

so the ODE is indeed exact and there is a solution of the form
. We have




With
, we have

so

Answer:
A = 49 in^2
Step-by-step explanation:
The perimeter of a square is found by
P = 4s where s is the side length
28 = 4s
Divide each side by 4
28/4 = 4s/4
7 =s
We want to find the area
A = s^2 where s is the side length
A = 7^2
A = 7^2
A = 49 in^2
Answer:
husky mix of germany shepherd
The difference of a number and seven is x - 7. So twice that, would be:
2(x - 7)
Since we can find that the graphs Y value is by a factor of 2 looking at the graph we can find
f(3)=8