heureka

Find v if the gradient between the points (5,2) and (11,v) is 3

$\begin{array}{|rcl|} \hline P_1 = \binom{5}{2} \qquad x_1 = 5 \qquad y_1 = 2\\ P_2 = \binom{11}{v} \qquad x_2 = 11 \qquad y_2 = v\\ \hline \text{gradient} &=& 3 \\ \text{gradient} &=& \frac{\Delta y}{\Delta x} \\ 3 &=& \frac{y_2-y_1}{x_2-x_1} \\ 3 &=& \frac{v-2}{11-5} \\ 3 &=& \frac{v-2}{6} \\ 3\cdot 6 &=& v-2 \\ 18 &=& v-2 \\ 18+2 &=& v \\ 20 &=& v \\ \mathbf{v} &\mathbf{=}& \mathbf{20} \\ \hline \end{array}$

A theatre charges $25 for adults and $15 for children. All 200 seats in the theatre were sold on opening night and the ticket sales totalled $4350. Calculate the number of adult tickets sold.

$\begin{array}{lrcll} x = \text{adults}\\ y = \text{children}\\ \\ \hline (1) & x + y &=& 200 \qquad & |\qquad -x \\ & y &=& 200 - x \\ \hline (2) & x\cdot 25 + y \cdot 15 &=& 4350 \qquad & |\qquad : 3 \\ & x\cdot 5 + y \cdot 3 &=& 870 \qquad & |\qquad y = 200 - x\\ & x\cdot 5 + ( 200 - x) \cdot 3 &=& 870 \\ & x\cdot 5 + 200 \cdot 3 - x \cdot 3 &=& 870 \\ & 5x + 200 \cdot 3 - 3x &=& 870 \\ & 5x + 600 - 3x &=& 870 \qquad & |\qquad -600 \\ & 5x - 3x &=& 870 -600 \\ & 2x &=& 270 \qquad & |\qquad : 2 \\ & \mathbf{ x } & \mathbf{=}& \mathbf{135}\\ \hline & y &=& 200 -x \qquad & |\qquad x = 135 \\ & y &=& 200-135 \\ & \mathbf{ y } &\mathbf{=}& \mathbf{65}\\ \hline \end{array}$

The number of adult tickets sold is 135.

Show that (2x-10)/(10x-1-5x) can be reduced to 10/5x

$\begin{array}{rcll} && \frac{2^x-10} {10^{x-1}-5^x } \qquad | \qquad \cdot \frac{2^x}{2^x}\\\\ &=& \frac{2^x\cdot (2^x-10)} {2^x(10^{x-1}-5^x) } \\\\ &=& \frac{2^x\cdot (2^x-10)} {2^x\cdot 10^{x-1}- 2^x\cdot 5^x } \\\\ &=& \frac{2^x\cdot (2^x-10)} {2^x\cdot 10^{x-1}- (2\cdot 5)^x } \\\\ &=& \frac{2^x\cdot (2^x-10)} {2^x\cdot 10^{x-1}- 10^x } \\\\ &=& \frac{2^x\cdot (2^x-10)} {2^x\cdot 10^{x-1}- 10\cdot 10^{x-1} } \\\\ &=& \frac{2^x\cdot (2^x-10)} { 10^{x-1} ( 2^x - 10 ) } \\\\ &=& \frac{2^x} { 10^{x-1} } \\\\ &=& \frac{2^x} { \frac{10^x}{10} } \\\\ &=& 10\cdot \frac{2^x}{10^x} \\\\ &=& 10\cdot \left(\frac{2}{10} \right)^x \\\\ &=& 10\cdot \left(\frac{1}{5} \right)^x \\\\ &=& 10\cdot \left(\frac{1}{5^x} \right) \\\\ &=& \dfrac{10}{5^x} \end{array}$

$\begin{array}{|rrcrrll|} \hline 1 &:97 &=& 0. &\text{ remainder }& 1 \\ 1\cdot 10 &:97 &=& 0 &\text{ remainder }& 10 \\ 10\cdot 10 &:97 &=& 1 &\text{ remainder }& 3 \\ 3\cdot 10 &:97 &=& 0 &\text{ remainder }& 30 \\ 30\cdot 10 &:97 &=& 3 &\text{ remainder }& 9 \\ 9\cdot 10 &:97 &=& 0 &\text{ remainder }& 90 \\ 90\cdot 10 &:97 &=& 9 &\text{ remainder }& 27 \\ 27\cdot 10 &:97 &=& 2 &\text{ remainder }& 76 \\ \dots\\ z\cdot 10 &:97 &=& A &\text{ remainder }& y \\ y\cdot 10 &:97 &=& 6 & \text{ remainder }& x \\ x\cdot 10 &:97 &=& 7 & \text{ remainder }& 1 & \qquad \text{ end of repeating part remainder } = 1\\ \hline \end{array}$

