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Is there any software by which I can can culatate eizen ventor of any martix? I would appreciate yoru help

Simplify the following trigonometric expression

Multiply cosθ/cscθ+1 by cscθ-1/cscθ-1

$\begin{array}{|rcll|} \hline && \left( \frac{ \cos(\phi) } {\csc(\phi)+1 } \right) \cdot \left( \frac{ \csc(\phi)-1 } { \csc(\phi)-1 } \right ) \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\csc(\phi)-1 \Big) } { \csc^2(\phi)-1 } \quad & | \quad \csc(\phi) = \frac{1}{\sin(\phi)} \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big) } { \frac{1}{\sin^2(\phi)}-1 } \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big) } { \frac{1-\sin^2(\phi)}{\sin^2(\phi)} } \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big)\cdot \sin^2(\phi) } { 1-\sin^2(\phi) } \quad & | \quad 1-\sin^2(\phi) = \cos^2(\phi) \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big)\cdot \sin^2(\phi) } { \cos^2(\phi) } \\\\ &=& \frac{ \Big(\frac{1}{\sin(\phi)}-1 \Big)\cdot \sin^2(\phi) } { \cos(\phi) } \\\\ &=& \frac{ \Big(\frac{1-\sin(\phi)}{\sin(\phi)}\Big)\cdot \sin^2(\phi) } { \cos(\phi) } \\\\ &=& \frac{ \Big(1-\sin(\phi)\Big)\cdot \sin(\phi) } { \cos(\phi) } \\\\ &=& \Big(1-\sin(\phi)\Big) \cdot \tan(\phi) \\\\ &=& \tan(\phi)-\tan(\phi)\cdot \sin(\phi) \\ \hline \end{array}$

n=1/1+2 + 1/1+2+3 + ...... + 1/1+2+3+....+2014 + 2/2015, n=?

$\begin{array}{|rcll|} \hline n &=& \frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4} + \ldots + \frac{1}{1+2+3+\ldots+2013} + \frac{1}{1+2+3+\ldots+2014} + \frac{2}{2015} \\ \hline \end{array}$

$\begin{array}{rcll} n &=& \frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4} + \ldots + \frac{1}{1+2+3+\ldots+2013} + \frac{1}{1+2+3+\ldots+2014} + \frac{2}{2015} \\ &=& \frac{1}{\left(\frac{1+2}{2}\right)\cdot 2} +\frac{1}{\left(\frac{1+3}{2}\right)\cdot 3} +\frac{1}{\left(\frac{1+4}{2}\right)\cdot 4} + \ldots + \frac{1}{\left(\frac{1+2013}{2}\right)\cdot 2013} + \frac{1}{\left(\frac{1+2014}{2}\right)\cdot 2014} + \frac{2}{2015} \\ &=& \frac{2}{2\cdot 3} +\frac{2}{3\cdot 4} +\frac{2}{4\cdot 5} + \ldots + \frac{2}{2013\cdot 2014} + \frac{2}{2014\cdot 2015} + \frac{2}{2015} \\ &=& 2\times \Big( \frac{1}{2015} + \frac{1}{2\cdot 3} +\frac{1}{3\cdot 4} +\frac{1}{4\cdot 5} + \ldots + \frac{1}{2013\cdot 2014} + \frac{1}{2014\cdot 2015} \Big)\\\\ && \mathbf{\frac{1}{a\cdot (a+1)} = \frac{1}{a} - \frac{1}{a+1}} \\\\ &=& 2\times \Big[ \frac{1}{2015} + \left(\frac{1}{2}-\frac{1}{3}\right) +\left(\frac{1}{3}-\frac{1}{4}\right) +\left(\frac{1}{4}-\frac{1}{5}\right) + \ldots \\ && + \left(\frac{1}{2013}-\frac{1}{2014}\right) + \left(\frac{1}{2014}-\frac{1}{2015}\right) \Big] \\\\ &=& 2\times \Big[ \frac{1}{2015} + \frac12+ \underbrace{\left(-\frac{1}{3}+\frac{1}{3}\right)}_{=0} +\underbrace{\left(-\frac{1}{4}+\frac{1}{4}\right) }_{=0} +\underbrace{\left(-\frac{1}{5}+\frac{1}{5}\right)}_{=0} + \ldots \\ && + \underbrace{\left(-\frac{1}{2013}+\frac{1}{2013}\right)}_{=0} + \underbrace{\left(-\frac{1}{2014}+\frac{1}{2014}\right)}_{=0} -\frac{1}{2015} \Big] \\\\ &=& 2\times \left(\frac{1}{2015}+ \frac12 -\frac{1}{2015} \right) \\\\ &=& 2\times \left(\frac12 \right) \\\\ &\mathbf{=}& \mathbf{1} \\ \end{array}$

