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"If three real numbers are in an arithmetic progression while their squares are in a geometric progression, find the sum of all the possible ratios of the geometric progression of the squares."

I get it thanks!! :"D

Thanks I understand it now! :D

I think so: 4! - 3! - 2! =24 - 6 - 2 =16

The question specified that the polynomial had rational coefficients.  The only way that could occur here is if the irrational terms in sqrt(3) and sqrt(2) were eliminated by complementary terms.  In general, an n'th order polynomial will have n roots (counting possible repeated ones).  Therefore with an odd degree polynomial with rational coefficients you couldn't have an odd number of roots that all contained irrational terms; the irrational ones would have to appear in complementary pairs. 

My attempt:

\(4! - 8 \\ = 24-8 \\ = 16\)

Ah I see thanks! Drawing it out helps but do you maybe also know a more convenient way to calculate this when I have a lot of numbers?

If you have four cards with the numbers 1,2,3 and 4
how many numbers can you make that are bigger than 2200?

\(\begin{array}{|rcll|} \hline 1234 & \\ 1243 & \\ 1324& \\ 1342& \\ 1423& \\ 1432& \\ \hline \end{array} \begin{array}{|rcll|} \hline 2134 & \\ 2143 & \\ 2314& \checkmark \\ 2341& \checkmark \\ 2413& \checkmark \\ 2431& \checkmark \\ \hline \end{array} \begin{array}{|rcll|} \hline 3124& \checkmark \\ 3142& \checkmark \\ 3214 & \checkmark \\ 3241 & \checkmark \\ 3412& \checkmark \\ 3421& \checkmark \\ \hline \end{array} \begin{array}{|rcll|} \hline 4123& \checkmark \\ 4132& \checkmark \\ 4213 & \checkmark \\ 4231 & \checkmark \\ 4312& \checkmark \\ 4321& \checkmark \\ \hline \end{array}\)

You can make 16 numbers that are bigger than 2200

You can post questions, just spell out what they are for.

Challenge questions that you know the answer to are good, just don't be deceptive about them.

Spell out that it is a challenge and you already know the answer.

In this instance it might also have been nice if you had spelt out that the answer was simple, requiring no algebra.

I always tell the kids that they need to give us as much info as they have if they want us to help them.

If they say an answer is wrong they should always state why they believe this. That goes for you too.

I answered your question thinking I was helping you. 

I would not have answered if I had understood it was a not a 'real' question.

However, it was nice to see your neat little answer. Using congruence and similar shapes is always interesting.

Hi! Thanks for the answer I really do appreciate it but if you can only use each number once how many would that be?

Sorry about that. You didn't mention it in the question. heureka's answer is correct.

No, I did not need the answer. I'm on this forum for fun, and to keep my mind occupied. 

But, if there's a problem, I wont post the questions anymore.  

If we list the multiples of 3 and 7 under 10 the results would look like this: 3, 6, 7, 9.
The sum of these multiples is 25.
Can you find the sum of the multiples of 3 and 7 under 1000?

\(\begin{array}{|rcll|} \hline 3,\ 6,\ 9,\ \ldots ,\ 999~ (=3*333) \\ 7,\ 14,\ 21,\ \ldots ,\ 994~ (=7*142) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{sum } &=& \dfrac{3+999}{2}*333 + \dfrac{7+994}{2}*142 \\\\ \text{sum } &=& \dfrac{1002}{2}*333 + \dfrac{1001}{2}*142 \\\\ \text{sum } &=& 166833 + 71071 \\\\ \mathbf{\text{sum }} &=& \mathbf{237904} \\ \hline \end{array}\)

A 48-foot rope is cut into three lengths. The second length of trope is twice the frist length. The third length of rope is one fifth the first length. What is the third length?

Hello Guest!

\(x+2x+\frac{x}{5}=48\\ \frac{5x+10x+x}{5}=48\\ 16x =48\cdot 5\\ x=15\)

\(The\ third\ length\ is \ \frac{15}{5}ft=3ft\)

  !

