The square root of 1,444 is: 38
± 38
√-48 + 35 + √ 25 + √-27 = √[(i2)(16)(3)] + 35 + 5 + √[(i2)(9)(3)] = 4i√3 + 38 + 9i√3 = 38 + (13√3)i
Assuming you mean consecutive numbers are integers, then: The integers will be the two integers between which the square root of 1406 lies. Calculating the square root of 1406 using the "long" division method: ________3__7 _____---------- ___3|_14_06 ________9 _______--- __67|__506 ________469 ________---- _________37 The square root of 1406 is 37.<something>, thus it lies between 37 and 38, so: 1406 = 37 × 38
what is the value of the 3 in 38
The square root of 38 = ± 6.1644146+sqrt2
38 is greater than the square root of 38
38 is not a perfect square. Its square root is a fraction and the square root of a perfect square is always an integer.
The square root of 38 = ± 6.1644146+sqrt2
The square root of 1,444 is: 38
± 38
6.164414003
No
Yes.
38 38X38=1444
± 6.2
√38 = 6.164414003 with the positive (whole number?) being 6.