heureka

könnte mir jemand dies nach t umstellen ich schaffe es einfach nicht 13.5=15*(1-e^(-t/0.3))

\(\begin{array}{|rcll|} \hline 13.5 &=& 15\cdot (~1-e^{ -\frac{t}{0.3}} ~) \quad & | \quad :15 \\ \frac{ 13.5}{15} &=& 1-e^{-\frac{t}{0.3}} \\ 0.9 &=& 1-e^{-\frac{t}{0.3}} \quad & | \quad -1 \\ 0.9 -1 &=& -e^{-\frac{t}{0.3}} \\ -0.1 &=& -e^{-\frac{t}{0.3}} \quad & | \quad \cdot(-1) \\ 0.1 &=& e^{-\frac{t}{0.3}} \\ 0.1 &=& \frac{1}{e^{\frac{t}{0.3}} } \quad & | \quad \cdot e^{\frac{t}{0.3}} \\ 0.1\cdot e^{\frac{t}{0.3}} &=& 1 \quad & | \quad \cdot 10 \\ e^{\frac{t}{0.3}} &=& 10 \quad & | \quad \ln() \text { auf beiden Seiten} \\ \ln( e^{\frac{t}{0.3}} ) &=& \ln(10) \\ \frac{t}{0.3} &=& \ln(10) \quad & | \quad \cdot 0.3 \\ t &=& \ln(10) \cdot 0.3 \\ t &=& 2.30258509299 \cdot 0.3 \\ \mathbf{ t } & \mathbf{=} & \mathbf{0.69077552790} \\ \hline \end{array}\)

Hi ! I have to show that
\(\sum \limits_{n=2}^{\infty} \left( \dfrac { \ln \left[(1+ \frac {1} {n})^n\cdot (n+1) \right]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \right)= \dfrac {1} {2\cdot \ln(2)}\)
Can anyone help ? Thank you in advance.

\(\begin{array}{|rcll|} \hline && \sum \limits_{n=2}^{\infty} \dfrac { \ln \left[(1+ \frac {1} {n})^n\cdot (n+1) \right]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { \ln \left[(\frac {n+1} {n})^n\cdot (n+1) \right]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { \ln \left[ \frac {(n+1)^n} {n^n}\cdot (n+1) \right]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { \ln [ \frac {(n+1)^{n+1}} {n^n} ]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { \ln \left[(n+1)^{n+1} \right]- \ln(n^n) } { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right] } \\\\ &=&\sum \limits_{n=2}^{\infty} \left( \dfrac { \ln \left[(n+1)^{n+1} \right] } { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right] } - \dfrac { \ln(n^n) } { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right] } \right) \\\\ &=&\sum \limits_{n=2}^{\infty} \left( \dfrac { 1 } { \ln(n^n) } - \dfrac { 1 } { \ln \left[(n+1)^{n+1} \right] } \right) \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln(n^n) } -\sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln \left[(n+1)^{n+1} \right] } \\\\ &=&\sum \limits_{n=1}^{\infty} \dfrac { 1 } { \ln[(n+1)^{n+1}] } -\sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln \left[(n+1)^{n+1} \right] } \\\\ &=& \dfrac{1}{\ln(2^2)} + \sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln[(n+1)^{n+1}] } -\sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln \left[(n+1)^{n+1} \right] } \\\\ &=& \dfrac{1}{\ln(2^2)} \\\\ &=& \dfrac{1}{2\cdot \ln(2)} \\ \hline \end{array}\)

sechs qadrat meter stoff kosten $36.
Wie viel geld muss man für 10 qadratmeter stoff bezahlen

\(\begin{array}{|rcll|} \hline && 10\ m^2 \cdot \frac{$36}{6\ m^2} \\ &=& $\ \frac{10\cdot 36}{6} \\ &=& $\ 10\cdot 6 \\ &=& $\ 60 \\ \hline \end{array}\)

Für 10 m² Stoff muss man $ 60 bezahlen.

The price of a scooter is € 2500.

Henk has not saved enough money to pay for the scooter.

He borrows € 1500 from his parents.

How many Euros has Henk for the scooter ?

