An object is launched on the moon at 16 feet per second (f/s) from a 1152-foot tall tower. The equation for the object's height s at time t seconds after launch is s(t) = -16t^2 + 16t + 1152, where s is in feet. When does the object strike the moon’s surface and why?

0.5seconds later

Step-by-step explanation:

Given the height modeled by the equation:

s(t) = -16t^2 + 16t + 1152

The velocity of the object at the moon surface is zero

ds(t)/dt = 0

ds(t)/dt = -32t + 16

0 = -32t + 16

32 t = 16

t = 16/32

t = 1/2

t = 0.5seconds

Hence the object strikes the moon surface 0.5seconds later

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