Solve -05^2+1/4-|-2^2-4|-(-11/2)^3*4/9

$$- 05 ^ { 2 } + \frac { 1 } { 4 } - | - 2 ^ { 2 } - 4 | - ( - 1 \frac { 1 } { 2 } ) ^ { 3 } \times \frac { 4 } { 9 }$$

Evaluate

$$-\frac{25}{4}=-6.25$$

Solution Steps

Calculate 5 to the power of 2 and get 25.

$$0\times 25+\frac{1}{4}-|-2^{2}-4|-\left(-\frac{1\times 2+1}{2}\right)^{3}\times \frac{4}{9}$$

Multiply 0 and 25 to get 0.

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

Add 0 and \frac{1}{4} to get \frac{1}{4}.

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

Calculate 2 to the power of 2 and get 4.

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

Subtract 4 from -4 to get -8.

$$\frac{1}{4}-|-8|-\left(-\frac{1\times 2+1}{2}\right)^{3}\times \frac{4}{9}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -8 is 8.

$$\frac{1}{4}-8-\left(-\frac{1\times 2+1}{2}\right)^{3}\times \frac{4}{9}$$

Subtract 8 from \frac{1}{4} to get -\frac{31}{4}.

$$-\frac{31}{4}-\left(-\frac{1\times 2+1}{2}\right)^{3}\times \frac{4}{9}$$

Multiply 1 and 2 to get 2.

$$-\frac{31}{4}-\left(-\frac{2+1}{2}\right)^{3}\times \frac{4}{9}$$

Add 2 and 1 to get 3.

$$-\frac{31}{4}-\left(-\frac{3}{2}\right)^{3}\times \frac{4}{9}$$

Calculate -\frac{3}{2} to the power of 3 and get -\frac{27}{8}.

$$-\frac{31}{4}-\left(-\frac{27}{8}\times \frac{4}{9}\right)$$

Multiply -\frac{27}{8} and \frac{4}{9} to get -\frac{3}{2}.

$$-\frac{31}{4}-\left(-\frac{3}{2}\right)$$

The opposite of -\frac{3}{2} is \frac{3}{2}.

$$-\frac{31}{4}+\frac{3}{2}$$

Add -\frac{31}{4} and \frac{3}{2} to get -\frac{25}{4}.

$$-\frac{25}{4}$$

Factor

$$-\frac{25}{4} = -6\frac{1}{4} = -6.25$$