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arsen [322]
3 years ago
12

Is this function in Slope-Intercept Form, Point Slope Form, or Standard Form?

Mathematics
2 answers:
Alexandra [31]3 years ago
6 0
It’s gonna be standard form
masya89 [10]3 years ago
4 0

Answer:

this is point slope form

Step-by-step explanation:

point slope form is y-y1=m(x-x1)

slope intercept is y=mx+b

standard form is Ax+By=c

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60

Step-by-step explanation:

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4 years ago
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
MATH1314-1043,1044 16wks SPR21
Brilliant_brown [7]

Step-by-step explanation:

how where is the graph cause I don't understand that

6 0
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