heureka

Find the largest negative integer x which satisfies the congruence

$34x+6\equiv 2\pmod {20}$.

$x = -6+10n \quad | \quad n \in \mathbb{Z}$

The largest negative integer x = - 6

Check:

$\begin{array}{|rcll|} \hline && 34\cdot(-6)+6 \pmod {20} \\ &\equiv & -204 + 6 \pmod {20} \\ &\equiv & -198 \pmod {20} \\ &\equiv & -18 \pmod {20} \\ &\equiv & -18+20 \pmod {20} \\ &\equiv & 2 \pmod {20} \\ \hline \end{array}$

Find the least positive integer x that satisfies 

$x+4609 \equiv 2104 \pmod{12}$.

$\begin{array}{|lrcll|} \hline & x+4609 &\equiv& 2104 \pmod{12} \\ & x+4609-2104 &=& n \cdot 12 \quad & | \quad n \in \mathbb{N} \\ & x+2505 &=& n \cdot 12 \\ & x&=& n \cdot 12 -2505 \\\\ \text{so } & n \cdot 12 &=& 2505 \\ & n &=& \uparrow \frac{2505}{12} \uparrow \\ & n &= & \uparrow 208.75 \uparrow \\ & n &= & 209 \\\\ & x&=& 209 \cdot 12 -2505 \\ & x&=& 2508 -2505 \\ & \mathbf{x} & \mathbf{=} & \mathbf{3} \\ \hline \end{array}$

Let B, C, and D be points on a circle. Let BC and the tangent to the circle at D intersect at A.

If AB = 4,

   AD = 8,

and AC is perperpendicular to AD,

then find CD.

$\begin{array}{|rcll|} \hline AD^2 &=& AB \cdot AC & \text{secant-tangent theorem} \\ \mathbf{AC} & \mathbf{=} & \mathbf{ \frac{AD^2}{AB} }\\\\ CD^2 &=& AD^2 + AC^2 & \text{pythagoras} \\ CD^2 &=& AD^2 + AC^2 & | \quad AC= \frac{AD^2}{AB} \\ CD^2 &=& AD^2 + \left(\frac{AD^2}{AB}\right)^2 \\ CD^2 &=& AD^2 + \frac{AD^4}{AB^2} \\ CD^2 &=& AD^2\cdot \left( 1+\frac{AD^2}{AB^2} \right) \\ CD^2 &=& AD^2\cdot \left[1+\left(\frac{AD}{AB}\right)^2 \right] \\ \mathbf{CD} & \mathbf{=} & \mathbf{AD \cdot \sqrt{1+\left(\frac{AD}{AB}\right)^2 } }\\ \hline \end{array}$

CD = ?

$\begin{array}{|rcll|} \hline \mathbf{CD} & \mathbf{=} & \mathbf{AD \cdot \sqrt{1+\left(\frac{AD}{AB}\right)^2 } } \quad | \quad AB=4 \quad AD=8 \\ CD & = & 8 \cdot \sqrt{1+\left(\frac{8}{4}\right)^2 } \\ CD & = & 8 \cdot \sqrt{1+ 2^2 } \\ CD & = & 8 \cdot \sqrt{5 } \\ CD & = & 17.88854382 \\ \hline \end{array}$

In the graph below, each grid line counts as one unit.

The line shown below passes through the point  (not shown on graph).

Find n .

The line passes through the points (0,3)  and (-7,-1)

 $\begin{array}{|rcll|} \hline \text{slope } = \frac{3-(-1)}{0-(-7)} &=& \frac{n-3}{1001-0} \\ \frac{3-(-1)}{0-(-7)} &=& \frac{n-3}{1001-0} \\ \frac{4}{7} &=& \frac{n-3}{1001} \\ 1001\cdot \frac{4}{7} &=& n-3 \\ 3+ 1001\cdot \frac{4}{7} &=& n \\ \mathbf{575} & \mathbf{=} & \mathbf{n} \\ \hline \end{array}$

bäcker sontag will seine pflaumenkuchen auf eine 96cm langen und 60cm breitenkuchenblech

in gleiche grossen quadratische Stücke schneiden
wie gross kann er die stücke höchstens machen

Es wird der größte gemeinsame Teiler von 60 und 96 gesucht.

