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Xóchitl pedals at a rate of [tex]\(18 \frac{km}{h}\)[/tex] on a stationary bike. Her friend Cowessess has already pedaled for 12 minutes at an average rate of [tex]\(10 \frac{km}{h}\)[/tex].

Assuming Cowessess's rate stays the same, how long would Xóchitl have to pedal to catch up to Cowessess's distance?
[tex]\[\square\][/tex] minutes


Sagot :

To determine how long Xóchitl needs to pedal to catch up to the distance that Cowessess has already covered, let’s break down the problem step by step.

### Step 1: Calculate the Distance Cowessess Has Traveled

Cowessess is pedaling at an average rate of [tex]\(10 \, \frac{km}{h}\)[/tex] for 12 minutes. First, we need to convert the time from minutes to hours:
[tex]\[ \text{Time} = \frac{12 \, \text{minutes}}{60 \, \text{minutes/hour}} = 0.2 \, \text{hours} \][/tex]

Next, calculate the distance. Distance traveled can be found using the formula:
[tex]\[ \text{Distance} = \text{Speed} \times \text{Time} \][/tex]
Plug in the values for Cowessess’s speed:
[tex]\[ \text{Distance}_{\text{Cowessess}} = 10 \, \frac{km}{h} \times 0.2 \, h = 2 \, km \][/tex]

### Step 2: Determine the Time Xóchitl Needs to Cover the Same Distance

Xóchitl needs to pedal the same distance to catch up. Her pedaling rate is [tex]\(18 \, \frac{km}{h}\)[/tex]. To find the time she needs to pedal, use the distance formula rearranged to solve for time:
[tex]\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \][/tex]
Using the distance Cowessess traveled and Xóchitl's speed:
[tex]\[ \text{Time}_{\text{Xóchitl}} = \frac{2 \, km}{18 \, \frac{km}{h}} = \frac{2}{18} \, h = \frac{1}{9} \, h \][/tex]

### Step 3: Convert Xóchitl’s Pedaling Time to Minutes

Xóchitl's time calculated is in hours. Convert this to minutes by multiplying by 60:
[tex]\[ \text{Time}_{\text{Xóchitl}} = \frac{1}{9} \, h \times 60 \, \frac{\text{minutes}}{h} = \frac{60}{9} \, \text{minutes} = 6.6667 \, \text{minutes} \][/tex]

### Conclusion

Therefore, Xóchitl would need to pedal for approximately [tex]\(6.67\)[/tex] minutes (or equivalently [tex]\(6 \frac{2}{3}\)[/tex] minutes) to catch up to the distance that Cowessess has already covered.
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