Question 795·Hard·Math

Read the axis labels at two grid intersections: the best-fit line passes through $(2,210)$ and $(10,390)$. These points are on the line, so they describe its predictions. The nearby dots are measured yields, not the line's predictions.

The slope is the change in predicted yield divided by the change in $x$. Type $m=\frac{390-210}{10-2}$ in Desmos. It shows $m=22.5$, so each one-unit increase in $x$ raises the predicted yield by $22.5$ kilograms per hectare.

At $x=2$, the line predicts $210$. Moving from $2$ to $x$ is a change of $x-2$, so multiply that change by the slope and add it to $210$. The linear model is:

Use $210$ from the line, not $220$ from the dot at $x=2$.

The target is a predicted yield of $465$ kilograms per hectare, so solve $f(x)=465$. Keep the slope in Desmos, define $f(x)=210+m(x-2)$, and type $f(x_1)\sim465$. The $\sim$ asks Desmos to find the input $x_1$; it reports $x_1\approx13.3333$. That is tens of kilograms per hectare, not bags.

Because $x_1$ counts tens of kilograms per hectare, multiply by $10$ to get kilograms per hectare. Multiply by $6.5$ hectares for the whole field, then divide by $25$ kilograms per bag. Type $\frac{10x_1(6.5)}{25}$; Desmos shows about $34.6667$ bags.

$34$ bags fall short of the required $34.6667$ bags. Since bags are whole items, round up when finding the minimum purchase that meets a target. The farmer should buy $35$ bags. Choice D.