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serg [7]
2 years ago
11

At the Kansas City airport a group of pilots for

Mathematics
1 answer:
Tju [1.3M]2 years ago
7 0

Answer:

35 is greater than the other options.Step-by-step explanation:

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Giving 10 points need urgently
larisa86 [58]

Answer:

{-2, - 1, 0, 1}

Step-by-step explanation:

The integer solutions will be {-2, - 1, 0, 1}

3 0
2 years ago
Read 2 more answers
Five small apples have about 390 calories.
Dvinal [7]
78 calories per apple :)
7 0
2 years ago
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An automotive repair center charges $45 for any part of the first hour of labor, and $25 for any part of each additional hour. W
IceJOKER [234]
The total cost is given by the equation:
C(t) = 45 + 25(h-1) where h is the number of hours worked.

We can check for each option in turn:

Option A:

Inequality 5 < x ≤ 6 means the hour is between 5 hours (not inclusive) to 6 hours (inclusive)
Let's take the number of hours = 5
C(5) = 45 + (5-1)×25 = 145
Let's take the number of hours = 6
Then substitute into C(6) = 45 + (6-1)×25 = 170
We can't take 145 because the value '5' was not inclusive.


Option B:
The inequality is 6 < x ≤ 7
We take number of hours = 6
C(6) = 25(6-1) + 45 = 170
We take number of hours = 7
Then C(7) = 25(7-1) + 45 = 195

Option C:
The inequality is 5 < x ≤ 6
Take the number of hours = 5
C(5) = 25(5-1) + 45 = 145
Take the number of hours = 6
C(6) = 25(6-1) + 45 = 170
We can't take the value 145 as '5' was not inclusive in the range, but we can take 170

Option D:
6 < x ≤ 7
25(6-1) + 45 < C(t) ≤ 25(7-1) + 45
170 < C(t) ≤ 195

Correct answer: C


7 0
3 years ago
600 centimeters__5 meters
uranmaximum [27]
600 centimeters > 5 meters
4 0
3 years ago
Bobby biked 12/3 hours on monday 21/3 hours on tuesday and 22/3 hours on wednesday. what is the total number of hours bobby spen
serg [7]
The fractions you're given can be simplified down, so you get (12 ÷ 3) 4 hours on Monday, (21 ÷ 3) 7 hours on Tuesday, and (22 ÷ 3) 7 1/3 hours on Wednesday. You can now add these up and get your answer as 18 1/3 hours, or 18.33.

I hope this helps!
8 0
3 years ago
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