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umka2103 [35]
3 years ago
7

How do you do these problems

Mathematics
1 answer:
avanturin [10]3 years ago
7 0
\cfrac{1}{3} =   \cfrac{2}{6}  =  \cfrac{3}{9}   =  \cfrac{6}{18}

\cfrac{4}{20} = \cfrac{3}{15} =  \cfrac{12}{60}  =  \cfrac{100}{500}

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The following integral requires a preliminary step such as long division or a change of variables before using the method of par
shtirl [24]

Division yields

\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}

Now for partial fractions: you're looking for constants <em>a</em>, <em>b</em>, and <em>c</em> such that

\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}

\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a

which gives <em>a</em> + <em>b</em> = 2, <em>c</em> = 0, and 2<em>a</em> = -7, so that <em>a</em> = -7/2 and <em>b</em> = 11/2. Then

\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}

Now, in the integral we get

\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx

The first two terms are trivial to integrate. For the third, substitute <em>y</em> = <em>x</em> ² + 2 and d<em>y</em> = 2<em>x</em> d<em>x</em> to get

\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}

7 0
3 years ago
2. Joyce waited three-quarters of an hour to
Evgen [1.6K]

Joyce waited 50 minutes waiting for the train

Step-by-step explanation:

divide 60 by 3 cause 3 quarters of an hour is 3/4 then you wanna take half an hour and add that (30 mins) getting 50 minutes

8 0
3 years ago
How do you do systems of equations ?
Arada [10]
There are 2 methods  you are taught in Middle School.

Elimination and Substitution.
8 0
4 years ago
Help help help pelesss please
blagie [28]

Answer:

its 1

Step-by-step explanation:

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8 0
3 years ago
Matt help plus<br> Nskdnisbeineijsbd
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Answer:

Step-by-step explanation:

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