heureka

4) 3√24 / √2   ?

  $$\boxed{\dfrac{3 *\sqrt{24} }{ \sqrt{2} }} \\ \\ =3 *\sqrt{ \dfrac{24 }{ 2} } \\ \\ =3 *\sqrt{ 12 } \\ \\ =3 *\sqrt{ 4*3} \\ \\ =3 *\underbrace{\sqrt{ 4}}_{2}*\sqrt{3} \\ \\ =3 *2*\sqrt{3} \\ \\ \boxed{=6\sqrt{3}}$$   

∫〖2x+1〗  ?

$$\int{(2x+1)\ dx } \\\\ = \int{2x\ dx } + \int{1\ dx }\quad | \quad 1=x^0 \;!\\\\ = 2\int{x^{\textcolor[rgb]{1,0,0}1} \ dx } + \int{x^{\textcolor[rgb]{1,0,0}0} \ dx }$$

$$\boxed { \int{x^{\textcolor[rgb]{1,0,0}{n}} \ dx } = { x^{\textcolor[rgb]{1,0,0}{n}+1} \over \textcolor[rgb]{1,0,0}{n}+1} }$$

$$\\= 2 * \left( \dfrac{ x^{\textcolor[rgb]{1,0,0}{1} +1} }{ \textcolor[rgb]{1,0,0}{1} +1 } \right)+ \dfrac{ x^{\textcolor[rgb]{1,0,0}{0} +1} }{ \textcolor[rgb]{1,0,0}{0} +1 } + c\\\\ =\not{2} \dfrac{ x^{2} }{ \not{2}} + \dfrac{ x^{1} }{1 } + c\\\\ = x^{2}+x + c$$

Hi Melody,

this is a arithmetic succession. Order is 3 ( So i hold 6 as a constant number in Delta 3 so the order is 3 ). I got the Polynom from Internet.

Beginning bei n= 0. If i set x = n+1 so the Polygon can written as:

$$p(x) \\ = (x-1)^3+3(x-1)^2+4(x-1)+2 \\ = x^3-3x^2+3x-1+3(x^2-2x+1)+4x-4+2 \\ = x^3-3x^2+3x-1+3x^2-6x+3+4x-4+2\\ = x^3-3x^2+3x^2+3x-6x+4x -1+3-4+2\\ = x^3+x \\ p(x)=x^3+x \quad | \quad x=1,2,3,... \quad Your\quad solution$$

        if     1  =  2

                2 = 10

                3 = 30

                4 = 68

        then 5 =   ?

$$\begin{array}{ccccc} n+1& p(n)& \Delta_1 & \Delta_2 & \Delta_3 \\ \\ 1&2& \\ &&8\\ 2&10&&12 \\ &&20&&\textcolor[rgb]{1,0,0}{6}\\ 3&30&&18&\downarrow \\ &&38&&(\textcolor[rgb]{1,0,0}{6})\\ 4&68&&(24) \\ &&(62)\\ 5&(130)& \end{array}$$

 5 =   130

Polynom

$$p(n) = n^3+3*n^2+4*n+2 \quad | \quad n = 0,1,2,3, ...$$

Cos (a+b) sin a=-3/5 sin b=-5/13 a & b in Q3  ?

$$\boxed{\cos(a+b)=\;?} \quad \sin{(a)} =-{3\over5} \qquad \sin {(b)}=-{5\over13}$$

$$\cos{(a+b)}=\cos{(a)}*\cos(b)-sin{(a)}*\sin{(b)}$$

cos(a)=?   and   cos(b)=?