12x^3+53x^2−34x+5=0

$\begin{array}{|l|rcll|} \hline & 12x^3+53x^2-34x+5 &=& 0 \\ \mathbf{ x_1 = -5} & 12\cdot (-5)^3 + 53\cdot (-5)^2 -34\cdot (-5) + 5 &=& 0 \\ & -1500+1325+170+5 &=& 0 \\ & 0 &=& 0\ \checkmark \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 12x^2-7x+1 &=& 0 \\ x &=& \frac{7\pm \sqrt{(-7)^2 - 4 \cdot 12 \cdot 1 } }{2\cdot 12} \\ &=& \frac{7\pm \sqrt{1} }{24} \\ &=& \frac{7\pm 1 }{24} \\\\ x_2 &=& \frac{7 + 1 }{24} \\ \mathbf{ x_2 } & \mathbf{=}& \mathbf{ \frac{1 }{3} } \\\\ x_3 &=& \frac{7 - 1 }{24} \\ \mathbf{ x_3 } & \mathbf{=}& \mathbf{\frac{1 }{4}} \\ \hline \end{array}$

$x_1 = -5 \\ x_2 = \frac{1 }{3} \\ x_3 = \frac{1 }{4}$

In the Small State Lottery,
three white balls are drawn at random from twenty balls labled 1-20 and
a blue Superball is drawn from ten balls labeled 21-30.
To win a prize you must match at least two of the white balls or match the blue Superball.
If you buy a ticket what is the probability that you would win a prize?

Let W be the probability of choosing 2 white balls out of 20 or 3 white balls out of 20 .

Let $\overline{\mathbf{W}}$be not W.

Let B be the probability of choosing one blue ball out of 10.

Let $\overline{\mathbf{B}}$ be not B

$\begin{array}{|rcll|} \hline W &=& \dfrac{\binom{3}{2}\binom{17}{1} + \binom{3}{3}\binom{17}{0} }{\binom{20}{3}} = \dfrac{13}{285} \\\\ \overline{W} &=& 1 - W = 1-\dfrac{13}{285} = \dfrac{272}{285} \\\\ B &=& \dfrac{\binom{1}{1}\binom{9}{0} }{\binom{10}{1}} = \dfrac{1}{10} \\\\ \overline{B} &=& 1 - B = 1 - \dfrac{1}{10} = \dfrac{9}{10} \\ \hline \end{array}$

$\begin{array}{|c|c|c|c|} \hline & \mathbf{B} & \mathbf{\overline{B}} \\ & \dfrac{1}{10} & \dfrac{9}{10} \\ \hline \mathbf{W} & W\cap B & W\cap \overline{B} \\ \dfrac{13}{285} & \dfrac{13}{285} \times \dfrac{1}{10} & \dfrac{13}{285} \times \dfrac{9}{10} \\ \hline \mathbf{\overline{W}} & \overline{W}\cap B & \overline{W}\cap \overline{B} \\ \dfrac{272}{285} & \dfrac{272}{285} \times \dfrac{1}{10} & \dfrac{272}{285} \times \dfrac{9}{10} \\ \hline \end{array}$

The probability of not a prize is
$\overline{W}\cap \overline{B} = \dfrac{272}{285} \times \dfrac{9}{10} \\$

The probability of a prize is
$\begin{array}{|rcll|} \hline && 1- \overline{W}\cap \overline{B} \\ &=& 1- \left(\frac{272}{285} \times \frac{9}{10} \right) \\ &=& \frac{285\cdot 10- 9\cdot 272}{285\cdot 10} \\ &=& \frac{402}{2850} \\ &=& \frac{67\cdot 6}{475\cdot 6} \\ &=& \frac{67}{475} \\ &=& 0.14105263158 \\ \hline \end{array}$

n=1/1+2 + 1/1+2+3 + ...... + 1/1+2+3+....+2014 + 2/2015, n=?

edited

continued fraction ?

