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Answer :
We start by determining how many minutes each person runs on their first three days.
For Shyla, she starts with 5 minutes on day 1 and then adds 4 minutes each day. Therefore:
[tex]$$
\begin{aligned}
\text{Day 1:} \quad & 5 \text{ minutes} \\
\text{Day 2:} \quad & 5 + 4 = 9 \text{ minutes} \\
\text{Day 3:} \quad & 5 + 4 + 4 = 13 \text{ minutes}
\end{aligned}
$$[/tex]
Thus, Shyla runs for 5, 9, and 13 minutes on days 1, 2, and 3 respectively.
For Diego, he starts with 10 minutes on day 1 and then adds 2 minutes each day. Thus:
[tex]$$
\begin{aligned}
\text{Day 1:} \quad & 10 \text{ minutes} \\
\text{Day 2:} \quad & 10 + 2 = 12 \text{ minutes} \\
\text{Day 3:} \quad & 10 + 2 + 2 = 14 \text{ minutes}
\end{aligned}
$$[/tex]
So, Diego runs for 10, 12, and 14 minutes on days 1, 2, and 3 respectively.
Next, to determine if there will ever be a day where Shyla runs more minutes than Diego, we can express the minutes each runs on day [tex]$n$[/tex] with formulas:
- For Shyla:
[tex]$$
S(n) = 5 + 4(n-1)
$$[/tex]
- For Diego:
[tex]$$
D(n) = 10 + 2(n-1)
$$[/tex]
To find when Shyla’s minutes exceed Diego’s minutes, we set up the inequality:
[tex]$$
5 + 4(n-1) > 10 + 2(n-1)
$$[/tex]
Now, simplify the inequality step by step:
1. Expand both sides:
[tex]$$
5 + 4n - 4 > 10 + 2n - 2
$$[/tex]
2. Simplify the constants on both sides:
[tex]$$
4n + 1 > 2n + 8
$$[/tex]
3. Subtract [tex]$2n$[/tex] from both sides:
[tex]$$
2n + 1 > 8
$$[/tex]
4. Subtract 1 from both sides:
[tex]$$
2n > 7
$$[/tex]
5. Divide by 2:
[tex]$$
n > 3.5
$$[/tex]
Since [tex]$n$[/tex] must be a whole number representing a day, the smallest integer greater than 3.5 is [tex]$n = 4$[/tex]. This means that starting on day 4, Shyla will run more minutes than Diego.
To verify, on day 4:
- Shyla runs:
[tex]$$
S(4) = 5 + 4(4-1) = 5 + 12 = 17 \text{ minutes}
$$[/tex]
- Diego runs:
[tex]$$
D(4) = 10 + 2(4-1) = 10 + 6 = 16 \text{ minutes}
$$[/tex]
Thus, on day 4, Shyla runs 17 minutes and Diego runs 16 minutes, so Shyla indeed runs more.
In summary:
- Shyla’s running minutes for the first three days are: 5 minutes, 9 minutes, and 13 minutes.
- Diego’s running minutes for the first three days are: 10 minutes, 12 minutes, and 14 minutes.
- There will be a day when Shyla runs more minutes than Diego. Specifically, starting on day 4, Shyla runs more minutes than Diego.
For Shyla, she starts with 5 minutes on day 1 and then adds 4 minutes each day. Therefore:
[tex]$$
\begin{aligned}
\text{Day 1:} \quad & 5 \text{ minutes} \\
\text{Day 2:} \quad & 5 + 4 = 9 \text{ minutes} \\
\text{Day 3:} \quad & 5 + 4 + 4 = 13 \text{ minutes}
\end{aligned}
$$[/tex]
Thus, Shyla runs for 5, 9, and 13 minutes on days 1, 2, and 3 respectively.
For Diego, he starts with 10 minutes on day 1 and then adds 2 minutes each day. Thus:
[tex]$$
\begin{aligned}
\text{Day 1:} \quad & 10 \text{ minutes} \\
\text{Day 2:} \quad & 10 + 2 = 12 \text{ minutes} \\
\text{Day 3:} \quad & 10 + 2 + 2 = 14 \text{ minutes}
\end{aligned}
$$[/tex]
So, Diego runs for 10, 12, and 14 minutes on days 1, 2, and 3 respectively.
Next, to determine if there will ever be a day where Shyla runs more minutes than Diego, we can express the minutes each runs on day [tex]$n$[/tex] with formulas:
- For Shyla:
[tex]$$
S(n) = 5 + 4(n-1)
$$[/tex]
- For Diego:
[tex]$$
D(n) = 10 + 2(n-1)
$$[/tex]
To find when Shyla’s minutes exceed Diego’s minutes, we set up the inequality:
[tex]$$
5 + 4(n-1) > 10 + 2(n-1)
$$[/tex]
Now, simplify the inequality step by step:
1. Expand both sides:
[tex]$$
5 + 4n - 4 > 10 + 2n - 2
$$[/tex]
2. Simplify the constants on both sides:
[tex]$$
4n + 1 > 2n + 8
$$[/tex]
3. Subtract [tex]$2n$[/tex] from both sides:
[tex]$$
2n + 1 > 8
$$[/tex]
4. Subtract 1 from both sides:
[tex]$$
2n > 7
$$[/tex]
5. Divide by 2:
[tex]$$
n > 3.5
$$[/tex]
Since [tex]$n$[/tex] must be a whole number representing a day, the smallest integer greater than 3.5 is [tex]$n = 4$[/tex]. This means that starting on day 4, Shyla will run more minutes than Diego.
To verify, on day 4:
- Shyla runs:
[tex]$$
S(4) = 5 + 4(4-1) = 5 + 12 = 17 \text{ minutes}
$$[/tex]
- Diego runs:
[tex]$$
D(4) = 10 + 2(4-1) = 10 + 6 = 16 \text{ minutes}
$$[/tex]
Thus, on day 4, Shyla runs 17 minutes and Diego runs 16 minutes, so Shyla indeed runs more.
In summary:
- Shyla’s running minutes for the first three days are: 5 minutes, 9 minutes, and 13 minutes.
- Diego’s running minutes for the first three days are: 10 minutes, 12 minutes, and 14 minutes.
- There will be a day when Shyla runs more minutes than Diego. Specifically, starting on day 4, Shyla runs more minutes than Diego.
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