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olga55 [171]
3 years ago
12

A farmer feeds her four horses 64 bales of hay each week. How many bales of hay would the farmer need in order to feed six horse

s for a week?
Mathematics
1 answer:
IrinaVladis [17]3 years ago
8 0

In this question, we're trying to find how many bales of hay the farmer needs to get in order to feed 6 horses

First, we would need to find how many bales of hay each horse eats.

Find this by dividing 64 by 4

64 ÷ 4 = 16

We now know that each horse eats 16 bales of hay

Now, multiply 16 by 6 in order to find how many bales of hay the farmer needs to get to feed 6 horses.

16 × 6 = 96

Therefore, the farmer needs to get 96 bales of hay in order to feed 6 horses.

Answer:

96 bales of hay

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Euclid’s root beer mug is shaped basically like a cylinder that is eight inches tall with a radius of three inches. Aristotle’s
olchik [2.2K]

The Euclid's mug holds the most root beer.

Step-by-step explanation:

Euclid's mug :

Height = 8 inches

Radius = 3 inches

Volume = π(r x r) h

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= 226.08 cubic inches

Aristotle's mug:

Height = 18 inches

Diameter = 4 inches

Radius = 2 inches

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The Euclid's mug holds the most root beer.

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Describe the graph of the equation x = 7. Is the equation a function?
DerKrebs [107]

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Air is being lost from a spherical balloon at a constant rate of 3/2cm3s-1. Find the rate at which the radius is decreasing at t
cricket20 [7]

Answer:

The rate at which the radius is decreasing when the radius is 6 cm is approximately 3.316 × 10⁻³ cm/s

Step-by-step explanation:

The rate at which air is being lost from the balloon = 3/2 cm³/s

The rate at which the radius is decreasing when the radius is 6 cm long is given as follows;

The rate at which air is being lost from the balloon = dV/dt = 3/2 cm³/s

dV/dt = dV/dr × dr/dt

Where;

dr/dt = The rate at which the radius is decreasing

dV/dr = d(4/3×π×r³)/dr = 4·π·r²

Therefore, we have;

dr/dt = (dV//dt)/(dV/dr) = (3/2 cm³/s)/(4·π·r²)

dr/dt = (3/2 cm³/s)/(4·π·r²)

When r = 6 cm, we have;

dr/dt = (3/2 cm³/s)/(4 × π × (6 cm)²) ≈ 3.316 × 10⁻³ cm/s

Therefore, the rate at which the radius is decreasing, dr/dt, when the radius is 6 cm long ≈ 3.316 × 10⁻³ cm/s.

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