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:
year
Step-by-step explanation:
![A=P\left ( 1+R/100\right )^{T}](https://tex.z-dn.net/?f=A%3DP%5Cleft%20%28%201%2BR%2F100%5Cright%20%29%5E%7BT%7D)
![A=Amount ,P=Principle , R=Rate , T=Time](https://tex.z-dn.net/?f=A%3DAmount%20%2CP%3DPrinciple%20%2C%20R%3DRate%20%2C%20T%3DTime)
CI=compound interest
![CI=A-P](https://tex.z-dn.net/?f=CI%3DA-P)
for each a and b
account earn each year by a=![45\$](https://tex.z-dn.net/?f=45%5C%24)
account b earns
%
find the time the value of b is more
![45](https://tex.z-dn.net/?f=45%3C400%5Cleft%20%28%201%2B5%2F100%20%5Cright%20%29%5E%7BT%7D-400)
![445](https://tex.z-dn.net/?f=445%3C%5Cleft%20400%5Cleft%20%28%201%2B5%2F100%20%5Cright%20%29%5E%7BT%7D)
![445/400](https://tex.z-dn.net/?f=445%2F400%3C%5Cleft%20%28%2021%2F20%20%5Cright%20%29%5E%7BT%7D)
![\ln \left ( 445/400 \right )](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%20%28%20445%2F400%20%5Cright%20%29%3CT%5Cln%20%5Cleft%20%28%2021%2F20%20%5Cright%20%29)
![.106609735](https://tex.z-dn.net/?f=.106609735%3C.0487901642%5Ctimes%20%20T)
![=2.1850](https://tex.z-dn.net/?f=%3D2.1850%3CT)
year
Answer:
ST = 4
Step-by-step explanation:
Answer:
3(6x+1)
Step-by-step explanation:
I hope its correct good luck best wishes