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3 years ago

An arrangement of objects where order is not important is a ____.

Mathematics

2 answers:

mina [271]3 years ago

5 0

Permutations is the answer ..

stiks02 [169]3 years ago

3 0

Permutations I think is correct

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This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

3 years ago

Pls help asap i will give brainerlist plus 15 points

mario62 [17]

Answer:

6/7

Step-by-step explanation:

4 0

3 years ago

( 7 − 1 9 ) ( 9 − 3 7 )

rusak2 [61]

Answer:

-12 × -28

=336.

<h2>Hope it helps you.</h2>

8 0

3 years ago

Sergio went to the library at 3:18 p.m. to do homework. He worked at the library for 57 minutes, then left to go home. What time

8_murik_8 [283]

Answer:

The time would be 4:15 I think

6 0

3 years ago

Is (x+3) a factor of P(x)=2X³+5X²+X+10 explain

anyanavicka [17]

Answer:

Doubt it because the answer has to be a factor of 10 but I may be wrong.

5 0

3 years ago