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Welcher Term besteht zwischen 1,12,33und 62

$$\\ \boxed{p(n) = a*n^3+b*n^2+c*n+d} \\
p(1)= 1 = a*1^3+b*1^2+c*1+d \\
p(2)= 12 = a*2^3+b*2^2+c*2+d \\
p(3)= 33 = a*3^3+b*3^2+c*3+d \\
p(4)= 62 = a*4^3+b*4^2+c*4+d \\
\hline
(1)\ a+b+c+d=1 \\
(2)\ 8a+4b+2c+d=12\\
(3)\ 27a+9b+3c+d=33\\
(4)\ 64a+16b+4c+d=62\\
\hline
I: (2)-(1): \ 7a+3b+c=11 \\
II:(4)-(3): \ 37a+7b+c+d=29\\
III: (3)-(2): \ 19a+5b+c=21\\
\hline$$

$$III-I: \ 12a+2b=10\\
II-III: \ 18a+2b =8 \\
18a-12a=8-10 \\
6a=-2\\
a=-\frac{1}{3}\\
-\frac{12}{3}+2b=10\\
2b=10+4\\
b=7\\
\hline
-\dfrac{7}{3}+21+c=11\\
c=-\dfrac{23}{3}\\
\hline
-\dfrac{1}{3}+7-\dfrac{23}{3}+d=1\\
-\dfrac{24}{3}+7+d=1 \\
d=2 \\
\hline$$

$$\\ \boxed{p(n) = -\frac{1}{3}*n^3+7*n^2-\frac{23}{3}*n+2} \\\\
p(1) = -8+7+2= 1 \quad \text{ okay!}\\
p(2) = 30-18 = 12 \quad \text{ okay!}\\
p(3) = 65-32 = 33 \quad \text{ okay!}\\
p(4) = 114-52= 62 \quad \text{ okay!}\\$$

What is a equation for this data. X=0,Y=2,X=1,Y=120,and X=2,Y=7200

$$\small{\text{
A parabola
$
y=ax^2+bx+c
$
is a equation. We need to calulate a, b and c.
}}$\\$
\small{\text{
We have $x_1 = 0$ and $y_1 =2 $ and calculate $
y_1=ax_1^2+bx_1+c \Rightarrow 2 = 0 + 0 + c \Rightarrow \textcolor[rgb]{1,0,0}{ c = 2 }
$
}}$\\$
\small{\text{
We also have $x_2 = 1$ and $y_2 =120 $ and calculate $
y_2=ax_2^2+bx_2+2 \Rightarrow 120 = a + b + 2
$
}}
$\\$
\small{\text{
$\Rightarrow a + b = 118 \quad (1)$
}}
$\\$
\small{\text{
Last but not least $x_3 = 2$ and $y_3 =7200 $ and calculate $
y_3=ax_3^2+bx_3+2 \Rightarrow 7200= a*4 + b*2 + 2 $
}}
$\\$
\small{\text{
$\Rightarrow 4a + 2b = 7198 \quad (2)$
}}$\\\\$
(1) \quad a + b = 118 \quad | \quad *2\\
(1) \quad 2a + 2b = 236 \\
(2) \quad 4a + 2b = 7198 \\
(2)-(1): 4a-2a = 2a = 7198 - 236 \\
\textcolor[rgb]{1,0,0}{a = 3481}\\
a+b= 118 \\
3481 + b = 118 \\
b= 118 - 3481 \\
\textcolor[rgb]{1,0,0}{b = -3363}\\
\small{\text{
The equation is \boxed{ $y = 3481x^2 - 3363x+2$ }
}}
$\\\\$
\small{\text{
Proof:
$ y(0) = 0-0+2 = 2 $ okay!
}}$\\$
\small{\text{
Proof:
$ y(1) = 3481-3363+2 = 120 $ okay!
}}$\\$
\small{\text{
Proof:
$ y(2) = 3481*4-3363*2+2 = 7200 $ okay!
}}$$

-29/50 as a decimal = -0.58

x + 9 = 18 + -2x

$$\begin{array}{rcl}
x + 9 &=& 18 + -2x \quad | \quad +2x \\
2x+x+9 &=& 18 \\
3x+9 &=& 18 \quad | \quad -9\\
3x &=& 18-9 \\
3x &=& 9 \quad | \quad :3\\
x &=& \frac{9}{3} \\
x &=& 3
\end{Array}

\text{Proof:} \\
3+9=18-2*3 \\
12 = 18-6 \\
12 = 12 \text{okay!}$$

1/2-what=-1 3/4

$$\begin{array}{rcl}
\frac{1}{2} -x &=& -1 \frac{3}{4} \quad | \quad *(-1)\\\\
-\frac{1}{2}+x &=& 1 \frac{3}{4} \quad | \quad +\frac{1}{2}\\\\
x &=& 1 \frac{3}{4}+\frac{1}{2}\\\\
x &=& \frac{4*1+3}{4}+\frac{1*2}{2*2}\\\\
x &=& \frac{7}{4}+\frac{2}{4}\\\\
x &=& \frac{9}{4}\\\\
x &=& \frac{8+1}{4}\\\\
x &=& \frac{8}{4}+\frac{1}{4}\\\\
x &=& 2+\frac{1}{4}\\\\
x &=& 2\frac{1}{4}
\end{array}$$

Hi,  Melody

excuse please

Hi Alan,

thank you very much.

