heureka

The size of a certain population of wild horses varies because of certain ecological pressures.
In 1999, the number of horses was 100.
In 2002, the population was down to 60,
but in 2005, the population was back up to 100.

Assume that 100 was the maximum number of horses and 60 was the minimum.
Write an equation that models the size of the population of horses,
in terms of what year it is,
starting with 1999.

\(\text{Let $x$ is the year}\)

\(\begin{array}{|rcll|} \hline f(x) &=& 20\sin\left(\frac{\pi}{3}(x-1999) + \frac{\pi}{2} \right ) + 80 \quad & | \quad \pi \mathrel{\hat{=}} 180^{\circ} \\\\ &=& 20\sin\left(\frac{180^{\circ}}{3}(x-1999) + \frac{180^{\circ}}{2} \right ) + 80 \\ \mathbf{f(x)} & \mathbf{=} & \mathbf{ 20\sin( 60^{\circ}(x-1999) + 90^{\circ} ) + 80 } \\ \hline \end{array}\)

The graph of
\(\huge{r = \dfrac{7}{3 \cos \theta + 2 \sin \theta}}\)
is a line. Find the slope of this line.

\(\begin{array}{|rcll|} \hline \mathbf{r(\theta)} &\mathbf{=}& \mathbf{\dfrac{7}{3 \cos \theta + 2 \sin \theta}} \\ \hline r(0) &=& \dfrac{7}{3 \cos(0) + 2 \sin(0)} \\\\ r(0) &=& \dfrac{7}{3} \\ \hline r(90) &=& \dfrac{7}{3 \cos(90) + 2 \sin(90)} \\\\ r(90) &=& \dfrac{7}{2} \\ \hline -m &=& \dfrac{r(90)}{r(0)} \\\\ &=& \dfrac{\dfrac{7}{2}}{ \dfrac{7}{3}} \\\\ -m &=& \dfrac{3}{2} \\\\ \mathbf{m} & \mathbf{=} & \mathbf{-\dfrac{3}{2}} \\ \hline \end{array}\)

Thank you, CPhill !

 Inverse?

If \(\large{f(x)=\dfrac{a}{x+2}}\), solve for the value of \(\large{a}\) so that \(\large{f(0)=f^{-1}(3a)}\).

\(\begin{array}{|rcll|} \hline f\left(f^{-1}(x) \right) &=& x \quad & | \quad x = 3a \\\\ f\left(f^{-1}(3a) \right) &=& 3a \quad & | \quad f^{-1}(3a) = f(0) \\\\ f\left(f(0) \right) &=& 3a \quad & | \quad f(0) = \dfrac{a}{0+2} = \dfrac{a}{2} \\\\ f\left(\dfrac{a}{2} \right) &=& 3a \quad & | \quad f\left(\dfrac{a}{2}\right) = \dfrac{a}{\dfrac{a}{2}+2} \\\\ \dfrac{a}{\dfrac{a}{2}+2} &=& 3a \\\\ \dfrac{1}{\dfrac{a}{2}+2} &=& 3 \\\\ \dfrac{a}{2}+2 &=& \dfrac{1}{3} \\\\ \dfrac{a}{2} &=& \dfrac{1}{3}-2 \\\\ \dfrac{a}{2} &=& -\dfrac{5}{3} \\\\ \mathbf{a} & \mathbf{=} & \mathbf{-\dfrac{10}{3}} \\\\ \hline \end{array}\)

Thank you, CPhill !

4. 

Given that \(AF=4\sqrt3\) and \(FC=5\sqrt{3}\), what is \(BC\)?

\(\text{Let $BC=x$}\\ \text{Let $AB=y$}\\ \text{Let $ED=z$}\\ \text{Let $BD=H$}\\ \text{Let $EF=h$}\\ \text{Let $AF=4\sqrt3$} \\ \text{Let $FC=5\sqrt{3}$}\\ \text{Let $AC=AF+FC=9\sqrt{3}$}\)

