heureka

A model of the eiffel tower that was purchased in a gift shop is 29.55 inches tall. the actual height of the eiffel tower is 985 feet, or 11,820 inches. what scale factor was used to make the model ?

$$\\\small{\text{ $ \dfrac{11820\ \mathrm{inches} } {29.55\ \mathrm{inches}} = 400 $ }}\\\\ \small{\text{ $ 1:400 $ }}$$

What is the nth term of this sequence?               3, 9, 27, 81, 243

$$\\a_{n+1} =3\cdot a_n \\ a_n = 3\cdot a_{n-1}\\$$

(1+3+5+⋯+3983)/1992

$$\small{\text{ $ \dfrac{ 1+3+5+\cdots +3983} {1992} = \ ? $ }}\\ \\ \small{\text{ The arithmetic Series $ 1+3+5+\cdots +3983 $ has $a_1 = 1$ and $a_n=3983$ and $d = 2$ }}\\ \small{\text{ The formula is $a_n = a_1+(n-1)\cdot d$ or $ 3983 = 1 + (n-1)\cdot 2 $ }}\\ \small{\text{ So $n = 1992 $ }}\\ \small{\text{ The sum $ = 1+3+5+\cdots +3983 = n \left( \dfrac{a_1+a_n}{2} \right) = 1992\cdot \left( \dfrac{1+3983}{2} \right) = 1992\cdot 1992 $ }}\\\\ \small{\text{ $ \dfrac{ 1+3+5+\cdots +3983} {1992} = \dfrac{1992\cdot 1992} {1992} = 1992$ }}\\ \\$$

What is 9 to the power of 999 ?

change of basis: 9 to 10

$$b_1^{ e_1 } = b_2^{ e_2 } \quad | \quad \ln() \\ e_1 \cdot \ln{(b_1)} = e_2 \cdot \ln{(b_2)} \\\\ e_2 = e_1 \cdot \dfrac{ \ln{(b_1)} } { \ln{(b_2)} } \\ \boxed{ b_1^{ e_1 } = b_2^{ e_1 \left( \cdot \dfrac{ \ln{(b_1)} } { \ln{(b_2)} } \right) }}\\\\ \small{\text{ $ b_1 = 9 \qquad e_ 1 = 999 \qquad b_2 = 10 $ }} \\\\ \small{\text{ $ 9^{999} = 10^{999 \left( \cdot \dfrac{ \ln{(9)} } { \ln{(10)} } \right) } $ }} \\\\ \small{\text{ $ 9^{999} = 10^{999 \cdot( 0.95424250944 ) } $ }} \\ \small{\text{ $ 9^{999} = 10^{953.288266930} $ }} \\ \small{\text{ $ 9^{999} = 10^{0.288266930}\cdot 10^{953} $ }} \\ \small{\text{ $ 9^{999} = 1.94207916858 \cdot 10^{953} $ }}$$

What is  (2+y)³ written out ?

$$\small{\text{ $(2+y)^3 = \ ? $}}\\ \small{\text{ $(2+y)^3 = \textcolor[rgb]{1,0,0}{1} \cdot 2^3 \cdot y^0 + \textcolor[rgb]{1,0,0}{3} \cdot 2^2 \cdot y^1 + \textcolor[rgb]{1,0,0}{3} \cdot 2^1 \cdot y^2 + \textcolor[rgb]{1,0,0}{1} \cdot 2^0 \cdot y^3 $}}\\ \small{\text{ $(2+y)^3 = 8\cdot y^0 + 12\cdot y^1 + 6 y^2 + y^3 $}}\\ \small{\text{ $(2+y)^3 = 8 + 12 y + 6 y^2 + y^3 $}}\\$$

sin-1(1.04) ?

If 1.04 is a complex number z:

