50PTS!
Describe the vertical asymptotes and holes for the graph of y = x-5/x^2 -1
2 answers:
Answer:
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Step-by-step explanation:
Ok
first of all, for q(x)/p(x)
if the degree of q(x) is less than the degree of p(x),then the horizontal assemtote is 0
then simplify
any factors you factored out is now a hole, remember them
to find the vertical assemtotes of a function, set the SIMPLIFIED denomenator equal to 0 and solve
so
y=(x-5)/(x^2-1)
q(x)<p(x)
horizontal assemtote is y=0
no factors to simplify so no holes
set denomenator to 0 to find vertical assemtote
x^2-1=0
(x-1)(x+1)=0
x-1=0
x=1
x+1=0
x=-1
the horizontal assemtotes are x=1 and -1
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Answer:
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Step-by-step explanation:
Volume of a cone:
V = πr²(h/3)
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Work:
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