Find the height, in feet of the rock after 10 seconds

Using the form: ax^2 + bx + c = y
y = h(x), c = 0

equation 1

x=1; y=58
a(1^2) + 1b = 58
a + b = 58

equation 2
x=2; y=112
a(2^2) + 2b = 112
4a + 2b = 112
:
Multiply the 1st equation by 2, subtract from the 2nd equation
a + b = 58 x 2 = 2a+2b = 116

subtract that from 2nd equation
4a +2b =112 – 2a+2b= 116 =

2a = -4

A = -4/2 = -2

Find b using: a + b = 58
-2 + b = 58
b = 58 + 2
b = 60

use equation: h(x) = -2x^2 + 60x

Find the height, in feet,of the rock after 10 seconds in the air.;
x=10
h(10) = -2(10^2) + 60(10)
h(10) = -2(100) + 600
h(10) = -200 + 600
h(10) = 400 ft

 The rock is 400 feet in the air after 10 seconds

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

< 3

step-by-step explanation:

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3