This must be on the moon as the acceleration due to gravity in this equation must be around 1/8 that on earth. :) Anyway...
h=-2t^2+9t+11
A)
dh/dt=-4t+9, when velocity, dh/dt=0, it is the maximum height reached
dh/dt=0 only when 4t=9, t=2.25 seconds
h(2.25)=21.125 ft (21 1/8 ft)
B)
As seen in A), the time of maximum height was 2.25 seconds after the squirrel jumped.
C)
The squirrel reaches the ground when h=0...
0=-2t^2+9t+11
-2t^2-2t+11t+11=0
-2t(t+1)+11(t+1)=0
(-2t+11)(t+1)=0, since t>0 for this problem...
-2t+11=0
-2t=-11
t=5.5 seconds.
Answer:
The solution of the system is (6, 9).
Step-by-step explanation:
Solve one of the variables and substitute the answer for the other variable.
Select one of the problems and solve for x.
3x+y=27
Subtract y from both sides.
3x+-y=27
Divide both sides by three.
x=1/3(-y + 27)
Multiply 1/3 by -y + 27.
x = -1/3y + 9
Replace -y/3 + 9 for x in the other problem 3x-2y=0.
3(-1/3y + 9) - 2y = 0
Multiply 3 by -y/3 + 9.
−y+27−2y=0
Add -y to -2y.
−3y+27=0
Subtract 27 from both sides of the problem.
−3y=−27
Divide both sides by −3.
y=9
Replace 9 for y in x=−1/3y+9.
x=−1/3*9+9
Multiply −1/3 times 9.
x=-3+9
Add 9 to -3.
x = 6.
x = 6, y = 9.