Question

Hello -- I could use some assistance solving the attached problems; slope intercept form. Thank you.Question 3

Answer 1

The slope-intercept form of an equation of a straight line is

`\begin{gathered} y=mx+b \\ \text{Where } \\ m\text{ is the slope and b is the y-intercept} \end{gathered}`

From the given table,

Taking two coordinates from the table,

`\begin{gathered} (x_1,y_1)\Rightarrow(0,100) \\ (x_2,y_2)\Rightarrow(1,105) \end{gathered}`

To find the equation of a staright line, the formula is

`(y-y_1)/(x-x_1)=(y_2-y_1)/(x_2-x_1)`

Sustituting for the coordinates into the formula above

`\begin{gathered} (y-y_1)/(x-x_1)=(y_2-y_1)/(x_2-x_1) \\ \frac{y-100_{}}{x-0}=\frac{105_{}-100_{}}{1-0} \\ \frac{y-100_{}}{x}=(5)/(1) \\ \text{Crossmultiply} \\ 1(y-100)=5* x \\ y-100=5x \\ y=5x+100 \end{gathered}`

Hence, the slope intercept form of the equation representing the data in the table is

`y=5x+100`