We solve A:

We have:

$\begin{array}{|rcl|} \hline \mathbf{x\cdot 10 }&\mathbf{:97} &\mathbf{=}& \mathbf{7} & \mathbf{\text{ remainder }}& \mathbf{1} \\ \text{ or }\\ x\cdot 10&&=& 7\cdot 97 & +& 1\\\\ x &=& \frac{ 7\cdot 97+1}{10} \\ x &=& \frac{ 680}{10} \\ \mathbf{x} & \mathbf{=}& \mathbf{68}\\ \hline \end{array}$

...and we have:

$\begin{array}{|rcl|} \hline \mathbf{y\cdot 10 }&\mathbf{:97} &\mathbf{=}& \mathbf{6} & \mathbf{\text{ remainder }}& \mathbf{x} \\ \text{ or }\\ y\cdot 10&&=& 6\cdot 97 & +& 68\\\\ y &=& \frac{ 6\cdot 97+68}{10} \\ y &=& \frac{ 650}{10} \\ \mathbf{y} & \mathbf{=}& \mathbf{65}\\ \hline \end{array}$

...and...

$\begin{array}{|rcl|} \hline \mathbf{z\cdot 10 }&\mathbf{:97} &\mathbf{=}& \mathbf{A} & \mathbf{\text{ remainder }}& \mathbf{y} \\ \text{ or }\\ z\cdot 10&&=& A\cdot 97 & +& 65\\\\ \hline \end{array}$

We have to solve the Diophantine equation  $10z -97A = 65.$

We can do it with a variation of the Euclidean algorithm, or we can do it with Euler's Method etc.

I will do it with Euler's theorem:

$\begin{array}{l} \text{In number theory, }\\ \text{Euler's theorem (also known as the Fermat–Euler theorem or Euler's totient theorem)}\\ \text{states that if } \mathbf{m} \text{ and } \mathbf{a} \text{ are coprime positive integers, then }\\ \hline a^{\varphi (m)} \equiv 1 \pmod{m} \qquad \text{, if }~ gcd(a,m)=1 \\\\ \text{or} \\ \\ a^{\varphi (m)} \pmod{m} \equiv 1 \qquad | \qquad \cdot \frac{1}{a}\\ \frac{ a^{\varphi (m)} } {a} \pmod{m} \equiv \frac{1}{a} = a^{-1}\\ a^{\varphi (m)-1} \pmod{m} \equiv \frac{1}{a} = a^{-1}\\ \begin{array}{|rcl|} \hline a^{-1} = \frac{1}{a} \equiv a^{\varphi (m)-1} \pmod{m} \qquad \text{, if }~ gcd(a,m)=1\\ \hline \end{array} \end{array}$

We need this Formula: $\begin{array}{|rcl|} \hline a^{-1} = \frac{1}{a} \equiv a^{\varphi (m)-1} \pmod{m} \qquad \text{, if }~ gcd(a,m)=1\\ \hline \end{array}$

an we need the Euler's Totient Function: $\varphi(10) = 10\cdot (1- \frac{1}{2} )\cdot (1- \frac{1}{5} ) = 4, \text{ because } ~ 10 = 2\cdot 5$

so we have:

$\begin{array}{rcl} 10z -97A &=& 65\\\\ 97A &=& 10\cdot z - 65 \qquad | \qquad \pmod{10} \\ 97A &\equiv& -65 \pmod{10} \\ 97A &\equiv& -65 +7\cdot 10 \pmod{10} \\ \mathbf{97A} & \mathbf{\equiv}& \mathbf{5 \pmod{10}}\\\\ 97\cdot A\cdot \frac{1}{97} & \equiv & \frac{5}{97} \pmod{10}\\ A & \equiv & 5\cdot \frac{1}{97} \pmod{10}\\\\ \frac{1}{97} \pmod{10} &\equiv& 97^{\varphi(10)-1} \pmod{10} \qquad \text{ see Formula above,}\\ &&\qquad \text{we can do it because, } ~gcd(97,10)=1\\ &\equiv& 97^{4-1} \pmod{10}\\ &\equiv& 97^3 \pmod{10}\\ &\equiv& 912673 \pmod{10}\\ &\equiv& 3 \pmod{10}\\\\ A & \equiv & 5\cdot \frac{1}{97} \pmod{10}\\\\ A & \equiv & 5\cdot 3 \pmod{10}\\ A & \equiv & 15 \pmod{10}\\ A & \equiv & 5 \pmod{10}\\ \mathbf{A} & \mathbf{=} & \mathbf{5} \end{array}$

What is 7 over Sin(44) times Sin(36)? I don't have a calculator that can do the over function and can't find one online...

