The temperature was 54 at 7:00 a.m. At 3:00 p.m. on the same day, the temperature was 82. What is the rate of change in temperat
1 answer:
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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)
We have
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)
We check with each options
'Or' represents the intersection of two graphs
'And' represents two separate graphs'
We have two separate shaded part in the given graph
So we ignore the options that has 'and' in between
LEts check first and second option
Simplify the first part and second part
multiply both sides by 2 .
x < 2 or 4x - 2 > = 26
solve 4x-2 > = 26
add 2 on both sides and then divide both sides by 4
4x >= 28
x >= 7
So solution is x<2 or x>=7 . that is the graph on number line
Lets check with second option
3x-3<3 or 2x+8>=22
add 3 on both sides
3x < 6
divide both sides by 3
so x< 2
2x+8>=22
subtract 8 on both sides
2x >= 30
divide both sides by 2
x >= 15
x<2 or x>=15 that does not satisfies the graph
So option A is correct