$\begin{array}{|rcll|} \hline n =\cfrac{1}{1+2+\cfrac{1}{1+2+3+\cfrac{1}{ 1+2+3+4+\ldots \cfrac{1}{1+2+3+4+\ldots + 2014 +\frac{2}{2015}}}}} \\ \hline \end{array}$

n = 0.31606040

y'+y=x

$\begin{array}{|l|rcll|} \hline & y'+y &=& x \\ & y' &=& x -y \\ \text{substitution} \ u=x-y & y' &=& u \\ & 1-y' &=& 1-u \quad & | \quad u' = 1-y'\\ & u' &=& 1-u \\ & \frac{du}{dx} &=& 1-u \\ & \frac{du}{1-u} &=& dx \\ & \int \frac{du}{1-u} &=& \int dx \\ & -\ln|1-u| &=& x + c_1 \\ & \ln|1-u| &=& -x - c_1 \\ & 1-u &=& e^{-x - c_1} = c_2e^{-x} \quad & \quad (c_2=e^{-c_1}) \\ & u &=& 1- c_2e^{-x}=1-ce^{-x} \quad & \quad (c_2=c) \\ \text{back substitution} \ x-y=u & x-y &=& 1-ce^{-x} \\\\ & \mathbf{ y }& \mathbf{=} & \mathbf{ce^{-x} +x-1} \\ \hline \end{array}$

proof y' + y = x

$\begin{array}{|rcll|} \hline \mathbf{ y }& \mathbf{=} & \mathbf{ce^{-x} +x-1} \\ \mathbf{ y' }& \mathbf{=} & \mathbf{-ce^{-x} +1 } \\ y'+y &=& -ce^{-x} +1 + ce^{-x} +x-1 \\ y'+y &=& x \ \checkmark \\ \hline \end{array}$

(3x^3+2x^2-25x-12)/(x+3)=?

$\dfrac{3x^3+2x^2-25x-12}{x+3}=\ ?$

The point (-3,-4) divides the line joining point A(-6,-7) and point B in the ratio 1:3.

Find the coordinates of B. 

$\text{Set } \vec{P} = \binom{-3}{-4} \\ \text{Set } \vec{A} = \binom{-6}{-7} \\ \text{Set } \vec{B} =\ ?$

Formula:

$\begin{array}{|rcll|} \hline \vec{P} = (1-\lambda)\vec{A}+\lambda\vec{B} \\ \hline \end{array}$

Ratio 1:3

$\begin{array}{|rcll|} \hline \lambda &=& \frac{1}{1+3} \\ &=& \frac14 \\ \hline \end{array}$

Solution for $\vec{B}$:

$\begin{array}{|rcll|} \hline \vec{P} &=& (1-\lambda)\vec{A}+\lambda\vec{B} \\ \lambda\vec{B} &=& \vec{P} - (1-\lambda)\vec{A} \\ \vec{B} &=& \frac{1}{\lambda} \cdot \Big( \vec{P} - (1-\lambda)\vec{A} \Big) \quad & | \quad \lambda &=& \frac14 \\ \vec{B} &=& \frac{1}{ \frac14 } \cdot \Big( \vec{P} - (1-\frac14)\vec{A} \Big) \\ \vec{B} &=& 4 \cdot ( \vec{P} - \frac34 \vec{A} ) \\ \vec{B} &=& 4 \vec{P} - 3\vec{A} \\ \vec{B} &=& 4 \binom{-3}{-4} - 3\binom{-6}{-7} \\ \vec{B} &=& 4 \binom{-3}{-4} + 3\binom{6}{7} \\ \vec{B} &=& \binom{-12}{-16} + \binom{18}{21} \\ \vec{B} &=& \binom{-12+18}{-16+21} \\ \mathbf{ \vec{B} } & \mathbf{=} & \mathbf{\dbinom{6}{5}} \\ \hline \end{array}$

Point B(6,5)

If a permutation is chosen at random from the letters "AAABBBCD",

what is the probability that it begins with at least 2 A's?
Round your answer to 6 decimal places as needed.

