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3 years ago

Answer number one please

Mathematics

2 answers:

Diano4ka-milaya [45]3 years ago

7 0

Answer:

16^3

Step-by-step explanation:

First, multiply the bases together. Then, add the exponent. Instead of adding the two exponents together, keep it the same.

tankabanditka [31]3 years ago

6 0

Answer:

-3 + 6 = -3

Step-by-step explanation:

add the terms -3 and 6

exponent law

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Martha's mom is making party bags for all of her guests. She has 24 green gumballs and 33 blue gumballs. If she splits them even

Dmitry_Shevchenko [17]

Answer:

3 bags

Step-by-step explanation:

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7 0

3 years ago

What is the equation for the perpendicular that passes through the point (-1,-2) ?

tangare [24]

Answer:

Perpendicular with respects to what?

You can't find it with only a point, you also need a slope to know which line is which.

There are an infinite number of lines that go through any point. Each has a slope, without slopes, we cant solve such a question.

It's line asking, how old is Jane's dog?

Who knows man? There's like thousands of Janes, which one is the one?

7 0

3 years ago

What is the multiplicative inverse of 1/16 ?

andrey2020 [161]

Answer:

16/1 I think

Step-by-step explanation:

8 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

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