what is 3.58333333333 as a fraction?
do you mean 3.58333333333..... or 3.58333333333
3.58333333333
well 3.58 as a fraction is 179/50
a=3.58333333333 ok do like this b=4.56 b456/100 or k=9.034 k=9034/1000 can you do this ?
all my answers are A: 4 7/18 B: 4 7/9 C: 2 7/9 D: 2 7/18
\[a=3.58333333333 =\frac{ 358333333333 }{10^{11} }\]
those answers , they aren't a=3.58333333333 may be you type the question wrong for example a=3.58333333333...
yes you type it wrong the original question must be to find a=3.58333333333... it is 3225/900=3.58333333333... i did check it
now lest begin by this correct question agree?
Try this: (assuming the 3's go on forever) x=3.58333333333... 100x=358.333333333... 10000x=35833.3333333.. 10000x-100x=9900x=35833.3333... - 358.33333... = 35833-358 = 35475. So x = 35475/9900 = 43/12 (divide numerator and denominator by 825). You could write it as 3 7/12. The choices you have are obviously all wrong, because neither of them is a number between 3 and 4, while 3.58333333333(...) is!
it says to multiply and simplify this: http://static.k12.com/bank_packages/files/media/mathml_7ffdad0085a802aaca0eb30b964ea95f3bf3013c_1.gif
suppose a=3.58333333333... multiply by 100 so 100a =358.33333333333333333333... *10 so 1000a=3583.33333333333333... now 1000a-100a=3583.33333333333333...-358.33333333333333... so 900a =3583-358 =3225 so a=3225/900=3583.33333333333333...
Just try to expand: multiply 2/3 with 1/3 and also 2/3 with 13/4. You then have to make the denominators the same... Do you know how to go on?
not really, i don't know much about fractions...
now we have a=3225/900 we can simplify to get the answer divide both 3225,900 by 3,then by 5,then by 5 a=43/12
\(\frac{2}{3}\cdot \frac{1}{3}+\frac{2}{3}\cdot \frac{13}{4}=\frac{2}{9}+\frac{26}{12}=\frac{2}{9}+\frac{13}{6}=...\) Now look f0r the Least Common Multiple (LCM) of 9 and 6: this is 18. That's your new denominator!
If you replace \(\frac{2}{9}\) with \(\frac{4}{18}\) and \(\frac{13}{6}\) with \(\frac{39}{18}\), you can now add the fractions: \(\frac{4}{18}+\frac{39}{18}=\frac{4+39}{18}=\frac{43}{18}=2\frac{7}{18}\).
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