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3 years ago

5

Half of the students in Max's class volunteer at the local community center. Fifteen students do not volunteer. If there are 12

boys in Max's class,how many girls are in his class?

2 answers:

alexdok [17]3 years ago

7 0

Answer: X=18

Step-by-step explanation:

let the numbers of girls be X

total in class = X + 12

number of students who do not volunteer

= (x + 12)/2

but this is 15

(x+12)/2 = 15

x+12 = 30

x = 18

So there are 18 girls in the class.

<h3>I hope this helped :) </h3>

xxTIMURxx [149]3 years ago

5 0

18
because 15 * 2 = 30; 30 - 12=18

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$\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}$

$\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}$

$\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}$

$\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}$

$\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1$

$\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}$

$\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x$

Step-by-step explanation:

Required

Simplify

Solving (1):

$\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}$

Factorize the numerator and the denominator

$\frac{x^2(x + 2) -9(x+2)}{x(x^2-x-6)}$

Factor out x+2 at the numerator

$\frac{(x^2 -9)(x+2)}{x(x^2-x-6)}$

Express x^2 - 9 as difference of two squares

$\frac{(x^2 -3^2)(x+2)}{x(x^2-x-6)}$

$\frac{(x -3)(x+3)(x+2)}{x(x^2-x-6)}$

Expand the denominator

$\frac{(x -3)(x+3)(x+2)}{x(x^2-3x+2x-6)}$

Factorize

$\frac{(x -3)(x+3)(x+2)}{x(x(x-3)+2(x-3))}$

$\frac{(x -3)(x+3)(x+2)}{x(x+2)(x-3)}$

Cancel out same factors

$\frac{(x+3)}{x}$

Hence:

$\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}$

Solving (2):

$\frac{3x^2 - 5x - 2}{x^3 - 2x^2}$

Expand the numerator and factorize the denominator

$\frac{3x^2 - 6x + x - 2}{x^2(x- 2)}$

Factorize the numerator

$\frac{3x(x - 2) + 1(x - 2)}{x^2(x- 2)}$

Factor out x - 2

$\frac{(3x + 1)(x - 2)}{x^2(x- 2)}$

Cancel out x - 2

$\frac{3x + 1}{x^2}$

Hence:

$\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}$

Solving (3):

$\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}$

Express x^2 - 9 as difference of two squares

$\frac{6 - 2x}{x^2 - 3^2} * \frac{15 + 5x}{4x}$

Factorize all:

$\frac{2(3 - x)}{(x- 3)(x+3)} * \frac{5(3 + x)}{2(2x)}$

Cancel out x + 3 and 3 + x

$\frac{2(3 - x)}{(x- 3)} * \frac{5}{2(2x)}$

$\frac{3 - x}{x- 3} * \frac{5}{2x}$

Express $3 - x$ as $-(x - 3)$

$\frac{-(x-3)}{x- 3} * \frac{5}{2x}\\$

$-1 * \frac{5}{2x}$

$-\frac{5}{2x}$

Hence:

$\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}$

Solving (4):

$\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x}$

Expand x^2 - 6x + 9 and factorize 5x - 15

$\frac{x^2 -3x -3x+ 9}{5(x - 3)} / \frac{5}{3-x}$

Factorize

$\frac{x(x -3) -3(x-3)}{5(x - 3)} / \frac{5}{3-x}$

$\frac{(x -3)(x-3)}{5(x - 3)} / \frac{5}{3-x}$

Cancel out x - 3

$\frac{(x -3)}{5} / \frac{5}{3-x}$

Change / to *

$\frac{(x -3)}{5} * \frac{3-x}{5}$

Express $3 - x$ as $-(x - 3)$

$\frac{(x -3)}{5} * \frac{-(x-3)}{5}$

$\frac{-(x-3)(x -3)}{5*5}$

$\frac{-(x-3)^2}{25}$

Hence:

$\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}$

Solving (5):

$\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}$

Factorize the numerator and expand the denominator

$\frac{x^2(x - 1) -1(x - 1)}{x^2 - x-x+1}$

Factor out x - 1 at the numerator and factorize the denominator

$\frac{(x^2 - 1)(x - 1)}{x(x -1)- 1(x-1)}$

Express x^2 - 1 as difference of two squares and factor out x - 1 at the denominator

$\frac{(x +1)(x-1)(x - 1)}{(x -1)(x-1)}$

$x +1$

Hence:

$\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1$

Solving (6):

$\frac{9x^2 + 3x}{6x^2}$

Factorize:

$\frac{3x(3x + 1)}{3x(2x)}$

Divide by 3x

$\frac{3x + 1}{2x}$

Hence:

$\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}$

Solving (7):

$\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x}$

Change / to *

$\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}$

Expand

$\frac{x^2-2x-x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}$

Factorize

$\frac{x(x-2)-1(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}$

$\frac{(x-1)(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}$

Cancel out x - 2 and x - 1

$\frac{1}{4x} * \frac{12x^2}{x} * \frac{x}{1}$

Cancel out x

$\frac{1}{4x} * \frac{12x^2}{1} * \frac{1}{1}$

$\frac{12x^2}{4x}$

$3x$

Hence:

$\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x$

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