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Three circles of radius 1 are externally tangent to each other and internally tangent to a larger circle.
What is the radius of the large circle?
Express your answer as a common fraction in simplest radical form.

Let Radius of the larger circle = R.
Let Radius of the three circles = r = 1.

Let Area of the triangle ABC = \(A_{ABC}\)
Let Area of the triangle AOB = \( A_{AOB}\)

Let AC = CB = BA = 2r
Let AO = OB = R-r

\(\mathbf{A_{ABC} = \ ?}\)

\(\begin{array}{|lrcll|} \hline (\text{Heron}) & s &=& \frac{2r+2r+2r}{2} \\ & &=& 3r \\ & A_{ABC}^2 &=& s(s-2r)(s-2r)(s-2r) \\ & A_{ABC}^2 &=& s(s-2r)^3 \quad & | \quad s = 3r \\ & A_{ABC}^2 &=& 3r(3r -2r)^3 \\ & A_{ABC}^2 &=& 3rr^3 \\ & A_{ABC}^2 &=& 3r^4 \\ & A_{ABC} &=& r^2 \sqrt{3} \\ \hline \end{array}\)

\(\mathbf{A_{AOB} = \ ?} \)

\(\begin{array}{|lrcll|} \hline & A_{AOB} &=& \frac{A_{ABC}}{3} \\ (1) & A_{AOB} &=& \frac{r^2 \sqrt{3}}{3} \\ \hline \end{array}\)

\(\mathbf{(\text{Heron})\ A_{AOB} = \ ?}\)

\(\begin{array}{|rcll|} \hline (\text{Heron}) & s &=& \frac{2r+(R-r)+(R-r)}{2} \\ & s &=& \frac{2R}{2} \\ & s &=& R \\ & A_{AOB}^2 &=& s(s-2r)[s-(R-r)][s-(R-r)] \quad & | \quad s = R \\ & A_{AOB}^2 &=& R(R-2r)[R-R+r)][R-R+r] \\ & A_{AOB}^2 &=& R(R-2r)r^2 \\ (2) & A_{AOB} &=&r\sqrt{R(R-2r)} \\ \hline \end{array}\)

(1) = (2)

\(\begin{array}{|rcll|} \hline \frac{r^2 \sqrt{3}}{3} &=& r\sqrt{R(R-2r)} \\ \frac{r \sqrt{3}}{3} &=& \sqrt{R(R-2r)} \quad & | \quad \text{square both sides}\\ \frac{r^2}{3} &=& R(R-2r) \\ R^2-2rR -\frac{r^2}{3} &=& 0 \quad & | \quad \cdot 3 \\ 3R^2-6rR - r^2 &=& 0 \\ \\ R &=& \frac{6r \pm \sqrt{36r^2-4\cdot 3\cdot (-r^2) } }{2\cdot 3} \\ R &=& \frac{6r \pm \sqrt{36r^2+12r^2 } }{6} \\ R &=& \frac{6r \pm \sqrt{48r^2 } }{6} \\ R &=& \frac{6r \pm 4r\sqrt{3} }{6} \\ R &=& r \pm \frac23 r \sqrt{3} \quad & | \quad R \gt 0 \\ R &=& r + \frac23 r \sqrt{3} \quad & | \quad r = 1 \\ \mathbf{R} & \mathbf{=} & \mathbf{1 + \frac23 \sqrt{3}} \\ \hline \end{array}\)

x=AB / 2

tan(APB/2) = 9 / 12 = x / y

y = (12/9) * x

x^2 +y^2 = 12^2

x^2 + [ (12/9) * x ]^2 =12^2

x = 36/5

AB = 2x

AB = 72/5 = 14.4

In the figure below, \( \angle E = 40^{\circ}\) and arc AB, arc BC, and arc CD all have equal length.
Find the measure of \(\angle ACD\), in degrees.

1. \(\mathbf{ \angle BCE =\ ? }\)

\(\begin{array}{|rcll|} \hline \angle BCE &=& 90^{\circ} - \frac{40^{\circ}}{2} \\ \mathbf{ \angle BCE } &\mathbf{ =}& \mathbf{ 70^{\circ}} \\ \hline \end{array} \)

2. \(\mathbf{ \angle BCA =\ ? }\)

\(\begin{array}{|rcll|} \hline \angle BCE + 2\cdot \angle BCA &=& 180^{\circ} \\ \angle BCA &=& \frac{180^{\circ}- \angle BCE} {2} \\ \angle BCA &=& 90^{\circ}- \frac{\angle BCE} {2} \quad & | \quad \angle BCE = 70^{\circ} \\ \angle BCA &=& 90^{\circ}- \frac{70^{\circ}} {2} \\ \angle BCA &=& 90^{\circ}- 35^{\circ} \\ \mathbf{ \angle BCA} &\mathbf{ =}& \mathbf{ 55^{\circ}} \\ \hline \end{array} \)

3. \(\mathbf{ \angle ACD =\ ? }\)

\(\begin{array}{|rcll|} \hline \angle ACD &=& \angle BCE - \angle BCA \\ \angle ACD &=& 70^{\circ} - 55^{\circ} \\ \mathbf{ \angle ACD} &\mathbf{ =}& \mathbf{ 15^{\circ} } \\ \hline \end{array}\)

A 4 digit number is equal to the digit sum of the number multiplied by 109.

