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Juli2301 [7.4K]
3 years ago
7

If the volume of a storage container is 6720 cubic inches, what is the height of the container

Mathematics
1 answer:
guajiro [1.7K]3 years ago
6 0
18.87 inches the cube root of 6720
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Solve the initial-value problem<br><br> y' = x^4 - \frac{1}{x}y, y(1) = 1.
natta225 [31]

The ODE is linear:

y'=x^4-\dfrac yx

y'+\dfrac yx=x^4

Multiplying both sides by x gives

xy'+y=x^5

Notice that the left side can be condensed as the derivative of a product:

(xy)'=x^5

Integrating both sides with respect to x yields

xy=\dfrac{x^6}6+C

\implies y(x)=\dfrac{x^5}6+\dfrac Cx

Since y(1)=1,

1=\dfrac16+C\implies C=\dfrac56

so that

\boxed{y(x)=\dfrac{x^5}6+\dfrac5{6x}}

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3 years ago
Geometry hw! Pls help!
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Answer:

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Step-by-step explanation:

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Every quadrilateral with opposite angles supplementary can be inscribed in a circle.TrueFalseIf a triangle is inscribed in a cir
solniwko [45]

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Answer 1: True

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What number is not a composite number 2,9,15,21?
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4 years ago
A standard number cube has 6 sides labeled 1 to 6. apu rolls a standard number cube 30 times. How many times can
Leviafan [203]

Answer:

The expected value of rolling a 5 or 6 is 10.

Step-by-step explanation:

The sample space of rolling a standard number cube is:

S = {1, 2, 3, 4, 5 and 6}

The cube is standard, this implies that each side has an equal probability of landing face-up.

So, the probability of all the six outcomes is same, i.e. \frac{1}{6}.

Now it is provided that Apu rolls the cube <em>n</em> = 30 times.

Let the random variable <em>X</em> represent the value on the face of cube.

The event of rolling a 5 and rolling a 6 are mutually exclusive, i.e. they cannot occur together.

So, P (X = 5 and X = 6) = 0.

Compute the probability of getting a 5 or 6 as follows:

P (X = 5 or X = 6) = P (X = 5) + P (X = 6) - P (X = 5 and X = 6)

                            = P (X = 5) + P (X = 6)

                            =\frac{1}{6}+\frac{1}{6}\\\\=\frac{2}{6}\\\\=\frac{1}{3}

Compute the expected value of rolling a 5 or 6 as follows:

E(X = 5\ \text{or}\ X = 6)=n\times P(X = 5\ \text{or}\ X = 6)

                               =30\times \frac{1}{3}\\\\=10

Thus, the expected value of rolling a 5 or 6 is 10.

8 0
3 years ago
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