Question

career math i unit 5 test linear functions and equations (a)
airplane pilot:
the graph below shows the rate at which fuel is burned when flying. use the amount of fuel used between hours 1.5 and 2.5 to make your slope calculation.
the y - axis is the distance measured in nm, and the x - axis is the time measured in hours.
write the equations for each line below.

1. high speed cruise: y=
2. long range cruise: y=

Explanation:

Step1: Recall slope - intercept form

The equation of a line is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. For lines passing through the origin ($b = 0$), the equation is $y=mx$. The slope $m=\frac{\Delta y}{\Delta x}$.

Step2: Find points for high - speed cruise

For high - speed cruise, we use the points at $x = 1.5$ and $x = 2.5$. Let's assume the distance (y - values) at $x = 1.5$ is $y_1$ and at $x = 2.5$ is $y_2$. From the graph, if we estimate the values (assuming we can read the graph accurately), say at $x = 1.5$ the distance for high - speed cruise is $y_1 = 575$ nm and at $x = 2.5$ the distance is $y_2=900$ nm. Then the slope $m_1=\frac{y_2 - y_1}{2.5 - 1.5}=\frac{900 - 575}{1}=325$. Since the line passes through the origin, the equation of the high - speed cruise line is $y = 325x$.

Step3: Find points for long - range cruise

For long - range cruise, using the points at $x = 1.5$ and $x = 2.5$. Let the distance (y - values) at $x = 1.5$ be $y_3$ and at $x = 2.5$ be $y_4$. If we estimate from the graph, say at $x = 1.5$ the distance for long - range cruise is $y_3 = 427$ nm and at $x = 2.5$ the distance is $y_4 = 687$ nm. Then the slope $m_2=\frac{y_4 - y_3}{2.5 - 1.5}=\frac{687 - 427}{1}=260$. Since the line passes through the origin, the equation of the long - range cruise line is $y = 260x$.

Answer:

1. $y = 325x$
2. $y = 260x$