Question

Decompose into partial fractions.
(4x+34)/(x^2 - 5x - 24)

Answer 1

\[\frac{4x+34}{x^2-5x-24}=\frac{4x+34}{(x-8)(x+3)}=\frac{A}{x-8}+\frac{B}{x+3}\]
\[4x+34=A(x+3)+B(x-8)=(A+B)x+3A-8B\]
\[\Rightarrow\quad A+B=4\,\,,\,\,3A-8B=34\quad\Rightarrow\quad A=6,B=-2\]
\[\frac{4x+34}{x^2-5x-24}=\frac{6}{x-8}+\frac{-2}{x+3}\]

Answer 2

Thanks nikvist...can u help me w/ another problem????

Answer 3

wait im not sure how you got a = 6 adn b = -2

Answer 4

nvr mind

Answer 5

where is another problem? :-)

Answer 6

Expand and simplify:

\[\sum_{n=0}^{4} n ^{2}/2\]

Answer 7

it is simple
\[\sum\limits_{n=0}^{4}\frac{n^2}{2}=\frac{1}{2}\sum\limits_{n=0}^{4}n^2=\frac{1}{2}(0^2+1^2+2^2+3^2+4^2)=\frac{1}{2}\cdot 30=15\]

Answer 8

i have another one

Answer 9

\[\sum_{n=1}^{3} 1/n!\]

Answer 10

im confused about teh factorial

Answer 11

\[\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}=\frac{1}{1}+\frac{1}{2\cdot 1}+\frac{1}{3\cdot 2\cdot 1}=1+\frac{1}{2}+\frac{1}{6}=\frac{5}{3}\]

Answer 12

About factorials. How would you simplify this expresion: 3(n+1)! / 5n!

Answer 13

do u distribute