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if x+y=5 and x-y=3 then xy= ?

$$\boxed{\\(x+y)^2=x^2+2xy+y^2 \qquad (x-y)^2=x^2-2xy+y^2} \\\\ \dfrac{(x+y)^2-(x-y)^2}{2}=\dfrac{2xy+2xy}{2}=2xy\\\\ 2xy=\dfrac{5^2-3^2}{2}\\\\ xy=\dfrac{5^2-3^2}{4} = \dfrac{25-9}{4}=\dfrac{16}{4}=4\\\\$$

Square and Multiply Algorithm:

$$64^{235} \mod{391}$$

$$\small{\text{ $ \begin{array}{lr} Exponent&z=1\\ &64 \mod 391 = 64 = b\\ \hline 235\ odd: &\underbrace{1*64}_{=z*b} \mod 391 = 64 = z\\ &(\underbrace{64}_{=b})^2 \mod 391 = 186 = b \\ :2 & \\ \hline 117\ odd: &\underbrace{64*186}_{=z*b} \mod 391 = 174 = z\\ &(\underbrace{186}_{=b})^2 \mod 391 = 188 = b \\ :2 & \\ \hline 58\ even: &(\underbrace{188}_{=b})^2 \mod 391 = 154 = b \\ :2 & \\ \hline 29\ odd: &\underbrace{174*154}_{=z*b} \mod 391 = 208 = z\\ &(\underbrace{154}_{=b})^2 \mod 391 = 256 = b \\ :2 & \\ \hline 14\ even: &(\underbrace{256}_{=b})^2 \mod 391 = 239= b \\ :2 & \\ \hline 7\ odd: &\underbrace{208*239}_{=z*b} \mod 391= 55= z\\ &(\underbrace{239}_{=b})^2 \mod 391 = 35= b \\ :2 & \\ \hline 3\ odd: &\underbrace{55*35}_{=z*b} \mod 391= 361 = z\\ &(\underbrace{35}_{=b})^2 \mod 391 = 52= b \\ :2 & \\ \hline 1\ odd: &\underbrace{361*52}_{=z*b} \mod 391= \textcolor[rgb]{1,0,0}{4} = z\\ &(\underbrace{52}_{=b})^2 \mod 391= 358= b \\ :2 & \\ 0\ end \end{array} $ }}$$

long modulo( long b, long x, long n) { // b = basic, x = exponent, n = modulo

   long z=1; b=b % n;

   while (x !=0) {

      if (x%2 !=0) z = (z*b)%n; // Exponent odd

      b =(b*b)% n; x = x/2;

  } // while

  return z;

}

see: For example:

See:

Square-and-Multiply-Algorithm with modulo for any $$\small{\text{$b^x\mod n$}}$$

f(x)= (x^2-3)/(x+4) ;

P(6/? )

Gleichung der Tangente in P an die Kurve ?

$$f(x)= \dfrac{(x^2-3) } { (x+4) } \qquad x_p = 6 \quad y_p = ? \\ y_p = f(x_p)=f(6)= \dfrac{(6^2-3) } { (6+4) } = \dfrac{33}{10}= 3.3\\ \boxed{y_p = 3.3} \\\\ f'(x)= \dfrac{ 2x(x+4)-(x^2-3)*1 } { (x+4)^2 }\\ m_{\small{\text{Tangente}}} =f'(x_p)= \small{\text{$ f'(6)=\dfrac{2*6(6+4)-(6^2-3)*1}{(6+4)^2 }=\dfrac{12*10-33}{10^2}=\dfrac{120-33}{100}=0.87$}}\\ \boxed{m_{\small{\text{Tangente}}}=0.87}\\\\ b_{\small{\text{Tangente}}}\ ?\\ y_p = m_{\small{\text{Tangente}}} * x_p+b_{\small{\text{Tangente}}}\\ b_{\small{\text{Tangente}}}= y_p-m_{\small{\text{Tangente}}} *x_p\\ b_{\small{\text{Tangente}}}=3.3-0.87*6\\ \boxed{b_{\small{\text{Tangente}}}=-1.92}\\\\ \small{\text{ Die Gleichung der Tangente am Punkt $P(6/3.3)$ lautet: }}\\ \boxed{y=0.87x-1.92}$$

What is the mean of 17, 18, 22, 15, 17, 16, 25, 21, and 20

The mean is 19:

$$\dfrac{17+ 18+ 22+ 15+ 17+ 16+ 25+ 21 +20}{9}= 19 \\\\ \dfrac{-3-2+2-5-3-4+5+1+0}{9}+20= 19$$

