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4 years ago

Please answer this correctly without making mistakes

Mathematics

1 answer:

IrinaVladis [17]4 years ago

5 0

Answer:

44 5/6 "the second one"

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Multiplying and dividing radical expressions and leaving them in factored form. I am trying to find the best and easy way to get

natka813 [3]

Answer:

$\frac{3x + 8}{36 - 2x} / \frac{27x^2 + 72x}{3x^2 - 27} = \frac{(x - 3)(x+3)}{6x(18 - x)}$

Step-by-step explanation:

Given

$\frac{3x + 8}{36 - 2x} / \frac{27x^2 + 72x}{3x^2 - 27}$

Required

Solve

Change / to *

$\frac{3x + 8}{36 - 2x} * \frac{3x^2 - 27}{27x^2 + 72x}$

Factor out 3

$\frac{3x + 8}{36 - 2x} * \frac{3(x^2 - 9)}{3(9x^2 + 24x)}$

$\frac{3x + 8}{36 - 2x} * \frac{(x^2 - 9)}{(9x^2 + 24x)}$

Factorize:

$\frac{3x + 8}{36 - 2x} * \frac{(x^2 - 9)}{3x(3x + 8)}$

Cancel out 3x + 8

$\frac{1}{36 - 2x} * \frac{(x^2 - 9)}{3x}$

Factorize:

$\frac{1}{2(18 - x)} * \frac{(x^2 - 9)}{3x}$

Combine

$\frac{x^2 - 9}{2*3x(18 - x)}$

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

Express the numerator as a difference of two squares

$\frac{(x - 3)(x+3)}{6x(18 - x)}$

Hence:

$\frac{3x + 8}{36 - 2x} / \frac{27x^2 + 72x}{3x^2 - 27} = \frac{(x - 3)(x+3)}{6x(18 - x)}$

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Answer:

Step-by-step explanation:

at least are key words for $\geq$ symbol

Area is L x W and since you have W its basically:

$33x\geq 165$

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