MaxWong

For 2nd question:

First find f(x):

\(f(x) = (g\cdot h)(x) = g(x) \cdot h(x) = (18^x)(14x^3)\)

Then find f '(x):

In this case we use product rule:

\(f'(x) = (18^x)'(14x^3) + (18^x)(14x^3)' = 14\ln 18 \cdot 18^x \cdot x^3 +18^x\cdot42x^2\)

Let us call the part before plus sign (1) and the part after plus sign (2)

Differentiate (1):

\((1)' = 14\ln 18 \left((18^x)'(x^3)+(18^x)(x^3)'\right)\\ =14\ln 18 (x^3\cdot18^x\ln 18+3\cdot18^x\cdot x^2)\)

Differentiate (2):

\((2)' = (18^x)'(42x^2) + (18^x)(42x^2)'\\ =42\ln 18\cdot18^x\cdot x^2+84\cdot18^x\cdot x\)

Add up (1)' and (2)'

\(\quad z(x) \\= (1)' + (2)'\\ =14\ln18(x^3\cdot 18^x\ln 18+3\cdot 18^x \cdot x^2) + 42\ln 18\cdot 18^x \cdot x^2 + 84 \cdot 18^x \cdot x\\ = 14x^218^x\ln 18(x + 3) + 42x \cdot 18^x(x\ln 18+2)\)

sin (-7 degrees) = -0.121869343405

sin (-7 radians) = -0.656986598719

\(2\times 2\times \cdot\cdot\cdot\times 2 \text{ (30 times)}\\ = 2^{30}\\ =1073741824\)

I don't need to go to school tomorrow :)

But the 1-day holiday is not for Thanksgiving :(

It is just for the next day of Sports Day and the school lets us take a rest because in Sports Day all students compete in sports and all will be tired......

Thanksgiving doesn't mean anything to our country :(

P.S.: Luckily I don't live in Mainland China or Christmas doesn't mean anything to our country...... I live in Hong Kong where Britain people ruled(but now not) so that we still have Christmas :)

370² = 136900 (normal) = 1.369 x 10^5 (scientific) 

\(\sqrt{\dfrac{1-\cos\frac{24}{25}}{2}}\\ =\cos\dfrac{12}{25}\)

It cannot be expressed in terms of pi :(

Exact value of sqrt(0.2)

\(\qquad\sqrt{0.2} \\=\sqrt{\dfrac{1}{5}}\\ =\dfrac{1}{\sqrt5}\\ =\dfrac{\sqrt5}{5}\)

\(\text{Q:Simplify }\dfrac{1+\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}\)

\(\quad\dfrac{1+\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}\\ =\dfrac{1}{\frac{\sqrt2}{2}}+\dfrac{\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}\\ =2\div \sqrt2 + 1\qquad\color{olive}\boxed{\color{red}2=(\sqrt2)^2}\\ = \sqrt2 + 1\)

\(\dfrac{519.88\text{ billion}}{2.37\text{ trillion}}\\ =\dfrac{519.88\times 10^9}{2.37 \times 10^{12}}\\ = \dfrac{519.88}{2.37\times 10^3}\\ =\dfrac{519.88}{2370}\\ \text{So the answer is approximately 1/4.}\)

Exact value approx. eq. 0.219

\(\sqrt{\dfrac{5}{8}-\dfrac{\sqrt5}{8}}\)??

it equals 

\(\sqrt{\left(\dfrac{\sqrt5-1}{2}\right)\left(\dfrac{\sqrt5}{4}\right)}\)

and the golden ratio = (sqrt5 - 1)/2, and golden ratio is often related to pentagons.

So maybe we can work that exact value out of pentagons? :)