heureka

So I have this question that I have no idea how to approach,
it was one of my math problems that belong in the category of factoring general quadratics,
but I didn't know how to factor it:
The equation  
$\large{y=-4.9t^2+42t+18.9 }$
describes the height (in meters) of a ball tossed up in the air at 42 meters per second from a height of 18.9 meters from the ground,
as a function of time in seconds.
In how many seconds will the ball hit the ground?

$\text{Hit the ground means $y=0\ !$}$

$\begin{array}{|rcll|} \hline -4.9t^2+42t+18.9 &=& 0 \quad | \quad \cdot(-1) \\ 4.9t^2-42t-18.9 &=& 0 \\\\ t &=& \dfrac{42\pm \sqrt{42^2-4\cdot 4.9 \cdot (-18.9) } } {2\cdot 4.9} \\ t &=& \dfrac{42\pm \sqrt{42^2+370.44 } } {9.8} \\ t &=& \dfrac{42\pm \sqrt{1764+370.44 } } {9.8} \\ t &=& \dfrac{42\pm \sqrt{2134.44 } } {9.8} \\ t &=& \dfrac{42\pm 46.2 } {9.8} \quad | \quad t>0\ !\\ t &=& \dfrac{42{\color{red}+} 46.2 } {9.8} \\\\ t &=& \dfrac{88.2 } {9.8} \\ \mathbf{ t } & \mathbf{=} & \mathbf{9\ \text{seconds} } \\ \hline \end{array}$

In 9 seconds the ball will hit the ground.

If $2ab = 12$, evaluate $8a^2b^2$.

$\begin{array}{|rcll|} \hline 2ab &=& 12 \quad | \quad : 2 \\ ab &=& \dfrac{12}{2} \\ \mathbf{ab} & \mathbf{=} & \mathbf{6} \\\\ 8a^2b^2 &=& 8 (ab)^2 \quad | \quad \mathbf{ab=6} \\ &=& 8\cdot (6)^2 \\ &=& 8\cdot 36 \\ &=& 288 \\ \hline \end{array}$

$8a^2b^2 = 288$

In acute triangle $ABC,\ \angle A = 68^\circ$.

Let be the circumcenter of triangle $ABC$.
Find, $\angle OBC$ in degrees.

$\text{Let $\angle OBC = {\color{red}x}$ }$

$\begin{array}{|rcll|} \hline 2{\color{red}x} + 2{\color{blue}y} + 2{\color{green}z} &=& 180^{\circ} \quad | \quad : 2 \\ {\color{red}x} + {\color{blue}y} + {\color{green}z} &=& 90^{\circ} \quad | \quad {\color{blue}y} + {\color{green}z} = \angle A \\ {\color{red}x} + A &=& 90^{\circ} \\ {\color{red}x} &=& 90^{\circ} - A \quad | \quad A = 68^{\circ} \\ {\color{red}x} &=& 90^{\circ} - 68^{\circ} \\ {\color{red}x} &=& 22^{\circ} \\ \hline \end{array}$

$\text{$\angle OBC$ in degrees is $22^{\circ}$}$

help
Let

$\large{f(x) = \dfrac{2x + 3}{kx - 2}}$.

Find all real numbers

$\mathbf{k}$

so that

$\large{f^{-1}(x) = f(x)}$.

discontinuity:

$\begin{array}{|rcll|} \hline 2x+3 &=& 0 \\ 2x &=& -3 \\ x &=& -\dfrac{3}{2} \\ \hline kx - 2 &\ne& 0 \\ kx &\ne& 2 \\ k &\ne& \dfrac{2}{x} \quad | \quad x = -\dfrac{3}{2} \\\\ k &\ne& \dfrac{2}{-\dfrac{3}{2}} \\\\ \mathbf{k} & \mathbf{\ne} & \mathbf{-\dfrac{4}{3}} \\ \hline \end{array}$

At $k = -\dfrac{4}{3} \Rightarrow x = -\dfrac{3}{2} \Rightarrow y = \dfrac{0}{0}$

help

Let 

$\large{S = \dfrac{1}{2^3} + \dfrac{1}{4^3} + \dfrac{1}{6^3} + \dotsb}$

and
$\large{T = \dfrac{1}{1^3} + \dfrac{1}{3^3} + \dfrac{1}{5^3} + \dotsb}$.
Find

$\large{\dfrac{S}{T}}$

1.

