A baseball is thrown in a parabolic arc. It's position above the ground at a given point in time can be represented by the quadratic function p (t) = 1/2gt + vot + po, where t > 0, g is -32 ft/sec/sec, vo is initial velocity, and po is its initial position above the ground. If the ball was thrown straight up at 16 ft/sec when it was 5 ft above the ground, how high did it go?

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval

the surface area is 124cm²

step-by-step explanation:

length: 4cm

width: 9cm

height:   2cm

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = = -

y = - x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - + c ⇒ c = -

y = - x - ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form