Well, there are various ways. The most simple way is to approximate the solution:
89 is between 9^2 and 10^2, therefore the first digit is 9
89 is between 9.4^2 and 9.5^2, therefore the second digit is 4
There is a second method. in which you use the Newton's method:
Consider the following equation: f(x)=x^2-89, which has sqrt[89] as its root.
Let x0 (the starting value) be 9.
This is the formula of Newton's method:
x_(n+1)= x_(n)- f(x_n)/f'(x_n)
x_(n+1)= x_(n)- (x_(n)^2-89)/2*(x_n)
Now we begin the approxiamton:
x_(2)= 9- (9^2-89)/(2*9)=9.4444
x_3=9.433
x_3=9.43398132
If you need digits, just use more iterations (The numbers of digits doubles for each iteration ).
The square root of 89 is halfway between the square root of 81 = 9 and the square root of 100 = 10. Rounded to two decimal places, the square root of 89 is equal to 9.43.
9.4
sqrt(356) = sqrt(4*89) = sqrt(4)*sqrt(89) = 2*sqrt(89)
The square root of 7921 is 89.
Providing it's a right angle triangle use Pythagoras' theorem: 252+402 = 2225 and the square root of this is about 47.16990566 meters The exact answer is 5 times the square root of 89
Rounded to five decimal places, sqrt(89) = ±9.43398The square root of 89 (written √89 or 891/2) = 9.434 (3dp)
The square root of 7921 is 89.
The square root of 89 is halfway between the square root of 81 = 9 and the square root of 100 = 10. Rounded to two decimal places, the square root of 89 is equal to 9.43.
9.433981132056603...
Find the factors of 89. Then put cancel out all of the factors that are perfect squares and put them outside of the root. Multiply the remaining factors under the root.
89
No
9.4
no.
An equation would be x = 890.5
9
sqrt(356) = sqrt(4*89) = sqrt(4)*sqrt(89) = 2*sqrt(89)