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3 years ago

Factor completely 8y^3 + 16y^2

Mathematics

2 answers:

Vlad1618 [11]3 years ago

3 0

Answer:

8y^2 (y+2)

Step-by-step explanation:

8y^3 + 16y^2

both terms contain an 8 and a y^2 so we can factor them out

8y^2 (y+2)

arlik [135]3 years ago

3 0

$8y^3=8y^2(y)\\\\16y^2=8y^2(2)\\\\8y^3+16y^2=8y^2(y)+8y^2(2)=8y^2(y+2)$

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Answer:

the price before taxes was $18,229.13

Step-by-step explanation:

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3 years ago

1 1/5 divided by 3/10

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3 years ago

What is the difference?

Serhud [2]

Answer:

The option $\frac{(x+5)(x+2)}{x^3-9x}$ is correct

The difference of the given expression is

$\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})=\frac{(x+5)(x+2)}{x^3-9x}$

Step-by-step explanation:

Given expression is $\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})$

To find the difference of the given expression as below :

$\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})$

$=\frac{2x+5}{x(x-3)}-(\frac{3x+5}{x(x^2-9)})-({\frac{x+1}{x^2-9})$

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

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

( using the formula $a^2-b^2=(a+b)(a-b)$ )

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

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$=\frac{x^2+7x+10}{x(x^2-9)}$ ( by factoring the quadratic polynomial )

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

Therefore $\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})=\frac{(x+5)(x+2)}{x^3-9x}$

Therefore the difference of the given expression is

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

Therefore option $\frac{(x+5)(x+2)}{x^3-9x}$ is correct

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3 years ago

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Answer:

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Step-by-step explanation:

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