$\displaystyle \frac {x^2 + 2x - 63}{7 - x} = \frac {(x + 9)(x - 7)}{-(x - 7)} = -(x + 9)$
the restriction is $\displaystyle x \ne 7$ since we have to consider the domain of the first function as well. in the original question, x = 7 makes the denominator zero
i use LaTex. see here
I hope you're not letting me do your homework for you... where's that suspicious smiley?same story here.
$\displaystyle \frac {5m^2 + 6m - 8}{5m^2 - 16m - 16} \cdot \frac {20m^2 + 16m}{4 - m^2} = \frac {(5m - 4) (m + 2)}{(m - 4) (5m + 4)} \cdot \frac {4m (5m + 4)}{(2 - m)(2 + m)}$
$\displaystyle = \frac {4m(5m - 4)}{(m - 4)(2 - m)}$
what do you think the restrictions are? remember, consider the original function as well
haha, yes you didOMG I DID IT!
not so hard was it?
m doesnt = 4 or -2 <---is that right?
and im stuck here, do i cancel or recipricate and muiltyply?
$\displaystyle
\frac {j^2+7j+12}{j^2+6j+9} \div \frac {j^2-16}{j^2+7j+12}
$
$\displaystyle
\frac {(j+3)(j+4)}{(j+3)(j+3)} \div \frac {(j-4)(j+4)}{(j+3)(j+4)}
$
what about the (2 - x) and the (5m + 4) in the denominators of the original fractions
multiply the first fraction by the reciprocal of the second fraction and see if anything cancels. if not, just multiply
and im stuck here, do i cancel or recipricate and muiltyply?
$\displaystyle
\frac {j^2+7j+12}{j^2+6j+9} \div \frac {j^2-16}{j^2+7j+12}
$
$\displaystyle
\frac {(j+3)(j+4)}{(j+3)(j+3)} \div \frac {(j-4)(j+4)}{(j+3)(j+4)}
$