Choose the answer based on the most efficient method as presented in the lesson.
2 answers:
You might be interested in
Answer:
see the procedure
Step-by-step explanation:
Looking at the graph we have
The graph represent a vertical parabola open upward
The vertex is a minimum
The vertex is the point (-4,-3)
The domain is the interval -----> (-∞,∞)
The Domain is all real numbers
The range is the interval ----> [-3,∞)
The range is all real numbers greater than or equal to -3
The graph is increasing in the interval (-4,∞)
The graph is decreasing in the interval (-∞,-4)
The minimum of the graph is y=-3 occurs at x=-4
Answer:
The correct answer is :
1. Line PQ (One line PQ).
Step-by-step explanation:
The first step to solve this question is to draw the plane A with the points P and Q lying on it.
We know that given two different points there is only one line that contains this two different points.
Let's analyze each option.
''2. Lines PQ and QP''
This option is wrong because there aren't two different lines. In fact it is only one line that can be named line PQ or line QP.
''3. The 2 lines PQ and QP plus another line that does not lie in plane A.''
This option is assuming that exist three lines that contain P and Q. This option is also wrong.
''1. Line PQ''
This option is correct. It will be clarify with the drawing I will attach.
''We can't name them all!''
This option is assuming that exist infinite lines that contain P and Q. This option is wrong.
In the drawing I call the line that contains P and Q as line L.
Given that P and Q lie in plane A necessarily the line L must lie on the plane A.
