48 points and the first answer is brainliest!
an object is launched upwards from an initial height of 4 feet above ground, with an initial velocity of 48 feet per second. if the object's position at time x is given by
f(x) = −16x^2 + vt + s, where v is initial velocity and s is initial height, which equation can be used to find the maximum height of the object after x seconds?
a) f(x) = −16(x − 3)^2 + 40
b) f(x) = −16(x − 1.5)^2 + 40
c) f(x) = −16(x − 1.5)^2 + 20
d) f(x) = −16(x − 1.5)^2 + 32

B)  f(x) =  -16(x - 1.5)² + 40

Step-by-step explanation:

The equation we have is: h(t) = -16x² + 48x + 4

The answers are all in vertex form, so we need to covert this equation into vertex form by completing the square.

Step 1:  factor out -16 from the first 2 terms...

-16(x² - 3x) + 4    

Step 2:  complete the square on this term...

-16(x² - 3x + (3/2)² - (3/2)²)

    -16(x² - 3x + (3/2)²) + 4 + 16(3/2)²

        -16(x - 3/2)² + 4 + 16(9/4)

             -16(x - 1.5)² + 4 + 4(9)

                   -16(x - 1.5)² + 40

So f(x) =  -16(x - 1.5)² + 40

its b i took the test

Step-by-step explanation:

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

its a straight line parallel to the time axis