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erik [133]
3 years ago
11

Evaluate the given integral by making an appropriate change of variables, where r is the rectangle enclosed by the lines x - y =

0, x - y = 7, x + y = 0, and x + y = 6.
Mathematics
1 answer:
vova2212 [387]3 years ago
6 0
\begin{cases}u=x-y\\v=x+y\end{cases}

\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\\\\dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}1&-1\\1&1\end{bmatrix}
\implies\det\mathbf J=2

The area of the region is then given by

\displaystyle\iint_R\mathrm dA=\int_{u=0}^{u=7}\int_{v=0}^{v=6}2\,\mathrm dv\,\mathrm du=84
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70℅ = 14
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70X = 1400

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3 years ago
A rectangular park has sidewalks on two adjacent sides. If it takes John 60 steps to
Sergeeva-Olga [200]

Answer:

Length = 80 m and breadth = 60 m If one moves along the two adjacent sides, one covers 80+60 = 140 m. Diagonal of the rectangle = √ ( length^2 + breadth^2) ...

5 0
2 years ago
In the inequality 6a+4b>10, what could be the possible value of a if b=2?
Arlecino [84]

We are given the following inequality:

6a+4b>10

If we replace b = 2, we get:

\begin{gathered} 6a+4(2)>10 \\ 6a+8>10 \end{gathered}

Now we solve for "a" first by subtracting 8 on both sides:

\begin{gathered} 6a+8-8>10-8 \\ 6a>2 \end{gathered}

Now we divide both sides by 6

\frac{6a}{6}>\frac{2}{6}

Simplifying:

a>\frac{1}{3}

Therefore, for b = 2, the possible values of "a" are those that are greater than 1/3

3 0
1 year ago
Find the quotient of these Complex Numbers. <br><br><br> (5-i) / (3+2i)
JulsSmile [24]

Let the given complex number

z = x + ix = \dfrac{5-i}{3+2i}

We have to find the standard form of complex number.

Solution:

∴ x + iy = \dfrac{5-i}{3+2i}

Rationalising numerator part of complex number, we get

x + iy = \dfrac{5-i}{3+2i}\times \dfrac{3-2i}{3-2i}

⇒ x + iy = \dfrac{(5-i)(3-2i)}{3^2-(2i)^2}

Using the algebraic identity:

(a + b)(a - b) = a^{2} - b^{2}

⇒ x + iy = \dfrac{15-10i-3i+2i^2}{9-4i^2}

⇒ x + iy = \dfrac{15-13i+2(-1)}{9-4(-1)} [ ∵ i^{2} =-1]

⇒ x + iy = \dfrac{15-2-13i}{9+4}

⇒ x + iy = \dfrac{13-13i}{13}

⇒ x + iy = \dfrac{13(1-i)}{13}

⇒ x + iy = 1 - i

Thus, the given complex number in standard form as "1 - i".

5 0
3 years ago
What percent of this grid is shaded
faltersainse [42]

Answer:

The 4th one cuz it’s a percent

Step-by-step explanation:

8 0
3 years ago
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