I have replaced 024 with 045

I found the answer in (a) is $${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{zy}}\right)}{{\mathtt{\,-\,}}\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{zy}}\right)}}$$ , how to solve (b)?

$$x=\dfrac{10zy-2y^2}{5zy} = 2-\dfrac{2}{5}*\dfrac{y}{z} \quad | \quad \dfrac{y}{z} = \dfrac{5}{3}\\\\
x= 2 - \frac{2}{3} = 1\frac{1}{3}$$

the number 1.45450450450450450450450450...... how can be written as an ordinary fraction ?

$$\\ \small{\text{
I.
$
1.45450450450450450450450450\dots = 1.4545 + 0.0000\overline{045} = \frac{14545} {10000} + 0.0000\overline{045} =\frac{2909}{2000} + 0.0000\overline{045}
$
}}$\\\\$
\small{\text{
II.
$
0.0000\overline{045} = 045*10^{-7} + 045*10^{-10} + 045*10^{-13} + 045*10^{-16} + \dots
$
}}
$\\$
\small{\text{
$
0.0000\overline{045} = (\underbrace{045*10^{-7}}_{=a}) + (045*10^{-7}) *10^{-3} + (045*10^{-7})*10^{-6} + (045*10^{-7})*10^{-9} + \dots
$
\textcolor[rgb]{1,0,0}{ this is a geometric series. }
}}
$\\$
\small{\text{
The sum of the geometric series is $ s=\frac{a}{1-r}$ we have $a=045*10^{-7}$ and $ r = \frac{ (045*10^{-7}) *10^{-3} }{ 045*10^{-7} } = \frac{ (045*10^{-7})*10^{-6} }{ (045*10^{-7}) *10^{-3} } = ... = 10^{-3}
$
}}
$\\$
\small{\text{
$
\begin{array}{rcl}
s &=& \frac{ 045*10^{-7} }{ 1- 10^{-3} } \\ \\
&=& \frac{10^3}{10^3}*\frac{ 045*10^{-7} }{ 1- 10^{-3} } \\\\
&=& \frac{45}{10^4(10^3-1)} \\\\
&=& \frac{45}{ 9990000 }
\end{Array}
$
}}$\\\\$
\small{\text{
III.
$
1.45450450450450450450450450\dots = 1.4545 + 0.0000\overline{045} = \frac{14545} {10000} + 0.0000\overline{045} =\frac{2909}{2000} + \frac{45}{ 9990000 }
$
}}$\\\\$
\small{\text{
$
1.45450450450450450450450450\dots
= \frac{2909}{2000} + \frac{45}{ 9990000 }
= \frac{2909}{2000} + \frac{1}{ 2220000 }
= \frac{3229}{2220}
$
}}$$

A sector has an angle of 107.9 and an arc length of 5.92m. Find its radius

arc length  b = 5.92 m

angle  $$\alpha \ensurement{^{\circ}}$$ = $$107.9\ensurement{^{\circ}}$$

radius r = ?

$$\boxed{b =r* \breve \alpha} \qquad \breve \alpha =\alpha \ensurement{^{\circ}} * \frac{2\pi}{360\ensurement{^{\circ}}} = \alpha \ensurement{^{\circ}} *\frac{\pi}{180\ensurement{^{\circ}}} \\\\
\begin{array}{rcl}
b &=& r* \breve \alpha \\
&=& r * \alpha \ensurement{^{\circ}} *\frac{\pi}{180\ensurement{^{\circ}}}
\end{array}\\\\\\
\boxed{r = \left( \frac{180\ensurement{^{\circ}} }{\pi} \right)*\frac{b} {\alpha\ensurement{^{\circ}}} = 57.2957795131*\frac{b} {\alpha\ensurement{^{\circ}}} }\\\\
\small{\text{
$
r = \left( \frac{180\ensurement{^{\circ}} }{\pi} \right)*\frac{5.92\ m}{107.9\ensurement{^{\circ}}} = 57.2957795131 * \frac{5.92\ m}{107.9\ensurement{^{\circ}}} = 3.14356825503\ m
$
}}$$

Radius is 3.14356825503 m