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3 years ago

What is the width of a rectangle with an area of 5/8 inches2 and a length of 1o inches

Mathematics

1 answer:

Nata [24]3 years ago

5 0

I hope this helps you

Area=length.width

Area=1.5/8

Area=5/8

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What is the square root of four hundred fifty six

4vir4ik [10]

Answer:

21.3541565041

Step-by-step explanation:

You can look it up online to easily find it. Thats how i found this answer.

Hope this helps!

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2 years ago

Read 2 more answers

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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2 years ago

Bob spends 30 hours in 4 weeks of gardening. How many hours does he garden in 5 weeks

Mademuasel [1]

Answer:

37.5 hrs

Step-by-step explanation:

You do cross multiplication hours on top and weeks on the bottom.

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3 years ago

In triangle ABC, what is the measure of angle C?

agasfer [191]

Depends on the angle of A and B

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3 years ago

I need help with this question

Delvig [45]

Answer:

Go download CameraMath on playstore that how

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2 years ago