These are all relatively prime....the  LCM  = 

3 * 5 * 7   =  

105

139  193  319  391  913  931  Total =  6 such triplets.

sinasinb = 1/2 [cos ( a-b) -cos(a+b)]

               = 1/2 * 12   ( cos 0 - cos pi/3)

                6  - 6 cos pi/3                            <============== Don't see that as an answer ....they MIGHT have intended  C.

P.S.    Those numbers in the chart are 'out of 100 '      so  example:    16.254   out of a hundred = .16254     

j          k

12        4
6         6
4       12

The center will be   x,0

the didttance from the center to the points is equal

use ditance formula for each point and equate them

(x-2)^2 + 3^2   = (x-3)^2 + 2^2

   -4x = -6x      x must = 0

center is at  0, 0

   calculate the distance to either of the points to find the radius

      3^2 + 2^2 = r^2

          13 = r^2                    then   r = sqrt (13)

Score 475 is   three standard deviations below the mean of 550   ( 25 x 3 )

  three standard deviations below 550 mean on exam B would be   550 - 3*100 = 250

First midpoint   1,3      second midpoint    3,-2

slope =   (3- -2)/(1-3) = .......

Are you sure that you want the "sum" or the "number" of  "free divisors of 54,000"?

54,000=(1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 36, 40, 45, 48, 50, 54, 60, 72, 75, 80, 90, 100, 108, 120, 125, 135, 144, 150, 180, 200, 216, 225, 240, 250, 270, 300, 360, 375, 400, 432, 450, 500, 540, 600, 675, 720, 750, 900, 1000, 1080, 1125, 1200, 1350, 1500, 1800, 2000, 2160, 2250, 2700, 3000, 3375, 3600, 4500, 5400, 6000, 6750, 9000, 10800, 13500, 18000, 27000, 54000)>>Total = 80 divisors.

(1, 4, 9, 16, 25, 36, 100, 144, 225, 400, 900, 3600) >Total = 12 - these are perfect squares included in the above 80 divisors.

You simply remove these 11 perfect squares from the 80 divisors and sum up what is left.

1 + ky = 3y     ( are you sure this is written properly?)

 1 + k(3) = 3(3)

    3k = 8

      k = 8/3

I'll let you finish:

P=30

The point   8, 3    is not on the line  x + 5y = 7

     the point    -8, 3     IS on the line

         plug into the second equation

                 16(-8)  + p(3) = Q

                   -128 + 3p = q           slope has to be 5 to be perpindicular  (the slope of the first line is  -1/5)

                       16x + py = q

                                  py = - 16x +q        so    -16/p = 5    p = -16/5

                     -128 - 16/5 (3) = q = -640/5 - 48/5 = -688/5 =Q=........

  p

hihihi, this line introduces the error:  "( its given that its not a square)"  after the second fold.   

on the contrary, it's given that after the second fold the figure is a square, specifically  

the problem states that:  "second rectangle (which is a square)"  after the second fold. 

A prime number can't be even.  Oh wait.  There is a single prime number that is even. 

This may not make any sense but you use simplifying by taking off zeros to make it more solvable.

1. there where too many zeros so i took away some and left 7 zeros so that you could still subtract properly. the new question is below

5 x 10,000,000 = 50,000,000 - 234,567 = 4,765,433 x 8 = 398,123,464

2. than you add on all the other zeros to get the answer below  

398,123,464,000,000,000,000

I hope this is helpfull! :D 

-Grace

A perfect square is of the form x^2 where x is a positive whole number

One less than a perfect square is    x^2 -1

\(x^2-1=(x-1)(x+1)\)

If this is a prime number then the smaller factor must be 1

x-1=1

x=2

x^2-1 = 3

Hence 3 is the only prime number that is followed by a perfect square.

The circle will have circumference 2.pi.3 = 6.pi so the given arc is 1/5 th of the way round so will have an angle at the centre equal to 2.pi/5.

Use the webcalc calculator on the Home page here.

As follows:

Why can't you do it?  Or at least some of it?

Finish solving this.

I only got one answer but maybe there are more.

\(|z-1|=|z+3|=|z-i|\\ |a+bi-1|=|a+bi+3|=|a+bi-i|\\ |(a-1)+bi|=|(a+3)+bi|=|a+(b-1)i|\\ (a-1)^2+b^2=(a+3)^2+b^2=a^2+(b-1)^2\\ (a-1)^2+b^2=(a+3)^2+b^2=a^2+(b-1)^2\\ etc \)

Hint

Put the points clockwise as   T Q R U E A C D

Start with the rectangle:

Where  TREC is the rectangle, and the other points are added on the sides afterwards

How do  Q and A need to be placed to make the description work?  

See if you can upload your picture.

Come back if this does not click with you.  Discuss what you have done.

B for number of big

S for number of small

S = B + a prime number

B+S=54

so

B+B+a prime = 54

2B + a prime = 54

2B is even

So can the prime be even or odd?

You can finish it.