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There should be: 6! / [2!.2!.2!] = 90 permutations. Here is a list of all of them !!

{1, 1, 2, 2, 3, 3} | {1, 1, 2, 3, 2, 3} | {1, 1, 2, 3, 3, 2} | {1, 1, 3, 2, 2, 3} | {1, 1, 3, 2, 3, 2} | {1, 1, 3, 3, 2, 2} | {1, 2, 1, 2, 3, 3} | {1, 2, 1, 3, 2, 3} | {1, 2, 1, 3, 3, 2} | {1, 2, 2, 1, 3, 3} | {1, 2, 2, 3, 1, 3} | {1, 2, 2, 3, 3, 1} | {1, 2, 3, 1, 2, 3} | {1, 2, 3, 1, 3, 2} | {1, 2, 3, 2, 1, 3} | {1, 2, 3, 2, 3, 1} | {1, 2, 3, 3, 1, 2} | {1, 2, 3, 3, 2, 1} | {1, 3, 1, 2, 2, 3} | {1, 3, 1, 2, 3, 2} | {1, 3, 1, 3, 2, 2} | {1, 3, 2, 1, 2, 3} | {1, 3, 2, 1, 3, 2} | {1, 3, 2, 2, 1, 3} | {1, 3, 2, 2, 3, 1} | {1, 3, 2, 3, 1, 2} | {1, 3, 2, 3, 2, 1} | {1, 3, 3, 1, 2, 2} | {1, 3, 3, 2, 1, 2} | {1, 3, 3, 2, 2, 1} | {2, 1, 1, 2, 3, 3} | {2, 1, 1, 3, 2, 3} | {2, 1, 1, 3, 3, 2} | {2, 1, 2, 1, 3, 3} | {2, 1, 2, 3, 1, 3} | {2, 1, 2, 3, 3, 1} | {2, 1, 3, 1, 2, 3} | {2, 1, 3, 1, 3, 2} | {2, 1, 3, 2, 1, 3} | {2, 1, 3, 2, 3, 1} | {2, 1, 3, 3, 1, 2} | {2, 1, 3, 3, 2, 1} | {2, 2, 1, 1, 3, 3} | {2, 2, 1, 3, 1, 3} | {2, 2, 1, 3, 3, 1} | {2, 2, 3, 1, 1, 3} | {2, 2, 3, 1, 3, 1} | {2, 2, 3, 3, 1, 1} | {2, 3, 1, 1, 2, 3} | {2, 3, 1, 1, 3, 2} | {2, 3, 1, 2, 1, 3} | {2, 3, 1, 2, 3, 1} | {2, 3, 1, 3, 1, 2} | {2, 3, 1, 3, 2, 1} | {2, 3, 2, 1, 1, 3} | {2, 3, 2, 1, 3, 1} | {2, 3, 2, 3, 1, 1} | {2, 3, 3, 1, 1, 2} | {2, 3, 3, 1, 2, 1} | {2, 3, 3, 2, 1, 1} | {3, 1, 1, 2, 2, 3} | {3, 1, 1, 2, 3, 2} | {3, 1, 1, 3, 2, 2} | {3, 1, 2, 1, 2, 3} | {3, 1, 2, 1, 3, 2} | {3, 1, 2, 2, 1, 3} | {3, 1, 2, 2, 3, 1} | {3, 1, 2, 3, 1, 2} | {3, 1, 2, 3, 2, 1} | {3, 1, 3, 1, 2, 2} | {3, 1, 3, 2, 1, 2} | {3, 1, 3, 2, 2, 1} | {3, 2, 1, 1, 2, 3} | {3, 2, 1, 1, 3, 2} | {3, 2, 1, 2, 1, 3} | {3, 2, 1, 2, 3, 1} | {3, 2, 1, 3, 1, 2} | {3, 2, 1, 3, 2, 1} | {3, 2, 2, 1, 1, 3} | {3, 2, 2, 1, 3, 1} | {3, 2, 2, 3, 1, 1} | {3, 2, 3, 1, 1, 2} | {3, 2, 3, 1, 2, 1} | {3, 2, 3, 2, 1, 1} | {3, 3, 1, 1, 2, 2} | {3, 3, 1, 2, 1, 2} | {3, 3, 1, 2, 2, 1} | {3, 3, 2, 1, 1, 2} | {3, 3, 2, 1, 2, 1} | {3, 3, 2, 2, 1, 1} (total: 90)

There are 4 ways to sum up to 9.

