![\bf \textit{sum of an infinite geometric serie}\\\\ \stackrel{for~~|r|\ \textless \ 1}{S=\sum\limits_{i=0}^{\infty}~a_1r^i\implies \cfrac{a_1}{1-r}}\qquad \begin{cases} a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ a_1=42\\ r=\frac{3}{4} \end{cases} \\\\\\ S=\cfrac{42}{1-\frac{3}{4}}\implies S=\cfrac{42}{\frac{1}{4}}\implies S=164](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bsum%20of%20an%20infinite%20geometric%20serie%7D%5C%5C%5C%5C%0A%5Cstackrel%7Bfor~~%7Cr%7C%5C%20%5Ctextless%20%5C%201%7D%7BS%3D%5Csum%5Climits_%7Bi%3D0%7D%5E%7B%5Cinfty%7D~a_1r%5Ei%5Cimplies%20%5Ccfrac%7Ba_1%7D%7B1-r%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Aa_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%0Ar%3D%5Ctextit%7Bcommon%20ratio%7D%5C%5C%0A----------%5C%5C%0Aa_1%3D42%5C%5C%0Ar%3D%5Cfrac%7B3%7D%7B4%7D%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AS%3D%5Ccfrac%7B42%7D%7B1-%5Cfrac%7B3%7D%7B4%7D%7D%5Cimplies%20S%3D%5Ccfrac%7B42%7D%7B%5Cfrac%7B1%7D%7B4%7D%7D%5Cimplies%20S%3D164)
bearing in mind that, the geometric sequence is "convergent" only when |r|<1, or namely "r" is a fraction between 0 and 1.
What midpoints are you wanting to measure?
Answer:
The point at (-7, -5) = a
The point at (9, 3) = b
The point at (-3, 7) = c
The "a" point of the triangle is 12 units away from the center point.
So, 12 x 1/4
=> 12/4
=> 3
So, the "a" point of the dilated figure is 3 units left from the center.
=> So, the dilated "a" point is at (2, -5)
The "b" point is 8/4 (= rise/run = y-axis / x-axis) from the center point.
=> 8/4 = 2
So, the "b" point of the dilated figure is 1 unit right and 2 units up from the center point.
=> So, the dilated "b" point is at (6, -3)
The "c" point is 12/8 units away from the center point.
=> 12/8 x 1/4
=> 3/2
So, the "c" point of the dilated figure is 3 units up and 2 units left from the center point.
=> So, the dilated "c" point is at (3, -2)
Answer:
z
=
![\frac{17+4w}{w-M}](https://tex.z-dn.net/?f=%5Cfrac%7B17%2B4w%7D%7Bw-M%7D)
Step-by-step explanation: