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farmergirl411
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
First you want to find the slope of teh graph from t=0 to t=6. Just use the slope formula.
I'm not quit sure how to do this problem so I need more info than that
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.
I understand how to the rest after the slope can you just help me with the slope
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.
so is the answer y=600x+5700
The slope should be a -6. Did you do 5700-2100 instead of 2100-5700?
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.
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
you just meany=-600x+57000 now good
m=(2100-5700)/(6-0)=-600
ok so then that is my answer right