Henk has € 2500 - € 1500 = € 1000

Welche Minimumseigenschaft hat das arithmetische Mittel?

what is the area of  A(0,3) B(4,-1) C(-2,1) D(2,-3)

\(\begin{array}{|rcll|} \hline \text{Let } \vec{A} &=& \binom{0}{3} \\ \text{Let } \vec{B} &=& \binom{4}{-1} \\ \text{Let } \vec{C} &=& \binom{-2}{1} \\ \text{Let } \vec{D} &=& \binom{2}{-3} \\ \hline \end{array}\)

Points clockwise:

\(\begin{array}{|rcll|} \hline 1. & \vec{A} &=& \binom{0}{3} \\ 2. & \vec{B} &=& \binom{4}{-1} \\ 3. & \vec{D} &=& \binom{2}{-3} \\ 4. & \vec{C} &=& \binom{-2}{1} \\ \hline \end{array}\)

Area:

\(\begin{array}{|rcll|} \hline 2A &=& |\vec{B} \times \vec{A}| + |\vec{D} \times \vec{B}| + |\vec{C} \times \vec{D}| + |\vec{A} \times \vec{C}| \\ 2A &=& |\binom{4}{-1} \times \binom{0}{3}| + |\binom{2}{-3} \times \binom{4}{-1}| + |\binom{-2}{1} \times \binom{2}{-3}| + |\binom{0}{3} \times \binom{-2}{1}| \\ 2A &=& 4\cdot 3-(-1)\cdot 0 +2(-1)-(-3)\cdot 4 +(-2)(-3)-1\cdot2 + 0\cdot 1 -3(-2) \\ 2A &=& 12-0 -2+12 +6-2 + 0 +6 \\ 2A &=& 32 \\ \mathbf{A} & \mathbf{=}& \mathbf{16} \\ \hline \end{array}\)

sinx-2sinxcosx=0

Formula:

\(\begin{array}{|rcll|} \hline 2\ sin(x) \cos(x) &=& \sin(2x)\\ \hline \end{array}\)

So:

\(\begin{array}{|rcll|} \hline 0 &=& \sin(x) - 2\cdot sin(x) \cos(x) \quad & | \quad 2\cdot sin(x) \cos(x) = \sin(2x) \\ 0 &=& \sin(x) - \sin(2x) \quad & | \quad + \sin(2x) \\ \sin(2x) &=& \sin(x) \\ \hline \end{array} \)

First result:

\(\begin{array}{|rcll|} \hline \sin(2x) &=& \sin(x) \quad & | \quad \arcsin() \text{ both sides } \\ 2x &=& x \quad & | \quad -x \\ 2x-x &=& 0 \\ \mathbf{ x } & \mathbf{=} & \mathbf{ 0 \pm 360^{\circ} \cdot n } \quad & | \quad n \in N \\ \hline \end{array}\)

Second result:

\(\begin{array}{|rcll|} \hline \sin(2x) &=& \sin(x) \quad & | \quad \sin(x) = \sin(180^{\circ}-x) \\ \sin(2x) &=& \sin(180^{\circ}-x) \quad & | \quad \arcsin() \text{ both sides } \\ 2x &=& 180^{\circ}-x \quad & | \quad +x \\ 2x+x &=& 180^{\circ} \\ 3x &=& 180^{\circ} \quad & | \quad : 3 \\ x &=& 60^{\circ} \\ \mathbf{ x } & \mathbf{=} & \mathbf{ 60^{\circ} \pm 360^{\circ} \cdot n } \quad & | \quad n \in N \\ \hline \end{array}\)

Third result:

\(\begin{array}{|rcll|} \hline \sin(2x) &=& \sin(x) \quad & | \quad -\sin(x) = \sin(180^{\circ}+x) \\ \sin(2x) &=& -\sin(180^{\circ}+x) \\ \sin(2x) &=& \sin(-180^{\circ}-x) \quad & | \quad \arcsin() \text{ both sides } \\ 2x &=& -180^{\circ}-x \quad & | \quad +x \\ 2x+x &=& -180^{\circ} \\ 3x &=& -180^{\circ} \quad & | \quad : 3 \\ x &=& -60^{\circ} \\ \mathbf{ x } & \mathbf{=} & \mathbf{ -60^{\circ} \pm 360^{\circ} \cdot n } \quad & | \quad n \in N \\ \hline \end{array} \)

1.

\(y = t\sin t\\ x = t\cos t\)

2.

\(y = t\cos t\\ x = t\sin t\)

An architect build a model of a 220-foot tall building he is designning.

The model is 25 inches tall and 10 inches wide.

How wide is the actual building ?

\(\begin{array}{|rcll|} \hline \frac{10\ in.}{25\ in.} &=& 0.4 \\\\ \frac{\text{wide}}{220\ foot} &=& 0.4 \\ \text{wide} &=& 0.4\cdot 220\ foot \\ \text{wide} &=& 88\ foot \\ \hline \end{array}\)

The building is 88 foot wide

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