[tex] \boxed{ \sf7 \frac{2}{3} }[/tex]
[tex] \it5 \frac{2}{5} -( - 2 \frac{4}{15} ) \\ [/tex]
First, we convert all mixed numbers into full fractions:
[tex]\it5 \frac{2}{5} = \frac{(5 \times 5 + 2)}{5} = \frac{27}{5} \\ [/tex]
[tex]\it2 \frac{4}{15} = \frac{(2 \times 15 + 4)}{15} = \frac{34}{15} \\ [/tex]
[tex] \: [/tex]
Now, we need to bring both to the same denominator, and because 5 is a factor or 15, we can bring both to fractions of 1/15, for wich we only need to transform the first fraction:
[tex]\it \frac{27}{5} = \frac{27}{5} \times \frac{3}{3} = \frac{81}{15} \\ [/tex]
So, the subtraction is actually
[tex] \it \frac{81}{15} - ( - \frac{34}{1} ) = \frac{81}{15} + \frac{34}{15} = \frac{115}{15} = \frac{23}{3} \\ [/tex]
[tex]\it \frac{23}{3} = 7 \\ [/tex] remainder [tex]2 = \frac{(7 \times 3 + 2)}{3} = 7 \frac{2}{3} \\ [/tex]
//@smokerz//
Answer:
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Verified answer
Answer:
[tex] \boxed{ \sf7 \frac{2}{3} }[/tex]
Step-by-step explanation:
[tex] \it5 \frac{2}{5} -( - 2 \frac{4}{15} ) \\ [/tex]
First, we convert all mixed numbers into full fractions:
[tex]\it5 \frac{2}{5} = \frac{(5 \times 5 + 2)}{5} = \frac{27}{5} \\ [/tex]
[tex]\it2 \frac{4}{15} = \frac{(2 \times 15 + 4)}{15} = \frac{34}{15} \\ [/tex]
[tex] \: [/tex]
Now, we need to bring both to the same denominator, and because 5 is a factor or 15, we can bring both to fractions of 1/15, for wich we only need to transform the first fraction:
[tex]\it \frac{27}{5} = \frac{27}{5} \times \frac{3}{3} = \frac{81}{15} \\ [/tex]
[tex] \: [/tex]
So, the subtraction is actually
[tex] \it \frac{81}{15} - ( - \frac{34}{1} ) = \frac{81}{15} + \frac{34}{15} = \frac{115}{15} = \frac{23}{3} \\ [/tex]
[tex]\it \frac{23}{3} = 7 \\ [/tex] remainder [tex]2 = \frac{(7 \times 3 + 2)}{3} = 7 \frac{2}{3} \\ [/tex]
[tex] \: [/tex]
//@smokerz//
Answer:
[tex] \boxed{ \sf7 \frac{2}{3} }[/tex]