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72%

This equation has infinite number of numerical solutions,

x = 1/179 (3590-571 y)

y = 1/571 (3590-179 x)

There are seven days in a week.  

63/7 = 9 weeks

Q^5 = 8sqrt2 = 8 (1.414)

Q^5 = 11.31    Then take the 5th root of both sides to get 'Q'  (see instructions below)

Q=1.6243

on Web2.0 calc enter    11.31 then the ' y checkmark x'   key,   then 5   then =

I am not sure what you mean here....      

m = 2.55 + n           2m=5n

hasnt that got old?

2....

Uhm what?

Because Ebony Gay is the No. 1 p**n

i give 0 fux

cuz it is deal with it b***h

are you blind or somthing? it is there twice! once at the start and once at the end!

its there!

121/64   x   4   =   (121)(4)/64  =  121/16

1549.51- 1549.51(.30) = 1084.66

3135993055.72

A clear cloudless day-time sky is bluebecause molecules in the air scatter blue light from the sun more than they scatter red light. When we look towards the sun at sunset, we see red and orange colours because the bluelight has been scattered out and away from the line of sight.

You give 14 apples to group 2 from group 1 now having 8 apples in group 1 and 28 apples in group 2. From group 2 with 28 apples give group 3 12 apples leaving group 2 with 16 apples group 1 with 8 apples and group 3 with 24 apples. Finally give group 1 8 apples from group 3 and you have every group with 16 apples! Hope this helped. :)

1.       2.      3.

22          14          12

8            28          12

8            16          24

16          16          16

I hope this is easier to understand.

Hi All,

I was intrigued by the answer Melody gave and would like to comment on it. See the approach of diving people into groups of same sizes is a pretty standard one. For example if you want ot divide a group of 8 people into pairs (i.e. all equal sizes) the standard approach shall be 8!/(((2!)^4)*4!), i.e divide 8! by 2!*2!*2!*2! and multiply the whole with 1/4! (since there are 4 groups of equal sizes) and it returns the same answer as 105. This can be used for anything. (3N as well). Say N people need to be distributed in 2 groups of 2 and 1 group of 3, then it is N!/2!*2!*3!*2! (i.e. 2! -> since 2 items belon to one group, 2! -> since 2 items belon to one group,3! -> since 2 items belon to one group, 2! -> since 2 groups are identical in size.) 

Now that concept out of the way, the thing that intrigued me is its application in geometrical figures, like the one discussed. I think it has happened because in such a scenario we fix one group namely AB and then form 3 paired groups, and since it is a square all the 8 arrangements will be identical, hence giving us the formula (2k-1)!/2^k-1*3!

i.e. (2k)!/(2k*((2!)^3)*3!)

I think that is the case. But definitely for sure such a formula wouldnt work out in other geometrical figures or if the numbers werent 8 for example, i.e we have 6 persons and there are 8 places. 

In a triangle ABC, AB = AC. D is the mid-point of BC, E is the foot of the perpendicular drawn from D to AC, and F is the mid-point of DE. Prove that AF is perpendicular to BE.

I. Definition:

\(\boxed{~ \begin{array}{lcl} \vec{a}= \vec{BC}\qquad |\vec{a}|= a \\ \vec{b}= \vec{AC}\qquad |\vec{b}|= b\qquad \vec{b}\cdot \vec{b} = b^2 \\ \quad \vec{a}\cdot \vec{b} = \frac{a}{2}\cdot a\\ \quad \vec{a}\cdot \vec{b} = \frac{a^2}{2}\\ \hline \vec{d}=\vec{AD}\\ \vec{d}=\vec{b}-\frac12 \vec{a} \\ \end{array} ~} \boxed{~ \begin{array}{lcl} \vec{f}=\vec{AE}\\ \vec{f}= \frac{ (\vec{b}\cdot\vec{d}) } {b^2} \vec{b}\\ \vec{f}= \frac{ (\vec{b}\cdot \left(\vec{b}-\frac12 \vec{a} \right) ) } {b^2} \vec{b}\\ \vec{f}= \frac{ (b^2 - \frac{ (\vec{a}\cdot \vec{b}) }{2} ) } {b^2} \vec{b}\\ \vec{f}= \frac{ (b^2 - \frac{ a^2 }{4} ) } {b^2} \vec{b}\\ \vec{f}= \left(1 - \frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \end{array} ~}\\\)

\(\boxed{~ \begin{array}{lcl} \vec{e}=\vec{ED}\\ \vec{e}=\vec{d}-\vec{f}\\ \vec{e}=\left( \vec{b}-\frac12 \vec{a} \right) -\vec{f}\\ \frac{\vec{e}}{2}=\left( \vec{b}-\frac12 \vec{a} \right)\frac12 -\frac12 \vec{f}\\ \vec{AF} = \frac{\vec{e}}{2}+\vec{f} = \left( \vec{b}-\frac12 \vec{a} \right)\frac12 -\frac12 \vec{f}+\vec{f}\\ \vec{AF} = \left( \vec{b}-\frac12 \vec{a} \right)\frac12 +\frac12 \vec{f}\\ \vec{AF} = \left( \vec{b}-\frac12 \vec{a} \right)\frac12 +\frac12 \left(1 - \frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \vec{AF} = \vec{b} \left[ \frac12 +\frac12 \cdot \left( 1 - \frac{ a^2 }{4b^2} \right) \right] -\frac14 \cdot \vec{a} \\ \vec{AF} = \vec{b} \left( \frac12 +\frac12 - \frac{ a^2 }{8b^2} \right) -\frac14 \cdot \vec{a} \\ \vec{AF} = \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) -\frac14 \cdot \vec{a} \\ \end{array} ~} \boxed{~ \begin{array}{lcl} \vec{BE} = \vec{a}-(\vec{b}-\vec{f})\\ \vec{BE} = \vec{a}-\vec{b}+\vec{f}\\ \vec{BE} = \vec{a}-\vec{b}+\left(1 - \frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \vec{BE} = \vec{a}-\vec{b}+ \vec{b} - \left(\frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \vec{BE} = \vec{a} - \left(\frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \end{array} ~}\)

