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2 years ago

I really need an answer as soon as possible. 10 pts worth it

Mathematics

1 answer:

viva [34]2 years ago

8 0

Answer:

A. Correct: When we plug in g(x) for the x in f(x), we get H(x).

B. Correct: When we plug in g(x) for the x in f(x), we get H(x).

C. Correct: When we plug in g(x) for the x in f(x), we get H(x).

D. Correct: When we plug in g(x) for the x in f(x), we get H(x).

Step-by-step explanation:

<em>Brainliest, please!</em>

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Estimate the perimeter, and also please show your work!

jeka57 [31]

Answer:

You can also just find the area and times that by 4

If you want I can do that, just spam tell me and I'll figure the area out and tell u the answer in comments.

Step-by-step explanation:

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1 year ago

Can someone help me in this plssss I really need it

kompoz [17]

Answer:

$\boxed{-3xy^{2}\sqrt [3] {2x^{2}}}$

Step-by-step explanation:

Your expression is

$\sqrt [3] {-54x^{5}y^{6}}$

Here's how I would simplify it.

$\begin{array}{rcll}\sqrt [3] {-54x^{5}y^{6}} & = & \sqrt [3] {(-1)^{3}\times 2 \times 27 \times x^{2} \times x^{3} \times y^{6}} & \text{Factored the cubes}\\& = & \sqrt [3] {(-1)^{3} \times 3^{3}\times x^{3} \times y^{6}\times 2 \times x^{2}} & \text{Grouped the cubes}\\\end{array}$

$\begin{array}{rcll}& = & \sqrt [3] {(-1)^{3} \times {3^{3}\times x^{3} \times y^{6}}} \times\sqrt [3] { 2 \times x^{2}} & \text{Separated the cubes}\\&=& \mathbf{-3xy^{2}\sqrt [3] {2x^{2}}} & \text{Took cube roots}\\\end{array}$

$\text{The simplified expression is $\boxed{\mathbf{-3xy^{2}\sqrt [3] {2x^{2}}}}$}$

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3 years ago