Answer:
The slope of a line that is perpendicular to the given line is 
Step-by-step explanation:
The equation of the line in Slope-Intercept form is:

Where "m" is the slope of the line and "b" is the y-intercept.
Solve for "y" from the equation of the line
:

You can observe that the slope of this line is:

By definition, the slopes of perpendicular lines are negative reciprocal, then, the slope of a line that is perpendicular to the give line, is

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The first question:3y=4x+7 (1)
-4x-4y=28 (2)
Let's plug in x = -4, -3, 3, and 4 and see the y-values :)
I have attached the table since it's hard to make a table with text :P
As you can see, when x = -4, that is when the y-values are equal. That means that is the solution to the system of equations. Your answer is
A) (-4, -3).
The second question:</span><span>-3x-y=-10 (1)
4x-4y=8 (2)
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When you graph both equations, you will see that they intersect at
D) (3, 1).
The third question:
We need to find the lines for revenues and expenses.
To find the line for revenues, make months the x-value and revenues the y-values. And find the equation. You should get y = 4,100x + 300.
To find the line for expenses, make months the x-value and expenses the y-values. And find the equation. You should get y = 1,900x + 19,990.
Now graph both solutions and see where they intersect. They intersect at approximately (8.95, 36995)
That would be during the month of
C) August.
$20,000 is between $15,000 and $49,999, so we'll use the interest rate of 6.5% (see row 3)
r = 6.5% = 6.5/100 = 0.065
We'll use the decimal form of the interest rate as it is most common for financial math problems.
P = 20,000 is the amount deposited
t = 1 year is the amount of time
We will plug those values into the formula
i = P*r*t
to get the following:
i = P*r*t
i = 20000*0.065*1
i = 1300
So Mark earns $1,300 in simple interest each year.
Answer:
Problem 2): 
which agrees with answer C listed.
Problem 3) : D = (-3, 6] and R = [-5, 7]
which agrees with answer D listed
Step-by-step explanation:
Problem 2)
The Domain is the set of real numbers in which the function (given by a graph in this case) is defined. We see from the graph that the line is defined for all x values between 0 and 240. Such set, expressed in "set builder notation" is:

Problem 3)
notice that the function contains information on the end points to specify which end-point should be included and which one should not. The one on the left (for x = -3 is an open dot, indicating that it should not be included in the function's definition, therefor the Domain starts at values of x strictly larger than -3. So we use the "parenthesis" delimiter in the interval notation for this end-point. On the other hand, the end point on the right is a solid dot, indicating that it should be included in the function's definition, then we use the "square bracket notation for that end-point when writing the Domain set in interval notation:
Domain = (-3, 6]
For the Range (the set of all those y-values connected to points in the Domain) we use the interval notation form:
Range = [-5, 7]
since there minimum y-value observed for the function is at -5 , and the maximum is at 7, with a continuum in between.