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lisov135 [29]
3 years ago
13

12 less than a number is the same as 5 times the same number increased by 4

Mathematics
1 answer:
Lina20 [59]3 years ago
7 0

Answer: 12<5x+4

Step-by-step explanation:

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Suppose that 10 &lt; n &lt; 11. A possible value for n is
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10 < n < 11 \\
or n ∈ ( 10 , 11 )
Possible value can be easily obtained by generating an arithmetic mean

n = \frac{10 + 11}{2}  = 10.5
Else we've infinite numbers between them ,
According to density property of real numbers , we can have any real number satisfying the given inequality under condition 10 < n < 11
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Simplify the expression. write the answer using scientific notation 4(6.2x10^11)
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= 2.48 × 10¹²
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3 years ago
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

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

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