Answer:
See explanation
Step-by-step explanation:
Consider triangles PTS and QTR. In these triangles,
- given;
- given;
- as vertical angles when lines PR and SQ intersect.
Thus, 
by AAS postulate.
Congruent triangles have congruent corresponding sides, so
Consider segments PR and QS:
![PR=PT+TR\ [\text{Segment addition postulate}]\\ \\QS=QT+TS\ [\text{Segment addition postulate}]\\ \\PT=QT\ [\text{Proven}]\\ \\ST=RT\ [\text{Given}]](https://tex.z-dn.net/?f=PR%3DPT%2BTR%5C%20%5B%5Ctext%7BSegment%20addition%20postulate%7D%5D%5C%5C%20%5C%5CQS%3DQT%2BTS%5C%20%5B%5Ctext%7BSegment%20addition%20postulate%7D%5D%5C%5C%20%5C%5CPT%3DQT%5C%20%5B%5Ctext%7BProven%7D%5D%5C%5C%20%5C%5CST%3DRT%5C%20%5B%5Ctext%7BGiven%7D%5D)
So,
![PR=SQ\ [\text{Substitution property}]](https://tex.z-dn.net/?f=PR%3DSQ%5C%20%5B%5Ctext%7BSubstitution%20property%7D%5D)
I JUST TOOK THE TEST AND THE ANSWER IS D
Option A. All the real values of x where x < -1
Procedure
Solve the inequality:
(x -3)(x+1)>0
That happens in two cases.
1) When both factors >0
x-3>0 and x+1>0
x>3 and x >-1
The intersection is x >3
2) When both factors <0
x-3<0 and x+1<0
x<3 and x<-1
the intersection is x<-1.
We have obtained that the function is positive for the intervals x < -1 and x > 3. But in one of those intervals the function is decresing and in the other is increasing.
You can recognize that the function given is a parabola and, because the coefficient of the quadratic term is positive, the parabola opens upward. Then the function is decreasing in the first interval and increasing in the second interval.
Answer:
Slope: 2
The slope how much y increases when x increases by 1.
Step-by-step explanation:
1. Find the difference between both y-coordinates. (6)
2. Find the difference between both x-coordinates. (3)
Divide the difference between y and x. y/x=2
The slope is 2.
