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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