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Science & Mathematics Mathematics

The number of gallons of water in a tank t minutes ofter the tank started to drain is Q(t)=200(30-t)^2....?

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hey can anyone help with this?? The number of gallons of water in a tank t minutes ofter the tank started to drain is Q(t)=200(30-t)^2. how fast is the water running out at the end of ten minutes.

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i suspect that you get the first derivative and then substitute in 10 for t... can anyone help? Follow

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Best Answer: Your suspicion is correct.

The number of gallons of water in a tank t minutes after the tank has started to drain is Q(T)=200(30-t)^2.? The number of gallons of water in a tank t minutes after the tank has started to drain isQ(t) = 200(30-t)^2?

Q'(t) = 200 * 2 * (-1) * (30 - t).....The (-1) is from the chain rule -> it is the derivative of 30 - t.

The number of gallons of water in a tank t minutes after the tank has started to drain is Q(t) = 250(35 - t)^2?

So Q'(t) = -400 * 20 = - 8,000 gallons per minute. Notice that this answer is negative because the quantity of water is decreasing.

If a tank holds 2000 gallons of water, which drains from the bottom of the tank in 80 minutes, then...?

The water is running out of the tank at a rate of 8,000 gallons per minute ...<<<... Answer If the use of the chain rule bothers you then you can simply multiply the terms of the equation for Q, getting:

Starting a fresh water tropical fish tank?

Q(t) = 200 * (900 - 60t + t^2) = 180,000 - 12,000t + 200t^2

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So Q'(t) = -12,000 + 400t = -12,000 + 4,000 = -8,000

Find the permutation that follows 526431 in the lexicographic ordering of permutations of{1,2,3,4,5,6}?

The tank is emptying at 8,000 gallons per minute...<<<... Same Answer as above!

Let f(x) be a function satisfying f'(X)=f(x) and f(0)=2 then òf(X)/3+4f(x) dx is?

Since both methods are valid the answers have to be the same. I prefer to use the chain rule and avoid some of the messier arithmetic. . Source(s): 4 years of experience teaching calculus

Find the maximum of the following function: h(x, y) = ln(x^32*y^16) given the constraints: 2x^2+y^2 = 8, x>0, y>0?

Gerry · 9 years ago 1

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Solution set to (5x-3)(4x+1)=0?

Q(t) = 200(900 - 60t + t²)

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Maths homework help?

= 200t² - 12,000t + 180,000

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We're asked how fast the water is running out. This is the rate of change in the volume of water with respect to time,

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or dQ/dt: A certain factory employs 60 workers . if 4 out of every 5 workers are married, how many married workers are employed by the factory?

Q'(t) = 400t - 12,000

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What adds to be 2 and multiplies to be -3?

At the end of 10 minutes,

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Q'(10) = 4,000 - 12,000 = -8,000 gal / min SOLUTION ? · 9 years ago 0

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Yup, Q'(t) is the rate the quantity of water is changing. To keep it clean, simplify before differentiating. Pamela · 9 years ago 0

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