Wildlife management personnel use predator-prey equations to model the populations of certain predators and their prey in the wild. Suppose the population M of a predator after t months is given by

M=750 + 125 sin (Pi/6)t

while the population N of its primary prey is given by

N=12250 + 3050 cos(Pi/6)t

Find the values of t, 0 is less than or equal to t which is less than 12, for which the predator population is 875. (There is one value of t). Find the values of t, 0 is less than or equal to t which is less than 12, for which the prey population is 10725. (There are two values of t).

**** I have already completed the predator equation and came up with t=2. In the prey equation, I have worked on it for a long time and cannot come up with 2 t values between 0 and 12. Help please!! Thank you.

For the predator M:



Subtract 750 from both sides, then divide by 125.

It should be straightforward from there.

Let me know what you get.

Do the same for the prey.

M=750 + 125 sin (Pi/6) t
875 = 750 +125 sin (pi/6) t
125 = 125 (1/2) t
125 = 62.5 t
t = 2

N= 12250 + 3050 cos(pi/6) t
10725 = 12250 + 3050 cos (pi/6) t
-1525 =3050 (rad3/2) t
****now where do I go??

You've got it pretty much done. Just divide by what's in front of the t and isolate the t.

The first one is correct.

I keep coming up with -rad3 over 3... AHHHHH

OK... I'll try again... here goes...

-1525 = 3050 (rad3/2) t
-1/2 = (rad3/2) t

-1 = rad3 (t)
-rad3/3 = t = less than 0!! AHHHH! What am I doing wrong? Is it simple algebra?

galactus, you are my last hope to whip me into shape here. Can you see what it is that I am doing wrong?

You are not doing it wrong. is the only solution to this equation if we set it equal to 10725 and solve for t.

Are you sure there is not a typo in the equation somewhere?