Answer:
B is your answer to this problem
Step-by-step explanation:
Answer:
Part 1) There are infinity locations for the point B
Part 2) see the explanation
Step-by-step explanation:
Part 1) How many possible locations are there for point B?
we know that
The equation of a line in point slope form is equal to

where


substitute

Convert to slope intercept form




Point B can be any point ( different from point A) that satisfies the linear equation
therefore
There are infinity locations for the point B
Part 2) Describes a method to location the point
To locate the point, one of the two coordinates must be known. The known coordinate is placed into the linear equation and the equation is solved to find the value of the missing coordinate
Example
Suppose that the x-coordinate of point B is 4
For x=4
substitute in the linear equation

so
The coordinates of point B is (4,10.5)
Answer:
B, C, D
Step-by-step explanation:
In this problem, the range is what the output, or y, can be. The origin, or the middie of the graph, is when x=0 and y=0. From the 10s on the screen, we can gather that 5 lines = a distance of 10 on the graph. Using this information, we can say
5 lines = distance of 10
divide both sides by 5 to find the distance for each line
1 line = distance of 2
The function goes from y=0 to three lines down, for a distance of 6. The range is therefore [-6,0] as all values from -6 to 0 on the y axis are included on the graph, including 0 and -6. In this range, -6, -2, and -1 are all included.
The length of OM is 4/3 the length of OJ, and the length of ON is 4/3 the length of OK. Thus, two pairs of sides from the triangles are proportional to each other.
Also, angle O from one of the triangles is equal to angle O of the other triangle because they are the same angle.
Thus, the two triangles are similar by SAS (side-angle-side) similarity theorem. This theorem is quite similar to the SAS congruence theorem.
To make a similarity statement, we just have to match corresponding parts when naming the triangle.
Similarity statement: ΔOJK~ΔOMN