The height of a small rock falling from the top of a 124-ft-tall building with an initial downward velocity of –30 ft/sec is modeled by the equation h(t) = –16t2 – 30t + 124, where t is the time in seconds. For which interval of time does the rock remain in the air?

0≤ t ≤2

Step-by-step explanation:

Hi, to solve this question the equation must be equal to 0 (h=0)

h(t) = –16t2 – 30t + 124

0 = –16t2 – 30t + 124

we have to apply the quadratic formula:

For: ax2+ bx + c

a=-16 ; b=-30; c=124

x =[ -b ± √b²-4ac] /2a

Replacing with the values given:

x =[ -(-30) ± √(-30)²-4(-16)124] /2(-16)

x = [ 30 ± √900 +7,936] /-32

x = [ 30± √8,836] /-32

x = [ 30 ± 94] /-32

Positive:

x = [ 30 + 94] /-32 = -3.875

Negative:

x =[ 30 - 94] /-32 = 2 seconds

Since time can't be negative, the rock remains in the air for 2 seconds.

0≤ t ≤2

Feel free to ask for more if needed or if you did not understand something.

38.5

step-by-step explanation:

lets say, first number = r

second number = 3r+7

total = 49

(r)+(3r+7)= 49

4r+7 = 49

4r = 42

r = 42/4

r= 10.5

3r+7 = 3(10.5)+7 = 38.5

hope this