(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:
7. r = -5
8. x = -1
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
Step-by-step explanation:
<u>Step 1: Define</u>
r + 2 - 8r = -3 - 8r
<u>Step 2: Solve for </u><em><u>r</u></em>
- Combine like terms: -7r + 2 = -3 - 8r
- Add 8r to both sides: r + 2 = -3
- Subtract 2 on both sides: r = -5
<u>Step 3: Check</u>
<em>Plug in r into the original equation to verify it's a solution.</em>
- Substitute in <em>r</em>: -5 + 2 - 8(-5) = -3 - 8(-5)
- Multiply: -5 + 2 + 40 = -3 + 40
- Add: -3 + 40 = -3 + 40
- Add: 37 = 37
Here we see that 37 does indeed equal 37.
∴ r = -5 is a solution of the equation.
<u>Step 4: Define equation</u>
-4x = x + 5
<u>Step 5: Solve for </u><em><u>x</u></em>
- Subtract <em>x</em> on both sides: -5x = 5
- Divide -5 on both sides: x = -1
<u>Step 6: Check</u>
<em>Plug in x into the original equation to verify it's a solution.</em>
- Substitute in <em>x</em>: -4(-1) = -1 + 5
- Multiply: 4 = -1 + 5
- Add: 4 = 4
Here we see that 4 does indeed equal 4.
∴ x = -1 is a solution of the equation.
1/ 1/5
1/1÷1/5
1/1×5/1
5/1=
5