      7     

–––––––    x   Sin(36)      =   x

 Sin(44)

When I try 7  :  Sin44 x Sin36 it always comes up as 5.884289931 but I know this is wrong, can you tell me a way to input this kind of calculation into a scientific calculator without the custom fraction thingy option?

gon or grad is a unit of measurement of an angle,
equivalent to$\frac{1}{400}$ of a turn, $\frac{9}{10}$ of a degree or $\frac{ \pi} {200}$ of a radian.

Symbol ^g or gon

$\begin{array}{rcll} \frac{7}{\sin(44^g)}\cdot \sin(36^g) &=& 5.88428993131 \end{array}$

Use degrees:

$\begin{array}{rcll} \frac{7}{\sin(44^{\circ})}\cdot \sin(36^{\circ}) &=& 5.92305072683 \end{array}$

What are the zeros of f(x)=x^3+3x^2-4x-12?

$\begin{array}{lrrl} x^3+3x^2-4x-12\\ & & 12 = 2\cdot 2 \cdot 3\\\\ (x-x_1)\cdot (x-x_2)\cdot (x-x_3) & \Rightarrow & -12 = (-x_1)\cdot(-x_2)\cdot(-x_3)& \\ & \Rightarrow & \boxed{ \begin{array}{rcll}12 = \underbrace{x_1}_{=2}\cdot \underbrace{x_2}_{=2} \cdot \underbrace{x_3}_{=3}\end{array} } & (\text{sign} \ ?)\\\\ (x-x_1)\cdot (x-x_2)\cdot (x-x_3) & & = x^2\cdot(-x_1)+x^2\cdot(-x_2)+x^2\cdot(-x_3) \\ & \Rightarrow & 3x^2= x^2\cdot(-x_1-x_2-x_3) \\ & & 3 = -(x_1+x_2+x_3) \\ & & -3 = x_1+x_2+x_3 \\\\ & & & \text{sum} \\ & & 2+2+3 & =7~ (\text{no})\\ & & 2+2-3 & =1~ (\text{no})\\ & & 2-2+3 & =3~ (\text{no})\\ & & 2-2-3 & =-3~ (\text{yes})\\ & & -2+2+3 & =3~ (\text{no})\\ & & -2+2-3 & =-3~ (\text{yes})\\ & & -2-2+3 & =-1~ (\text{no})\\ & & -2-2-3 & =-7~ (\text{no}) \end{array}$

$\boxed{ \begin{array}{lrrl} x_1 = 2\\ x_2 = -2\\ x_3 = -3\\ x^3+3x^2-4x-12 = (x-2)\cdot(x+2)\cdot(x+3) \end{array} }$

Hallo asinus,

ich verwende LaTeX, siehe oben beim Sigma-Zeichen.

und

Viel Spaß damit.

Ein Wurzelzeichen schreibt man mit \sqrt{2}, eine Wurzel 3 mit \sqrt[3]{2}...

Eine neue Zeile mit "\\"

Willst du an den Latex-Code, dann rechte Maustaste, im Kontextmenü "Show Math As" -> Tex Commands

Es erscheint ein Fenster mit dem Latex-Code, den man mit Drag & Drop entnehmen kann:

$\sqrt{2}\\ \sqrt[3]{2}$

451. zahl der Formel f(n)= f (n−1)∗2 + f (n−2) + 2;

Ich denke es handelt sich um eine rekursive Folge. Man könnte auch schreiben:

$a_n = a_{n-1}\cdot 2 + a_{n-2} + 2$

Leider fehlen zur Berechnung von $a_{451}$ die Anfangswerte $a_1$ und $a_2$

Die Fibonacci-Folge wäre zum Beispiel:

Die Werte wären dann:

Larry has 4-cent stamps and 9-cent stamps, which he can combine to produce various amounts of postage. For example, he can make 40 cents by using four 9-cent stamps and a 4-cent stamp, or by using ten 4-cent stamps. However, there are some amounts of postage he can't make exactly, such as 10 cents. {nl} {nl} What is the largest number of cents that Larry cannot make exactly from a combination of 4- and/or 9-cent stamps? {nl} {nl} Explain how you know your answer is correct. (You should explain two things: why Larry can't make the amount of your answer, and why he can make any bigger amount.)

