Determine the distance between −18.26 and −2.45 on a number line.(1 point)
Responses
−20.71
20.71
−15.81
15.81
The distance between -18.26 and -2.45 on a number line can be found by subtracting the two numbers:
-2.45 - (-18.26) = -2.45 + 18.26 = 15.81
So, the distance between -18.26 and -2.45 is 15.81 units.
The correct response is:
15.81
Subtract 24−(−15)
To subtract a negative number, we can change it to addition. So, we have:
24 - (-15) = 24 + 15
Adding the numbers, we get:
24 + 15 = 39
Therefore, the result of 24 - (-15) is 39.
What is 17 4/9 + 5/9 − 8 − 10?
To simplify the expression 17 4/9 + 5/9 - 8 - 10, we first add the two fractions:
17 4/9 + 5/9 = 17 + 4/9 + 5/9 = 17 + (4+5)/9 = 17 + 9/9 = 17 + 1 = 18
Next, we subtract 8 and 10 from the result:
18 - 8 - 10 = 10 - 10 = 0
Therefore, the value of the expression 17 4/9 + 5/9 - 8 - 10 is 0.
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 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)
To find the change in temperature during the three hours, we need to subtract the initial temperature from the final temperature.
Let's call the initial temperature at 9 a.m. "T1" and the final temperature at noon "T2."
T1 = -5.8 degrees Celsius
T2 = -1.6 degrees Celsius
The change in the temperature, ΔT, can be found by subtracting T1 from T2:
ΔT = T2 - T1
Substituting the values of T1 and T2:
ΔT = (-1.6) - (-5.8)
ΔT = -1.6 + 5.8
ΔT = 4.2
The change in temperature during these three hours is 4.2 degrees Celsius.
This means that the temperature increased by 4.2 degrees Celsius from 9 a.m. to noon.