Applications of Slope and Intercept

1. D. \(2\) percent per minute

The rate at which the battery was charged is equivalent to the rate of change of this relationship. In linear relationships, the rate of change is represented by the& slope of the line. We can calculate this slope from any two points on the line.

Two points whose coordinates are clearly visible from the graph are \((0, 40)\) and \((5, 50)\).

Now, to find the slope, let's take the ratio of the corresponding differences in the \(y\)-values and the \(x\)-values:

\(\frac{50-40}{5-0}=\frac{10}{5}=2\)

The slope of the line is \(2\), which means the battery was charged at a rate of \(2\) percent per minute.

2. \(10\) degrees Celsius

To find the temperature that corresponds to \(2\) minutes, we need to look for the point on the graph where Time is \(2\).

The point we are looking for is \((2,10)\), which means that after \(2\) minutes, the pizza's temperature was \(10\) degrees Celsius.

3. B. \(\frac{1}{4}\) minute

To find the duration that corresponds to a draining of \(18\) liters of water, we need to find the relationship's rate of change. In linear relationships, the rate of change is represented by the slope of the line. We can calculate this slope from any two points on the line.

Two points whose coordinates are clearly visible from the graph are \((0,360)\) and \((2.5,180)\).

Now, to find the slope, let's take the ratio of the corresponding differences in the \(y\)-values and the \(x\)-values:

\(\frac{180-360}{2.5-0}=\frac{-180}{2.5}=-72\)

The slope of the line is \(-72\), which means the rate of change is \(72\) liters per minute. So \(18\) liters of water were drained every \(\frac{18}{72}=\frac{1}{4} \) minutes.

4. D. \(0.6\) per kilometer

The rate at which Karl's truck consumed fuel is equivalent to the rate of change of this relationship. In linear relationships, the rate of change is represented by the slope of the line. We can calculate this slope from any two points on the line.

Two points whose coordinates are clearly visible from the graph are \((250, 350)\) and \((500, 200)\).

Now, to find the slope, let's take the ratio of the corresponding differences in the \(y\)-values and the \(x\)-values:

\(\frac{200-350}{500-250}=\frac{-150}{250}=-0.6\)

The slope of the line is \(-0.6\), which means Karl's truck consumed its fuel at a rate of \(0.60\) liters per kilometer.