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A group of adults and kids went to see a movie. Tickets cost $7.00 each for adults and $4.50 each for kids, and the group paid $64.00 in total. There were 4 fewer adults than kids in the group.
Find the number of adults and kids in the group.

$\begin{array}{lrcl} a = \text{adults}\\ k = \text{kids}\\ \hline (1) & a\cdot $7.00 + k\cdot $4.50 &=& $64.00\\ (2) & a &=& k-4 \\ \hline & a\cdot $7.00 + k\cdot $4.50 &=& $64.00 \qquad | \qquad a &=& k-4\\ & (k-4)\cdot $7.00 + k\cdot $4.50 &=& $64.00\\ & k\cdot $7.00 -4\cdot $7.00 + k\cdot $4.50 &=& $64.00\\ & k\cdot $7.00 + k\cdot $4.50 -4\cdot $7.00&=& $64.00\\ & k\cdot $11.5 -4\cdot $7.00&=& $64.00 \\ & k\cdot $11.5 - $28.00&=& $64.00 \qquad | \qquad + $28.00\\ & k\cdot $11.5 &=& $64.00 + $28.00\\ & k\cdot $11.5 &=& $92.00 \qquad | \qquad : $11.5\\ & k &=& \frac{ $92.00}{ $11.5 } \\ & \mathbf{k} &\mathbf{=}& \mathbf{8} \\ \hline & a &=& k-4 \qquad | \qquad k = 8\\ & a &=& 8-4 \\ & \mathbf{a} &\mathbf{=}& \mathbf{4} \\ \end{array}$

The number of adults is 4

and the number of kids in the group is 8

what is the unit digit in 2^2015

$\begin{array}{rcll} 2^{2015} \pmod {10}\\ &\equiv& 2^{13\cdot 155} \pmod {10} \qquad \boxed{~2^{13} \pmod {10} \equiv 8192 \pmod {10}\equiv 2 \pmod {10}~} \\ &\equiv& \left(2^{13}\right)^{155} \pmod {10} \qquad | \qquad 2^{13} \pmod {10}\equiv 2 \pmod {10}\\ &\equiv& 2^{155} \pmod {10}\\ &\equiv& 2^{13\cdot 11+12} \pmod {10}\\ &\equiv& 2^{13\cdot 11}\cdot 2^{12} \pmod {10}\\ &\equiv& \left(2^{13}\right)^{11}\cdot 2^{12} \pmod {10} \qquad | \qquad 2^{13} \pmod {10}\equiv 2 \pmod {10}\\ &\equiv& 2^{11}\cdot 2^{12} \pmod {10}\\ &\equiv& 2^{11+12} \pmod {10}\\ &\equiv& 2^{23} \pmod {10}\\ &\equiv& 2^{13+10} \pmod {10}\\ &\equiv& 2^{13}\cdot 2^{10} \pmod {10} \qquad | \qquad 2^{13} \pmod {10}\equiv 2 \pmod {10}\\ &\equiv& 2\cdot 2^{10} \pmod {10}\\ &\equiv& 2^{11} \pmod {10} \\ &\equiv& 2048\pmod {10}\\ &\equiv& 8\pmod {10}\\ \mathbf{2^{2015} \pmod {10}} & \mathbf{\equiv} & \mathbf{8\pmod {10}}\\ \end{array}$

4:5=6:x ?

$\begin{array}{rcll} 4:5 &=& 6:x \qquad \text{or}\\ \frac45 &=& \frac6x \qquad \updownarrow\\ \frac54 &=& \frac{x}6 \qquad | \qquad \cdot 6\\ 6\cdot \frac54 &=& x \\ x &=& 6\cdot \frac54 \\ x &=& 3\cdot \frac52 \\ x &=& \frac{15}2 \\ \mathbf{x} &\mathbf{=}& \mathbf{7.5} \\ \end{array}$

|2x+5| = 3x+4

$\begin{array}{rcl} |2x+5| &=& 3x+4 \qquad | \qquad ()^2\\ (2x+5)^2 &=& (3x+4)^2 \\ 4x^2+20x+25 &=& 9x^2+24x+16\\ 5x^2+4x-9 &=& 0 \\\\ \boxed{~ x = {-b \pm \sqrt{b^2-4ac} \over 2a} ~}\\ 5x^2+4x-9 &=& 0 \\ x_{1,2} &=& {-4 \pm \sqrt{16-4\cdot 5 \cdot(-9)} \over 2\cdot 5}\\ x_{1,2} &=& {-4 \pm \sqrt{16+180} \over 10}\\ x_{1,2} &=& {-4 \pm 14 \over 10}\\\\ x_1 &=& {-4 +14 \over 10}\\ x_1 &=& {10 \over 10}\\ \mathbf{x_1} &\mathbf{=}& \mathbf{1}\\ x_2 &=& {-4 -14 \over 10}\\ x_2 &=& {-18 \over 10}\\ \mathbf{x_2} &\mathbf{=}& \mathbf{-1.8} \qquad \text{no solution}\\ \end{array}$

Rotate 90 counter clockwise about the origin 

$\begin{array}{rcl} \dbinom{x_p}{y_p}\cdot \begin{pmatrix} \cos{(90)} & \sin{(90)}\\ - \sin{(90)} & \cos{(90)} \\ \end{pmatrix}\\\\ = \dbinom{x_p}{y_p}\cdot \begin{pmatrix} 0&1\\ -1& 0\\ \end{pmatrix}\\\\ =\dbinom{-y_p}{x_p} \end{array}$

Determine where f(x) intersects the x-axis ( f(x) =0) .

f(x)=x^2+5x+6

$\begin{array}{rcl} &x^2+5x+6 &=& 0\\ \text{Factorisation }& (x+2)(x+3) &=& 0\\ &x_1 &=& -2\\ &x_2 &=& -3 \end{array}$

Determine where f(x) intersects the x-axis ( f(x) = 0 ).

f(x)=x^2+9x+18

$\begin{array}{rcl} &x^2+9x+18 &=& 0\\ \text{Factorisation }& (x+3)(x+6) &=& 0\\ &x_1 &=& -3\\ &x_2 &=& -6 \end{array}$

can you put brackets in this equation to make it correct: 24/6-2^2-1^2=9

$(~(~24/6-2 ~)^2-1~)^2=9$

frequency

$440\ \text{Hz} \cdot 2^{\frac{12}{12}} = 880\ \text{Hz}$

How can I describe a function in which its value of 1 doubles the functions value, but greater than 1 does nothing

$2\cdot f(x) *\left[ \frac{1~-~1 \pmod x}{x} \right] \qquad x \ge 1$