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Let A, B, C, and D be points on a circle such that AB = 11 and CD = 19.
Point P is on segment AB with AP = 6, and Q is on segment CD with CQ = 7.
The line through P and Q intersects the circle at X and Y. If PQ = 27, find XY

\(\begin{array}{|rcll|} \hline \text{Let AB } &=& 11 \\ \text{Let AP } &=& 6 \\ \text{Let PB } &=& \text{ AB } - \text{ AP } \\ &=& 11 - 6 \\ &=& 5 \\\\ \text{Let CD } &=& 19 \\ \text{Let CQ } &=& 7 \\ \text{Let QD } &=& \text{ CD } - \text{ CQ } \\ &=& 19 - 7 \\ &=& 12 \\\\ \text{Let PQ } &=& 27 \\\\ \text{Let XP } &=& x \\ \text{Let YQ } &=& y \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & \text{AP } \cdot \text{PB } &=& x \cdot \text{PY } \quad & | \quad \text{PY } = y + \text{ PQ } \\ & 6 \cdot 5 &=& x \cdot ( y + \text{ PQ } ) \\ & 30 &=& x \cdot ( y + 27 ) \\ & x &=& \frac{30}{y + 27} \\\\ (2) & \text{CQ } \cdot \text{QD } &=& y \cdot \text{QX } \quad & | \quad \text{QX } = x + \text{ PQ } \\ & 7 \cdot 12 &=& y \cdot ( x + \text{ PQ } ) \\ & 84 &=& y \cdot ( x + 27 ) \quad & | \quad x = \frac{30}{y + 27} \\ & 84 &=& y \cdot ( \frac{30}{y + 27} + 27 ) \\ & 84 &=& y \cdot \left( \frac{30}{y + 27} + 27 \right) \\ & 84 &=& y \cdot \left( \frac{30 +27\cdot (y + 27 ) }{y + 27} \right) \\ & 84 &=& y \cdot \left( \frac{30 +27y +27^2 }{y + 27} \right) \\ & 84 &=& y \cdot \left( \frac{759 +27y}{y + 27} \right) \\ & 84\cdot(y + 27) &=& y \cdot ( 759 +27y ) \\ & 84y + 84\cdot 27 &=& 759y +27y^2 \\ & 27y^2 +675y - 84\cdot 27 &=& 0 \quad & | \quad : 27 \\ & y^2 + 25y - 84 &=& 0 \\\\ & y_{1,2} &=& \frac{-25 \pm \sqrt{25^2 -4 \cdot(-84) } } {2} \\ & y_{1,2} &=& \frac{-25 \pm 31 } {2} \quad & | \quad y > 0!\\ & y &=& \frac{-25 + 31 } {2} \\ & \mathbf{y} &\mathbf{=}& \mathbf{3} \\\\ & x &=& \frac{30}{y + 27} \quad & | \quad y = 3 \\ & x &=& \frac{30}{3 + 27} \\ & x &=& \frac{30}{30} \\ & \mathbf{x} &\mathbf{=}& \mathbf{1} \\\\ & \text{XY } &=& x + \text{ PQ } + y \\ & \text{XY } &=& 1 + 27 + 3 \\\\ & \mathbf{ \text{XY } } & \mathbf{=} & \mathbf{31} \\ \hline \end{array} \)

sum of the numbers starting from 14 to 180?

\(\small{ \begin{array}{|rcll|} \hline && 14+15+16 +\cdots +178 +179+180 \\ &=& (14-13+13) + (15-13+13) + (16-13+13) +\cdots + (178-13+13) + (179-13+13) + (180-13+13) \\ &=& (1+13) + (2+13) + (3+13) +\cdots + (165+13) + (166+13) + (167+13) \\ &=& 167\cdot 13 + (1+2+3+4+\cdots + 165+166+167)\\ &=& 167\cdot 13 + \frac{1+167}{2}\cdot 167 \\ &=& 2171 + 14028 \\ &=& 16199 \\ \hline \end{array} } \)

When you subtract one square number from another the answer is 35. What are the two square numbers?

