heureka

 find the sum of 7 terms of 75,-15,3,-3/5

$$\small{
\text{ratio} = -\dfrac15 = -0.2 \qquad a_1 = 75
}\\\\
\small{
\text{geometric sequence: } a_n = a_1 \cdot r^{n-1}
}\\\\
\begin{array}{rclclcr}
\small{a_1 &=& 75 \cdot r^0 &=& 75 }\\ \\
\small{a_2 &=& 75 \cdot r^1 &=& 75 \cdot \left( -\dfrac15 \right)^1 &=& -15 }\\\\
\small{a_3 &=& 75 \cdot r^2 &=& 75 \cdot \left( -\dfrac15 \right)^2 &=& 3 }\\\\
\small{a_4 &=& 75 \cdot r^3 &=& 75 \cdot \left( -\dfrac15 \right)^3 &=& -0.6 }\\\\
\small{a_5 &=& 75 \cdot r^4 &=& 75 \cdot \left( -\dfrac15 \right)^4 &=& 0.12 }\\\\
\small{a_6 &=& 75 \cdot r^5 &=& 75 \cdot \left( -\dfrac15 \right)^5 &=& -0.024 }\\\\
\small{a_7 &=& 75 \cdot r^6 &=& 75 \cdot \left( -\dfrac15 \right)^6 &=& 0.0048 }\\\\
\end{array}\\\\
\small{
\mathrm{sum}_7 = 75-15+3-0.6+0.12-0.024+0.0048= 62.5008
}\\\\
\small{
\mathrm{sum}_7 = a_1 \cdot \left( \dfrac{ 1-r^7} {1-r} \right)
= 75\cdot \left( \dfrac{ 1-(-0.2)^7} {1-(-0.2)} \right)
=75\cdot \left( \dfrac{ 1+0.2^7} {1.2)} \right)
=62.5008
}$$

$$\boxed{~~
\small{
\mathrm{sum}_n = a_1 \cdot \left( \dfrac{ 1-r^n} {1-r} \right)
}
~~}$$

Sogar null hoch null ist eins ?

Warum ist jede Zahl hoch null immer eins ?

$$1 = \dfrac{ a^1 }{ a^1 } = a^1\cdot a^{-1} = a^{1-1} = a^0$$

$$0^0 \text{ ist unbestimmt! }$$

Berechne den winkel gamma und delta wenn alpha 45 und beta 75 ist.

Wenn es sich um ein Sehnenviereck handelt, das heißt, das Viereck liegt in einem Kreis, dann

berechnen sich die Winkel gamma und delta wie folgt:

alpha + gamma = 180,   45 + gamma = 180,  also gamma = 180 - 45, gamma = 135

beta + delta = 180,  75 + delta = 180,  also delta =180 - 75, delta = 105

irene had a total of 1686 red,blue and green balloons for sale.the ratio of red to blue was 2:3.after irene sold 3/4 blue and 1/2 red.she had 922 balloons left.hiw many blue balloons does irene have at first ?

 $$\small{\text{$
\begin{array}{lrcl}
(1) & r + b + g &=& 1686 \\\\
(2) & \dfrac{r}{b} &=& \dfrac23 \qquad \text{ so } \qquad r = \dfrac23 \cdot b\\\\
\hline
\\
(2) \text{ in } (1): \quad & \dfrac23 \cdot b + b+ g &=& 1686 \\\\
(I) & \dfrac53 \cdot b + g &=& 1686 \\\\
\hline
\\
(3) & \dfrac14 \cdot b + \dfrac12 \cdot r +g &=& 922\\\\
(2) \text{ in } (3): \quad & \dfrac14 \cdot b +\dfrac12 \cdot \dfrac23 \cdot b + g &=& 922 \\\\
& \dfrac14 \cdot b +\dfrac13 \cdot b + g &=& 922 \\\\
& \dfrac3{12} \cdot b +\dfrac4{12} \cdot b + g &=& 922 \\\\
(II) & \dfrac7{12} \cdot b + g &=& 922 \\\\
\hline
\\
(I)-(II): & \dfrac53 \cdot b - \dfrac7{12} \cdot b &=& 1686 - 922 \\\\
& \dfrac53 \cdot b - \dfrac7{12} \cdot b &=& 764 \\\\
& \dfrac{20}{12} \cdot b - \dfrac7{12} \cdot b &=& 764 \\\\
& \dfrac{13}{12} \cdot b &=& 764 \\\\
& b &=& \dfrac{ 764\cdot 12 } {13} \\\\
&\mathbf{ b }& \mathbf{=}& \mathbf{705}
\end{array}
$}}$$

Irene does  have at first 705 blue balloons.

check: 176 ( blue ) + 235 ( red ) + 511 ( green ) = 922

          705 ( blue )  + 470 ( red ) + 511 ( green ) = 1686

          176 / 705 = 1 / 4

          235 / 470 = 1 / 2

          705 * ( 2 / 3 ) = 470

Two math students erect a sun "shade" on the beach (similar to a lean too). This "shade" is rectangular in shape with dimensions 1.5 m long and 2 m wide, and it makes an angle of 60° with the ground. When the sun's rays cast down on this "shade" there will be an area of shade made on the ground (also a rectangle). What is the area of shade that the students will have to sit in at 12 noon ? (that is, what is the projection of the shade onto the ground)? (Assume the sun’s rays are shining directly down).

$$\vec{a}=1.5\cdot
\begin{pmatrix} \cos{ (60\ensurement{^{\circ}} ) \\ \sin{ (60\ensurement{^{\circ}} ) }\\ \end{pmatrix}\qquad
\vec{b} = \begin{pmatrix} 0\\2 \end{pmatrix}\\\\
\begin{array}{lll}
\text{area}=\left|\vec{a}_{projection}\times\vec{b} \right|
& \qquad \vec{a}_{projection} = \left(\vec{a}\cdot \vec{e}_x\right)\cdot \vec{e}_x
& \qquad \vec{e}_x = \begin{pmatrix} 1\\0 \end{pmatrix}\\\\
& \qquad \vec{a}_{projection} =\left( 1.5\cdot \begin{pmatrix} \cos{ (60\ensurement{^{\circ}} ) \\ \sin{ (60\ensurement{^{\circ}} ) }\\ \end{pmatrix}\cdot \begin{pmatrix} 1\\0 \end{pmatrix}\right)\cdot \begin{pmatrix} 1\\0 \end{pmatrix} \\\\
& \qquad \vec{a}_{projection} = \left( 1.5\cdot \cos{(60\ensurement{^{\circ}} )} +0\right) \cdot \begin{pmatrix} 1\\0 \end{pmatrix} \\\\
& \qquad \vec{a}_{projection} = \begin{pmatrix} 1.5\cdot \cos{(60\ensurement{^{\circ}} )}\\0 \end{pmatrix} \\\\
\text{area}=\left|\begin{pmatrix} 1.5\cdot \cos{(60\ensurement{^{\circ}} )}\\0 \end{pmatrix}\times\begin{pmatrix} 0\\2 \end{pmatrix} \right| \\\\
\text{area}=2\cdot 1.5\cdot \cos{ (60\ensurement{^{\circ}} ) }
\qquad | \qquad \cos{ (60\ensurement{^{\circ}} ) } = \dfrac{1}{2}\\\\
\text{area}=2\cdot 1.5\cdot \dfrac{1}{2}\\\\
\mathbf{\text{area}=1.5 ~\mathrm{m^2}}\\\\
\end{array}$$

what is 5 C 4 ?

$$\small{
\boxed{~~ \binom{n}k=\binom{n}{n-k}\qquad \text{ and } \qquad \binom{n}1 = n ~~}\qquad
_5C_4 = \binom54 = \binom51 = 5
}$$

Was ist die Ableitung von x*(x+1)^-1

1. Methode:

$$f(x) = x\cdot(x+1)^{-1}$$

$$\boxed{
~~f(x)=u\cdot v \qquad f'(x) = u\cdot v' + u'\cdot v ~~
}$$

$$\begin{array}{rclrcl}
u &=& x \quad & \quad v &=& (x+1)^{-1} \\
u' &=& 1 \quad & \quad v' &=& (-1)(x+1)^{-2}\cdot 1= -(x+1)^{-2}\\\\
\end{array}\\\\
\begin{array}{rclrcl}
f'(x) &=& x \cdot [-1(x+1)^{-2}] + 1\cdot (x+1)^{-1} \\
f'(x) &=& -x \cdot (x+1)^{-2} + (x+1)^{-1} \\ \\
f'(x) &=& \dfrac{ -x } { (x+1)^2 } + \dfrac{ 1 } { x+1 } \\ \\
f'(x) &=& \dfrac{ -x } { (x+1)^2 } + \dfrac{ 1 } { x+1 } \left(\dfrac{x+1}{x+1} \right) \\ \\
f'(x) &=& \dfrac{ -x +x+1} { (x+1)^2 } \\ \\
\mathbf{f'(x)} &\mathbf{=}& \mathbf{\dfrac{1} { (x+1)^2 }} \\ \\
\end{array}$$

