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.

Given:

Current revenue = $6000

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) < 6000

Since 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,  for 5 or more numbers of $5 fee increases, the revenue from fees will actually be less than its current value.

It would be 15 m how i got the answer: (20x20) + b^2 = (25x25)400 + b^2 = 625625-400=225square that and get 15hope this : )