Agym has 200 members who pay $30 per month for unlimited use of the gym’s equipment. a survey of the members indicates that for each $5 increase in the monthly fee, the gym will lose 20 members. this means that the revenue r from fees, which is currently $6000 per month, will become r(f)= -100f^2+400f+6000, where f is a whole number of $5 fee increases. write and solve a quadratic inequality to answer the questions: for what numbers of $5 fee increases will the revenue from fees actually be less than its current value? write down all your work and submit an image of the work to go along with your answer.  

Current revenue from fees = $6000

We are given the formula, R(f)= -100f² + 400f + 6000
where f represents the number of $5 fee increases

The question is asking us to find f, when 6000 > R(f)
⇒ 6000 > -100f² + 400f + 6000

Subtract 6000 from both sides of the inequality
6000 - 6000 > -100f² + 400f + 6000 - 6000
0 > -100f² + 400f

Add 100f² to both sides of the inequality
⇒ 0 + 100f² > -100f² + 100f² + 400f
100f² > 400f

Divide the inequality by 100
100f²/100 > 400f/100
f² > 4f

Divide both sides of the inequality by f
f²/f > 4f/f
f > 4

⇒ f ≥ 5
We therefore conclude that, for 5 or more numbers of $5 increment in the fees, the revenue from fees will actually be less than its current value.

A= b/2 because u simplying both sides of equation