Lets say we have
P(x)/q(x)
vertical assymtotes are in the form x=something, not y=0
y=0 are horizontal assemtotes
so verticall assymtotes
reduce the fraction
set the denomenator equal to zero
those values that make the deomenator zero are the vertical assymtotes
the horizontal assymtote
when the degree of P(x)<q(x), then HA=0
when the degree of P(x)=q(x), then divide the leading coefient of P(x) by the leading coeficnet of q(x)
example, f(x)=(2x^2-3x+3)/(9x^2-93x+993), then HA is 2/9
ok so for vertical assymtote example
f(x)=x/(x^2+5x+6)
the VA's are at x=-3 and x=-2
horizontal assymtote
make degree same
f(x)=(3x^2-4)/(8x^2+9x),
the HA is 3/8
hope I helped, read the whole thing then ask eusiton
Yea no solution is your answer
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Step-by-step explanation:
2:45+30 min=3:15
3:15+12 min = 3:27
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each of his parents are 42 years old.
Step-by-step explanation:
Answer:
Problem1= a Problem 2=a
Step-by-step explanation:
Hope this helps