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x^a=y^b=k and x^c=y^d=t, then which of the following is true.

a) ac=bd

b) ad=bc

c) a/d=c/b

d) a^c=b^d

e) a+c=b+d

$x^a=y^b \\ x^c=y^d$

$\begin{array}{|rcll|} \hline x^a &=& y^b \quad & | \quad \text{exponentiate both sides with } \frac{c}{a} \\ \displaystyle x^{a\cdot\frac{c}{a}} &=& \displaystyle y^{b\cdot\frac{c}{a}} \\ x^{c} &=& \displaystyle y^{b\cdot\frac{c}{a}}\quad & | \quad x^{c} = y^d \\ y^d &=& \displaystyle y^{b\cdot\frac{c}{a}} \\ \Rightarrow d &=&\displaystyle b\cdot\frac{c}{a} \\ \mathbf{ad} & \mathbf{=} & \mathbf{bc} \\ \hline \end{array}$

The values of the four variables a, b, c, and d are 9, 11, 13, and 15, though not necessarily in that order.

What is the number of possible values of the expression ab+bc+cd+da?

Let a = 9
Let b = 11
Let c = 13
Let d = 15

$\small{ \begin{array}{|r|c|r|r|r|r|rcl|} \hline & \text{All permutations} & & & & & \\ & \text{of a,b,c,d} & A & B & C & D & AB+BC+CD+DA &=& (B+D)(A+C) \\ \hline 1. & abcd & 9 & 11 & 13 & 15 & (11+15)(9+13) = 26*22 &=& 572 \\ 2. & abdc & 9 & 11 & 15 & 13 & (11+13)(9+15) = 24*24 &=& \qquad 576 \\ 3. & acbd & 9 & 13 & 11 & 15 & (13+15)(9+11) = 28*20 &=& \qquad \qquad 560 \\ 4. & acdb & 9 & 13 & 15 & 11 & (13+11)(9+15) = 24*24 &=& \qquad 576 \\ 5. & adcb & 9 & 15 & 13 & 11 & (15+11)(9+13) = 26*22 &=& 572 \\ 6. & adbc & 9 & 15 & 11 & 13 & (15+13)(9+11) = 28*20 &=& \qquad \qquad 560 \\ 7. & bacd & 11 & 9 & 13 & 15 & (9+15)(11+13) = 24*24 &=& \qquad 576 \\ 8. & badc & 11 & 9 & 15 & 13 & (9+13)(11+15) = 22*26 &=& 572 \\ 9. & bcad & 11 & 13 & 9 & 15 & (13+15)(11+9) = 28*20 &=& \qquad \qquad 560 \\ 10. & bcda & 11 & 13 & 15 & 9 & (13+9)(11+15) = 22*26 &=& 572 \\ 11. & bdca & 11 & 15 & 13 & 9 & (15+9)(11+13) = 24*24 &=& \qquad 576 \\ 12. & bdac & 11 & 15 & 9 & 13 & (15+13)(11+9) = 28*20 &=& \qquad \qquad 560 \\ 13. & cbad & 13 & 11 & 9 & 15 & (11+15)(13+9) = 26*22 &=& 572 \\ 14. & cbda & 13 & 11 & 15 & 9 & (11+9)(13+15) = 20*28 &=& \qquad \qquad 560 \\ 15. & cabd & 13 & 9 & 11 & 15 & (9+15)(13+11) = 24*24 &=& \qquad 576 \\ 16. & cadb & 13 & 9 & 15 & 11 & (9+11)(13+15) = 20*28 &=& \qquad \qquad 560 \\ 17. & cdab & 13 & 15 & 9 & 11 & (15+11)(13+9) = 26*22 &=& 572 \\ 18. & cdba & 13 & 15 & 11 & 9 & (15+9)(13+11) = 24*24 &=& \qquad 576 \\ 19. & dbca & 15 & 11 & 13 & 9 & (11+9)(15+13) = 20*28 &=& \qquad \qquad 560 \\ 20. & dbac & 15 & 11 & 9 & 13 & (11+13)(15+9) = 24*24 &=& \qquad 576 \\ 21. & dcba & 15 & 13 & 11 & 9 & (13+9)(15+11) = 22*26 &=& 572 \\ 22. & dcab & 15 & 13 & 9 & 11 & (13+11)(15+9) = 24*24 &=& \qquad 576 \\ 23. & dacb & 15 & 9 & 13 & 11 & (9+11)(15+13) = 20*28 &=& \qquad \qquad 560 \\ 24. & dabc & 15 & 9 & 11 & 13 & (9+13)(15+11) = 22*26 &=& 572 \\ \hline \end{array} }$

