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Gemiola [76]
3 years ago
12

A bucket that weighs 4lband a rope of negligible weight are used todraw water from a well that is 80ftdeep. The bucket is filled

with 40lbof waterand is pulled up at a rate of 2ft/s, but water leaks out of a hole in the bucketat a rate of 0.2lb/s. Find the work done in pulling the bucket to the top ofthe well.
Mathematics
1 answer:
liq [111]3 years ago
8 0

Answer:

Workdone = 3200 lb.ft

Step-by-step explanation:

We are told that the bucket is filled with 40 lb of water but water leaks out of a hole in the bucket at a rate of 0.2lb/s

Thus,

Weight of water at any given time (t) would be;

w(t) = 40 - 0.2t - - - - (1)

We are told the bucket is pulled up at a rate of 2ft/s.

Thus, height at time (t); y = 0 + 2t = 2t

Since y = 2t,

Then,t = y/2

Put y/2 for t in eq 1

Thus; w(y) = 40 - 0.2(y/2)

w(y) = 40 - 0.1y

Now, at y = 80 ft, we have;

w(80) = 40 - 0.1(80)

w(80) = 40 - 8 = 32 lb

Since 32 lbs are left, it means there is always water in the bucket.

Thus, work done is;

W = 80,0[∫(Total weight).dy]

W = 80,0[∫[(weight of rope) + (weight of bucket) + (weight of water)]dy]

W = 80,0[∫[0 + 4 + 40 - 0.1y]dy]

Integrating, we have;

W = [44y - y²/20] at boundary of 80 and 0

So,

W = [44(80) - 80²/20] - [0 - 0²/20]

W = 3200 lb.ft

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Answer:

P(X= k) = (1-p)^k-1.p

Step-by-step explanation:

Given that the number of trials is

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Given p = success,

1 - p = failure

Hence the distribution is described as: Pr ( FFFF.....FS)

Pr(X= k) = (1-p)(1-p)(1-p)....(1-p)p

Pr((X=k) = (1 - p)^ (k-1) .p

Since N<=k

Pr (X =k) = p(1-p)^k-1, k= 1,2,...k

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0, elsewhere.

Given p= 0.2, k= 3,

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3 years ago
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3 years ago
Find the value of k for which the line y = kx + 6 is a tangent to the curve x^2 + y^2 – 10x + 8y = 84
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Answer:

k = 1/2

Step-by-step explanation:

input y = kx +6 into the equation of the curve x² + y² – 10x + 8y = 84

x² + (kx + 6)² - 10x + 8(kx + 6) = 84

expand:

x² + k²x² + 12kx + 36 - 10x + 8kx + 48 = 84

simplify by collecting like terms:

x² + k²x² + 20kx - 10x + 84 = 84

subtract 84 on both sides to bring it to the left:

x² + k²x² + 20kx - 10x + 84 - 84 = 0

x² + k²x² + 20kx - 10x = 0

factorise out x:

x²(1 + k²) + x(20k - 10) = 0

using the discriminant b² - 4ac where b is 20k - 10, a is 1 + k² and c is 0, substitute them in the formula b² - 4ac:

b² - 4ac

(20k - 10)² - 4(1 + k²)(0) = 0

the part highlighted in bold is gone because it's all multiplied by 0, so we are left with (20k - 10)² = 0

(20k - 10)² is the same as

(20k - 10)(20k - 10)

equate both to 0

20k - 10 = 0 and 20k - 10 = 0

add 10 on both sides

20k = 10 and 20k = 10

divide 20 on both sides

k = 10/20 and k = 10/20 which are both the same

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2 years ago
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<h3>Answer: 6 Frames</h3>

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Profit is the money you make.

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If you sold five frames, you would make: (8.50 x 5 = $42.50)

Note: $42.50 <u>is not</u> greater than your expenses, so you haven't make any profit yet.

If you sold six frames, you would make: (8.50 x 6 = $51.00)

Note: $51.00 <u>is</u> greater than your expenses, so you made some profit.

<h3>Additional Information</h3>

To find the money you made or your profit, simply subtract the money you received from your expenses.

$51.00 - $46.00 = $5.00

You only made <u>$5.00</u>, but to make a larger profit, you have to sell more frames.

<h3>Conclusion</h3>

I hope my explanation provided excellent aid. Oh, and you're welcome.

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