$$\textstyle{ \cos{(a)}=\sqrt{1-\sin^2{(a)}}= \sqrt{1-({3\over5})^2}= {\sqrt{5^2-3^2}\over5}={\sqrt{16}\over5}={\pm4\over5}=\pm{4\over5} }$$

$$\textstyle{ \cos{(b)}=\sqrt{1-\sin^2{(b)}}= \sqrt{1-({5\over13})^2}= {\sqrt{13^2-5^2}\over13}={\sqrt{144}\over13}={\pm12\over13}=\pm{12\over13} }$$

$$\cos{(a+b)}=\pm ({4\over5}) \times ({12\over13}) - (-{3\over5}) \times (-{5\over13})$$

$$\cos{(a+b)}=\pm ({4\over5}) \times ({12\over13}) - ({3\over5}) \times ({5\over13})$$

$$\cos{(a+b)}=({\pm(4*12)\over5*13}) - ({3*5\over5*13})$$

$$\cos{(a+b)}={\pm(4*12)-3*5\over5*13}$$

$$\cos{(a+b)}={\pm48-15\over65}$$

$$\text{1.) }\cos{(a+b)}={48-15\over65}={33\over65}=0.50769230769$$

$$\text{2.) }\cos{(a+b)}={-48-15\over65}=-{63\over65}=-0.96923076923$$

2 sin 120 cos 120    ?

$$\\\sin{(\alpha+\beta)}= \sin{(\alpha)} * \cos{(\beta)} + \cos{(\alpha)} * \sin{(\beta)} \\ \sin{(\alpha+\alpha)}= \sin{(\alpha)} * \cos{(\alpha)} + \cos{(\alpha)} * \sin{(\alpha)}\\\\ \boxed{ \sin{(2*\alpha)}= 2*\sin{(\alpha)} * \cos{(\alpha)}}\\\\ \sin{(2*120\ensurement{^{\circ}} )}= 2\sin{(120\ensurement{^{\circ}} )} \cos{(120\ensurement{^{\circ}} )}= \sin{(240 \ensurement{^{\circ}} ) =-0.86602540378$$

$$\\ \sin{(240 \ensurement{^{\circ}} ) = \sin{(360\ensurement{^{\circ}-120 \ensurement{^{\circ} } ) = -\sin{(120 \ensurement{^{\circ} } ) =-\frac{\sqrt{3} } {2} =-0.86602540378$$

3/15 divided by 7/10  ?

$$\begin{array}{rcl} \dfrac { \frac {3}{15}} {\textcolor[rgb]{1,0,0}{\frac {7}{10} }} = \frac {3}{15}\times\textcolor[rgb]{1,0,0}{\frac {10}{7}} = \frac {3*\textcolor[rgb]{1,0,0}{10 }}{15*\textcolor[rgb]{1,0,0}{7}} = \frac {3*\textcolor[rgb]{1,0,0}{2*5 }}{3*5*\textcolor[rgb]{1,0,0}{7}} = \frac {3*\textcolor[rgb]{1,0,0}{5*2 }}{3*5*\textcolor[rgb]{1,0,0}{7}} =\frac {\textcolor[rgb]{1,0,0}{2}}{\textcolor[rgb]{1,0,0}{7}}\end{array} =0.28571428571$$

3x + y = -4

$$\Rightarrow x=-{1\over3}y-{4\over3}$$  

$$\\\text{Y axis Segment } = -\frac{4}{3} \\\\ \text{Gradient to the Y axis} = -\frac{1}{3} \quad \text{(+ clockwise; - counterclockwise)}$$

wenn 80 Reiskörner 2 gramm wiegen,wieviel wiegen dann 80 Trillionen Reiskörner ,bitte mit rechenweg.

  Eine Million hoch 3 = 1 Trillion = 10^18 = 1.000.000.000.000.000.000

$$\dfrac { 2 \; Gramm } { 80\; Reisk\ddot{o}rner } * 80 \; Trillionen \; Reisk\ddot{o}rner \\\\ =2 \; Trillionen \; Gramm * \dfrac{80 \; Reisk\ddot{o}rner}{80 \; Reisk\ddot{o}rner} \quad | \quad \dfrac{80 Reisk\ddot{o}rner}{80 Reisk\ddot{o}rner}=1\\\\ =2 \; Trillionen \; Gramm \\\\ =2 *10^{18}g \\\\ =2 \;Eg \quad | \quad Eg = Exa \;Gramm$$