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https://socratic.org/questions/five-numbers-have-an-average-of-90-four-of-the-numbers-are-87-93-89-and-85-what- | 1,723,069,117,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00517.warc.gz | 409,488,914 | 6,174 | # Five numbers have an average of 90. Four of the numbers are 87, 93, 89, and 85. What is the remaining number?
Jan 19, 2017
Remaining number is $96$
#### Explanation:
Average of given set of numbers is their sum divided by their count.
Here we have five numbers and as their average is $90$,
their sum must be $5 \times 90 = 450$
Bur four of them add up to $87 + 93 + 89 + 85 = 354$
Hence, remaining number is $450 - 354 = 96$ | 131 | 435 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-33 | latest | en | 0.967253 | 131 |
https://cracku.in/10-a-cube-of-side-12-cm-is-painted-red-on-all-the-fac-x-cat-1996 | 1,716,862,688,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00525.warc.gz | 149,478,222 | 23,688 | Question 10
# A cube of side 12 cm is painted red on all the faces and then cut into smaller cubes, each of side 3 cm. What is the total number of smaller cubes having none of their faces painted?
Solution
Total number of cubes will be = $$\frac{12 \times 12 \times 12}{3 \times 3 \times 3}$$ = 64
Cubes with 3 sides painted = cubes at corner sides = 8
Cubes with 2 sides painted = cubes on lines = 24
Cubes with 1 sides painted = cubes on middle of surface = 24
So cubes with no side painted = 64 - (8+24+24) = 8 | 155 | 516 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-22 | latest | en | 0.914671 | 142 |
https://socratic.org/questions/how-do-you-solve-8-2x-is-greater-than-or-equal-to-4 | 1,638,103,153,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358520.50/warc/CC-MAIN-20211128103924-20211128133924-00075.warc.gz | 491,534,131 | 5,777 | # How do you solve 8-2x is greater than or equal to -4?
## When are you supposed to flip the greater than or equal to sign?
##### 1 Answer
Mar 31, 2018
$x \le 6$
#### Explanation:
$8 - 2 x \ge - 4$ is our equation
To solve for the inequality you do it normally like you would for an equation, although if you multiply or divide by a negative number you flip the inequality
$- 2 x \ge - 12$
Now we have to divide both sides by $- 2$ so we will flip the inequality
$x \le 6$ | 141 | 481 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-49 | latest | en | 0.909142 | 145 |
https://socratic.org/questions/beth-has-900-in-the-bank-every-time-she-rides-the-bus-she-spends-2-how-do-you-wr | 1,581,879,431,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141396.22/warc/CC-MAIN-20200216182139-20200216212139-00519.warc.gz | 564,294,427 | 6,309 | # Beth has $900 in the bank. Every time she rides the bus she spends$2. How do you write an inequality that Beth can use to see how many times she can ride the bus and still have more than $500 in the bank? ##### 1 Answer This is the answer for (b) bus rides she takes. $500 < 900 - 2 b$#### Explanation: $500 < 900 - 2 b$$500 - 900 < - 2 b$$- 400 < - 2 b$$\frac{- 400}{- 2} > b$We reverse the signs above because we are dividing by a negative number. $200 > b\$
She can take less than 200(a lot of) bus rides. | 158 | 511 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-10 | longest | en | 0.926667 | 149 |
https://socratic.org/questions/how-do-you-find-the-gcf-of-24-and-42 | 1,555,634,778,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526923.39/warc/CC-MAIN-20190419001419-20190419023419-00412.warc.gz | 569,945,231 | 5,744 | # How do you find the GCF of 24 and 42?
Jan 9, 2017
$6$
#### Explanation:
The Greatest Common Factor is the highest number that can be used to divide the two given numbers. It is easily found by writing down the factors of the two numbers and selecting the common one(s) which are the highest.
In the given examples, the factors of the two numbers are as follows:
$24 : 2 , 2 , 2 , 3$
$42 : 2 , 3 , 7$
Since there are two numbers common between the two sets of factors, the GCF is:
$2 \times 3 = 6$ | 144 | 507 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2019-18 | latest | en | 0.934767 | 151 |
https://socratic.org/questions/how-do-you-solve-7x-2x-15 | 1,566,608,538,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319155.91/warc/CC-MAIN-20190823235136-20190824021136-00221.warc.gz | 634,332,311 | 5,787 | # How do you solve 7x = 2x + 15?
Jun 2, 2017
$x = 3$
#### Explanation:
$7 x = 2 x + 15$
We can move $2 x$ to the other side, changing it from positive to negative.
$7 x - 2 x = 15$
$5 x = 15$
Now we can work backwards. The last thing that happened to $x$ is that it was multiplied by $5$ to get to $15$, so in working backwards, we will do the opposite to $15$ to get $x$.
x= 15 ÷ 5
color(blue)(x = 3 | 148 | 410 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-35 | latest | en | 0.936504 | 153 |
https://socratic.org/questions/stella-makes-2-royalties-on-a-book-she-wrote-how-much-money-did-her-book-earn-in | 1,586,373,860,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371821680.80/warc/CC-MAIN-20200408170717-20200408201217-00282.warc.gz | 680,212,473 | 6,077 | Stella makes 2% royalties on a book she wrote. How much money did her book earn in sales last year if she made $53,000 in royalties? 1 Answer S=($53000)/.02=$2,650,000 Explanation: We can take the book sales ($S$), multiply it by the royalty percentage (%), and see how much she makes in royalties ($R$): S xx % = R Now let's substitute in what we know: S xx 2% =$53,000
To find $S$ (book sales) we now divide both sides by 2% (which is the same as 0.02):
S=($53000)/.02=$2,650,000 | 151 | 483 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2020-16 | longest | en | 0.961586 | 154 |
https://socratic.org/questions/mary-sold-half-of-her-comic-books-and-then-bought-8-more-she-now-has-18-how-many#619526 | 1,721,484,304,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515164.46/warc/CC-MAIN-20240720113754-20240720143754-00459.warc.gz | 479,301,903 | 5,937 | # Mary sold half of her comic books and then bought 8 more. she now has 18. how many did she begin with?
May 24, 2018
#### Explanation:
If she bought 8 more and in total she have 18 before she buy those 8 books she had 10 so $10 \cdot 2 = 20$ because she sold half of her books that means she had double the amount of 10.
May 24, 2018
20
#### Explanation:
Work backwards from the final count.
Final count $\to 18$
This was after adding eight more $18 - 8 = 10$
This was after selling half the original amount
original amount $\to 2 \times 10 = 20$ | 158 | 558 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-30 | latest | en | 0.988775 | 158 |
https://socratic.org/questions/how-do-you-simplify-65-5 | 1,601,368,609,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401632671.79/warc/CC-MAIN-20200929060555-20200929090555-00523.warc.gz | 542,537,141 | 5,818 | # How do you simplify 65 + -5?
Jan 30, 2016
$65 - 5 = 60$
#### Explanation:
re white as $\left(65\right) + \left(- 5\right)$
The plus sign can be considers as the action 'Put with then count' from where you are on the number line towards the right. (+ve direction).
The minus sign can be considered as 'change direction of count'
So the + is first telling you to count to the right on the number line.
The we have the minus which is saying "now change the direction of count on the number line.
Put all this together and you end up with:
$65 - 5 = 60$ | 152 | 560 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2020-40 | latest | en | 0.907422 | 162 |
https://socratic.org/questions/the-sum-of-two-times-a-number-and-5-is-45-what-is-the-number | 1,638,588,265,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362923.11/warc/CC-MAIN-20211204003045-20211204033045-00033.warc.gz | 563,046,530 | 6,170 | # The sum of two times a number and 5 is 45 what is the number?
Nov 21, 2016
20
#### Explanation:
The sum: +
Two times a number: 2x
Let $x$ be the unknown number that $2$ is being multiplied by
The sum of 2 times a number and 5:
$2 x + 5 = 45$
Perform the opposite operation on the left side of the equation by subtracting $5$ on both sides of the equation
$2 x = 40$
Isolate $x$ by performing the opposite operation, which is division of $2$, on both sides of the equation
$x = 20$
The unknown number is $20$ | 153 | 522 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-49 | latest | en | 0.940712 | 164 |
https://socratic.org/questions/what-is-the-cost-of-each-eraser-if-you-bought-a-magazine-for-5-and-four-erasers- | 1,638,137,005,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358591.95/warc/CC-MAIN-20211128194436-20211128224436-00390.warc.gz | 594,674,130 | 5,961 | # What is the cost of each eraser if you bought a magazine for $5 and four erasers for a total of$25?
Sep 28, 2014
The buyer bought four erasers and a magazine for $25. The cost of each magazine is$5
The cost of an eraser is x dollars
The cost of 4 erasers is 4x
Cost of the Erasers + Cost of the Magazine = Total Cost
4x + $5 =$25
Subtract 5 from both sides to isolate the term with the variable.
4x + 5 -5 = 25 -5
4x = 20
Divide by 4 on both sides to isolate the variable.
x = $\frac{20}{4}$ = \$5 | 162 | 510 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-49 | latest | en | 0.926245 | 168 |
https://socratic.org/questions/how-do-you-solve-4-7x-5x | 1,638,098,370,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358520.50/warc/CC-MAIN-20211128103924-20211128133924-00296.warc.gz | 488,830,930 | 5,959 | # How do you solve 4 - 7x = 5x?
Apr 11, 2018
$x = \frac{1}{3}$
#### Explanation:
First, you'll want to combine like terms by bringing the $x$'s together. This can be done by adding $7 x$ to both sides.
$4 - 7 x = 5 x$
to get
$4 = 12 x$
To solve for $x$, you must divide both sides of the equation by the coefficient of $x$ which is $12$ and then simplify.
$x = \frac{4}{12}$
$x = \frac{1}{3}$
$x$ is therefore equal to $\frac{1}{3}$. | 156 | 444 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-49 | latest | en | 0.928288 | 169 |
https://api-project-1022638073839.appspot.com/questions/a-bag-contains-orange-blocks-and-pink-blocks-the-ratio-of-orange-block-to-pink-b | 1,718,515,131,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861643.92/warc/CC-MAIN-20240616043719-20240616073719-00057.warc.gz | 87,637,999 | 6,316 | # A bag contains orange blocks and pink blocks. The ratio of orange block to pink blocks is 2 to 11. If 121 pink blocks are in the bag, how many orange blocks are in the bag?
Sep 15, 2017
$22$ orange blocks.
#### Explanation:
So, there are $2$ orange blocks for every $11$ pink blocks.
Let's denote the amount of orange blocks as $x$.
$\frac{121}{11} = \frac{x}{2}$
Multiplying both sides by $22$, we get
$242 = 11 x$
Dividing both sides by $11$, we arrive at our answer:
$\frac{242}{11} = \frac{11 x}{11} \to 22 = x$ | 162 | 521 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.909884 | 169 |
https://socratic.org/questions/how-do-you-solve-4-6x-1-7-3x-2 | 1,601,578,644,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402131986.91/warc/CC-MAIN-20201001174918-20201001204918-00736.warc.gz | 583,604,448 | 6,136 | # How do you solve 4( 6x - 1) = 7( 3x + 2)?
Mar 11, 2018
Distribute and then combine.
