heureka

the expression (3/2x+1)(3/2x-1)-(3/2x-1)^2 is equivalent to?

\(\begin{array}{ll} \left( \frac32x+1 \right) \left( \frac32x-1 \right) - \left( \frac32x-1 \right)^2 \\\\ = \left( \frac32x+1 \right) \left( \frac32x-1 \right) - \left( \frac32x-1 \right)\left( \frac32x-1 \right) \qquad & | \qquad \text{factorise } \left( \frac32x-1 \right) \\\\ = \left( \frac32x-1 \right) \left[ \left( \frac32x+1 \right) - \left( \frac32x-1 \right) \right]\\\\ = \left( \frac32x-1 \right) \left( \frac32x+1 - \frac32x+1 \right)\\\\ = \left( \frac32x-1 \right) \left( 2 \right)\\\\ = 2 \left( \frac32x-1 \right) \qquad & | \qquad \text{expand} \\\\ = 2\cdot \frac32x-2\cdot 1 \\\\ = 3x-2 \\\\ \mathbf{ \left( \frac32x+1 \right) \left( \frac32x-1 \right) - \left( \frac32x-1 \right)^2 } \mathbf{ = 3x-2 } \end{array}\)

tan^2(30)= 1/cos^2(30) - 1 explain why. 

\(\begin{array}{rcl} \text{Use the Pythagorean identity } \sin^2{(\theta)} + \cos^2(\theta) =1 \\ \end{array}\)

\(\begin{array}{rcl} \sin^2{(30^{\circ})} + \cos^2(30^{\circ}) &=& 1 \qquad & | \qquad : \cos^2(30^{\circ})\\\\ \frac{\sin^2{(30^{\circ})} } {\cos^2(30^{\circ})} + \frac{ \cos^2(30^{\circ}) } {\cos^2(30^{\circ}) } &=& \frac{1} { \cos^2{(30^{\circ})} } \qquad & \frac{\sin^2{(30^{\circ})} } {\cos^2(30^{\circ})} = \tan^2{(30^{\circ})}\\\\ \tan^2{(30^{\circ})} + 1 &=& \frac{1} { \cos^2{(30^{\circ})} } \qquad & | \qquad -1\\\\ \mathbf{\tan^2{(30^{\circ})} }& \mathbf{=} & \mathbf{ \frac{1} { \cos^2{(30^{\circ})} } -1} \end{array}\)

What square is the product of four consecutive odd integers ?

\(\begin{array}{rcl} (2n-3)(2n-1)(2n+1)(2n+3) &=& x^2 \\ (2n-3)(2n+3)(2n-1)(2n+1) &=& x^2 \\ (4n^2-3^2)(4n^2-1) &=& x^2 \\ 16n^4-40n^2+3^2 &=& x^2 \qquad n = 0\\ 3^2 &=& x^2 \\ x &=& 3 \\\\ (2\cdot 0-3)(2\cdot 0-1)(2\cdot 0+1)(2\cdot 0+3) &=& 3^2 \\ (-3)\cdot (-1)\cdot (1)\cdot (3) &=& 3^2 \\ \end{array}\)

Frau Dambietz will ihr  Ferienziel in 4,5 Stunden erreichen. Sie geht von einer Durchschnittsgeschwindigkeit von 120km/h aus. Aufgrund von Staus und  dichten Verkehr schafft sie nur 90km/h wie viele Stunden kommt sie später an ?

s = Weg

v = Geschwindigkeit

t = Zeit

Weg-Zeit-Gesetz: \(s = v \cdot t\)

Streckenfahrt ohne Verzögerung : \(\begin{array}{lcll} s&=& v_1\cdot t_1 \qquad v_1 = 120\ \mathrm{\frac{km}{h}} \qquad t_1 = 4,5 \ \mathrm{h}\\ \end{array} \)

Streckenfahrt mit Staus:  \(\begin{array}{lcll} s&=& v_2\cdot t_2 \qquad v_2 = 90\ \mathrm{\frac{km}{h}} \qquad t_2 = ? \ \mathrm{h}\\ \end{array} \)

Die Gesamtstrecke s bleibt gleich:

\(\begin{array}{rlcll} s = & v_1\cdot t_1&=& v_2\cdot t_2\\ & v_1\cdot t_1&=& v_2\cdot t_2 \qquad : v_2\\ & \frac{ v_1\cdot t_1 } { v_2 } &=& t_2\\ & t_2 &=& \frac{ v_1\cdot t_1 } { v_2 } \\ & t_2 &=& \frac{ 120\ \mathrm{\frac{km}{h}} \cdot 4,5 \ \mathrm{h} } { 90\ \mathrm{\frac{km}{h}} } \\ & t_2 &=& \frac{ 120\cdot 4,5 } { 90 } \ \mathrm{h} \\ & t_2 &=& 6 \ \mathrm{h} \\ \end{array}\)

Wie viele Stunden kommt sie später an ?

Die Zeitdifferenz beträgt:  \(6 \ \mathrm{h}-4,5 \ \mathrm{h} = 1,5 \ \mathrm{h}\)

Sie kommt 1,5 h später an.

Summe einer Geometrische Reihe

\(\begin{array}{rcll} s &=& 1 &+& a + a^2+a^3+a^4+a^5+\ \cdots \ a^{n-2} + a^{n-1} \\ - a\cdot s &=& && a + a^2+a^3+a^4+a^5+\ \cdots \ a^{n-2} + a^{n-1} +a^n\\ \hline s - a\cdot s &=& 1 - a^n \\ (1-a)s &=& 1 - a^n \\ s &=& \frac{ 1 - a^n }{1-a} \\ s &=& \frac{ a^n -1 }{a-1} \\ \end{array}\)

2^0 =1 =1

2^0+2^1 =1 + 2 =3

2^0+2^1+2^2 =1 + 2 + 4 =7

2^0+2^1+2^2+2^3 =1 + 2 + 4 + 8 =15

2^0+2^1+2^2+2^3+2^4 =1 + 2 + 4 + 8 + 16 =31

2^0+2^1+2^2+2^3+2^4+2^5 =1 + 2 + 4 + 8 + 16 + 32 =63

Geometrische Reihe a = 2, Bildungsgesetz:

 \(\begin{array}{rcll} s &=& \frac{ 2^n -1 }{2-1}\\\\ s &=& \frac{ 2^n -1 }{1}\\\\ \mathbf{s} & \mathbf{=} & \mathbf{2^n -1} \\ \\ \end{array}\\ \begin{array}{lcll} n=1: & s=2^1-1 &=& 1 \\ n=2: & s=2^2-1 &=& 3 \\ n=3: & s=2^3-1 &=& 7 \\ \cdots \\ n=6: & s=2^6-1 &=& 63 \\ \end{array} \)

3^0 =1 =1

3^0+3^1 =1 + 3 =4

3^0+3^1+3^2 =1 + 3 + 9 =13

3^0+3^1+3^2+3^3 =1 + 3 + 9 + 27 =40

3^0+3^1+3^2+3^3+3^4 =1 + 3 + 9 + 27 + 81 =121

3^0+3^1+3^2+3^3+3^4+3^5 =1 + 3 + 9 + 27 + 81 + 243 =364

Geometrische Reihe a = 3, Bildungsgesetz:

\(\begin{array}{rcll} s &=& \frac{ 3^n -1 }{3-1}\\\\ \mathbf{ s }& \mathbf{=} & \mathbf{ \frac{ 3^n -1 }{2} }\\ \\ \end{array}\\ \begin{array}{lcll} n=1: & s= \frac{ 3^1-1 }{2} &=& 1 \\ n=2: & s= \frac{ 3^2-1 }{2} &=& 4 \\ n=3: & s= \frac{ 3^3-1 }{2} &=& 13 \\ \cdots \\ n=6: & s= \frac{ 3^6-1 }{2} &=& 364 \\ \end{array}\)

Skydivers use an instrument called an altimeter to determine their height above Earths surface. An altimeter measures atmospheric pressure and converts it to altitude based on the relationship between pressure and altitude. One model for atmospheric pressure P (in kilo pascals, kPa) as a function of altitude a (in kilometers) is P=100e^(-a/8).  

A. Since an altimeter measures pressure directly, pressure is the independent variable for an altimeter.   Rewrite the model P=100e^(-a/8) so that it gives altitude as a function of pressure. 

\(\small{ \begin{array}{rcll} P &=& 100 e^{\frac{-a}{8}} \qquad & | \qquad : 100 \\\\ \frac{P}{100} &=& e^{\frac{-a}{8}} \qquad & | \qquad \ln{} \\\\ \ln{ ( \frac{P}{100} ) } &=& \ln{ ( e^{\frac{-a}{8}} ) } \\ \ln{ ( \frac{P}{100} ) } &=& \frac{-a}{8} \qquad & | \qquad \cdot (-1) \\\\ -\ln{ ( \frac{P}{100} ) } &=& \frac{a}{8} \\\\ \frac{a}{8} &=& -\ln{ ( \frac{P}{100} ) } \\\\ \frac{a}{8} &=& 0 -\ln{ ( \frac{P}{100} ) } \qquad & | \qquad \ln{(1)} = 0 \\\\ \frac{a}{8} &=& \ln{(1)} -\ln{ ( \frac{P}{100} ) } \\\\ \frac{a}{8} &=& \ln{ \left( \frac{1} { \frac{P}{100} } \right) } \\\\ \frac{a}{8} &=& \ln{ ( \frac{100}{P} ) } \qquad & | \qquad \cdot 8 \\\\ \mathbf{ a } &\mathbf{=}& \mathbf{ 8\cdot \ln{ \left( \frac{100}{P} \right) } }\\ \end{array} }\)

