1. Domain.
We have
in the denominator, so:
![x^2-2x-3\neq0\\\\(x^2-2x+1)-4\neq0\\\\(x-1)^2-4\neq0\\\\(x-1)^2-2^2\neq0\qquad\qquad[\text{use }a^2-b^2=(a-b)(a+b)]\\\\(x-1-2)(x-1+2)\neq0\\\\ (x-3)(x+1)\neq0\\\\\boxed{x\neq3\qquad\wedge\qquad x\neq-1}](https://tex.z-dn.net/?f=x%5E2-2x-3%5Cneq0%5C%5C%5C%5C%28x%5E2-2x%2B1%29-4%5Cneq0%5C%5C%5C%5C%28x-1%29%5E2-4%5Cneq0%5C%5C%5C%5C%28x-1%29%5E2-2%5E2%5Cneq0%5Cqquad%5Cqquad%5B%5Ctext%7Buse%20%7Da%5E2-b%5E2%3D%28a-b%29%28a%2Bb%29%5D%5C%5C%5C%5C%28x-1-2%29%28x-1%2B2%29%5Cneq0%5C%5C%5C%5C%0A%28x-3%29%28x%2B1%29%5Cneq0%5C%5C%5C%5C%5Cboxed%7Bx%5Cneq3%5Cqquad%5Cwedge%5Cqquad%20x%5Cneq-1%7D)
So there is a hole or an asymptote at x = 3 and x = -1 and we know, that answer B) is wrong.
2. Asymptotes:
We have only one asymptote at x = -1 (and hole at x = 3), thus the correct answer is A)
Answer:
C
Expand the bracets then simplfy and then factorise
Answer:
D
Step-by-step explanation:
You would have to eliminate A because the students have not yet been to other school districts especially if it's every third. You would also have to eliminate B because well, there really isnt any other place to eat but the lunchroom because otherwise it might pose a problem, and with C it can be difficult but again, if we were to try to change something the time they take really has no impact. D would have to be the correct answer because every third student will have various favorite lunch items and will ultimately tell the school that the gross macaroni that is like...unedible is bad and needs to be taken out...immediatley.
Answer:
the world would be filled with people
we would all be dead
Step-by-step explanation:
Answer:1+3x+9+x+8+2x= 6x+9
so =3(x+3)
Step-by-step explanation:
