194
194
The square root of fraction 20 over 25: = √20 ...√25 = √4(5) ......5 = 2√5 .....5 or 2 square root of 5 over 5
The square root of 8 over 25 is irrational, and real.
13. The figures provided (including the answer) satisfy Pythagoras' Theorem. a2 = b2 + c2 a2 = 52 + 122 = 25 + 144 = 169 Then a = √169 = 13
194
194
That depends where you are placing the brackets. sqrt(169 - 25) = sqrt(144) = 12, however, sqrt(169) - 25 = 13 - 25 = -12.
If the 13 was the long side (hypoteneuse), the other side would be the square root of 13²-5² = square root of 169-25 = square root of 144 = 12. (according to Pythagoras). If the 13 was a short side, the other is square root of 13² + 5² = square root of 169+25 = square root of 194, or about 13.93
The square root of fraction 20 over 25: = √20 ...√25 = √4(5) ......5 = 2√5 .....5 or 2 square root of 5 over 5
Well, honey, to find the square root of 625 over 36, you first find the square root of the numerator (625) which is 25, and the square root of the denominator (36) which is 6. So, the square root of 625 over 36 is 25 over 6. Math doesn't have to be as complicated as your ex's drama, darling.
The square root of 8 over 25 is irrational, and real.
the square root of 122+52 (144 + 25 = 169) which is 13.
√4225 = √(25)(169) = √(5^2)(13^2) = 5 x 13 = 65
13. The figures provided (including the answer) satisfy Pythagoras' Theorem. a2 = b2 + c2 a2 = 52 + 122 = 25 + 144 = 169 Then a = √169 = 13
1.6
The square root of 25 is 5.