B. To check the function in part a, use the fact that atmospheric pressure at Earths surface is about 100kPa. 

\(\small{ \begin{array}{rcll} a & = & 8\cdot \ln{ \left( \frac{100}{P} \right) } \qquad P = 100\ \mathrm{kPa} \\\\ a & = & 8\cdot \ln{ \left( \frac{100}{100} \right) } \\\\ a & = & 8\cdot \ln{ ( 1 ) } \qquad & | \qquad \ln{(1)} = 0 \\\\ a & = & 8\cdot 0 \\\\ \mathbf{ a } & \mathbf{=} & \mathbf{0\ \mathrm{km} } \end{array} }\)

C. Suppose a skydiver deploys the parachute when the altimeter measures 87kPa. Use the function in part a to determine the skydivers altitude. Give your answer in both kilometers and feet (1 kilometer=3281 feet).

\(\small{ \begin{array}{rcll} a & = & 8\cdot \ln{ \left( \frac{100}{P} \right) } \qquad P = 87\ \mathrm{kPa} \\\\ a & = & 8\cdot \ln{ \left( \frac{100}{87} \right) } \\\\ a & = & 8\cdot \ln{ ( 1.14942528736 ) } \\\\ a & = & 8\cdot 0.13926206733 \\\\ \mathbf{ a } & \mathbf{=} & \mathbf{1.11409653867\ \mathrm{km} } \\\\ a & = & 1.11409653867\ \mathrm{km}\cdot \frac{3281\ \mathrm{ft} }{ 1\ \mathrm{km}}\\\\ \mathbf{ a } & \mathbf{=} & \mathbf{3655.35074337\ \mathrm{feet} } \\\\ \end{array} } \)

Mr.Thompson is on a diet. He currently weighs 260 pounds. He loses 4 pounds per month. After how many months will Mr. Thompson reach his goal weight of 220 pounds? in standard form and slope intercept form

x = month

slope intercept form: \(y= 260-4x\)

standard form: \(4x + y= 260\)

\(\begin{array}{rcl} 220 &=& 260 - 4x \\ 4x &=& 260-220\\ 4x &=& 40 \qquad | :4 \\ x &=& \frac{40}{4} \\ \mathbf{x} & \mathbf{=} & \mathbf{10} \end{array}\)

Mr. Thompson reach his goal weight of 220 in 10 month

Lass f eine Polynom Funktion mit Echtkoeffizienten sein. Wenn -7, 9, -8 +7i und 2-2i die Nullstellen von f sind, was ist der kleinst moeglichste Grad in der Standart Form?

\(\small{ \begin{array}{lcll} \text{Sind } z_1, \ \cdots \ , z_n \text{ die Lösungen der Gleichung } z^n + a_{n – 1}z^{n – 1}+ \ \cdots \ + a_1 \cdot z + a_0 = 0 \text{ mit } a_{n – 1}, \ \cdots \ , \ a_0 \in C,\\ \text{so gilt: } (z – z_1) \cdot (z – z_2) \cdot (z – z_3) \cdot \ \cdots \ \cdot (z – z_n) = z^n + a_{n – 1} z^{n – 1} + \ \cdots \ + a_1 \cdot z + a_0\\\\ \text{Ist } z = a + b · i \text{ eine Lösung einer Polynomgleichung } a_n z^n + a_{n – 1}z^{n – 1} + \ \cdots \ + a_0 = 0 \text{ mit reellen}\\ \text{Koeffizienten, so ist auch die konjugiert komplexe Zahl } z^* = a – b · i \text{ eine Lösung der Gleichung. }\\\\ f(x) = (x+7)\cdot (x-9)\cdot [x-(-8+7i)]\cdot [x-(-8-7i)] \cdot [x-(2-2i)]\cdot [x-(2+2i)] \\ = x^{\color{red}6} +10\ x^5-30\ x^4-1194\ x^3-2039\ x^2+18604\ x-56952 = 0 \end{array} }\)

Der kleinst mögliche Grad in der Standard Form ist 6.

Siehe auch 7.5.2 (Gleichungen höheren Grads, Fundamentalsatz der Algebra) in: