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3!*4!=6*24= 144

Both x and y are positive real numbers,

and the point (x,y) lies on or above both of the lines having equations \(2x + 5y = 10\) and \(3x + 4y = 12\).

What is the least possible value of \(8x + 13y\)?

\(\begin{array}{|lrcll|} \hline & 2x + 5y &=& 10 \\ & 3x + 4y &=& 12 \\ & 3x + 4y &=& 12 \\ \hline \text{sum} & 8x + 13y &=& 10+12+12 \\ & \mathbf{8x + 13y} &=& \mathbf{34} \\ \hline \end{array}\)

(b)

In the expansion of \(\left(1 + x\right)^n\), three consecutive coefficients are in the ratio \(1:7:35\).
Find the positive integer \(n\).

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k + 1}} &=& \dfrac{n!}{(k+1)!(n-k-1)!} \quad &| \quad k!(k+1)=(k+1)! \\\\ &=& \dfrac{1}{k+1} *\dfrac{n!}{k!(n-k-1)!} \quad &| \quad (n-k-1)!(n-k)=(n-k)! \\\\ &=& \dfrac{n-k}{k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{n-k}{k+1} \dbinom{n}{k} } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dbinom{n}{k - 1} : \dbinom{n}{k} : \dbinom{n}{k + 1} &=& 1 : 7 : 35 \\\\ &&\boxed{ \mathbf{\dbinom{n}{k - 1} = \dfrac{k}{n-k+1} \dbinom{n}{k} }\\ ~\\ \mathbf{\dbinom{n}{k + 1} = \dfrac{n-k}{k+1} \dbinom{n}{k} } } \\\\ \dfrac{k}{n-k+1} \dbinom{n}{k} : \dbinom{n}{k} : \dfrac{n-k}{k+1} \dbinom{n}{k} &=& 1 : 7 : 35 \\ \hline \dfrac{\dbinom{n}{k}}{\dfrac{k}{n-k+1} \dbinom{n}{k}} &=& \dfrac{7}{1} \\\\ \dfrac{n-k+1}{k} &=& 7 \\\\ n-k+1 &=& 7k \\ \mathbf{n} &=& \mathbf{8k -1} \qquad (1) \\ \hline \dfrac{\dfrac{n-k}{k+1} \dbinom{n}{k} } {\dbinom{n}{k}} &=& \dfrac{35}{7} \\\\ \dfrac{n-k}{k+1} &=& 5 \\\\ 5(k+1) &=& n-k \\ 5k+5 &=& n-k \\ 6k &=& n-5 \\ \mathbf{k} &=& \mathbf{ \dfrac{n-5}{6} } \qquad (2) \\ \hline n &=& 8k-1 \quad | \quad k=\dfrac{n-5}{6} \\ n &=& 8 \dfrac{(n-5)}{6}-1 \\ n &=& \dfrac{4}{3}(n-5)-1 \\ & & \dotsb \\ \mathbf{n} &=& \mathbf{23} \\\\ k &=& \dfrac{n-5}{6} \quad | \quad n=23 \\\\ k &=& \dfrac{23-5}{6} \\ & & \dotsb \\ \mathbf{k} &=& \mathbf{3} \\ \hline \dbinom{n}{k - 1} : \dbinom{n}{k} : \dbinom{n}{k + 1} &=& 1 : 7 : 35 \\\\ \dbinom{23}{2} : \dbinom{23}{3} : \dbinom{23}{4} &=& 1 : 7 : 35 \\\\ 253 : 1771 : 8855 &=& 1 : 7 : 35 \\ \hline \end{array}\)

You can solve this one in your head in less then a minute.

