Solve sqrt{8}sqrt{3}-2lceil-5rceil+-{left(1/3right)}^2

$$\sqrt{ 8 } \sqrt{ 3 } -2 \lceil -5 \rceil +- { \left( \frac{ 1 }{ 3 } \right) }^{ 2 }$$

Evaluate

$$2\sqrt{6}+\frac{89}{9}\approx 14.787868374$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$2\sqrt{6}-2\lceil -5\rceil -\left(\frac{1}{3}\right)^{2}$$

The ceiling of a real number a is the smallest integer number greater than or equal to a. The ceiling of -5 is -5.

$$2\sqrt{6}-2\left(-5\right)-\left(\frac{1}{3}\right)^{2}$$

Multiply 2 and -5 to get -10.

$$2\sqrt{6}-\left(-10\right)-\left(\frac{1}{3}\right)^{2}$$

The opposite of -10 is 10.

$$2\sqrt{6}+10-\left(\frac{1}{3}\right)^{2}$$

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$2\sqrt{6}+10-\frac{1}{9}$$

Convert 10 to fraction \frac{90}{9}.

$$2\sqrt{6}+\frac{90}{9}-\frac{1}{9}$$

Since \frac{90}{9} and \frac{1}{9} have the same denominator, subtract them by subtracting their numerators.

$$2\sqrt{6}+\frac{90-1}{9}$$

Subtract 1 from 90 to get 89.

$$2\sqrt{6}+\frac{89}{9}$$

Factor

$$\frac{18 \sqrt{6} + 89}{9} = 14.787868374455245$$