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tatiyna
3 years ago
12

What is the conjugate of 12 - sqrt 5

Mathematics
2 answers:
Lyrx [107]3 years ago
7 0
12+sqrt 5, is your answer, hoped this helped, 
Nataliya [291]3 years ago
5 0

Answer:

the conjugate of  12-\sqrt{5} is  12+\sqrt{5}

Step-by-step explanation:

the conjugate of 12 - sqrt 5

In general , the conjugate of a+bi is a-bi

to get the we change the sign in the middle

Change the negative sign to positive sign before the square root

the conjugate of  12-\sqrt{5} is  12+\sqrt{5}

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Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices
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Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt

Where P, Q are scalar functions

We want to compute

\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths \large C_1,C_2,C_3,C_4 which represents the sides of the rectangle.

\large C_1 is the line segment from (0,0) to (5,0)

\large C_2 is the line segment from (5,0) to (5,1)

\large C_3 is the line segment from (5,1) to (0,1)

\large C_4 is the line segment from (0,1) to (0,0)

Then

\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize \large C_1 as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

\large \displaystyle\int_{C_1}xydx+x^2dy=0

 Now the second integral. We parametrize \large C_2 as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25

The third integral. We parametrize \large C_3 as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2

The fourth integral. We parametrize \large C_4 as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

\large \displaystyle\int_{C_4}xydx+x^2dy=0

So

\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2

Now, let us compute the value using Green's theorem.

According with this theorem

\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx

where A is the interior of the rectangle.

so A={(x,y) |  0≤ x≤ 5,  0≤ y≤ 1}

We have

\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x

so

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