Answer:
the total amount of water supplier per hour to the region within a circle of radius R=110 ( that is from distance r, 0<r<110)
![W(R) = 2\pi [1-(R+1)e^{-R}]](https://tex.z-dn.net/?f=W%28R%29%20%3D%202%5Cpi%20%5B1-%28R%2B1%29e%5E%7B-R%7D%5D)
Step-by-step explanation:
if f(r) describes the water supplied at a distance r , the total amount supplied inside a region that goes from 0 until the circle of radius R, is the sum of all f(r) values from 0 until R, that is the integral value over these limits.
The formula deduction can be found in the attached picture
There is an "r" that multiplies e^-r as result of changing from rectangular coordinates to polar ones.(dx*dy --> r*dr*da)
The two angles form a straight line which is equal to 180 degrees. This makes the angles supplemtary.
To find x, add the two angles together to equal 180:
7x + x+20 = 180
Combine like terms:
8x + 20 = 180
Subtract 20 from both sides:
8x = 160
Divide both sides by 8:
X = 20
Number of ribbons that can be cut = 6 ÷ 1/2
Number of ribbons that can be cut = 6 x 2/1
Number of ribbons that can be cut = 12
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Answer: 12 1/2 foot pieces can be cut from 6 feet ribbon
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Answer: A. 
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
So to find the rate of the area:
Using 
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:
The rate at which the area is increasing is given by expression 
.
B. At t = 5, rate is:
At 5 seconds, the area is expanded at a rate of 1.76 cm/s.
12/5 - 3/10 = 24-3 /10 = 21/10
OR 2.1 in decimal form
