heureka

Math(Input=Result) Error!

$${binom}{\left({\left({\frac{{\mathtt{3}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}\right)}^{{\mathtt{10}}}\right)} = {binom}{\left(\tiny\text{Error: loop}\right)}$$

$$\left(\dfrac{3}{x^2}-4x^3\right)^{10}\\$$

I think the answer is:

$$10Cr\times \left(\dfrac{3}{x^2}\right)^{\textcolor[rgb]{1,0,0}{10-r}}\times (-4x^3)^{\textcolor[rgb]{1,0,0}{r}}\\ =10Cr\times 3^{10-r} \times (-4)^r \times x^{-2(10-r)} \times x^{3r}\\ =10Cr\times 3^{10-r} \times (-4)^r \times x^{-2(10-r)+3r}\\ =10Cr\times 3^{10-r} \times (-4)^r \times x^{-20+2r+3r}\\ =10Cr\times 3^{10-r} \times (-4)^r \times x^{-20+5r}$$

so -20+5r=0  -> r = 4

the term we want is

$$10C\textcolor[rgb]{1,0,0}{4}\times 3^6\times (-4)^4 \times x^0$$

(1) -8x - 12y = 36  ->   8x + 12y = -36

(2)  5y -   6x = 41  ->  -6x +   5y =  41

$$\boxed{\begin{array}{rcrcr} 8x & + & 12y &=& -36 \\ -6x & + & 5y &=& 41 \end{array}}$$

\boxed{ $$\begin{array}{rcrcr}  8x & + & 12y &=& -36 \\ -6x & + & 5y &=&  41  \end{array}$$ }

  $$\\x= \frac{\begin{vmatrix} -36 & 12 \\ 41 & 5 \end{vmatrix}}{\begin{vmatrix} 8& 12 \\ -6 & 5 \end{vmatrix}} =\frac{-36*5-41*12}{8*5-(-6)*12} =\frac{-180-492}{40+72} =\frac{-672}{112}=-6\\ \boxed{x=-6} \\y= \frac{\begin{vmatrix} 8 & -36 \\ -6 & 41 \end{vmatrix}}{\begin{vmatrix} 8& 12 \\ -6 & 5 \end{vmatrix}} =\frac{8*41-(-6)(-36)}{8*5-(-6)*12} =\frac{328-216}{40+72} =\frac{112}{112}=1\\ \boxed{y=1}$$

$${\mathtt{9}}{m}{\mathtt{\,\times\,}}{\mathtt{10}}{m}{\mathtt{\,\times\,}}{\mathtt{30}}{m}{\mathtt{\,\times\,}}{\mathtt{9}}{m} = {\mathtt{24\,300}}\left[{m}{\mathtt{\,\times\,}}{m}{\mathtt{\,\times\,}}{m}{\mathtt{\,\times\,}}{m}\right]$$  = $$24300[m^4]$$

3-th square of (-125):

$$\\\sqrt[3]{-125}=\sqrt[3]{125}\times\sqrt[3]{-1}=(5)* \left\{ e^{i\frac{1}{3}(\pi+0*\pi)}, e^{i\frac{1}{3}(\pi+2*\pi)}, e^{i\frac{1}{3}(\pi+4*\pi)} \right\}\\ \sqrt[3]{-125}=(5)* \left\{ e^{i\frac{1}{3}\pi}, e^{i\frac{3}{3}\pi}, e^{i\frac{5}{3}\pi} \right\}\\ \sqrt[3]{-125}=(5)* \left\{ e^{i\frac{1}{3}\pi}, e^{i\pi}, e^{i\frac{5}{3}\pi} \right\}\\ \boxed{e^{i\pi}=-1}\\ \sqrt[3]{-125}=(5)* \left\{ e^{i\frac{1}{3}\pi}, -1, e^{i\frac{5}{3}\pi} \right\}\\ \\ \sqrt[3]{-125}=(5)* e^{i\frac{1}{3}\pi} \quad \text{complex number} \\\\ \sqrt[3]{-125}=(5)*(-1)=-5\quad \text{real number }\\\\ \sqrt[3]{-125}=(5)* e^{i\frac{5}{3}\pi} \quad \text{complex number} \\$$

The "Julian Date" of 1.2.2014 0:0:0 o'clock is 2456689.50000

The "Julian Date" of 1.2.2014 12:0:0 is 2456690.00000

The "Julian Date"

see:

http://aa.usno.navy.mil/data/docs/JulianDate.php

The Julian date for CE 2014 February 1 00:00:00.0 UT is JD 2456689.500000

The Julian date for CE 2014 February 1 12:00:00.0 UT is JD 2456690.000000 

The Julian date for CE 2014 May 29 12:00:00.0 UT is JD 2456807.000000 

$$b = \text{ arc length }= 12cm$$

$$\alpha = 20 \ensuremath{^\circ}$$

r = radius  ?

