Let x and y be your two consecutive whole numbers
x < sqrt(142) < y
x^2 < 142 < y
So, we are looking for x and y such that x^2 < 142 and y^2 > 142.
The closest squared number to 142 is 144 = 12^2.
Next is 11^2 = 121.
11 and 12 are consecutive.
11^2 = 121 < 142 < 144 = 12^2
Thus, 11 and 12 are your numbers
Answer:
D is the answer im happy to help you any time
4×6817=27256 so it is 27 thousand
Answer:
The range is from 125 to 133.
Step-by-step explanation:
That's the lowest and highest the weights go, therefore the range is from 125 and anything in between up to 133.
Answer:
348
Step-by-step explanation:
6 8 10 22 75 78 81 354
subtract the highest number to the lowest number