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:
Step-by-step explanation:
Problem One (left panel)
<em><u>Question A</u></em>
- The y intercept happens when x = 0
- That being said, the y intercept is 50. It was moving when the timing began.
<em><u>Question B</u></em>
The rate of change = (56 - 52)/(3 - 1) = 4/2 = 2 miles / hour^2 (you have a slight acceleration.
<em><u>Question C</u></em>
- 60 = a + (n-1)d
- 60 = 50 + (n - 1)*2
- 10/2 = (n - 1)*2/2
- 5 = n - 1
- 6 = n
The way I have done it the domain is n from 1 to 6
Question 2 (Right Panel)
<em><u>Question A</u></em>
The equation for the table is f(x) = 3x - 3 which was derived simply by putting all three points into y = ax + b and solving.
- f(0) = ax + b
- -3 = a*0) + b
- b = - 3
- So far what you have is
- f(x) = ax - 3
- f(-1) = a*(-1) - 3 but we know (f(-1)) = -6
- - 6 = a(-1) - 3 add 3 to both sides
- -6 +3 = a(-1) -3 + 3
- -3 = a*(-1) Divide by - 1
- a = 3
- f(x) = 3x - 3 Answer for f(x)
- The slope of f(x) = the coefficient in front of the x
- f(x) has a slope of 3
- g(x) has a slope of 4
<em><u>Part B</u></em>
- f(x) has a y intercept of - 3
- g(x) has a y intercept of -5
- f(x) has the greater y intercept.
- -3 > - 5
The Greatest Common factor of 14 and 18 is 2. Here is my work ~
Answer:The particles in a solid are tightly packed and locked in place. Although we cannot see it or feel it, the particles are moving = vibrating in place.
