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2 years ago

10

A book sold 33,100 copies in its first month of release. Suppose this represents 7.8% of the number of copies sold to date. How

many copies have been sold to date?
Round your answer to the nearest whole number.

1 answer:

matrenka [14]2 years ago

3 0

33100 = 7.8% of total
33100 = .078 x total (change from percent to decimal)
33100/.078 = total (divide both sides by decimal)

424358.9 = total

Round to 424,359 copies.

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

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

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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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2 years ago