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

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: