(10+2x)(8+2x) = 120
4x^2 + 36x + 80 = 120
x^2 + 9x + 20 = 30
x^2 + 9x - 10 = 0
(x+10)(x-1) = 0
x = 1 inch dimension
Domain: (-♾,2)U(2,♾), {x | x ≠2}
Range: (-♾,3) U(3,♾), {y | y ≠ 3}
Sorry that’s all I can help with
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.
Mean, x_bar = 1518
Standard deviation, sigma = 325
Range required: 1550 ≤ X ≤ 1575
Z = (X - x_bar)/sigma
Z1 = (1550-1518)/325 ≈ 0.1
Z2 = (1575-1518)/325 ≈ 0.18
From Z tables,
P(Z1) = 0.5398
P(Z2) = 0.5714
P(1550≤X≤1575) = P(Z2) - P(Z1) = 0.5714 - 0.5398 = 0.0316
The correct answer is C.