$$\\ \small{\text{ $ \boxed{ \sin^{-1}(z) = -i\cdot \log\left(\sqrt{1-z^2}+i\cdot z \right) \qquad $Complex logarithm $\log{(z)} = \ln{(|z|)}+i\cdot \varphi } $ }}\\\\ \small{\text{We have $z = 1.04$}}\\ \small{\text{ \sin^{-1}(z)=sin^{-1}(1.04) = -i\cdot \log\left(\sqrt{1-(1.04)^2}+i\cdot 1.04 \right) $}}\\ \small{\text{ sin^{-1}(1.04) = -i\cdot \log\left(\sqrt{1-(1.04)^2}+i\cdot 1.04 \right) $}}\\ \small{\text{ sin^{-1}(1.04) = -i\cdot \log\left(\sqrt{ -0.0816 }+i\cdot 1.04 \right) $}}\\ \small{\text{ sin^{-1}(1.04) = -i\cdot \log\left(\sqrt{0.0816}\cdot \sqrt{-1}+i\cdot 1.04 \right) $}}\\ \small{\text{ sin^{-1}(1.04) = -i\cdot \log\left(\sqrt{0.0816}\cdot i+i\cdot 1.04 \right) $}}\\ \small{\text{ sin^{-1}(1.04) = -i\cdot \log\left(0.28565713714\cdot i+i\cdot 1.04 \right) $}}\\ \small{\text{ sin^{-1}(1.04) = -i\cdot \log\left(0.28565713714\cdot i+i\cdot 1.04 \right) $}}\\ \small{\text{ sin^{-1}(1.04) = -i\cdot \mathbf{ \log{\left( 1.32565713714 \cdot i \right)} } $}}\\\\ \small{\text{ a+b\cdot i = \mathbf{1.32565713714 \cdot i } \qquad a= 0 $ and $ b = 1.32565713714 $}}\\ \small{\text{ $\textcolor[rgb]{0,0,1}{|1.32565713714 \cdot i|} = \sqrt{a^2+b^2} = \sqrt{0^2+1.32565713714^2} = \textcolor[rgb]{0,0,1}{1.32565713714} $}}\\ \small{\text{ $\varphi = \tan^{-1}{ \left(\dfrac{1.32565713714}{0} \right) } \quad \Rightarrow \quad \varphi = \textcolor[rgb]{1,0,0}{\dfrac{\pi}{2}} $}} \\\\ \small{\text{ $\log{\left( 1.32565713714 \cdot i \right)} = \ln{ ( \textcolor[rgb]{0,0,1}{ 1.32565713714 } ) } + i \cdot \textcolor[rgb]{1,0,0}{ \dfrac{\pi}{2} } $}}\\\\ \small{\text{ sin^{-1}(1.04) = -i\cdot \left( \ln{ ( 1.32565713714 ) } + i \cdot \dfrac{\pi}{2} \right) \quad \ln{ ( 1.32565713714 ) } = 0.28190828905 $}}\\ \small{\text{ sin^{-1}(1.04) = -i\cdot \left( 0.28190828905 + i \cdot \dfrac{\pi}{2} \right) $}}\\\\ \small{\text{ sin^{-1}(1.04) = -i\cdot 0.28190828905 - i^2 \cdot \dfrac{\pi}{2} \right) \quad | \quad \boxed{i^2=-1} $}}$$

$$\\ \small{\text{ $ sin^{-1}(1.04) = \dfrac{\pi}{2} - 0.28190828905 \cdot i $}}\\ \small{\text{ $ sin^{-1}(1.04) = 1.57079632679 - 0.28190828905 \cdot i $}}\\$$

$$\small{\text{ $ \boxed{ \dfrac{4i^6+3i } {1-2i} -(9+4i) =\ ? } $ }}\\\\ \small{\text{ $ i^2 = -1 \qquad i^4 = i^2 \cdot i^2 = (-1)\cdot (-1) = 1 \qquad i^6 = i^2 \cdot i^2 \cdot i^2 = (-1)\cdot (-1) \cdot (-1) = -1 $ }}\\$$

we set

$$\small{\text{$i^6 = (-1)^3 = -1$}}$$

so we have:

$$\small{\text{ $ \dfrac{4i^6+3i } {1-2i} -(9+4i) \quad | \quad \boxed{i^6=-1} $ }}\\\\ \small{\text{ $ =\dfrac{-4+3i } {1-2i} -(9+4i) $ }}\\\\ \small{\text{ $ =\dfrac{ (-4+3i) \codt(1+2i) } { (1-2i)\codt(1+2i) } -(9+4i) \quad | \quad \boxed{(a-ib)\cdot (a+ib) = a^2+b^2} $ }}\\\\ \small{\text{ $ =\dfrac{ (-4+3i) \codt(1+2i) } { 1^2+2^2 } -(9+4i) $ }}\\\\ \small{\text{ $ =\dfrac{ (-4+3i) \codt(1+2i) } { 5 } -(9+4i) $ }}\\\\ \small{\text{ $ =\dfrac{ (-4+3i) \codt(1+2i) } { 5 } -5\cdot\dfrac{ (9+4i) }{5} $ }}\\\\ \small{\text{ $ =\dfrac{ (-4+3i) \codt(1+2i)-5\codt(9+4i) } { 5 } $ }}\\\\ \small{\text{ $ =\dfrac{ -4-8i+3i+6i^2 -45 -20i } { 5 } \quad | \quad \boxed{i^2=-1} $ }}\\\\ \small{\text{ $ =\dfrac{ -4 -8i + 3i +6\cdot (-1) -45 -20i } { 5 } $ }}\\\\ \small{\text{ $ =\dfrac{ -4 -8i + 3i -6 -45 -20i } { 5 } \quad $ we put together }}\\\\ \small{\text{ $ =\dfrac{ -55 -25i } { 5 } $ }}\\\\ \small{\text{ $ =-11 -5i $ }}$$

In einem Raum gibt es 8 Lampen, die man unabhängig voneinander ein- und ausschalten kann. Wie viele Beleuchtungsarten gibt es, wenn

a) genau 5 Lampen brennen sollen,

$$\small{\text{ $ \left(\begin{array}{c}8\\5\end{array}\right) =\dfrac{8\ !}{5\ ! \cdot (8-5)\ ! } =\dfrac{8\ !}{5\ ! \cdot 3\ ! } =\dfrac{5\ !\cdot 6 \cdot 7 \cdot 8}{5\ ! \cdot 3\ ! } =\dfrac{ 6 \cdot 7 \cdot 8}{ 3\ ! } =\dfrac{ 6 \cdot 7 \cdot 8}{ 6 } = 7 \cdot 8 = 56 $ }}$$

b) mindestens 5 Lampen brennen sollen?

