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3 years ago

Can you please help me with #39 please!!?? Please explain part b. Please!!!!

Mathematics

1 answer:

zysi [14]3 years ago

4 0

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Can someone solve what is n/n^3

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Answer:

Alex = No B.

Step-by-step explanation:

1) Use Quotient Rule: $\frac{x^{a} }{x^{b} } =x^{a-b}$

$n^{1-3}$

2) Simplify 1 - 3 = 2

$n^{-2}$

3) Use Negative Power Rule: $x^{-a} =\frac{1}{x^{a} }$

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2 years ago

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Option B is correct

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4 years ago

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HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH

8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

$4+\sqrt{6x}$

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

$4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}$

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

$\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}$

Thank you

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