     1. 36 x AAABBBCD
     2. 36 x AAABBBDC
     3. 36 x AAABBCBD
     4. 36 x AAABBCDB
     5. 36 x AAABBDBC
     6. 36 x AAABBDCB
     7. 36 x AAABCBBD
     8. 36 x AAABCBDB
     9. 36 x AAABCDBB
    10. 36 x AAABDBBC
    11. 36 x AAABDBCB
    12. 36 x AAABDCBB
    13. 36 x AAACBBBD
    14. 36 x AAACBBDB
    15. 36 x AAACBDBB
    16. 36 x AAACDBBB
    17. 36 x AAADBBBC
    18. 36 x AAADBBCB
    19. 36 x AAADBCBB
    20. 36 x AAADCBBB
    21. 36 x AABABBCD
    22. 36 x AABABBDC
    23. 36 x AABABCBD
    24. 36 x AABABCDB
    25. 36 x AABABDBC
    26. 36 x AABABDCB
    27. 36 x AABACBBD
    28. 36 x AABACBDB
    29. 36 x AABACDBB
    30. 36 x AABADBBC
    31. 36 x AABADBCB
    32. 36 x AABADCBB
    33. 36 x AABBABCD
    34. 36 x AABBABDC
    35. 36 x AABBACBD
    36. 36 x AABBACDB
    37. 36 x AABBADBC
    38. 36 x AABBADCB
    39. 36 x AABBBACD
    40. 36 x AABBBADC
    41. 36 x AABBBCAD
    42. 36 x AABBBCDA
    43. 36 x AABBBDAC
    44. 36 x AABBBDCA
    45. 36 x AABBCABD
    46. 36 x AABBCADB
    47. 36 x AABBCBAD
    48. 36 x AABBCBDA
    49. 36 x AABBCDAB
    50. 36 x AABBCDBA
    51. 36 x AABBDABC
    52. 36 x AABBDACB
    53. 36 x AABBDBAC
    54. 36 x AABBDBCA
    55. 36 x AABBDCAB
    56. 36 x AABBDCBA
    57. 36 x AABCABBD
    58. 36 x AABCABDB
    59. 36 x AABCADBB
    60. 36 x AABCBABD
    61. 36 x AABCBADB
    62. 36 x AABCBBAD
    63. 36 x AABCBBDA
    64. 36 x AABCBDAB
    65. 36 x AABCBDBA
    66. 36 x AABCDABB
    67. 36 x AABCDBAB
    68. 36 x AABCDBBA
    69. 36 x AABDABBC
    70. 36 x AABDABCB
    71. 36 x AABDACBB
    72. 36 x AABDBABC
    73. 36 x AABDBACB
    74. 36 x AABDBBAC
    75. 36 x AABDBBCA
    76. 36 x AABDBCAB
    77. 36 x AABDBCBA
    78. 36 x AABDCABB
    79. 36 x AABDCBAB
    80. 36 x AABDCBBA
    81. 36 x AACABBBD
    82. 36 x AACABBDB
    83. 36 x AACABDBB
    84. 36 x AACADBBB
    85. 36 x AACBABBD
    86. 36 x AACBABDB
    87. 36 x AACBADBB
    88. 36 x AACBBABD
    89. 36 x AACBBADB
    90. 36 x AACBBBAD
    91. 36 x AACBBBDA
    92. 36 x AACBBDAB
    93. 36 x AACBBDBA
    94. 36 x AACBDABB
    95. 36 x AACBDBAB
    96. 36 x AACBDBBA
    97. 36 x AACDABBB
    98. 36 x AACDBABB
    99. 36 x AACDBBAB
   100. 36 x AACDBBBA
   101. 36 x AADABBBC
   102. 36 x AADABBCB
   103. 36 x AADABCBB
   104. 36 x AADACBBB
   105. 36 x AADBABBC
   106. 36 x AADBABCB
   107. 36 x AADBACBB
   108. 36 x AADBBABC
   109. 36 x AADBBACB
   110. 36 x AADBBBAC
   111. 36 x AADBBBCA
   112. 36 x AADBBCAB
   113. 36 x AADBBCBA
   114. 36 x AADBCABB
   115. 36 x AADBCBAB
   116. 36 x AADBCBBA
   117. 36 x AADCABBB
   118. 36 x AADCBABB
   119. 36 x AADCBBAB
   120. 36 x AADCBBBA
 
      4320 x AA $\ldots$

The probability that "AAABBBCD" begins with at least 2 A's is
$\frac{4320}{8!} =\frac{4320}{40320} =0.10714285714 =10.7143 \%$