heureka

How do I simplify the equation 3y^2-3 over y^2-1 ?

$$\dfrac{3y^2-3 }{ y^2-1}\\\\ =\dfrac{3\cdot(y^2-1) }{ y^2-1}\\\\ =3$$

what is the ones digit of 9^40 ?

$$9^{40}\mod 10\\ =( \underbrace{9^2}_{=1\mod 10} )^{20}\mod 10 \qquad | \qquad 9^2\mod 10 =\ 81 \mod 10 =\ 1 \\\\ =1^{20}\mod 10 \qquad\qquad | \quad 1^{20} = 1 \\ =1\mod 10$$

Perform the indicated operations and simplify the expression.

$$\small{\text{ $ \left( x^{\frac{1}{4}} + 1 \right) \cdot \textcolor[rgb]{1,0,0}{\left( \dfrac{1}{4}x^{-\frac{1}{4}} \right)} - \left( x^{\frac{1}{4}} - 1 \right) \cdot \textcolor[rgb]{1,0,0}{\left( \dfrac{1}{4}x^{-\frac{1}{4}} \right)} $}}\\ \small{\text{ $ = \textcolor[rgb]{1,0,0}{\left( \dfrac{1}{4}x^{-\frac{1}{4}} \right)} \cdot \left[ \left( x^{\frac{1}{4}} + 1 \right) - \left( x^{\frac{1}{4}} - 1 \right) \right] $}}\\\\ \small{\text{ $ = \left( \dfrac{1}{4}x^{-\frac{1}{4}} \right) \cdot \left( \not{x^{\frac{1}{4}}} + 1 - \not{x^{\frac{1}{4}}} + 1 \right) $}}\\\\ \small{\text{ $ =\left( \dfrac{2}{4}x^{-\frac{1}{4}} \right) $}}\\\\ \small{\text{ $ =\left( \dfrac{1}{2}x^{-\frac{1}{4}} \right) $}}$$

Hallo gandalf the green,

Omis Bedingungen sind richtig, bzw. habe ich auch.

Nur die Gleichung lautet bei mir anders:

$$}}\\\\\\ \small{\text{Die Gleichung m\"u \ss te lauten: $ \boxed{y = x^3 - x^2 - 5x + 5} $ }}$$

Eine GRF 3. Grades hat einen Hochpunkt H(-1|8). Bei x=1 hat sie die Gerade mit der Funktionsgleichng g(x)=-4x+4 als Tangente. Bestimmen Sie f(x). Ich finde die 4. Bedingung nicht

Gesucht ist die Gleichung:  $$y=ax^3+bx^2+cx+d$$

1. Bedingung: Hochpunkt H(-1|8) :

$$\small{\text{ $ 8=a(-1)^3+b(-1)^2+c(-1)+d\\ \qquad \boxed{8=-a+b-c+d} \quad (1) $ }}$$

2. Bedingung: Punkt ( x=1, y=0 )  g(1) = -4*(1) + 4 = 0 = f(1)

$$\small{\text{ $ 0=a(1)^3+b(1)^2+c(1)+d=0\\ \qquad \boxed{0=a+b+c+d} \quad (2) $ }}$$

Die Bedingungen zur Ableitung :  $$y'=3ax^2+2bx+c$$

3. Bedingung: f'(1) = -4     g'(1) = f'(1) = -4

$$\small{\text{ $ -4=3a(1)^2+2b(1)+c\\ \qquad \boxed{-4=3a+2b+c} \quad (3) $ }}$$

4. Bedingung:  f'(-1)= 0 existiert ein Hochpunkt

$$\small{\text{ $ 0=3a(-1)^2+2b(-1)+c\\ \qquad \boxed{0=3a-2b+c} \quad (4) $ }}$$

Alle Bedingungen in der Übersicht als Gleichungssystem:

$$\small{\text{ $ \begin{array}{lrcl} (1) & 8 &=& - a +b - c +d \\ (2) & 0 &=& a + b + c +d \\ (3) &-4 &=& 3a + 2b + c \\ (4) & 0 &=& 3a-2b+c \\ \hline (1)+(2)& 8 &=& 2b+2d \quad | \quad :2 \\ (6) & 4 &=& b+d \\ \hline (3)+(4)& -4 &=& 6a + 2c \quad | \quad :2 \\ (5) & -2 &=& 3a+c \\ \hline (2) & 0 &=& a + (b+d) + c \quad | \quad b+d = 4\\ & 0 &=& a + 4 + c \\ & a &=& -4-c \\ \hline (5) & 3a &=& -2 -c \quad | \quad a = -4-c \\ & 3(-4-c) &=& -2-c \\ & -12-3c &=& -2-c \\ & 2c &=& -10 \\ & \textcolor[rgb]{1,0,0}{c} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{-5} \\ \hline & a &=& -4 -c \quad | \quad c = -5 \\ & a &=& -4 -(-5) \\ & a &=& -4 + 5 \\ & \textcolor[rgb]{1,0,0}{a} & \textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{1} \\ \hline (3) & 2b &=& -4-3a-c \quad | \quad c = -5 \quad a = 1\\ & 2b &=& -4-3+5 \\ & 2b &=& -7 + 5 \\ & 2b &=& -2 \\ & \textcolor[rgb]{1,0,0}{b} & \textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{-1} \\ \hline (6) & d &=& 4-b \quad | \quad b = -1 \\ & d &=& 4 - (-1) \\ & d &=& 4+1 \\ & \textcolor[rgb]{1,0,0}{d} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{5} \\ \hline \end{array} $ }}\\\\\\ \small{\text{Die Gleichung lautet: $ \boxed{y = x^3 - x^2 - 5x + 5} $ }}$$

