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what is the question?
yes find the vertical asymptotes if any
set denominator = 0 and solve:\[x(x+1)=0\] \[x=0\] or \[x=-1\] so there are two.
need the horizontal one as well? degree of numerator is 1,degree of denominator is 2, and since 2 > 1 it is \[y=0\]
gotta make sure that the bottom doesnt cancel the top; if it does, that would be a hole; but this problem is good....
oh so the bases never changes when re-writing??
not in that case :)
if they are different you cannot do it.
k thanks ...was always confused about that..lol teachers dont kno what they are talking about..lol
sort of its a new thing in orlando ..its a college readiness class that is basically preparing you to take pre-calc in college
i amristre can u help me after her
orlando eh... im over near brooksville
perhaps :) hypers are easy
amristre can u help me after her
*15 all of it.... 6x = 5x +30 x = 30 maybe?
amristre can u help me after
60/5 = 30/3 +2 6 = 10+2 .... gonna have to see what I did lol
i can yes...
2x x --- = --- + 2 5 3 2x = 5x/3 +10 2x = 5x+30/3 6x = 5x +30 x = 30 ..... 2(30) 30 ---- = --- + 2 5 3 2(6) = 10 + 2 12 = 12 ; well it worked that time lol.. my dbl check must have been off
60/5 = 12 lol.....
answer is either -+30 or -+60..but thanks lol
x = 30 fer sure
ok her is my impossible problem...lol
x^2+7x+10 times x^2+6x ---------- ----------- multiply x^2+8x+12 x^2+13x+40
factor to see what cancels... 2,5,x,6 ------ 6,2,8,5 ,,x, --- = x/(x+8) if I see it right ,,8,
but can you walk me through it so i can no how to do it myself