A projectile is thrown upward so that its distance above the ground after t seconds is h=-121t^2 +504t. After how many seconds does it reach it maximum height?

2.08secs

Step-by-step explanation:

Given the equation modeled by the height as h=-121t^2 +504t

The velocity of the object at the max height is zero, hence;

v = dh/dt = 0

dh/dt = -2(121)t + 504

dh/dt = -242t + 504

0 = -242t + 504

242t = 504

t = 504/242

t = 2.08secs

Hence the projectile reaches its maximum height after 2.08secs

b is 38 degrees.

a is 52 degrees.

this is because if b is 38 and a is 14 degrees more than b than 38+14=52 and 52+38=90 degrees. hope this .

final total: 63

step-by-step explanation:

60 times sales tax

60 times 0.05

3

60 + 3

63