Please help me....trying to get my HS diploma, i did not graduate :(
A ball is thrown in air and it's height, h(t) in feet, at any time, t in seconds, is represented by the equation h(t)=−t2+7t. When is the ball higher than 10 feet off the ground?


Please Help Metrying To Get My HS Diploma I Did Not Graduate A Ball Is Thrown In Air And Its Height Ht In Feet At Any Time T In Seconds Is Represented By The Eq class=

Sagot :

Answer:

A.

Step-by-step explanation:

so, the equation is

h(t) = -t² + 7t

so, we need to find the solutions for t (the time when the ball is exactly 10 ft in the air). there had to be 2 solutions, as the ball first goes up passing the 10 ft height, and then comes back down again, passing the 10 ft mark a second time. and between these 2 times the ball is higher (but not equal, so, we can only use < or > as inequality signs) than 10 ft.

10 = -t² + 7t

-t² + 7t - 10 = 0

the generation solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = t

a = -1

b = 7

c = -10

t = (-7 ± sqrt(7² - 4×-1×-10))/(2×-1) =

= (-7 ± sqrt(49 - 40))/-2 = (-7 ± sqrt(9))/-2

t1 = (-7 + 3)/-2 = -4/-2 = 2 seconds

t2 = (-7 - 3)/-2 = -10/-2 = 5 seconds

so, between 2 and 5 seconds airtime the ball is higher than 10 ft.

and remember : HIGHER THAN.

so, we cannot use any equality (like <= or >=).

t must be higher than 2 and lower than 5 :

2 < t < 5