how to represent 0.47 bar. over 7 on p/q form???
Answers
Answered by
4
0.47 bar we can represent like this
0.47+0.0047+0.000047.....................
so it is in G.P ( Geometric progression)
so
a = 0.47
r = T2/T1 = 0.0047/0.47 = 0.01
S(infinity) = ?
S(infinity) = a/(1-r)
S(infinity) = 0.47/(1-0.01)
S(infinity) = 0.47/0.99
S(infinity) = (47/100)/(99/100)
S(infinity) = 47/100×100/99
S(infinity) = 47/99
0.47 bar in p/q form is 47/99
0.47+0.0047+0.000047.....................
so it is in G.P ( Geometric progression)
so
a = 0.47
r = T2/T1 = 0.0047/0.47 = 0.01
S(infinity) = ?
S(infinity) = a/(1-r)
S(infinity) = 0.47/(1-0.01)
S(infinity) = 0.47/0.99
S(infinity) = (47/100)/(99/100)
S(infinity) = 47/100×100/99
S(infinity) = 47/99
0.47 bar in p/q form is 47/99
Answered by
0
Let A=0.47bar
Then 100A=47.47bar
So 100A-A=47
Therefore 99A=47
A=47÷99
Then 100A=47.47bar
So 100A-A=47
Therefore 99A=47
A=47÷99
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