There are 5 ways to sum up to 8. 

There are 36 different rolls on the die, but 9 ways to die :D

1/4 chance that you will die when you roll the die. (Okay I will stop)

The probability that A, B, or C will walk the plank is 

1/4 * 3 or 75%

Arghhhh!!

Hope this helped!!!

Supermanaccz

I am a little confused on the notations, can you please use latex? Then I can attempt the problem 

1)

983 should appear on both of their lists, since Anna is collecting numbers with different(non-repeating) digits and Oleg is collecting prime numbers. This is the largest such number up to 1023 that meets both conditions.

Disjoint events cannot happen at the same time. In other words, they are mutually exclusive. Put in formal terms, events A and B are disjoint if their intersection is zero: ... Disjoint events are disjointed, or not connected. Another way of looking at disjoint events are that they have no outcomes in common

A compound event is one in which there is more than one possible outcome.

Determining the probability of a compound event involves finding the sum of the probabilities of the individual events and, if necessary, removing any overlapping probabilities.

I hope this helped, 

Supermanaccz

Idenpendent events are events that do not effect each other. 

The occurring of A does not affect the outcome of B.

For example, 

rolling a 6 on a die and flipping a head on a coin. 

These events do not affect each other.

I hope this helped,

supermanaccz

I hope this helped :)

If by surfaces you mean area, here you go:

For the first shape, the area is:

17 * 15 which is 255.

For the second shape, we first need to find the height. 

To find the height we can use the Pythagorean theorem. 

5^2 - 3^2 = x^2 —————>  x = 4.

The area is 5 *4 ÷ 2 which also is 10.

The area for the third shape is:

17 *18÷2 which is 153.

To find the area for the final shape, we also need to find the height using the Pythagorean theorem. 

13^2 - 5^2 = x^2 ——————> x = 12

7*12 = 84. 

The area of the shapes in order is:

225, 10, 153, and 84.

Thanks so much for your answer Heaureka, that's some impressive Latex!

ok then...i guess i can

Can you translate it into "plaintext"??

The diagonals of a trapezoid are perpendicular and have lengths 8 and 10.

Find the length of the median of the trapezoid.

Area A = ?

\(\begin{array}{|rcll|} \hline 2A &=& (8-y)x + xy + (10-x)y+(10-x)(8-y) \\ 2A &=& 8x-yx+xy+10y-xy+80-10y-8x+xy \\ 2A &=& 80 \\ \mathbf{A} & \mathbf{=} & \mathbf{40} \\ \hline \end{array}\)

h = ?

\(\begin{array}{|rclrcl|} \hline \sin(B) &=& \dfrac{h}{8} & \sin(A) &=& \dfrac{h}{10} \quad & | \quad B = 90^{\circ}-A \\ \sin(90^{\circ}-A)^2+ \sin(A)^2 &=& \dfrac{h^2}{8^2} +\dfrac{h^2}{10^2} \\ \cos(A)^2+ \sin(A)^2 &=& \dfrac{h^2}{8^2} +\dfrac{h^2}{10^2} \\ 1 &=& \dfrac{h^2}{8^2} +\dfrac{h^2}{10^2} \\ \dfrac{h^2}{8^2} +\dfrac{h^2}{10^2} &=& 1 \\ h^2\left( \dfrac{1}{8^2} + \dfrac{1}{10^2} \right) &=& 1 \\ h^2\left( \dfrac{164}{8^210^2} \right) &=& 1 \\ h &=& \dfrac{8\cdot 10}{\sqrt{164}} \\ h &=& \dfrac{8\cdot 10}{\sqrt{4\cdot 41}} \\ h &=& \dfrac{8\cdot 10}{2\sqrt{41}} \\ \mathbf{ h }&\mathbf{=}& \mathbf{\dfrac{40}{\sqrt{41}} } \\ \hline \end{array}\)

median m = ?

