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Why can't you just take the square root, get +/- 2i and then take the square root again?

Glad you were able to figure it out Omi !

Make the MIDDLE number  '2x+1  '

    this will make ODD numbers for all of x values

  the smaller number next to it would be    2x+1    -2 

     the larger number next to it would be  2x+1  +2      these two numbers sum to 24   +     2x+1

        2x+1   -2     + 2x+1   +2   = 24  + 2x+1

        4x +2= 25   + 2x+1

         2x =23

          x = 11.5

So the Middle one is    2 x +1   = 2 (11.5) +1 = 24     ...

Whoa....this is not possible      the smallest   and the largest are two odd numbers that sum to an EVEN number

                                                          adding 24 to the MIDDLE ODD number will result in an odd number !    Not even...

Here are the solution:

If   f(x) = g(x)    Just set them equal

7      =      x^2+2x -8       subtract 7 from both sides

0 = x^2 + 2x - 15             now either factor  in your head....or use the quadratic formula     a = 1 b = 2  c = -15

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) 

(x+5)(x-3) = 0       so x = -5   or 3    

You just need to find the smallest number that can be divided by 2    or  5    or   6    evenly

   The Least Common Multiple

there are various ways to do this   

start with 5     multiples of 5   are    5   10  15   20  25  30   35    etc     see any number in htere that is also divisible by 2   and  6 ?

I'll show you the second line....you should be able to do the rest

Seond line says x = 1     f(x)  =x^2 -6x+8      this equals   (1)^2 -6(1) +8 = 3

                                      g(x) = x -2              this equals    1 -2               =-1      fill in the chart

Do the same for all of the other x values given       Looks like you should find TWO value of x   where   f(x)  and g(x) are equal.....GO !

How many pounds of dried apples and peaches will be used?

Find the solution to the sytem:

Sum of all the internal angles = 720°

Hence:  35 + (35+a) + (35+2a) + (35+3a) + (35+4a)  (35+5a) = 720

Solve for a, then calculate 35+5a.

A brute force and ignorance program gives the following results:

17   there are 6538 ways

19   there are 12117 ways

23   there are 22967 ways

29   there are 15267 ways

31   there are 9142 ways

37   there are 462 ways

total = 66493 ways

Hence probability = 66493/6⁷   (or  0.2375 approximately)

4.5*a=16.2

is the same as

a=16.2/4.5

a=3.6 

Just ask if you need further elaboration.  :)

Cheers

It's 35% of 21 000 seats which is calculated as 0.35*21 000= 7 350 fewer tickets sold. 

 

You will be making hanging flower baskets. The plants you have picked out are blooming annuals and non-blooming annuals. The blooming annuals cost $3.20 each and non-blooming annuals cost $1.50 each. You bought a total of 24 plants for $49.6. Write a linear system of equations that you can use to find how many of each type of plant you bought.

Let the variable b represent the number of hours spent babysitting.

Then, the number of hours spent lifeguarding = b + 2.

Amount of money made:  9·b + 17·(b + 2)  =  138

Simplifying:                       9b + 17b + 34  =  138

                                                  26b + 34  =  138

                                                           26b  =  104

                                                               b  =  4      <---  hours spent babysitting

                                                         b + 2  =  6      <---  hours spent lifeguarding

If the problem is "r divided by 8 = 5", then  r/8 = 5     --->     multiply both sides by 8     --->     5x8  =  40.

If the problem is "r^8  = 5", then  r⁸ = 5     --->     take the 8^th root of both sides     --->     5^1/8  =  1.222844...

Love the geogebra Melody.

In answer to your question, it's a standard angle property of a circle.

Take a diameter from Q across the circle to T on the other side.

Angle QPT will be right angle,since it's based on the diameter, and then angle PTQ will equal the angle between PQ and the tangent, ( subtract from 90 twice). 

Then, all angles based on the same arc, PQ, are equal etc.

In triangle PQR, PQ=15 and QR=25 The circumcircle of triangle PQR is constructed.