Find the biggest numbber satisfied the above condition

\(\begin{array}{|rccccl|} \hline 1. & 1 & 0 & 9 & 0 \\ 2. & 1 & 3 & 0 & 8 \\ 3. & 1 & 4 & 1 & 7 \\ 4. & 1 & 5 & 2 & 6 \\ 5. & 1 & 6 & 3 & 5 \\ 6. & 1 & 7 & 4 & 4 \\ 7. & 1 & 8 & 5 & 3 \\ 8. & 1 & 9 & 6 & 2 \\ 9. & 2 & 2 & 8 & 9 \\ 10. & \mathbf{2} & \mathbf{3} & \mathbf{9} & \mathbf{8} & \text{max} \\ \hline \end{array} \)

Bitte um Kommentar

Nehmen wir mal eine beliebige Ausgangslage von "0 0 0 0" an:

Wie viele Möglichkeiten gibt es ein beliebiges Rad zu verdrehen (nach den Vorgaben).

Verdrehung z.B. des ersten Rades:

1. Möglichkeit: Eine Zahl weiter: 1 0 0 0

2. Möglichkeit: Zwei Zahlen weiter: 2 0 0 0

3. Möglichkeit: Eine Zahl zurück: 9 0 0 0

4. Möglichkeit: Zwei Zahlen zurück: 8 0 0 0

Für ein Rad ergeben sich insgesamt 4 Möglichkeiten.

Für zwei Räder ergeben sich 4x4 = 16 Möglichkeiten.

Es gibt 6 Möglichkeiten 2 unterschiedliche Räder zu kombinieren.
\(\begin{array}{|lcccc|} \hline & 0 & 0 & 0 & 0 \\ \hline 1. & x & x & 0 & 0 \\ 2. & x & 0 & x & 0 \\ 3. & x & 0 & 0 & x \\ 4. & 0 & x & x & 0 \\ 5. & 0 & x & 0 & x \\ 6. & 0 & 0 & x & x \\ \hline \end{array}\)

Also gibt es insgesamt 16 x 6 = 96 Möglichkeiten.

Solve for X

x+1-2sqrt(x+4)=0

\(\begin{array}{|rcll|} \hline x+1-2\sqrt{x+4} &=& 0 \quad & | \quad +2\sqrt{x+4} \\ x+1 &=& 2\sqrt{x+4} \quad & | \quad \text{square both sides} \\ (x+1)^2 &=& 4(x+4) \\ x^2+2x+1 &=& 4x+16 \\ x^2+2x+1 &=& 4x+16 \quad & | \quad -4x \\ x^2+2x-4x+1 &=& 16 \\ x^2-2x+1 &=& 16 \quad & | \quad - 16 \\ x^2-2x+1-16 &=& 0 \\ x^2-2x-15 &=& 0 \\ (x-5)(x+3) &=& 0 \\\\ x_1 = 5 & \text{ or } & x_2 = -3 \\ \hline \end{array} \)

Proof:

\(x_1 = 5:\)
\(\begin{array}{|rcll|} \hline 5+1-2\sqrt{5+4} &\overset{?}{=}& 0 \\ 6-2\sqrt{9} &\overset{?}{=}& 0 \\ 6-2\cdot 3 &\overset{?}{=}& 0 \\ 6-6 &\overset{?}{=}& 0 \\ 0 &\overset{!}{=}& 0 \\\\ x = 5 \text{ is a solution } \\ \hline \end{array} \)

\(x_2 = -3\)
\(\begin{array}{|rcll|} \hline -3+1-2\sqrt{-3+4} &\overset{?}{=}& 0 \\ -2-2\sqrt{1} &\overset{?}{=}& 0 \\ -2-2\cdot 1 &\overset{?}{=}& 0 \\ -2-2 &\overset{?}{=}& 0 \\ -4 &\ne & 0 \\\\ x = -3 \text{ is not a solution } \\ \hline \end{array}\)

The only solution is x = 5

Factor and express the answer as a quotient with positive exponents.

x^(-2)-4x^(-5)+3x^(-8)

\(\begin{array}{|rcll|} \hline && x^{-2}-4x^{-5}+3x^{-8} \\ &=& x^{-8}\left( \frac{x^{-2}} {x^{-8}} - \frac{4x^{-5}} {x^{-8}} +3 \frac{ x^{-8} } {x^{-8}} \right) \\ &=& x^{-8} ( x^{-2} x^{8} - 4x^{-5} x^{8} +3 ) \\ &=& x^{-8} ( x^{-2+8} - 4x^{-5+8} +3 \\ &=& x^{-8} ( x^{6} - 4x^{3} +3 ) \\ &=& x^{-8} (x^3-1)(x^3-3) \\ &=& \dfrac{(x^3-1)(x^3-3)}{x^{8}} \\ \hline \end{array}\)

Given that
\(f(2)=5\)
and \(f^{-1}(x+4)=2f^{-1}(x)+1\)
for all \(x\),
find \(f^{-1}(17)\).

