heureka

$x^2 \cdot (2x + x^2) \\ (x^2y - xy^{-1}) \cdot x^{-1}y \\ (2x^{n-1} + 2x^{n+1} - 2x^{n}) \cdot 2x^{1-n} \\ (25a^{m-1} + 24a^{m-t}) : 7a^{m} \\$

1.

$\begin{array}{rcll} x^2 \cdot (2x + x^2) &=& x^2 \cdot 2x + x^2 \cdot x^2 \\ &=& 2x^3 + x^4 \\ \end{array}$

2.

$\begin{array}{rcll} (x^2y - xy^{-1}) \cdot x^{-1}y &=& x^2y \cdot x^{-1}y - xy^{-1} \cdot x^{-1}y \\ &=& x^{2-1}y^2 - x^{1-1}y^{1-1} \\ &=& x^{1}y^2 - x^{0}y^{0} \qquad & \qquad x^0 = 1 \qquad y^0 = 1\\ &=& xy^2 - 1\\ \end{array}$

3.

$\begin{array}{rcll} (2x^{n-1} + 2x^{n+1} - 2x^{n}) \cdot 2x^{1-n} &=& 2x^{n-1}\cdot 2x^{1-n} + 2x^{n+1}\cdot 2x^{1-n} - 2x^{n} \cdot 2x^{1-n}\\ &=& 2\cdot 2\cdot x^{n-1+1-n} + 2\cdot 2\cdot x^{n+1+1-n} - 2\cdot 2\cdot x^{n+1-n}\\ &=& 4\cdot x^{0} + 4\cdot x^{2} - 4\cdot x^{1} \qquad & \qquad x^0 = 1 \\ &=& 4\cdot 1 + 4\cdot x^{2} - 4\cdot x^1 \\ &=& 4 + 4\cdot x^{2} - 4\cdot x \\ &=& 4\cdot( x^{2} - x + 1) \\ \end{array}$

4.

$\begin{array}{rcll} \dfrac{25a^{m-1} + 24a^{m-t}} { 7a^{m} } &=& \dfrac{25a^{m-1}}{ 7a^{m} } + \dfrac{24a^{m-t}} { 7a^{m} }\\\\ &=& \dfrac{25}{7} \cdot \dfrac{a^{m-1}}{ a^{m} } + \dfrac{24}{7} \cdot \dfrac{a^{m-t}} { a^{m} }\\\\ &=& \dfrac{25}{7} \cdot a^{m-1-m} + \dfrac{24}{7} \cdot a^{m-t-m}\\\\ &=& \dfrac{25}{7} \cdot a^{-1} + \dfrac{24}{7} \cdot a^{-t} \\\\ &=& \dfrac{25}{7a} + \dfrac{24}{7a^t}\\\\ &=& \dfrac17 \cdot \left( \dfrac{25}{a} + \dfrac{24}{a^t} \right) \end{array}$

hab mal eine Frage. Und zwar bin ich mir unsicher ob ich die wurzel aus 4-2x in der kettenregel machen kann.. bitte um hilfe

Ja.

$\begin{array}{rcll} y &=& \sqrt{4-2x} \\ y &=& (4-2x)^{\frac12} \\ y'&=& \frac12\cdot (4-2x)^{\frac12-1}\cdot (-2) \\ y'&=& \frac12\cdot (4-2x)^{-\frac12}\cdot (-2) \\ y'&=& \frac{-2}{2}\cdot (4-2x)^{-\frac12} \\ y'&=& - (4-2x)^{-\frac12} \\ \mathbf{y'}&\mathbf{=}&\mathbf{ - \frac{1}{\sqrt{4-2x} } }\\ \end{array}$

Wie löst man Betragsungleichungen?

Könnt ihr mir anhand folgender aufgabe zeigen, wie es geht?

|2x-1| ≥ 2|x+3| in R

$\begin{array}{rcll} |2x-1| &\ge& 2\cdot |x+3| \qquad & \qquad (\text{quadriere beide Seiten})\\ (2x-1)^2 &\ge& 2^2\cdot (x+3)^2 \qquad & \qquad (\text{ausklammern})\\ 4x^2-4x+1 &\ge& 4\cdot(x^2+6x+9) \\ 4x^2-4x+1 &\ge& 4x^2+24x+36 \qquad & \qquad -4x^2\\ -4x+1 &\ge& 24x+36 \qquad & \qquad -24x\\ -28x+1 &\ge& 36 \qquad & \qquad -1\\ -28x &\ge& 35 \qquad & \qquad :7\\ -4x &\ge& 5 \qquad & \qquad :(-4) \quad (~\ge \text{ wird zu } \le~)\\ x &\le& -\frac{5}{4} \\ \mathbf{x} &\mathbf{\le}& \mathbf{ -1,25 } \\ \end{array}$

5+_x6-3=14 ?

