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galina1969 [7]
3 years ago
15

Tommy has a pet monkey. Every day, his monkey eats 4 apples in the morning. The monkey also eats two bananas for every banana th

at tommy eats. Write an equation to describe this situation where x is the number of bananas tommy eats and y is the total number of peices of fruit the monkey eats
Mathematics
2 answers:
Licemer1 [7]3 years ago
4 0

Answer:

Step-by-step explanation:

Let x represent the number of bananas that Tommy eats.

Let y represent the total number of pieces of fruit that the monkey eats.

Every day, his monkey eats 4 apples in the morning. The monkey also eats two bananas for every banana that tommy eats. This means that the number of bananas that the monkey eats is 2x. The expression for the total number of fruits eaten by the monkey becomes

y = 2x + 4

jeka57 [31]3 years ago
3 0

Answer:

Step-by-step explanation:

Y=2X+4

Where y is the the total no of fruits the monkey eats and x is the no of bananas the Tommy eats... 4 represents the 4 apples eaten every morning by the monkry

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2 years ago
The function f, where f(m) represents the amount of water, in gallons, remaining in a bathtub after m minutes of draining at a c
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kvasek [131]
<h3>Given</h3>
  • a cone of height 0.4 m and diameter 0.3 m
  • filling at the rate 0.004 m³/s
  • fill height of 0.2 m at the time of interest
<h3>Find</h3>
  • the rate of change of fill height at the time of interest
<h3>Solution</h3>

The cone is filled to half its depth at the time of interest, so the surface area of the filled portion will be (1/2)² times the surface area of the top of the cone. The filled portion has an area of

... A = (1/4)(π/4)d² = (π/16)(0.3 m)² = 0.09π/16 m²

This area multiplied by the rate of change of fill height (dh/dt) will give the rate of change of volume.

... (0.09π/16 m²)×dh/dt = dV/dt = 0.004 m³/s

Dividing by the coefficient of dh/dt, we get

... dh/dt = 0.004·16/(0.09π) m/s

... dh/dt = 32/(45π) m/s ≈ 0.22635 m/s

_____

You can also write an equation for the filled volume in terms of the filled height, then differentiate and solve for dh/dt. When you do, you find the relation between rates of change of height and area are as described above. We have taken a "shortcut" based on the knowledge gained from solving it this way. (No arithmetic operations are saved. We only avoid the process of taking the derivative.)

Note that the cone dimensions mean the radius is 3/8 of the height.

V = (1/3)πr²h = (1/3)π(3/8·h)²·h = 3π/64·h³

dV/dt = 9π/64·h²·dh/dt

.004 = 9π/64·0.2²·dh/dt . . . substitute the given values

dh/dt = .004·64/(.04·9·π) = 32/(45π)

7 0
3 years ago
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