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Solve the following word problem. The air temperature at 9 a.m. is −5.8 degrees Celsius. The air temperature at noon is −1.6 de...Asked by man i love latinasssss
Solve the following word problem. The air temperature at 9 a.m. is −5.8 − 5.8 degrees Celsius. The air temperature at noon is −1.6 − 1.6 degrees Celsius. What is the change in the temperature during these three hours? Write and solve an equation to show your answer. Then explain what your answer means.(4 points)
Answers
Answered by
man i love latinasssss
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Answered by
man i love latinasssss
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Answered by
GPT-4o mini
To find the change in temperature from 9 a.m. to noon, we can set up the equation using the temperatures at these times.
Let:
- \( T_1 = -5.8 \) °C (temperature at 9 a.m.)
- \( T_2 = -1.6 \) °C (temperature at noon)
The change in temperature (\( \Delta T \)) can be calculated using the formula:
\[
\Delta T = T_2 - T_1
\]
Substituting the values we have:
\[
\Delta T = -1.6 - (-5.8)
\]
This can be simplified as follows:
\[
\Delta T = -1.6 + 5.8
\]
Now, performing the addition:
\[
\Delta T = 4.2
\]
Thus, the change in temperature during these three hours is \( 4.2 \) degrees Celsius.
### Explanation of the Answer:
The answer of \( 4.2 \) °C means that the temperature increased by 4.2 degrees from 9 a.m. to noon. Initially, the temperature was quite cold at −5.8 °C, but by noon, it had warmed up to −1.6 °C. This indicates a significant warming trend in the air temperature over the course of three hours.
Let:
- \( T_1 = -5.8 \) °C (temperature at 9 a.m.)
- \( T_2 = -1.6 \) °C (temperature at noon)
The change in temperature (\( \Delta T \)) can be calculated using the formula:
\[
\Delta T = T_2 - T_1
\]
Substituting the values we have:
\[
\Delta T = -1.6 - (-5.8)
\]
This can be simplified as follows:
\[
\Delta T = -1.6 + 5.8
\]
Now, performing the addition:
\[
\Delta T = 4.2
\]
Thus, the change in temperature during these three hours is \( 4.2 \) degrees Celsius.
### Explanation of the Answer:
The answer of \( 4.2 \) °C means that the temperature increased by 4.2 degrees from 9 a.m. to noon. Initially, the temperature was quite cold at −5.8 °C, but by noon, it had warmed up to −1.6 °C. This indicates a significant warming trend in the air temperature over the course of three hours.
Answered by
GPT-4o mini
To find the change in temperature from 9 a.m. to noon:
Let:
- \( T_1 = -5.8 \) °C (9 a.m.)
- \( T_2 = -1.6 \) °C (noon)
The change in temperature (\( \Delta T \)) is calculated as follows:
\[
\Delta T = T_2 - T_1 = -1.6 - (-5.8) = -1.6 + 5.8 = 4.2
\]
### Result:
The change in temperature is \( 4.2 \) °C.
### Explanation:
This means the temperature increased by 4.2 degrees from −5.8 °C at 9 a.m. to −1.6 °C at noon, indicating a warming trend over three hours.
Let:
- \( T_1 = -5.8 \) °C (9 a.m.)
- \( T_2 = -1.6 \) °C (noon)
The change in temperature (\( \Delta T \)) is calculated as follows:
\[
\Delta T = T_2 - T_1 = -1.6 - (-5.8) = -1.6 + 5.8 = 4.2
\]
### Result:
The change in temperature is \( 4.2 \) °C.
### Explanation:
This means the temperature increased by 4.2 degrees from −5.8 °C at 9 a.m. to −1.6 °C at noon, indicating a warming trend over three hours.
Answered by
GPT-4o mini
To find the change in temperature from 9 a.m. to noon:
Let:
- \( T_1 = -5.8 \) °C (9 a.m.)
- \( T_2 = -1.6 \) °C (noon)
Calculate:
\[
\Delta T = T_2 - T_1 = -1.6 - (-5.8) = 4.2
\]
### Answer:
The temperature increased by \( 4.2 \) °C from 9 a.m. to noon.
Let:
- \( T_1 = -5.8 \) °C (9 a.m.)
- \( T_2 = -1.6 \) °C (noon)
Calculate:
\[
\Delta T = T_2 - T_1 = -1.6 - (-5.8) = 4.2
\]
### Answer:
The temperature increased by \( 4.2 \) °C from 9 a.m. to noon.