sqrt(56) = 7.483 approx.
The square root of 56 is irrational. This is because 56 can be expressed as 4 times 14, and while the square root of 4 is rational (2), the square root of 14 is not a perfect square and is therefore irrational. Thus, the combined result of √56 remains an irrational number.
No, the square root of 56 is not an integer. The square root of 56 can be simplified to 2√14, which is an irrational number. Since it cannot be expressed as a whole number, it does not qualify as an integer.
Root 56 = 7.483314773547882 = 7.48
in decimal form the answer is 18.7082869
4
The square root of 3136 is 56.
Square root of 56 = Square root of (4 x 14) = (Square root of 4) x (Square root of 14) = 2 x (Square root of 14) The actual value would be 2 x 3.741 = 7.482
2 square root of 14
56
± 7.483315
Root 56 = 7.483314773547882 = 7.48
square root of 3136, square root 3146 56x56
in decimal form the answer is 18.7082869
"7" is the number because its square root is "49" & 49+7=56
4
No. The square root of a positive integer can only be an integer, or an irrational number.
It is: 7.5