$\begin{array}{rcll} 5+ \text{_} \times 6 - 3 &=& 14 \\ 5+ x \times 6 - 3 &=& 14 \quad & | \quad + 3\\ 5+ x \times 6 &=& 14 + 3\\ 5+ x \times 6 &=& 17 \quad & | \quad -5\\ x \times 6 &=& 12 \quad & | \quad :6 \\ x &=& \frac{12}{6} \\ \mathbf{x} &\mathbf{=}& \mathbf{2} \\ \end{array}$

Probe:

$\begin{array}{rcll} 5+ 2 \times 6 - 3 &=& 14 \\ 5+ 12 - 3 &=& 14 \\ 17 - 3 &=& 14 \\ 14 &=& 14 \\ \end{array}$

(sqrt(3)-4+sqrt(5))/((3*sqrt(15))+15-(12*sqrt(5))

$\begin{array}{rcll} \frac{ \sqrt{3}-4+\sqrt{5} } { 3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} } &=& \frac{ \sqrt{3}-4+\sqrt{5} } { 3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} } \cdot \frac{ \sqrt{45} }{ \sqrt{45} } \\\\ &=& \frac{ \sqrt{3}\sqrt{45}-4\sqrt{45}+\sqrt{5}\sqrt{45} } { (3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} )\cdot \sqrt{45} } \quad & | \quad \small{\sqrt{5}\sqrt{45} = \sqrt{225} = 15}\\\\ &=& \frac{ \sqrt{3}\sqrt{45}-4\sqrt{45}+15 } { (3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} )\cdot \sqrt{45} } \quad & | \quad \small{ \sqrt{45} = \sqrt{9\cdot 5} = 3\sqrt{5} }\\\\ &=& \frac{ \sqrt{3}\cdot 3\sqrt{5}-4\cdot3\sqrt{5}+15 } { (3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} )\cdot \sqrt{45} } \\\\ &=& \frac{ 3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} } { (3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} )\cdot \sqrt{45} } \\\\ &=& \frac{1 } { \sqrt{45} } \\\\ \mathbf{\frac{ \sqrt{3}-4+\sqrt{5} } { 3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} } }&\mathbf{=}& \mathbf{\frac{1 } { \sqrt{45} } } \\\\ &\mathbf{=}& \mathbf{0.149071198499986 } \end{array}$

Let the series  $a_{n+1} =1+\sum{a_n}+a_n$ if $a_2=7$ what is $a_5$ ?

$\begin{array}{rcll} a_{n+1} &=& 1+\sum{a_n}+a_n \\\\ a_2 &=& 1 + a_1 + a_1\\ \mathbf{a_2} &\mathbf{=}& \mathbf{1 + 2a_1} \qquad & | \qquad a_2 = 7\\ 7 &=& 1 + 2a_1 \\ 6 &=& 2a_1 \\ 3 &=& a_1 \\ \mathbf{a_1} &\mathbf{=}& \mathbf{3}\\ \end{array}$

$\begin{array}{rcll} a_3 &=& 1 + a_1 + a_2 + a_2\\ a_3 &=& 1 + a_1 + 2a_2 \qquad & | \quad \small{a_2 = 1+2a_1}\\ a_3 &=& 1 + a_1 + 2\cdot(1+2a_1)\\ a_3 &=& 1 + a_1 + 2 + 4a_1\\ \mathbf{a_3} &\mathbf{=}& \mathbf{3 + 5a_1}\qquad & | \qquad a_1 = 3\\ a_3 &=& 3 + 5\cdot 3\\ \mathbf{a_3} &\mathbf{=}& \mathbf{18}\\ \end{array}$

$\begin{array}{rcll} a_4 &=& 1 + a_1 + a_2 + a_3 + a_3\\ a_4 &=& 1 + a_1 + a_2 +2a_3 \qquad & | \quad \small{a_2 = 1+2a_1 \quad a_3 = 3 + 5a_1}\\ a_4 &=& 1 + a_1 + 1+2a_1+ 2\cdot( 3 + 5a_1)\\ a_4 &=& 1 + a_1 + 1+2a_1+ 6 + 10a_1\\ \mathbf{a_4} &\mathbf{=}& \mathbf{8 + 13a_1}\qquad & | \qquad a_1 = 3\\ a_4 &=& 8 + 13\cdot 3\\ \mathbf{a_4} &\mathbf{=}& \mathbf{47}\\ \end{array}$

$\begin{array}{rcll} a_5 &=& 1 + a_1 + a_2 + a_3 + a_4 + a_4\\ a_5 &=& 1 + a_1 + a_2 + a_3 + 2a_4 \qquad & | \quad \small{a_2 = 1+2a_1 \quad a_3 = 3 + 5a_1 \quad a_4 = 8 + 13a_1}\\ a_5 &=& 1 + a_1 + 1+2a_1+ 3 + 5a_1 +2\cdot (8 + 13a_1)\\ a_5 &=& 1 + a_1 + 1+2a_1+ 3 + 5a_1 + 16 + 26a_1\\ \mathbf{a_5} &\mathbf{=}& \mathbf{21 + 34a_1}\qquad & | \qquad a_1 = 3\\ a_5 &=& 21 + 34\cdot 3\\ \mathbf{a_5} &\mathbf{=}& \mathbf{123}\\ \end{array}$

Find the solutions of the following system. Then graph the system of equations to confirm your algebraic solutions.