Die Zahl 96 hat 12 Teiler: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
Die Zahl 60 hat 12 Teiler: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Der größte gemeinsame Teiler von 60 und 96 ist die Zahl 12.

Die Kuchenstücke können höchstens 12cm x 12cm sein.

Find the value of x that makes each statement true

sin (x/3 + 10) = cos x

1. x₁ = ?

$\begin{array}{|rcll|} \hline \sin(\frac{x}{3}+10^{\circ}) &=& \cos(x) \quad & | \quad \cos(x) = \sin(90^{\circ}-x) \\ \sin(\frac{x}{3}+10^{\circ}) &=& \sin(90^{\circ}-x) \\ \frac{x}{3}+10^{\circ} &=& 90^{\circ}-x \\ x+\frac{x}{3}+10^{\circ} &=& 90^{\circ} \\ x+\frac{x}{3} &=& 80^{\circ} \\ x\cdot (1+\frac{1}{3}) &=& 80^{\circ} \\ x\cdot (\frac{4}{3}) &=& 80^{\circ} \\ x &=& 80^{\circ}\cdot \frac{3}{4} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{60^{\circ}} \\ \hline \end{array}$

2. x₂ = ?

$\begin{array}{|rcll|} \hline \sin(\frac{x}{3}+10^{\circ}) &=& \cos(x) \quad & | \quad \cos(x) = \cos(-x) \\ \sin(\frac{x}{3}+10^{\circ}) &=& \cos(-x) \quad & | \quad \cos(-x) = \sin(90^{\circ}-(-x)) \\ \sin(\frac{x}{3}+10^{\circ}) &=& \sin(90^{\circ}-(-x)) \\ \sin(\frac{x}{3}+10^{\circ}) &=& \sin(90^{\circ}+x) \\ \frac{x}{3}+10^{\circ} &=& 90^{\circ}+x \\ x-\frac{x}{3} &=& -80^{\circ} \\ x\cdot (1-\frac{1}{3}) &=& -80^{\circ} \\ x\cdot (\frac{2}{3}) &=& -80^{\circ} \\ x &=& -80^{\circ}\cdot \frac{3}{2} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-120^{\circ} } \\ \hline \end{array}$

identity

Formula:

$\begin{array}{|rcll|} \hline \sec(t) &=& \frac{1}{\cos(t)} \\ \tan(t) &=& \frac{\sin(t)}{\cos(t)} \\ \csc(t) &=& \frac{1}{\sin(t)} \\ \sin^2(t) &=& 1-\cos^2(t) \\ \hline \end{array}$

$\begin{array}{rcll} && \dfrac{ \tan(t) } {\sec(t)-\cos(t)} \\\\ &=& \dfrac{ \frac{\sin(t)}{\cos(t)} } {\frac{1}{\cos(t)}-\cos(t)} \\\\ &=& \dfrac{\sin(t)}{\cos(t)} \cdot \left( \frac{1} {\frac{1}{\cos(t)}-\cos(t)} \right) \\\\ &=& \dfrac{\sin(t)} {\frac{\cos(t)}{\cos(t)}-\cos(t)\cos(t)} \\\\ &=& \dfrac{\sin(t)} {1-\cos^2(t)} \\\\ &=& \dfrac{\sin(t)} {\sin^2(t)} \\\\ &=& \dfrac{1} {\sin(t)} \\\\ &=& \csc(t) \\ \end{array}$

Boncic's Identity:

ei x pi + ei x tau = 0

$\begin{array}{rcll} e^{i\pi}+ e^{i\tau} &\stackrel{?}=& 0 \quad &| \quad \tau = 2=pi \\ e^{i\pi}+ e^{i2\pi} &\stackrel{?}=& 0 \\ e^{i\pi}+ (e^{i\pi})^2 &\stackrel{?}=& 0 \quad &| \quad e^{i\pi} = -1\ (\text{Euler's Identity}) \\ -1+ (-1)^2 &\stackrel{?}=& 0 \\ -1+ 1 &=& 0\ \checkmark \\ \end{array}$

