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xeze [42]
3 years ago
6

An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the ref

inery to storage tanks located on the south bank of the river 7 km east of the refinery. The cost of laying pipe is $400,000/km over land to a point P on the north bank and $800,000/km under the river to the tanks. To minimize the cost of the pipeline, how far from the refinery should P be located? (Round your answer to two decimal places.)
Mathematics
1 answer:
Deffense [45]3 years ago
4 0
To minimize the cost, we take the straight distance from the refinery to the other side of the river as 2 km. Also, the 7 km will be the distance that has to be traveled by the pipeline in land. The total cost, C, is therefore,
               total cost =   (2 km)($800,000/km) + (7 km)($400,000 /km)
                    total cost = $4,400,000
Thus, the total cost of the pipeline is approximately $4,400,000.00. 
 
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3 years ago
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21

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2 years ago
There are 9 performers who are to present their acts at a variety show. How many different ways are there to schedule their appe
nika2105 [10]
For this type of question it is easiest to think about it in terms of a simple multiplication. The way we can do this is by working out how many possibilities there are for each position on the program. This is as follows:
For the first act, there are 9 possible performers we can pick
For the second act, because we have already assigned a performer to the first act, there are only 8 possible performers
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This goes on and on until we reach the final act, where we have used all the performers except one.
Therefore we can say:
For the first act there are 9 possibilities
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If we want to write this mathematically we can just say that there are:
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3 0
3 years ago
StartLayout enlarged left-brace 1st row 4 x minus 5 y = 5 2nd row negative 0.08 x + 0.10 y = 0.10 EndLayout
FinnZ [79.3K]

Answer:

x = 1.25

y = 0

Step-by-step explanation:

Given

4x - 5y = 5

0.08x + 0.10y = 0.10

Required

Determine the solution

Make x the subject of formula in: 4x - 5y = 5

4x  = 5 + 5y

Divide both sides by 4

\frac{4x}{4}  = \frac{5 + 5y}{4}

x = \frac{5 + 5y}{4}

Substitute x = \frac{5 + 5y}{4} in 0.08x + 0.10y = 0.10

0.08(\frac{5 + 5y}{4}) + 0.10y = 0.10

Solve the fraction

0.02(5 + 5y) + 0.10y = 0.10

Open the bracket

0.1 + 0.1y + 0.10y = 0.10

0.1 + 0.2y = 0.10

Subtract 0.1 from both sides

0.1-0.1 + 0.2y = 0.10 - 0.1

0.2y = 0

Divide both sides by 0.2

y =0

Substitute 0 for y in x = \frac{5 + 5y}{4}

x = \frac{5 + 5*0}{4}

x = \frac{5 + 0}{4}

x = \frac{5 }{4}

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5 0
3 years ago
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IRINA_888 [86]

~~~~~~ \textit{Compound Interest Earned Amount \underline{in 18 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$1300\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &18 \end{cases}

A=1300\left(1+\frac{0.06}{1}\right)^{1\cdot 18}\implies A=1300(1.06)^{18}\implies A\approx 3710.64 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y = 1300(1.06)^x~\hfill

7 0
2 years ago
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