Question

In the forrest the population of gray squirrels oscillates between a high of 5700 on January 1 (t=0)and a low of 2100 in July 1(t=6),
Find a function for the population p(t)
of grey squirrels in terms of time t months

Answer 1

First you want to find the slope of teh graph from t=0 to t=6. Just use the slope formula.

Answer 2

I'm not quit sure how to do this problem so I need more info than that

Answer 3

After that, apply the slope to the formula, y=mx+b where m equals the slope. To solve for b, plug in either the first set of coordinates, (0,5700) or the second one (6,2100). If you get the right slope, the b value will be equal either way.

Now plug in m and b in y=mx+b and you're done.

Answer 4

I understand how to the rest after the slope can you just help me with the slope

Answer 5

The slope formula is \[m= (y _{2}-y _{1})/(x _{2}-x _{1})\]

The x value is the time and the y value is the population since the population varies with time. As such, time is teh independent variable, x.

Answer 6

so is the answer y=600x+5700

Answer 7

The slope should be a -6. Did you do 5700-2100 instead of 2100-5700?

Answer 8

It is importable to be very careful with this. y2 and x2 mean the second end of teh slope, right of the first end, y1 and x1. 5700 is at t=0 while 2100 is at t=6 so 2100 is y2 value.

Answer 9

I didi that and you get 3600. sO then you take 3600/6=600 and m=600. Then you do5700=600(0)+b. and b=5700

Answer 10

you just meany=-600x+57000 now good

Answer 11

That's right

Answer 12

m=(2100-5700)/(6-0)=-600

Answer 13

ok so then that is my answer right

Answer 14

Yeah

Answer 15

p(t)=-600x+5700

Answer 16

Correct

Answer 17

thanks

Answer 18

You're welcome