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:
Step-by-step explanation:
Use this formula,
where n is the amount of even intergers, a is the starting term, d is the common difference.
- the amount of even intergers from 2 to 200 is 100
- The starting number is 2
- The common difference is 2
N + (3n - 15) = 101 is the equation, the answer to the equation is 29
The anwser is C!!!!!!!!!!!!!!!!!!!!!! ik this because i calculated it my self
Answer:
-88
Step-by-step explanation:
Because when you all ad it down you get -96 and then -8= -88
