heureka

Can you simplfy the following please

$$w=\frac{-1}{-1/2cos(theta)^2+(7/16)}
=
\dfrac{-1}{-\dfrac{1}{2} \cos{
( \theta^2 )}
+\dfrac{7}{16}
}
=
\dfrac{-2}
{
\dfrac{7}{8}
-
\cos{
( \theta^2 )}
}
=
\dfrac{2}
{
\cos{
( \theta^2 )}
-
\dfrac{7}{8}
}$$

Wie komme ich bei der Berechnung des Wertes einer europäischen Option von d1 auf N(d1)? Sprich mit welcher Funktion komme ich von -0.66626 auf eine standardnormale Verteilung von 0.74738 ?

Die Normalverteilung läßt sich nicht auf eine elementare Formel zurückführen, deshalb ist hier mit einer Reihenentwicklung zu rechnen:

Gegeben ist d1 = -0.66626 und gesucht ist N(-d1) = 0.74738!

$$N(-d1) = \frac{1}{2}
\left\{
1+\frac{2}{\sqrt{\pi}}
\left[
\frac{1}{1*0!}\left( \frac{0.66626 }{\sqrt{2}}\right)^1
-\frac{1}{3*1!}\left( \frac{0.66626 }{\sqrt{2}}\right)^3
+\frac{1}{5*2!}\left( \frac{0.66626 }{\sqrt{2}}\right)^5
-\frac{1}{7*3!}\left( \frac{0.66626 }{\sqrt{2}}\right)^7
+\frac{1}{9*4!}\left( \frac{0.66626 }{\sqrt{2}}\right)^9
-\frac{1}{11*5!}\left( \frac{0.66626 }{\sqrt{2}}\right)^{11}
\ \pm ...
\right]
\right\}$$

$$N(0.66626) = \frac{1}{2}
\left\{
1+\frac{2}{\sqrt{\pi}}
\left(
0.74111686404
-0.03485499086
+0.00232083205
-0.00012264558
+0.00000529304
-0.00000019224
\ \pm ...
\right)
\right\}$$

$$N(0.66626) = \frac{1}{2}
\left\{
1+\frac{2}{\sqrt{\pi}}
\left(
0.43846526044
\ \pm ...
\right)
\right\}$$

$$N(0.66626) = \frac{1}{2}
\left\{
1+0.49475506538
\right\}$$

$$N(0.66626) = 0.747378$$

Find the value of y so that the points (-5,10) and (3,y) lie on a line with slope -3/8

$$slope = \frac{\Delta y}{\Delta x} = \frac
{y-10}
{3-(-5)}
=
\frac
{y-10}
{8}
\\\\
\text{also slope } = -\frac{3}{8}$$

$$so\ we\ have \quad
\frac{y-10}{8} =
-\frac{3}{8} \qquad or\ \quad y-10=-3$$

the value of y = 7

if the frequency of an electromagnetic raditaion is 7.5x10^10Hz what is the wavelength ?

$$Hz = \frac{1}{s}$$

$$\text{speed of light: } c = 299 792 458 \ \frac{ m }{ s }
= 0,0299792458 * 10^{10}\ \frac{ m }{ s }$$

$$\text{frequency : } f = 7,5 * 10^{10}\ \frac{ 1 }{ s }$$

$$\text{wavelength: } \lambda = \dfrac{ c }{ f }
=
\frac { 0,0299792458 * 10^{10}\ \frac{ m }{ s } }
{7,5 * 10^{10}\ \frac{ 1 }{ s } }
=
\frac { 0,0299792458 } { 7,5 } \ m =0.00399723277\ m$$

the wavelength is 0.00399723277 m   or   $$3.99723277 * 10^{-3} \ m$$

Find the LCM of 8, 3, 4 ?

Least Common Multiple, you need the prime factorization:

$$\mbox{prime factorization} \\\\
\begin{array}{rccccc}
8 =& 2^{\textcolor[rgb]{1,0,0}{3}}\ *& 3^0\\
3 =& 2^0 \ *& 3^{\textcolor[rgb]{1,0,0}{1}}\\
4 = & 2^2 \ *& 3^0\\
\hline
\mbox{in each case the biggest exponent}
& 2^{\textcolor[rgb]{1,0,0}{3}}
\ *& 3^{\textcolor[rgb]{1,0,0}{1}}
&= 24 = LCM \\
\end{array}$$

LCM of 8, 3, 4

24= 3 * 8

24= 8 * 3

24= 6 * 4

find the intercepts and use them to graph the equation

$$\begin{array}{rcl}
y & = & 3x+3 \quad | \quad -3x\\ \\
y-3x &=& 3 \quad | \quad :3\\ \\
\frac{y-3x}{3} &=&1\\\\
\frac{y}{3} + \frac{-3x}{3} &=&1\\\\
\frac{y}{3} + \frac{-x}{1} &=&1\\\\
\frac{y}{3} + \frac{x}{-1} &=&1\\\\
\frac{x}{-1} + \frac{y}{3} &=&1
\end{array}$$

$$\boxed{\dfrac{x}{ x_{intercept} } + \dfrac{y}{ y_{intercept} } &=&1 }\\\\
x_{intercept} = -1\\
y_{intercept} = 3$$