\(\begin{array}{|rcll|} \hline A &=& \dfrac{a+c}{2} \cdot h \quad & | \quad m = \dfrac{a+c}{2}\\\\ A &=& m \cdot h \\\\ m &=& \dfrac{A}{h} \\\\ m &=& \dfrac{40}{ \dfrac{40}{\sqrt{41} } } \\\\ m &=& \dfrac{40}{40}\cdot \sqrt{41} \\\\ m &=& \sqrt{41} \\\\ \mathbf{ m} &\mathbf{ =} & \mathbf{ 6.40312423743} \\ \hline \end{array} \)

Sorry for the confusion, I am working with the definition that a trapezoid is a quadrilateral in which exactly one pair of opposite sides are parallel.

HI Melody;

Yes, it is a rhombus. It is not different. A rhombus is a special parallelogram and a parallelogram is a special trapezoid.

The diagonals should be perpendicular to each other.

Hi Omi, it is good to see you here. 

But..... that is a rhombus, I do not think it is the shape that  AConfusedGuy meant :)

The diagonals of a trapezoid are perpendicular and have lengths 8 and 10. Find the length of the median of the trapezoid.

Trapezoid means different things in different countries.

I am assuming that you mean a 4 sided figure where 2 of the sides are parallel. (I would call this a trapezium)

I am wondering if you are referring to what I would call an isoscles trapezium where the 2 non parallel sides are equal in length?

How many subsets of the set $\{1,2,3,4,\ldots,10\}$ have no two consecutive numbers as members?

\(\begin{array}{|l|cr|r|} \hline & &&\text{okay} \\ \hline \binom{10}{0}\times \ 0-\text{elements-subsets } &=& 1 & 1 \\ \binom{10}{1}\times \ 1-\text{elements-subsets } &=& 10 & 10 \\ \binom{10}{2}\times \ 2-\text{elements-subsets } &=& 45 & 36 \\ \binom{10}{3}\times \ 3-\text{elements-subsets } &=& 120 & 56 \\ \binom{10}{4}\times \ 4-\text{elements-subsets } &=& 210 & 35 \\ \binom{10}{5}\times \ 5-\text{elements-subsets } &=& 252 & 6 \\ \binom{10}{6}\times \ 6-\text{elements-subsets } &=& 210 & 0 \\ \binom{10}{7}\times \ 7-\text{elements-subsets } &=& 120 & 0 \\ \binom{10}{8}\times \ 8-\text{elements-subsets } &=& 45 & 0 \\ \binom{10}{9}\times \ 9-\text{elements-subsets } &=& 10 & 0 \\ \binom{10}{10}\times \ 10-\text{elements-subsets } &=& 1 & 0 \\ \hline \text{sum } &=& & 144 \\ \hline \end{array}\)

The subsets:

   1. { }

   2. {1}
   3. {2}
   4. {3}
   5. {4}
   6. {5}
   7. {6}
   8. {7}
   9. {8}
  10. {9}
  11. {10}

