Solve sqrt{16}*sqrt[3]{-27/64}+sqrt{-2^2}

$$\sqrt{ 16 } \times \sqrt[ 3 ]{ - \frac{ 27 }{ 64 } } + \sqrt{ { -2 }^{ 2 } }$$

Evaluate

$$-1$$

Solution Steps

Calculate the square root of 16 and get 4.

$$4\sqrt[3]{-\frac{27}{64}}+\sqrt{\left(-2\right)^{2}}$$

Calculate \sqrt[3]{-\frac{27}{64}} and get -\frac{3}{4}.

$$4\left(-\frac{3}{4}\right)+\sqrt{\left(-2\right)^{2}}$$

Multiply 4 and -\frac{3}{4} to get -3.

$$-3+\sqrt{\left(-2\right)^{2}}$$

Calculate -2 to the power of 2 and get 4.

$$-3+\sqrt{4}$$

Calculate the square root of 4 and get 2.

$$-3+2$$

Add -3 and 2 to get -1.

$$-1$$

Factor

$$-1$$