For this case we have that by definition, the equation of a line in the slope-intersection form is given by:

Where:
m: It's the slope
b: It is the cut-off point with the y axis
On the other hand we have that if two lines are perpendicular, then the product of their slopes is -1. So:

The given line is:

So we have:

We find 

So, a line perpendicular to the one given is of the form:

We substitute the given point to find "b":

Finally we have:

In point-slope form we have:

ANswer:

Answer:
The limit of this function does not exist.
Step-by-step explanation:


To find the limit of this function you always need to evaluate the one-sided limits. In mathematical language the limit exists if

and the limit does not exist if

Evaluate the one-sided limits.
The left-hand limit

The right-hand limit

Because the limits are not the same the limit does not exist.
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>
Answer:-18
-6
0
3
9
Step-by-step explanation:
115% is the same as 1.15
some number = x
115% of some number is 120
1.15 of x is 120
1.15 * x = 120
x = 104.3478
<span>x ≈ 104.3</span>