MaxWong

Do 2.5 * 10^(12) for \(2.5\times10^{12}\)

\(3\pi d = 12\pi\\ 3\pi d = 4(3\pi)\\ d = 4.\)

\(-95\div 9\\ =-\dfrac{95}{9}\\ =-10\dfrac{5}{9}\)

I assume that the $42.25 is associated with the cleaning of the park. LOL.

$42.25 / 6.5hrs =$6.5/hour <---- ElectricPavlov you calculated wrongly......

\((-5)\times 7 - 6\times (-2)\\ =-(5\times 7)+(6\times 2)\\ = -35 + 12\\ = 12 - 35\\ = -23\)

The expression 

= 2.5(10 cos 36 + 4 sin 35 - 3 cos 75)

= \(\dfrac{5(10\sin 54 + 4\sin 35 - 3\sin 15)}{2}\)

= \(\dfrac{5\left(10(\frac{1+\sqrt5}{4})+4((\sqrt[3]{-\frac{\sqrt6+\sqrt2}{32}+(\frac{i}{32})(2\sqrt{2-\sqrt3})})-\sqrt[3]{-\frac{\sqrt6+\sqrt2}{32}-(\frac{i}{32})(2\sqrt{2-\sqrt3})})-3(\dfrac{\sqrt3-1}{2\sqrt2})\right)}{2}\)

Upper part is

\(5\left(10(\frac{1+\sqrt5}{4})+4((\sqrt[3]{-\frac{\sqrt6+\sqrt2}{32}+(\frac{i}{32})(2\sqrt{2-\sqrt3})})-\\\sqrt[3]{-\frac{\sqrt6+\sqrt2}{32}-(\frac{i}{32})(2\sqrt{2-\sqrt3})})-3(\dfrac{\sqrt3-1}{2\sqrt2})\right)\)

I know this is complicated but it is completely real.

The question in the title:

\(\sqrt{\dfrac{3}{5}}\\ =\dfrac{\sqrt3}{\sqrt5}\leftarrow\text{Distributive property of square roots}\\ = \dfrac{\sqrt3}{\sqrt5}\times\dfrac{\sqrt5}{\sqrt5}\\ =\dfrac{\sqrt3 \times \sqrt5}{(\sqrt5)^2}\\ =\dfrac{\sqrt{15}}{5}\leftarrow\text{Multiplicative property of square roots}\)

\(\quad\displaystyle\mathtt{P}^{16}_3\\ = 3360\)

Use the calculator!!