Given:
Current revenue = $6000
<span>R(f)= -100f^2 + 400f + 6000
where f is a whole number of $5 fee increases
We are told to find f, when R(f) </span><span>< 6000
Since </span>R(f)= -100f^2+400f+6000
R(f) < 6000 ⇒ -100f^2+400f+6000 < 6000
Subtract 6000 from both sides
-100f^2 + 400f + 6000 - 6000 < 6000 - 6000
-100f^2 + 400f < 0
⇒ 400f - 100f^2 < 0
Divide the equation by 100
400f/100 - 100f^2/100 < 0/100
4f - f^2 < 0
Add f^2 to both sides of the equation
4f - f^2 + f^2 < 0 + f^2
4f < f^2
Divide both sides by f
4f/f < (f^2)/f
4 < f
⇒ f > 4
⇒ f ≥ 5
Therefore, <span> for 5 or more numbers of $5 fee increases, the revenue from fees will actually be less than its current value.</span>
Answer:
i believe its D
Step-by-step explanation:
i hope that's right if its not I'm sorry
24 + 11 = 35 + 11 = 46 + 11 = 57 + 11 = 68 +11 = 79 + 11 = 90 + 11 = 101
17 + 12 = 29 + 12 = 41 + 12 = 53 + 12 = 65 + 12 = 77 + 12 = 89 + 12 = 101
$101, 7 months
