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Vikentia [17]
3 years ago
11

The fuel cost per hour for running a ship is approximately one half the cube of the speed (measured in knots) plus additional fi

xed costs of $216 per hour. Find the most economical speed to run the ship for a 500 M (nautical mile) trip. Note: Assume that there are no major disturbances, such as heavy tides or stormy seas
Mathematics
1 answer:
andrezito [222]3 years ago
7 0

Answer:

6 knots

Step-by-step explanation:

Let the speed be v knots

then time taken to cover 500 M = 500 / v hrs

fuel consumption /hr = 216 + 0.5v^3

let F be the fuel consumption for trip

= [500/v][216 + 0.5v^3]

= 500[216/v + 0.5v^2]

dF/dv = 500[ - 216/v^2 + v]

d^2F/d^2v = 500[432/v^3 + 1] , i.e. +ve

so setting dF/dv will give a minima

500[ -216/v^2 + v] = 0

or v = 216/v^2

or v^3 = 216

solving, we get v = [216]^(1/3) = 6 knots

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OLga [1]

Answer:

Each sweater costs $8.30

Step-by-step explanation:

j = jeans

s = sweater

3j + 3s = 105.69

3(26.93) + 3s = 105.69

80.79 + 3s = 105.69

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3s = 24.9

/3   /3

s = 8.3 ($8.30)

Check:

3j + 3s = 105.69

3(26.93) + 3(8.3) = 105.69

80.79 + 24.9 = 105.69

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Hope this helps!

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2 years ago
Rick, Maya, and Tom work in a call center. On a particular day, Rick worked 5 and ½ hours, Maya worked 6.1 hours, and Tom worked
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Hope I helped! :)

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4 years ago
Paulo's economics course requires two papers ? ?one long and one short ? ?throughout the semester. The number of pages, l l, in
mixer [17]

Answer:

27 pages.

Step-by-step explanation:

Let l be the number of pages in the long paper and s be the number of pages in the short paper.

We have been given that the total number of pages for both papers is 40. We can represent this information in an equation as:

l+s=40...(1)

We are also told that the number of pages in the long paper is one more than two times the number of pages in the short paper. We can represent this information in an equation as:

l=2s+1...(2)

We will use substitution method to solve system of linear equations.

From equation (1) we will get,

s=40-l

Upon substituting this value in equation (2) we will get,

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Upon adding 2l to both sides of our equation we will get,

l+2l=80-2l+2l+1

l+2l=80+1

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Let us divide both sides of our equation by 3 we will get,

\frac{3l}{3}=\frac{81}{3}

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