. A rocket carrying fireworks is launched from a hill above a lake. The rocket will fall into lake after exploding at its maximum height. The rocket's height above the surface of the lake is given by h(t) = −16t 2 + 64t + 80, where h is in feet and t is in seconds

Step-by-step explanation:

Since we are not told what to find, we can as well find the time taken by the rocket to reach its maximum height as shown

The rocket reaches its maximum height when the velocity is zero

v = dh/dt

v = -32t+64

Since v = 0 at maximum height

0 = -32t+64

32t = 64

t = 64/32

t = 2secs

Hence the rocket reaches the maximum height after 2secs

d.192

step-by-step explanation:

12*16=192

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752