Solve sqrt{1.6*62.5}-|sqrt[3]{-1125/64}|+sqrt{(-0.2)^4}*(-5)^3

$$\sqrt { 1.6 \times 62.5 } - | \sqrt[ 3 ] { - 11 \frac { 25 } { 64 } } | + \sqrt { ( - 0.2 ) ^ { 4 } } \times ( - 5 ) ^ { 3 }$$

Evaluate

$$2.75$$

Solution Steps

Multiply 1.6 and 62.5 to get 100.

$$\sqrt{100}-|\sqrt[3]{-\frac{11\times 64+25}{64}}|+\sqrt{\left(-0.2\right)^{4}}\left(-5\right)^{3}$$

Calculate the square root of 100 and get 10.

$$10-|\sqrt[3]{-\frac{11\times 64+25}{64}}|+\sqrt{\left(-0.2\right)^{4}}\left(-5\right)^{3}$$

Multiply 11 and 64 to get 704.

$$10-|\sqrt[3]{-\frac{704+25}{64}}|+\sqrt{\left(-0.2\right)^{4}}\left(-5\right)^{3}$$

Add 704 and 25 to get 729.

$$10-|\sqrt[3]{-\frac{729}{64}}|+\sqrt{\left(-0.2\right)^{4}}\left(-5\right)^{3}$$

Calculate \sqrt[3]{-\frac{729}{64}} and get -\frac{9}{4}.

$$10-|-\frac{9}{4}|+\sqrt{\left(-0.2\right)^{4}}\left(-5\right)^{3}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{9}{4} is \frac{9}{4}.

$$10-\frac{9}{4}+\sqrt{\left(-0.2\right)^{4}}\left(-5\right)^{3}$$

Subtract \frac{9}{4} from 10 to get \frac{31}{4}.

$$\frac{31}{4}+\sqrt{\left(-0.2\right)^{4}}\left(-5\right)^{3}$$

Calculate -0.2 to the power of 4 and get 0.0016.

$$\frac{31}{4}+\sqrt{0.0016}\left(-5\right)^{3}$$

Calculate the square root of 0.0016 and get 0.04.

$$\frac{31}{4}+0.04\left(-5\right)^{3}$$

Calculate -5 to the power of 3 and get -125.

$$\frac{31}{4}+0.04\left(-125\right)$$

Multiply 0.04 and -125 to get -5.

$$\frac{31}{4}-5$$

Subtract 5 from \frac{31}{4} to get \frac{11}{4}.

$$\frac{11}{4}$$

Factor

$$\frac{11}{2 ^ {2}} = 2\frac{3}{4} = 2.75$$