The tangent to the circumcircle at Q is drawn, and the line through P parallel to this tangent intersects QR at S.

Find RS.

\(\text{Let $C$ is the center of the circle } \\ \text{Let $PS=TS=h$ } \\ \text{Let $RS=x$ } \\ \text{Let $QS=25-x$ } \)

\(\begin{array}{|lrcll|} \hline 1. & \left(25-x\right)^2+h^2 &=& 15^2 \qquad \text{or}\qquad \mathbf{ h^2 = 15^2-\left(25-x\right)^2} \\\\ 2. & x(25-x) &=& h * h \qquad \text{Intersecting chords theorem} \\ & x(25-x) &=& h^2 \\ & x(25-x) &=& 15^2-\left(25-x\right)^2 \\ & 25x-x^2 &=& 15^2-( 25^2-50x+x^2) \\ & 25x-x^2 &=& 15^2-25^2+50x-x^2 \\ & 25x &=& 15^2-25^2+50x \\ & 25^2-15^2 &=& 25x \\ & 25x &=& 25^2-15^2 \\ & 25x &=& 400 \\ & \mathbf{x} &=& \mathbf{16} \\ \hline \end{array}\)

The number of ways is C(11,3) + C(10,2) + C(9,1) = 219.

I do not see any short way of calculating this.

Are you meant to be writing a little computer program to do it?

Thanks Guest,

I am just trying to make sense of your answer.

Why is   \(\angle QRP = \alpha\)        ?

Yes you are right guest.

I fixed an error in mine and now it always stays at 16.

I wss see if I can work out how to give a proper link to what i have created.

I now think Guest #4 below is correct.  There is a unique answer.  Because that answer is independent of the shape of the triangle it is legitimate to choose an easy one (a right-angled one) on which to do the calculation:

RS is indeed 16.

I think that the answer is unique, and I think that it is 16.

Using Guest's diagram, the triangles QPR and QSP are similar.

The angle between QP and the tangent, call it alpha, is equal to the angle QPS and also the angle QRP.

Similarly the angle between QR and the tangent is equal to the angle QSP and also the angle QPR.

So, QP/QS = QR/QP, 

QS = QP.QP/QR =225/25 = 9, so SR = 16.

That fits the approximate values that Melody found by drawing.

Alan seems to chosen the particular case where angle PQR is a right angle.

Using the angles I called alpha earlier, from the triangle PQR, tan(alpha) = 15/25 = 3/5.

From the triangle PQS, tan(alpha) = QS/15, so QS = 45/5 = 9, as above.

The base diameter of a right circular cone is \(18~\text{cm}\).
If the lateral area is \(516.4~\text{cm}^2\), find its volume in cubic centimeters

\(\text{Let the radius of the circle $ r = 9~\text{cm} $ } \\ \text{Let the lateral area $ F_\text{lateral} = 516.4~\text{cm}^2$ } \\ \text{Let the circle area $ F_\text{circle} = \pi r^2 =81\pi= 254.469\ldots~\text{cm}^2$ }\)

volume:

\(\begin{array}{|rcll|} \hline \mathbf{V_{\text{circular cone}}} &=& \mathbf{\dfrac{r}{3}*\sqrt{F_\text{lateral}^2 - F_\text{circle}^2}} \\\\ &=& \dfrac{9~\text{cm}}{3}*\sqrt{\left(516.4~\text{cm}^2\right)^2 - \left(254.469\ldots~\text{cm}^2\right)^2} \\\\ &=& 3~\text{cm}*\sqrt{266668.96~\text{cm}^4 - 64,754.4744755~\text{cm}^4} \\ &=& 3~\text{cm}*\sqrt{201,914.485524~\text{cm}^4} \\ &=& 3~\text{cm}*449.348957409~\text{cm}^2 \\ &=& 1348.04687223~\text{cm}^3 \\ \mathbf{ \mathbf{V_{\text{circular cone}}} } &=& \mathbf{1348.0~\text{cm}^3} \\ \hline \end{array} \)