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Art [367]
3 years ago
9

Based on the graph which statement is true

Mathematics
1 answer:
Dahasolnce [82]3 years ago
6 0

Answer:

D. He needs 6 cups of flour for 3 batches

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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.

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3 years ago
What is the answer of 6xy + (-9xy)
kati45 [8]

Answer:

-3xy

Step-by-step explanation:

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A is the answer:)))))))))))))))))))))))
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John ordered medium French fries $1.89,2 hot dogs $2.19,and an iced tea $1.35. All the menu items are subject to a sales tax of
tatuchka [14]

$4.18 would be his change back

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Identify the a b and c values of the following quadratic expression 4x^2+5x=4
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A is the 4

B is the 5

and C is also 4.


A is always the number in front of x^2. B is always the number i front of the normal x, and C is the number that is on the other side of the equal sign!

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