                    A = 4*6 + 4*15    

By Vieta

In the form   ax^2 + bx + c

The product of the roots =   c /a   =     (2m) / 3

So

2 * 3  =  2m / 3

6  =  2m  / 3

18  = 2m

m  = 9

Look at the graph here :  https://www.desmos.com/calculator/vsoqo96xtr

The  least possible value  of  8x + 13y  will occur at  the  intersection of  these  lines

So  we have that

2x + 5y  =10

3x + 4y  =  12          multiply the first eauation through by 4  and the second  by -5

8x + 20y  = 40

-15x - 20y = -60      add these

-7x  = -20

x = 20/7 =  the x coordinate of the  intersection of these lines

And  using 2x + 5y = 10    we can find the y  coordinate  as

2(20/7) + 5y  =10

40/7 + 5y = 70/7

5y=   70/7 - 40/7

5y = 30/7       divide both sides by 5

y = 6/7

So.....the minimum value of  8x + 13y  =

8(20/7) + 13(6/7)  =

[160  + 78 ]  / 7  =

238 / 7

34

(a)

Simplify \(\dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k - 1}} &=& \dfrac{n!}{(k-1)!(n-k+1)!} \quad &| \quad (k-1)!k=k!~ \text{or}~(k-1)!= \dfrac{k!}{k} \\\\ &=& k *\dfrac{n!}{k!(n-k+1)!} \quad &| \quad (n-k)!(n-k+1)=(n-k+1)! \\\\ &=& \dfrac{k}{n-k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{k}{n-k+1} \dbinom{n}{k} } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}} \\\\ &=& \dfrac{\dbinom{n}{k}}{\dfrac{k}{n-k+1} \dbinom{n}{k} } \\\\ &=& \dfrac{n-k+1}{k} * \dfrac{\dbinom{n}{k}}{\dbinom{n}{k} } \\\\ &=& \mathbf{\dfrac{n-k+1}{k}} \\ \hline \end{array}\)

8/15

Thank you, CPhill !

Very nice, heureka    !!!!

These form a circle with a radius  of  5

And one complete  graph is traced out   from t = 0  to t  = pi

So.....

Speed  =   Distance Traveled   /Time      ....so....

Speed  =     

2pi (5)

_____  =     10

   pi

We can  also prove this with Calculus

dx/ dt  =   -10 cos (-2t)           dy/ dt   =   -10 sin (2t)

Note that  -10cos(-2t)   =  -10cos(2t)

The speed is given by

sqrt  [   (dx/dt)^2   + (dy/dt)^2  [  =

sqrt  [ (-10cos(2t))^2 +  (-10 sin (2t))^2  ]  =

sqrt  [  (-10)^2   (sin^2(2t)  + cos^2(2t)) ]   =

sqrt  (100 *  1 ] =

sqrt [ 100 ]  =

10

Let \(\omega\) be a complex number such that \(\omega^5 = 1\) and \(\omega \neq 1\).
\(\dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^2}{1 + \omega^4} + \dfrac{\omega^3}{1 + \omega} + \dfrac{\omega^4}{1 + \omega^3}\).

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^2}{1 + \omega^4} + \dfrac{\omega^3}{1 + \omega} + \dfrac{\omega^4}{1 + \omega^3} } \quad | \quad \omega^4 = \dfrac{\omega^5}{\omega}=\dfrac{1}{\omega} \\\\ &=& \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^2}{1 + \dfrac{1}{\omega}} + \dfrac{\omega^3}{1 + \omega} + \dfrac{\omega^4}{1 + \omega^3} \\\\ &=& \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^3}{1 + \omega} + \dfrac{\omega^3}{1 + \omega} + \dfrac{\omega^4}{1 + \omega^3} \\\\ &=& \dfrac{2\omega^3}{1 + \omega} + \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^4}{1 + \omega^3} \quad | \quad \omega^3 = \dfrac{\omega^5}{\omega^2}=\dfrac{1}{\omega^2} \\\\ &=& \dfrac{2\omega^3}{1 + \omega} + \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^4}{1 + \dfrac{1}{\omega^2}} \\\\ &=& \dfrac{2\omega^3}{1 + \omega} + \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^6}{1 + \omega^2} \quad | \quad \omega^6 = \omega^5 \omega=\omega \\\\ &=& \dfrac{2\omega^3}{1 + \omega} + \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega}{1 + \omega^2} \\\\ &=& \dfrac{2\omega^3}{1 + \omega} + \dfrac{2\omega}{1 + \omega^2} \\\\ &=& 2\left( \dfrac{ \omega^3}{1 + \omega} + \dfrac{ \omega}{1 + \omega^2} \right) \\\\ &=& 2\left( \dfrac{ \omega^3(1 + \omega^2) + (1 + \omega)\omega}{(1 + \omega)(1 + \omega^2) } \right) \\\\ &=& 2\left( \dfrac{ \omega^5 + \omega + \omega^2 + \omega^3 }{ 1 + \omega + \omega^2 + \omega^3 } \right) \quad | \quad \omega^5 = 1 \\\\ &=& 2\left( \dfrac{ 1 + \omega + \omega^2 + \omega^3 }{ 1 + \omega + \omega^2 + \omega^3 } \right) \\\\ &=& \mathbf{2} \\ \hline \end{array}\)

