Answer :
Let's review the problem step-by-step and identify where Omyra made errors in her calculations.
### Step-by-Step Solution:
1. Determine the speeds and time:
- Joseph jogs at 8 kilometers per hour.
- Isabelle rides at 12 kilometers per hour.
- Both are traveling for 15 hours.
2. Calculate the distance traveled by each:
- Joseph's distance: [tex]\( 8 \, \text{km/h} \times 15 \, \text{hours} = 120 \, \text{km} \)[/tex].
- Isabelle's distance: [tex]\( 12 \, \text{km/h} \times 15 \, \text{hours} = 180 \, \text{km} \)[/tex].
3. Apply the Pythagorean theorem to find the distance between Joseph and Isabelle:
- They are traveling on perpendicular paths (north and west), forming a right-angled triangle.
- Let the distances traveled by Joseph and Isabelle be the legs of the triangle, a and b, respectively:
[tex]\[ a = 120 \, \text{km} \quad \text{and} \quad b = 180 \, \text{km} \][/tex]
- The distance apart (hypotenuse, c) is found using:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
[tex]\[ c^2 = 120^2 + 180^2 \][/tex]
[tex]\[ c^2 = 14400 + 32400 \][/tex]
[tex]\[ c^2 = 46800 \][/tex]
4. Solve for c:
- Take the square root of both sides to find the hypotenuse:
[tex]\[ c = \sqrt{46800} \approx 216.33 \, \text{km} \][/tex]
### Review of Omyra's Work:
Let's compare this correct solution to Omyra's work:
[tex]\[ \begin{aligned} 8^2+12^2 & -a^2 \\ 64+24 & =a^2 \\ 88 & =a^2 \end{aligned} \][/tex]
Here are Omyra's errors:
1. Error in Squaring the Speeds:
- The squares should be [tex]\(8^2 = 64\)[/tex] and [tex]\(12^2 = 144\)[/tex], not [tex]\(64 + 24\)[/tex].
2. Error in Adding the Squares:
- Omyra incorrectly added the squared values as [tex]\(64 + 24 = 88\)[/tex]. The correct sum is [tex]\(64 + 144 = 208 \, \text{(if calculating speeds after 1 hour only, which she did wrong here)}\)[/tex].
### Corrected Calculations:
- Correct speeds and distances after 15 hours squared:
[tex]\[ 120^2 = 14400 \, \text{km}^2 \quad \text{and} \quad 180^2 = 32400 \, \text{km}^2 \][/tex]
- Correct sum:
[tex]\[ 14400 + 32400 = 46800 \, \text{km}^2 \][/tex]
- Distance between Joseph and Isabelle:
[tex]\[ c = \sqrt{46800} \approx 216.33 \, \text{km} \][/tex]
To sum up, the correct results lead to the following errors:
- Error in Basic Squaring:
[tex]\(8^2\)[/tex] should equal 64 and [tex]\(12^2\)[/tex] should equal 144.
- Error in Summation:
Sum [tex]\(64\)[/tex] and [tex]\(144\)[/tex] correctly rather than the false [tex]\( 64 + 24 = 88 \)[/tex].
By identifying these key mistakes, we can see why her result was inaccurate.
### Step-by-Step Solution:
1. Determine the speeds and time:
- Joseph jogs at 8 kilometers per hour.
- Isabelle rides at 12 kilometers per hour.
- Both are traveling for 15 hours.
2. Calculate the distance traveled by each:
- Joseph's distance: [tex]\( 8 \, \text{km/h} \times 15 \, \text{hours} = 120 \, \text{km} \)[/tex].
- Isabelle's distance: [tex]\( 12 \, \text{km/h} \times 15 \, \text{hours} = 180 \, \text{km} \)[/tex].
3. Apply the Pythagorean theorem to find the distance between Joseph and Isabelle:
- They are traveling on perpendicular paths (north and west), forming a right-angled triangle.
- Let the distances traveled by Joseph and Isabelle be the legs of the triangle, a and b, respectively:
[tex]\[ a = 120 \, \text{km} \quad \text{and} \quad b = 180 \, \text{km} \][/tex]
- The distance apart (hypotenuse, c) is found using:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
[tex]\[ c^2 = 120^2 + 180^2 \][/tex]
[tex]\[ c^2 = 14400 + 32400 \][/tex]
[tex]\[ c^2 = 46800 \][/tex]
4. Solve for c:
- Take the square root of both sides to find the hypotenuse:
[tex]\[ c = \sqrt{46800} \approx 216.33 \, \text{km} \][/tex]
### Review of Omyra's Work:
Let's compare this correct solution to Omyra's work:
[tex]\[ \begin{aligned} 8^2+12^2 & -a^2 \\ 64+24 & =a^2 \\ 88 & =a^2 \end{aligned} \][/tex]
Here are Omyra's errors:
1. Error in Squaring the Speeds:
- The squares should be [tex]\(8^2 = 64\)[/tex] and [tex]\(12^2 = 144\)[/tex], not [tex]\(64 + 24\)[/tex].
2. Error in Adding the Squares:
- Omyra incorrectly added the squared values as [tex]\(64 + 24 = 88\)[/tex]. The correct sum is [tex]\(64 + 144 = 208 \, \text{(if calculating speeds after 1 hour only, which she did wrong here)}\)[/tex].
### Corrected Calculations:
- Correct speeds and distances after 15 hours squared:
[tex]\[ 120^2 = 14400 \, \text{km}^2 \quad \text{and} \quad 180^2 = 32400 \, \text{km}^2 \][/tex]
- Correct sum:
[tex]\[ 14400 + 32400 = 46800 \, \text{km}^2 \][/tex]
- Distance between Joseph and Isabelle:
[tex]\[ c = \sqrt{46800} \approx 216.33 \, \text{km} \][/tex]
To sum up, the correct results lead to the following errors:
- Error in Basic Squaring:
[tex]\(8^2\)[/tex] should equal 64 and [tex]\(12^2\)[/tex] should equal 144.
- Error in Summation:
Sum [tex]\(64\)[/tex] and [tex]\(144\)[/tex] correctly rather than the false [tex]\( 64 + 24 = 88 \)[/tex].
By identifying these key mistakes, we can see why her result was inaccurate.
