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Ostrovityanka [42]
3 years ago
12

Please help (20 points)

Mathematics
1 answer:
Evgesh-ka [11]3 years ago
8 0

35.\\V=\dfrac{\pi r^2h}{3}\ \ \ \ |\cdot3\\\\3V=\pi r^2h\ \ \ \ |:\pi r^2\\\\\boxed{h=\dfrac{3V}{\pi r^2}}

36.\\V=150\ m^3\\r=5\ m\\h=?\\\\\text{substitute}\\\\h=\dfrac{3\cdot150}{\pi\cdot5^2}=\dfrac{450}{25\pi}=\dfrac{18}{\pi}\approx\dfrac{18}{3}=6\ m\\\\Answer:\ The\ height\ is\ greater\ than\ a\ radius.

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Read 2 more answers
3. At noon Lan is in his blue mini-van 300 km north of Makenna in her Bugatti. Lan heads south
LiRa [457]

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

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2 years ago
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