Answer:
No because 21 is close to it.
Step-by-step explanation:
To be considered an outlier, it would need to be a farther distance such as the distance from 28 to 21.
Hope this helps!!! PLZ MARK BRAINLIEST!!!
A) I included a graph, look below.
B)
Input the y in x = y + 3.
x = (-4x - 3) + 3
x = -4x + 0
Add 4x to both sides.
5x = 0
Divide both sides by 5.
x = 0
Input that x value in y = -4x - 3
y = -4(0) - 3
y = 0 - 3
y = -3
(0, -3)
C)
Convert both equations to Standard Form.
x = y + 3
Subtract y from both sides.
x - y = 3
y = -4x - 3
Add 4x to both sides.
4x + y = -3
Add the equations together.
4x + y = -3
x - y = 3
equals
5x = 0
Divide both sides by 5.
x = 0
Input that into one of the original equations.
0 = y + 3
Subtract 3 from both sides.
-3 = y
(0, -3)
Answer: A. ![A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}](https://tex.z-dn.net/?f=A%27%28t%29%3D%5Cfrac%7B4356%5Cpi%7D%7B%28t%2B11%29%5E%7B3%7D%7D-%5Cfrac%7B527076%5Cpi%7D%7B%28t%2B11%29%5E%7B5%7D%7D)
B. A'(5) = 1.76 cm/s
Step-by-step explanation: <u>Rate</u> <u>of</u> <u>change</u> measures the slope of a curve at a certain instant, therefore, rate is the derivative.
A. Area of a circle is given by
![A=\pi.r^{2}](https://tex.z-dn.net/?f=A%3D%5Cpi.r%5E%7B2%7D)
So to find the rate of the area:
![\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%3D%5Cfrac%7BdA%7D%7Bdr%7D.%5Cfrac%7Bdr%7D%7Bdt%7D)
![\frac{dA}{dr} =2.\pi.r](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdr%7D%20%3D2.%5Cpi.r)
Using ![r(t)=3-\frac{363}{(t+11)^{2}}](https://tex.z-dn.net/?f=r%28t%29%3D3-%5Cfrac%7B363%7D%7B%28t%2B11%29%5E%7B2%7D%7D)
![\frac{dr}{dt}=\frac{726}{(t+11)^{3}}](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D%3D%5Cfrac%7B726%7D%7B%28t%2B11%29%5E%7B3%7D%7D)
Then
![\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%3D2.%5Cpi.r.%5B%5Cfrac%7B726%7D%7B%28t%2B11%29%5E%7B3%7D%7D%5D)
![\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%3D2.%5Cpi.%5B3-%5Cfrac%7B363%7D%7B%28t%2B11%29%5E%7B2%7D%7D%5D.%5Cfrac%7B726%7D%7B%28t%2B11%29%5E%7B3%7D%7D)
Multipying and simplifying:
![\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%3D%5Cfrac%7B4356%5Cpi%7D%7B%28t%2B11%29%5E%7B3%7D%7D%20-%5Cfrac%7B527076%5Cpi%7D%7B%28t%2B11%29%5E%7B5%7D%7D)
The rate at which the area is increasing is given by expression
.
B. At t = 5, rate is:
![A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}](https://tex.z-dn.net/?f=A%27%285%29%3D%5Cfrac%7B4356%5Cpi%7D%7B%285%2B11%29%5E%7B3%7D%7D%20-%5Cfrac%7B527076%5Cpi%7D%7B%285%2B11%29%5E%7B5%7D%7D)
![A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}](https://tex.z-dn.net/?f=A%27%285%29%3D%5Cfrac%7B4356%5Cpi%7D%7B4096%7D%20-%5Cfrac%7B527076%5Cpi%7D%7B1048576%7D)
![A'(5)=\frac{2408693760\pi}{4294967296}](https://tex.z-dn.net/?f=A%27%285%29%3D%5Cfrac%7B2408693760%5Cpi%7D%7B4294967296%7D)
![A'(5)=1.76268](https://tex.z-dn.net/?f=A%27%285%29%3D1.76268)
At 5 seconds, the area is expanded at a rate of 1.76 cm/s.
Answer:
I think its 3p not sure but u can try that