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Yuki888 [10]
1 year ago
11

8th grade geometry part 3, please explain the steps :)))

Mathematics
1 answer:
Illusion [34]1 year ago
7 0

Answer: Volume = 97.6

Step-by-step explanation:

Volume of the cone first

V = 1/3 h π r²

<u>h = 0.7</u>

<u>r = 2</u>

V = 1/3 × 0.7 × 2² × 3.14

V = 1/3 × 0.7 × 4 × 3.14

V = 1/3 × 8.792

V = 2.93

Then find the volume of the cylinder if the cone was there.

V = π r² h

<u>r = 2</u>

<u>h = 8 </u>

V = 3.14 × 2² × 8

V = 3.14 × 4 × 8

V = 3.14 × 32

V = 100.48

Now you subtract the volume of the cone from the cylinder.

100.48 - 2.93 = 97.55

Rounded to the nearest tenth is 97.6

Hope this was helpful, I know it is a lot to look at but I tried my best.

Also, can I get brainliest? I worked really hard on this.

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Step-by-step explanation:

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From the graph attached,

Function is defined for all real values of x.

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On y-axis values of the function vary from -2 (excluding 2) to positive infinity.

Therefore, range of the function will be (-2, ∞).

2). Average rate of change of a function in the interval (a, b) is defined by,

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By using this expression we can find the average rate of change in the given interval.

Please give the correct interval for which the average rate of change is to be calculated.

8 0
2 years ago
The cross country team ran 1.797 hours yesterday and 1.56 hours today. Each member of the team ran 3.4 miles every hour. How man
nignag [31]

Answer:

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Step-by-step explanation:

To find the answer to this question, multiply the miles per hour by the amount of hours they ran.

1.797 (hours they ran yesterday) * 3.4 (miles per hour)

7 0
3 years ago
Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x
djverab [1.8K]

I'll assume the ODE is

y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}

Solve the homogeneous ODE,

y'' - 3y' + 2y = 0

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y = C_1 e^x + C_2 e^{2x}

For nonhomogeneous ODE (1),

y'' - 3y' + 2y = e^x

consider the ansatz particular solution

y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x

Substituting this into (1) gives

a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1

For the nonhomogeneous ODE (2),

y'' - 3y' + 2y = e^{2x}

take the ansatz

y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}

Substitute (2) into the ODE to get

b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1

Lastly, for the nonhomogeneous ODE (3)

y'' - 3y' + 2y = e^{-x}

take the ansatz

y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}

and solve for c.

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6 0
1 year ago
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igor_vitrenko [27]

Answer:

$7,000 at a rate of 7% and $21,000 at a rate of 14%.

Step-by-step explanation:

Let x be amount invested at 7% and y be amount invested at 14%.

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x+y=28,000...(1)

The interest earned at 7% in one year would be 0.07x and interest earned at 14% in one year would be 0.14x.

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Form equation (1), we will get:

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Therefore, an amount of $7,000 was invested at a rate of 14%.

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