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2 years ago

10

The variables x and y vary directly. Use the values to find the constant of proportionality, k. Then write an equation that rela

tes x and y.
y=72; x=3

2 answers:

den301095 [7]2 years ago

7 0

Y/X=? Basically saying that 72 divided by 3 is what

Travka [436]2 years ago

3 0

y = kx

Replace y and x to solve for k:

72 = 3k

Divide both sides by 3:

k = 24

Now replace k with 24:

y = 24x

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A barrel of crude oil contains about 5.61 cubic feet of oil. How many barrels of oil contained in 1 mile (5280 feet) of a pipeli

8_murik_8 [283]

1)volume of the pipeline

The pipeline is a cylinder, therefore;
Volume (cylinder)=πr²h
r=radius
h=height of the cylinder

diameter=6 in*(1 ft / 12 in)=0.5 ft
raius=diameter / 2=0.5 ft / 2=0.25 ft.
height=5280 ft

Volume (pipeline)=π(0.25 ft)²(5280 ft)=330π ft³≈1036.73 ft³.

2) we calculate the number of barrel
1 mile of oil in this pipeline is 330π ft³ of oil.

1 barrel of crude------------------5.61 ft³
x----------------------------------330π ft³

x=(1 barrel*330π ft³) / 5.61 ft³=184.8 barrels

3) we calculate the price.

1 barrel---------------$100
184.8 barrels---------- x

x=(184.8 barrels * $100) / 1 barrel=$18,480

Solution: ≈$18,480

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2 years ago

What is the greatest common factor of the pair of numbers?

geniusboy [140]

The answer is 66.
66 * 3 = 198
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3 0

2 years ago

Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):

$y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1 \right)} \, dx$

$y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx$

$y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C$

$C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0$

$C = 0$

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

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Answer:

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2 years ago

Read 2 more answers

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2 years ago

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