Well....we can generate a quadratic for this

Let the  polynomial be

y = ax^2 + bx + c

Since  (0, 4)   is in the graph, then  c  = 4

And using the other two points we  have this system

13  = a(6)^2  + b(6) + 4

 -12 = a(9)^2  + b(9) + 4        simplify

36a + 6b  = 9

81a + 9b  = - 16         

Divide the   first equation through by 6

6a + b = 3/2

b = (3/2) - 6a          sub this into the second equation for  b   and we have that

81a +9(3/2 -6a)  = -16

81a + 27/2 - 54a   = -16

27a  =    -16 - 27/2

27a  =  [ -32 - 27 ] / 2

27a  = [ -59 ] /2    

a =  [ -59/27] / 2

a = -59/54

And

b  =  (3/2)  - 6 (-59/54 )

b = (3/2) + 59/9

b = 145/18

So.....the polynomial is

-(59/54)x^2  + (145/18)x +  4 

Here's a graph  :  https://www.desmos.com/calculator/jrc6i4llx4

The  M  can occupy  any one of six positions

The  rest  of the letters  can  be arranged in   6!/2  =  360 ways

We must divide by 2  because the A's  are not  distinct

So....the  number of "words"  without  M at the end  = 

6 * 360  = 

2160

By the Pythagorean Theorem the height of the large triangle  is given by

sqrt  [ 10^2 - 6^2 ]   = sqrt  [100 - 36 ]  =  sqrt (64)   =8

So  the  area  of the large triangle  is given by

(1/2) (base) (height)  =

(1/2) ( 6+15) (8)  =

(1/2) (21) (8)  =

4 * 21  =

84 units^2

The answer is -2.

By the theory of cyclotomic polynomials, the smallest such n is 24.

Let  angle  BAD  = y  = angle DAC

And since AB = Ad, then angles ABD  and ADB  are equal

So....let angles  ABD, ADB  =  x

Since angle ADB is vertical to angle CDE, then  angle  CDE  = x

And since CD  = CE,   then  angle CDE  = angle DEC  =  x

So  triangles ABD  and CDE  are similar by AA congruency

So angle DCE  = y

And by the exterior angle theorem, angle ABD  = angle DAC + angle ACD

So

x   = y + angle ACD

x - y   = angle ACD

So  angle ACE  =  angle DCE +  angle ACD

So   angle ACE  = y  + (x - y)    =  x

So   angle  DEC  = x = angle AEC

And angle ACE  = x

So   angle AEC  = angle ACE  means that triangle AEC is isosceles

It's interesting that the sum of the internal angles of the shape OXYZ is equal to the angle XYZ.

                           17.5° +65° + 40° = 122.5°     

Point Q is chosen at random inside equilateral triangle XYZ. Find the probability that Q is closer to the center of the triangle than to X, Y or Z. (In other words, let O be the center of the triangle. Find the probability that OQ is shorter than all of QX, QY and QZ)

See  the following image ;

We have an equilateral  triangle with a side of 2  and a height of sqrt (3)

O is the point of intersection of the angle bisectors  of the triangle and BY = BO = DO

So.....BY  is 1/3  of the height of  triangle  XYZ

And triangle AYC  is similar to triangle XYZ......so the area of  triangle AYC  =  (1/3)^2  = (1/9) that of triangle XYZ

And Q will be closer to Y than it is to O  when Q is located inside of triangle AYC

So.....using symmetry, we will have three such smaller triangles, so their total area = 3(1/9)   = 1/3 that of triangle XYZ

So......the probability that  Q will be closer to O  than to any of the  vertices of triangle XYZ   =  2/3

5 

The coefficient of 5 tells us that the speed of the particle is 5.

The number of possible seatings is 4!*4! = 576.

ok thank you! i finally understand these problems now..