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
Answer:
i think it would be 87% i have been working on ths for a mnet and it got confusing for a sec i may be rong but i am pretty shure that this is the answer
btw i cant spell
Step-by-step explanation:
Answer:
The number next to a variable is called the Coefficient of the variable.
P = 2(L + W)
L = W + 5
A = 4P + 2
P = 2(W + 5 + W)
P = 2(2W + 5)
P = 4W + 10
A = 4P + 2
A = 4(4W + 10) + 2
A = 16W + 42
A = L * W
A = W(W + 5)
A = W^2 + 5W
W^2 + 5W = 16W + 42
W^2 + 5W - 16W - 42 = 0
W^2 - 11W - 42 = 0
(W + 3)(W - 14) = 0
W - 14 = 0
W = 14 <==
L = W + 5
L = 14 + 5
L = 19 <==
P = 2(19 + 14)
P = 2(33)
P = 66
A = L * W
A = 19 * 14
A = 266
answer : length = 19, width = 14....perimeter = 66....area = 266