The number of possible values of the expression ab+bc+cd+da=(b+d)(a+c) is 3

The values are 560, 572, and 576

One question.....you must be able to use numbers that have 3 leading 0s as possible four digit numbers????

Because....had I picked 9999 originally and then 9998....you would have had to use 0001 to get the first sum of 9999

And if I had picked 9997 as my last number.....you would have had to use 0002   to get the second 9999 sum......

Is this correct ???

Yes, You were ​very close, spot on infact!!!

I learned this in 3rd grade and it took me a week to figure it out!

HAHAHA!!!!!...I know it will come to 29997  !!!

I think I have the trick....although I'm not sure

You ask the person to pick a 4 digit number

Then...you add  9999 * 2   =  19998  to that number

This will be the "future number"

When they pick the next number.....you make sure that the number you pick and  their second number sum to 9999

When they pick their final number......you do the same thing......you make sure that your final number and their final number sum to 9999

So....the future number is just   their first number +  9999 * 2

Am I close ???

(I have to go very soon but I will be back tomorrow)

YES!!

You got it!!

Now add up all of the 4 digit numbers

Hahaha!! and I choose 

8888

I know the next number you're going to pick.....8888  !!!

OK

1111

Now you once more

Okay now I choose one, 

6666

OK

3333

Haha that just happened to happen!! I didn't even notice!

Yes, that is true, if I was not able to choose a number it would not work.

Now you pick another number randomly.

You will probably figure it out but here it is!

You future number will be 29997.

​​​

Ah....thanks, hectictar.....I forgot about the word "tangent"    !!!!!

[-81 / 625]   /   [243 / 3125]    =

[-81 / 625 ]  *  [ 3125 / 243 ]          swap denominators

[-81 / 243 ]  *  [ 3125  / 625 ] 

[ -81 / ( 81 * 3 ) ]  *  [  (625 * 5)  / 625 ]

[  -1 / 3 ] *  [ 5 / 1 ]  =

-5 / 3

[ -81 / 625 ] / [ 243 / 3125 ]

                                                     Invert the second fraction and multiply.

=   [ -81 / 625 ] * [ 3125 / 243 ]

=   [ -81 * 3125 ] / [ 625 * 243 ]

=   -253125 / 151875

=   -5/3

≈   -1.667

y = 2x^2  + kx +  6

y  = 4 - x

Set these equal

4 - x  = 2x^2 + kx + 6          rearrange as

2x^2 + (k + 1)x  +  2  =   0

If there is one solution, the discriminant must   = 0

So we have that

(k + 1)^2  - 4(2) (2)  = 0

(k + 1) ^2   - 16  =  0

( k + 1)^2   = 16        take the negative root

k + 1   =   -4

k  =   - 5

thx that was correct.

thx that was correct.

3. Let f(x) = px + q, where p and q are real numbers. Find p+q if $f(f(f(x))) = 8x + 21.

f(f(x)) =  p (px + q)  + q    =    p^2x + pq + q

So

f ( f ( f(x)))  =  f  [   p^2x + pq + q ]  =

p [ p^2x + pq + q ] + q   =   8x + 21

[p^3x]  + [ p^2q + pq + q ] =  8x + 21

This must mean that

(p^3)x  = 8x   ⇒    p^3  = 8   ⇒  p  = 2

And it must also mean that

p^2q+ pq  + q  =  21           and using p = 2, we have that

2^2 q  + 2q +  q   = 21

4q  +  3q  = 21

7q  = 21

q  = 3

So....   p  +  q   =   2  +  3   =   5