3 years ago

Which expression would be easier to simplify if you used the associative property to change the grouping?

Mathematics

2 answers:

vlabodo [156]3 years ago

6 0

Answer:

Step-by-step explanation:

BlackZzzverrR [31]3 years ago

5 0

It well be C for your answer

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Can you prove that def=hgf? Justify your answer.

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3 years ago

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tester [92]

Answer:

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Step-by-step explanation:

We can use ratios to solve

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$(-2x^3+5x^2-4x+8)-(-2x^3+2x-3)\\\\=-2x^3+5x^2-4x+8-(-2x^3)-2x-(-3)\\\\=-2x^3+5x^2-4x+8+2x^3-2x+3\\\\=(-2x^3+2x^3)+5x^2+(-4x-2x)+(8+3)\\\\=5x^2-6x+11$

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3 years ago

Please help me Algebra 1

Oduvanchick [21]

Answer:

a. $\sqrt{x^n}$ = $x^\frac{n}{2}$

b. $\sqrt{x^n} = x^\frac{(n-1)}{2}\sqrt{x}$

Step-by-step explanation:

a) When n is even, then it is divisible by 2. Because of this, you can write:

$\sqrt{x^n} = (x^n)^\frac{1}{2}$

$\sqrt{x^n} = x^\frac{n}{2}$

b) When n is odd, then n - 1 is even. This would make it divisible by 2, and there would be a remainder of 1, so we can write:

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

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$\sqrt{x^n} = x^\frac{(n-1)}{2}$ × $x^\frac{1}{2}$

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

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3 years ago

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Answer: 64 i believe

Hope this helps :)

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3 years ago