This solution explains how to find the sum of two repeating decimals, $0.\overline{23}$ and $0.\overline{22}$.
Repeating decimals can be accurately represented as fractions. A repeating decimal of the form $0.\overline{ab}$ is equal to the fraction $\frac{ab}{99}$.
Now, add the fractional representations:
$ \frac{23}{99} + \frac{22}{99} = \frac{23 + 22}{99} = \frac{45}{99} $Convert the resulting fraction $\frac{45}{99}$ back into a repeating decimal. Since the denominator is 99, the numerator represents the repeating part.
$ \frac{45}{99} = 0.454545... = 0.\overline{45} $Therefore, the value of $0.\overline{23} + 0.\overline{22}$ is $0.\overline{45}$.
What will the value of the following be (correct to three decimal points)?
$160.342 - 32.124$
Arrange the following numbers in their increasing order.
(1) $-0.96$
(2) $0.83$
(3) $0.24$
(4) $-0.64$
(5) $0.58$
Simplify the given expression.
$9 \times 0.9 \times 0.09 \times 0.009 \times \frac{1}{0.3} \times \frac{1}{0.03} \times \frac{1}{0.003}$
What is the result when 0.129129129… is converted to a fraction?
Which of the following statement(s) is/are correct?
I. (3/11) > 0.3
II. (7/8) > 0.86
The value of \(1.\overline{3}+0.\overline{69}-0.5\overline{23}\) is equal to:
The value of \(0.\bar 4 + 0.5 \bar 9 - 0.4 \overline{23}\) is equal to:
If 19 × 23 = 437, then find the value of (190 × 0.023).