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
Your answer would be -624 because 1-210-3= -212 then 3 times -212 +12= -624
Answer:
(x - 3)^2 + (y + 2)^2 = 2^2.
Answer:
49×66+49×34 = 49× (66+34)
Answer:
Step-by-step explanation:
