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mezya [45]
3 years ago
7

What is the value of ry' when x = 5 and y = 10?

Mathematics
2 answers:
Dafna11 [192]3 years ago
7 0

Answer:

1250

Step-by-step explanation:

X^3 • Y^1

Since X=5 & Y=10,

Substitute

5^3 •10^1

simplify

5*5*5*10=1250

dlinn [17]3 years ago
5 0

Answer:

135

Step-by-step explanation:

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How do you solve - (4n + 5y + 6)
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-4n + 5y + 6

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Distribute the negative

−4n + 5y + 6

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What is the slope of the line containing the points (1,-1) and (3,3)
marishachu [46]

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4 0
3 years ago
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

5 0
3 years ago
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