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nordsb [41]
2 years ago
5

Is Figure B a scale copy of Figure A?

Mathematics
1 answer:
Lisa [10]2 years ago
7 0

Answer:

oae yo msg 12 bestfriend lai send gara

Yadi 4 jana ko answer autai vaya ma really timima tyo khubi xa .......

Q- what is good in me??

1 Eyes

2 smile

3 style

4 Attiude

5 Voice

6 heart

7 looks

Mali pani send gara la........ i am waiting.

Step-by-step explanation:

yes it is a scale copy of Figure A

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Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2
exis [7]

The Jacobian for this transformation is

J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}

with determinant |J| = 12, hence the area element becomes

dA = dx\,dy = 12 \, du\,dv

Then the integral becomes

\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv

where R' is the unit circle,

\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1

so that

\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}

Then

\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}

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F(x) = x5 – 8x4 + 16x3?
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Answer:

Step-by-step explanation:

The first guy got it backwards

Thre first solution is 4

The second solution is 0

6 0
3 years ago
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What expression is equivalent to 4x-5-6x+7?
lesantik [10]

Answer:

-2x + 2

Step-by-step explanation:

you didn't post choices but combining like terms you get

-2x + 2

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2 years ago
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