heureka

What does PIE equal?

\(\pi = 4\cdot \arctan(1) \qquad \text{angle in rad}\)

Triangle ABC has a right angle at angle B. Legs AB and CB are extended past point B to points D and E, respectively, such that angle EAC = angle ACD = 90 degrees. Prove that EB * BD = AB * BC.

Below is the diagram representing the described triangle.

Geometric mean theorem:

\(\begin{array}{|lrcll|} \hline (1) & AB^2 &=& EB \cdot BC \\ (2) & BC^2 &=& AB\cdot BD \\ \hline (1) \cdot (2): & AB^2\cdot BC^2 &=& EB \cdot BC\cdot AB\cdot BD \quad & | \quad : (AB\cdot BC) \\ & AB\cdot BC &=& EB\cdot BD \\ \hline \end{array}\)

Help.

Geometric mean theorem:

\(\begin{array}{|rcll|} \hline 5^2 &=& 12\cdot x \\ x &=& \frac{25}{12} \\ \hline \end{array}\)

Cathetus Theorem:

\(\begin{array}{|rcll|} \hline y^2 &=& x\cdot (x+12) \\ y^2 &=& \frac{25}{12}\cdot (\frac{25}{12}+12) \\ y^2 &=& \frac{25}{12}\cdot (\frac{25+144}{12}) \\ y^2 &=& \frac{25}{12}\cdot (\frac{169}{12}) \\ y^2 &=& \frac{5^2}{12^2}\cdot 13^2 \\ y &=& \frac{5 }{12 }\cdot 13 \\ y &=& \frac{65}{12 }\\ \hline \end{array}\)

scientific notation 0.5401

\(0.5401 = 5.401\cdot 10^{-1}\)

if given the numbers 8,3,2,and 22 how do i find 34

\(\begin{array}{|rcll|} \hline 34 &=& (22+3-8)\cdot 2 \\ \hline \end{array} \)

please answer and simplify squareroot of -2 times square root of -12. Remember i factor.

\(\sqrt{-2}\cdot \sqrt{-12} = \sqrt{(-2)\cdot(-12)}=\sqrt{24}=\sqrt{4\cdot 6}=\sqrt{4}\cdot \sqrt{6}=2\cdot \sqrt{6}=4.89897948557\)

The graph:

*Find the equation of the tangent to the circle above that passes through the beginning of axis O (0,0)

1. Center and radius of the circle:

\(\begin{array}{|rcll|} \hline x^2+y^2-6x-2y+8 &=& 0 \\ x^2-6x+y^2-2y &=& -8 \\ (x-\frac{6}{2})^2-3^2 +(y-\frac{2}{2})^2 -1 &=& -8 \\ (x-3)^2 +(y-1)^2 &=& 9+1-8 \\ (x-3)^2 +(y-1)^2 &=& 2 \\\\ Center = C(3,1) && Radius\ r= \sqrt{2} \\ \hline \end{array}\)

\(OC = \sqrt{x_c^2+y_c^2} =\sqrt{3^2+1^2}= \sqrt{10}\)

\(\begin{array}{|rcll|} \hline \tan(\alpha) &=& \frac{y_c}{x_c} \\ &=& \frac{1}{3} \\ \tan(\beta) &=& \frac{r}{\sqrt{OC^2-r^2}} \\ &=& \frac{\sqrt{2}}{\sqrt{10-2}} \\ &=& \frac{1}{2} \\ \hline \end{array} \)

2. Tangent above slope m:

\(\begin{array}{|rcll|} \hline m=\tan{(\alpha+\beta)} &=& \frac{\tan(\alpha)+\tan(\beta)} {1-\tan(\alpha)\tan(\beta)} \\ m &=& \frac{\frac{1}{3}+\frac{1}{2}} {1-\frac{1}{3}\cdot \frac{1}{2}} \\ m &=& 1 \\\\ y &=& mx +b \quad & | \quad m = 1 \qquad b = 0 \\ \mathbf{y} &\mathbf{=}& \mathbf{x} \\ \hline \end{array} \)

3. Tangent below slope m:

\(\begin{array}{|rcll|} \hline m=\tan{(\alpha-\beta)} &=& \frac{\tan(\alpha)+\tan(\beta)} {1+\tan(\alpha)\tan(\beta)} \\ m &=& \frac{\frac{1}{3}-\frac{1}{2}} {1+\frac{1}{3}\cdot \frac{1}{2}} \\ m &=& -\frac{1}{7} \\\\ y &=& mx +b \quad & | \quad m = -\frac{1}{7} \qquad b = 0 \\ \mathbf{y} &\mathbf{=}& \mathbf{-\frac{1}{7}\cdot x} \\ \hline \end{array}\)

factoring 5y^2-80

\(\begin{array}{|rcll|} \hline && 5y^2-80 \\ &=& 5y^2-4^2\cdot 5 \\ &=& 5\cdot(y^2-4^2) \\ &=& 5\cdot(y-4)\cdot (y+4) \\ \hline \end{array} \)

what is 3004 in binary?

\(\begin{array}{|rclclcl|} \hline 3004 &:& 2 &=& 2\cdot 1502 &+& 0 \\ 1502 &:& 2 &=& 2\cdot 751 &+& 0 \\ 751 &:& 2 &=& 2\cdot 375 &+& 1 \\ 375 &:& 2 &=& 2\cdot 187 &+& 1 \\ 187 &:& 2 &=& 2\cdot 93 &+& 1 \\ 93 &:& 2 &=& 2\cdot 46 &+& 1 \\ 46 &:& 2 &=& 2\cdot 23 &+& 0 \\ 23 &:& 2 &=& 2\cdot 11 &+& 1 \\ 11 &:& 2 &=& 2\cdot 5 &+& 1 \\ 5 &:& 2 &=& 2\cdot 2 &+& 1 \\ 2 &:& 2 &=& 2\cdot 1 &+& 0 \\ 1 &:& 2 &=& 2\cdot 0 &+& 1 \\ \hline \end{array}\)

\(3004_{10}=101110111100_2\)