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spin [16.1K]
3 years ago
7

If 23/8, 11/2, 9/6, and 15/7 are placed in order from least to greatest, which would be first?

Mathematics
1 answer:
yarga [219]3 years ago
8 0

Answer:

\frac{9}{6}

Step-by-step explanation:

The given fractions are;

\frac{23}{8},\frac{11}{2},\frac{9}{6},\frac{15}{7}

The least common denominator is 168.


We express all the fractions in equivalent form with the LCD as the denominator.


\frac{23}{8}=\frac{21\times23}{21\times8}=\frac{483}{168}


\frac{11}{2}=\frac{84\times11}{84\times2}=\frac{924}{168}


\frac{9}{6}=\frac{28\times9}{28\times6}=\frac{252}{168}


\frac{15}{7}=\frac{24\times15}{24\times7}=\frac{360}{168}


We now compare the equivalent forms and arrange from the least to the greatest.

\frac{252}{168}\:

This implies that,

\frac{9}{6}\:

Therefore the first would be \frac{9}{6}.







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