  12. {1, 3}
  13. {1, 4}
  14. {1, 5}
  15. {1, 6}
  16. {1, 7}
  17. {1, 8}
  18. {1, 9}
  19. {1, 10}
  20. {2, 4}
  21. {2, 5}
  22. {2, 6}
  23. {2, 7}
  24. {2, 8}
  25. {2, 9}
  26. {2, 10}
  27. {3, 5}
  28. {3, 6}
  29. {3, 7}
  30. {3, 8}
  31. {3, 9}
  32. {3, 10}
  33. {4, 6}
  34. {4, 7}
  35. {4, 8}
  36. {4, 9}
  37. {4, 10}
  38. {5, 7}
  39. {5, 8}
  40. {5, 9}
  41. {5, 10}
  42. {6, 8}
  43. {6, 9}
  44. {6, 10}
  45. {7, 9}
  46. {7, 10}
  47. {8, 10}

  48. {1, 3, 5}
  49. {1, 3, 6}
  50. {1, 3, 7}
  51. {1, 3, 8}
  52. {1, 3, 9}
  53. {1, 3, 10}
  54. {1, 4, 6}
  55. {1, 4, 7}
  56. {1, 4, 8}
  57. {1, 4, 9}
  58. {1, 4, 10}
  59. {1, 5, 7}
  60. {1, 5, 8}
  61. {1, 5, 9}
  62. {1, 5, 10}
  63. {1, 6, 8}
  64. {1, 6, 9}
  65. {1, 6, 10}
  66. {1, 7, 9}
  67. {1, 7, 10}
  68. {1, 8, 10}
  69. {2, 4, 6}
  70. {2, 4, 7}
  71. {2, 4, 8}
  72. {2, 4, 9}
  73. {2, 4, 10}
  74. {2, 5, 7}
  75. {2, 5, 8}
  76. {2, 5, 9}
  77. {2, 5, 10}
  78. {2, 6, 8}
  79. {2, 6, 9}
  80. {2, 6, 10}
  81. {2, 7, 9}
  82. {2, 7, 10}
  83. {2, 8, 10}
  84. {3, 5, 7}
  85. {3, 5, 8}
  86. {3, 5, 9}
  87. {3, 5, 10}
  88. {3, 6, 8}
  89. {3, 6, 9}
  90. {3, 6, 10}
  91. {3, 7, 9}
  92. {3, 7, 10}
  93. {3, 8, 10}
  94. {4, 6, 8}
  95. {4, 6, 9}
  96. {4, 6, 10}
  97. {4, 7, 9}
  98. {4, 7, 10}
  99. {4, 8, 10}
 100. {5, 7, 9}
 101. {5, 7, 10}
 102. {5, 8, 10}
 103. {6, 8, 10}

 104. {1, 3, 5, 7}
 105. {1, 3, 5, 8}
 106. {1, 3, 5, 9}
 107. {1, 3, 5, 10}
 108. {1, 3, 6, 8}
 109. {1, 3, 6, 9}
 110. {1, 3, 6, 10}
 111. {1, 3, 7, 9}
 112. {1, 3, 7, 10}
 113. {1, 3, 8, 10}
 114. {1, 4, 6, 8}
 115. {1, 4, 6, 9}
 116. {1, 4, 6, 10}
 117. {1, 4, 7, 9}
 118. {1, 4, 7, 10}
 119. {1, 4, 8, 10}
 120. {1, 5, 7, 9}
 121. {1, 5, 7, 10}
 122. {1, 5, 8, 10}
 123. {1, 6, 8, 10}
 124. {2, 4, 6, 8}
 125. {2, 4, 6, 9}
 126. {2, 4, 6, 10}
 127. {2, 4, 7, 9}
 128. {2, 4, 7, 10}
 129. {2, 4, 8, 10}
 130. {2, 5, 7, 9}
 131. {2, 5, 7, 10}
 132. {2, 5, 8, 10}
 133. {2, 6, 8, 10}
 134. {3, 5, 7, 9}
 135. {3, 5, 7, 10}
 136. {3, 5, 8, 10}
 137. {3, 6, 8, 10}
 138. {4, 6, 8, 10}

