The perpendicular bisector goes through the midpoint.
What is the midpoint of the 2 points?
We simply add up the x points and divide by 2. Then we add up the y points and divide by 2.
So,
Midpoint is:
![(\frac{3-9}{2},\frac{-1+5}{2})=(-3,2)](https://tex.z-dn.net/?f=%28%5Cfrac%7B3-9%7D%7B2%7D%2C%5Cfrac%7B-1%2B5%7D%7B2%7D%29%3D%28-3%2C2%29)
Also, the perpendicular bisector's slope is the negative reciprocal of the line's slope.
The slope of the line is change in y points divided by change in x points.
Slope =
![\frac{5--1}{-9-3}=\frac{5+1}{-12}=\frac{6}{-12}=-\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B5--1%7D%7B-9-3%7D%3D%5Cfrac%7B5%2B1%7D%7B-12%7D%3D%5Cfrac%7B6%7D%7B-12%7D%3D-%5Cfrac%7B1%7D%7B2%7D)
The slope (m) of the perpendicular line (negative reciprocal) is basically:
![2](https://tex.z-dn.net/?f=2)
Equation of line is:
![y-y_1=m(x-x_1)](https://tex.z-dn.net/?f=y-y_1%3Dm%28x-x_1%29)
m is the slope (what we got "2")
x1 and y1 are the respective point where it passes through (which is the midpoint, which is (-3,2)
So, equation of perpendicular bisector is:
C 15 x 12 =$180 plus $8 shipping is $188.00
Step-by-step explanation:
Volume of a sphere is
![\frac{4}{3} \pi {r}^{3}](https://tex.z-dn.net/?f=%20%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%7Br%7D%5E%7B3%7D%20)
plug in 70 ft for the radius to get 1,436,755 cubic ft then divide that by 4 since the prompt describes a quarter sphere, giving you a tank volume of 359,189 cubic ft.
Volume of a cylinder is
![\pi {r}^{2} h](https://tex.z-dn.net/?f=%5Cpi%20%7Br%7D%5E%7B2%7D%20h)
from the second prompt we know the tanks are identical since they are said to be congruent which means the hight and radius are the same. Plug in 120 f for h and 35 ft for r to get the volume of a cylinder and then divide that by 2 to get the volume of a single tank. the volume of both tanks would be
![\pi {35}^{2} (120) = 461,814](https://tex.z-dn.net/?f=%5Cpi%20%7B35%7D%5E%7B2%7D%20%28120%29%20%3D%20461%2C814)
To show tank model is a sixth of the original so you divide the radius by 6 and perform the same calculation as in the first portion of the problem ro get the volume a quarter sphere
![\frac{4}{3} \pi {( \frac{70}{6} )}^{3} = 6,652 \: {ft}^{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%7B%28%20%5Cfrac%7B70%7D%7B6%7D%20%29%7D%5E%7B3%7D%20%20%3D%206%2C652%20%5C%3A%20%20%7Bft%7D%5E%7B3%7D%20)
To get the percentage of the model relative to the original you divide the models volume by the full scale volume
![\frac{6652}{1436755} \times 100 = 0.46](https://tex.z-dn.net/?f=%20%5Cfrac%7B6652%7D%7B1436755%7D%20%5Ctimes%20100%20%20%3D%200.46)
which is less than 1% of the full scale volume.
1. x=50 2. x=4 hope this helps:)