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:
Your answers are correct. However, the instructions say to write the formula, and in my class you would write A= bh ÷ 2.
Step-by-step explanation:
However, the instructions say to write the formula, and in my class you would write A= bh ÷ 2. Also, you may want to write A= for every line of math that you do. If your class doesn't do that, then disregard that. :)
|x - 1| > 4 ⇔ x - 1 > 4 or x - 1 < -4 |add 1 to both sides
x > 5 or x < -3
Answer: c. {x| x < -3 or x > 5}.
The length of the other side of the yard is 16 feet
work: 160 divided by 10
hope it helps
H = 5 + ut - 16t^2
When the balls falls back to the ground, h = 0
0 = 5 + 3.2u - 16(3.2)^2
0 = 5 + 3.2u - 163.84
0 = 3.2u - 158.84
3.2u = 158.84
u = 158.84/3.2 = 49.6 ft/s
Required equation is
h = 5 + 49.6t - 16t^2
