Solve 2*(3/8)^2-3/2*3/8+7/10

$$2 \times ( \frac { 3 } { 8 } ) ^ { 2 } - \frac { 3 } { 2 } \times \frac { 3 } { 8 } + \frac { 7 } { 10 }$$

Evaluate

$$\frac{67}{160}=0.41875$$

Solution Steps

Calculate \frac{3}{8} to the power of 2 and get \frac{9}{64}.

$$2\times \frac{9}{64}-\frac{3}{2}\times \frac{3}{8}+\frac{7}{10}$$

Express 2\times \frac{9}{64} as a single fraction.

$$\frac{2\times 9}{64}-\frac{3}{2}\times \frac{3}{8}+\frac{7}{10}$$

Multiply 2 and 9 to get 18.

$$\frac{18}{64}-\frac{3}{2}\times \frac{3}{8}+\frac{7}{10}$$

Reduce the fraction \frac{18}{64} to lowest terms by extracting and canceling out 2.

$$\frac{9}{32}-\frac{3}{2}\times \frac{3}{8}+\frac{7}{10}$$

Multiply \frac{3}{2} times \frac{3}{8} by multiplying numerator times numerator and denominator times denominator.

$$\frac{9}{32}-\frac{3\times 3}{2\times 8}+\frac{7}{10}$$

Do the multiplications in the fraction \frac{3\times 3}{2\times 8}.

$$\frac{9}{32}-\frac{9}{16}+\frac{7}{10}$$

Least common multiple of 32 and 16 is 32. Convert \frac{9}{32} and \frac{9}{16} to fractions with denominator 32.

$$\frac{9}{32}-\frac{18}{32}+\frac{7}{10}$$

Since \frac{9}{32} and \frac{18}{32} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9-18}{32}+\frac{7}{10}$$

Subtract 18 from 9 to get -9.

$$-\frac{9}{32}+\frac{7}{10}$$

Least common multiple of 32 and 10 is 160. Convert -\frac{9}{32} and \frac{7}{10} to fractions with denominator 160.

$$-\frac{45}{160}+\frac{112}{160}$$

Since -\frac{45}{160} and \frac{112}{160} have the same denominator, add them by adding their numerators.

$$\frac{-45+112}{160}$$

Add -45 and 112 to get 67.

$$\frac{67}{160}$$

Factor

$$\frac{67}{2 ^ {5} \cdot 5} = 0.41875$$