heureka

Hallo,

wie klammere ich eine Funktion mit Wurzel aus?(Bitte mit Rechenweg)

\(\begin{array}{|lrcll|} \hline && \frac23\cdot \left( -\sqrt{\frac23} \right) -2\cdot \left( -\sqrt{\frac23} \right) \\ &=& \left( -\sqrt{\frac23} \right)\cdot \Big( \frac23 -2 \Big) \\ &=& \left( -\sqrt{\frac23} \right)\cdot \Big( \frac23 -\frac63 \Big) \\ &=& \left( -\sqrt{\frac23} \right)\cdot \Big( \frac{2-6}{3} \Big) \\ &=& \left( -\sqrt{\frac23} \right)\cdot \Big( -\frac{4}{3} \Big) \\ &=& \left( \sqrt{\frac23} \right)\cdot \Big( \frac{4}{3} \Big) \\ &=& \frac{4}{3}\cdot \sqrt{\frac23} \\ \hline \end{array} \)

Help? I don't know how to do this question

Tangent Secant Theorem

\(\overline{AD}^2 = \overline{AB} \cdot (\overline{AB} + \overline{BC}) \\ \overline{AD}^2 = 16 \cdot (16 +9) \\ \overline{AD}^2 = 16 \cdot 25 \\ \overline{AD} = 4 \cdot 5 \\ \overline{AD} = 20 \)

first 3 terms of arithmetic sequence

where the 21st term is -38 and the 50th term is -125

Let
\(\begin{array}{ll} x=21 & t_x=t_{21}=-38 \\ y=50 & t_y=t_{50}=-125 \\ z=1 & t_z=t_1= ? \\ \end{array} \)

\(\begin{array}{|lrcll|} \hline \text{AP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ &t_{1} &=& (-38)\cdot (\frac{1-50}{21-50}) + (-125)\cdot (\frac{21-1}{21-50}) \\ &t_{1} &=& (-38)\cdot (\frac{-49}{-29}) + (-125)\cdot (\frac{20}{-29}) \\ &t_{1} &=& -38\cdot (\frac{49}{29}) + 125\cdot (\frac{20}{29}) \\ &t_{1} &=& \frac{-38\cdot 49+ 125\cdot 20}{29} \\ &t_{1} &=& \frac{638}{29} \\ & \mathbf{ t_{1} } & \mathbf{=} & \mathbf{22} \\ \hline \end{array}\)

Let
\(\begin{array}{ll} x=21 & t_x=t_{21}=-38 \\ y=50 & t_y=t_{50}=-125 \\ z=2 & t_z=t_2= ? \\ \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{AP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ &t_{2} &=& (-38)\cdot (\frac{2-50}{21-50}) + (-125)\cdot (\frac{21-2}{21-50}) \\ &t_{2} &=& (-38)\cdot (\frac{-48}{-29}) + (-125)\cdot (\frac{19}{-29}) \\ &t_{2} &=& -38\cdot (\frac{48}{29}) + 125\cdot (\frac{19}{29}) \\ &t_{2} &=& \frac{-38\cdot 48+ 125\cdot 19}{29} \\ &t_{2} &=& \frac{551}{29} \\ & \mathbf{ t_{2} } & \mathbf{=} & \mathbf{19} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline d&=&t_2-t_1 \\ d&=&19-22 \\ d&=&-3 \\\\ t_3 &=& t_2 +d \\ t_3 &=& 19-3 \\ \mathbf{ t_{3} } & \mathbf{=} & \mathbf{16} \\ \hline \end{array}\)

The first 3 terms of arithmetic sequence: 22, 19, 16, ...

a5 = 7, a8 = 56. Find a11

\(\begin{array}{|lrcll|} \hline \text{AP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ \text{GP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x^{\frac{z-y}{x-y}} \cdot t_y^{\frac{x-z}{x-y}} } \\ \hline \end{array} \)

Let

\(\begin{array}{ll} x= 5 & t_x=t_5=7 \\ y=8 & t_y=t_8=56 \\ z=11 & t_z=t_{11}=? \\ \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{AP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ &t_{11} &=& 7\cdot (\frac{11-8}{5-8}) + 56\cdot (\frac{5-11}{5-8}) \\ &t_{11} &=& 7\cdot (\frac{3}{-3}) + 56\cdot (\frac{-6}{-3}) \\ &t_{11} &=& 7\cdot (-1) + 56\cdot 2 \\ &t_{11} &=& -7 + 112 \\ & \mathbf{ t_{11} } & \mathbf{=} & \mathbf{105} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \text{GP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x^{\frac{z-y}{x-y}} \cdot t_y^{\frac{x-z}{x-y}} } \\ &t_{11} &=& 7^{\frac{11-8}{5-8}} \cdot 56^{\frac{5-11}{5-8}} \\ &t_{11} &=& 7^{-1} \cdot 56^{2} \\ &t_{11} &=& \frac{3136}{7} \\ & \mathbf{ t_{11} } & \mathbf{=} & \mathbf{448} \\ \hline \end{array} \)

sin(arcsin x + arccos x)

i)
\(\begin{array}{lrcll} & \cos{(\varphi)} &=& x \\ \text{or}& \quad \varphi &=& \arccos{(x)}\\ \end{array} \)

ii)
\(\begin{array}{lrclcl} &\sin{(90^\circ-\varphi)} &=& \cos{(\varphi)} &=& x \\ \text{or}& \quad 90^\circ-\varphi && &=& \arcsin{(x)}\\ \end{array} \)

iii)

\(\begin{array}{rcll} && \sin\Big(\arcsin(x)+\arccos(x)\Big) \\ &=& \sin(90^\circ-\varphi+\varphi) \\ &=& \sin(90^\circ)\\ &=& 1\\ \end{array}\)

If sin(A)=cos(A) was is the angle measurement?