#### Explanation:
You would have to distribute the $4$ into $6 x - 1$, making it
$24 x - 4$
Doing the same on the other side, you have
$21 x + 14$
$24 x - 4 = 21 x + 14$
You have to combine like terms by subtracting and adding from each side. If you subtract $21 x$ from each side and add $4$ to each side, you will end up with the equation
$3 x = 18$
Divide both sides by $3$ and you get
$x = 6$ | 169 | 490 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2020-40 | latest | en | 0.905778 | 169 |
https://socratic.org/questions/on-a-50-question-test-a-student-answered-5-questions-incorrectly-what-percent-di | 1,718,751,259,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861794.76/warc/CC-MAIN-20240618203026-20240618233026-00156.warc.gz | 478,748,088 | 6,029 | # On a 50 question test, a student answered 5 questions incorrectly. What percent did the student answer correctly?
Feb 2, 2016
If $5$ questions were answered incorrectly, $45$ questions were answered correctly and the percentage answered correctly is 45/50 xx 100 = 90%
#### Explanation:
To find the percentage of questions answered correctly (50 of which $5$ were answered wrong so $45$ were answered right), we must divide the number of correct answers by the total number of questions.
In this case that would be $\frac{45}{50}$.
To convert that decimal fraction to a percentage, the answer is multiplied by $100$.
$\frac{45}{50} = 0.9$ and $0.9 \times 100$ equals 90%. | 169 | 680 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.966194 | 171 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-7-algebraic-fractions-7-2-multiplying-and-dividing-algebraic-fractions-problem-set-7-2-page-285/5 | 1,537,731,691,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159744.52/warc/CC-MAIN-20180923193039-20180923213439-00073.warc.gz | 757,726,955 | 13,016 | ## Elementary Algebra
We can never divide fractions. Rather, we must flip the numerator and the denominator for the second fraction listed and then multiply. When we multiply fractions, we multiply the numerators of the fractions by each other and the denominators of the fractions by each other. Thus, we obtain that this expression is equal to: $\frac{17 \times -9}{19 \times 9}=-153/171$ In order to simplify fractions, we want to cancel out anything that is in both the numerator and the denominator of the fraction. In order to do this, we look for common factors for the values in the numerator and the denominator of the fraction. 153 and 171 have a common factor of 9, so we simplify: $-153/171=-(17*9)/(9*19)=-17/19$ | 174 | 726 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2018-39 | longest | en | 0.901392 | 172 |
https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-1-section-1-1-introduction-to-algebra-variables-and-mathematical-models-exercise-set-page-11/71 | 1,539,668,611,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510019.12/warc/CC-MAIN-20181016051435-20181016072935-00088.warc.gz | 942,955,211 | 14,310 | ## Introductory Algebra for College Students (7th Edition)
$4(x +5) = 33$.
First identify the task. In this case, it is to express "the product of four and a number increased by five is thirty-three" as an equation. We are told to assume the number is "x". A method to solving this problem is to divide the given statement into sub parts and then combine each of these parts into an equation. "The product of four and a number" can be expressed as "$4x$". "increased by five" can be expressed as "$+5$". "Is equal to three" reminds us to connect the two sides with an equals sign, "$=33$". By putting these together, the answer is $4(x +5) = 33$. | 167 | 647 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2018-43 | longest | en | 0.966009 | 172 |
https://socratic.org/questions/suppose-you-are-starting-an-office-cleaning-service-you-have-spent-315-on-equipm | 1,600,937,800,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00437.warc.gz | 551,417,959 | 6,252 | # Suppose you are starting an office-cleaning service. You have spent $315 on equipment. To clean an office, you use$4 worth of supplies. You charge $25 per office. How many offices must you clean to break even? ##### 1 Answer Jul 12, 2018 Number of offices to be cleaned to cover the cost of equipment $= 15$#### Explanation: Cost of equipment =$.315
Cost of supplies =$.4 Charge per office =$.25
Number of offices to be cleaned to cover the cost of equipment $= x$
Then -
$25 x - 4 x = 315$
$21 x = 315$
$x = \frac{315}{21} = 15$
Number of offices to be cleaned to cover the cost of equipment $= 15$ | 167 | 606 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-40 | latest | en | 0.93355 | 172 |
https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-8-test-page-621/10 | 1,571,624,369,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987751039.81/warc/CC-MAIN-20191021020335-20191021043835-00026.warc.gz | 912,113,380 | 12,244 | ## Introductory Algebra for College Students (7th Edition)
$$5\sqrt {2}$$
According to the product rule for square roots, the square root of two radicals that are multiplied together is the product of the two radicands. In this problem, we first multiply the two radicands together to get one radical expression: $$\sqrt {10} \times \sqrt {5} = \sqrt {50}$$ We can now simplify by using the product rule for square roots to split the radical into two component radicals. We want to divide the radical expression so that the radicand is the product of one perfect square and another factor: $$\sqrt {50} = \sqrt {25 \times 2}$$ Now we take the square root of $25$, which is $5$: $$5\sqrt {2}$$ | 175 | 693 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-43 | latest | en | 0.919217 | 172 |
https://api-project-1022638073839.appspot.com/questions/if-kate-triples-her-height-she-will-be-as-tall-as-if-she-added-147-5-inches-to-h | 1,718,791,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00756.warc.gz | 83,043,074 | 6,420 | # If Kate triples her height, she will be as tall as if she added 147.5 inches to half her height. How tall is Kate?
Jun 2, 2018
$\text{59 inches}$.
#### Explanation:
Let's call Kate's height $x$. We're going to set this up as an equation. Since three times her height is equal to $147.5$ inches plus half her height, this can be written as:
$3 x = 147.5 + 0.5 x$
Here $x$ is her height in inches. By subtracting $0.5 x$ on both sides, we get:
$2.5 x = 147.5$
Divide both sides by $2.5$
$x = 59$ | 162 | 504 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.941563 | 172 |
https://socratic.org/questions/samantha-s-notebook-has-14-pages-more-than-three-times-as-much-as-lisa-s-noteboo | 1,638,094,182,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358480.10/warc/CC-MAIN-20211128073830-20211128103830-00223.warc.gz | 588,135,449 | 6,199 | # Samantha's notebook has 14 pages more than three times as much as Lisa's notebook. If Samantha's notebook has 260 pages, how many pages are there in Lisa's notebook?
##### 1 Answer
Nov 29, 2016
Lisa's notebook has $82$ pages.
#### Explanation:
If we represent the number of pages in Lisa's notebook as $x$, then the number of pages in Samantha's notebook will be $\left(3 x + 14\right)$, which we know to be $260$. Hence:
$3 x + 14 = 260$
Subtract $14$ from each side.
$3 x = 246$
Divide each side by $3$.
$x = 82$ which is the number of pages in Lisa's notebook. | 163 | 574 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-49 | latest | en | 0.92675 | 173 |
https://socratic.org/questions/how-many-0-12-pencils-can-mr-velazquez-buy-for-4-80 | 1,582,850,484,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146907.86/warc/CC-MAIN-20200227221724-20200228011724-00093.warc.gz | 495,000,380 | 5,941 | # How many $0.12 pencils can Mr. Velazquez buy for$4.80?
Mr. Velazquez can buy $40$ pencils
$0.12 p = 4.80$
Where $p$ is the number of pencils which cost $0.12 per pencil and $4.80$is the amount of money that Mr. Velazquez has to spend. So we solve the equation for $p$$0.12 p = 4.8$Divide $0.12$from both sides: $\frac{\cancel{0.12}}{\cancel{\textcolor{red}{0.12}}} p = \frac{4.80}{\textcolor{red}{0.12}}$$p = 40$So he can buy $40\$ pencils exactly | 167 | 450 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-10 | latest | en | 0.910631 | 173 |
https://socratic.org/questions/what-is-the-rule-of-divisibility-by-7#249488 | 1,721,872,484,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.61/warc/CC-MAIN-20240724232540-20240725022540-00511.warc.gz | 457,398,569 | 5,917 | # What is the rule of divisibility by 7?
Apr 4, 2016
#### Explanation:
Starting from right mark off the digits in groups of threes (just as we do when we put commas in large numbers).
Now add up alternate group of numbers and find the difference between the two. If the difference is divisible by $7$, entire number is divisible by $7$.
For example $123456789113$, these are grouped as $123$, $456$, $789$ and $113$
and $123 + 789 = 912$ and $456 + 113 = 569$.
As difference between $912 - 569 = 343$
As $343$ is divisible by $7$, $123456789113$ is divisible by $7$ | 162 | 573 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-30 | latest | en | 0.924752 | 173 |
https://socratic.org/questions/sara-can-lift-a-10-pound-weight-by-applying-30-pounds-of-force-to-a-lever-if-the | 1,582,847,423,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146907.86/warc/CC-MAIN-20200227221724-20200228011724-00109.warc.gz | 500,501,522 | 5,900 | # Sara can lift a 10-pound weight by applying 30 pounds of force to a lever. If the amount of force Sara needs to apply to a lever is proportional to the weight she can lift, how much force would she need to apply in order to lift a 25-pound weight?
Jun 30, 2018
$75$ pounds of force to the lever.
#### Explanation:
We know that the force she applies to the lever is directly proportional to the weight she can lift.
So, we get:
$\frac{30}{10} = \frac{{F}_{{l}_{2}}}{25}$
$3 = \frac{{F}_{{l}_{2}}}{25}$
$\therefore {F}_{{l}_{2}} = 3 \cdot 25$
$= 75$ | 169 | 558 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-10 | latest | en | 0.917856 | 175 |
https://socratic.org/questions/how-do-you-solve-16-m-3-5-3 | 1,576,407,069,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541307813.73/warc/CC-MAIN-20191215094447-20191215122447-00119.warc.gz | 546,231,939 | 6,028 | # How do you solve 16 m + 3 5 = 3?
Oct 18, 2016
This problem is solved by isolating the variable, so that we determine that $m = - 2$.
#### Explanation:
1. Subtract $35$ from both sides. This gets the $16$m by itself. It is necessary to perform this operation on both sides to maintain the balance of the equation.
$16 m + 35 - 35 = 3 - 35$
2. Divide both sides by $16$. This gets the $m$ by itself. Again, it is necessary to perform this on both sides so that the equation stays balanced.
$\frac{16 m}{16} = \frac{3 - 35}{16}$
3. Simplify.
$m = - 2$ | 172 | 555 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-51 | longest | en | 0.932611 | 175 |
https://socratic.org/questions/on-a-50-question-test-a-student-answered-5-questions-incorrectly-what-percent-di#220383 | 1,723,400,723,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008125.69/warc/CC-MAIN-20240811172916-20240811202916-00397.warc.gz | 416,501,637 | 6,083 | # On a 50 question test, a student answered 5 questions incorrectly. What percent did the student answer correctly?
##### 1 Answer
Feb 2, 2016
If $5$ questions were answered incorrectly, $45$ questions were answered correctly and the percentage answered correctly is 45/50 xx 100 = 90%
#### Explanation:
To find the percentage of questions answered correctly (50 of which $5$ were answered wrong so $45$ were answered right), we must divide the number of correct answers by the total number of questions.
In this case that would be $\frac{45}{50}$.
To convert that decimal fraction to a percentage, the answer is multiplied by $100$.
$\frac{45}{50} = 0.9$ and $0.9 \times 100$ equals 90%. | 174 | 695 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-33 | latest | en | 0.962892 | 175 |
https://api-project-1022638073839.appspot.com/questions/what-is-the-equation-of-the-line-that-goes-through-5-6-and-is-perpendicular-to-y | 1,590,368,692,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347387155.10/warc/CC-MAIN-20200525001747-20200525031747-00519.warc.gz | 248,625,337 | 6,073 | # What is the equation of the line that goes through (5, -6) and is perpendicular to y=9?