$\mathbf{S+T =\ ?}$

$\begin{array}{|rcll|} \hline S+T &=& \left(\dfrac{1}{2^3} + \dfrac{1}{4^3} + \dfrac{1}{6^3} + \dotsb \right) +\left(\dfrac{1}{1^3} + \dfrac{1}{3^3} + \dfrac{1}{5^3} + \dotsb \right) \\ S+T &=& \color{red}{\dfrac{1}{1^3} + \dfrac{1}{2^3} + \dfrac{1}{3^3} + \dfrac{1}{4^3} + \dfrac{1}{5^3} + \dfrac{1}{6^3} + \dotsb} \\ \hline \end{array}$

2.

$\mathbf{\dfrac{S}{T} =\ ?}$

$\begin{array}{|rcll|} \hline S &=& \dfrac{1}{2^3} + \dfrac{1}{4^3} + \dfrac{1}{6^3}+ \dfrac{1}{8^3}+ \dfrac{1}{10^3} + \dotsb \\ &=& \dfrac{1}{(2\cdot 1)^3} + \dfrac{1}{(2\cdot 2)^3} + \dfrac{1}{(2\cdot 3)^3}+ \dfrac{1}{(2\cdot 4)^3}+ \dfrac{1}{(2\cdot 5)^3} + \dotsb \\ &=& \dfrac{1}{2^3 1^3} + \dfrac{1}{2^3 2^3} + \dfrac{1}{2^3 3^3}+ \dfrac{1}{2^3 4^3}+ \dfrac{1}{2^3 5^3} + \dotsb \\ &=& \dfrac{1}{2^3} \underbrace{ \left(\color{red}{\dfrac{1}{1^3} + \dfrac{1}{2^3} + \dfrac{1}{3^3}+ \dfrac{1}{4^3}+ \dfrac{1}{5^3} + \dotsb} \right) }_{=S+T} \\ &=& \dfrac{1}{2^3}(S+T) \\\\ S &=& \dfrac{1}{8}(S+T) \quad | \quad \cdot 8 \\ 8S &=& S+T \quad | \quad -S \\ 7S &=& T \quad | \quad :7 \\ S &=& \dfrac{T}{7} \quad | \quad:T \\ \mathbf{\dfrac{S}{T}} & \mathbf{=} & \mathbf{\dfrac{1}{7}} \\ \hline \end{array}$

What is the area of the composite figure whose vertices have the following coordinates?

(−2, −2) , (4, −2) , (5, 1) , (2, 3) , (−1, 1)

$\begin{array}{|r|r|r|r|r|} \hline \text{Point} & x & y & \\ \hline 1 & -2 & -2 &\\ & & & (-2)(-2) - 4(-2) & = 12 \\ 2 & 4 & -2 & \\ & & & 4\cdot 1 - 5 (-2) & = 14 \\ 3 & 5 & 1 &\\ & & & 5\cdot 3 - 2 (1) & = 13 \\ 4 & 2 & 3 &\\ & & & 2\cdot 1 - (-1)3 & = 5 \\ 5 & -1 & 1 &\\ & & & (-1)(-2) - (-2)1 & = 4 \\ 1 & -2 & -2 & \\ \hline & & & & \text{sum} = 48 \\ & & & & \text{area of the composite figure } = \dfrac{\text{sum}}{2} = \dfrac{\text{48}}{2} = \mathbf{24} \\ \hline \end{array}$

An ant moves on the following lattice, beginning at the dot labeled A. 

Each minute he moves to one of the dots neighboring the dot he was at, choosing from among its neighbors at random.

What is the probability that after 5 minutes he is at the dot labeled B ?

If

$\large{ f(x)=\dfrac{a}{x+2}}$,

solve for the value of $\mathbf{a}$ so that

$\large{f(0)=f^{-1}(3a)}$.

