Question

Estimate the value of the irrational number (0.6289731...)2. Up to how many decimal digits is the estimation correct? 4 5 6 7

Answer 1

Answer:

7 decimal digits

Step-by-step explanation:

Because it ws rounded up to 0.6289731

and the known digits after the decimal are 7 in numbers.

0.6290 ......4 decimal digits

0.62897..... 5 decimal digits

0.628973.... 6 decimal digits

0.628973 ..... 7 decimal digits

Answer 2

Answer:

0.927

Step-by-step explanation:

The value can be estimated using the binomial theorem for expanding numbers. The theorem states:

For an expansion

$(1+x)^{n} = 1+ \frac{n}{1}x+\frac{n(n-1)}{1*2} x^{2} + ...$

Now, let's take the value 0.6289731. In the expression, let 0.6289731 be equal to $( 1 - 0.0371)$

We can use the binomial expansion above:

$(1-0.0371)^{2} = 1+ (-)\frac{(2)}{1} (0.0371) + \frac{2(2-1)}{1*2} (-0.0371)^{2} + ...$ up to three terms

This gives 1-0.0734+0.00137641 = 0.927 up to 3 decimal places