Alle Beleuchungsarten( inkl. keine Leuchte brennt! ) =

$$\\ \small{\text{ $ \left(\begin{array}{c}8\\0\end{array}\right) +\left(\begin{array}{c}8\\1\end{array}\right) +\left(\begin{array}{c}8\\2\end{array}\right) +\left(\begin{array}{c}8\\3\end{array}\right) +\left(\begin{array}{c}8\\4\end{array}\right) +\left(\begin{array}{c}8\\5\end{array}\right) +\left(\begin{array}{c}8\\6\end{array}\right) +\left(\begin{array}{c}8\\7\end{array}\right) +\left(\begin{array}{c}8\\8\end{array}\right) =2^8 $ }}\\\\ \small{\text{mindestens 5 Leuchten brennen $=2^8 - \left(\begin{array}{c}8\\0\end{array}\right) -\left(\begin{array}{c}8\\1\end{array}\right) -\left(\begin{array}{c}8\\2\end{array}\right) -\left(\begin{array}{c}8\\3\end{array}\right) -\left(\begin{array}{c}8\\4\end{array}\right) $ }}\\\\ \small{\text{ $ \left(\begin{array}{c}8\\ \textcolor[rgb]{1,0,0}{0}\end{array}\right) = 1 $ (kein Leuchte brennt) }}\\ \small{\text{ $ \left(\begin{array}{c}8\\ \textcolor[rgb]{1,0,0}{1}\end{array}\right) = 8 $ (eine Leuchte brennt) }}\\ \small{\text{ $ \left(\begin{array}{c}8\\ \textcolor[rgb]{1,0,0}{2}\end{array}\right) = 28 $ (zwei Leuchten brennen) }}\\ \small{\text{ $ \left(\begin{array}{c}8\\ \textcolor[rgb]{1,0,0}{3}\end{array}\right) = 56 $ (drei Leuchten brennen) }}\\ \small{\text{ $ \left(\begin{array}{c}8\\ \textcolor[rgb]{1,0,0}{4}\end{array}\right) = 70 $ (vier Leuchten brennen) }}\\\\ \small{\text{mindestens 5 Leuchten brennen $=2^8 -1-8-28-56-70 $}}\\ \small{\text{$=256-1-8-28-56-70=93 $ }}$$

Find the indefinite integral of the function f(x)=(4sqrt(x)-3/2)(2/x+x^2)

$$\\ \small{\text{ $f(x)= \left(4\sqrt{x}-\frac{3}{2}\right)\cdot \left(\frac{2}{x}+x^2\right) $}}\\\\ \small{\text{ $f(x)=4\sqrt{x} \cdot \frac{2}{x} + 4\sqrt{x} \cdot x^2 - \frac{3}{2} \cdot \frac{2}{x} - \frac{3}{2} \cdot x^2 $}}\\\\ \small{\text{ $f(x)=8 x^{\frac{1}{2}} \cdot x^{-1} + 4 x^{\frac{1}{2}} \cdot x^2 - 3 \cdot x^{-1} - \frac{3}{2} \cdot x^2 $}}\\\\ \boxed{ \small{\text{ $f(x)=8 x^{-\frac{1}{2}} + 4 x^{\frac{5}{2}} - 3 x^{-1} - \frac{3}{2} \cdot x^2 $}}}\\\\ \int{(8 x^{-\frac{1}{2}} + 4 x^{\frac{5}{2}} - 3 x^{-1} - \frac{3}{2} \cdot x^2 )}\ dx \\\\ \small{\text{ $ = 8\int{(x^{-\frac{1}{2}} ) }\ dx + 4\int{( x^{\frac{5}{2}} ) }\ dx - 3\int{( x^{-1} ) }\ dx - \frac{3}{2} \int{(x^2 ) }\ dx $ }}\\ \boxed{\int{(x^{-\frac{1}{2}} ) }\ dx = 2\sqrt{x} +c \qquad \int{(x^{ \frac{5}{2}} ) }\ dx = \frac{2}{7}x^3\sqrt{x} +c } \\ \boxed{\int{(x^{-1}) }\ dx = \ln{(x)} +c \qquad \int{(x^{2}) }\ dx = \frac{x^3}{3} +c } \\\\\\ \small{\text{ $ \int{(8 x^{-\frac{1}{2}} + 4 x^{\frac{5}{2}} - 3 x^{-1} - \frac{3}{2} \cdot x^2 )}\ dx =16\sqrt{x} + \frac{8}{7}x^3\sqrt{x} -3\ln{(x)} -\frac{x^3}{2} +c $ }}$$