$$\\\mathbf{How\ to\ calcultate\ the\ \textit{Euler phi function} \ \phi(n):}\\ $ We have the prime factorization of n = p_1\cdot p_2\cdot p_3 \cdots\\ \phi(n) = n \cdot (1-\frac{1}{p_1})\cdot (1-\frac{1}{p_2}) \cdot (1-\frac{1}{p_3}) \cdots$$

$$\\\mathbf{Example\ 1: n = 6 } $\\ The prime factorization of 6 = 2 * 3 = p_1*p_2 \\ \phi(6) = 6 \cdot (1-\frac{1}{2}) \cdot (1-\frac{1}{3}) \\ \phi(6) = 6 \cdot \frac{1}{2} \cdot \frac{2}{3} \\ \phi(6) = \frac{6}{3} \\ \phi(6) = 2$$

$$\\\mathbf{Example\ 2: n = 9 } $\\ The prime factorization of 9 = 3^2 = p_1^2 \\ \phi(9) = 9 \cdot (1-\frac{1}{3}) \\ \phi(9) = 9 \cdot \frac{2}{3} \\ \phi(6) = 3 \cdot 2 \\ \phi(6) = 6$$

$$\\\mathbf{Example\ 3: n = 7 } $\\ The prime factorization of 7 = 7 = p_1 \qquad 7 $ is a prime number!$ \\ \phi(7) = 7 \cdot (1-\frac{1}{7}) \\ \phi(7) = 7 \cdot \frac{6}{7} \\ \phi(7) = 6$$

$$\\\mathbf{Example\ 4: n = 11 } $\\ The prime factorization of 11 = 11 = p_1 \qquad 11 $ is a prime number!$ \\ \phi(11) = 11 \cdot (1-\frac{1}{11}) \\ \phi(11) = 11 \cdot \frac{10}{11} \\ \phi(11) = 10$$

$$\boxed{\text{ In general $ \phi(p) = p-1 $, if p is a prime number }}\\\\ \begin{array}{lr} p = 2: &\phi(2) = 1 \qquad =(2-1)\\ p = 3: &\phi(3) = 2 \qquad =(3-1)\\ p = 5: &\phi(5) = 4 \qquad =(5-1)\\ p = 7: &\phi(7) = 6 \qquad =(7-1)\\ p = 11: &\phi(11) = 10 \qquad =(11-1)\\ p = 13: &\phi(13) = 12 \qquad =(13-1)\\ \cdots & \phi(p) = p-1 \end{array}$$

The first 99 values of the Phi function are:

	+0	+1	+2	+3	+4	+5	+6	+7	+8	+9
0+		1	1	2	2	4	2	6	4	6
10+	4	10	4	12	6	8	8	16	6	18
20+	8	12	10	22	8	20	12	18	12	28
30+	8	30	16	20	16	24	12	36	18	24
40+	16	40	12	42	20	24	22	46	16	42
50+	20	32	24	52	18	40	24	36	28	58
60+	16	60	30	36	32	48	20	66	32	44
70+	24	70	24	72	36	40	36	60	24	78
80+	32	54	40	82	24	64	42	56	40	88
90+	24	72	44	60	46	72	32	96	42	60

Hello Melody,

$$\\\mathbf{De{finition:}}\\ \begin{text} L{et} $ n \ge 1$ be an integer. Then we de{fine} the \\ \textit{Euler phi function} $\phi$ by\\ $\phi(n)=$ the number of positive integers \\ less than $n$ that are relatively prime to $n$ \end{text}$$

$$\\\mathbf{Relatively Prime:}\\ $ \begin{text} Describes two numbers for which\\ the only common factor is 1. \\ In other words, relatively prime numbers have\\ a greatest common factor $(gcf)$ of $1$. \\ For example, $6$ and $35$ are relatively prime $(gcf = 1)$.\\ The numers $6$ and $8$ are not relatively prime $(gcf = 2)$. \end{text}$$

$$\\\mathbf{Example\ 1: n = 6 } $ and n is not a prime number $\\ \samll{\text{ $ \begin{array}{rr} \textcolor[rgb]{1,0,0}{1}& gcf(6,1) = \textcolor[rgb]{1,0,0}{1}\\ 2& gcf(6,2) = 2\\ 3& gcf(6,3) = 3\\ 4& gcf(6,4) = 2\\ \textcolor[rgb]{1,0,0}{5}& gcf(6,5) = \textcolor[rgb]{1,0,0}{1}\\ 6& gcf(6,6) = 6\\ \end{array} $ }}\\ 6 \text{ has } \textcolor[rgb]{0,0,1}{2} \text{ relative primes } 1 \text{ and } 5 \text { see the red color. So } \phi(6) = \textcolor[rgb]{0,0,1}{2}$$