\(\begin{array}{|lrcll|} \hline & \cos{(\alpha)} = \dfrac{H}{x} &=& \dfrac{z}{H} \\ (1) & \mathbf{H^2} & \mathbf{=} & \mathbf{xz} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline & 2A = H\cdot AC &=& xy \\ (2) & \mathbf{H} & \mathbf{=} & \mathbf{\dfrac{xy}{AC}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & x^2+y^2 &=& (AC)^2 \\ (3) & \mathbf{y^2} & \mathbf{=} & \mathbf{(AC)^2-x^2} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & \cos{(\alpha)} &=& \dfrac{h}{z} \\ & \tan{(\alpha)} &=& \dfrac{h}{AF} \\ & h = z\cdot \cos{(\alpha)} &=& AF\cdot \tan{(\alpha)} \\ & z &=& AF \cdot \dfrac{\sin{(\alpha)}} {\cos^2{(\alpha)}} \\\\ &&& \sin{(\alpha)} = \dfrac{x}{AC} \\ &&& \cos^2{(\alpha)} = 1-\sin^2{(\alpha)} =\dfrac{(AC)^2-x^2}{(AC)^2} \\\\ (4) & \mathbf{ z } & \mathbf{=} & \mathbf{ AF\cdot \dfrac{AC\cdot x}{(AC)^2-x^2} } \\ \hline \end{array}\)

We substitute in (1):

\(\begin{array}{|rcll|} \hline \mathbf{H^2} & \mathbf{=} & \mathbf{xz} \quad & | \quad \mathbf{H} \mathbf{=} \mathbf{\dfrac{xy}{AC}} \\ \dfrac{x^2y^2}{(AC)^2} &=& xz \quad & | \quad \mathbf{ z } \mathbf{=} \mathbf{ AF\cdot \dfrac{AC\cdot x}{(AC)^2-x^2} } \\ \dfrac{x^2y^2}{(AC)^2} &=& \dfrac{AF\cdot AC\cdot x^2}{(AC)^2-x^2} \\ \dfrac{y^2}{(AC)^2} &=& \dfrac{AF\cdot AC}{(AC)^2-x^2} \quad & | \quad \mathbf{y^2} \mathbf{=} \mathbf{(AC)^2-x^2} \\ \dfrac{(AC)^2-x^2}{(AC)^2} &=& \dfrac{AF\cdot AC}{(AC)^2-x^2} \\ \left((AC)^2-x^2\right)^2 &=& AF\cdot (AC)^3 \quad & | \quad \text{sqrt both sides} \\ (AC)^2-x^2 &=& AC\cdot \sqrt{AF\cdot AC} \\ x^2 &=& (AC)^2 - AC\cdot \sqrt{AF\cdot AC} \\ x &=& \sqrt{ (AC)^2 - AC\cdot \sqrt{AF\cdot AC} } \\ \mathbf{x} &\mathbf{=}& \mathbf{\sqrt{ AC\cdot \left( AC - \sqrt{AF\cdot AC} \right)} } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{x} &\mathbf{=}& \mathbf{\sqrt{ AC\cdot \left( AC - \sqrt{AF\cdot AC} \right)} } \quad & | \quad AC = 9\sqrt{3} \qquad AF = 4\sqrt{3} \\ x &=& \sqrt{ 9\sqrt{3}\cdot \left(9\sqrt{3} - \sqrt{108} \right)} \\ x &=& \sqrt{ 243- 9\sqrt{3}\cdot \sqrt{108} } \\ x &=& \sqrt{ 243- 9\sqrt{324} } \\ x &=& \sqrt{ 243- 9\cdot 18 } \\ x &=& \sqrt{ 243- 162 } \\ x &=& \sqrt{ 81 } \\ \mathbf{x} &\mathbf{=}& \mathbf{9} \\ \hline \end{array} \)

BC  is 9

For a certain natural number n,

n^2 gives a remainder of 4 when divided by 5, and

n^3 gives a remainder of 2 when divided by 5.

What remainder does n give when divided by 5?

\(\begin{array}{|lrcll|} \hline (1) & n^2 & \equiv & 4 \pmod{5} \\ (2) & n^3 & \equiv & 2 \pmod{5} \\ \\ \hline \\ \dfrac{(2)}{(1)}: & \dfrac{n^3}{n^2} & \equiv & \dfrac{2}{4} \pmod{5} \\\\ &n & \equiv & \dfrac{1}{2} \pmod{5} \\\\ &\mathbf{n} & \mathbf{\equiv} & \mathbf{2^{-1} \pmod{5}} \\ \hline \end{array} \)

Modular multiplicative inverse by Euler:

\(\text{According to Euler's theorem, if $a$ is coprime to $m$, that is, $gcd(a, m) = 1$, then} \\ {\displaystyle a^{\phi (m)}\equiv 1{\pmod {m}},} \\ \text{where ${\displaystyle \phi }$ is Euler's totient function. } \)

The modular multiplicative inverse can be found directly:

\({\displaystyle a^{\phi (m)-1}\equiv a^{-1}{\pmod {m}}.}\\ \text{In the special case where $m$ is a prime, ${\displaystyle \phi (m)=m-1}$ and a modular inverse is given by } \\ {\displaystyle a^{-1}\equiv a^{m-2}{\pmod {m}}.}\)

\(\text{So $2$ is coprime to $5$ }: \)

\(\begin{array}{|rcll|} \hline 2^{-1} & \equiv & 2^{5-2} \pmod{5} \\ & \equiv & 2^{3} \pmod{5} \\ & \equiv & 8 \pmod{5} \\ & \equiv & 8-5 \pmod{5} \\ & \equiv & 3 \pmod{5} \\ \hline \end{array}\)

\( \begin{array}{|rcll|} \hline &\mathbf{n} & \mathbf{\equiv} & \mathbf{2^{-1} \pmod{5}} \\ &\mathbf{n} & \mathbf{\equiv} & \mathbf{3 \pmod{5}} \\ \hline \end{array}\)

inverse matrices

(b)

Hence find the inverse, \(A_n^{-1} \) of \(A_n\) in terms of \(n\)

Formula inverse matrice:

\(\begin{array}{|rcll|} \hline A &=& e\cdot \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \\\\ A^{-1} &=& \dfrac{e\cdot \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix}}{ det(A)} \qquad det(A) = e^2\cdot(a\cdot d - c\cdot b) \\\\ && \mathbf{ \boxed{ A^{-1} = \dfrac{ \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix}}{e\cdot(a\cdot d - c\cdot b)} } } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A_n &=& 5^{n-1} \begin{pmatrix} 2n+5 & -2n \\ 2n & 5-2n \\ \end{pmatrix} \\\\ A_n^{-1} &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } {5^{n-1}\cdot \Big((2n+5)(5-2n) - (2n)(-2n) \Big) } \\\\ &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } {5^{n-1}\cdot (25-4n^2+4n^2 ) } \\\\ &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } { 5^{n-1}\cdot 5^2 } \\\\ &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } { 5^{n-1+2} } \\\\ &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } { 5^{n+1} } \\\\ \mathbf{A_n^{-1}} & \mathbf{=} & \mathbf{ 5^{-n-1} \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } \\ \hline \end{array}\)

2)
Given that m and n are positive integers such that m=6 (mod 9) and n=0 (mod 9),
what is the largest integer that mn is necessarily divisible by?

\(\begin{array}{|rcll|} \hline m & \equiv & 6\pmod{9} \\ \text{or} \\ m &=& 6 + z\cdot 9 ,\quad z \in \mathbb{Z} \\\\ n & \equiv & 0\pmod{9} \\ \text{or} \\ n &=& 0 + z_2 \cdot 9, \quad z_2 \in \mathbb{Z} \\\\ \hline \\ mn &=& (6 + z\cdot 9)\cdot z_2 \cdot 9 \\ mn &=& 3\cdot(2 + z\cdot 3)\cdot z_2 \cdot 9 \\ \mathbf{mn} &\mathbf{=}& \mathbf{{\color{red}27}\cdot(2 + 3z)\cdot z_2} \\ \hline \end{array}\)

The largest integer is 27

1) 

We have positive integers a, b and c such that a>b>c.

When a, b and c are divided by 19, the remainders are 4, 2 and 18 respectively.

When the number 2a+b-c is divided by 19, what is the remainder?

\(\begin{array}{|rcll|} \hline a & \equiv & 4\pmod{19} \quad & | \quad \cdot 2 \\ \mathbf{2a} & \mathbf{\equiv} & \mathbf{8\pmod{19}} \\\\ \mathbf{b} & \mathbf{\equiv} & \mathbf{2\pmod{19}} \\\\ c & \equiv & 18\pmod{19} \\ c & \equiv & 18-19\pmod{19} \\ \mathbf{c} & \mathbf{\equiv} & \mathbf{-1\pmod{19} } \\ \\ \hline \\ 2a+b-c & \equiv & 8+2-(-1) \pmod{19} \\ \mathbf{2a+b-c} & \mathbf{\equiv} & \mathbf{{\color{red}11} \pmod{19}} \\ \hline \end{array}\)

The remainder is 11