Given: 0.3454545... Assuming the number is a recurring decimal that repeats "45" forever, we try to find ea equivalent fraction.
Let A=0.3, and B=0.0454545..., so 0.3454545=A+B.
Change both decimals,A and B, into fraction forms separately.
A=0.3=3/10, and B can be changed in ways shown below:
Sol.1: B=0.0454545...=0.045+0.00045+0.0000045+...
=45(0.001+0.00001+0.0000001+...)=45x0.0010101...
While, 1/99=0.010101..., and 1/990=0.00101010...,
so B=45x1/990=5/110. Thus A+B=3/10 + 5/110=38/110=19/55
The answer is: 0.3454545...=19/15
Sol.2: 1000B=45.4545..., and 10B=0.4545...,
so 1000B-10B=45, that is 990B=45.
We have: B=45/990=5/110 Thus, A+B=3/10 + 5/110=19/55
The answer is: 0.34545...=19/55
Sol.3: B=0.045+0.00045+0.0000045=45/10^3(1 + 1/10^2 + 1/10^4 + 1/10^6...)
Use the formula of sum,S, for infinite geometric series, where the first term is a and the common ratio is r (lrl<1): S=a/(1-r) Here, a=1 and r=1/10^2
So, B=45/10^3 x 1/(1-1/10^2)=45/10^3 x 10^2/(10^2-1)=45/990=5/110
Thus, A+B=3/10+5/110=19/55
The answer is: 0.34545...=19/55