64^235 mod 391

$$\small{\text{ $ \begin{array}{ll} 64^{235} \mod 391\\ =(64^5)^{47}\mod 391 &\quad | \quad (64^5) \mod 391 = 302 \\ = (302)^{47}\mod 391 \\ = 302^2 *(302)^{45}\mod 391 \\ = 302^2 *(302^3)^{15}\mod 391 &\quad | \quad (302^3) \mod 391 = 4 \\ = 302^2*4^{15}\mod 391\\ = 302^2*(4^3)^{5} \mod 391 &\quad | \quad 4^3= 64 \\ =302^2*(64)^5 \mod 391 &\quad | \quad (64^5) \mod 391 = 302 \\ =302^2*302 \mod 391\\ =302^3 \mod 391 &\quad | \quad (302^3) \mod 391 = 4 \\ = 4 \mod 391 \end{array} $ }}$$

y=_x+().   x 3,4,5,6,8 and y 4,5,6,10,15

$$\small{\text{ Determine the least squares regression line for (n=5): $ \qquad \begin{array}{rrr} & x & y \\ 1 & 3 & 4 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \\ 4 & 6 & 10 \\ 5 & 8 & 15 \\ \end{array} $ }}\\ \sum{X} = 3 + 4 + 5 + 6 + 8 = 26 \qquad \overline{X} = \frac{26}{5}=5.2\\ \sum{Y} = 4 + 5 + 6 + 10 + 15 = 40 \qquad \overline{Y} = \frac{40}{5}=8\\ \\ \small{\text{ $ \sum{ (X-\overline{X})^2 } = (3-5.2)^2 + (4-5.2)^2 + (5-5.2)^2 + (6-5.2)^2 + (8-5.2)^2 = 14.8 $ }}\\ \small{\text{ $ \sum{(Y-\overline{Y})^2} = (4-8)^2 + (5-8)^2 + (6-8)^2 + (10-8)^2 + (15-8)^2 =82 $ }}\\ \small{\text{ $ \sum{(X-\overline{X})(Y-\overline{Y})} = (3-5.2)(4-8) + (4-5.2)(5-8) + (5-5.2)(6-8) + (6-5.2)(10-8) + (8-5.2) (15-8) = 34 $ }}\\\\ \small{\text{ Standard Deviation in x und y: }}\\ \small{\text{ $ \quad SD_x=\sqrt{ \dfrac{ \sum{ (X-\overline{X})^2 } } {n-1} } = \sqrt{ \dfrac{ 14.8 }{ 5-1 } } = 1.92353840617 $ }}\\ \small{\text{ $ \quad SD_y=\sqrt{ \dfrac{ \sum{ (Y-\overline{Y})^2 } } {n-1} } == \sqrt{ \dfrac{ 82 }{ 5-1 } } = 4.52769256907 $ }} }}\\\\ \small{\text{ Correlation Coefficient, r : }}\\ \small{\text{ $ r = \dfrac{ \sum (X-\overline{X}) (Y-\overline{Y}) } { \sqrt{ \sum (X-\overline{X})^2 } \sqrt{ \sum (Y-\overline{Y})^2 } } = \dfrac{ 34 } { \sqrt{ 14.8 } \sqrt{ 82 } } = 0.97598048329 $ }}\\\\$$

$$$\\\\$ \small{\text{ PREDICTED $Y = a + b X $, }}\\ \small{\text{ where the slope $b$ and intercept $a$ are calculated in the following order: }}\\ \small{\text{ $ b = r \dfrac{ \mbox{SD}_Y}{\mbox{SD}_X} =0.97598048329 *\dfrac{4.52769256907 }{1.92353840617}=2.29729729730 $ }}\\ \\ \small{\text{ $ a = \overline{Y} - b \overline{X} = 8 - 2.29729729730* 5.2 = -3.94594594597 $ }}\\\\ \small{\text{ The regression line is }} \small{\text $ \boxed{ \text{Predicted }\quad Y = a + bX = -3.94594594597+ (2.29729729730 )X $ }}$$

How he made this !

$$\int {\frac{1}{\sqrt{4-(x-3)^2}} }\ dx =\int \frac{1}{\sqrt{4[ 1- \left( \frac{x-3}{2} \right)^2} ] }\ dx =\frac{1}{2}\int \frac{1}{\sqrt{1- \left( \frac{x-3}{2} \right)^2}}\ dx\\\\ \small{\text{We substitute: $ \frac{x-3}{2}=\sin{(u)}, $ so we have $ \frac{1}{2}\ dx=\cos{(u)}\ du $ or $ \ dx=2\cos{(u)}\ du $ }}\\ =\frac{1}{2}\int \frac{2\cos{(u)}}{\sqrt{1- \sin^2{(u)}}}\ du =\frac{1}{2}\int \frac{2\cos{(u)}}{\cos{(u)}}}\ du =\int\ du=u\\ \small{\text{We substitute back: $ u=\arcsin{ ( \frac{x-3}{2} ) } $ }}\\\\ \int {\frac{1}{\sqrt{4-(x-3)^2}} }\ dx =\sin^{-1}{ ( \frac{x-3}{2} ) }$$