For a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height, y, and time, x , when the projectile reaches the ground

Solution

We know that the x coordinate of a quadratic function is given by:

Vx= -b/2a

And the y coordinate correspond to the maximum value of y.

Then the best options are C and D but the best option is:

D) The maximum height is a y coordinate of the vertex of the quadratic function, which occurs when x = -b/2a

The projectile reaches the ground when the height is zero. The time when this occurs is the x-intercept of the zero of the function that is farthest to the right.

7 0

10 months ago

Please find the answer

Serggg [28]

Answer:

x^3 + 1/x^3 = 488

Step-by-step explanation:

x^2 + 1/x^2 = 62
x^2 + 1/x^2 + 2 = 64
( adding 2 in both sides )
(x + 1/x ) ^2 = 64
x + 1/x = 8

now,

( x+ 1/x ) ^ 3 = 512
x^3 + 1/x^3 + 3 × x × 1/x ( x + 1/x )
x^3 + 1/x^3 + 3 ( 8 )
( since x + 1/x = 8 )
x^3 + 1/x^3 + 24 = 512
x^3 + 1/x^3 = 488

hence, we got x^3 + 1/x^3 = 488

8 0

3 years ago