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:
-5.04199038
Step-by-step explanation:
(y2-y1)/(x2-x1)
Plus, mthway can double check
Answer:
A
Step-by-step explanation:
The given angles are vertical and thus congruent, so
3x = x + 40 ( subtract x from both sides )
2x = 40 ( divide both sides by 2 )
x = 20 → A
6y = - x + 8
y = - 1/6x + 4/3
y intercept = 4/3
Olivia needs 15 yards of ribbon.
2.5x18=45 feet
divide 45ft by 3 to see how many yards that would be.
45/3=15
Hope that I helped
