#1**+1 **

x = t + 3 ⇒ t = x - 3

y = 2t^2 - 4

y = 2 ( x - 3)^2 - 4

y = 2 (x^2 -6x + 9) - 4

y = 2x^2 - 12x + 14 correct !!!!

When

t = -4, x = -1

t = 0 , x = 3

GOOD JOB !!!!!

CPhill May 4, 2020

#2**0 **

Thank you! could you possibly look at another post I made earlier that says

Sam attempts a 42-yard field goal in a football game. For his attempt to be a success, the football needs to pass through the uprights and over the crossbar that is 10 feet above the ground.

Sam kicks the ball from the ground with an initial velocity of 70 feet per second, at an angle of 50° with the horizontal.

Which statement is true of Sam's attempt?

The kick is good! The football clears the crossbar by approximately 6 feet.

The kick is good! The ball clears the crossbar by nearly 2 feet.

The kick is good! The ball clears the crossbar by nearly 15 feet.

The kick is not successful. The ball hits the ground short of the crossbar.

Guest May 4, 2020

#3**+1 **

We can use this equation

y = -16t^2 + V_{0}sin (A) *t + H_{0}

Where V_{0} = initial velocity

H_{0} = initial height = 0

A = the launch angle

t = time

42 yds = 126 ft

The time it takes for the kick to reach the crossbar can be found as

126 = 70cos(70°) * t

t = 126 / [70 cos (50°)] = 2,8 s

So

To find the height at the end we have

y = -16 (2.8)^2 + (70)sin (50°)*(2.8)

y = 24.87 ft

It clears the bar by 14.87 ft ≈ 15 ft

CPhill May 4, 2020