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
2=6
4=8
6=10
8=12
I added +4 for each
Answer:
x = 22.66
Step-by-step explanation:
If you type in both the coordinates it’ll give you the answer on google. That’s how I did this.