\(\begin{array}{rclrr} a^2 - b^2 &=& 35 \\ \underbrace{(a-b)}_{=5}\cdot \underbrace{(a+b)}_{=7} &=& 35 = 5\cdot 7 \\ \end{array}\)

\(\begin{array}{lrcl|r|r} & && & I. & II. \\ \hline & a-b &=& 5 & \\ & && & + & - \\ & a+b &=& 7 & \\ \hline \end{array} \)

\(\begin{array}{lrcl} I. \\ & 2a &=& 7+5 \\ & 2a &=& 12 \\ & a &=& 6 \\\\ II.\\ & 2b &=& 7-5\\ & 2b &=& 2 \\ & b &=& 1 \end{array}\)

\(\begin{array}{|rcll|} \hline 6^2 - 1^2 &=& 35 \\ \hline \end{array}\)

(〖(-9)〗^3×〖(-12)〗^4×〖(10)〗^(-2))/(〖(15)〗^(-2)×〖18〗^2 )

\(\begin{array}{|rcll|} \hline && \displaystyle \frac{ (-9)^3 \cdot (-12)^4 \cdot 10^{-2} } { 15^{-2} \cdot 18^2 } \\\\ &=& \displaystyle \frac{ (-9)^3 \cdot (-12)^4 \cdot 10^{-2} } { 15^{-2} \cdot 18^2 } \\\\ &=& \displaystyle \frac{ (-9)^3 \cdot (-12)^4 \cdot 15^{2} } { 10^{2} \cdot 18^2 } \\\\ &=& \displaystyle \frac{ (-9)^3 \cdot 12^4 \cdot 15^{2} } { 10^{2} \cdot 18^2 } \\\\ &=& \displaystyle -\frac{ 9^3 \cdot 12^4 \cdot 15^{2} } { 10^{2} \cdot 18^2 } \\\\ &=& \displaystyle -\frac{ 729\cdot 20736 \cdot 225 } { 100 \cdot 324 } \\\\ &=& \displaystyle -\frac{ 729\cdot 20736 \cdot 9\cdot 25 } { 4\cdot 25 \cdot 324 } \\\\ &=& \displaystyle -\frac{ 729\cdot 20736 \cdot 9 } { 4 \cdot 324 } \\\\ &=& \displaystyle -\frac{ 729\cdot 81\cdot 256 \cdot 9 } { 4 \cdot 81\cdot 4 } \\\\ &=& \displaystyle -\frac{ 729\cdot 256 \cdot 9 } { 4 \cdot 4 } \\\\ &=& \displaystyle -\frac{ 729\cdot 256 \cdot 9 } { 16 } \\\\ &=& \displaystyle -\frac{ 729\cdot 16\cdot 16 \cdot 9 } { 16 } \\\\ &=& \displaystyle - 729\cdot 16 \cdot 9 \\\\ &=& \displaystyle - 104976 \\ \hline \end{array} \)

(x^2+3)^2  auflösen ?     x^4+9 ist ja schwachsinn

\(\begin{array}{|rcll|} \hline && (x^2+3)^2 \\ &=& (x^2+3)\cdot (x^2+3) \\ &=& x^2\cdot (x^2+3) + 3 \cdot (x^2+3) \\ &=& x^2\cdot x^2 + x^2\cdot 3 + 3 \cdot (x^2+3) \\ &=& x^4 + 3x^2 + 3 \cdot (x^2+3) \\ &=& x^4 + 3x^2 + 3 \cdot x^2 + 3 \cdot 3 \\ &=& x^4 + 3x^2 + 3x^2 + 9 \\ &=& x^4 + 2\cdot 3x^2 + 9 \quad & | \quad \text{ 1. Binom } (a+b)^2 = a^2 + 2ab + b^2 !\\ &=& x^4 + 6x^2 + 9 \\ \hline \end{array}\)

for every 45 points there is 30 points how much is one point

\(\begin{array}{|r|r|} \hline \text{points} & \text{points} \\ \hline 45 & 30 \\ 90 & 60 \\ 135 & 90 \\ \cdots & \cdots \\ x & x\cdot \frac{30}{45} \\ 1 & 1\cdot \frac{30}{45} \\ \hline \end{array} \)

would the square root of -12 be equal to 12i?

