9y^-5/3y^2r^-3 what is the answer in a fraction

Mathematics

1 answer:

allsm [11]3 years ago

8 0

Answer:

3r^3/y^7.

Step-by-step explanation:

9y^-5 / 3y^2r^-3

9 / 3 = 3

y ^-5 / y^2 = y^-7 = 1 / 7y

1 / r^-3 = r^3

So the answer is 3r^3/y^7.

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Find the midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9)

JulsSmile [24]

The midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9) is $\left(\frac{-9}{2}, \frac{-3}{2}\right)$

<u>Solution:</u>

Given, two points are (-6, 6) and (-3, -9)

We have to find the midpoint of the segment formed by the given points.

The midpoint of a segment formed by $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { and }\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ is given by:

$\text { Mid point } \mathrm{m}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

$\text { Here in our problem, } x_{1}=-6, y_{1}=6, x_{2}=-3 \text { and } y_{2}=-9$

Plugging in the values in formula, we get,

$\begin{array}{l}{m=\left(\frac{-6+(-3)}{2}, \frac{6+(-9)}{2}\right)=\left(\frac{-6-3}{2}, \frac{6-9}{2}\right)} \\\\ {=\left(\frac{-9}{2}, \frac{-3}{2}\right)}\end{array}$