2x-y=-3

4(x-2)+(y-1)=36

$\begin{array}{rcll} (1) & 2x-y&=&-3 \qquad & | \qquad \cdot (-1)\\ & -2x+y &=& 3 \qquad & | \qquad +2x\\ & y &=& 3 +2x\\\\ (2) & 4(x-2)+(y-1) &=& 36 \\ & 4x-8+y-1 &=& 36 \\ & 4x+y-9 &=& 36 \\ & 4x+y-9 &=& 36 \qquad & | \qquad +9\\ & 4x+y &=& 36+9\\ & 4x+y &=& 45 \qquad & | \qquad -4x\\ & y &=& 45-4x\\\\ x: & 3 +2x&=& 45-4x \qquad & | \qquad +4x\\ & 3 +6x&=& 45\qquad & | \qquad -3\\ & 6x&=& 42 \qquad & | \qquad :6\\ & \mathbf{ x }& \mathbf{=} & \mathbf{7}\\\\ y: & y &=& 3 +2x \qquad & | \qquad x=7\\ & y &=& 3 +2\cdot 7\\ & y &=& 3 +14\\ & \mathbf{y} &\mathbf{=}& \mathbf{17} \end{array}$

What do the symbols mean Heureka ?

Symbolic notation in Text notation:

(P⋁¬Q)⋀(¬P⋁Q)⋀(¬P⋁¬Q)  = ( P or (not Q)  ) and (  (not P) or Q ) and ( (not P) or ( not Q )  )

Symbolic notation in Text notation:

¬P⋀¬Q = ( not P ) and ( not Q )

⋁  = or

⋀  = and

¬   = not

¬P  = not P

¬Q  = not Q

P ⋁ Q = P or Q

P ⋀ Q = P and Q

I need a truth table for this but i dont know how to do it..

(P⋁¬Q)⋀(¬P⋁Q)⋀(¬P⋁¬Q) 

(P⋁¬Q)⋀(¬P⋁Q)⋀(¬P⋁¬Q)  = ¬P⋀¬Q

A parabola has its focus at (–2, 7) and its vertex at (–2, –2). What is the equation of the parabolas directrix?

parabola: $y=ax^2+bx+c$

focus: $(x_f = -2,\ y_f = 7)$

vertex:  $(x_v = -2,\ y_v = -2)$

directrix: $y = c - \frac{b^2+1}{4a}$

$\boxed{~ \begin{array}{rclrcl} x_f &=& -\frac{b}{2a} & y_f &=& c - \frac{b^2-1}{4a} \\ x_v &=& -\frac{b}{2a} & y_v &=& c - \frac{b^2}{4a} \\ \end{array} ~}$

1. a?

$\begin{array}{rcll} y_f = c - \frac{b^2-1}{4a} &=& c - \frac{b^2}{4a} +\frac{1}{4a} \\ y_v &=& c - \frac{b^2}{4a} \\\\ y_f-y_v &=& \frac{1}{4a} \\ a &=& \frac{1}{4\cdot (y_f-y_v)} \\ a &=& \frac{1}{4\cdot (7-(-2))} \\ a &=& \frac{1}{4\cdot 9} \\ \mathbf{a} &\mathbf{=}& \mathbf{\frac{1}{36}} \\ \end{array}$

2. b?

$\begin{array}{rcl} x_v &=& -\frac{b}{2a} \\ \frac{b}{2a} &=& -x_v \\ b &=& -2\cdot a\cdot x_v \\ b &=& -2\cdot \frac{1}{36} \cdot (-2) \\ b &=& 4\cdot \frac{1}{36} \\ \mathbf{b} &\mathbf{=}& \mathbf{\frac{1}{9}} \\ \end{array}$

3. c?

$\begin{array}{rcl} y_v &=& c - \frac{b^2}{4a} \\ \frac{b^2}{4a} + y_v &=& c \\ c &=& \frac{b^2}{4a} + y_v \\ c &=& \frac{(\frac{1}{9})^2}{4\cdot (\frac{1}{36})} + (-2) \\ c &=& \frac{1}{9}-2 \\ \mathbf{c} &\mathbf{=}& \mathbf{-\frac{17}{9}} \\ \end{array}$

The equation of the parabola is $y=\frac{1}{36}x^2+\frac{1}{9}x-\frac{17}{9}$

4. directrix

$\begin{array}{rcl} y &=& c - \frac{b^2+1}{4a} \\ y &=& -\frac{17}{9} - \frac{(\frac{1}{9})^2+1}{4\cdot \frac{1}{36} } \\ y &=& -\frac{17}{9} - 9\cdot (\frac{1}{81}+1) \\ y &=& -\frac{17}{9} -\frac19 - 9 \\ y &=& -2 - 9 \\ \mathbf{y} &\mathbf{=}& \mathbf{-11} \\ \end{array}$

The equation of the parabolas directrix is y = -11