Solve (7/5-3/8-1/2)^2:(7/2*frac{3}{20})+frac{3}{10}*frac{23}{2^2}=

$$( \frac { 7 } { 5 } - \frac { 3 } { 8 } - \frac { 1 } { 2 } ) ^ { 2 } : ( \frac { 7 } { 2 } \cdot \frac { 3 } { 20 } ) + \frac { 3 } { 10 } \cdot \frac { 23 } { 2 ^ { 2 } } =$$

Evaluate

$$\frac{9}{4}=2.25$$

Solution Steps

Subtract \frac{3}{8} from \frac{7}{5} to get \frac{41}{40}.

$$\frac{\left(\frac{41}{40}-\frac{1}{2}\right)^{2}}{\frac{7}{2}\times \frac{3}{20}}+\frac{3}{10}\times \frac{23}{2^{2}}$$

Subtract \frac{1}{2} from \frac{41}{40} to get \frac{21}{40}.

$$\frac{\left(\frac{21}{40}\right)^{2}}{\frac{7}{2}\times \frac{3}{20}}+\frac{3}{10}\times \frac{23}{2^{2}}$$

Calculate \frac{21}{40} to the power of 2 and get \frac{441}{1600}.

$$\frac{\frac{441}{1600}}{\frac{7}{2}\times \frac{3}{20}}+\frac{3}{10}\times \frac{23}{2^{2}}$$

Multiply \frac{7}{2} and \frac{3}{20} to get \frac{21}{40}.

$$\frac{\frac{441}{1600}}{\frac{21}{40}}+\frac{3}{10}\times \frac{23}{2^{2}}$$

Divide \frac{441}{1600} by \frac{21}{40} by multiplying \frac{441}{1600} by the reciprocal of \frac{21}{40}.

$$\frac{441}{1600}\times \frac{40}{21}+\frac{3}{10}\times \frac{23}{2^{2}}$$

Multiply \frac{441}{1600} and \frac{40}{21} to get \frac{21}{40}.

$$\frac{21}{40}+\frac{3}{10}\times \frac{23}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{21}{40}+\frac{3}{10}\times \frac{23}{4}$$

Multiply \frac{3}{10} and \frac{23}{4} to get \frac{69}{40}.

$$\frac{21}{40}+\frac{69}{40}$$

Add \frac{21}{40} and \frac{69}{40} to get \frac{9}{4}.

$$\frac{9}{4}$$

Factor

$$\frac{3 ^ {2}}{2 ^ {2}} = 2\frac{1}{4} = 2.25$$