Answer:
1) P(x) =14400x+10300 2)y=-4x+19 3)[tex]y=\frac{-2x}{3} -14[/tex] 4) x=-11
(Please check the pictures for better understanding)
Step-by-step explanation:
1) Firstly, let's organize the data. Let the dependent variable x be the years, and the Range P(x) the Profit. So arranging these linear functions y=ax+b and plugging the coordinates for each point (1, 24700) and (4, 67900). Goes:
24,700=x+b
67,900=4x +b
1.1)Solving for b, this Linear System of Equation by the Addition Method goes:
[tex]24,700=x+b*(-4)\\67,900=4x+b\\----------\\-98,800=-4x-4b\\67,900=4x+b\\ -----------\\-30900=-3b\\\frac{3b}{3} =\frac{30900}{3}\\ b=10300[/tex]
1.2) Now, let's find the slope based on those two points (1, 24700) and (4, 67900) and using the formula to find the slope:
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
[tex]m=\frac{67900-24700}{4-1}\\ m=\frac{43200}{3}\\ m=14400[/tex]
So now we can write this Linear Equation for the Profit:
P(x) =14400x+10300
2) An equation of the line perpendicular to 4y=x-8 has a slope equal to [tex]m_{2}[/tex]=-[tex]\frac{-1}{m_{1}}[/tex] the line. In this case, an equation perpendicular to 4y=x-8 .
Rearranging the given equation of the line 4y=x-8 goes:
[tex]y=\frac{x}{4} -\frac{8}{4} \\ y=\frac{x}{4}-2[/tex]
2.1) The slope of this new line is equivalent to m=-4 since -4 is opposite inverse value for the slope of its perpendicular line (1/4).
If the line passes through the point (6,-5), then this point already belongs to this line.
Then, we can write.
-5=-4(6)+b
-5=-24+b
-5+24=-24+24+b
b=19
Now we can write the function:
y=-4x+19
3) Firstly, let's rearrange the equation. Into a General Equation of the Line y=ax+b
2x+3y=5
-2x+2x+3y=5-2x
3y=5-2x
[tex]\frac{3y}{3} =\frac{5}{3} -\frac{2x}{3}[/tex]
[tex]y=\frac{-2x}{3}+\frac{5}{3}[/tex]
3.2) Let's find out b, the linear coefficient by plugging the coordinates of the point (-3,-8).
Also, Parallel lines have the same value for the slope such as [tex]m_{1}=m_{2}[/tex]
-8=-3(-2/3)+b
-8=2+b
-8-2=2-2+b
b=-10
Then the equation of this parallel line is:
[tex]y=\frac{-2x}{3} -10[/tex]
4) Now on this last question, we have to find an equation of the line through the point (-11,-4) perpendicular to another line y=15
Suppose the slope of the first line is [tex]m_{1}[/tex] and the value of the slope of its perpendicular is [tex]m_{2}=\frac{-1}{m_{1}}[/tex]
y=15 is a horizontal line with slope equals zero, we can rewrite it as y=0x+15.
Then the slope is zero for horizontal lines.
A perpendicular line to a horizontal line is a vertical one.
In addition to this, a vertical line passes through only one point in the x-axis. Then, there is no horizontal variation. This leads the formula of a slope to an undefinition for Real Set.
[tex]m=\frac{y_{2}-y_{1}}{0}[/tex]
The only perpendicular line which passes through the point (-11,-4) to y=15 is the line x=-11.
