a ball is thrown from a height of 4 feet with an intial upward velocity of 40 ft per second. The ball's height h (in feet) after t seconds is given by the following. h=4+40t-16t^(2 )Find the values of t for which the ball's height is 26 feet.

Step-by-step explanation:

you need o solve 26=4+40t-16t^2

the equation becomes:

the equation becomes:-16t^2+40t+4-26=0

the equation becomes:-16t^2+40t+-22=0 or8t^2-20t+11=0

8t^2-20t+11=0, a=8 , b=-20 c=11 the discriminant =b^2-4ac=(-20)^2-4x8x11=48

t1=(20-squarootof48)/16 =13/16 =0.81 seconds t2=27/16=1.68 second I rounded my answers

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}