2. Methode:

$$f(x) = \dfrac{ x } { x+1 }$$

$$\boxed{
~~f(x)=\dfrac{ u } { v } \qquad
f'(x) = \dfrac{ u } { v } \left(\dfrac{ u' } { u }-\dfrac{ v' } { v }
\right)~~
}$$

$$\begin{array}{rclrcl}
u &=& x \quad & \quad v &=& x+1 \\
u' &=& 1 \quad & \quad v' &=& 1\\\\
\end{array}\\\\
\begin{array}{rclrcl}
f'(x) &=& \dfrac{ x } { x+1 } \left(\dfrac{ 1 } { x }-\dfrac{ 1 } {x+1} \right)\\\\
f'(x) &=& \dfrac{ x } { x+1 } \left(\dfrac{ x+1-x } { x(x+1) } \right)\\\\
f'(x) &=& \dfrac{ x } { x+1 } \left(\dfrac{ 1} { x(x+1) } \right)\\\\
\mathbf{f'(x)} &\mathbf{=}& \mathbf{\dfrac{1} { (x+1)^2 }} \\ \\
\end{array}$$

$$\small{\text{$
\begin{array}{lcl}
$Let $ a, b, c, $ and $ n $ be positive integers. If $
a + b + c = 19 \cdot 97 $ and $ \\
\left[a + n = b - n = \dfrac{c}{n}\right],
$ compute the value of $ a .
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{lrrrcl}
& a+b+c=19\cdot 97 \\
\\
\hline
\\
(1)& a+n=k \\
(2) & b-n=k &\qquad \qquad (1)+(2): & a+b&=& 2\cdot k\\
(3) & \dfrac{c}{n}=k &\qquad \qquad $so$ & c &=& k\cdot n\\\\
& & & a+b+c &=& 2\cdot k + k\cdot n=19\cdot 97
\\
\hline
\\
\end{array}
$}}\\\\
\small{\text{$
\begin{array}{rrclrcl}
& 2\cdot k + k\cdot n &=& 19\cdot 97\\\\
& k\cdot(2+n)&=& 19\cdot 97\\\\
I. & \textcolor[rgb]{1,0,0}{k}\cdot\textcolor[rgb]{0,0,1}{(2+n)}&=& \textcolor[rgb]{1,0,0}{19}\cdot \textcolor[rgb]{0,0,1}{97}\\\\
& \underline{k=19} && \underline{2+n = 97 }& \qquad \Rightarrow
\qquad n&=& 95\\
& && & a+n&=& k\\
& && & a+95&=& 19\\
& && & a&=& -76 ~$ negative! $\\ \\
II. & \textcolor[rgb]{1,0,0}{k}\cdot\textcolor[rgb]{0,0,1}{(2+n)}&=& \textcolor[rgb]{1,0,0}{97}\cdot \textcolor[rgb]{0,0,1}{19}\\\\
& \underline{k=97} && \underline{2+n = 19 }& \qquad \Rightarrow
\qquad n&=& 17\\
& && & a+n&=& k\\
& && & a+17&=& 97\\
& && & a&=& 80 ~$ okay! $\\
\end{array}
$}}$$

The Smith family has 13 children, all born exactly 2 years apart. How old will the second-youngest child be when the oldest child is five times the age of the youngest child?

$$\small{\text{$
\begin{array}{rcl}
$arithmetical sequence $
~
\underbrace{c_1}_{child ~1} ~~
\underbrace{(c_1+1\cdot 2)}_{child ~2} ~~
\underbrace{(c_1+2\cdot 2)}_{child ~3} ~~
\underbrace{(c_1+3\cdot 2)}_{child ~4} ~~
\cdots\\\\
c_n=c_1+(n-1)\cdot2\\\\
c_{13}=c_1+12\cdot2\\
c_{13}=c_1+24\\
\end{array}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
c_{13}=c_1+24 = 5\cdot c_1\\
c_1+24 = 5\cdot c_1\\
4\cdot c_1 = 24 \\
\mathbf{c_1 = 6} \\\\
c_2 = c_1+2\\
c_2 = 6+2\\
\mathbf{c_2 = 8}
\end{array}
$}}$$

The second-youngest child is 8 years old