G1.03: Examples 4-7

Answer: Since it is linear, [latex]C=mp+b[/latex], where [latex]m=0.027[/latex] and [latex]b=4.50[/latex]. Thus [latex]C=0.027p+4.50[/latex] is the formula.

Answer: Let [latex]m=1.35[/latex] and [latex]({{x}_{0}},{{y}_{0}})=(5,40)[/latex]. Then [latex-display]\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,y-{{y}_{0}}=m(x-{{x}_{0}})\\&\,\,\,\,\,\,\,\,\,\,\,\,y-40=1.35(x-5)\\&\,\,\,\,\,\,\,\,\,\,\,\,y-40=1.35x-6.75\\&\,\,y-40+40=1.35x-6.75+40\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=1.35x+33.25\end{align}[/latex-display] Thus the equation of the line is [latex]y=1.35x+33.25[/latex]. Check: We can look at the formula and read the slope as 1.35, so that checks. To see whether the given point is on it, evaluate the formula at [latex]x=5[/latex] and see whether it gives the correct y-value. [latex]y=1.35(5)+33.25=6.75+33.25=40[/latex]. This checks!

1. Choose the appropriate variable to be the output variable and call it y. Then call the input variable x.
2. Write two points as [latex]({{x}_{1}},{{y}_{1}})[/latex]and [latex]({{x}_{2}},{{y}_{2}})[/latex].
3. Use the two points to compute the slope. Call it m.
4. Pick one of the points (either is fine) and call it [latex]({{x}_{0}},{{y}_{0}})[/latex].
5. Plug those values into this equation. [latex](y-{{y}_{0}})=m(x-{{x}_{0}})[/latex]
6. Solve for y. That gives the equation of the line.
7. If different letters are needed besides x and y for the input and output variables, replace the x and y in the formula with those different letters.

Answer: Let [latex]({{x}_{1}},{{y}_{1}})[/latex]= (2,6) and = (4,11). [latex-display]m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\,\,\,\,\,=\,\,\,\frac{11-6}{4-2}\,\,=\frac{5}{2}\,\,\,\,\,=\,\,\,2.5[/latex-display] Now, to use the point-slope form, I will use [latex]({{x}_{0}},{{y}_{0}})[/latex]= (2,6) to get [latex]\begin{align}&(y-{{y}_{0}})=m(x-{{x}_{0}})\\&(y-6)=2.5(x-2)\\&y-6=2.5x-5\\&y\,\,\,\,\,=2.5x\,\,+\,\,1\\\end{align}[/latex] So the equation of the line is [latex]y=2.5x+1[/latex], the slope is 2.5, the y-intercept is 1. Check: Use the formula we derived to determine the output value for [latex]x=2[/latex] and see if the given point is on the line.   [latex]y=2.5x+1=2.5(2)+1=5+1=6[/latex]. This tells us that (2,6) is on the line. Use the formula we derived to determine the output value for [latex]x=4[/latex] and see if the second given point is on the line.   [latex]y=2.5x+1=2.5(4)+1=10+1=11[/latex]. This tells us that (4,11) is on the line. So this checks.

1. Find the formula to predict the oatmeal needed (called M) from the flour needed (F.)
2. Interpret the slope.
3. Interpret the y-intercept.

Answer: a. Here, the output variable (y) is M and the input variable (x) is F. So the two points given are (0,0) and (3, 2). Using the slope formula, [latex]m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\,\,\,\,\,=\,\,\,\frac{2-0}{3-0}\,\,=\frac{2}{3}\,\,\,\,\,=\,\,\,0.667[/latex]cups of oatmeal per one cup of flour. Then, to use the point-slope form of the line, we’ll use and the first point, which is (0,0). [latex-display]\begin{align}&(y-{{y}_{0}})=m(x-{{x}_{0}})\\&(y-0)=0.667(x-0)\\&\,\,\,\,\,\,\,\,\,\,\,y=0.667x\\&\,\,\,\,\,\,\,\,\,\,\,y\,=0.667x+0\\\end{align}[/latex-display] So the formula is [latex]M=0.667\cdot{F}[/latex], so the slope is 0.667 and the y-intercept is 0.

1. Interpret the slope: For every 1 additional cup of flour, we should use 0.667 additional cups of oatmeal.
2. Interpret the intercept: If we use 0 cups of flour, then we will use 0 cups of oatmeal.

Licenses & Attributions