In a geometric progression of real numbers, the sum of the first two terms is 7,

and the sum of the first six term is 91. 

Find the sum of the first four terms.

\(\begin{array}{|rclrcl|} \hline \mathbf{s_4 = a \dfrac{1-r^4}{1-r}} && \mathbf{s_2 = a \dfrac{1-r^2}{1-r}} \\ \hline \\ \dfrac{s_4}{s_2} &=& a \dfrac{1-r^4}{1-r} \above 1pt a \dfrac{1-r^2}{1-r} \\\\ \dfrac{s_4}{s_2} &=& \dfrac{(1-r^4)}{1-r^2} \\\\ \dfrac{s_4}{s_2} &=& \dfrac{(1-r^2)(1+r^2)}{1-r^2} \\\\ \dfrac{s_4}{s_2} &=& 1+r^2 \\\\ s_4 &=& s_2(1+r^2) \quad | \quad s_2 = 7 \\\\ \mathbf{s_4} &=& \mathbf{7(1+r^2)} \qquad (1) \\ \hline \end{array} \begin{array}{|rclrcl|} \hline \mathbf{s_2} &=& \mathbf{a+ ar} \\\\ s_2 &=& a(1+r) \quad | \quad s_2 = 7 \\\\ 7 &=& a(1+r) \\\\ \mathbf{a} &=& \mathbf{\dfrac{7}{1+r} } \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s_6 &=& S_2 + ar^2+ar^3+ar^4+ar^5 \quad | \quad s_2 = 7, s_6 = 91 \\ 91 &=& 7 + ar^2+ar^3+ar^4+ar^5 \\ 84 &=& ar^2+ar^3+ar^4+ar^5 \\ 84 &=& ar^2 (1 +r+r^2+r^3) \quad | \quad \mathbf{a=\dfrac{7}{1+r} } \\\\ 84 &=& \dfrac{7r^2(1 +r+r^2+r^3)}{1+r} \quad | \quad 1 +r+r^2+r^3 = \dfrac{1-r^4}{1-r} \\\\ 84 &=& \dfrac{7r^2\dfrac{1-r^4}{1-r}}{1+r} \\\\ 84 &=& \dfrac{7r^2(1-r^4)}{(1-r)(1+r)} \\\\ 84 &=& \dfrac{7r^2(1-r^4)}{(1-r^2)} \\\\ 84 &=& \dfrac{7r^2(1-r^2)(1+r^2)}{(1-r^2)} \\\\ 84 &=& 7r^2(1+r^2) \quad | \quad : 7 \\\\ 12 &=& r^2(1+r^2) \\\\ 12 &=& r^2 + r^4 \\\\ r^4+r^2 -12 &=& 0 \\\\ r^2 &=& \dfrac{-1\pm \sqrt{1-4*(-12)} }{2} \\ r^2 &=& \dfrac{-1\pm \sqrt{49} }{2} \\ r^2 &=& \dfrac{-1\pm 7 }{2} \\ \\ r^2 &=& \dfrac{-1+7 }{2} \\ r^2 &=& \dfrac{6 }{2} \\ r^2 &=& \mathbf{3} \\\\ r^2 &=& \dfrac{-1- 7 }{2} \\ r^2 &=& \dfrac{-8 }{2} \\ r^2 &=& -4 \qquad \text{r is a complex number! No solution} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{s_4} &=& \mathbf{7(1+r^2)} \quad | \quad r^2 = 3 \\ s_4 &=& 7(1+3) \\ s_4 &=& 7*4 \\ \mathbf{s_4} &=& \mathbf{28} \\ \hline \end{array}\)

The sum of the first four terms is 28

Yes, that works too.

Did you actually want that question answered or was it just put there as a fun challenge question?

If it was just a fun challenge question that you already knew how to do then you should have said so from the very beginning.

The length of a vegetable garden is 6 feet longer than its width. If the area of the garden is 135 square feet, find its dimensions.

