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kari74 [83]
2 years ago
15

1. Use precise mathematical language to justify and explain each mathematical process.

Mathematics
1 answer:
iris [78.8K]2 years ago
6 0

Mathematical processes are the steps used to determine the result of operations

  • The perimeter of the rectangle is 10x^2 - 14x + 28
  • The area of the rectangle is 5x^3 +17x^2  -31x + 45

The dimension of the rectangle is given as:

Length = 5x^2 - 8x + 9

Width = x + 5

<u>(1) The perimeter of the triangle</u>

This is the sum of the side lengths of the rectangle.

So, we have:

Perimeter = 2 \times (Length + Width)

This gives

Perimeter = 2 \times (5x^2 - 8x + 9 + x + 5)

Collect like terms

Perimeter = 2 \times (5x^2 - 7x + 14)

Open brackets

Perimeter = 10x^2 - 14x + 28

Hence, the perimeter of the rectangle is 10x^2 - 14x + 28

<u>(2) The area of the triangle</u>

This is the product of the dimensions of the rectangle.

So, we have:

Area = Length \times  Width

This gives

Area = (5x^2 - 8x + 9) \times  (x + 5)

Open brackets

Area = 5x^3 - 8x^2 + 9x +25x^2 - 40x + 45

Collect like terms

Area = 5x^3 - 8x^2 +25x^2+ 9x  - 40x + 45

Area = 5x^3 +17x^2  -31x + 45

Hence, the area of the rectangle is 5x^3 +17x^2  -31x + 45

Read more about areas and perimeters at:

brainly.com/question/15656646

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Lady_Fox [76]

For the first integral, z = π/4 is a pole of order 3 and lies inside the contour |z| = 1. Compute the residue:

\displaystyle \mathrm{Res}\left(\frac{e^z\cos(z)}{\left(z-\frac\pi4\right)^3}, z=\frac\pi4\right) = \lim_{z\to\frac\pi4}\frac1{(3-1)!} \frac{d^{3-1}}{dz^{3-1}}\left[e^z\cos(z)\right]

We have

\dfrac{d^2}{dz^2}[e^z\cos(z)] = -2e^z \sin(z)

and so

\displaystyle \mathrm{Res}\left(\frac{e^z\cos(z)}{\left(z-\frac\pi4\right)^3}, z=\frac\pi4\right) = - \lim_{z\to\frac\pi4} e^z \sin(z) = -\frac{e^{\pi/4}}{\sqrt2}

Then by the residue theorem,

\displaystyle \int_C \frac{e^z\cos(z)}{\left(z-\frac\pi4\right)^3} \, dz = 2\pi j \left(-\frac{e^{\pi/4}}{\sqrt2}\right) = \boxed{-\sqrt2\,\pi e^{\pi/4} j}

For the second integral, z = 2j and z = j/2 are both poles of order 2. The second poles lies inside the rectangle, so just compute the residue there as usual:

\displaystyle \mathrm{Res}\left(\frac{\cosh(2z)}{(z-2j)^2\left(z-\frac j2\right)^2}, z=\frac j2\right) = \lim_{z\to\frac j2}\frac1{(2-1)!} \frac{d^{2-1}}{dz^{2-1}}\left[\frac{\cosh(2z)}{(z-2j)^2}\right] = \frac{16\cos(1)-24\sin(1)}{27}j

The other pole lies on the rectangle itself, and I'm not so sure how to handle it... You may be able to deform the contour and consider a principal value integral around the pole at z = 2j. The details elude me at the moment, however.

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I'm pretty sure this is it because there are only 2 like terms so you can't combine anything with 4x.
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