since the diameter of the base of the cylinder is 6 feet, then its radius is half that, or 3 feet.
![\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=9 \end{cases}\implies V=\pi (3)^2(9)\implies V=81\pi](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cylinder%7D%5C%5C%5C%5C%0AV%3D%5Cpi%20r%5E2%20h~~%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0Ah%3Dheight%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Ar%3D3%5C%5C%0Ah%3D9%0A%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Cpi%20%283%29%5E2%289%29%5Cimplies%20V%3D81%5Cpi)
Among all these no., see the first no. that lies before or left side of decimal point, here... 2 is common so next check the no. After decimal...
2.0, 2.2, 2.4, 2 is considered as 2.0 and then we have 2.0...
Let's not take 2.0s and let's take 2.2 and 2.4 which is greater than 2.0.
Next see the second no. That comes after decimal
We have 2.24 and 2.4 which can be taken as 2.40... The no.s after decimal,
24 is greater than 40 so 2.4 is greater...
Answer:
<u>m = 0</u>, y = -6
Step-by-step explanation:
using point slope form :
![y - y_{1} = m(x - x_{1})](https://tex.z-dn.net/?f=y%20-%20y_%7B1%7D%20%3D%20m%28x%20-%20x_%7B1%7D%29)
y-y1=m(x-x1) ; initial point (x1,y1) = (4,-6), final point (x,y) = (7,-6).
one substituted, it should look like this:
-6--6 = m(7-4),
-6+6= m(7-4)
0 = m(7-4)
0 = 3m
0/3 = 3/3m
0 = m
m = 0/3 = 0
Answer:
![y=(x-2)^2-1](https://tex.z-dn.net/?f=y%3D%28x-2%29%5E2-1)
Step-by-step explanation:
<u>Complete the square</u>
<u />![y=x^2-4x+3\\\\y+1=x^2-4x+3+1\\\\y+1=x^2-4x+4\\\\y+1=(x-2)^2\\\\y=(x-2)^2-1](https://tex.z-dn.net/?f=y%3Dx%5E2-4x%2B3%5C%5C%5C%5Cy%2B1%3Dx%5E2-4x%2B3%2B1%5C%5C%5C%5Cy%2B1%3Dx%5E2-4x%2B4%5C%5C%5C%5Cy%2B1%3D%28x-2%29%5E2%5C%5C%5C%5Cy%3D%28x-2%29%5E2-1)