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2 years ago

Circle D is centered at (-3, 3) with a radius of 5 and circle E is centered at (3,-3) with a radiu

Mathematics

1 answer:

bezimeni [28]2 years ago

4 0

I think “c” is the correct answer. I am not sure

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Graph the system by hand.

weqwewe [10]

Answer:

dont khan academy

Step-by-step explanation:

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why

cheet

7 0

2 years ago

Indeed help with this plz

gizmo_the_mogwai [7]

Answer:

x=\frac{6-y-z}{3}

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

6 6/9 divided by 1 5/7

makvit [3.9K]

Exact Form:
28/9
Decimal Form:
3.1
Mixed Number Form:
3 1/9

3 0

2 years ago

Someone please help?

Digiron [165]

The answer to your question is,

$3.84, you take $69.20 and divide it by 18 and there is your answer!

-Mabel <3

4 0

3 years ago

Find the following integral

ololo11 [35]

There's nothing preventing us from computing one integral at a time:

$\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2$

$\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2$

$\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx$

Expand the integrand completely:

$x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x$

Then

$\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}$

4 0

2 years ago