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Sagot :
To determine the equation that represents the change in altitude of the airplane from a given table of data, we follow these steps:
1. Identify the given data:
[tex]\[ \begin{array}{|c|c|} \hline \text{Minutes (m)} & \text{Altitude (a in ft)} \\ \hline 1 & 35,000 \\ \hline 2 & 31,000 \\ \hline 3 & 27,000 \\ \hline 4 & 23,000 \\ \hline \end{array} \][/tex]
2. Determine the change in altitude per minute:
- The altitude decreases uniformly, so we can calculate the difference between any two consecutive altitudes. Let's use the first two data points.
[tex]\[ \Delta a = \text{Altitude at minute 1} - \text{Altitude at minute 2} = 35,000 - 31,000 = 4,000 \text{ ft} \][/tex]
- This means the airplane descends 4,000 feet each minute.
3. Formulate the general equation:
- The general form of the linear equation is:
[tex]\[ a = \text{slope} \times m + \text{intercept} \][/tex]
- The slope (rate of change) is calculated as -4,000 since the airplane is descending:
[tex]\[ \text{slope} = -4,000 \text{ ft/min} \][/tex]
4. Calculate the y-intercept (initial altitude when m = 0):
- We know the altitude at minute 1:
[tex]\[ \text{Altitude at minute } 1 = -4,000 \times 1 + \text{intercept} \][/tex]
[tex]\[ 35,000 = -4,000 \times 1 + \text{intercept} \][/tex]
Solving for the intercept:
[tex]\[ \text{intercept} = 35,000 + 4,000 = 39,000 \text{ ft} \][/tex]
5. Write the final equation:
- Plugging the slope and the intercept into the equation form:
[tex]\[ a = -4,000m + 39,000 \][/tex]
However, the correct intercept based on given data from the result (31000), the proper intercept would be:
6. Verification: Let's double-check the intercept we calculated using substitution:
[tex]\[\begin{aligned} a &= -4,000m + 31,000 . \end{aligned}] Therefore, the refined equation actually becomes: \[ a = -4000m + 35000. \][/tex]
Finally, this confirms the initial intercept calculation being correct.
1. Identify the given data:
[tex]\[ \begin{array}{|c|c|} \hline \text{Minutes (m)} & \text{Altitude (a in ft)} \\ \hline 1 & 35,000 \\ \hline 2 & 31,000 \\ \hline 3 & 27,000 \\ \hline 4 & 23,000 \\ \hline \end{array} \][/tex]
2. Determine the change in altitude per minute:
- The altitude decreases uniformly, so we can calculate the difference between any two consecutive altitudes. Let's use the first two data points.
[tex]\[ \Delta a = \text{Altitude at minute 1} - \text{Altitude at minute 2} = 35,000 - 31,000 = 4,000 \text{ ft} \][/tex]
- This means the airplane descends 4,000 feet each minute.
3. Formulate the general equation:
- The general form of the linear equation is:
[tex]\[ a = \text{slope} \times m + \text{intercept} \][/tex]
- The slope (rate of change) is calculated as -4,000 since the airplane is descending:
[tex]\[ \text{slope} = -4,000 \text{ ft/min} \][/tex]
4. Calculate the y-intercept (initial altitude when m = 0):
- We know the altitude at minute 1:
[tex]\[ \text{Altitude at minute } 1 = -4,000 \times 1 + \text{intercept} \][/tex]
[tex]\[ 35,000 = -4,000 \times 1 + \text{intercept} \][/tex]
Solving for the intercept:
[tex]\[ \text{intercept} = 35,000 + 4,000 = 39,000 \text{ ft} \][/tex]
5. Write the final equation:
- Plugging the slope and the intercept into the equation form:
[tex]\[ a = -4,000m + 39,000 \][/tex]
However, the correct intercept based on given data from the result (31000), the proper intercept would be:
6. Verification: Let's double-check the intercept we calculated using substitution:
[tex]\[\begin{aligned} a &= -4,000m + 31,000 . \end{aligned}] Therefore, the refined equation actually becomes: \[ a = -4000m + 35000. \][/tex]
Finally, this confirms the initial intercept calculation being correct.
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