Julissa is running a 10-kilometer race at a constant pace. After running for 18 minutes, she completes 2 kilometers. After running for 54 minutes, she completes 6 kilometers. Her trainer writes an equation letting [tex]t[/tex], the time in minutes, represent the independent variable and [tex]k[/tex], the number of kilometers, represent the dependent variable.

Which equation can be used to represent [tex]k[/tex], the number of kilometers Julissa runs in [tex]t[/tex] minutes?

A. [tex]k - 2 = \frac{1}{9}(t - 18)[/tex]
B. [tex]k - 18 = \frac{1}{9}(t - 2)[/tex]
C. [tex]k - 2 = 9(t - 18)[/tex]
D. [tex]k - 18 = 9(t - 2)[/tex]



Answer :

To determine which equation represents the number of kilometers [tex]\( k \)[/tex] Julissa runs in [tex]\( t \)[/tex] minutes, we start by noting two observations from the problem:

1. After 18 minutes, Julissa has run 2 kilometers.
2. After 54 minutes, Julissa has run 6 kilometers.

From these observations, we can find her constant running pace.

First, we calculate the change in time (Δt) and the change in distance (Δk):

[tex]\[ \Delta t = 54 \text{ minutes} - 18 \text{ minutes} = 36 \text{ minutes} \][/tex]

[tex]\[ \Delta k = 6 \text{ kilometers} - 2 \text{ kilometers} = 4 \text{ kilometers} \][/tex]

Next, we determine her running pace, which is the distance per unit time. The slope (m), or her running pace in kilometers per minute, is:

[tex]\[ m = \frac{\Delta k}{\Delta t} = \frac{4 \text{ kilometers}}{36 \text{ minutes}} = \frac{1}{9} \text{ kilometers per minute} \][/tex]

We use the point-slope form of a linear equation [tex]\( k - k_1 = m (t - t_1) \)[/tex], where [tex]\( t_1 \)[/tex] and [tex]\( k_1 \)[/tex] are given points on the line, and [tex]\( m \)[/tex] is the slope. Here, we can use the point (18, 2):

[tex]\[ k - 2 = \frac{1}{9}(t - 18) \][/tex]

So, the equation that represents the number of kilometers [tex]\( k \)[/tex] Julissa runs in [tex]\( t \)[/tex] minutes is:

[tex]\[ k - 2 = \frac{1}{9}(t - 18) \][/tex]

Thus, the correct equation is:

[tex]\(\boxed{k-2=\frac{1}{9}(t-18)}\)[/tex]