The square root of 9604 is 98.
√98 = 7√2 = 9.9
square root of (2 ) square root of (3 ) square root of (5 ) square root of (6 ) square root of (7 ) square root of (8 ) square root of (9 ) square root of (10 ) " e " " pi "
square root 2 times square root 3 times square root 8
We use the property of square roots that says the square root of (ab)=square root (a) multiplied by square root of b So square root (4x)=square root (4) mutiplies by square root of x =2(square root (x)) 2sqrt(x)
98 is 49 times 2 so the square root of 98 is the square root of 49 times the square root of 2. But the square root of 49 is 7 , so the answer is 7 times the square root of 2.
No because the square root of 98 is 7 times the square root of 2
The simplified square root of 98 is: ±9.9 or ±10
No. If the square root of 98 was an integer, 98 would be a perfect square.
The square root of 9604 is 98.
√98 ~= 9.899
49
49
It is 98 because 98*98 = 9604
9.899494936611665It is: 7 times square root of 2± 9.899495
2.83 to the nearest hundredth.
Expressed as a surd, sqrt(98) + sqrt(2) = 8 sqrt(2).