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 $0.\overline{45}$.
Simplify the given expression.
$9 \times 0.9 \times 0.09 \times 0.009 \times \frac{1}{0.3} \times \frac{1}{0.03} \times \frac{1}{0.003}$
Arrange the following numbers in their increasing order.
(1) $-0.96$
(2) $0.83$
(3) $0.24$
(4) $-0.64$
(5) $0.58$
The value of \(\frac{1}{4} + \frac{{[{{(20.35)}^2} - {{(8.35)}^2}] \times 0.0175}}{{{{(1.05)}^2} + (1.05)(27.65)}}\) is:
The value of \(0.4\overline 6 + 0.7\overline {23} - 0.3\overline 9 \times 0.\overline 7 \) is:
The value of \(\frac{48.3\times[(4.95)^2+4.95\times13.25]}{[(12.55)^2-(5.65)^2]\times19.8} \) is:
Find the value of (1.6) 3 - (0.9) 3 - (0.7) 3.
What is the value of x, if \(5\left( {1 - \frac{x}{5}} \right) - (5 - x) - \frac{1}{{200}}{\rm{of (20 - x) = 0}}{\rm{.08}}\) ?