Solve (sqrt{3})^2-pi^0+sqrt{3}cos30^circ

$$( \sqrt { 3 } ) ^ { 2 } - \pi ^ { 0 } + \sqrt { 3 } \cos 30 ^ { \circ }$$

Evaluate

$$\frac{7}{2}=3.5$$

Short Solution Steps

The square of \sqrt{3} is 3.

$$3-\pi ^{0}+\sqrt{3}\cos(30)$$

Calculate \pi to the power of 0 and get 1.

$$3-1+\sqrt{3}\cos(30)$$

Subtract 1 from 3 to get 2.

$$2+\sqrt{3}\cos(30)$$

Get the value of \cos(30) from trigonometric values table.

$$2+\sqrt{3}\times \frac{\sqrt{3}}{2}$$

Express \sqrt{3}\times \frac{\sqrt{3}}{2} as a single fraction.

$$2+\frac{\sqrt{3}\sqrt{3}}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{2}{2}.

$$\frac{2\times 2}{2}+\frac{\sqrt{3}\sqrt{3}}{2}$$

Since \frac{2\times 2}{2} and \frac{\sqrt{3}\sqrt{3}}{2} have the same denominator, add them by adding their numerators.

$$\frac{2\times 2+\sqrt{3}\sqrt{3}}{2}$$

Do the multiplications in 2\times 2+\sqrt{3}\sqrt{3}.

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

Do the calculations in 4+3.

$$\frac{7}{2}$$