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:

750+125sin(\frac{\pi}{6})t=875

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. \frac{-\sqrt{3}}{3} 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?