Answer :
The explicit rules for Hayden and Nate are y = 12 + 2 * (x - 1) and [tex]y = 12 \times (1 + 0.1)^{(\text{x} - 1)}[/tex]
Hayden studied for 26 hours and Nate for 23.4 hours
Total study hours for each boy over 15 weeks are 390 and 381.6
The week number when Nate catches up to Hayden is 1
A. Explicit rules for Hayden and Nate:
Let the week number be x
So, we have
For Hayden:
y = 12 + 2 * (x - 1)
For Nate:
[tex]y = 12 \times (1 + 0.1)^{(\text{x} - 1)}[/tex]
B. Number of hours each student studied in week 8:
Here, we set x = 8
So, we have
For Hayden:
[tex]\[ \text{y} = 12 + 2 \times (8 - 1) = 12 + 2 \times 7 = 12 + 14 = 26 \text{ hours} \][/tex]
For Nate:
[tex]\[ \text{y} = 12 \times (1 + 0.1)^{(8 - 1)} = 12 \times (1.1)^7 \][/tex]
[tex]\[ \text{y} \approx 12 \times 1.948717 = 23.4\text{ hours} \][/tex]
C. Total study hours for each boy over 15 weeks:
For Hayden: we have
[tex]\[ \text{Total} = \sum_{n=1}^{15} (12 + 2 \times (n - 1)) \][/tex]
Using the sum of AP series, we have
[tex]\[ \text{Total} = \dfrac{15}{2}(2 \times 12 + (15 - 1) \times 2)[/tex]
Evaluate
[tex]\[ \text{Total} = 390[/tex]
For Nate:
[tex]\[ \text{Total} = \sum_{n=1}^{15} (12 \times (1 + 0.1)^{n - 1}) \][/tex]
[tex]\[ \text{Total} = \sum_{n=1}^{15} (12 \times (1 .1)^{n - 1}) \][/tex]
Using the sum of GP, we have
[tex]\[ \text{Total} = \dfrac{12 \times (1 .1)^{15} - 1}{1.1 - 1} \][/tex]
Evaluate
[tex]\[ \text{Total} = \dfrac{12 \times 3.18}{14} \][/tex]
[tex]\[ \text{Total} = 381.6[/tex]
D. Week when Nate catches up to Hayden in study hours:
This is represented as
[tex](12 \times (1.1)^{x - 1}) = (12 + 2 \times (x - 1))[/tex]
Solving graphically, we have
x = 1
This mean sthat
The week number when Nate catches up to Hayden is 1
A. Hayden: [tex]\(S_H(n) = 12 + 2(n-;1)\)[/tex] Nate: [tex]\(S_N(n) = 12(1 + 0.10)^{n-1}\)[/tex]
B. Hayden: 26 hours ; Nate: approximately 25.94 hours.
C. Hayden: 600 hours ; Nate: approximately 1168.46 hours.
D. Nate catches up to Hayden in week 16.
A.
1. For Hayden: The explicit rule for Hayden's study time can be described as [tex]\(S_H(n) = 12 + 2(n-1)\),[/tex] where (n) represents the week number.
2. For Nate: The explicit rule for Nate's study time can be described as
[tex]\(S_N(n) = 12(1 + 0.10)^{n-1}\),[/tex] where (n) represents the week number.
B. To find out how many hours each student studied in week 8:
1. For Hayden: [tex]\(S_H(8) = 12 + 2(8-1) = 12 + 14 = 26\)[/tex] hours.
2. For Nate: [tex]\(S_N(8) = 12(1 + 0.10)^{8-1} = 12(1.10)^7 \approx 25.94\)[/tex] hours.
C. To find out how many hours each boy studied in the entire 15 weeks:
1. For Hayden [tex]: \(S_H(15) = 12 + 2(15-1) = 12 + 28 = 40\)[/tex] hours per week. Therefore, [tex]\(40 \times 15 = 600\)[/tex] hours in total.
2. For Nate: [tex]\(S_N(15) = 12(1 + 0.10)^{15-1} = 12(1.10)^{14} \approx 77.90\)[/tex] hours per week. Therefore, [tex]\(77.90 \times 15 \approx 1168.46\)[/tex] hours in total.
D. To find out in which week Nate will catch up to Hayden in the number of hours spent on studying:
We set [tex](S_N(n) = S_H(n)\)[/tex] and solve for (n):
[tex]\[12(1.10)^{n-1} = 12 + 2(n-1)\]\[1.10^{n-1} = \frac{12 + 2(n-1)}{12}\]\[1.10^{n-1} = 1 + \frac{n-1}{6}\]\[1.10^{n-1} = \frac{n + 5}{6}\][/tex]
Taking the natural log of both sides:
[tex]\[n-1 = \log_{1.10}\left(\frac{n+5}{6}\right)\]\[n = \log_{1.10}\left(\frac{n+5}{6}\right) + 1\][/tex]
Using numerical methods or software, we find [tex]\(n \approx 15.72\).[/tex]
Since (n) must be an integer, Nate will catch up to Hayden in week 16.
