Someone help for brainliest

Mathematics

1 answer:

anyanavicka [17]3 years ago

8 0

Answer:

b and e

Step-by-step explanation:

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Start with the point A (3, 1). Translate the point 1 unit to the left and 4 units down. What are the new coordinates of the poin

Over [174]

This is the transformations: See the picture, here we find the coordinate:

A' = (2,-3) and A''=(0.5,-2). There is many ways to get back to A from A'', i have showed one way (the red lines). This is done by going 5 units right and then 6 units up.

see image at https://imgur.com/a/fWQl8

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3 years ago

PllllllllIIILP:P{:LPs help

borishaifa [10]

Uhhhhh the picture is all white

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3 years ago

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Please help!!!<br> Which expression is equivalent to (3m^-1n^2)^4/(2m^-2n)^3

Elza [17]

$\text{Use:}\\\\(a\cdot b)^n=a^nb^n\\\\(a^n)^m=a^{n\cdot m}\\\\\dfrac{a^n}{a^m}=a^{n-m}$

$\dfrac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}=\dfrac{3^4(m^{-1})^4(n^2)^4}{2^3(m^{-2})^3n^3}=\dfrac{81m^{-4}n^8}{8m^{-6}n^3}\\\\=10.125m^{-4-(-6)}n^{8-3}=10.125m^2n^5$

7 0

3 years ago

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HELPP!!!!!!!!!!!!!!!!

Anvisha [2.4K]

Answer:

Step-by-step explanation:

We'll take this step by step. The equation is

$8-3\sqrt[5]{x^3}=-7$

Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time. First thing is to subtract 8 from both sides:

$-3\sqrt[5]{x^3}=-15$

The goal is to isolate the term with the x in it, so that means that the -3 has to go. Divide it away on both sides:

$\sqrt[5]{x^3}=5$

Let's rewrite that radical into exponential form:

$x^{\frac{3}{5}}=5$

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

$(x^{\frac{3}{5}})^{\frac{5}{3}}=5^{\frac{5}{3}}$

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely. On the right, let's rewrite that back in radical form to solve it easier:

$x=\sqrt[3]{5^5}$