 139. {1, 3, 5, 7, 9}
 140. {1, 3, 5, 7, 10}
 141. {1, 3, 5, 8, 10}
 142. {1, 3, 6, 8, 10}
 143. {1, 4, 6, 8, 10}
 144. {2, 4, 6, 8, 10}

Find $$\sum_{k = 0}^{10} (k + 2) \cdot 3 \cdot 2^k.$$

\(\displaystyle \sum \limits_{k = 0}^{10} (k + 2) \cdot 3 \cdot 2^k = 33 \cdot 2^{11} = 67584\)

If $f(x) = \frac{4x+1}{3}$ what is the value of $\left[f^{-1}(1)\right]^{-1}$?

\(\begin{array}{|rcll|} \hline f(x) &=& \dfrac{4x+1}{3} \\ \hline y &=& \dfrac{4x+1}{3} \\\\ 3y &=& 4x+1 \\\\ 3y-1 &=& 4x \\\\ x &=& \dfrac{3y-1}{4} \quad & | \quad x \leftrightarrow y \\\\ y &=& \dfrac{3x-1}{4} \\\\ f^{-1}(x) &=& \dfrac{3x-1}{4} \\ \hline f^{-1}(1) &=& \dfrac{3\cdot 1-1}{4} \\\\ &=& \dfrac{2}{4} \\\\ &=& \dfrac{1}{2} \\ \hline \left[f^{-1}(1)\right]^{-1} &=& \left(\dfrac{1}{2}\right)^{-1} \\\\ &=& \dfrac{1}{ \left(\dfrac{1}{2}\right)^{1}} \\\\ &=& \dfrac{1}{ \dfrac{1}{2} } \\\\ &=& 1\cdot \dfrac{2}{1} \\\\ \mathbf{ \left[f^{-1}(1)\right]^{-1}} &\mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

Calculate $$ S = \sum_{k=1}^n \frac 1{k(k+1)(k+2)}. $$

\(\begin{array}{|lcll|} \hline S_n = \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{1}{2 \cdot 3 \cdot 4} + \dfrac{1}{3 \cdot 4 \cdot 5} + \dfrac{1}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)} \\ \hline \end{array}\)

Formula:

\(\begin{array}{|lcll|} \hline \text{in general}:\ \dfrac{1}{n(n+d)} = \dfrac{1}{d}\left(\dfrac{1}{n}- \dfrac{1}{n+d} \right) \\ \hline \\ \begin{array}{lrcll} \text{we need}: & \dfrac{1}{(n+1)(n+2)} &=& \dfrac{1}{n+1}-\dfrac{1}{n+2} \\\\ & \dfrac{1}{n(n+1)} &=& \dfrac{1}{n}-\dfrac{1}{n+1} \\\\ & \dfrac{1}{n(n+2)} &=& \dfrac{1}{2} \left( \dfrac{1}{n}-\dfrac{1}{n+2} \right) \\ \end{array} \\ \hline \end{array}\)

we rearrange:

\(\begin{array}{|rcll|} \hline \dfrac{1}{n \cdot (n+1) \cdot (n+2)} \\\\ &=& \dfrac{1}{n}\times \dfrac{1}{(n+1) \cdot (n+2)} \\\\ &=& \dfrac{1}{n}\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{n}\times \dfrac{1}{n+1} - \dfrac{1}{n}\times \dfrac{1}{n+2} \\\\ &=& \left(\dfrac{1}{n}-\dfrac{1}{n+1} \right)- \dfrac{1}{2} \times \left(\dfrac{1}{n} -\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{n} - \dfrac{1}{n+1} -\dfrac{1}{2n} + \dfrac{1}{2(n+2)} \\\\ \mathbf{\dfrac{1}{n \cdot (n+1) \cdot (n+2)} } & \mathbf{=} & \mathbf{ \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)} } \\ \hline \end{array}\)

telescoping series:

