48cm2 is the answer because
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.
D= # of dimes
q= # of quarters
QUANTITY EQUATION:
d + q= 64
COST EQUATION:
0.10d + 0.25q= $9.25
STEP 1:
multiply quantity equation by -0.10 to be able to eliminate the d term in step 2
(-0.10)(d + q)= (-0.10)(64)
-0.10d - 0.10q= -6.40
STEP 2:
add equation from step 1 to cost equation to eliminate the d term and solve for q
Add
0.10d + 0.25q= $9.25
-0.10d - 0.10q= -6.40
0.15q= 2.85
divide both sides by 0.15
q= 19 quarters
STEP 3:
substitute q value in step 2 into either original equation to find d value
d + q= 64
d + 19= 64
subtract 19 from both sides
d= 45 dimes
CHECK:
0.10d + 0.25q= $9.25
0.10(45) + 0.25(19)= 9.25
4.50 + 4.75= 9.25
9.25= 9.25
ANSWER: There are 45 dimes and 19 quarters.
Hope this helps! :)
This can solved using the cosine law which is:
c² = a² + b² - 2ab cos θ
Using the values given from the problem
6² = b² + b² - 2bb cos 112.62
And solving for b
36 = 2b² - 2b² cos 112.62
b = 3.6
The answer is the 3rd option.
