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NeX [460]
3 years ago
9

Simplify and reduce to the lowest terms.

itle=" \frac{-7}{12} " alt=" \frac{-7}{12} " align="absmiddle" class="latex-formula"> ÷ \frac{2}{3} =
Mathematics
1 answer:
Paladinen [302]3 years ago
6 0
When you divide two fractions, you're actually multiplying one of them by the reciprocal of the other. First, find the reciprocal of the second fraction by flipping it upside down. Then, multiply it by the first fraction. (Numerator x numerator and denominator x denominator)

\frac{-7}{12} ÷ \frac{2}{3}   Replace the second fraction with it's reciprocal
\frac{-7}{12} x \frac{3}{2}   Multiply (-7 x 3 and 12 x 2)
\frac{-21}{24} Both 21 and 24 are divisible by three, so divide them by 3
\frac{-7}{8}
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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}}

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2 years ago
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When the following fraction is reduced, what will be the exponent on the m? 27mn^3/56m^6n
Oksi-84 [34.3K]
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5 0
2 years ago
Read 2 more answers
Circumference of a circle
solmaris [256]

Answer:

50.27 units

Step-by-step explanation:

The formula for the circumference of a circle is:

C = \pid

where, d = diameter

Now plug in the values as shown below:

C = \pi16

C = 50.27

3 0
3 years ago
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