MaxWong

\(\text{Number of students} = \dfrac{143-8}{5} = 27\)

Make some changes to your diagram.

Now AF = 1, angle AFB = 90 degrees, and the angle at the bottom right corner is called G.

\(\quad\cos(a+b)\\ =\dfrac{\text{AC}}{\text{AB}}\\ =\dfrac{\text{AG} - \text{GC}}{\text{AB}}\\ =\dfrac{\text{AG}}{\frac{1}{\cos b}} - \dfrac{\text{GC}}{\frac{1}{\cos b}}\\ =\cos a \cos b - \tan b \sin a \cos b\\ = \cos a \cos b - \sin a \sin b\)

Let coordinates of C be \(\begin{pmatrix} 4\\ 2\\ 9 \end{pmatrix} + \lambda \begin{pmatrix}-3\\-1\\2\end{pmatrix}\) as the line passes through C.

Rewrite the coordinates of C: \(\begin{pmatrix}4-3\lambda\\2-\lambda\\9+2\lambda\end{pmatrix}\).

The dot product of \(\vec{\text{BA}}\) and \(\vec{\text{BC}}\) is 0. (Why?)

\(\begin{pmatrix}-2\\4\\6\end{pmatrix} \begin{pmatrix}-2-3\lambda&&4-\lambda&&6+2\lambda\end{pmatrix} = 0\\ -2(-2-3\lambda) + 4(4-\lambda) + 6(6+2\lambda) = 0\\ 4 + 6\lambda + 16 - 4\lambda + 36 + 12 \lambda = 0\\ 14\lambda +56 = 0\\ \lambda = -4\)

So the coordinates of C is \(\begin{pmatrix}16\\6\\1\end{pmatrix}\).

It is an indeterminate form.

You can say, because 0 * a = 0, 0/0 = 0(1/0) = 0.

You can say, because 1/0 = infinity, 0/0 = infinity.

You can say, because a/a = 1, 0/0 = 1.

You get different values if you do it differently, so you can't really determine its value. That's why it is indeterminate.

\(\sec \theta + \tan \theta = p\\ \sec^2 \theta - \tan^2 \theta = p (\sec \theta - \tan \theta)\\ \sec \theta - \tan \theta = \dfrac{1}{p}\\ 2 \sec \theta = p + \dfrac{1}{p}\\ \sec \theta = \dfrac{p^2 + 1}{2p}\\ \cos \theta = \dfrac{2p}{p^2 + 1}\\ \sin \theta = \sqrt{1 - \left(\dfrac{2p}{p^2 +1}\right)^2} = \dfrac{1-p^2}{1+p^2}\\ \color{red}\boxed{\csc \theta = \dfrac{1+p^2}{1-p^2} }\)

\(\text{To prove:} \displaystyle \sum_{cyc(a,b,c)} \dfrac{a^3}{b^2 (5a+2b)} \geq \dfrac{3}{7}\\ \text{By Cauchy-Schwarz inequality: } \displaystyle \sum_{cyc(a,b,c)} \dfrac{a^3}{b^2 (5a+2b)} \sum_{cyc(a,b,c)}ab^2 (5a+2b) \geq (a^2+b^2+c^2)^2\\ \text{If we can show that } \displaystyle \sum_{cyc(a,b,c)} \dfrac{a^3}{b^2 (5a+2b)} \geq \dfrac{(a^2+b^2+c^2)^2}{\displaystyle\sum_{cyc(a,b,c)}ab^2 (5a+2b)} \geq \dfrac{3}{7}\text{, then we are done.}\\ \text{That means we now need to prove } \displaystyle 3 \sum_{cyc(a,b,c)} ab^2 (5a+2b) \leq 7(a^2+b^2+c^2)^2.\\ \text{We can easily see that: }3 \displaystyle\sum_{cyc(a,b,c)} ab^2 (5a+2b) = 15 \sum_{cyc(a,b,c)} a^2 b^2 + 6\sum_{cyc(a,b,c)}ab^3\\ \text{By AM-GM inequality: } \displaystyle3\sum_{cyc(a,b,c)} ab^2 (5a+2b) \geq 3\left(15 \left(a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}}\right) + 6\left(a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}}\right)\right) = 63 a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}}\\ \text{In other words: }\displaystyle3\sum_{cyc(a,b,c)} ab^2 (5a + 2b) \geq 63a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}}\\ \text{Which means: } 63a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}} \leq 7(a^2+b^2+c^2)^2\text{, which is true.} \)

I don't know if this is correct or not. This is too challenging for a beginner in proving inequalities...

If AC was the hypotenuse, then AB = 30/sin(45^o) = 15 √2 

Solute: Mean Green Brand Degreaser, Bleach

Solvent: Water.

You dilute bleach and degreaser with water, so water is a solvent, bleach and the degreaser are solutes.

\(\left(\dfrac{1}{2}\right)^{\dfrac{5000}{60}} \approx \\ 0.000\;000\;000\;000\;000\;000\;000\;000\;082\;066\;710\;908\;398\;\\ \;\;\;484\;932\;599\;748\;169\;573\;032\;246\;267\;286\;055\;990\;653\;\\ \;\;\;700\;864\;138\;695\;608\;921\;461\;002\;162\;199\;233\;793\;883\;\\ \;\;\;026\;545\;529\;934\;157\;752\;325\;784\;886\;826\;150\;426\;386\;\\ \;\;\;495\;457\;067\;186\;335\;978\;132\;830\;029\;967\;467\;146\;205\;\\\)

This is a very nice approximation :)

Use LaTeX. The icon should look like this: \(\Sigma \;\;\text{LaTeX}\).

Fractions: \dfrac{numerator}{denominator}. Example: \dfrac{3}{5} \(\rightarrow \dfrac{3}{5}\).

Boxes: \boxed{(Your equation)}.

Example: 
\boxed{\displaystyle\int^1_0 f(x) \mathbb dx = \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\dfrac{1}{n}f\left(\dfrac{k}{n}\right)}

\(\boxed{\displaystyle\int^1_0 f(x) \mathbb dx = \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\dfrac{1}{n}f\left(\dfrac{k}{n}\right)}\).

For more about how to write math expressions in LaTeX, please refer to my LaTeX post.

Link: https://web2.0calc.com/questions/a-post-of-latex

For advanced LaTeX skills, refer to this much older post by Melody: https://web2.0calc.com/questions/latex