As we have to solve for b, it means we have to isolate b
So first of all subtract 
from both sides
So we get
Now we need b , but its b^2 , so we can take square root on both sides.
taking square roots on both sides
Hence option B is correct
(a) The "average value" of a function over an interval [a,b] is defined to be
(1/(b-a)) times the integral of f from the limits x= a to x = b.
Now S = 200(5 - 9/(2+t))
The average value of S during the first year (from t = 0 months to t = 12 months) is then:
(1/12) times the integral of 200(5 - 9/(2+t)) from t = 0 to t = 12
or 200/12 times the integral of (5 - 9/(2+t)) from t= 0 to t = 12
This equals 200/12 * (5t -9ln(2+t))
Evaluating this with the limits t= 0 to t = 12 gives:
708.113 units., which is the average value of S(t) during the first year.
(b). We need to find S'(t), and then equate this with the average value.
Now S'(t) = 1800/(t+2)^2
So you're left with solving 1800/(t+2)^2 = 708.113
<span>I'll leave that to you</span>
Answer:
(a) 8.15
(b) 12.92
Step-by-step explanation:
Given: P = $3000, r = 0.085
Where
A is the Amount
P is the Principal
r is the rate
t is the time
(a) For the amount to double, A = 2 × P
A = 2 × $3000
A = $6000
Take 
of both sides
But 
∴ 
t = 8.15
(b) For the amount to double, A = 3 × P
A = 3 × $3000
A = $9000
Take of both sides
But
∴ 
t = 12.92
Answer:
The inverse relation G^(-1) is not a function. Why not? Because the y value y = 3 is paired up with more than one x value (x = 5, x = 2). The inverse relation G^(-1) is the set shown below
{(3,5), (3,2), (4,6)}
All I've done is swap the (x,y) values for each ordered pair to form the inverse relation. As you can see, x = 3 leads to multiple y value outputs which is why this relation is not a function. So in short, the answer is choice C. To have the inverse relation be a function, each value in the original domain must map to exactly one value in the range only. However that doesn't happen as the domain values map to an overlapping y value (y = 3).
Choice B would be the correct answer. I hope that helps! :D