II. Prove that AF is perpendicular to BE.

\(\boxed{~ \begin{array}{lcl} \vec{AF} = -\frac14 \cdot \vec{a} + \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) \\ \vec{BE} = \vec{a} - \vec{b} \left(\frac{ a^2 }{4b^2} \right)\\ \text{Perpendicular, if } \ (~\vec{AF}\cdot \vec{BE}~) = 0 \\ \hline \\ (~\vec{AF}\cdot \vec{BE}~) = \left ( -\frac14 \cdot \vec{a} + \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) \right) \cdot \left( \vec{a} - \vec{b} \left(\frac{ a^2 }{4b^2} \right) \right) \\ (~\vec{AF}\cdot \vec{BE}~) = -\frac14 \cdot \vec{a} \cdot \vec{a} + \frac14 \cdot \vec{a} \cdot \vec{b} \left(\frac{ a^2 }{4b^2} \right) + \vec{b} \cdot \vec{a} \left( 1 - \frac{ a^2 }{8b^2} \right) - \vec{b}\cdot \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) \left(\frac{ a^2 }{4b^2} \right)\\ \qquad \vec{a} \cdot \vec{a} = a^2 \qquad \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} =\frac{a^2}{2} \qquad \vec{b}\cdot \vec{b}=b^2 \\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4} + \frac14 \cdot \frac{a^2}{2} \left(\frac{ a^2 }{4b^2} \right) + \frac{a^2}{2} \left( 1 - \frac{ a^2 }{8b^2} \right) - b^2 \left( 1 - \frac{ a^2 }{8b^2} \right) \left(\frac{ a^2 }{4b^2} \right)\\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4} + \frac{a^4}{32b^2} + \frac{a^2}{2} - \frac{ a^4 }{16b^2} - \left( 1 - \frac{ a^2 }{8b^2} \right) \left(\frac{ a^2 }{4} \right)\\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4} + \frac{a^4}{32b^2} + \frac{a^2}{2} - \frac{ a^4 }{16b^2} -\frac{ a^2 }{4}+ \frac{ a^4 }{32b^2} \\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4}-\frac{ a^2 }{4}+ \frac{a^2}{2} + \frac{a^4}{32b^2} + \frac{ a^4 }{32b^2} - \frac{ a^4 }{16b^2} \\ \color{red }(~\vec{AF}\cdot \vec{BE}~) \color{black }= -\frac{ a^2 }{2}+ \frac{a^2}{2} + \frac{a^4}{16b^2} - \frac{ a^4 }{16b^2} \color{red }= 0\\ \end{array} ~}\)

cos x cos 30

\(cos30 cos x=\frac{\sqrt3 cosx}{2}\)

sin(x+15 degree) = 3 cos( x-15 degree)

\(\small{ \boxed{~ \text{Formula: }\quad \begin{array}{lcl} \sin{ (x+y) } &=& \sin{(x)}\cdot \cos{(y)} + \sin{(y)} \cdot \cos{(x)} \\ \cos{ (x-y) } &=& \cos{(x)}\cdot \cos{(y)} + \sin{(x)} \cdot \sin{(y)} \\ \end{array} ~}\\ \begin{array}{rcl} \sin{ ( x + 15^{\circ} ) } &=& 3 \cos{ ( x - 15^{\circ} ) } \\ \sin{(x)}\cdot \cos{(15^{\circ})} + \sin{(15^{\circ})} \cdot \cos{(x)} &=& 3 \cdot [~ \cos{(x)}\cdot \cos{( 15^{\circ})} + \sin{(x)} \cdot \sin{( 15^{\circ})} ~] \\ \sin{(x)}\cdot \cos{(15^{\circ})} + \sin{(15^{\circ})} \cdot \cos{(x)} &=& 3 \cdot \cos{(x)}\cdot \cos{( 15^{\circ})} + 3 \cdot \sin{(x)} \cdot \sin{( 15^{\circ})} \quad | \quad : \cos{(x)} \quad x\ne 90^{\circ}\\ \tan{(x)}\cdot \cos{(15^{\circ})} + \sin{(15^{\circ})} &=& 3 \cdot \cos{( 15^{\circ})} + 3 \cdot \tan{(x)} \cdot \sin{( 15^{\circ})} \\ \tan{(x)}\cdot \cos{(15^{\circ})} - 3 \cdot \tan{(x)} \cdot \sin{( 15^{\circ})} &=& 3 \cdot \cos{( 15^{\circ})} - \sin{(15^{\circ})} \qquad | \qquad : \cos{(15^{\circ})} \\ \tan{(x)} - 3 \cdot \tan{(x)} \cdot \tan{( 15^{\circ})} &=& 3 - \tan{(15^{\circ})} \\ \tan{(x)}\cdot \left[~ 1 - 3 \cdot \tan{( 15^{\circ})} ~ \right] &=& 3 - \tan{(15^{\circ})} \\ \tan{(x)} &=& \frac{ 3 - \tan{(15^{\circ})} } { 1 - 3 \cdot \tan{( 15^{\circ})} } \\ \tan{(x)} &=& \frac{ 2.7320508076 } { 0.1961524227 } \\ \tan{(x)} &=& 13.9282032303 \\ \mathbf{ x } & \mathbf{=} & \mathbf{ 85.8933946491^{\circ} \pm k\cdot 180^{\circ} \qquad k \in Z } \end{array} }\)