I. How do you find the area of a figure by only using points?

For n points, the points must be oriented:

 $$\boxed{2A= \sum \limits_{i=1}^n(y_i+y_{i+1})(x_i-x_{i+1}) \quad x_{n+1} = x_1 \quad y_{n+1} = y_1}$$ [Carl Friedrich Gauß]

For example (-5,3) (-1,2) (0,6), the points are oriented!

$$\begin{array}{|c|r|r|}
\hline
Point & x & y \\
\hline
1 &-5&3 \\
2 &-1&2 \\
3 &0&6 \\
\hline
\end{array}$$ 

$$\begin{array}{rcl}
2A & = & (y_1+y_2)(x_1-x_2)+(y_2+y_3)(x_2-x_3)+(y_3+y_1)(x_3-x_1) \\
2A & = & (3+2)(-5+1)+(2+6)(-1-0)+(6+3)(0+5)\\
2A & = & (5)(-4)+(8)(-1)+(9)(5)\\
2A & = & -20 - 8 + 45\\
2A & = & 17\\
A & = & 8.5
\end{array}$$

II. How do you find the Perimeter of a figure by only using points?

$$\boxed{Perimeter = \sum \limits_{i=1}^n
\sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2}\quad x_{n+1} = x_1 \quad y_{n+1} = y_1}
}$$

Example:

$$\begin{array}{|c|r|r|}
\hline
Point & x & y \\
\hline
1 &-5&3 \\
2 &-1&2 \\
3 &0&6 \\
\hline
\end{array}$$

$$\begin{array}{rcl}
Perimeter & = & \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}+ \sqrt{(x_3-x_2)^2+(y_3-y_2)^2}+ \sqrt{(x_1-x_3)^2+(y_1-y_3)^2} \\

& = & \sqrt{(-1+5)^2+(2-3)^2}+ \sqrt{(0+1)^2+(6-2)^2}+ \sqrt{(-5-0)^2+(3-6)^2} \\
& = & \sqrt{(4)^2+(-1)^2}+ \sqrt{(1)^2+(4)^2}+ \sqrt{(-5)^2+(-3)^2} \\
& = & \sqrt{16+1}+ \sqrt{1+16}+ \sqrt{25+9} \\
& = & \sqrt{17}+ \sqrt{17}+ \sqrt{34} \\
Perimeter & = & = 14.0771631461
\end{array}$$

If the x-intercept is 4,what could the y-intercept be ?

$$\text{x-intercept} = -\dfrac{
\text{y-intercept}
}{m}
\qquad \text{or} \qquad \text{y-intercept} = -4m$$

The digits of a two-digit number differ by 3 . If the digits are interchanged and the resulting number is added to the original number, we get 143 . What can be he original number ?

digit 1 = a

digit 2 = b

a - b = 3

Number 1   ab    is a*10 + b

Number 2   ba    is b*10 + a

Number 1 + Number 2 =  (a*10 + b) + (b*10 + a) = 143

                                                 10(a+b) + (a+b) = 143

                                                              11(a+b) = 143

                                                                   (a+b) = 143/11 = 13

I. a + b = 13

II.  a - b = 3

----------------

I+II: 2a = 16  or  a = 8

I-II:  2b = 10  or  b = 5

The Number is 85 or 58

Hallo asinus,

es gibt hier eine Methode, die Gleichung $$\boxed{h^3 - 25,5h^2 + 1102,950 = 0}$$ zu Fuß aufzulösen:

$$\boxed{h^3 - 25,5h^2 + 1102,950 = 0 \qquad b=-25,5 \quad d=1102,95 }$$

$$\boxed{
B=-\frac{b^2}{3}=-216,75 \qquad D=\frac{2b^3}{27}+d=-125,3 }$$

$$\boxed{a=\sqrt{-\frac{4}{3}B} = 17}$$

$$\boxed{q=\frac{D}{a^3}= -0,02550376552}$$

$$\boxed{4q=-0,10201506208}$$ 

Ist 4q<1, erhalten wir drei reelle Lösungen:

$$\boxed{\phi=arcsin(4q)=-5,85521855948\ensurement{^{\circ}}}$$

$$\boxed{
h_1=a\sin(\frac{1}{3}\phi)-\frac{b}{3}=17*(-0,03405769325) + 8,5=7,92101921469}$$

$$\boxed{
h_2=a\sin(\frac{1}{3}(\phi+360\ensurement{^{\circ}}))-\frac{b}{3}=17*(0,88255184178) + 8,5=23,5033813103}$$

$$\boxed{
h_3=a\sin(\frac{1}{3}(\phi+720\ensurement{^{\circ}}))-\frac{b}{3}=17*(-0,84849414853) + 8,5=-5,92440052504}$$