Answer:
sorry can't help hope someone gets it
Ok
y=-0.04x^2+8.3x+4.3
when the rocket reaches the ground (when height=0, ie when y=0), then the rocket will land, find the x coordinate
set y=0
0=-0.04x^2+8.3x+4.3
use quadratic formula
if you have ax^2+bx+c=0, then
x=

a=-0.04
b=8.3
c=4.3
x=

x=208.017 or -0.516785
xrepresents horizontal distance
you cannot have a negative horizontal distance unless you fired and theh wind blew it backwards
therefor x=280.017 is the answer
208.02 m
11! = 39,916,800
The erased digits are 1, 0, 0.
9514 1404 393
Answer:
(x, y) = (8, 2)
Step-by-step explanation:
The relevant equations are ...
3x -y = 22
x +2y = 12
__
We can eliminate y by adding twice the first equation to the second.
2(3x -y) +(x +2y) = 2(22) +(12)
7x = 56
x = 8
Substituting into the first equation gives ...
3(8) -y = 22
y = 24 -22 = 2
The first number is 8; the second number is 2.
The answer is .... C. d>0
Hope it helps !!!!