Respuesta :
We can start by modeling the teacher's tree height.
First, we define as time x=0 as the time when the teacher plant the 24 inches tree.
Then, y(0) = 24. This is the y-intercept (b=24), as it is the value of y when x=0.
This tree grows 10 inches per year. This is the slope when y is expressed in inches and x is expressed in years. Then, m=10.
We can write the equation for the height of the teacher's tree as:
[tex]\begin{gathered} y=mx+b \\ y=10x+24 \end{gathered}[/tex]Our tree is planted at x=5 (five years after the teacher plant the tree) and, as it is planted from seed, the height at this time x=5 is y(5)=0.
The slope is equal to the growth rate, that is 18.5 inches a year. Then, m=18.5.
As we know one point of the line and the slope, we can write the equation in slope-point form and then re-arrange it into slop-intercept form:
[tex]\begin{gathered} y-y_0=m(x-x_0) \\ y-0=18.5(x-5) \\ y=18.5x-18.5\cdot5 \\ y=18.5x-92.5 \end{gathered}[/tex]We then can find when the two trees reach the same height by equalizing both expressions for the heights:
[tex]\begin{gathered} y_{\text{teacher}}=y_{me} \\ 10x+24=18.5x-92.5 \\ 24+92.5=18.5x-10x \\ 116.5=8.5x \\ x=\frac{116.5}{8.5} \\ x\approx13.7 \end{gathered}[/tex]We can calculate the height of the trees when they have the same height by relacing x with 13.7 in anny of the 2 equations:
[tex]y(13.7)=10(13.7)+24=137+24=161[/tex]We can find the x-intercept by finding the values of x for y=0.
In our case, we know that at the moment of planting, x=5, the height is y=0, so the x-intercept is x=5.
In the case of the teacher, we have to calculate:
[tex]\begin{gathered} y=0=10x+24 \\ 10x=-24 \\ x=-\frac{24}{10} \\ x=-2.4 \end{gathered}[/tex]The x-intercept is x=-2.4.
We then can fill the table as:
Teacher:
Intercepts: x-intercept = -2.4, y-intercept = 24.
Slope: m = 10
Equation: y = 10x+24
Me:
Intercepts: x-intercept = 5, y-intercept = -92.5
Slope: m = 18.5
Equation: y=18.5x-92.5
Answer: the trees reach the same height, y=161 in., at time x=13.7 years.
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