Answer:
The correct option is A.
Step-by-step explanation:
Domain:
The expression in the denominator is x^2-2x-3
x² - 2x-3 ≠0
-3 = +1 -4
(x²-2x+1)-4 ≠0
(x²-2x+1)=(x-1)²
(x-1)² - (2)² ≠0
∴a²-b² =(a-b)(a+b)
(x-1-2)(x-1+2) ≠0
(x-3)(x+1) ≠0
x≠3 for all x≠ -1
So there is a hole at x=3 and an asymptote at x= -1, so Option B is wrong
Asymptote:
x-3/x^2-2x-3
We know that denominator is equal to (x-3)(x+1)
x-3/(x-3)(x+1)
x-3 will be cancelled out by x-3
1/x+1
We have asymptote at x=-1 and hole at x=3, therefore the correct option is A....
Answer:
we really can't tell because we are not able to use a protactor and a ruler
Answer:
f = .39
Step-by-step explanation:
d (v) = 2.15 v^2 / (64.4 f)
v = 40
d = 138
substitute these in
138 = 2.15 (40)^2 / (64.4 *f)
simplify
138 = 2.15 (*1600) / (64.4 *f)
138 = 3440 / (64.4 *f)
multiply each side by 64.4f
138 * 64.4 f = 3440
8887.2 f = 3440
divide by 8887.2
8887.2 f /8887.2= 3440/8887.2
f=.38707
to the nearest hundredth
f = .39
Your answer is f(y) > 0
A is your answer
Answer:
a
Step-by-step explanation:
