. A fruit company recently released a new applesauce. By the end of its first​ year, profits on this product amounted to $24,700.  The anticipated profit for the end of the fourth year is $67,900.  After the first​ year, the ratio of change in time to change in profit is constant. Let x be years and P be profit in dollars.P(x)=2. Find an equation of the line. Write the equation using function notation.Through (6,−5)​; perpendicular to 4y=x-83. Find an equation of the line. Write the equation using function notation.Through (-3,-8) ; parallel to 2x+3y=54. Find the equation of the lineThrough (-11,-4) perpendicular to y=15

Accepted Solution

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+b67,900=4x +b1.1)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+103002) 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+bb=19Now we can write the function:y=-4x+193) Firstly, let's rearrange the equation. Into a General Equation of the Line y=ax+b2x+3y=5-2x+2x+3y=5-2x3y=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+bb=-10Then 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=15Suppose 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.