The maximum value can be determined by taking the derivative of the function.
(dh/dt) [h(t)] = h'(x) = -9.8t + 6
Set h'(x) = 0 to find the critical point
-9.8t + 6 = 0
-9.8t = -6
t = 6/9.8
Plug the time back into the function to find the height.
h(6/9.8) = -4.9(6/9.8)^2 + 6(6/9.8) + .6
= 2.4
And I don't understand your second question.
Answer:
(-3, 4)
Step-by-step explanation:
First, add 3 to point B's x coordinate:
-6 + 3 = -3
Then, add 1 to the y coordinate:
3 + 1
= 4
So, the coordinates of pre image point A are (-3, 4)
Answer:
60
Step-by-step explanation:
A= bh
A = 6 x 10
A = 60
12+2+8+6+57=85
85/5= 17
mean=17