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3 years ago

15

What is the prime factorization for 160 using exponents?

2 answers:

Fynjy0 [20]3 years ago

8 0

The prime factorization of 160 = 25•5.

bija089 [108]3 years ago

3 0

Answer:

$2^5*5$

Step-by-step explanation:

<u>Step 1: Prime Factorization</u>

<u /> $160$ → $2 * 2 * 2 * 2 * 2 * 5$

$2 * 2 * 2 * 2 * 2 * 5$ → $2^5*5$

Answer: $2^5*5$

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2 years ago

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Answer:

<u><em>Part A:</em></u> D. $y = (x + 1)(x+3)\\$

<u><em>Part B:</em></u> C. $3x(x-3) = 0\\$

Step-by-step explanation:

For part A) we just have to plug in 0 for x and solve for y until we find the equation that says 3 is the value for y when x is 0. For purposes of speeding up the process the correct answer is D. I will show how to check for it now.

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Now plug in 0 for x.

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Now plug in x for 0 and solve:

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3 years ago

Can you show me the steps to factor out 8x^3 - 27y^3?

Travka [436]

$(2x-3y)(4x^2+6xy+9y^2)$ is the factored form of $8x^3-27y^3$

<em><u>Solution:</u></em>

We have to factor the given expression

$8x^3-27y^3$

We can use the algebraic identity to factor the given expression

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Lets make the given expression in form of $a^3-b^3$

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Now apply the algebraic property in above expression

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Therefore,

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$(2x)^3-(3y)^3=(2x-3y)(4x^2+6xy+9y^2)$

Thus the given expression is factored out

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3 years ago