Feb 28, 2017
See the entire solution explanation below:
#### Explanation:
$y = 9$ is a vertical line because it has a value of $9$ for each and every value of $x$.
Therefore, a line perpendicular to the will be a horizontal line and $x$ will have the same value for each and every value of $y$.
The equation for a horizontal line is $x = a$.
In this case we are given the point $\left(5 , - 6\right)$ which has a value of $5$ for $x$.
Therefore, the equation for the line in this problem is:
$x = 5$ | 166 | 598 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-24 | latest | en | 0.907396 | 176 |
https://braingenie.ck12.org/skills/103388/learn | 1,603,305,035,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107877420.17/warc/CC-MAIN-20201021180646-20201021210646-00441.warc.gz | 206,419,198 | 1,807 | ### Sample Problem
If the length of a rectangle increases by 2 cm, its area will increase by 12 cm2. If the width increases by 3 cm, its area will increase by 21cm2. What is the original area of the rectangle? cm2
#### Solution
By increasing the length by 2 cm, we are basically adding a rectangle that is 2 cm by the width of the original rectangle. Since the area of the added rectangle is 12, the width of the original rectangle is therefore 12 ÷ 2 = 6 cm. By increasing the width by 3 cm, we are basically adding a rectangle that is 3 cm by the length of the original rectangle. Since the area of this added rectangle is 21, the length is 21 ÷ 3 = 7cm. Thus the original rectangle has an area of 7 × 6 = 42 cm2. | 191 | 718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2020-45 | latest | en | 0.928852 | 176 |
https://algebra-class-ecourse.com/question/you-pay-35-each-month-for-satellite-tv-plus-4-50-for-every-movie-you-purchase-write-and-solve-an-11176597-42/ | 1,643,306,939,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305277.88/warc/CC-MAIN-20220127163150-20220127193150-00513.warc.gz | 153,524,038 | 14,161 | ## You pay \$35 each month for satellite TV, plus \$4.50 for every movie you purchase. Write and solve an equation to find the number of movies p
Question
You pay \$35 each month for satellite TV, plus \$4.50 for every movie you purchase. Write and solve an equation to find the number of movies purchased during a month with a total bill of \$48.50
0
1. We know that \$35 is for satellite TV, so we can subtract that from the total.
48.50 – 35 = 13.50
Now we divide what’s left of the total by the cost of a movie.
13.50 ÷ 4.50 = 3
3 movies were purchased.
Hope I could help!
2. So 3 movies were purchased | 166 | 616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-05 | latest | en | 0.958409 | 177 |
https://socratic.org/questions/how-do-you-solve-2x-5x-12 | 1,582,199,006,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144722.77/warc/CC-MAIN-20200220100914-20200220130914-00098.warc.gz | 575,128,037 | 6,196 | # How do you solve 2x = 5x -12?
Jun 21, 2018
$x = 4$
#### Explanation:
we get by adding $12$ on both sides
$12 + 2 x = 5 x$
subtracting $2 x$ on both sides
$12 = 3 x$
so $x = 4$
Aug 9, 2018
$x = 4$
#### Explanation:
First solve for $x$ by subtracting $5 x$ from both sides:
$- 3 x = - 12$
To make things a little bit nicer, multiply both sides by $- 1$:
$3 x = 12$
Then solve for $x$ by dividing both sides by $3$:
$x = 4$ | 166 | 437 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2020-10 | longest | en | 0.900181 | 177 |
https://socratic.org/questions/how-do-you-solve-14-x-3x-2 | 1,582,873,763,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147054.34/warc/CC-MAIN-20200228043124-20200228073124-00155.warc.gz | 543,334,080 | 6,155 | # How do you solve 14-x=3x-2?
Jul 8, 2016
$x = 4$
#### Explanation:
For problems like this you want to get the variable on one side of the equation and your constants ( a number on its own) on the other side.
Put your variables on the right side of the equation by adding x on both sides. You should end up with:
$14 = 4 x - 2$
Add 2 on both sides of the equation so the constants are on the left side leaving 4x on the right and 16 on the left:
$16 = 4 x$
Divide both sides by 4 to solve for x:
$\frac{16}{4} = \frac{4 x}{4}$
$x = 4$ | 167 | 545 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-10 | longest | en | 0.901275 | 178 |
https://socratic.org/questions/when-a-number-is-decreased-by-20-of-itself-the-result-is-20-what-is-the-number#416567 | 1,656,619,615,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103877410.46/warc/CC-MAIN-20220630183616-20220630213616-00428.warc.gz | 557,655,145 | 5,768 | # When a number is decreased by 20% of itself, the result is 20. What is the number?
May 2, 2017
25
#### Explanation:
Since the number is reduced by 20% we have 80% left. That 80% is equal to 20. We can use cross multiplication to relate the two.
80% is equal to $\frac{80}{100} = \frac{8}{10} = \frac{4}{5}$
We know that 20 is 80% of something so we have $\frac{20}{x}$
These two should be equal so we can do the following
$\frac{4}{5} = \frac{20}{x}$
Using cross multiplication we get
$4 x = 100$
Solving we get
$x = 25$ | 176 | 530 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-27 | latest | en | 0.924482 | 178 |
https://socratic.org/questions/how-do-you-solve-5x-7-10#563116 | 1,642,856,387,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303845.33/warc/CC-MAIN-20220122103819-20220122133819-00038.warc.gz | 574,727,541 | 5,927 | # How do you solve 5x+7=10?
Mar 3, 2018
$x = \frac{3}{5}$ or $0.6$
#### Explanation:
$5 x + 7 = 10$
You need to subtract seven from both sides.
$5 x + 7 - 7 = 10 - 7$
The two sevens on the left side cancel out and if you subtract seven from ten then you are left with this:
$5 x = 3$
You then need to divide both sides by five so that the five on the left side becomes a one (which you don't need to write), and divide the three on the right side by five as well.
So therefore,
$x = \frac{3}{5} = 0.6$ | 166 | 512 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-05 | latest | en | 0.923578 | 179 |
https://www.univerkov.com/the-sides-of-the-rectangle-are-5-and-12-cm-will-find-its-diagonals/ | 1,686,117,522,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653608.76/warc/CC-MAIN-20230607042751-20230607072751-00428.warc.gz | 1,155,492,594 | 6,213 | # The sides of the rectangle are 5 and 12 cm. Will find its diagonals.
We are given a rectangle with sides 5 cm and 12 cm. Let its length be 12 cm and width 5 cm. We need to find its diagonal. We can draw a diagonal and consider one of the resulting triangles. Since all the corners of the rectangle are straight, then the triangle will turn out to be straight. We know two of his legs, which means he can find the hypotenuse according to the rule: the square of the hypotenuse is equal to the sum of the squares of the legs.
Let’s find the sum of the squares of the legs:
5 ^ 2 + 12 ^ 2 = 25 + 144 = 169.
So our hypotenuse will be equal to the root of 169. And the root of 169 is 13. | 186 | 685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-23 | latest | en | 0.944331 | 179 |
https://socratic.org/questions/how-do-you-solve-16x-4-3x-3 | 1,582,197,090,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144722.77/warc/CC-MAIN-20200220100914-20200220130914-00252.warc.gz | 580,440,535 | 6,252 | # How do you solve 16x + 4= 3x - 3?
May 25, 2018
$x = - \frac{7}{13}$
#### Explanation:
You can start the equation off by trying to move all the numbers with variables to the left. To achieve this, you can subtract $3 x$ from the right and subtract it from the left. If you did it right, your equation would look like this:
$13 x + 4 = - 3$
Then, to make the variable $x$ alone on the left side, you can subtract $4$ from both sides to get
$13 x = - 7$
Then you would divide both sides by $13$ to get x to become one number which is
$x = - \frac{7}{13}$ | 171 | 562 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-10 | longest | en | 0.93578 | 179 |
https://socratic.org/questions/omar-wants-to-buy-chairs-for-his-new-office-each-chair-costs-12-and-there-is-a-f | 1,638,419,111,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00017.warc.gz | 594,928,043 | 6,230 | # Omar wants to buy chairs for his new office. Each chair costs $12 and there is a flat delivery fee of$10. If he has $80, how many chairs can he buy? ##### 1 Answer Oct 27, 2016 This is of the form $a x + b = c$#### Explanation: $a =$price per chair ($12)
$x =$number of chairs
$b =$delivery fee ($10) Now the problem can be filled in: $12 \cdot x + 10 = 80 \to$(subtract 10 from both sides) $12 \cdot x = 70 \to$(divide by 12) $x = \frac{70}{12} = 5 \frac{10}{12}$So he can buy 5 chairs and have$10 left. He is \$2 short for the 6th chair. | 180 | 541 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-49 | latest | en | 0.947965 | 180 |
https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-8-cumulative-review-exercises-page-621/24 | 1,571,805,607,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987829458.93/warc/CC-MAIN-20191023043257-20191023070757-00207.warc.gz | 893,491,997 | 12,349 | ## Introductory Algebra for College Students (7th Edition)
The number is $43$.
First, we want to translate this word problem into an algebraic equation that we can solve. Let's look at the different pieces of the statement: "Seven subtracted from" means to subtract $7$ from another number. "Five times a number" means to multiply $5$ by an unknown number $x$. "Is $208$" means "$= 208$." Now we can put these parts together to form an algebraic equation: $$5x - 7 = 208$$ To solve this equation, we add $7$ to each side to isolate the variable on one side of the equation and the constants on the other: $$5x = 215$$ Now we divide by $5$ on each side of the equation to solve for $x$: $$x = 43$$ | 183 | 697 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-43 | latest | en | 0.910193 | 180 |
https://socratic.org/questions/how-do-you-find-the-next-two-terms-of-the-geometric-sequence-20-30-45#606424 | 1,638,118,382,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358570.48/warc/CC-MAIN-20211128164634-20211128194634-00156.warc.gz | 606,379,699 | 5,891 | # How do you find the next two terms of the geometric sequence 20, 30, 45,...?
May 3, 2018
$67.5$ and $101.25$
#### Explanation:
In a geometric sequence the ratio of a term with its immediately preceding term is always constant and is known as constant ratio. Therefore subsequent terms can be obtained by multiplying the term by common ratio.
Here in $20 , 30 , 45 , \ldots \ldots$, the ratio is $\frac{30}{20} = \frac{45}{30} = \frac{3}{2} = 1.5$.
Hence next i.e. fourth term in the series is $45 \times 1.5 = 67.5$
and fifth term is $67.5 \times 1.5 = 101.25$ | 181 | 569 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-49 | latest | en | 0.941894 | 180 |
https://socratic.org/questions/how-do-you-solve-5b-2-17 | 1,638,140,936,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00403.warc.gz | 601,557,502 | 5,962 | # How do you solve 5b + 2 = 17?
Jul 27, 2015
$b = 3$
#### Explanation:
We will group the numbers with each other and leave the the variable $b$ alone on one side of the equation.
This means that you will have $+ 2$ to be next to the 17, but because the 17 is on the other side of the equation, $+ 2$ will become $- 2$.
