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Sagot :
To find the [tex]\( x \)[/tex]-intercept (the time it takes to reach the school) given the data provided, we need to find the time at which the distance from the school is zero.
Given data:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (in minutes) } x & \text{Distance (in meters) } f(x) \\ \hline 0 & 36 \\ 3 & 32 \\ 6 & 28 \\ 9 & 24 \\ 12 & 20 \\ \hline \end{array} \][/tex]
Let’s use the method of linear interpolation to estimate the time at which the distance would be zero.
### Step-by-Step Solution:
1. Identify the Two Points for Interpolation:
We can observe that the distance is decreasing linearly. To find the [tex]\( x \)[/tex]-intercept, use the two end points in the given table:
- At [tex]\( x = 0 \)[/tex], the distance [tex]\( f(0) = 36 \)[/tex] meters.
- At [tex]\( x = 12 \)[/tex], the distance [tex]\( f(12) = 20 \)[/tex] meters.
2. Determine the Slope of the Line:
The slope [tex]\( m \)[/tex] of the line can be found using the two points [tex]\((x_1, y_1) = (0, 36)\)[/tex] and [tex]\((x_2, y_2) = (12, 20)\)[/tex].
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{20 - 36}{12 - 0} = \frac{-16}{12} = -\frac{4}{3} \][/tex]
3. Equation of the Line:
Using the point-slope form of the line equation [tex]\( y - y_1 = m(x - x_1) \)[/tex]:
[tex]\[ f(x) - 36 = -\frac{4}{3} (x - 0) \][/tex]
Simplifying, we get:
[tex]\[ f(x) = -\frac{4}{3} x + 36 \][/tex]
4. Find the [tex]\( x \)[/tex]-Intercept:
The [tex]\( x \)[/tex]-intercept occurs where [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 0 = -\frac{4}{3} x + 36 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -\frac{4}{3} x = -36 \implies x = 36 \times \frac{3}{4} = 27 \][/tex]
### Interpretation:
- The [tex]\( x \)[/tex]-intercept is [tex]\( x = 27 \)[/tex] minutes.
- Meaning: It takes approximately 27 minutes to reach the school.
Therefore, the point [tex]\( (27, 0) \)[/tex] indicates the time it takes to reach the school, where the distance becomes zero.
Given data:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (in minutes) } x & \text{Distance (in meters) } f(x) \\ \hline 0 & 36 \\ 3 & 32 \\ 6 & 28 \\ 9 & 24 \\ 12 & 20 \\ \hline \end{array} \][/tex]
Let’s use the method of linear interpolation to estimate the time at which the distance would be zero.
### Step-by-Step Solution:
1. Identify the Two Points for Interpolation:
We can observe that the distance is decreasing linearly. To find the [tex]\( x \)[/tex]-intercept, use the two end points in the given table:
- At [tex]\( x = 0 \)[/tex], the distance [tex]\( f(0) = 36 \)[/tex] meters.
- At [tex]\( x = 12 \)[/tex], the distance [tex]\( f(12) = 20 \)[/tex] meters.
2. Determine the Slope of the Line:
The slope [tex]\( m \)[/tex] of the line can be found using the two points [tex]\((x_1, y_1) = (0, 36)\)[/tex] and [tex]\((x_2, y_2) = (12, 20)\)[/tex].
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{20 - 36}{12 - 0} = \frac{-16}{12} = -\frac{4}{3} \][/tex]
3. Equation of the Line:
Using the point-slope form of the line equation [tex]\( y - y_1 = m(x - x_1) \)[/tex]:
[tex]\[ f(x) - 36 = -\frac{4}{3} (x - 0) \][/tex]
Simplifying, we get:
[tex]\[ f(x) = -\frac{4}{3} x + 36 \][/tex]
4. Find the [tex]\( x \)[/tex]-Intercept:
The [tex]\( x \)[/tex]-intercept occurs where [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 0 = -\frac{4}{3} x + 36 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -\frac{4}{3} x = -36 \implies x = 36 \times \frac{3}{4} = 27 \][/tex]
### Interpretation:
- The [tex]\( x \)[/tex]-intercept is [tex]\( x = 27 \)[/tex] minutes.
- Meaning: It takes approximately 27 minutes to reach the school.
Therefore, the point [tex]\( (27, 0) \)[/tex] indicates the time it takes to reach the school, where the distance becomes zero.
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