Hello Guest!

\((w+6ft)\times w =135\ ft^2\\ w^2+6w-135=0\\ w=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\\ w=-3\pm\sqrt{3^2+135}\\ w=-3\pm 12\)

\(w=9\ ft\\ l=9\ ft+6\ ft=15\ ft\)

  !

Thank you, CPhill !

f(x)    =  x^3 * x^(-2) * x ^(1/2)  =  x ^(3 - 2 + 1/2)  =  x ^( 1 + 1/2)  =  x^(3/2)

 l(x)    = 3x^2  [  2  - x^(-1) ]  + 3x   - (2x)^2

           =  6x^2  -  3x^2*x^(-1)  + 3x   - 4x^2

            =  6x^2  - 3x  + 3x  -  4x^2

             =  2x^2

                               x √x                                         x * x^(1/2) * x^(3)

 h(x)  =            __________________    =             ______________   =

                         (x^(-1))^(-2)  * x^(-3)                              x^2 

                                           x^ (1 + 1/2 + 3)              x ^(4 + 1/2)

                                         ______________  =      ___________  =   x^(4 + 1/2 - 2)  =

                                                      x^2                         x^2

                                                 x^(2 + 1/2)  = 

                                                   x^(5/2)

                  x^5  *  x^ (5/4)                x^5  * x^(5/4)            x^5 * x^(5/4)

k(x)   =     _____________  =           ____________  =   ____________   =   x^5

                     √ [x^(5/2)]                    (x^(5/2)(^(1/2)            x^(5/4)

Elegant response, CPhill!

1. The tangent function is negative in quadrant IV .

tan  = y /x     x is positive, y is negative in Q4....so.....true

2. The cotangent function is positive in quadrant III .

cot  = x/y       x,y are both positive here...so....true

3. The cosine function is positive in quadrant I .

All trig functions are positive in Q1

4.The cosecant function is positive in quadrant III .

csc  = r/y    y is negative here....r is always positive....so....false

5. The sine function is negative in quadrant II .

sine  = y /r    y,r  are positve in Q2....so....false

Pure  concentrate =  100%  = 1

20% concentrate  =.20

Let x  be the amount of pure concentrate

So

1x  + .20 (10) =  .50 (10 + x)    simplify

x  + 2  =  5  + .5x      rearrange as

x - .5x  =  5 -2

.5x = 3    divide both sides by .5

x =  6 L  of pure concentrate

f(f(x))  =  3

From the graph we can see that

f(-4)   = 3      and    f(2)  = 3     and   f(5)  = 3

So....this must mean that

f (x)   = -4                     and     f (x)  = 2                      and      f(x)  = 5

x =  not on graph                      x = -2  and 0                         x  =  3

Check

f ( f(-2))  =  f ( 2)  =  3

f( f(0))    =  f(2)   =  3

f(f(3)) = f(5)   = 3

So......the values of  x are   -2, 0 and 3

What about this...?     56 / 4 = 14   

I believe  that  this  follows something known as "Snell's Law"

I'm  assuming  that the building is 30ft  from the teenager

Let the distance that the teenager is from the mirror  = D

Let the distance that the mirror is from the  mirror  =30 - D

Let the building height  =  H

The angle of incidence  = the angle of reflection....this means that

tan (70) = 5 / D      (1)       and     tan (70) = H / (30 - D)    (2)

Rearranging (1)  we get that   D = 5 /tan (70)

Sub this into (2)  and we have that

tan (70)   =   H / [ 30  - 5/tan(70) ]

[ 30 -5 / tan (70) ] * tan (70)  = H  ≈  77.4 ft   =building height

Using similar triangles

D / 5  =  (30 - D) / 77.4     cross-multiply

77.4D = 5 (30 - D)

77.4 D   =150 - 5D

82.4D  =150

D = 150 / 82.4 ≈  1.82  ft = distance teenager is from the  mitrror