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2 years ago

13

What is the measure of a base angle of an isosceles triangle if the vertex angle measure 38 and the two congruent sides each mea

sure 21 units ?

1 answer:

MissTica2 years ago

7 0

38+21 Units= 59 units .

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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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1 year ago

5. Adriana gives half of her t-shirts to charity. Then

fenix001 [56]

I believe 22. Because 11+3=14, but without that, you would add 11 to 11, which gives you 22

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1 year ago

Please help 3(2x+1)+2(x+4)

Naya [18.7K]

Answer: 8x+11

Step-by-step explanation:

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1 year ago

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leonid [27]

1650
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Hope this will be helpful

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2 years ago

Read the following poem:

pentagon [3]

Answer:

7, I believe. I have to fill in the extra character space, so I hope you have a nice day :)

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