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:
the value of x is 2.
Step-by-step explanation:
5(3x - 4) = 2(2x + 1)
15x - 20 = 4x + 2
15x - 4x = 2 + 20
11x = 22
x = 22/11
x = 2
Answer:
The answer is x = -13/4
Step-by-step explanation:
-13/4 + 2 / -13/4 +3 = 5
