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AB  = AP  + BP

CD = CP + DP

By the  intersecting chord theorem

AP *  BP  = CP * DP

3 * BP  =  8 *  DP

BP / DP  = 8 / 3

This problem is alcumus and homework on another website.

H = hardcover books

Paperback books   = twice the harcover books  = 2H

Original number in the  collection  = H + 2H  =  3H

Books  checked out  = H   + x(3H/9)   where x is the number of members checking out the paperback books

Paperbacks returned  = 2(3H /9)

Paperbacks not returned  = (x - 2)(3H/9)

So  we have this equation

Books in collection  = [ books returned ] + [ books not returned ]

3H  = [2 (3H/9)] + [ H + (x - 2)(3H/9) ]     simplify

3H =  [ H(2/3) ] + [ H + (H)(x - 2)/3 ]     divide through  by H

3  = (2/3) + 1 + (x - 2)/3

3  = (2/3) + 1 + (x - 2)/3      multiply through by 3

9 = 2 + 3 + x - 2

9 = 3 + x

9 - 3   = x

6   = x

You need to make 5 calls.....one to Member 7427  and  (x - 2)  = (6 - 2)  = 4 more to those who checked out the unreturned paperbacks 

Thank you to everyone who replied. You answers are appreciated!

Thanks

That's a pretty neat method, Alan....I haven't seen it before.....!!!

Top  left figure....we have  a hexagon....it it composed of 12 equal right triangles.....

each  has  legs  of  45/2  and 45 / [2√3] units.....each triangle has an area  of  (1/2) (product of the legs)...so we have

Area  =

12 * (1/2) * (45/2) * (45 / [2√3 ] =

6 * 45^2 / [4√3] =

2025√3 / 2 units^2  ≈

1753.7  units^2

The next three figures are  equilateral triangles.....their areas  =  [√3/4] * side^2

I'll calculate the area of the one with a side length of   20√3....you should be able to do the other two

Area  = [√3/4 ] [20√3]^2   =  [√3/ 4] [ 1200]  = 300√3  units^2 ≈ 519.6 units^2

The last two figures are also equilateral triangles  with circumradiuses  of 14 and 28...the area of either  is given by  3* (1/2)(circumradius)^2 * √3/2

The area of the  one with the circumradius of 14  =

3*(1/2) (14)^2 * √3/2  =

(3/4)(196)√3  =

147√3  units^2 ≈

254.6 units^2

You should be able to calculate the area of the one with the circumradius of 28 based on this

 ∑(n, 1, 1007, sqrt(n))=21 319.2574907101 
So, 21,319 is a prime number!

I'm not entirely sure the following is exactly the Tschirnhaus transformation method, but it's as close as I can get!

P(short)  =  1  - P(tall)  =  1  - .15   =  .85  = 85%

Eh....I'm pretty sjure that you could have done the same thing, Coldplay....

CPhill you don't need to compose an answer, the question has already been answered previously on this site.

This question has already been answered on the site.

https://web2.0calc.com/questions/question_139

CPhill said it better than I could've. Hope it helped!

( pq^3)^3 (4p^2q)^2

________________                using  (a^m)^n   =  a^(m * n)....so we have

     (2pq^2)^3

(p^3 * q^9) (4^2 * p^4 * q^2 )

_______________________

  ( 2^3 * p^3 * q^6)

( p^3 * q^9 * 16  * p^4  * q^2)

________________________         using  a^m * a^n  =  a^(m + n)

   (8 * p^3 * q^6)

16 *p^7 * q^11

_____________           using    a^m / a^n   =  a^(m - n)

     8 * p^3 * q^6 

2 * p^4 * q^5          just as Coldplay found   !!!!!

Last  one

x  ≤ -4   this means  ( -inf, 4 ]

or          

x  ≥  5    this means  [ 5, inf )

So  we  have

(-inf, 4 ] or [5, inf )

The answer is I belive 2p^4 * q^5.

I am working on a clear step by step solution,

Second one

y  + 3x ≥ 2    ⇒  y  ≥ -3x + 2

y + 4 ≥ (1/2)x   ⇒   y ≥ (1/2)x - 4

Look at the graph  on the bottom left...note that   (2,0)  is a point in the shaded region

So...replacing these into the inequalities, we have that

0 ≥  -3(2) + 2              and          0 ≥ (1/2)(2) - 4

0 ≥ -6 + 2                                    0 ≥ 1 - 4

0 ≥ -4   true                                 0 ≥  -3    true

So.....this graph is the correct one

Thanks everybody!!!

For the first one...let's pick a point in the shaded region...I'll choose  (-1, 6)

This point should satisfy all of the inequalities  in any answer

Look at the third  answer....we have that

6 - 6  ≤ -11 (-1)         6  + 3(-1) ≥  -4           -2(-1)  + 3(6)   > 15                       

0  ≤ 11   true             6  - 3  ≥ -4                     2    +  18   >   15

                                     3  ≥  -4   true                20  >  15     true

So.....this answer is the correct one

Oh, I didn't notice the reply button, oops, but thanks.

Nice, I knew there were far more than just 1 possible answer. Thanks Melody (and anyone else who answers). Also, I don't know how to set it as the best answer, though I'm at school now, not home, so I'm probably not able to do that.

x = (1/4)y + a               y  = (1/4)x + b

4x = y + 4a                  4y = x + 4b

y = (4x - 4a)                x  = (4y - 4b)

y = 4 ( x - a)                x = 4 ( y - b)

2 = 4( 1 - a)                 1  = 4 (2 - b)

(2/4)  = 1 - a               (1/4)  = 2 - b

(1/2)  = 1  - a                b = 2 - 1/4

a  = 1  -1/2                   b = 8/4 - 1/4

a  = 1/2                         b = 7/4

So...a +  b  =  1/2 + 7/4  =  2/4 + 7/4   =  9/4   

If you are having trouble understanding what I have done I shall try to compare it to base 10

Remember:

In base 10

\(23606 = 2*10^4+3*10^3+6*10^2+0*10+6\)

or

\(110100\\ = 1*10^5+1*10^4+0+1*10^2+0+0 \\ = 10^5+10^4+0+10^2+0+0\)

Yes the second answer is definitely the correct one. 

I am glad you gave it a go guest one, that is the best way to learn.  I hope you see this continuation.

Good work guest 2   

I shall also try to explain.

When the base-b number 11011_b is multiplied by b-1, then 1001_b is added, what is the result (written in base b)?

\(11011_b=b^4+b^3+0+b+1\\ (b-1)*(b^4+b^3+0+b+1)\\ =b^5+b^4+0+b^2+b\\ \quad\;\;-(b^4+b^3+0+b+1)\\ =b^5+0-b^3+b^2+0-1\\ Now\\ 1001_b=b^3+0+0+1\\~\\ \;\;\;(b^5+0-b^3+b^2+0-1)\\ \;\;\;+\qquad\;\;(b^3+0+0+1)\\ =b^5+0+0+b^2+0+0\\ =100100_b \)