$$\\\mathbf{Example\ 2: n = 9 } $ and n is not a prime number $\\ \samll{\text{ $ \begin{array}{rr} \textcolor[rgb]{1,0,0}{1}& gcf(9,1) = \textcolor[rgb]{1,0,0}{1}\\ \textcolor[rgb]{1,0,0}{2}& gcf(9,2) = \textcolor[rgb]{1,0,0}{1}\\ 3& gcf(9,3) = 3\\ \textcolor[rgb]{1,0,0}{4}& gcf(9,4) = \textcolor[rgb]{1,0,0}{1}\\ \textcolor[rgb]{1,0,0}{5}& gcf(9,5) = \textcolor[rgb]{1,0,0}{1}\\ 6& gcf(9,6) = 3\\ \textcolor[rgb]{1,0,0}{7}& gcf(9,7) = \textcolor[rgb]{1,0,0}{1}\\ \textcolor[rgb]{1,0,0}{8}& gcf(9,8) = \textcolor[rgb]{1,0,0}{1}\\ 9& gcf(9,9) = 9\\ \end{array} $ }}\\ 9 \text{ has } \textcolor[rgb]{0,0,1}{6} \text{ relative primes } 1,2,4,5,7,8 \text { see the red color. So } \phi(9) = \textcolor[rgb]{0,0,1}{6}$$

$$\\\mathbf{Example\ 3: n = 7 } $ and n is a prime number $\\ \samll{\text{ $ \begin{array}{rr} \textcolor[rgb]{1,0,0}{1}& gcf(7,1) = \textcolor[rgb]{1,0,0}{1}\\ \textcolor[rgb]{1,0,0}{2}& gcf(7,2) = \textcolor[rgb]{1,0,0}{1}\\ \textcolor[rgb]{1,0,0}{3}& gcf(7,3) = \textcolor[rgb]{1,0,0}{1}\\ \textcolor[rgb]{1,0,0}{4}& gcf(7,4) = \textcolor[rgb]{1,0,0}{1}\\ \textcolor[rgb]{1,0,0}{5}& gcf(7,5) = \textcolor[rgb]{1,0,0}{1}\\ \textcolor[rgb]{1,0,0}{6}& gcf(7,6) = \textcolor[rgb]{1,0,0}{1}\\ 7& gcf(7,7) = 7\\ \end{array} $ }}\\ \small{ 7 \text{ has } 7-1=\textcolor[rgb]{0,0,1}{6} \text{ relative primes } 1 \text{ until } 6 \text { see the red color. So } \phi(7) = \textcolor[rgb]{0,0,1}{6}.}\\ \text{ So } \phi \text{ of a prime number } \phi(p) = p-1$$

In a triangle ABC, a,b,c are 3 opposite side of each angle A,B,C, if a^2-c^2=2b, and SinA*CosC=3*CosA*SinC, find b

$$\boxed{ (1)\quad a^2-c^2=2b\qquad (2) \quad SinA*CosC=3*CosA*SinC}$$

$$\small{\text{ sine rule: }} \frac{sinA}{a}=\frac{sinC}{c} \quad \Rightarrow \quad sinA=\frac{a}{c}\cdot sinC \\ (2):\qquad \frac{a}{c}\cdot sinC\cdot cosC = 3cosA\cdot sinC\Rightarrow \boxed{ a\cdot cosC=3c\cdot cosA }\\ \small{\text{ in every triangle: }}b=a\cdot cosC+c\cdot cosA\Rightarrow\boxed{ c\cdot cosA = b - a\cdot cosC }\\ a\cdot cosC=3 (b - a\cdot cosC)\\ 4\cdot a\cdot cosC = 3b\\ \boxed{ a\cdot cosC = \frac{3} {4}\cdot b }\\\\ \small{\text{ law of cosines: }} c^2=a^2+b^2-2ac\cdot cosC\\ (1): \qquad a^2-c^2=2ab\cdot cosC-b^2 =2b\\ 2ab\cdot cosC-b^2 =2b \quad | \quad :b \\ 2\cdot a\cdot cosC-b =2 \quad | \quad a\cdot cosC = \frac{3}{4}\cdot b \\ 2\cdot \frac{3}{4}\cdot b - b =2\\ \frac{3}{2}\cdot b - b =2 \quad | \quad \cdot 2 \\ 3b-2b=4\\ b=4$$

what is Φ6?

$$\!\ \varphi(6) = 2. \qquad 1 \text{ and } 5 \\\\ 6 = 2 \cdot 3\\ \!\ \varphi(6)}=6\cdot (1-\frac{1}{2})( 1-\frac{1}{3} )}=2$$