All figures can be divided into 4 groups.

Group 1: 4n

Group 2: 4n+1

Group 3: 4n+2

Group 4: 4n+3

With 4-cent stamps we get Group 1. (Beginning with n=1 we have 4, 8, 12, 16, 32, ...)

With one 9-cent(4*2+1) we can start Group 2. (Beginning with n=2 we have 9 ) and continuing 4*3 + 1(=13) with one 4-cent, 4*4 + 1(=17) with two 4-cent, 4*5+1(=21) with three 4-cent etc.

With two 9-cent(4*4+2) we can start Group 3. (Beginning with n=4 we have 18 ) and continuing 4*5 + 2(=22) with one 4-cent, 4*6 + 2(=26) with two 4-cent, 4*7+2 (=30) with three 4-cent etc.

With three 9-cent(4*6+3) we can start Group 4. (Beginning with n=6 we have 27 ) and continuing 4*7 + 3(=31) with one 4-cent, 4*8 + 3(=35) with two 4-cent, 4*9+3 (=39) with three 4-cent etc.

$\begin{array}{|r|r|r|r|r|} \hline \text{All Numbers} \\ \hline \text{Group 1} &\text{Group 2} &\text{Group 3} &\text{Group 4} \\ 4n+1 & 4n+2 & 4n+3 & 4n \\ \hline 1 & 2 & 3 & 4\checkmark \\ & & & & +4 \\ 5 & 6 & 7 & 8\checkmark \\ & & & & +4 \\ 9\checkmark & 10 & 11 & 12\checkmark \\ & & & & +4 \\ 13\checkmark & 14 & 15 & 16\checkmark \\ & & & & +4 \\ 17\checkmark & 18\checkmark & 19 & 20\checkmark \\ & & & & +4 \\ 21\checkmark & 22\checkmark & {\color{red}23} & 24\checkmark \\ & & & & +4 \\ 25\checkmark & 26\checkmark & 27\checkmark & 28\checkmark \\ & & & & +4 \\ 29\checkmark & 30\checkmark & 31\checkmark & 32\checkmark \\ & & & & +4 \\ 33\checkmark & 34\checkmark & 35\checkmark & 36\checkmark \\ & & & & +4 \\ 37\checkmark & 38\checkmark & 29\checkmark & 40\checkmark \\ & & & & +4 \\ \dots\checkmark & \dots\checkmark & \dots\checkmark & \dots\checkmark \\ \hline \end{array}$

the largest number of cents that Larry cannot make exactly from a combination of 4- and/or 9-cent stamps is 23.

ua/ue=1/sqrt(R^2(2pi*f*c)^2+1 wie stelle ich die formel nach f um ?

$\begin{array}{rcll} \frac{U_a}{U_e} &=& \frac{1}{ \sqrt{ R^2\cdot(2\pi\cdot f \cdot C )^2 + 1 } } \\\\ \frac{U_a}{U_e} &=& \frac{1}{ \sqrt{1+ (2\pi\cdot f \cdot R \cdot C )^2 } } \\\\ \frac{U_e}{U_a} &=& \sqrt{1+ (2\pi\cdot f \cdot R \cdot C )^2 } \qquad & | \qquad \text{quadriere beide Seiten} \\\\ \left( \frac{U_e}{U_a} \right)^2 &=& 1+ (2\pi\cdot f \cdot R \cdot C )^2 \\ 1+ (2\pi\cdot f \cdot R \cdot C )^2 &=& \left( \frac{U_e}{U_a} \right)^2 \qquad & | \qquad -1\\ (2\pi\cdot f \cdot R \cdot C )^2 &=& \left( \frac{U_e}{U_a} \right)^2 -1 \qquad & | \qquad \sqrt{} \\ 2\pi\cdot f \cdot R \cdot C &=& \sqrt{ \left( \frac{U_e}{U_a} \right)^2 -1 } \\ ( 2\pi\cdot R \cdot C) \cdot f &=& \sqrt{ \left( \frac{U_e}{U_a} \right)^2 -1 } \qquad & | \qquad : ( 2\pi\cdot R \cdot C)\\\\ \mathbf{ f } & \mathbf{=} & \mathbf{ \dfrac{ \sqrt{ \left( \frac{U_e}{U_a} \right)^2 -1 } } { 2\pi\cdot R \cdot C } }\\ \end{array}$