The equation $y = -4.9t^2 + 42t + 18.9$ describes the height (in meters) of a ball tossed up in the air at 42 meters per second from a height of 18.9 meters from the ground, as a function of time in seconds. In how many seconds will the ball hit the ground?

9 seconds

Step-by-step explanation:

y = -4.9t² +42t +18.9

if the ball will be on the ground then y =0 and we need to find t by solving the quadratic equation

-4.9t² +42t +18.9 = 0

t= (-b±√b²-4ac)/2a, quadratic formula

t= {-42±√[42²-4*(-4.9)(18.9)] }/ 2(-4.9) , use calculator to solve

t= (-42 ±√2134.44)/ -9.8, solve the square root

t= (-42 ±46.2)/-9.8

one t = (-42 - 46.2)/-9.8 = - 88.2/ -9.8 = 9 seconds

another t = (-42 + 46.2)/-9.8 = 4.2 / -9.8 ≈ - 0.43 seconds reject this answer

answer: 5 17/33 % (to the tenth)

step-by-step explanation:

he guessed: 34.5 pounds

actual weight is 32.7

there for he is off:

34.5 -32.7 = 1.8

32.7 is the 100%

1.8 represents %? of 32.7

therefore:

32.7   - 100%

1.8     -     x%

x= 1.8*100%/32.7% = 18*100%/327%

x=1800%/327%   = (5 165/327)%

to the tenth:

5 17/33 %

we would need a picture of the quadrilateral in order to answer the question. also, it seems that the question is missing some components.