Question 66
What is the value of $$(9\div30)^2\times2.4+0.3\ of\ 12\times(1-0.3)^2+9\times(0.3)^2$$?
Solution
= $$\left(\frac{3}{10}\right)^2\times2.4+0.3\ of\ 12\times(0.7)^2+9\times0.09$$
= $$0.09\times2.4+3.6\times0.49+0.81$$
= $$0.216+1.764+0.81$$
= $$0.216+1.764+0.81$$
= 2.79
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