$$\boxed{b=2\pi r\frac{\alpha\ensuremath{^\circ}}{360\ensuremath{^\circ}}}$$

$$\\\\ \Rightarrow r = \dfrac{b*180\ensuremath{^\circ}}{\pi*\alpha\ensuremath{^\circ}} \\\\ r = \dfrac{12\;cm*180\ensuremath{^\circ}}{\pi*20\ensuremath{^\circ}}=34,377\;cm$$

The radius is 34.377 cm

The answer is C=11, B=19 and L=70

1. 5C+2B+0.10L = 100    -> 50C+20B+L=1000
2. C+B+L=100

$$\boxed{(1) - (2) : 19B+49C=900 }$$

19B+49C=900

49C=900 mod 19

49C-2*19C=(900-47*19) mod 19

49C-38C=11C=7 mod 19

11C+19z=7

19z=7 mod 11

19z-1*11z=(7-0*11) mod 11

8z=7 mod 11

8z+11u=7

11u=7 mod 8

11u - 1*8u=(7-0*8) mod 8

3u=7 mod 8

3u+8v=7

8v=7 mod 3

8v-2*3v=(7-2*3) mod 3

2v=1 mod 3

2v+3w=1

3w=1 mod 2

3w - 1*2w=(1-0*2) mod 2

$$\boxed{w=1 \bmod 2 }$$

back:

w=1-2g    g is a integer

v:

2v+3w=1

2v+3(1-2g)=1

-> v=-1+3g

u:

3u+8v=7

3u+8(-1+3g)=7

-> u=5-8g

z:

8z+11u=7

8z+11(5-8g)=7

-> z=-6+11g

C:

11C+19z=7

11C+19(-6+11g)=7

-> C=11-19g

B:

19B+49C=900

19B+49(11-19g)=900

-> B=19+49g

L:

C+B+L=100

L=100-C-B

L=100-(11-19g)-(19+49g)

L=100-11+19g-19-49g

L=70-30g

The result is:

$$\boxed{C=11-19g\quad B=19+49g \quad L=70-30g}$$

C can't be negative so g must be 0:

g=0

C=11-19*0=11

B=19+49*0 =19

L=70-30*0 =70

$$\\\textcolor[rgb]{1,0,0}{C=11\quad B=19\quad L=70}$$

11*$5.00 + 19*$2.00 + 70*$0.10 = $55+$38+$7=$100

11+19+70=100

-3x^2+12x-9  ?

$$\\-3x^2+12x-9 \quad | \quad : -3 \\ \boxed{x^2-4x+3=0}\\ x^2-4x \text{ } \textcolor[rgb]{1,0,0}{ + 2^2 - 2^2}+3=0\\\\ \left(x-2\right)^2-4+3=0\\\\ \left(x-2\right)^2-1=0\\\\ \left(x-2\right)^2=1\quad | \quad \pm\sqrt{} \\\\ x-2=\pm1\quad |\quad +2\\$$

$$\\x=\pm1+2\\ x_1=1+2=3 \quad\text{and}\quad x_2=-1+2=1\\ \boxed{x_1=3 \quad\text{and}\quad x_2=1}$$

$$\boxed{(-1 + i)^7 \quad? }$$

z=x+iy  $$\boxed{z=-1+i} \quad \Rightarrow x=-1, y=\textcolor[rgb]{1,0,0}{+}1$$

$$r=\sqrt{x^2+y^2}=\sqrt{(-1)^2+1^2}=\sqrt{2}$$

$$\phi = sign(y)*\cos^{-1}{\left(\frac{x}{r}\right)} =\textcolor[rgb]{1,0,0}{+}\cos^{-1}(\frac{-1}{\sqrt{2}}) =\cos^{-1}(-\frac{\sqrt{2}}{2})=\frac{3}{4}\pi$$

$$\boxed{z=r*e^{i\phi}}$$

$$\\z=\sqrt2*e^{i*\frac{3}{4}\pi} \\ z^7=\left(\sqrt2*e^{i*\frac{3}{4}\pi}\right)^7 \\ z^7=\left(\sqrt2\right)^7*\left(e^{i*\frac{3}{4}\pi}\right)^7 \\ z^7=\left(\sqrt2\right)^7*e^{i*\frac{7*3}{4}\pi} \\ z^7=\left(\sqrt2\right)^7*e^{i*\frac{21}{4}\pi} = \left(\sqrt2\right)^7*e^{i\left(4\pi+\pi+\frac{1}{4}\pi\textcolor[rgb]{1,0,0}{-4\pi}}\right)\\$$