$\begin{array}{|rcll|} \hline f\Big( f^{-1}(x) \Big) &=& x \quad | \quad x = 3a \\ f\Big( f^{-1}(3a) \Big) &=& 3a \quad | \quad f^{-1}(3a)=f(0) \\ f\Big( f(0) \Big) &=& 3a \quad | \quad f(0) = \dfrac{a}{0+2} = \dfrac{a}{2} \\ f\left( \dfrac{a}{2} \right) &=& 3a \quad | \quad f\left(\dfrac{a}{2} \right) = \dfrac{a}{\dfrac{a}{2}+2} = \dfrac{2a}{4+a} \\ \dfrac{2a}{4+a} &=& 3a \quad | \quad :a \\\\ \dfrac{2 }{4+a} &=& 3 \\\\ 4+a &=& \dfrac{2}{3} \\\\ a &=& \dfrac{2}{3} -4 \\\\ a &=& \dfrac{2}{3} -\dfrac{12}{3} \\\\ \mathbf{ a } &\mathbf{=}&\mathbf{ -\dfrac{10}{3}} \\ \hline \end{array}$

Prove algebraically that the square of any odd number is always 1 more than a multiple of 8.

Let n stand for any integer in your working.

Let k be a non-negative integer.

$\begin{array}{|rcll|} \hline n^2 &=& (2k-1)^2 \quad | \quad 2k-1 \text{ is any odd number } k > 0,\ k\in \mathbb{N} \\ &=& 4k^2-4k+1 \\ &=& 4k(k-1)+1 \quad | \quad \text{The product of two consecutive integers must be even so let k(k-1)=2m } \\ &=& 4\cdot 2m +1 \\ &=& 8m +1 \\ \mathbf{n^2} & \mathbf{=} & \mathbf{8m +1 \equiv 1 \pmod{8}} \\ \hline \end{array}$

Geometry

As shown in the diagram, $\dfrac{BD}{DC}=2$ , $\dfrac{CE}{EA}=3$, and $\dfrac{AF}{FB}=4$. Find $\dfrac{[DEF]}{[ABC]}$.

$\begin{array}{|rcll|} \hline [ABC] &=& \dfrac{4EA\cdot3DC}{2}\sin(C) \\ \mathbf{[ABC]} &\mathbf{=}& \mathbf{6 EA\cdot DC \sin(C)} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline [AFE] &=& \dfrac{1EA\cdot4FB}{2}\sin(A) \quad | \quad \sin(A)=\dfrac{3DC}{5FB}\sin(C) \\ [AFE] &=& \dfrac{1EA\cdot4FB}{2}\cdot \dfrac{3DC}{5FB}\sin(C) \\ \mathbf{[AFE]} &\mathbf{=}& \mathbf{ \dfrac{6}{5} EA\cdot DC \sin(C)} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline [FBD] &=& \dfrac{2DC\cdot 1FB}{2}\sin(B) \quad | \quad \sin(B)=\dfrac{4EA}{5FB}\sin(C) \\ [FBD] &=& \dfrac{2DC\cdot 1FB}{2}\cdot \dfrac{4EA}{5FB}\sin(C) \\ \mathbf{[FBD]} &\mathbf{=}& \mathbf{ \dfrac{4}{5} EA\cdot DC \sin(C)} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline [EDC] &=& \dfrac{1DC\cdot 3EA}{2}\sin(C) \\ \mathbf{[EDC]} &\mathbf{=}& \mathbf{ \dfrac{3}{2} EA\cdot DC \sin(C)} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline [AFE]+[FBD]+[EDC]+[DEF] &=& [ABC] \\ [DEF] &=& [ABC] -([AFE]+[FBD]+[EDC]) \quad | \quad : [ABC] \\\\ \dfrac{[DEF]}{[ABC]} &=& \dfrac{[ABC] -([AFE]+[FBD]+[EDC])} {[ABC]} \\\\ \dfrac{[DEF]}{[ABC]} &=&1 -\dfrac{[AFE]+[FBD]+[EDC]} {[ABC]} \\\\ \dfrac{[DEF]}{[ABC]} &=&\small{1 -\dfrac{\dfrac{6}{5} EA\cdot DC \sin(C)+\dfrac{4}{5} EA\cdot DC \sin(C)+\dfrac{3}{2} EA\cdot DC \sin(C)} {6 EA\cdot DC \sin(C)} }\\\\ \dfrac{[DEF]}{[ABC]} &=&1 -\dfrac{\dfrac{35}{10}} {6 } \\\\ \dfrac{[DEF]}{[ABC]} &=&1 - \dfrac{7}{12} \\\\ \mathbf{\dfrac{[DEF]}{[ABC]}} &\mathbf{=}& \mathbf{\dfrac{5}{12}} \\ \hline \end{array}$