\(\small{ \begin{array}{rcl|rcl|rcl} \sin(A)&=&\cos(A) \\ \cos(A)-\sin(A) &=& 0 & \sin(45^{\circ}-A) &=& \sin(45^{\circ})\cdot\cos(A)- \cos(45^{\circ})\cdot\sin(A) & \sin(45^{\circ})=\cos(45^{\circ}) = \frac{\sqrt{2}}{2} \\ & & & \sin(45^{\circ}-A) &=& \frac{\sqrt{2}}{2}\cdot\cos(A)- \frac{\sqrt{2}}{2}\cdot\sin(A) \\ & & & \sin(45^{\circ}-A) &=& \frac{\sqrt{2}}{2}\Big(\cdot\cos(A)- \sin(A) \Big) \\ \frac{2}{\sqrt{2}}\cdot \sin(45^{\circ}-A)&=& 0 & \frac{2}{\sqrt{2}}\cdot \sin(45^{\circ}-A) &=& \cos(A)- \sin(A) \\ \sin(45^{\circ}-A)&=& 0 \\ 45^{\circ}-A &=& \arcsin(0) \\ 45^{\circ}-A &=& 0\pm n\cdot 360^{\circ} \qquad n \in \mathbb{N} \\ \mathbf{A} & \mathbf{=} & \mathbf{45^{\circ} \pm n\cdot 360^{\circ}} \\\\ \sin(45^{\circ}-A)=\sin(180^{\circ}-(45^{\circ}-A) ) &=& 0 \\ 180^{\circ}-(45^{\circ}-A) &=& \arcsin(0) \\ 180^{\circ}-45^{\circ}+A &=& \arcsin(0) \\ 135^{\circ} + A &=& 0\pm n\cdot 360^{\circ} \qquad n \in \mathbb{N} \\\\ A &=& -135^{\circ} \pm n\cdot 360^{\circ} \\ A &=& -135^{\circ} + 360^{\circ} \pm n\cdot 360^{\circ} \\ \mathbf{A} & \mathbf{=} & \mathbf{225^{\circ} \pm n\cdot 360^{\circ}} \\ \end{array} } \)

What is the distance to the earth’s horizon

Tangent Secant Theorem

What is the distance to the earth’s horizon from point P?

Round to the nearest tenth. 

r = 3959 mi

\(x^2 = 15.6 \cdot (15.6 + 2\cdot 3959) \\ x = 351.8\ \text{mi}\)

A skydiver is at an altitude of 1.4 mi above the earth’s surface.

From the skydiver’s viewpoint, what is the distance to the horizon?

Round to the nearest tenth.

r = 3959 mi

\(x^2 = 1.4\cdot (1.4 + 2\cdot 3959) \\ x = 105.3\ \text{mi}\)

Michael is 12 years older than Brandon. Seventeen years ago,

Michael was 4 times as old as Brandon.

How old is Brandon now?

Let m = Michael
Let b = Brandon

\(\begin{array}{|lrcll|} \hline (1) & m &=& b+12 \\\\ (2) & m-17 &=& 4\cdot (b-17) \quad & | \quad m=b+12 \\ & b+12-17 &=& 4\cdot (b-17) \\ & b+5 &=& 4b- 68 \\ & 3b&=& 63 \\ & \mathbf{b} & \mathbf{=} & \mathbf{21} \\\\ (1) & m &=& b+12 \quad & | \quad b=21 \\ & m &=& 21+12 \\ & \mathbf{m} & \mathbf{=} & \mathbf{33} \\ \hline \end{array} \)

Michael is 33 and Brandon is 21

Was gibt -119 in Binär

Mit Zweier Komplement und in 8-Bit?

Methode 1:

Vorzeichen ignorieren und ins Binärsystem umrechnen: 119₍₁₀₎ = 01110111₍₂₎

Invertieren: Not[ 01110111] = 10001000

Eins addieren: 10001000 + 00000001 = 10001001

−119₍₁₀₎ = 10001001₍₂₎  (Mit Zweier Komplement und in 8-Bit)

Methode 2:

Ist \({\displaystyle x} \) eine negative Zahl, so errechnet sich \({\displaystyle x} \) in Zweierkomplementdarstellung ( \({\displaystyle x_{z}}\) ) mit  \({\displaystyle n}\) Stellen wie folgt:

\(\begin{array}{rcll} x_z &=& 2^n -|x| \qquad n = 8 Bit\\ x_z &=& 2^8- |-119| \\ x_z &=& 256- 119 \\ x_z &=& 137 \\ \end{array} \)

137₍₁₀₎ = 10001001₍₂₎  (Mit Zweier Komplement und in 8-Bit)

sin v = 0,5

Decide the value of: sin 2v

\(\begin{array}{rcl|rcl|rcl} \sin(2v) &=& 2\cdot \sin(v)\cdot \cos(v) & \cos(v) &=& \sqrt{1-\sin^2(v)} & \sin(v) = \frac12 \\ & & & &=& \sqrt{1-\left(\frac12 \right)^2} \\ & & & &=& \sqrt{1- \frac14} \\ & & & &=& \sqrt{\frac34} \\ & & & &=& \frac{\sqrt{3}}{2} \\ &=& 2\cdot \frac12\cdot \frac{\sqrt{3}}{2} & &=& \frac{\sqrt{3}}{2} \\ &=& \frac{\sqrt{3}}{2} \\ \end{array} \)

\(\sin(2v) = \frac{\sqrt{3}}{2}\)