What to did on this .paper thank you for your. please help me

Mathematics

1 answer:

matrenka [14]3 years ago

3 0

So, to solve the questions on your paper, you need to first have all of the bottom numbers on the fractions similar numbers. So if you have the numbers 1/4 and 4/8, you need to get them to both be the same number. Therefore, 1/4 should become 2/8, because you doubled the bottom number (you also then need to do the same to the top number). You don't need to change 4/8, because it's bottom number is already 8. Then, you can take 4/8 and 1/8 and add them. To add them, you just add the top numbers. The top numbers make 5, so then you add the denominator, and make the number 5/8. Since you can't simplify 5/8, just keep it. But if it was 4/8, you could simplify it to 1/2. You can do all of this same to all of the numbers on your paper. :D If you are subtracting, subtract only the top numbers, but if you need to multiply or divide, multiply/divide both numerator AND denominator. :D I hope I helped! If I didn't, please tell me, or just delete my answer. :D

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Help me please with numbers 1-8 I'm stuck!!

siniylev [52]

Hopefully this helps, and sorry I didn't do 8 I'm limited on time

5 0

3 years ago

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

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