$5 b = 17 - 2$
$5 b = 15$
Therefore,
$b = \frac{15}{5}$
now divide both sides of the equation by $5$ to get
$\frac{\cancel{5} \cdot b}{\cancel{5}} = \frac{15}{5}$
$b = 3$ | 176 | 502 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-49 | latest | en | 0.906221 | 180 |
https://braingenie.ck12.org/skills/103385/learn | 1,603,553,308,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107883636.39/warc/CC-MAIN-20201024135444-20201024165444-00566.warc.gz | 249,831,348 | 1,951 | ### Sample Problem
Look at the figure below. The shaded area is 53 cm2. What is the area of the square?
cm2
#### Solution
If we want to get the area of the square, we have to get the relationship of the square’s area and the shaded area. We do this by ‘dissecting’ the figure into triangles and a rectangle by drawing lines parallel to the square as shown.
We see that there are rectangles that are formed on the four corners of the square. In each rectangle on the four corners, we see that the half is shaded and the other half is white. Thus the shaded area contains the middle rectangle, making it 3 × 2 = 6 cm bigger than the white one. Therefore, the area of the square is (53 - 6) × 2 + 6 = 100 cm2. | 178 | 711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2020-45 | latest | en | 0.956163 | 180 |
https://educationexpert.net/english/1748058.html | 1,631,980,165,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00109.warc.gz | 270,725,282 | 7,025 | 16 October, 05:03
# The third grade class collected 148 books to donate to the library the fourth grade class collected 175books the students need pack the books into boxes each box hold 9 books how many boxes do they need to hold in all
0
1. 16 October, 06:33
0
They need 36 boxes.
Explanation:
First, we need the calculate the sum of the books collected. That means we have 148 + 175 = 323 books in total. Now we have to find the number of boxes needed. Since in each box they can put 9 books, we need to divide the total number of books by the number of books per box.
So they will need 323 : 9 = 35.9 boxes
Basically this means that they will need 36 boxes although one of them won't be completely filled with books. | 185 | 726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-39 | latest | en | 0.947904 | 181 |
https://socratic.org/questions/the-mean-of-4-numbers-is-5-and-the-mean-of-3-different-numbers-is-12-what-is-the#382699 | 1,709,026,915,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00361.warc.gz | 540,157,718 | 6,125 | # The mean of 4 numbers is 5 and the mean of 3 different numbers is 12. What is the mean of the 7 numbers together?
8
#### Explanation:
The mean of a set of numbers is the sum of the numbers over the count of the set (the number of values).
We have a set of four numbers and the mean is 5. We can see that the sum of the values is 20:
$\frac{20}{4} = 5$
We have another set of three numbers whose mean is 12. We can write that as:
$\frac{36}{3} = 12$
To find the mean of the seven numbers together, we can add the values together and divide by 7:
$\frac{20 + 36}{7} = \frac{56}{7} = 8$ | 176 | 594 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-10 | latest | en | 0.903465 | 182 |
https://socratic.org/questions/59b7406db72cff30bb0c7bd4#474309 | 1,716,681,270,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058858.81/warc/CC-MAIN-20240525222734-20240526012734-00763.warc.gz | 468,655,260 | 5,838 | # Question #c7bd4
Sep 12, 2017
you will need 7 pizzas
#### Explanation:
Multiply the number of people by two to find out how many pieces of pizza is needed
$35 \times 2 = 70$
Divide 70 by 10 to find the amount of pizzas $70 \div 10 = 7$
This means that 7 pizzas are needed for the party!
Sep 12, 2017
7
#### Explanation:
This is a problem of equating two expressions – one for the number of pieces in a pizza, and one for the number expected to be eaten.
Pieces eaten will be $35 \times 2 = 70$. so, if each pizza has 10 slices, the number needed is $\frac{70}{10} = 7$. | 172 | 582 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-22 | latest | en | 0.917168 | 182 |
https://socratic.org/questions/four-less-than-the-product-of-a-number-and-8-is-twice-the-sum-of-the-number-and-#593069 | 1,696,100,211,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510707.90/warc/CC-MAIN-20230930181852-20230930211852-00824.warc.gz | 565,955,089 | 6,289 | # Four less than the product of a number and 8 is twice the sum of the number and 10. What is the number?
Apr 12, 2018
$x = 4$
#### Explanation:
Four less than the product of a number and $8$:
$8 x - 4$
is twice the sum of the number and $10$, so
$8 x - 4 = 2 \left(x + 10\right)$
Simplify.
$8 x - 4 = 2 x + 20$
Add $4$ to both sides, which is
$8 x = 2 x + 24$
Subtract $2 x$ from both sides:
$6 x = 24$
Divide $6$ from both sides to isolate the $x$
$x = 4$ | 172 | 472 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2023-40 | latest | en | 0.900092 | 183 |
https://api-project-1022638073839.appspot.com/questions/how-do-you-solve-5x-7-28 | 1,600,695,688,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400201699.38/warc/CC-MAIN-20200921112601-20200921142601-00759.warc.gz | 270,772,657 | 6,183 | # How do you solve –5x – 7 < 28?
Aug 13, 2015
Get rid of the constant first, then the coefficient.
#### Explanation:
Let's look at the equation.
$- 5 x - 7 < 28$
We want to solve this inequality step by step.
First, get rid of the constant. In this case, it would be the $7$. We can do this by adding 7 to both sides.
$- 5 x < 35$
Looks a lot easier now! We just multiply both sides by $- 5$ to get the answer.
There's a catch, though. When you multiply both sides of an inequality by a negative coefficient, you have to flip the inequality. So,
$- 5 x < 35$
becomes,
$x > - 7$ | 177 | 589 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2020-40 | latest | en | 0.932101 | 184 |
https://socratic.org/questions/coach-eddie-ordered-400-baseball-gloves-30-the-gloves-were-left-handed-how-many- | 1,586,303,739,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371806302.78/warc/CC-MAIN-20200407214925-20200408005425-00340.warc.gz | 700,407,715 | 6,116 | # Coach Eddie ordered 400 baseball gloves, 30% the gloves were left-handed How many left of handed gloves Eddie?
Nov 15, 2016
There will be 120 left handed gloves.
#### Explanation:
In this problem we are looking for 30% of 400.
"Percent" or "%" means "out of 100" or "per 100", Therefore 30% can be written as $\frac{30}{100}$.
When dealing with percents the word "of" means "times" or "to multiply".
Finally, lets call the number we are looking for "n".
Putting this altogether we can write this equation and solve for $n$:
$n = \frac{30}{100} \times 400$
$n = \frac{12000}{100}$
$n = 120$ | 171 | 602 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-16 | longest | en | 0.902962 | 184 |
https://socratic.org/questions/eight-more-than-four-times-a-number-is-the-same-as-two-less-than-five-times-the- | 1,638,179,103,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358702.43/warc/CC-MAIN-20211129074202-20211129104202-00417.warc.gz | 598,417,843 | 6,151 | # Eight more than four times a number is the same as two less than five times the number. What is the number?
Dec 10, 2016
The number is 10
#### Explanation:
First, let's call the number we are looking for $n$.
Eight more than four times a number can be written as:
$4 n + 8$
Two less than five times the number can be written as:
$5 n - 2$
Because these are the same we can write the following equation and solve for $n$:
$4 n + 8 = 5 n - 2$
$4 n - 4 n + 8 + 2 = 5 n - 4 n - 2 + 2$
$0 + 8 + 2 = 5 n - 4 n - 0$
$10 = n$ | 181 | 532 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-49 | latest | en | 0.906037 | 184 |
https://jm.edugain.com/questions/From-a-deck-of-cards-Jeff-took-out-one-card-at-random-What-is-the-probability-that-he-got-a-prime-number | 1,632,329,392,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057371.69/warc/CC-MAIN-20210922163121-20210922193121-00648.warc.gz | 374,883,863 | 12,216 | From a deck of cards, Jeff took out one card at random. What is the probability that he got a prime number?
4 13
Step by Step Explanation:
1. There are 52 cards in a deck, with 13 cards of each suit (diamond, club, heart, and spade).
2. 13 cards of a suit are the numbers from 2 to 10, 3 face cards and 1 ace card.
3. From the numbers 2 to 10, only 2, 3, 5, and 7 are prime. So there are four prime numbered cards per suit, which gives a total of 16 prime numbered cards.
4. So, there are 16 prime numbered cards out of a total of 52 cards.
So, the probability that we pick a prime numbered card =
Number of prime numbered cards Total number of cards
=
16 52
=
4 13 | 194 | 671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-39 | latest | en | 0.963473 | 184 |
https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-3-section-3-4-the-slope-intercept-form-of-the-equation-of-a-line-exercise-set-page-253/79 | 1,571,201,470,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986664662.15/warc/CC-MAIN-20191016041344-20191016064844-00503.warc.gz | 931,176,398 | 12,478 | ## Introductory Algebra for College Students (7th Edition)
$$x = 8$$
To solve this equation, we first want to get rid of the fractions. We do this by multiplying each term by the least common denominator, which, in this case, is $4$: $$4(\frac{x}{2}) + 4(7) = 4(13) - 4(\frac{x}{4})$$ We divide out common factors to get rid of the fractions: $$2x + 28 = 52 - x$$ Add $x$ to both sides to isolate the variable on one side of the equation: $$3x + 28 = 52$$ Subtract $28$ on both sides to move the constants to one side of the equation: $$3x = 24$$ Divide by $3$ on both sides of the equation to solve for $x$: $$x = 8$$ | 188 | 619 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-43 | latest | en | 0.915033 | 185 |
https://socratic.org/questions/how-do-you-find-the-least-common-multiple-of-15-20 | 1,624,058,957,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643354.47/warc/CC-MAIN-20210618230338-20210619020338-00416.warc.gz | 467,667,816 | 5,814 | # How do you find the Least common multiple of 15, 20?
LCM = 60
#### Explanation:
Let's first find the prime factors of these two numbers:
$15 = 3 \times 5$
$20 = 2 \times 2 \times 5$
The LCM will have all the prime factors from the two numbers but we can drop replicates across the numbers we're doing this for. In this example:
From 15, we get both the 3 and the 5
From 20, we get two 2s - the 5 is already accounted for in the 5 from the 15.
The LCM is:
$2 \times 2 \times 3 \times 5 = 4 \times 15 = 60$
$15 \times 2 \times 2 = 60$
$20 \times 3 = 60$ | 190 | 563 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-25 | latest | en | 0.90444 | 185 |
https://www.vedantu.com/question-answer/arun-bought-a-car-for-rs-350000-the-next-year-class-10-maths-icse-5ee1d950d6b8da423ca73c49 | 1,618,939,753,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039476006.77/warc/CC-MAIN-20210420152755-20210420182755-00627.warc.gz | 1,192,885,943 | 85,644 | Question
# Arun bought a car for Rs. 350,000. The next year, the price went up to Rs. 370,000. What was the percentage of price increased?
Hint: In this question Arun bought a car for some amount and later after certain duration of time the price of that car went up, so the percentage increase in amount of car will be increment in price divided by original price multiplied with 100 so, use this basic concept to reach the solution of the problem.
So, increment in the price is $= 370,000 - 350,000 = Rs.{\text{ }}20,000$
Therefore percentage of price increased $= \dfrac{{20,000}}{{350,000}} \times 100$
$= \dfrac{2}{{35}} \times 100 = \dfrac{{40}}{7} = 5.714$% | 189 | 667 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-17 | longest | en | 0.938584 | 185 |
https://api-project-1022638073839.appspot.com/questions/edwards-mother-in-57-years-old-her-age-3-years-more-than-twice-edwards-what-is-e | 1,719,244,590,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00275.warc.gz | 76,185,900 | 6,454 | # Edwards mother in 57 years old her age 3 years more than twice Edwards what is Edwards age?
Mar 7, 2018
$27$ years old.
