Answer:
Neither perpendicular nor parallel.
Step-by-step explanation:
To find the answer, you just need to find the slopes of each line.
The slope of the line with points at (-6, 5) and (-2, 7) would be (7 - 5) / (-2 + 6) = 2 / 4 = 1/2.
The slope of the line with points at (4, 2) and (6, 6) would be (6 - 2) / (6 - 4) = 4 / 2 = 2.
They are most definitely not the same line.
They are not perpendicular, either, since although they are [eek I forgot the term for it] opposites, they are both positive. To be perpendicular, one slope must be negative and the other positive.
They are not parallel, since the slopes are not the same.
So, the relationship between the lines are that they are neither parallel nor perpendicular.
Hope this helps!
Answer:
The graph is shown below.
Step-by-step explanation:
The given piecewise function is
![f(x)=\begin{cases}-x & \text{ if } x](https://tex.z-dn.net/?f=f%28x%29%3D%5Cbegin%7Bcases%7D-x%20%26%20%5Ctext%7B%20if%20%7D%20x%3C0%20%5C%5C%20x%5E2%20%26%20%5Ctext%7B%20if%20%7D%20x%5Cgeq%200%20%5Cend%7Bcases%7D)
From the given piecewise function, it is clear that
for x<0.
Put different values of x in f(x) where x<0 and find the values of f(x) to make a table of values as shown below.
Table of values:
x f(x)
-3 3
-2 2
-1 1
Now, plot the points (-4,4),(-3,3),(-2,2),(-1,1) and join them by straight line. There is an open circle at x=0 because 0 is not included in this function.
From the given piecewise function, it is clear that
for
.
Table of values:
x f(x)
0 0
1 1
2 4
Plot the points (0,0), (1,1), (2,4) and join them by a curve. There is closed circle at 0 because 0 is included in this function.
The graph is shown below.
Hello!
The answer is 6.25.
The functions whose graphs include the point (16, 4) is y = x−12.
Option: D.
<u>Step-by-step explanation:</u>
To find the function that passes through the point (16,4).
Substitute the point in all the equations and equate the final values to find the correct equation.
A. y=2x.
Substitute (16,4).
4=2(16).
4≠32.
Thus the equation doesn't contains the point (16,4).
B. y=
.
Substitute (16,4).
4=
.
4≠ 256.
Thus the equation doesn't contains the point (16,4)
C. y=x+12.
4=16+12.
4≠28.
Thus the equation doesn't contains the point (16,4)
D. y=x−12.
4=16-12.
4=4.
Thus the equation contains the point (16,4)
E. y=14x.
4=14(16).
4≠224.
Thus the equation doesn't contains the point (16,4)
∴The functions whose graphs include the point (16, 4) is y = x−12.