Answer:
1st integer=x
2nd integer=x+1
3rd integer=x+2
4th integer=x+3
2(x +(x+2))= 3(x+1) +11
solve
2 (2x +2) = 3x + 3 + 11
4x + 4 = 3x + 14
-3x-4 =-3x -4
x=10
1st integer=10
2nd integer=11
3rd integer=12
4th integer=13
Step-by-step explanation:
hopefully this helped
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....