#### Explanation:
If we consider that $x$ is Edward's age, twice of his age is $2 x$ and 3 more than that, $2 x + 3$, is equal to his mom's age. so we can form an algebraic equation as follows.
$2 x + 3 = 57$
We should take $+ 3$ to the other side of the equation
$2 x + 3 - 3 = 57 - 3$
$2 x = 54$
Isolate $x$: take $2$ to the other side of the equation
$\frac{2 x}{2} = \frac{54}{2}$
$x = 27$ | 182 | 526 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.943132 | 185 |
https://socratic.org/questions/how-do-you-solve-2-5x-37 | 1,638,050,402,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358233.7/warc/CC-MAIN-20211127193525-20211127223525-00293.warc.gz | 596,293,550 | 6,023 | # How do you solve 2 + 5x < 37?
Jul 28, 2015
Subtract 2 from both sides, then divide both sides by 5 to obtain $x < 7$
#### Explanation:
$2 + 5 x < 37$ (Initial form)
$2 + 5 x - 2 < 37 - 2$ (Subtract 2 from both sides to isolate $5 x$ on one side of the inequality)
$5 x < 35$ (Result of subtraction)
$\frac{5 x}{5} < \frac{35}{5}$ (Divide both sides by 5)
(Note: If instead we were dividing by a negative number, then the direction of the inequality would flip. Here we are dividing by a positive, though, so that won't occur)
$x < 7$ (Result of division) | 185 | 565 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2021-49 | latest | en | 0.910125 | 186 |
https://answers.everydaycalculation.com/subtract-fractions/4-42-minus-5-40 | 1,566,544,722,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318011.89/warc/CC-MAIN-20190823062005-20190823084005-00029.warc.gz | 369,662,153 | 2,825 | Solutions by everydaycalculation.com
## Subtract 5/40 from 4/42
4/42 - 5/40 is -5/168.
#### Steps for subtracting fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 42 and 40 is 840
2. For the 1st fraction, since 42 × 20 = 840,
4/42 = 4 × 20/42 × 20 = 80/840
3. Likewise, for the 2nd fraction, since 40 × 21 = 840,
5/40 = 5 × 21/40 × 21 = 105/840
4. Subtract the two fractions:
80/840 - 105/840 = 80 - 105/840 = -25/840
5. After reducing the fraction, the answer is -5/168
- | 195 | 514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2019-35 | latest | en | 0.90214 | 186 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-2-solving-equations-2-1-solving-one-step-equations-practice-and-problem-solving-exercises-page-85/11 | 1,720,830,258,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00171.warc.gz | 649,284,301 | 15,273 | ## Algebra 1
$n = 19$
$27 + n = 46$ In order to solve this algebraic expression, we will need to eliminate the constants from the left hand side in order to equate it with the variable. To achieve this, we join the constants together by subtracting $27$ from both sides as shown below: $27 + n - 27 = 46 - 27$ Simplify to get: $n = 19$ This means that the algebraic equation has a possible solution at $n = 19$ Testing: To test the truth of our statement, we replace for $n$ with $19$ in the original equation as shown below: $27 + n = 46$ $27 + 19 = 46$ Join like terms together to get: $46 = 46$ It is true that 46 = 46. This is a proof that the statement $n = 19$ is also true. | 200 | 681 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-30 | latest | en | 0.918617 | 186 |
https://socratic.org/questions/a-waiter-received-a-5-25-tip-if-the-check-totaled-25-what-percentage-was-receive | 1,718,888,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00402.warc.gz | 484,556,472 | 5,898 | # A waiter received a $5.25 tip. If the check totaled$25, what percentage was received?
Nov 14, 2016
This was a 21% tip.
#### Explanation:
The percentage of the tip, let's call it $p$, is calculated by dividing the amount of the tip, let's call it $t$ by the total amount of the bill, let's call it $b$ and multiplying it by 100. The formula for this is:
$p = \frac{t}{b} \cdot 100$
Because we know the amount of the tip and the amount of the bill we can substitute these numbers and solve for $p$:
p = ($5.25)/($25.00) * 100
p = ($525.00)/($25.00)
$p = 21$ | 174 | 566 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-26 | latest | en | 0.941808 | 186 |
https://socratic.org/questions/sara-has-16-red-flowers-yellow-flowers-she-wants-bouquets-with-the-same-number-a#326726 | 1,642,362,327,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300010.26/warc/CC-MAIN-20220116180715-20220116210715-00109.warc.gz | 597,265,206 | 6,194 | # Sara has 16 red flowers and 24 yellow flowers. She wants to make bouquets with the same number of each color flower in each bouquet. What is the greatest number of bouquets she can make?
Oct 26, 2016
She will be able to make 8 bouquets:
Each bouquet will have 2 red flowers and 3 yellow flowers.
#### Explanation:
Sara obviously wants to use all the flowers so she does not any left over. She needs to find a number which divides into 16 and 24,
This is just an indirect way of using the HCF of 16 and 24, which is 8.
$16 = 2 \times 8$
$24 = 3 \times 8$
She will be able to make 8 bouquets:
Each bouquet will have 2 red flowers and 3 yellow flowers. | 180 | 660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-05 | latest | en | 0.961949 | 187 |
https://socratic.org/questions/to-get-a-b-in-math-alexandria-pappas-must-average-80-on-five-tests-scores-on-the-1 | 1,638,769,840,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.39/warc/CC-MAIN-20211206042636-20211206072636-00097.warc.gz | 598,066,583 | 6,247 | # To get a B in math, Alexandria Pappas must average 80 on five tests scores on the first four tests were 84. 77, 84, and 75. What is the lowest score that she can get on the last test and still get a B?
Oct 19, 2016
The final grade must be 83
#### Explanation:
To think through this logically, she wants to get a average of 80 over 5 tests. $5 \cdot 80$ = 400. We want the total of all the test grades to equal 400.
Sum up what you know to get a base.
$84 + 77 + 84 + 75$ = 317
now we want to know what will add to 317 to get 400.
$400 - 317$ = 83
So her final grade must be at least an 83 to get an average of 80 or higher. | 196 | 633 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-49 | latest | en | 0.95074 | 187 |
https://socratic.org/questions/how-do-you-solve-and-graph-f-6-5-and-f-4-2 | 1,580,250,354,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251783342.96/warc/CC-MAIN-20200128215526-20200129005526-00019.warc.gz | 669,249,278 | 6,212 | # How do you solve and graph f-6<5 and f-4>=2?
Mathematically, this is expressed as $6 \le f < 11$
#### Explanation:
We first need to solve each part of this problem:
$f - 6 < 5 \implies f < 11$
$f - 4 \ge 2 \implies f \ge 6$
The graph, then, will consist of a line along the $f$-axis (probably looking very much like a number line) and will be a solid dot at 6 (to indicate that 6 is part of the solution), a hollow dot at 11 (to indicate that all the points up to but not including 11 are part of the solution), and a solid line connecting the two dots.
Mathematically, this is expressed as $6 \le f < 11$ | 183 | 614 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2020-05 | longest | en | 0.947694 | 188 |
https://cracku.in/4-two-valves-a-and-b-can-fill-a-sump-in-37frac12-min-x-rrb-alp-13th-aug-2018-shift-2 | 1,656,496,960,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00150.warc.gz | 243,027,971 | 23,034 | Question 4
# Two valves A and B can fill a sump in $$37\frac{1}{2}$$ minutes and 45 minutes respectively. Both valves are opened. The sump will be filled in just 30 minutes, if valve B is turned off after:
Solution
Let's Solve with LCM method.
LCM of ( $$\frac{75}{2}$$ and 45) = 225, assume this is total unit of work.
Thus, efficiency of A = $$\frac{225}{75/2}$$= 6 units/min and B = $$\frac{225}{45}$$ = 5 units/min.
Since A works for 30 mins, he will finish = 6 x 30 = 180 units.
Remaining Work = 225 - 180Â
= 45 units to be completed by B in = $$\frac{45}{5}$$ = 9 mins. | 195 | 583 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-27 | latest | en | 0.905371 | 188 |
https://socratic.org/questions/a-designer-needs-to-create-perfectly-circular-necklaces-the-necklaces-each-need- | 1,566,607,902,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319155.91/warc/CC-MAIN-20190823235136-20190824021136-00432.warc.gz | 636,029,117 | 5,851 | A designer needs to create perfectly circular necklaces. The necklaces each need to have a radius of 10 cm. What is the largest number of necklaces that can be made from 1000 cm of wire?
May 30, 2017
The designer can make a maximum of $15$ necklaces.
Explanation:
We can find this out by dividing the total amount of wire by the circumference of one necklace.
$C = 2 \pi R$
$C = 2 \times 10 \pi$
$C = 20 \pi$
$C = 62.832 \text{cm}$
Now we can find out the amount of necklaces that the designer can make.
1000 ÷ 62.832 = 15.915
$\therefore$ The designer can make a maximum of $15$ necklaces. | 176 | 601 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2019-35 | latest | en | 0.913558 | 188 |
https://socratic.org/questions/3-out-of-8-students-are-athletes-if-there-are-1280-students-at-the-school-how-ma | 1,596,900,444,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737883.59/warc/CC-MAIN-20200808135620-20200808165620-00309.warc.gz | 502,170,154 | 5,958 | # 3 out of 8 students are athletes. If there are 1280 students at the school, how many are not athletes?
Oct 21, 2016
$800$ students at the school are not athletes.
#### Explanation:
If 3 out of 8 students are athletes ($\frac{3}{8} \times s t u \mathrm{de} n t s$), it stands to reason that 5 out of 8 students are not athletes ($\frac{5}{8} \times s t u \mathrm{de} n t s$).
If we represent the number of students who are not athletes by $x$ and we know the total number of students to be $1280$, we can write an equation:
$x = \frac{5}{8} \times 1280$
$x = 5 \times 160$
$\therefore x = 800$ | 189 | 602 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-34 | latest | en | 0.902302 | 189 |
https://socratic.org/questions/how-do-you-simplify-sqrt-49 | 1,575,654,814,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540490743.16/warc/CC-MAIN-20191206173152-20191206201152-00398.warc.gz | 561,036,317 | 6,095 | # How do you simplify sqrt(49)?
May 14, 2018
$\sqrt{49} = 7$
#### Explanation:
As $7 \cdot 7 = 49$ it follows that 7 is the square root of 49.
Therefore $\sqrt{49} = 7$
Since we don't have $\pm$ in front of the square root sign, we are only asked for the positive value.
May 14, 2018
$\sqrt{49} = 7$
#### Explanation:
$\sqrt{49}$
The radical symbol over a number means to take the square root of that number, or to find the number that can be multiplied by itself to get the number under the radical.
In this case it is $7$, since $7 \times 7 = 49$
The negative root is not asked for in this case. | 180 | 609 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-51 | latest | en | 0.902375 | 190 |
https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-2-review-exercises-page-204/46 | 1,571,700,717,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987795253.70/warc/CC-MAIN-20191021221245-20191022004745-00406.warc.gz | 913,819,612 | 12,343 | ## Introductory Algebra for College Students (7th Edition)
The number is $10$.