\(\begin{array}{|rcll|} \hline S_n &=& \mathbf{\dfrac{1}{2}} &\mathbf{-}& \mathbf{\dfrac{1}{2}} &\color{red}+& \color{red}\dfrac{1}{6} \\\\ &\mathbf{+}& \mathbf{\dfrac{1}{4}} &\color{red}-& \color{red}\dfrac{1}{3} &\color{blue}+& \color{blue}\dfrac{1}{8} \\\\ &\color{red}+& \color{red}\dfrac{1}{6} &\color{blue}-& \color{blue}\dfrac{1}{4} &\color{red}+& \color{red}\dfrac{1}{10} \\\\ &\color{blue}+& \color{blue}\dfrac{1}{8} &\color{red}-& \color{red}\dfrac{1}{5} &\color{green}+& \color{green}\dfrac{1}{12} \\\\ && \ldots \\\\ &+\color{red}& \color{red}\dfrac{1}{2(n-2)} &\color{green}-& \color{green}\dfrac{1}{n-1} &\color{red}+& \color{red}\dfrac{1}{2n} \\\\ &\color{green}+& \color{green}\dfrac{1}{2(n-1)} &\color{red}-& \color{red}\dfrac{1}{n} &\mathbf{+}& \mathbf{\dfrac{1}{2(n+1)}} \\\\ &\color{red}+& \color{red}\dfrac{1}{2n} &\mathbf{-}& \mathbf{\dfrac{1}{n+1}} &\mathbf{+}& \mathbf{\dfrac{1}{2(n+2)}} \\ \hline \end{array}\)

The part of each term cancelling with part of the next two diagonal terms:
Example:

\(\begin{array}{|lcll|} \hline \dfrac{1}{6}-\dfrac{1}{3}+\dfrac{1}{6} = 0 \\\\ \dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{8} = 0 \\\\ \dfrac{1}{10}-\dfrac{1}{5}+\dfrac{1}{10} = 0 \\\\ \ldots \\\\ \dfrac{1}{2n}-\dfrac{1}{n} + \dfrac{1}{2n} = 0 \\ \hline \end{array}\)

So \(S_n\) is, we have all black terms left :

\(\begin{array}{|rcll|} \hline S_n &=& \dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4} + \dfrac{1}{2(n+1)} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2(n+1)} + \dfrac{1}{2(n+2)} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2}\left( \dfrac{1}{n+1} - \dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2}\left( \dfrac{1}{(n+1)(n+2)} \right) \\\\ \mathbf{S_n} &\mathbf{=}& \mathbf{\dfrac{1}{4} - \dfrac{1}{2(n+1)(n+2)} } \\ \hline \end{array}\)

\(\displaystyle \mathbf{ S_n = \sum \limits_{k=1 }^{n} \dfrac 1{k(k+1)(k+2)} = \dfrac{1}{4} - \dfrac{1}{2(n+1)(n+2) } }\)

Rectangle ABCD is folded along BD, and point C lands on C1. BC1 and AD intersect at point E.

AB=5, AD=10. What is the length of DE?

\(\text{Let $AB=DC=b$ } \\ \text{Let $AD=BC=a$ } \\ \text{Let $DE=BE'=E'C'=x$ } \\ \text{Let $BD=d$ } \)

\(\begin{array}{|rcll|} \hline d^2 &=& a\cdot 2x \\\\ x &=& \dfrac{d^2}{a} \quad & | \quad d^2 = a^2+b^2 \\ &\mathbf{x } &\mathbf{=}& \mathbf{\dfrac{a^2+b^2}{2a}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a&=&10 \\ b &=& 5 \\\\ ED=x &=& \dfrac{5^2+10^2}{2\cdot 10} \\ x &=& \dfrac{125}{20} \\ x &=& \dfrac{25}{4} \\ \mathbf{x } &\mathbf{=}& \mathbf{6.25} \\ \hline \end{array}\)

It is used to refer to the four quarters that are made when you devide the coordinate plane by the x and y axis