identities in Quadrant I

$\begin{array}{|rcll|} \hline \sin(x) &=& \sqrt{1-\cos^2(x)} \quad & | \quad \cos(x) = \frac{2}{5} \sqrt{6} \\ &=& \sqrt{1-\left(\frac{2}{5} \sqrt{6}\right)^2} \\ &=& \sqrt{1- \frac{4}{25} \cdot 6 } \\ &=& \sqrt{1- \frac{24}{25} } \\ &=& \sqrt{\frac{25-24}{25} } \\ &=& \sqrt{\frac{1}{25} } \\ \sin(x) &=& \frac{1}{5} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \sin(2x) &=& 2\cdot \sin(x)\cdot \cos(x) \\ &=& 2\cdot \frac{1}{5}\cdot \frac{2}{5} \sqrt{6} \\ &=& \frac{4}{25}\cdot \sqrt{6} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \cos(2x) &=& 1-2\cdot \sin^2(x) \\ &=& 1-2\cdot \left( \frac{1}{5} \right)^2 \\ &=& 1- \frac{2}{25} \\ &=& \frac{25-2}{25} \\ &=& \frac{23}{25} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \tan(2x) &=& \frac{\sin(2x)}{\cos(2x)} \\ &=& \frac{\frac{4}{25}\cdot \sqrt{6}} {\frac{23}{25}} \\ &=& \frac{4}{23}\cdot \sqrt{6}\\ \hline \end{array}$

Decompose v into two vectors v1 and v2,

where v1 is parallel to w and v2 is orthogonal to w.

v = i - j, w = i + 2j

$\vec{v} = \binom{1}{-1}\\ \vec{w} = \binom{1}{2}$

$\begin{array}{rcll} \vec{v_1} &=& \lambda \cdot \vec{w} \\ \vec{v_2} &=& \mu \cdot \vec{w_\perp} \\ \hline \vec{v}=\vec{v_1}+\vec{v_2} &=& \lambda \cdot \vec{w} + \mu \cdot \vec{w_\perp} \\ \vec{v} &=& \lambda \cdot \vec{w} + \mu \cdot \vec{w_\perp} \quad &| \quad \cdot \vec{w} \\ \vec{v}\cdot \vec{w} &=& \lambda \cdot \vec{w}\cdot \vec{w} + \mu \cdot \vec{w_\perp} \cdot \vec{w} \quad &| \quad \vec{w_\perp} \cdot \vec{w} = 0 \quad \vec{w}\cdot \vec{w} = w^2 = 1^2+2^2 = 5\\ \vec{v}\cdot \vec{w} &=& \lambda \cdot w^2 +0 \\ \vec{v}\cdot \vec{w} &=& \lambda \cdot w^2 \quad &| \quad : w^2 \\ \lambda &=& \frac{ \vec{v}\cdot \vec{w} } {w^2} \\ \mathbf{ \vec{v_1} } &\mathbf{=}&\mathbf { \left( \frac{ \vec{v}\cdot \vec{w} } {w^2} \right) \cdot \vec{w} } \\ \mathbf{ \vec{v_2} } &\mathbf{=}&\mathbf { \vec{v} - \vec{v1} }\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{ \vec{v_1} } &\mathbf{=}&\mathbf { \left( \frac{ \vec{v}\cdot \vec{w} } {w^2} \right) \cdot \vec{w} } \\ & = & \left( \frac{ \binom{1}{-1}\cdot \binom{1}{2} } {5} \right) \cdot \binom{1}{2} \\ & = & \left( \frac{ 1-2 } {5} \right) \cdot \binom{1}{2} \\ & = & \left( \frac{ -1 } {5} \right) \cdot \binom{1}{2} \\ \mathbf{ \vec{v_1} } &\mathbf{=}&\mathbf { \binom{-0.2}{-0.4} } \\\\ \mathbf{ \vec{v_2} } &\mathbf{=}&\mathbf { \binom{1}{-1} - \vec{v1} }\\ & = & \binom{1}{-1} - \binom{-0.2}{-0.4} \\ & = & \binom{1}{-1} + \binom{0.2}{0.4} \\ & = & \binom{1+0.2}{-1+0.4} \\ \mathbf{ \vec{v_2} } &\mathbf{=}&\mathbf { \binom{1.2}{-0.6} } \\ \hline \end{array}$