Answer: 3.675 seconds
Step-by-step explanation:
Hi, when the object hits the ground, h=0:
h=−16t^2+48.6t+37.5
0=−16t^2+48.6t+37.5
We have to apply the quadratic formula:
For: ax2+ bx + c
x =[ -b ± √b²-4ac] /2a
Replacing with the values given:
a=-16 ; b=48.6; c=37.5
x =[ -(48.6) ± √(-48.6)²-4(-16)37.5] /2(-16)
x = [ -48.6 ± √ 4,761.96] /-32
x = [ -48.6 ± 69] /-32
Positive:
x = [ -48.6 + 69] /-32 = -0.6375
Negative:
x = [ -48.6 - 69] /-32 = 3.675 seconds (seconds can't be negative)
Feel free to ask for more if needed or if you did not understand something.
Divide the volume by the two measures. 728 ÷ 8 = 91 ÷ 6.5 = 14 so, the shoe box is we inches long.
Answer:
The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h
Step-by-step explanation:
At noon the location of Lan = 300 km north of Makenna
Lan's direction = South
Lan's speed = 60 km/h
Makenna's direction and speed = West at 75 km/h
The distance Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km
The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km
The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km
Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.
By Pythagoras' theorem, we have;
s² = x² + y²
The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61
ds²/dt = dx²/dt + dy²/dt
2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt
2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900
ds/dt = 900/(2×30·√61) ≈ 1.92
The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h
