At the time the rocket hits the ground h=0, given that h=-16t²+320t+32
when h=0, our equation will be:
-16t²+320t+32=0
solving the above by completing square method we proceed as follows;
-16t²+320t+32=0
divide though by -16 we get
t²-20t-2=0
t²-20t=2
but
c=(-b/2a)^2
c=(20/2)^2
c=100
hence:
t²-20t+100=100+2
(t-10)(t-10)=102
√(t-10)²=√102
t-10=√102
hence
t=10+/-√102
t~20.1 or -0.1
since it must have taken long, then the answer is 20.1 sec
H(t) = f(t) - g(t)
= 500(1.2)^t - 380(1.15)^t
Taking out the greatest common factor, which is 20:
= 20[(25)(1.2)^t - (19)(1.15)^t]
This is the third choice.
Note that you cannot subtract the bases of the exponents, for example (1.2^t - 1.15^t) cannot be simplified into something like 0.05^t.
The slope intercept form is y=mx+b. m being the rate of the slope (rise over run) so in this case 2/1, or simply 2. b is the y intercept, or where a line passes through the y intercept, in this case it is -1.
360 - 250 = 110
m<1 = 110/2 = 55
answer
55