First, let's translate this statement into an algebraic equation we can solve. Let's first break it into its component parts. "$6$ times a number" is $6x$. "Decreased by $20$" means to subtract $20$. "Is $4$ times the number" means $= 4x$. Now, let's put the components together to make an equation: $$6x - 20 = 4x$$ Add $20$ to each side to move the constant to one side of the equation: $$6x = 4x + 20$$ Subtract $4x$ from each side to isolate variables to the other side of the equation: $$2x = 20$$ Divide both sides by $2$ to solve for $x$: $$x = 10$$ The number is $10$. | 191 | 656 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-43 | latest | en | 0.922557 | 191 |
https://socratic.org/questions/in-math-class-60-percent-of-the-students-are-males-there-are-30-students-in-the- | 1,586,374,471,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371821680.80/warc/CC-MAIN-20200408170717-20200408201217-00276.warc.gz | 682,177,663 | 6,113 | # In math class, 60 percent of the students are males. There are 30 students in the class. How many students are males?
Dec 12, 2016
18 students are males.
#### Explanation:
This question can be rewritten as: What is 60% of 30?
"Percent" or "%" means "out of 100" or "per 100", Therefore 60% can be written as $\frac{60}{100}$.
When dealing with percents the word "of" means "times" or "to multiply".
Finally, lets call the number we are looking for "n".
Putting this altogether we can write this equation and solve for $n$ while keeping the equation balanced:
$n = \frac{60}{100} \times 30$
$n = \frac{1800}{100}$
$n = 18$ | 178 | 634 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2020-16 | longest | en | 0.905788 | 191 |
https://socratic.org/questions/how-many-ways-can-you-arrange-the-letters-in-the-word-factor | 1,713,571,303,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817463.60/warc/CC-MAIN-20240419234422-20240420024422-00845.warc.gz | 471,118,460 | 6,008 | # How many ways can you arrange the letters in the word factor?
May 24, 2018
6! = 720 ways.
#### Explanation:
There are 6 choices for which letter goes first.
There are 5 choices left for which letter goes second.
There are 4 choices left for which letter goes third.
...
There is 1 choice left for which letter goes sixth.
The total number of ways to arrange all these letters in a row is thr product of all these numbers of choices.
$6 \times 5 \times 4 \times 3 \times 2 \times 1$
$= 720$
This number can also be written as 6!
### Note:
This works because all the letters in "factor" are unique. If there are duplicates, we would need to divide our answer by the number of duplicate words created due to each duplicated letter. | 183 | 739 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-18 | latest | en | 0.930776 | 191 |
https://socratic.org/questions/what-is-2x-7-18 | 1,638,433,090,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00503.warc.gz | 588,514,175 | 6,067 | # What is 2x+7=18?
Jun 8, 2018
$x = 5.5$
#### Explanation:
First, you subtract $7$ from both sides. You are left with
$2 x = 11$
Then, you divide $2$ from both sides and you get your answer. The answer is
$x = 5.5$
Jun 9, 2018
$x = \frac{11}{2}$
#### Explanation:
We want to solve for $x$ here. Let's start by subtracting $7$ from both sides. Doing this, we now have
$2 x = 11$
Lastly, dividing both sides by $2$, we get
$x = \frac{11}{2}$, which can be alternatively written as $5.5$.
Hope this helps! | 176 | 517 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-49 | latest | en | 0.933722 | 192 |
https://socratic.org/questions/michelle-scored-48-of-their-team-s-points-their-team-had-a-total-of-50-points-ho | 1,718,994,554,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862132.50/warc/CC-MAIN-20240621160500-20240621190500-00307.warc.gz | 474,727,529 | 5,981 | # Michelle scored 48% of their team's points their team had a total of 50 points, how many points did she score?
Nov 18, 2016
Michelle scored 24 points.
#### Explanation:
Another way to state this problem is what is 48% of 50.
"Percent" or "%" means "out of 100" or "per 100", Therefore 48% can be written as $\frac{48}{100}$.
When dealing with percents the word "of" means "times" or "to multiply".
Finally, lets call the number we are looking for "n".
Putting this altogether we can write this equation and solve for $n$ while keeping the equation balanced:
$n = \frac{48}{100} \times 50$
$n = \frac{2400}{100}$
$n = 24$ | 178 | 633 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-26 | latest | en | 0.944869 | 192 |
https://socratic.org/questions/a-piece-of-cardboard-is-20-m-long-and-9-m-wide-joyce-cuts-square-pieces-from-it-#609528 | 1,716,804,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00777.warc.gz | 461,774,494 | 5,869 | A piece of cardboard is 20 m long and 9 m wide. Joyce cuts square pieces from it. If each square piece has a side of 2 m, how many pieces would Joyce get?
May 8, 2018
40 squares
Explanation:
The length is 20 m so divide 20 by 2 to see how many squares you can fit along its length.
$\frac{20}{2} = 10$
The width is 9 m so if you divide 9 by 2 this will tell you how many squares across the width.
$\frac{9}{2} = 4.5$
You can fit 10 squares along its length and 4 squares across its width. So if you can fit 10 rows, 4 times that would be 40 squares
$10 \times 4 = 40$
There would be a length 1 m by 10 m of waste.
Hope this is clearer. | 196 | 644 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-22 | latest | en | 0.95651 | 192 |
https://unitconverter.io/months/years/5425 | 1,716,699,989,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058868.12/warc/CC-MAIN-20240526043700-20240526073700-00328.warc.gz | 500,390,706 | 4,501 | # 5,425 months to years
to
5,425 Months = 452.08 Years = 452 Years and 1 Month
# How to Convert 5,425 Months to Years
Let's understand how to change 5,425 months into years.
## 1. Convert Months to Years
To change months to years, divide the number of months by 12 (since there are 12 months in 1 year).
In this case: 5,425 months ÷ 12 = 452 years
## 2. Determine Remaining Months
The remainder of this division represents the months that don't make up a full year. We use this remainder to find out the remaining months.
In this case: The remainder when dividing 5,425 by 12 is 1 month.
## 3. Final Answer
Therefore, 5,425 months is equal to 452 years and 1 month. | 191 | 677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-22 | latest | en | 0.910455 | 192 |
https://englishlangkan.com/question/mr-osei-and-his-wife-are-both-teachers-and-earn-520-and-450-per-month-respectively-find-the-rati-12918961-39/ | 1,631,975,158,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056476.66/warc/CC-MAIN-20210918123546-20210918153546-00718.warc.gz | 297,192,760 | 14,599 | ## Mr.Osei and his wife are both teachers and earn \$520 and \$450 per month respectively.Find the ratio of Mr.Osei’s icome to that of his wife
Question
Mr.Osei and his wife are both teachers and earn \$520 and \$450 per month respectively.Find the ratio of Mr.Osei’s icome to that of his wife.
in progress 0
4 days 2021-09-15T00:51:51+00:00 1 Answer 0
\$52 to \$45
Step-by-step explanation:
First put their income into a ratio:
520 to 450
Now you can put it into a fraction and then simplify it:
520/450
Divide each by 10
520/10 = 52
450/10 = 45
So 52/45 | 172 | 569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2021-39 | latest | en | 0.960265 | 192 |
https://www.tutorialspoint.com/sides-of-two-similar-triangles-are-in-the-ratio-4-9-areas-of-these-triangles-are-in-the-ratio-a-2-3-b-4-9-c-81-16-d-16-81 | 1,685,840,681,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00517.warc.gz | 1,133,639,908 | 11,345 | # Sides of two similar triangles are in the ratio 4:9. Areas of these triangles are in the ratio(a) 2:3(b) 4:9(c) 81:16(d) 16:81
Given:
Sides of two similar triangles are in the ratio 4:9.
To do:
We have to find the ratio of the areas of these triangles.
Solution:
We know that,
If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
Therefore,
Ratio of the areas of triangles $=(\frac{4}{9})^{2}$
$=\frac{16}{81}$
The ratio of the areas of the given triangles is $16:81$.
Tutorialspoint
Simply Easy Learning | 172 | 615 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-23 | latest | en | 0.902968 | 192 |
http://answers.everydaycalculation.com/compare-fractions/3-5-and-9-8 | 1,529,653,080,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864364.38/warc/CC-MAIN-20180622065204-20180622085204-00312.warc.gz | 19,618,786 | 2,788 | Solutions by everydaycalculation.com
## Compare 3/5 and 9/8
1st number: 3/5, 2nd number: 1 1/8
3/5 is smaller than 9/8
#### Steps for comparing fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 5 and 8 is 40
2. For the 1st fraction, since 5 × 8 = 40,
3/5 = 3 × 8/5 × 8 = 24/40
3. Likewise, for the 2nd fraction, since 8 × 5 = 40,
9/8 = 9 × 5/8 × 5 = 45/40
4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction
5. 24/40 < 45/40 or 3/5 < 9/8
and | 212 | 538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2018-26 | latest | en | 0.908552 | 194 |
https://socratic.org/questions/how-do-you-find-the-slope-and-y-intercept-for-the-line-y-7 | 1,723,457,432,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00409.warc.gz | 425,972,781 | 6,045 | # How do you find the slope and y-intercept for the line y=-7?
Sep 23, 2015
For the equation $y = - 7$
- the slope is $0$
- the y-intercept is $- 7$
#### Explanation:
The slope is defined as the change in $y$ for a corresponding change in $x$
Since the value of $y$ does not change (i.e. its change is $0$) no matter by how much $x$ changes,
the slope is $0$
The y-intercept is the value of $y$ when $x = 0$.
$y = - 7$ for all values of $x$
and therefore the y-intercept is $- 7$
Note that the equation $y = - 7$ is a vertical line, 7 units below the X-axis. | 183 | 564 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-33 | latest | en | 0.913745 | 194 |
https://socratic.org/questions/in-a-certain-area-there-are-35-houses-to-55-businesses-how-do-you-write-the-rati | 1,601,204,169,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400274441.60/warc/CC-MAIN-20200927085848-20200927115848-00706.warc.gz | 611,458,987 | 6,065 | # In a certain area, there are 35 houses to 55 businesses. How do you write the ratio of houses to businesses as a fraction in simplest form. Then explain its meaning?
Nov 4, 2016
There are $\frac{7}{11}$ times as many houses as businesses.
#### Explanation:
$\text{houses":"businesses} = 35 : 55$
This ratio can also be written in fractional form as:
$\textcolor{w h i t e}{\text{XXX}} \frac{35}{55}$
which can be simplified (by dividing each of the numerator and denominator by $5$)
$\textcolor{w h i t e}{\text{XXX}} = \frac{7}{11}$
If you divided the businesses up into (5 groups of) $11$ and then evenly assigned houses among those groups, each group would have $7$ houses. | 190 | 684 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-40 | latest | en | 0.947383 | 195 |
https://socratic.org/questions/58afcbd8b72cff5ee8748b31 | 1,544,929,693,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827252.87/warc/CC-MAIN-20181216025802-20181216051802-00252.warc.gz | 722,608,355 | 6,234 | # A class has 50 women and 82 men. If a student is picked at random, what's the probability of picking a woman?
$\frac{50}{132} = \frac{25}{66}$
#### Explanation:
There is a class of 50 women and 82 men, so 132 students all told. If we pick one person at random, what is the probability that person will be a woman?
Probability is a ratio, with the numerator being the number of ways we can satisfy a condition (or set of conditions) and the denominator being the number of ways we can get results.
In this case, there are 50 picks that will satisfy the condition of the pick being a woman and 132 total picks, so we have:
$\frac{50}{132}$
which can be reduced down:
$\frac{50}{132} = \frac{25}{66}$ | 184 | 707 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2018-51 | longest | en | 0.925719 | 195 |
https://www.esaral.com/q/sonal-attended-her-school-on-204-days-in-a-full-year-64695 | 1,719,266,123,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00209.warc.gz | 666,863,784 | 11,423 | # Sonal attended her school on 204 days in a full year.
