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3 years ago

what is the name of the quadrilateral with 2 pairs of eqaual sides and 2 pairs of opposite angles that are equal?

Mathematics

1 answer:

klasskru [66]3 years ago

8 0

A rhombus i think is what you are talking about

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In shop a crunchy peanut butter costs $325 for 250g

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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

hichkok12 [17]

Answer:

P is exactly 3km east from the oil refinery.

Step-by-step explanation:

Let's d be the distance in km from the oil refinery to point P. So the horizontal distance from P to the storage is 3 - d and the vertical distance is 2. Hence the diagonal distance is:

$\sqrt{(3 - d)^2 + 2^2} = \sqrt{(3 - d)^2 + 4}$

So the cost of laying pipe under water with this distance is

$800000\sqrt{(3 - d)^2 + 4}$

And the cost of laying pipe over land from the refinery to point P is 400000d. Hence the total cost:

$800000\sqrt{(3 - d)^2 + 4} + 400000d$

We can find the minimum value of this by taking the 1st derivative and set it to 0

$800000\frac{2*0.5*(3-d)(-1)}{\sqrt{(3 - d)^2 + 4}} + 400000 = 0$

We can move the first term over to the right hand side and divide both sides by 400000

$1 = 2\frac{3 - d}{\sqrt{(3 - d)^2 + 4}}$

$\sqrt{(3 - d)^2 + 4} = 6 - 2d$

From here we can square up both sides

$(3 - d)^2 + 4 = (6 - 2d)^2$

$9 - 6d + d^2 + 4 = 36 - 24d + 4d^2$

$3d^2-18d+27 = 0$

$d^2 - 6d + 9 = 0$

$(d - 3)^2 = 0$

$d -3 = 0$

d = 3

So the cost of pipeline is minimum when P is exactly 3km east from the oil refinery.

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3 years ago