Solve (61)/(3)\div(3)/(5)\times(19)/(9)+(6)/(3)

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Question

$$\frac{ 61 }{ 3 } \div \frac{ 3 }{ 5 } \times \frac{ 19 }{ 9 } + \frac{ 6 }{ 3 }$$

Evaluate

$\frac{5957}{81}\approx 73.543209877$

Solution Steps

Divide $\frac{61}{3}$ by $\frac{3}{5}$ by multiplying $\frac{61}{3}$ by the reciprocal of $\frac{3}{5}$.

$$\frac{61}{3}\times \frac{5}{3}\times \frac{19}{9}+\frac{6}{3}$$

Multiply $\frac{61}{3}$ times $\frac{5}{3}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{61\times 5}{3\times 3}\times \frac{19}{9}+\frac{6}{3}$$

Do the multiplications in the fraction $\frac{61\times 5}{3\times 3}$.

$$\frac{305}{9}\times \frac{19}{9}+\frac{6}{3}$$

Multiply $\frac{305}{9}$ times $\frac{19}{9}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{305\times 19}{9\times 9}+\frac{6}{3}$$

Do the multiplications in the fraction $\frac{305\times 19}{9\times 9}$.

$$\frac{5795}{81}+\frac{6}{3}$$

Divide $6$ by $3$ to get $2$.

$$\frac{5795}{81}+2$$

Convert $2$ to fraction $\frac{162}{81}$.

$$\frac{5795}{81}+\frac{162}{81}$$

Since $\frac{5795}{81}$ and $\frac{162}{81}$ have the same denominator, add them by adding their numerators.

$$\frac{5795+162}{81}$$

Add $5795$ and $162$ to get $5957$.

$$\frac{5957}{81}$$

Show Solution

Factor

$\frac{7 \cdot 23 \cdot 37}{3 ^ {4}} = 73\frac{44}{81} = 73.54320987654322$