Hey there! I'm happy to help!
To find the x value of the midpoint, you add the x values and divide by 2.
-5+4=-1
-1/2=-1/2
For the y value, you add the y values and divide by 2.
5+6=11
11/2=5 1/2
So, the midpoint is (-1/2, 5 1/2).
Have a wonderful day! :D
Answer: B.
Y = 4x
Step-by-step explanation:
Given that the side of a shape is x
The perimeter of the shape is y
Perimeter of a rectangle can be expressed as:
P = 2L + 2W
Where L = length and W = width
But if the shape is a square, and the side of the shape is x, then, the perimeter will be
P = 4x
Where perimeter P = Y
Therefore, perimeter Y will be;
Y = 4x
Therefore, the correct answer is B which is:
Y = 4x

so, as you can see above, the common ratio r = 1/2, now, what term is +4 anyway?


so is the 8th term, then, let's find the Sum of the first 8 terms.

![\bf S_8=512\left[ \cfrac{1-\left( \frac{1}{2} \right)^8}{1-\frac{1}{2}} \right]\implies S_8=512\left(\cfrac{1-\frac{1}{256}}{\frac{1}{2}} \right)\implies S_8=512\left(\cfrac{\frac{255}{256}}{\frac{1}{2}} \right)\\\\\\S_8=512\cdot \cfrac{255}{128}\implies S_8=1020](https://tex.z-dn.net/?f=%20%5Cbf%20S_8%3D512%5Cleft%5B%20%5Ccfrac%7B1-%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5E8%7D%7B1-%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cright%5D%5Cimplies%20S_8%3D512%5Cleft%28%5Ccfrac%7B1-%5Cfrac%7B1%7D%7B256%7D%7D%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%20%5Cright%29%5Cimplies%20S_8%3D512%5Cleft%28%5Ccfrac%7B%5Cfrac%7B255%7D%7B256%7D%7D%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%20%5Cright%29%5C%5C%5C%5C%5C%5CS_8%3D512%5Ccdot%20%5Ccfrac%7B255%7D%7B128%7D%5Cimplies%20S_8%3D1020%20)