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Julli [10]
4 years ago
8

Suppose that you are the power company and must provide power to a lighthouse on a rock one mile off shore. The nearest pole to

run the line from is 2 miles up the shore from the line straight out to the lighthouse. If the line costs $1000/mile on land and $1500/mile on water, find the path for the line that minimizes the total cost. Your answer should state how far from the pole your line leaves the shore and goes to the lighthouse. Show that this is a minimum and find the cost for the path you have chosen.
Mathematics
1 answer:
Levart [38]4 years ago
6 0
If we let x as the length of the line on line and y on water, and C as the total c ost, then we have the equation:
C = 1000x + 1500y

The figure formed by the scenario is a right triangle, so we have:
y² = 1² + (2 - x)²
Solving for y,
y = √(1 + 4 - 4x + x²)
y = √(5 - 4x + x²)

Subsituting:
C = 1000x + 1500√(5 - 4x + x²)
The minimum length of the line can be solved by taking the derivative of the equation and equating it to 0. So,
1000 + 1500(0.5)(-4 +2x)/√(5 - 4x + x²) = 0
x = 1.33 miles
From the equation: y = √(5 - 4x + x²)
y = 1.20 miles
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Answer:

17x² + 2x – 3

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(9x² + 3x – 5) + (8x² – x + 2) →

[Group like terms]

(9x² + 8x²) + (3x – x) + (2 – 5) →

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3 years ago
A number cube is rolled. What is the probability that a number greater than 4 will be rolled?
jek_recluse [69]

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1/6

Step-by-step explanation:

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2 years ago
Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion do
V125BC [204]

Answer:

Step-by-step explanation:

It is given that

\Delta=\begin{vmatrix}3&0&3\\2 &3&3\\0 &4&-2\end{vmatrix}

By cofactor expansion across the first row, we get

\Delta=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}

\Delta=3\left[(-1)^{1+1}\begin{vmatrix}3&3\\4&-2\end{vmatrix}\right]+0\left[(-1)^{1+2}\begin{vmatrix}2&3\\0&-2\end{vmatrix}\right]+3\left[(-1)^{1+3}\begin{vmatrix}2&3\\0&4\end{vmatrix}\right]

\Delta=3\left[-18\right]+0\left[(-1)(-4)\right]+3\left[8\right]

\Delta=-54+0+24

\Delta=-30

Therefore, the value of determinant is -30.

By cofactor expansion across the second column, we get

\Delta=a_{12}C_{12}+a_{22}C_{22}+a_{32}C_{32}

\Delta=0\left[(-1)^{2+1}\begin{vmatrix}2&3\\0&-2\end{vmatrix}\right]+3\left[(-1)^{2+2}\begin{vmatrix}3&3\\0&-2\end{vmatrix}\right]+4\left[(-1)^{3+2}\begin{vmatrix}3&3\\2&3\end{vmatrix}\right]

\Delta=0\left[(-1)(-4)\right]+3\left[(-6)\right]+4\left[(-1)3\right]

\Delta=-18-12

\Delta=-30

Therefore, the value of determinant is -30.

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4 years ago
A caterer charges $120 for 15 people to cater a party and $7 per person after that.
Nitella [24]

Answer

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Step-by-step explanation:

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4 years ago
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Answer:

I think C

Step-by-step explanation:

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