Question:
Sonal attended her school on 204 days in a full year. If her attendance is 85%, find the number of days on which the school was opened.
Solution:
Let $x$ be the number of days the school was opened.
Number of days Sonal attended school $=204$ days
Percentage of her attendance $=85 \%$ of $\mathrm{x}$
$=\left(\mathrm{x} \times \frac{85}{100}\right)$
$=\frac{85 \mathrm{x}}{100}$
Now, $\frac{85 x}{100}=204$
$\Rightarrow x=\left(204 \times \frac{100}{85}\right)$
$\Rightarrow x=240$
$\therefore$ The school was opened for 240 day. | 177 | 608 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.915507 | 195 |
brilliant-team.com | 1,713,461,103,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817222.1/warc/CC-MAIN-20240418160034-20240418190034-00797.warc.gz | 131,653,170 | 6,852 | Q:
# tire rim has a diameter of 15 in. What is the circumference of the tire rim? Use 3.14 for pi
Accepted Solution
A:
The meaning of circumference is to find the distance around a circle or in other words the perimeter of the circle. The circumference of a circle can be calculate by using the formula C=2πr or C=πd.
In this exercise is given that diameter of a tire rim is 15 inches, and it is asked to find its circumference using 3.14 as the value of pi. In this problem the diameter of the circle is given, so in this case the formula C=πd would be used, where d represents the diameter.
$$C= \pi d$$
$$C=(3.14)(15 in)$$
$$C=47.1 in$$
The circumference of the tire rim is 47.1 inches. | 189 | 694 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-18 | latest | en | 0.90135 | 195 |
https://socratic.org/questions/what-are-the-dimensions-of-a-garden-if-anne-wants-to-put-a-fence-around-her-rose#115599 | 1,713,349,762,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817146.37/warc/CC-MAIN-20240417075330-20240417105330-00586.warc.gz | 495,157,576 | 6,183 | # What are the dimensions of a garden if Anne wants to put a fence around her rose bed that is one-and-a-half times as long as it is wide and uses 50 feet of fencing?
Dec 25, 2014
Let's start by calling the smaller side $x$.
The longer side will then be $1.5 x$
Total length of fence is then two times the smaller side plus two times the longer side, and this must be 50. Or in formula:
$2 \cdot x + 2 \cdot 1.5 x = 5 x$
$5 x = 50$
divide both sides by 5:
$x = 10$
Conclusion:
Short side is 10 feet
Long side is 15 feet
Challenge
Try and work this out if one of the short sides of the rose bed is against the house, so it doesn't need a fence there. | 188 | 654 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-18 | latest | en | 0.95698 | 196 |
https://socratic.org/questions/a-shoe-store-costs-1800-dollars-a-month-to-operate-the-average-wholesale-cost-of#578028 | 1,642,729,909,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00659.warc.gz | 601,079,335 | 6,329 | # A shoe store costs $1800 dollars a month to operate. The average wholesale cost of each pair of shoes is$25, and the average price of each pair of shoes is $65. How many pairs of shoes must the store sell each month to break even? ##### 1 Answer Mar 20, 2018 The store has to sell $45$pairs of shoes. #### Explanation: The store has a base cost of $1800, the cost per pair of shoes is $25. Each pair of shoes is sold for $65, therefore the profit per pair of shoes is
$65 -$25 = $40 The formula for calculating the amount that needs to be sold would look like this; $40 x = 1800$To determine the value of $x$, we take this formula; $x = \frac{1800}{40}$$x = 45$Therefore, the store needs to sell $45\$ pairs of shoes to break even. | 200 | 734 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-05 | latest | en | 0.917689 | 196 |
https://socratic.org/questions/how-do-you-evaluate-p-2-5-when-p-14 | 1,575,772,668,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540504338.31/warc/CC-MAIN-20191208021121-20191208045121-00495.warc.gz | 563,128,373 | 6,011 | # How do you evaluate p/2-5 when p=14?
Nov 23, 2016
2
#### Explanation:
Begin by substituting $14$ in for $p$ because that is its value:
$\frac{14}{2} - 5$
P.E.M.D.A.S
Parenthesis, Exponents, Multiply, Divide, Add, Subtract
Skip the first 3 steps (P.E.M.) because that doesn't apply in this problem
So, the first step in this problem is to Divide
*Remember that every fraction is just division
$\frac{14}{2} = 7$
So, the problem now looks like:
$7 - 5$
The next step would be to Add, but that doesn't apply, so subtract
$7 - 5 = 2$ | 171 | 546 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2019-51 | latest | en | 0.907554 | 196 |
https://socratic.org/questions/a-total-of-30-people-made-cake-for-a-bake-sale-each-person-made-2-cakes-mrs-lari-1 | 1,582,850,196,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146907.86/warc/CC-MAIN-20200227221724-20200228011724-00157.warc.gz | 498,607,496 | 5,969 | # A total of 30 people made cake for a bake sale .Each person made 2 cakes Mrs.Larios found 20% of these cakes were chocolate.How many chocolate cakes were made for the sale?
Jun 3, 2018
12 of the cakes were chokolate.
#### Explanation:
As 30 people all made 2 cakes each, there were
$2 \cdot 30 = 60$ cakes in all.
To find out 20% of this the easiest is to change it into a fraction this time, as 20%=20/100=1/5
20% of the 60 cakes, therefore, is
$\frac{60}{5} = 12$ cakes
(you could also take:
20% of $60 = \frac{20}{100}$ of $60 = \frac{20}{100} \cdot 60 = 12$
Conclusion: 12 of the cakes were chokolate. | 198 | 613 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-10 | longest | en | 0.986166 | 197 |
https://unitconverter.io/months/years/9903 | 1,709,496,280,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00033.warc.gz | 582,047,646 | 4,535 | # 9,903 months to years
to
9,903 Months = 825.25 Years = 825 Years and 3 Months
# How to Convert 9,903 Months to Years
Let's understand how to change 9,903 months into years.
## 1. Convert Months to Years
To change months to years, divide the number of months by 12 (since there are 12 months in 1 year).
In this case: 9,903 months ÷ 12 = 825 years
## 2. Determine Remaining Months
The remainder of this division represents the months that don't make up a full year. We use this remainder to find out the remaining months.
In this case: The remainder when dividing 9,903 by 12 is 3 months.
## 3. Final Answer
Therefore, 9,903 months is equal to 825 years and 3 months. | 191 | 680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-10 | latest | en | 0.915833 | 197 |
https://braingenie.ck12.org/skills/102192/learn | 1,590,372,183,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347387155.10/warc/CC-MAIN-20200525001747-20200525031747-00321.warc.gz | 288,398,315 | 1,834 | ### Sample Problem
There were a certain number of students in both Class A and Class B. After 10% of the students in Class A were transferred into Class B, the number of students in Class B was increased by 20%. If there are 30 students in Class B now, how many students were originally in Class A?
students
#### Solution
For Class B
The number of students was increased by 20% to 30, then its original number of students is 100% and the final number is 100% + 20% = 120%.
Use the formula: Whole = Part ÷ Percent
The original number of students in Class B: 30 ÷ 120% = 25
Thus, the transferred students: 30 – 25 = 5
For Class A,
10% of its original students is 5.
Thus, the original number of students in Class A: 5 ÷ 10% = 50 | 199 | 737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-24 | latest | en | 0.984166 | 197 |
https://socratic.org/questions/what-percent-is-24-of-96 | 1,544,913,176,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00205.warc.gz | 742,152,544 | 5,832 | # What percent is 24 of 96?
24/96=0.25=25%
#### Explanation:
We know that 50% of a number is the same as $\frac{1}{2}$ of the number. And so what is $\frac{1}{2}$ of 96?
• $\frac{1}{2} \times 96 = 48$, which is the same as
• $0.5 \times 96 = 48$, which is the same as
• 50%xx96=48
Let's work with that last one for a minute. We can divide both sides by 96 to get:
50%=48/96
What we want to know is What percent of 96 is 24? Let's put the 24 in for the 48, and then do $\frac{24}{96}$ to find the percentage:
24/96=0.25=25% | 193 | 532 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2018-51 | latest | en | 0.948415 | 197 |
https://prep4gmat.com/zh-hans/qod-problem-solving-screws-and-nails-2/ | 1,685,457,004,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645810.57/warc/CC-MAIN-20230530131531-20230530161531-00653.warc.gz | 520,999,012 | 9,792 | Problem Solving — Screws and Nails
Explanation
Let S be the number of screws, N the number of nails, and H the number of hooks that Joseph bought.
Simplifying by using cents instead of dollars, we have:
(a) 1N + 2H + 3S = 32
We also know the following:
(b) N + H + S = 18
(c) N = 2S
Substituting (c) into both (a) and (b) gives us:
(a) 5S + 2H = 32
(b) 3S + H = 18
To solve the above equations, we can multiply (b) by 2 and subtract (a):
6S + 2H - 5S - 2H = 36 - 32
Thus, S = 4, N = 8, and H = 6.
The correct choice is B. | 199 | 527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-23 | longest | en | 0.919862 | 197 |
https://socratic.org/questions/how-do-you-solve-4n-12-12-4n | 1,576,471,536,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541317967.94/warc/CC-MAIN-20191216041840-20191216065840-00111.warc.gz | 542,177,822 | 6,054 | # How do you solve 4n - 12 = 12 - 4n?
Jul 8, 2016
$n = 3$
#### Explanation:
$4 n - 12 = 12 - 4 n$
This is our given equation. Add $4 n$ to both sides
$8 n - 12 = 12$
We end up with all variable terms on one side. Now add $12$ to both sides to get the variable term by itself on that side
$8 n = 24$
Now all we have to do is divide out the eight from both sides to get a solution for $n$.
$n = 3$
You can check your work (as always) and plug in $3$ for all variables $n$. You end up with $0 = 0$, which is true, which also indicates we did our work correctly and $3$ is a correct solution. | 192 | 599 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-51 | longest | en | 0.945259 | 198 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-2-solving-equations-2-1-solving-one-step-equations-practice-and-problem-solving-exercises-page-85/10 | 1,721,715,146,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518014.29/warc/CC-MAIN-20240723041947-20240723071947-00170.warc.gz | 678,207,112 | 15,177 | ## Algebra 1
$x = 4$
$6 = x + 2$ In order to solve this algebraic expression, we need to join the constants together on the left hand side. We do not move the coefficient of the variable since we have only one unknown. To calculate this, we subtract 2 from both sides of the equation in order to eliminate the constants from the right hand side as shown: $6 - 2 = x + 2 - 2$ Simplify to get: $4 = x$ Rearranging it, we have: $x = 4$ This means that the expression has a possible solution at $x = 4$ If we need to test the answer, we replace for x with 4 in the original equation as below: $6 = x + 2$ $6 = 4 + 2$ Simplify to get: $6 = 6$ 6 is equal to 6. This shows that the solution $x = 4$ is also correct. | 212 | 709 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-30 | latest | en | 0.91703 | 198 |
https://api-project-1022638073839.appspot.com/questions/what-is-the-sale-price-on-a-150-item-that-is-on-sale-for-25-off | 1,586,503,276,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371886991.92/warc/CC-MAIN-20200410043735-20200410074235-00143.warc.gz | 337,969,015 | 6,416 | #### Explanation:
So the original price is $150 and the discount is 25%, right?. So just get the original price, multiply by the discount and divide by $100$to find 25%. 150xx25%=3750/100=37.50. So now that you know the amount that you saved, just subtract $37.50$from $150$: $150 - 37.50$and you'll get $112.50 as the sale price.
