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azamat
4 years ago
5

Given a circle with radius of 2, which is the degree measure of an arc whose length is 1/2 ?

Mathematics
1 answer:
Finger [1]4 years ago
3 0
180 degrees. It would be half of the circle.
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Add or subtract.<br><br> a.<br> 5 + 12 =<br> 14
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a. Factorize the denominator:

\dfrac{x+14}{x^2-2x-8}=\dfrac{x+14}{(x-4)(x+2)}

Then we're looking for a,b such that

\dfrac{x+14}{x^2-2x-8}=\dfrac a{x-4}+\dfrac b{x+2}

\implies x+14=a(x+2)+b(x-4)

If x=4, then 18=6a\implies a=3; if x=-2, then 12=-6b\implies b=-2. So we have

\dfrac{x+14}{x^2-2x-8}=\dfrac3{x-4}-\dfrac2{x+2}

as required.

b. Same setup as in (a):

\dfrac{-3x^2+5x+6}{x^3+x^2}=\dfrac{-3x^2+5x+6}{x^2(x+1)}

We want to find a,b,c such that

\dfrac{-3x^2+5x+6}{x^2(x+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac c{x+1}

Quick aside: for the second term, since the denominator has degree 2, we should be looking for another constant b' such that the numerator of the second term is b'x+b. We always want the polynomial in the numerator to have degree 1 less than the degree of the denominator. But we would end up determining b'=0 anyway.

\implies-3x^2+5x+6=ax(x+1)+b(x+1)+cx^2

If x=0, then b=6; if x=-1, then c=-2. Expanding everything on the right then gives

-3x^2+5x+6=ax^2+ax+bx+b+cx^2=(a-2)x^2+(a+6)x+6

which tells us a-2=-3 and a+6=5; in both cases, we get a=-1. Then

\dfrac{-3x^2+5x+6}{x^2(x+1)}=-\dfrac1x+\dfrac6{x^2}-\dfrac2{x+1}

as required.

5 0
4 years ago
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