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Sagot :
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]
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]
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