An object is launched with a velocity at 64 feet per second from a platform 30 feet high. the function h(t)= -16t2+64t+30 represents the function where t is time in seconds, and h is height in feet. what is the maximum height of the object ever reaches? a) 92 ftb) 94 ftc) 2 ftd) 94 ftwhen does it reach the maximum height? a) 2 secb) 94 secc) -2 secd) 3 sec

To find the maximum or minimum value of a function, we can find the derivative of the function, set it equal to 0, and solve for the critical points.

H'(t) = -32t + 64

Now find the critical numbers:

-32t + 64 = 0

-32t = -64

t = 2 seconds

Since H(t) has a negative leading coefficient, we know that it opens downward. This means that the critical point is a maximum value rather than a minimum. If we weren't sure, we could check by plugging in a value for t slightly less and slighter greater than t=2 into H'(t):

H'(1) = 32

H'(3) = -32

As you can see, the rate of change of the object's height goes from increasing to decreasing, meaning the critical point at t=2 is a maximum.

To find the height, plug t=2 into H(t):

H(2) = -16(2)^2 +64(2) + 30 = 94

The answer is 94 ft at 2 sec.

yes

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!