Arock is launched from a cannon. its height, h(x), can be represented by a quadratic funtion in terms of time, x, in seconds. after 1 secound, the rock is 58 feet in the air; after 2 seconds, it is 112 feet in the air. find the height, in feet of the rock after 10 seconds in the air! (im on plato)

This is the quadratic function:

h(x)=ax²+bx+c

We have two points:

(1,58)

(2,112)

Now, we calculate this quadratic funtion.

we assume that h(0)=0

Therefore:

a(0)²+b(0)+c=0

c=0

(1,58)

a(1)²+b(1)=58

a+b=58  (1)

(2,112)

a(2)²+b(2)=112

4a+2b=112

2a+b=56  (1)

With the equations (1) and (2) we make a system of equations:

a+b=58

2a+b=56

we can solve this system of equations by reduction method.

-(a+b=58)

2a+b=56

a=-2

-2(a+b=58)

 2a+b=56

      -b=-60  ⇒  b=60

The function is:

h(x)=ax²+bx+c

h(x)=-2x²+60x

Now find the height, in feet, of the rock after 10 seconds in the air.

h(10)=-2(10)²+60(10)

h(10)=-200+600=400

400 ft.

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation: