An object is launched at 19.6 meters per second (m/s) from a 58.8 meter tall platform. The equation for the object's height s at t time seconds after the launch is s(t)=−4.9t2+19.6t+58.8 s
(
t
)
=
-
4
.
9
t
2
+
19
.
6
t
+
58
.
8
When does the object strike the ground?

$12. that is my answer.

answer: -2+√x,-2-√x

i hope this , i don't know if you're looking for a pair or not.

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3