\(\begin{array}{|rcll|} \hline && \sqrt{-12} \\ &=& \sqrt{(-1)\cdot 12 } \\ &=& \sqrt{12\cdot (-1) } \\ &=& \sqrt{12}\cdot \sqrt{ -1 } \quad & | \quad \sqrt{ -1 } = i \\ &=& \sqrt{12}\cdot i \\ \hline \end{array} \)

given for one ticket, probability to win is 1/9, what is the probability og winning twice if we bought 10 tickets

\(\begin{array}{|rcll|} \hline && \displaystyle \binom{10}{2} \cdot \left(\frac{1}{9} \right)^2 \cdot \left(\frac{8}{9}\right)^8 \\\\ &=& \displaystyle \frac{10}{2}\cdot \frac{9}{1} \cdot \frac{8^8}{9^{10}} \\\\ &=& \displaystyle \frac{45\cdot 8^8}{9^{10}} \\\\ &=& \displaystyle \frac{754\, 974\, 720}{3\, 486\, 784\, 401} \\\\ &=& 0.21652463507 \qquad (21.65\ \%) \\ \hline \end{array}\)

what is the magnitude of j^(5/8)? please give an explanation.

\(\begin{array}{|rcll|} \hline j^{(\frac{5}{8})} = \sqrt[8]{j^5} = \sqrt[8]{j}= \sqrt[8]{z}\quad & | \quad j^5 = j^2\cdot j^2 \cdot j = (-1)\cdot(-1) \cdot j = 1\cdot j = j \\ \hline \end{array}\)

Formula:

\(\begin{array}{|lcll|} \hline z &=& a +bj \qquad |z| = \sqrt{a^2+b^2} \qquad |z| = \text{ magnitude of z } \\ \hline \end{array}\)

\(\begin{array}{|lcll|} \hline z &=& j \\ &=& 0+1\cdot j \qquad a = 0 \qquad b = 1 \\\\ |z| &=& \sqrt{0^2+1^2} \\ &=& \sqrt{1} \\ |z| &=& 1 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \sqrt[8]{z} &=&\sqrt[8]{j} \\ &=& \sqrt[8]{|z|} \cdot \left( \cos( \frac{\varphi}{8} ) + i\cdot \sin(\frac{\varphi}{8}) \right) \\ &=& \sqrt[8]{1} \cdot \left( \cos( \frac{\varphi}{8} ) + i\cdot \sin(\frac{\varphi}{8}) \right) \\ &=& 1 \cdot \left( \cos( \frac{\varphi}{8} ) + i\cdot \sin(\frac{\varphi}{8}) \right) \\ \hline \end{array} \)

The magnitude of \(j^{\frac{5}{8}} = j^{\frac{1}{8}} = \sqrt[8]{j}\) is:

\(\begin{array}{rcll} \sqrt[8]{|z|} &=& \sqrt[8]{1} \\ &=& 1 \\ \end{array}\)

I have a physics question if anybody can help me out.

At 1m away a politiction produces a sound intensity at 80 dB in a park without obstables.

Approx what is the sound intensity at 30 meters away? I'm considering using 1 over R formula,

however it doesn't give an answer in dB.

Sound I be using logarithms?

Formula:

\(\begin{array}{|rrcll|} \hline Let & I_1 &=& 80\ dB \\ Let & r_1 &=& 1\ m \\ Let & I_2 &=& ? \\ Let & r_2 &=& 30\ m \\\\ & I_2 &=& I_1 - 10 \cdot |~ \log( (\frac{d_1}{d_2})^2 ) ~| \qquad \text{ sound intensity }\\\\ & I_2 &=& 80\ dB - 10 \cdot |~\log( (\frac{1\ m}{30\ m})^2 )~| \\ & I_2 &=& 80\ dB - 10 \cdot |~\log( 0.00111111111 )~| \\ & I_2 &=& 80\ dB - 10 \cdot |~(-2.95424250944)~| \\ & I_2 &=& 80\ dB - 10 \cdot 2.95424250944 \\ & I_2 &=& 80\ dB - 29.5424250944 \\\\ & \mathbf{ I_2 } & \mathbf{ = } & \mathbf{ 50.4575749056\ dB } \\ \hline \end{array} \)