Solve -2/3+4sqrt{1/16}*149

$$\frac{ -2 }{ 3 } +4 \sqrt{ \frac{ 1 }{ 16 } } \times 149$$

Evaluate

$$\frac{445}{3}\approx 148.333333333$$

Solution Steps

Fraction \frac{-2}{3} can be rewritten as -\frac{2}{3} by extracting the negative sign.

$$-\frac{2}{3}+4\times 149\sqrt{\frac{1}{16}}$$

Multiply 4 and 149 to get 596.

$$-\frac{2}{3}+596\sqrt{\frac{1}{16}}$$

Rewrite the square root of the division \frac{1}{16} as the division of square roots \frac{\sqrt{1}}{\sqrt{16}}. Take the square root of both numerator and denominator.

$$-\frac{2}{3}+596\times \frac{1}{4}$$

Multiply 596 and \frac{1}{4} to get \frac{596}{4}.

$$-\frac{2}{3}+\frac{596}{4}$$

Divide 596 by 4 to get 149.

$$-\frac{2}{3}+149$$

Convert 149 to fraction \frac{447}{3}.

$$-\frac{2}{3}+\frac{447}{3}$$

Since -\frac{2}{3} and \frac{447}{3} have the same denominator, add them by adding their numerators.

$$\frac{-2+447}{3}$$

Add -2 and 447 to get 445.

$$\frac{445}{3}$$

Factor

$$\frac{5 \cdot 89}{3} = 148\frac{1}{3} = 148.33333333333334$$