Question

\(\left(\frac{1}{2}\right)^2+\left(-\frac{3}{4}\right)-\frac{\sqrt{9}}{12}\)

Answer 1

\(\left(\frac{1}{2}\right)^2+\left(-\frac{3}{4}\right)-\frac{\sqrt{9}}{12}=\frac{1}{4}-\frac{3}{4}-\frac{3}{12}=\frac{-3}{4}\)

Answer 2

\(\left(\frac{1}{2}\right)^2+\left(-\frac{3}{4}\right)-\frac{\sqrt{9}}{12}\)

\(=\frac{1}{4}+\left(-\frac{3}{4}\right)-\frac{3}{12}\)

\(=\left(-\frac{2}{4}\right)-\frac{3}{12}\)

\(=\left(-\frac{6}{12}\right)-\frac{3}{12}\)

\(=\left(-\frac{6}{12}\right)+\left(-\frac{3}{12}\right)=-\frac{9}{12}=-\frac{3}{4}\)

Answer 3

${\left(\frac{1}{2}\right)}^2+\left(-\frac{3}{4}\right)-\frac{\sqrt{9}}{12}$

= \(\frac{1}{4}+\left(-\frac{3}{4}\right)-\frac{\sqrt{9}}{12}\)

= \(\frac{1-3}{4}-\frac{\sqrt{9}}{12}\)

=\(\frac{-2}{4}-\frac{\sqrt{9}}{12}\)

=\(\frac{-6}{12}-\frac{3}{12}\)

=\(\frac{-6-3}{12}\)

=\(\frac{-3}{4}\)

Answer 4

\(\left(\frac{1}{2}\right)^2+\left(-\frac{3}{4}\right)-\frac{\sqrt{9}}{12}=\frac{2}{4}-\frac{3}{4}-\frac{1}{4}\)

\(=\frac{2-3-1}{4}=\frac{-2}{4}=\frac{-1}{2}\)

Answer 5

\(\left(\dfrac{1}{2}\right)^2+\left(-\dfrac{3}{4}\right)-\dfrac{\sqrt{9}}{12}\)

\(=\dfrac{1}{4}-\dfrac{3}{4}-\dfrac{3}{12}\)

\(=\dfrac{-3}{4}\)