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1 year ago

11

GEOMETRY)

1 answer:

evablogger [386]1 year ago

8 0

Answer: 130

Step-by-step explanation:

Let the measure of arc DB be x.

Then,

$80=\frac{x+30}{2}\\\\160=x+30\\\\x=130$

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Use integration by parts to find the integrals in Exercise.<br> ∫^3_0 3-x/3e^x dx.

Viefleur [7K]

Answer:

8.733046.

Step-by-step explanation:

We have been given a definite integral $\int _0^3\:3-\frac{x}{3e^x}dx$ . We are asked to find the value of the given integral using integration by parts.

Using sum rule of integrals, we will get:

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx$

We will use Integration by parts formula to solve our given problem.

$\int\ vdv=uv-\int\ vdu$

Let $u=x$ and $v'=\frac{1}{e^x}$ .

Now, we need to find du and v using these values as shown below:

$\frac{du}{dx}=\frac{d}{dx}(x)$

$\frac{du}{dx}=1$

$du=1dx$

$du=dx$

$v'=\frac{1}{e^x}$

$v=-\frac{1}{e^x}$

Substituting our given values in integration by parts formula, we will get:

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(x*(-\frac{1}{e^x})-\int _0^3(-\frac{1}{e^x})dx)$

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx=3x-\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

Compute the boundaries:

$3(3)-\frac{1}{3}(-\frac{3}{e^3}- (\frac{1}{e^3}))=9+\frac{4}{3e^3}=9.06638$

$3(0)-\frac{1}{3}(-\frac{0}{e^0}- (\frac{1}{e^0}))=0-(-\frac{1}{3})=\frac{1}{3}$

$9.06638-\frac{1}{3}=8.733046$

Therefore, the value of the given integral would be 8.733046.

6 0

3 years ago

What is the equation of the circle with center (–2, 5) that passes through the point (–3, 8)?

marta [7]

(X-(-2))2+(y-5)2=10

Is the third

4 0

2 years ago

Elevator 1 moved up 15 feet from the ground level. Its position is labeled as +15. Elevator 2 moved down 6 feet from the ground

sergey [27]

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Elevator 2 moved down 6 feet from the ground level hence its position is labelled as $-6$

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3 years ago

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stich3 [128]

Answer:

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6 0

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