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Helga [31]
4 years ago
15

I need help with this problem please ​

Mathematics
1 answer:
Gwar [14]4 years ago
4 0

Answer:

  (a) reduction

  (b) 1/2

Step-by-step explanation:

The image figure A'B'C'D' is smaller than the original figure ABCD, so the dilation is a <em>reduction</em>.

Each of the points A'B'C'D' is half as far from the origin as the original points ABCD, so the scale factor is 1/2.

_____

Draw lines C'C and D'D. You will see they meet at the origin, which is the center of dilation. Then look at how far the points are along those lines. C' is one grid square diagonal along the line; C is 2 grid square diagonals along that line, so is twice as far from the origin. That is, C' is 1/2 the distance of C, so represents a reduction by a scale factor of 1/2.

The same distance considerations are observed along the line D'D. The point D' is the diagonal of a 2x1 rectangle from the origin (A distance of √5.) The point D is the diagonal of a 4x2 rectangle from the origin, so is twice as far. Once again D' is 1/2 the distance of D, so represents a reduction by a factor of 1/2.

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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:

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

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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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