Apr 7, 2018
Apr 7, 2018
$112.50 #### Explanation: Instead of finding the amount of reduction for the sale price, we can work with the percentages. If a discount of 25% is being given, it means that the sale price is 75% of the original price. 100%-25% = 75% Find 75% of $150
$\frac{75}{100} \times 150$
=\$112.50 | 201 | 649 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 22, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-16 | latest | en | 0.92044 | 198 |
https://socratic.org/questions/how-do-you-write-3x-4-3x-2-in-standard-form-1 | 1,653,779,927,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00384.warc.gz | 601,798,244 | 5,997 | # How do you write 3x=4-3x^2 in standard form?
Mar 28, 2018
$- 3 {x}^{2} - 3 x + 4 = 0$
#### Explanation:
Standard forms of polynomials are in the form $a {x}^{2} + b x + c = 0$. The key realizations are:
a) We want to set the polynomial equal to zero
b) The exponents should descend from left to right
So, let's subtract $3 x$ from both sides to set the equation equal to zero. We get:
$4 - 3 {x}^{2} - 3 x = 0$
This still isn't in standard form, since the exponents aren't in descending order. We can rewrite this as
$- 3 {x}^{2} - 3 x + 4 = 0$
Which would be our final answer in standard form. | 205 | 606 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-21 | latest | en | 0.90347 | 199 |
https://socratic.org/questions/lea-wants-to-put-a-fence-around-her-garden-her-garden-measures-14-feet-by-15-fee | 1,638,276,169,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358973.70/warc/CC-MAIN-20211130110936-20211130140936-00240.warc.gz | 582,387,619 | 6,192 | # Lea wants to put a fence around her garden. Her garden measures 14 feet by 15 feet. She has 50 feet of fencing. How many more feet of fencing does Lea need to put a fence around her garden?
Jan 19, 2017
Lea needs $8$ more feet of fencing.
#### Explanation:
Presuming the garden to be rectangular, we can find out the perimeter by the formula $P = 2 \left(l + b\right)$, where P=Perimeter, l=length and b=breadth.
$P = 2 \left(14 + 15\right)$
$P = 2 \left(29\right)$
$P = 58$
Since the perimeter is $58$ feet and Lea has $50$ feet of fencing, she will need:
$58 - 50 = 8$ more feet of fencing. | 189 | 603 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2021-49 | latest | en | 0.951377 | 199 |
https://socratic.org/questions/how-do-you-write-137-5-as-a-fraction | 1,624,529,617,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488552937.93/warc/CC-MAIN-20210624075940-20210624105940-00363.warc.gz | 472,602,491 | 5,992 | # How do you write 137.5% as a fraction?
Jan 20, 2016
$\frac{11}{8}$
#### Explanation:
We can write a percentage as a fraction by taking the percentage and dividing it by 100. In this case:
137.5% = 137.5/100
We want to simplify our fraction now, and the first step is to get rid of the decimals by multiplying both top and bottom by $2$:
$= \frac{275}{200}$
By looking at this fraction, you might notice that both $275$ and $200$ are divisible by $5$, so let's divide both by $5$
$= \frac{55}{40}$
We notice again that $55$ and $40$ are divisible by $5$, so we divide both by $5$ again
$= \frac{11}{8}$ | 188 | 614 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-25 | latest | en | 0.931126 | 199 |
https://socratic.org/questions/how-do-you-find-the-amount-of-sales-tax-for-a-board-game-that-costs-25-00-at-a-7 | 1,586,302,510,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371806302.78/warc/CC-MAIN-20200407214925-20200408005425-00337.warc.gz | 694,973,930 | 6,156 | # How do you find the amount of sales tax for a board game that costs $25.00 at a 7% tax rate? ##### 1 Answer Dec 4, 2016 7% sales tax on a board game costing$25.00 is $1.75 #### Explanation: What we are looking for in this problem is what is 7% of$25.00.
"Percent" or "%" means "out of 100" or "per 100", Therefore 7% can be written as $\frac{7}{100}$.
When dealing with percents the word "of" means "times" or "to multiply".
Finally, lets call the number we are looking for "n".
Putting this altogether we can write this equation and solve for $n$ while keeping the equation balanced:
n = 7/100 xx $25.00 n =$175.00/100
n = \$1.75 | 193 | 638 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2020-16 | longest | en | 0.937291 | 199 |
https://socratic.org/questions/5a2f281911ef6b1bee5261e4 | 1,566,608,299,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319155.91/warc/CC-MAIN-20190823235136-20190824021136-00054.warc.gz | 653,033,046 | 5,992 | # Question #261e4
Dec 12, 2017
Converting the words to an algebraic expression yields $3 x - 6$.
#### Explanation:
Let's break it down.
"6 less than . . . " means we’re subtracting $6$.
“ . . . the product of 3 and a number.” means 3 and a number ($x$) are being multiplied.
Combining both expressions yields: $3 x - 6$.
Dec 12, 2017
$3 x - 6$
#### Explanation:
The “ product of 3 and a number “ may be written as $3 x$ and “six less than the product “ can therefore be written as $3 x - 6$
Dec 12, 2017
Product of 3 and a number $= 3 \times x$
6 less $= 3 x - 6$
$3 x - 6$ | 198 | 586 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2019-35 | latest | en | 0.910786 | 199 |
https://socratic.org/questions/the-number-of-students-at-franklin-high-school-increased-from-840-to-910-over-a- | 1,722,806,823,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00593.warc.gz | 424,400,341 | 6,270 | # The number of students at Franklin High School increased from 840 to 910 over a 5-year period. What was the percent of increase?
Oct 25, 2016
The percent of increase in students is 8.33%.
#### Explanation:
The increase in numbers of students at Franklin High School over a five-year period was $910 - 840$ which is $70$.
To find the percent increase, let us consider the percent value as $x$. Since we are calculating the percent increase from $840$ and the increase is $70$, we can write the equation:
$840 \times \frac{x}{100} = 70$
Multiplying both sides by $\frac{100}{840}$, we get:
$x = 70 \times \frac{100}{840}$
$x = \frac{7000}{840}$
$x = 8.33$ | 184 | 665 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-33 | latest | en | 0.935671 | 200 |
http://thibaultlanxade.com/general/samantha-has-a-shelf-that-is-95-over-4-inches-wide-how-many-books-can-samantha-arrange-on-the-shelf-if-each-book-is-5-over-4-inches-thick-29-because-the-number-of-books-is-95-over-4-multiplied-by-5-over-4-28-because-the-number-of-books-is-95-o | 1,726,178,598,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00718.warc.gz | 28,014,920 | 7,441 | Q:
# Samantha has a shelf that is 95 over 4 inches wide. How many books can Samantha arrange on the shelf if each book is 5 over 4 inches thick? 29, because the number of books is 95 over 4 multiplied by 5 over 4 28, because the number of books is 95 over 4 divided by 4 over 5 19, because the number of books is 95 over 4 divided by 5 over 4 17, because the number of books is 95 over 4 multiplied by 4 over 5
Accepted Solution
A:
namely, how many times does 5/4 go into 95/4? well, let's just divide,
$$\bf \cfrac{\quad \frac{95}{4}\quad }{\frac{5}{4}}\implies \cfrac{95}{\underline{4}}\cdot \cfrac{\underline{4}}{5}\implies \cfrac{95}{5}\implies 19$$ | 222 | 659 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-38 | latest | en | 0.906657 | 200 |
https://www.gradesaver.com/textbooks/math/geometry/geometry-common-core-15th-edition/chapter-5-relationships-within-triangles-chapter-test-page-345/13 | 1,713,115,466,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00192.warc.gz | 754,916,220 | 13,902 | ## Geometry: Common Core (15th Edition)
$x = 7$
One angle in the two triangles is marked congruent. The adjacent angles are also congruent because they are both right angles, and right angles are congruent. We have two congruent triangles. One side is congruent because it is shared by both triangles. Therefore, these two triangles are congruent by AAS. Corresponding parts of congruent triangles are congruent; therefore, we can set the two sides whose expressions are given equal to one another: $5x - 8 = 2x + 13$ Subtract $2x$ from each side of the equation to move variables to the left side of the equation: $3x - 8 = 13$ Add $8$ to each side of the equation to move constants to the right side of the equation: $3x = 21$ Divide each side by $3$ to solve for $x$: $x = 7$ | 201 | 779 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-18 | latest | en | 0.933154 | 200 |
https://socratic.org/questions/how-do-you-graph-the-inequality-x-4 | 1,585,765,257,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505826.39/warc/CC-MAIN-20200401161832-20200401191832-00380.warc.gz | 709,706,164 | 6,248 | # How do you graph the inequality x>4?
May 9, 2017
On a number line graph:
Place an open circle at $4$ and an arrow pointing to the right.
#### Explanation:
$x$ is greater than $4$ which means that $4$ can not be part of the solution, that is why it is an open circle and not a closed circle.
$x$ is greater than $4$ means that any number larger than $4$ is part of the answer so even $4.00000000000001$ is part of the answer. so the line starts at the open circle and points to the right.
If you are using a set of axes you will have a vertical dashed line at $x = 4$ and shade the area to the right of the line. graph{x>4 [-1.813, 8.187, -2.44, 2.56]} | 192 | 660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-16 | latest | en | 0.935093 | 200 |
https://socratic.org/questions/yasmine-is-having-a-birthday-party-with-snacks-and-activities-for-her-guests-at- | 1,579,650,815,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606226.29/warc/CC-MAIN-20200121222429-20200122011429-00377.warc.gz | 677,293,758 | 6,001 | # Yasmine is having a birthday party with snacks and activities for her guests. At one table, five people are sharing three-quarters of a pizza. what equal-sized portion of the whole pizza will each of the five people receive?
Jan 11, 2017
$\frac{3}{20}$
#### Explanation:
On offer is $\frac{3}{4}$ pizza
Each person is given a slice that that is the same size as all the rest.
So the $\frac{3}{4}$ pizza is split into 5 equally sized parts.
Thus each of the 5 slices is $\frac{1}{5} \times \frac{3}{4} = \frac{1 \times 3}{5 \times 4} = \frac{3}{20} {\textcolor{w h i t e}{}}^{\text{ths}} \textcolor{w h i t e}{.}$ of a whole pizza | 195 | 637 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2020-05 | longest | en | 0.902087 | 200 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-2-solving-equations-2-1-solving-one-step-equations-practice-and-problem-solving-exercises-page-86/67 | 1,534,578,345,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213405.46/warc/CC-MAIN-20180818060150-20180818080150-00063.warc.gz | 857,947,136 | 15,033 | ## Algebra 1
In order to solve this algebraic expression, we must isolate the variable. A variable is the letter in the problem (for instance, in the equation 2x=10, x is the variable). The section header tells us to solve the equation using “multiplication or division,” so we know that the only things we should do to solve this problem are multiplication and division. Because multiplication and division are inverse operations (they cancel each other out), we know we must divide when a number is being multiplied by the variable and multiply when the variable is being divided by a number. In the equation -2 ½ =t/10, we must get rid of the 10, which the variable is being divided by, in order to solve. Thus, we multiply both sides by 10 to find that t=-25. We plug in -25 for t in the equation, and we get $-2.5=-25/10$. This is true, so the solution checks. | 205 | 866 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2018-34 | longest | en | 0.938087 | 200 |
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