You are playing basketball with your friends. The height (in feet) of the ball above the ground t seconds after a shot is released from your hand is modeled by the function f(t)= -16t^2 + 32t + 5.2. What is the value of t when the ball is released from your hand ?

2.15 seconds

Step-by-step explanation:

Given the equation of the height (in feet) of the ball above the ground t seconds after a shot is released from your hand modeled by the function

f(t)= -16t² + 32t + 5.2 where t is in seconds.

Note that the value of the height at the point the ball is released from the hand is 0 feet.

To get value of t when the ball is released from your hand, we will substitute f(t) = 0 into the modeled equation as shown;

f(t)= -16t² + 32t + 5.2

-16t² + 32t + 5.2 = 0

Multiply through by 10

-160t² + 320t + 52 = 0

Dividing through by -4

40t² - 80t - 13 = 0

t = 80±√80²-4(40)(-13)/2(40)

t = 80±√6400+2080/80

t = 80±√8480/80

t = 80±92.09/80

t = 80+92.09/80

t = 172.09/80

t = 2.15 secs

Hence, the value of t when the ball is released from your hand is approximately 2.15seconds.

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

its kind of blurry. can you put a clearer pic?

step-by-step explanation: