heureka

Find all values of x such that 

\(\dfrac{x}{x+4} = -\dfrac{9}{x+3}\)

\dfrac{x}{x+4} = -\dfrac{9}{x+3}.
If you find more than one value, then list your solutions in increasing order, separated by commas.

\(\text{ $x\ne -4$ and $x\ne -3$ }\)

\(\begin{array}{|rcll|} \hline \dfrac{x}{x+4} &=& -\dfrac{9}{x+3} \quad & | \quad \cdot (x+3) \\\\ \dfrac{x(x+3)}{x+4} &=& -9 \quad & | \quad \cdot (x+4) \\\\ x(x+3) &=& -9(x+4) \\\\ x^2+3x &=& -9x -36 \quad & | \quad +9x \\\\ x^2+12x &=& -36 \quad & | \quad +36 \\\\ x^2+12x+36 &=&0 \\\\ x &=& \dfrac{-12\pm \sqrt{12^2 -4\cdot 36 } }{2} \\\\ x &=& \dfrac{-12\pm \sqrt{144 -144 } }{2} \\\\ x &=& \dfrac{-12\pm \sqrt{0} }{2} \\\\ x &=& \dfrac{-12\pm 0 }{2} \\\\ x &=& \dfrac{-12}{2} \\\\ \mathbf{x} &\mathbf{=}& \mathbf{-6} \\ \hline \end{array}\)

What values of x satisfy |x - 4| + |x + 4| <= 10 ?

Express your answer in interval notation.

\( \huge |x + 4| + |x - 4| \le 10\)

Here we have two different amounts.

To dissolve them, we must make a double case distinction.

Usually we did this one after the other.
For reasons of space, we start with the first case in which the content of the left amount
is greater than or equal to zero.

\(\begin{array}{|rcll|} \hline & \underline{x\ge -4:}& \\\\ & x+4 + |x-4 | \le 10 \\ \\ \underline{\text{for } x\ge 4:} && \underline{\text{for } x \lt 4: } \\\\ x+4+x-4 \le 10 && x+4-(x-4) \le 10 \\ 2x \le 10 && 8 \le 10 \\ \boxed{ x \le 5 } && \\ \hline \end{array} \)

The other case was \(x \lt -4\).
In this area the content of the right amount is
always negative. A further case distinction is therefore not necessary here.

\(\begin{array}{|rcll|} \hline & \underline{x \lt -4:} \\\\ & -(x+4) - (x-4) \le 10 \\ \\ & -x-4-x+4 \le 10 \\ & -2x \le 10 \quad & \quad :(-2)\\ & x \ge \dfrac{10}{-2} \\ & \boxed{x \ge -5} \\ \hline \end{array} \)

Values of x in interval notation: [-5 , 5  ]

The operation \(*\) is defined for non-zero a andb  as follows: 

\(a * b = \frac{1}{a} + \frac{1}{b}. \)

If \(a+b=13\) and \(a \times b=25\), 

what is the value of  \(a*b\)  ?

Express your answer as a common fraction.

\(\begin{array}{|rcll|} \hline a+b &=& 13 \\ a\times b &=& 25 \\\\ \dfrac{a+b}{a\times b}&=& \dfrac{13}{25} \\\\ \dfrac{a}{a\times b} + \dfrac{b}{a\times b} &=& \dfrac{13}{25} \\\\ \dfrac{1}{b} + \dfrac{1}{a} &=& \dfrac{13}{25} \\\\ \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{13}{25} \quad & | \quad \dfrac{1}{a} + \dfrac{1}{b} = a*b \\\\ \mathbf{a*b} & \mathbf{=} & \mathbf{\dfrac{13}{25}} \\ \hline \end{array}\)

Modulo inverses
Find
\(24^{-1} \pmod{11^2}\).
That is, find the residue b  for which \(24b \equiv 1\pmod{11^2}\).
Express your answer as an integer from 0 to , inclusive.

\(\begin{array}{|rcll|} \hline & 24\cdot \underbrace{24^{-1}}_{=b} \equiv 1 \pmod{11^2} \\ \hline \end{array}\)

\(\begin{array}{rcll} \text{Let} \\ & 24\cdot 5 = 120 \equiv -1 \pmod{11^2} \\ \end{array} \)

\(\begin{array}{llcll} \text{square this equation: } \\ & (24\cdot 5)\cdot(24\cdot 5) &\equiv& (-1)\cdot(-1) \pmod{11^2} \\ & 24\cdot (5^2\cdot 24) &\equiv& 1 \pmod{11^2} \\ & 24\cdot 600 &\equiv& 1 \pmod{11^2} \quad | \quad 600 \equiv 116 \pmod{11^2} \\ & 24\cdot \underbrace{116}_{=(24)^{-1}=b} &\equiv& 1 \pmod{11^2} \\ \end{array} \)

\(\text{So $24^{-1} \pmod{11^2}=\boxed{116}$ is the multiplicative inverse to $24$ modulo $11^2$}.\)

Define
\(f(x)=\dfrac{1+x}{1-x}\)
and
\(g(x)=\dfrac{-2}{x+1}\).
Find the value of
\(\large{g(f(g(f(\ldots g(f(12)) \ldots ))))}\)
where there are 16 compositions of the functions g, and f,
alternating between the two.

\(\begin{array}{|rcll|} \hline f(x) &=& \dfrac{x+1}{1-x} \\\\ g(f(x)) = \dfrac{-2}{\dfrac{x+1}{1-x}+1} &&=& x-1 \\\\ f(g(f(x))) = \dfrac{x-1+1}{1-(x-1)} &=& \dfrac{x+0}{2-x} \\\\ g(f(g(f(x)))) = \dfrac{-2}{\dfrac{x}{2-x}+1}& &=&x-2 \\\\ f(g(f(g(f(x))))) = \dfrac{x-2+1}{1-(x-2)} &=& \dfrac{x-1}{3-x} \\\\ g(f(g(f(g(f(x))))))=\dfrac{-2}{\dfrac{x-1}{3-x}+1} & &=& x-3 \\\\ f(g(f(g(f(g(f(x))))))) = \dfrac{x-3+1}{1-(x-3)} &=& \dfrac{x-2}{4-x} \\\\ g(f(g(f(g(f(g(f(x))))))))=\dfrac{-2}{\dfrac{x-2}{4-x}+1} & &=& x-4 \\\\ \ldots \\ \hline \end{array}\)

\(\text{When there are $\mathbf{16}$ compositions of the functions $g$, and $f$, alternating between the two.} \)

\(\begin{array}{|rcll|} \hline g(f(g(f(\ldots g(f(12)) \ldots )))) &=& x-16 \quad & | \quad x = 12 \\ &=& 12-16 \\ &=& -4 \\ \hline \end{array}\)

I have the following terms of an arithmetic sequence:

\(\frac{1}{2}, x-1, 3x, \ldots\)

$\frac{1}{2}, x-1, 3x, \ldots$.

Solve for \(x\).

Formula of an arithmetic sequence:

\(\begin{array}{|rcll|} \hline \begin{vmatrix} a_i & a_j & a_k \\ i & j & k \\ 1 & 1 & 1 & \\ \end{vmatrix} = 0 \\\\ a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\ \hline \end{array} \)

\(\text{Set $i=1$, $\ j=2$, $\ k=3$ } \\ \text{Set $a_i=a_1=\frac12$, $\ a_j=a_2=x-1$, $\ a_k=a_3=3x$ } \)

\(\begin{array}{|rcll|} \hline a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\\\ \frac12(2-3)+(x-1)(3-1)+3x(1-2) &=& 0 \\ \frac12(-1)+(x-1)(2)+3x(-1) &=& 0 \\ -\frac12+2x-2-3x &=& 0 \\ -\frac52-x &=& 0 \\ \frac52+x &=& 0 \\ x &=& -\frac52 \\ \mathbf{x} &\mathbf{=}& \mathbf{-2.5} \\ \hline \end{array} \)

I assume,

how to convert 82 mg to lb.

