Solve 36:(51/5-7)+(1/2)^3*04^2

$$36 : ( 5 \frac { 1 } { 5 } - 7 ) + ( \frac { 1 } { 2 } ) ^ { 3 } \cdot 04 ^ { 2 }$$

Evaluate

$$-20$$

Solution Steps

Multiply 5 and 5 to get 25.

$$\frac{36}{\frac{25+1}{5}-7}+\left(\frac{1}{2}\right)^{3}\times 0\times 4^{2}$$

Add 25 and 1 to get 26.

$$\frac{36}{\frac{26}{5}-7}+\left(\frac{1}{2}\right)^{3}\times 0\times 4^{2}$$

Convert 7 to fraction \frac{35}{5}.

$$\frac{36}{\frac{26}{5}-\frac{35}{5}}+\left(\frac{1}{2}\right)^{3}\times 0\times 4^{2}$$

Since \frac{26}{5} and \frac{35}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{36}{\frac{26-35}{5}}+\left(\frac{1}{2}\right)^{3}\times 0\times 4^{2}$$

Subtract 35 from 26 to get -9.

$$\frac{36}{-\frac{9}{5}}+\left(\frac{1}{2}\right)^{3}\times 0\times 4^{2}$$

Divide 36 by -\frac{9}{5} by multiplying 36 by the reciprocal of -\frac{9}{5}.

$$36\left(-\frac{5}{9}\right)+\left(\frac{1}{2}\right)^{3}\times 0\times 4^{2}$$

Express 36\left(-\frac{5}{9}\right) as a single fraction.

$$\frac{36\left(-5\right)}{9}+\left(\frac{1}{2}\right)^{3}\times 0\times 4^{2}$$

Multiply 36 and -5 to get -180.

$$\frac{-180}{9}+\left(\frac{1}{2}\right)^{3}\times 0\times 4^{2}$$

Divide -180 by 9 to get -20.

$$-20+\left(\frac{1}{2}\right)^{3}\times 0\times 4^{2}$$

Calculate \frac{1}{2} to the power of 3 and get \frac{1}{8}.

$$-20+\frac{1}{8}\times 0\times 4^{2}$$

Multiply \frac{1}{8} and 0 to get 0.

$$-20+0\times 4^{2}$$

Calculate 4 to the power of 2 and get 16.

$$-20+0\times 16$$

Multiply 0 and 16 to get 0.

$$-20+0$$

Add -20 and 0 to get -20.

$$-20$$

Factor

$$-20$$