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)

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