Solve sqrt{25}+sqrt[3]{1/8}-sqrt{(-3)^2}-|-2|

$$\sqrt { 25 } + \sqrt[ 3 ] { \frac { 1 } { 8 } } - \sqrt { ( - 3 ) ^ { 2 } } - | - 2 |$$

Evaluate

$$\frac{1}{2}=0.5$$

Solution Steps

Calculate the square root of 25 and get 5.

$$5+\sqrt[3]{\frac{1}{8}}-\sqrt{\left(-3\right)^{2}}-|-2|$$

Calculate \sqrt[3]{\frac{1}{8}} and get \frac{1}{2}.

$$5+\frac{1}{2}-\sqrt{\left(-3\right)^{2}}-|-2|$$

Add 5 and \frac{1}{2} to get \frac{11}{2}.

$$\frac{11}{2}-\sqrt{\left(-3\right)^{2}}-|-2|$$

Calculate -3 to the power of 2 and get 9.

$$\frac{11}{2}-\sqrt{9}-|-2|$$

Calculate the square root of 9 and get 3.

$$\frac{11}{2}-3-|-2|$$

Subtract 3 from \frac{11}{2} to get \frac{5}{2}.

$$\frac{5}{2}-|-2|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -2 is 2.

$$\frac{5}{2}-2$$

Subtract 2 from \frac{5}{2} to get \frac{1}{2}.

$$\frac{1}{2}$$

Factor

$$\frac{1}{2} = 0.5$$