Question #53fde

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Answer 1 · 2016-05-23T20:20:13

See explanation

This is in a simplified form

Other forms are

$\frac{67}{4} \leftarrow$ $67/4 larr$ Improper fraction

$16.75 \leftarrow$ $16.75 larr$ decimal

'~~~~~~~~~ The teaching bit ~~~~~~~~~~~~~~~

Method to derive $\frac{67}{4}$ $67/4$

Write as $16 \frac{3}{4}$ $16 3/4$ as $16 + \frac{3}{4}$ $16+3/4$

If you multiply by 1 you do not change the value or the way it looks.

If you multiply I but in the form of $1 = \frac{4}{4}$ $1=4/4$ then you do not change the value but you do change the way it looks.

Multiply $16$ $16$ by 1 but in the form of $1 = \frac{4}{4}$ $1=4/4$

$16 \times \frac{4}{4} = \frac{16 \times 4}{4} = \frac{64}{4}$ $16xx4/4 =(16xx4)/4 = 64/4$

So you can write $16 + \frac{3}{4}$ $16+3/4$ as $\frac{64}{4} + \frac{3}{4} = \frac{67}{4}$ $64/4+3/4 = 67/4$

Answer 2 · 2016-05-31T21:36:55

If seen as $16^{\frac{3}{4}}$ $16^(3/4)$ it simplifies to $2^{3} = 8$ $2^3 = 8$

I wonder if the question was not meant to be $16^{\frac{3}{4}}$ $16^(3/4)$ ?

This is more likely because there is a calculation involved.
However, it would not be 'solve' rather determine or evaluate.

$16^{\frac{3}{4}}$ $16^(3/4)$ can also be written as ${\sqrt[4]{16}}^{3}$ $root4(16)^3$

This simplifies to $2^{3} = 8$ $2^3 = 8$