Question

Express 0.48 in p/q form​

Answer 1

Answer:

0.48 in p/q form is $\frac{12}{15}$

Step-by-step explanation:

Given digit is 0.48.

Now,

$0.48 \\ = \frac{48}{100} \\ = \frac{12}{25}$

So, fraction form of 0.48 is $\frac{12}{25}$

Therefore, p/q form of 0.48 is $\frac{12}{15}$

where p is 12 and 15.

This is a problem of fraction part of Algebra.

Some important Algebra formulas.

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )</p><p>$

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