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3 years ago

Simplify: 20x + 27y + 38y + 29x

Mathematics

2 answers:

Drupady [299]3 years ago

8 0

Answer:

=49x+65y

Step-by-step explanation:

Xelga [282]3 years ago

7 0

Answer:

49x+65y

Step-by-step explanation:

Only the x can be added to the x and only the y can be added to the y.

So 20+29=49 then add an x

27+38=65 then add a y

Always put your answer in alphabetical order.

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The question is Stephen read for 34 minutes on Monday and 45 minutes on Tuesday. Use breaking apart to find how many minutes Ste

user100 [1]

Breaking apart means that you just take the tens from each number, and the ones from each number, then add them up.

The tens from your numbers are:
30 from 34
40 from 45

The ones from your numbers are:
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Now, you have to add up the tens and ones separately.
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3 years ago

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Ilia_Sergeevich [38]

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Step-by-step explanation:

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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

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

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2 years ago