Lupe tosses a ball up to Quyen, waiting at a third-story window, with an initial velocity of 30 feet per second. She releases the ball from a height of 6 feet. The equation h = –16t2 + 30t + 6 represents the height h of the ball after t seconds. If the ball must reach a height of 25 feet for Quyen to catch it, does the ball reach Quyen? Explain. (Hint: Substitute 25 for h and use the discriminant.)

-316 is the discriminant and thus the ball doesn't reach Quyen because a negative discriminant means that it's not a real possibility

Step-by-step explanation:

We are given that the equation of the the height h of the ball after t seconds is;

h = –16t² + 30t + 6

Now, we want to know If the ball reaches Quyen after it must have reached a height of 25 feet.

Thus, h = 25ft

So,

25 = –16t² + 30t + 6

Subtract 25 from both sides to give;

–16t² + 30t + 6 - 25 = 0

–16t² + 30t - 19 = 0

Using quadratic equation, we have;

t = [-(30) ± √(30² - (4 x -16 x - 19)]/(2 x 30)

t = [-30 ± √(-316)]/60

We are told to use the discriminant which is -316 from above.

So, No, the ball doesn't reach Quyen because a negative discriminant means that it's not a real possibility

answer: f true

step-by-step explanation:

7. d. 64

8. i. 32

9. c. 125

10. j. none of the above

11. d. 81

12. i. 216

13. e. none of the above

14. g. 240

15. e. none of the above

16. g. 32

17.   e. none of the above