\(f^{-1}\) is the inverse of f

Formula:
\(\begin{array}{|rcll|} \hline y &=&f(x) \\ x &=& f^{-1}(y) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline f(2) &=& 5 \quad & | \quad x= 2 \qquad y = 5 \\ x &=& f^{-1}(y) \\ 2 &=& f^{-1}(5) \\\\ \mathbf{f^{-1}(5)} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & f^{-1}(17) &=& 2f^{-1}(13) + 1 \quad & | \quad 17 = x + 4 \qquad x = 13 \\ (2) & f^{-1}(13) &=& 2f^{-1}(9) + 1 \quad & | \quad 13 = x + 4 \qquad x = 9 \\ (3) & f^{-1}(9) &=& 2f^{-1}(5) + 1 \quad & | \quad 9 = x + 4 \qquad x = 5 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (3) & f^{-1}(9) &=& 2f^{-1}(5) + 1 \quad & | \quad \mathbf{f^{-1}(5)= 2} \\ & f^{-1}(9) &=& 2\cdot 2 + 1 \\ & f^{-1}(9) &=& 5 \\\\ (2) & f^{-1}(13) &=& 2f^{-1}(9) + 1 \quad & | \quad \mathbf{f^{-1}(9)= 5} \\ & f^{-1}(13) &=& 2\cdot 5 + 1 \\ & f^{-1}(13) &=& 11 \\\\ (1) & f^{-1}(17) &=& 2f^{-1}(13) + 1 \quad & | \quad \mathbf{f^{-1}(13)= 11} \\ & f^{-1}(17) &=& 2\cdot 11 + 1 \\ & \mathbf{f^{-1}(17)} &\mathbf{=}& \mathbf{23} \\ \hline \end{array}\)

question

\(\text{For complex numbers } \\ z, \text{let } \left[f(z) = \left\{ \begin{array}{cl} z^{2}&\text{ if }z\text{ is not real}, \\ z+2 &\text{ if }z\text{ is real}. \end{array} \right.\right] \\ \text{Find } f(i)+f(1)+f(-1)+f(-i).\)

\(\begin{array}{|l|clcrc|l|} \hline z = i && f(i) = z^2 = i^2 &=& -1 && \text{(z is not real)} \\ z = 1 && f(1) = z+2 = 1+2 &=& 3 && \text{(z is real)} \\ z = -1 && f(-1) = z+2 = -1+2 &=& 1 && \text{(z is real)} \\ z = -i && f(-i) = z^2 = (-i)^2 = i^2 &=& -1 && \text{(z is not real)} \\ \hline && f(i)+f(1)+f(-1)+f(-i) = -1+3+1-1 = 2 \\ \hline \end{array} \)

Guten Tag,

ich möchte die Magnus Tetens Formel nach T umstellen. Ich habe die fertig umgestellte Formel bereits,

komme aber wenn ich die selber umstellen möchte nicht soweit.

\(p_s(T) =610,7*e^{\left( \dfrac{ 17,27*T}{237,3+T } \right) }\)

\(\small{ \begin{array}{|rcll|} \hline p_s(T) &=& 610,7*e^{\left( \dfrac{ 17,27*T}{237,3+T } \right) } \quad & | \quad : 610,7 \\\\ \dfrac{p_s(T)}{610,7} &=& e^{\left( \dfrac{ 17,27*T}{237,3+T } \right) } \quad & | \quad \ln{()} \\\\ \ln\left( \dfrac{p_s(T)}{610,7} \right) &=& \dfrac{ 17,27*T}{237,3+T } \quad & | \quad *(237,3+T) \\\\ \ln\left( \dfrac{p_s(T)}{610,7} \right)*(237,3+T) &=& 17,27*T \\\\ \ln\left( \dfrac{p_s(T)}{610,7} \right)*237,3 + T* \ln\left( \dfrac{p_s(T)}{610,7} \right) &=& 17,27*T \quad & | \quad - T* \ln\left( \dfrac{p_s(T)}{610,7} \right) \\\\ \ln\left( \dfrac{p_s(T)}{610,7} \right)*237,3 &=& 17,27*T - T* \ln\left( \dfrac{p_s(T)}{610,7} \right) \\\\ \ln\left( \dfrac{p_s(T)}{610,7} \right)*237,3 &=& T*\left[ 17,27 - \ln\left( \dfrac{p_s(T)}{610,7} \right) \right] \quad & | \quad : 17,27 - \ln\left( \dfrac{p_s(T)}{610,7} \right) \\\\ \dfrac{\ln\left( \dfrac{p_s(T)}{610,7} \right)*237,3} {17,27 - \ln\left( \dfrac{p_s(T)}{610,7} \right) } &=& T \\ \hline \end{array} }\)