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:
n ÷ 2
Step-by-step explanation:
if you insert a 12 for 'n', the corresponding value below would be 12÷2, or 6
this follows the pattern in the table
45/50 and 46/50 could be one of the options of 4.5/5 and 4.6/5
20,000,000+400,000+80,000+4,000+100+60+3
Answer:
the answer is C. 20/27 cubic cm.
Step-by-step explanation:
V = LWH
length 5(1/3 cm)
width 2(1/3 cm)
height 2(1/3 cm)
V = (5/3 cm) (2/3 cm) (2/3 cm)
V = 20/27 cubic cm.