\(\begin{array}{|rcll|} \hline && \mathbf{83\ \text{mg}} \\\\ &=& 83\ \text{mg} \cdot \dfrac{ 1\ \text{kg} } { 1000000\ \text{mg} } \cdot \dfrac{ 2.2046226218\ \text{lb} } { 1\ \text{kg} } \\\\ &=& \dfrac{ 83\cdot 2.2046226218 } { 1000000 } \ \text{lb} \\\\ &\mathbf{=}& \mathbf{0.00018298368 \ \text{lb} } \\ \hline \end{array}\)

2.
Convert 373 milliseconds to units of megashakes.
One "shake" is equal to 10 nanoseconds[ns].

\(\begin{array}{|rcll|} \hline && \mathbf{373\ \text{ms}} \\\\ &=& 373\ \text{ms} \cdot \dfrac{ 1000000\ \text{ns} }{ 1\ \text{ms}} \cdot \dfrac{ 1 \ \text{shake} }{ 10\ \text{ns}} \cdot \dfrac{ 1 \ \text{megashake} }{ 1000000\ \text{shake}} \\\\ &=& \dfrac{ 373\cdot 1000000 }{ 10\cdot 1000000 }\ \text{megashake} \\\\ &=& \dfrac{ 373 }{ 10 }\ \text{megashake} \\\\ &\mathbf{=}& \mathbf{37.3\ \text{megashake}} \\ \hline \end{array} \)

1.

In Africa one "su" is 1.66 meters.

The average height of a human is 5 feet and 7 inches.

How to convert this height to units of "su" ?

feet to inches:

\(\begin{array}{|rcll|} \hline && 5\ \text{feet} + 7\ \text{inches} \\\\ &=& 5\ \text{feet} \cdot \dfrac{12\ \text{inches}}{1\ \text{feet}} + 7\ \text{inches} \\\\ &=& 60\ \text{inches} + 7\ \text{inches} \\\\ &=& 67\ \text{inches} \\\\ \hline \end{array} \)

inches to "su":

\(\begin{array}{|rcll|} \hline && \mathbf{67\ \text{inches}} \\\\ &=& 67\ \text{inches} \cdot \dfrac{ 2.54\ \text{cm} }{ 1\ \text{inch}} \cdot \dfrac{ 1\ \text{meter} }{ 100\ \text{cm}} \cdot \dfrac{ 1\ \text{su} }{ 1.66\ \text{meter}} \\\\ &=& \dfrac{67\cdot 2.54}{ 100\cdot 1.66} \ \text{su} \\\\ &=& \dfrac{170.18}{ 166} \ \text{su} \\\\ &\mathbf{=}& \mathbf{1.02518072289 \ \text{su}} \\ \hline \end{array}\)

There are 5 black balls and 5 white balls in a box. A person picks 5 balls from the box at random.
What is the probability that exactly 3 of 5 balls are white?

Hypergeometric distribution:
\(N\) is the population size \(= 10\),
\(K\) is the number of succeess states in the population \(= 5\),
\(n\) is the number of draws(i.e. quantitiy drawn in each trial) \(= 5\),
\(k\) is the number of abserved successes \(= 3\).

\(\begin{array}{|rcll|} \hline && \dfrac{ \binom{K}{k} \binom{ N-K}{n-k} }{ \binom{N}{n} } \\\\ &=& \dfrac{ \binom{5}{3}_{\text{white}} \binom{5}{2}_{\text{black}} } { \binom{10}{5}_{\text{all}} } \quad & | \quad \binom{5}{3}=\binom{5}{5-3}=\binom{5}{2} \\\\ &=& \dfrac{ \binom{5}{2} \binom{5}{2} } { \binom{10}{5} } \quad & | \quad \binom{5}{2}= \frac52\cdot \frac41 = 10 \\\\ &=& \dfrac{ 10\cdot 10 } { \binom{10}{5} } \quad & | \quad \binom{10}{5}=\frac{10}{5}\cdot \frac94 \cdot \frac83\cdot \frac72 \cdot \frac61 = 252 \\\\ &=& \dfrac{ 100 } { 252 } \\\\ &=& \dfrac{ 50 } { 126 } \\\\ &=& \dfrac{ 25 } { 63 } \\\\ &=& 0.39682539683\quad (=39.7\ \%) \\ \hline \end{array}\)