21/5 + 4/3 - 1/2 what is the answer
1 answer:
You might be interested in
Answer:
459
Step-by-step explanation:
The "long division method" algorithm for square root makes use of the relation described by the square of a binomial.
(a +b)² = a² +2ab +b² = a² +b(2a +b)
<h3>Steps</h3>The value for which the root is desired is written with digits marked off in pairs either side of the decimal point.
The initial digit of the root is the integer part of the square root of the most-significant pair. Here that is floor(√21) = 4. This is shown in the "quotient" spot above the leftmost pair. The square of this value is subtracted, and the next pair brought down for consideration. Here, that means the next "dividend" is 506.
The next "divisor" will be 2 times the "quotient" so far, with space left for a least-significant digit. Here, that means 506 will be divided by 80 + some digits. As in regular long division, determining the missing digit involves a certain amount of "guess and check." We find that the greatest value 'b' that will give b(80+b) ≤ 506 is b=5. This is the next "quotient" digit and is placed above the "dividend" pair 06. The product 5(85) = 425 is subtracted from 506, and the next "dividend" pair is appended to the result. This makes the next "dividend" equal to 8181.
As in the previous step, the next "divisor is 2 times the quotient so far: 2×45 = 90, with space left for the least significant digit. 8181 will be divided by 900-something with a "quotient" of 9. So, we subtract the product 9(909) = 8181 from the "dividend" 8181 to get the next "dividend." That result is zero, so we're finished.
The root found here is 459.
__
<em>Additional comment</em>
In practice, roots are often computed using iterative methods, with some function providing a "starter value" for the iteration. Some iterative methods can nearly double the number of good significant digits in the root at each iteration.
Using this "long division method," each "iteration" adds a single significant digit to the root. Its advantage is that it always works, and is generally suitable for finding roots by hand. Once the number of root digits begins to get large, the "divisor" starts to be unwieldy.
