Recent content by Burjam

By combine them, do you mean take a Fn+1 in the second equation as Fn+1 = Fn(1+0.014) - 119 and then try to solve for a and C?

How will adding the c to both sides eliminate the Fn+1? None of the problems I have done or have examples of with infinite series so far have anything like this, so I don't really have anything to go by.

I don't know how to write this equation without Fn being in terms of Fn+1 or Fn-1.

Fn+1 = Fn(1 + 0.014) - 119?

Homework Statement A fishery manager knows that her fish population naturally increases at a rate of 1.4% per month, while 119 fish are harvested each month. Let Fn be the fish population after the nth month, where F0 = 4500 fish. Assume that that process continues indefinitely. Use the...

Haha sorry I did this while really tired... my bad.

I miss a square root? Where does a square root come into play? None of the sides of the triangle should have a square root in them.

Homework Statement Under #3 Homework Equations Trig identities The Attempt at a Solution The picture attached is my attempt. The square in the upper upper left is the problem and the one in the lower right is my solution. I'm seeing that I'm getting the wrong answer, but not how.

That is correct. And thank you, I was able to work out the solution. I don't know why, but I kept insisting to myself that this simplified integral needed a u substitution when in reality all I had to do was simplify it further with some trig identities. Sometimes when I do problems in a rush I...

Homework Statement ∫(√(64 - x^2)) / x dx I must solve this using a sin substitution. Homework Equations x = 8sinΘ dx = 8cosΘ dΘ Θ = arcsin(x/8) Pythagorean Identities The Attempt at a Solution (After substitution) = ∫8cosΘ * (√(64 - 64sin^2Θ)) / 8sinΘ dΘ = ∫(cosΘ * (√(64(1 - sin^2Θ))) /...

Ok so I've become aware that I need to solve for the constant of proportionality k. dp/dt = k(20-p) The price rises $5 in 6 months, so dp/dt=5/6 when p=10 5/6=k(20-10) k=1/12 dp/dt=(20-p)/12 From here, I get what appears to be a faulty equation: p = 20 - 10e^t/12 The reason I say this is...

Ok so solving for k. The water level dropped 1 inch in an hour at L=35 so 1 = -k√35 k=-1/√35 Does that look right?

I was thinking of the other problem I posted. I think this is the same problem with that one too.

Ok so (dp/dt)/20-p = k then dp/dt = k(20-p) The problem is there are no conditions I can use in this problem that allow me to solve for k. I have conditions for p, but none for dp/dt.

Sorry for the double post, but actually it looks like the sign doesn't even matter. Can someone please give me a little more insight? I'm pretty sure at this point it has to do with my initial setup.