$$z^7=\left(\sqrt2\right)^7*e^{i\left(\pi+\frac{1}{4}\pi}\right)\\$$

$$\boxed{e^{i\phi}=\cos{\phi}+i\sin{\phi}}$$

$$\\z^7=\left(\sqrt2\right)^7 \left[ \cos{(\pi+\frac{\pi}{4} )}+i*sin{(\pi+\frac{\pi}{4})} \right]\\ z^7=\left(\sqrt2\right)^7 \left[ \underbrace{\cos{\pi}}_{-1}\cos{\frac{\pi}{4}} - \underbrace{\sin{\pi}}_0\sin{\frac{\pi}{4}} +i\left( \underbrace{\sin{\pi}}_0\cos{\frac{\pi}{4}}+\underbrace{\cos{\pi}}_{-1}\sin{\frac{\pi}{4}} \right) \right]\\ z^7=\left(\sqrt2\right)^7 \left( -\cos{\frac{\pi}{4}}-i\sin{\frac{\pi}{4}}\right)$$

$$\sin{\frac{\pi}{4}}=\cos{\frac{\pi}{4}}=\frac{\sqrt{2} }{2}\quad\\ z^7=\left(\sqrt2\right)^7 \left( -\frac{\sqrt{2}}{2}-i{\frac{\sqrt{2}}{2}\right)\\ z^7=\left(\sqrt2\right)^7\frac{\sqrt{2}}{2}\left( -1-i\right)\\ z^7=\frac{\left(\sqrt{2}\right)^8}{2}\left(-1-i\right)$$

$$\\z^7=\frac{2^4}{2}\left(-1-i\right)\\\\ z^7=2^3\left(-1-i\right)\\ z^7=8\left(-1-i\right)\\ z^7=-8-8i\\$$

$$\boxed{(-1+i)^7=-8-8i}$$

$$\boxed{2^{\frac{x}{2}} * 4^{\frac{x}{6}}* \left[(\frac{1}{8})^{\frac{1}{x}}\right]^{\frac{1}{6}}=2^2*2^{\frac{1}{3}} }\\\\ \Rightarrow 2^{\frac{x}{2}} *2^{2*\frac{x}{6}}*2^{-3*\frac{1}{x}*\frac{1}{6}}=2^{2+\frac{1}{3}}\\\\ \Rightarrow 2^{\frac{x}{2}} *2^{\frac{x}{3}}*2^{\left(-\frac{1}{2x}\right)}=2^{\frac{7}{3}}\\\\ \Rightarrow 2^{ \left(\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}\right) }=2^{\frac{7}{3}} \quad | \quad ln\\\\ \Rightarrow { \left(\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}\right)*\ln(2) }=\frac{7}{3}*\ln(2)\\\\ \Rightarrow { \frac{x}{2}+\frac{x}{3}-\frac{1}{2x}\right }=\frac{7}{3}\\\\$$

$$\\\Rightarrow { \frac{3x}{3x}*\frac{x}{2}+\frac{2x}{2x}*\frac{x}{3} -\frac{3}{3}*\frac{1}{2x}\right }=\frac{7}{3}\\\\ \Rightarrow \frac{3x^2+2x^2-3}{6x}=\frac{7}{3} \quad | \quad *6x\\\\ \Rightarrow 5x^2-3=6x*\frac{7}{3}\\ \Rightarrow 5x^2-3=14x\\$$

$$\\\Rightarrow 5x^2-14x-3=0\\\\ \Rightarrow x^2-\frac{14}{5}x-\frac{3}{5}=0\\\\ \Rightarrow x_{1,2}=\frac{14}{2*5}\pm\sqrt{\frac{14*14}{(2*5)*(2*5)}+\frac{3}{5}}\\\\ \Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{14*14}{(10)*(10)}+\frac{20}{20}*\frac{3}{5}}\\\\ \Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{196}{100}+\frac{60}{100}}\\\\ \Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{196+60}{100}}\\\\ \Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{256}{100}}\\\\ \Rightarrow x_{1,2}=\frac{14}{10}\pm\frac{16}{10}\\\\ \textcolor[rgb]{1,0,0}{x_1=}\frac{14}{10}+\frac{16}{10}=\frac{30}{10}=\textcolor[rgb]{1,0,0}{3}\\\\ \textcolor[rgb]{1,0,0}{x_2=}\frac{14}{10}-\frac{16}{10}=-\frac{2}{10}=\textcolor[rgb]{1,0,0}{-\frac{1}{5}}\\\\$$