You would divide the rate
First, let's convert each line to slope-intercept form to better see the slopes.
Isolate the y variable for each equation.
2x + 6y = -12
Subtract 2x from both sides.
6y = -12 - 2x
Divide both sides by 6.
y = -2 - 1/3x
Rearrange.
y = -1/3x - 2
Line b:
2y = 3x - 10
Divide both sides by 2.
y = 1.5x - 5
Line c:
3x - 2y = -4
Add 2y to both sides.
3x = -4 + 2y
Add 4 to both sides.
2y = 3x + 4
Divide both sides by 2.
y = 1.5x + 2
Now, let's compare our new equations:
Line a: y = -1/3x - 2
Line b: y = 1.5x - 5
Line c: y = 1.5x + 2
Now, the rule for parallel and perpendicular lines is as follows:
For two lines to be parallel, they must have equal slopes.
For two lines to be perpendicular, one must have the negative reciprocal of the other.
In this case, line b and c are parallel, and they have the same slope, but different y-intercepts.
However, none of the lines are perpendicular, as -1/3x is not the negative reciprocal of 1.5x, or 3/2x.
<h3><u>B and C are parallel, no perpendicular lines.</u></h3>
4.8 * 0 is the scientific notation
I shall have a conversation with you, if you don’t mind
Answer:
Therefore, Michael concludes option C)
C)
Step-by-step explanation:
Given:
1. DG = 3 and the area of square DEFG is 9.
2. AG = 4 and the area of square GHIA is 16.
3. DA = 5 and the area of square ABCD is 25.
So we have,

Now Add DG² and AG² we get

Which is also called as Pythagoras theorem i.e

Therefore, Michael concludes option C)
C)