Use the two rows of the table: how much does the cost increase when $k$ increases from $300$ to $700$? Divide the change in cost by the change in $k$ to get the rate per kWh.

Once you know the rate per kWh (the slope), plug that and one of the $(k, C)$ pairs from the table into $C = mk + b$ to solve for the fixed monthly fee $b$.

After you find the slope and the fixed fee, compare them to the options. Also try plugging $k = 300$ and $k = 700$ into each option to see which one matches both given costs.

In Desmos, make a table with the two rows $(300, 54)$ and $(700, 94)$. Think of $x$ as $k$ and $y$ as $C$ so the points show the given usages and costs.

Below the table, type each option as an equation using $x$ for $k$ and $y$ for $C$, for example: y = 0.1x + 30, y = 0.15x + 9, y = 0.1x + 24, and y = x/4 + 54. Make sure all the lines are visible.

Look at which line passes exactly through both data points from the table. The correct equation is the one whose graph goes through both $(300, 54)$ and $(700, 94)$ without missing either point.

Because the utility charges a fixed monthly fee plus a usage fee that depends on $k$, the relationship between $C$ and $k$ is linear.

We can write it in the form

where:

• $m$ is the cost per kilowatt-hour (the slope), and
• $b$ is the fixed monthly fee (the $y$-intercept).

From the table, the two data points are $(300, 54)$ and $(700, 94)$, where $k$ is like $x$ and $C$ is like $y$.

The slope $m$ is:

So the cost per kWh is $0.10$, and the equation so far is

Now use one of the points from the table to find $b$.

Using $(300, 54)$, substitute $k = 300$ and $C = 54$ into

That gives the equation

Next, solve this equation for $b$.},{

Compute $0.1 \cdot 300$:

So the equation from the previous step becomes

Subtract $30$ from both sides:

Substitute this back into the linear form $C = 0.1k + b$ to get the final equation: