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:
14
Step-by-step explanation:
Let's find (f-g)(x)
Now let's plug in x=2, so to find (f-g)(2)
Hello there, I am unable to use the “less than” symbol. Sorry :( .
1/5x "less than" 8
x “Less than” 40
Prime number is 37,41.
Therefore x is 37.
Answer:
see below (I hope this helps!)
Step-by-step explanation:
Square roots are usually in the form a√b where a and b are any rational number.
