Therate of changeis [tex]\boxed{\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ i}}{{\mathbf{n}}^{\mathbf{2}}}}[/tex] per day.
Further explanation:
It is given that 9 days ago, the area covered by a mold and a piece of bread is [tex]3{\text{ i}}{{\text{n}}^2}[/tex].
Consider days as [tex]x[/tex] and the area covered by a mold and a piece of breadas [tex]y[/tex].
Since, the initial value that is first day is consider as 0 and the areacovered by a mold and a piece of bread is [tex]3{\text{ i}}{{\text{n}}^2}[/tex] so the value of [tex]{x_1}[/tex] is [tex]0[/tex] and the value of [tex]{y_1}[/tex] is [tex]3[/tex].
Now, after 9 days ,the area covered by a mold and a piece of bread is [tex]9{\text{ i}}{{\text{n}}^2}[/tex] so the value of [tex]{x_2}[/tex] is [tex]9[/tex] and the value of [tex]{y_2}[/tex] is [tex]9[/tex].
The rate of change is defined as the ratio of the change in [tex]y[/tex] with respect to change in [tex]x[/tex] .
[tex]{\text{Rate of change }}= \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}{\text{}}[/tex] ......(1)
Now, substitute [tex]0[/tex] for [tex]{x_1}[/tex] , [tex]9[/tex] for [tex]{x_2}[/tex], [tex]3[/tex] for [tex]{y_1}[/tex] and [tex]9[/tex] for [tex]{y_2}[/tex] in equation (1) to obtain the rate of change.
[tex]\begin{aligned}{\text{Rate of change }}&= \frac{{9 - 3}}{{9 - 0}}\\ &= \frac{6}{9}\\&=\frac{2}{3}\\\end{aligned}[/tex]
Therefore, rate of change is [tex]\dfrac{2}{3}{\text{ i}}{{\text{n}}^2}[/tex].
Thus, the rate of change is [tex]\boxed{\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ i}}{{\mathbf{n}}^{\mathbf{2}}}}[/tex] per day.
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Answer Details:
Grade: Junior High School
Subject: Mathematics
Chapter: Surface Area and Volumes
Keywords: Surface area, linear equation, system of linear equations in two variables, largest printed area, derivative, differentiation, mold and piece of bread
