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Area  of  pizza  with 12 in diameter  =  pi (12//2)^2  =  pi (6)^2  =   36pi  in^2

So each person eats  18pi in ^2

Then  4 people  must  eat    4 (18 pi)  =  72 pi  in^2

So....we need to solve this

72pi  =  3 (pi * r^2)     divide both sides by 3 pi

24  = r^2   take the positive root

sqrt (24)  = r

So....the diameter =  2r  =  2qrt (24)  = 2 *2sqrt (6)  =   4sqrt (6)  in 

Well explained! :D

No prob  !!!!

The first  summation implies  that

[a1 + a21] [21/2]  = 693

[ a1 + a21] [10.5]  = 693       divide both sides  by  10.5

a1  + a21  =  66

So   we have 10 pairs of terms  each summing to  66

So....the 11th term must be   693 - 10(66)  =  693 - 660 = 33

And the  second summation  sums  the   odd tems  of  the series

So....

(a1 + a21)  =  66

(a3 + a19) =   66

(a5 +  a17) =  66

(a7 + a15) =   66

(a9 + a13) =   66

  a11         =   33

So....the  second sum  =   5(66) + 33   =   363

Never mind. Sorry!

Thanks!

It said it was wrong.

Let N be the  number of books she started with 

To Fred she sells  2 books  and then 1/5 of what is left  =    2 + (1/5)(N - 2)    books

To Joan....she sells  6  books and what remains

What remains  is  what Verna started with  - what she sells to Fred  - 6 books she sold to Joan

So she sells Joan  6 + (1/5)(N - [2 + (1/5)(N-2)]  - 6)

So  we have this inequality

2 + (1/5)(N - 2)  >  6 + (1/5)(N - (2 + (1/5)(N - 2)) - 6)    mutiply through by 5

10  + (N - 2)  >  30 +  (N - (2 + (1/5)(N - 2)) - 6)

N + 8  >   30 + ( N - (2 + (1/5)N - 2/5) - 6)

N + 8 > 30 + (N  - 2 - (1/5)N  + 2/5  - 6)

N + 8 >  30  + (4/5)N - 38/5      multiply through by 5

5N + 40  > 150 + 4N - 38

5N + 40 > 150 + 4N  - 38

N > 150 - 38 - 40

N > 72

Then...she had  to  have at  least 73 books in her collection

Check....if she started with just 72 books

She sells  to Fred    2 + (1/5)(72 - 2)  =  2 + 14  =  16 books

She has  56 books left

She sels to Joan   6  + (1/5) (56 -6)  =  6 + 10 = 16 books

But....she sells to Fred more books.....so....N must be greater than 72

There are only four matrices that work.

The number of solutions is 13770.

The solutions are 3, -3, 7, -7.

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Shouldn't this post be off-topic?

CPhill already answered this question an hour ago when you posted it before.

Thanks! That's very nice of you.

Thanks!

Oh, okay, thanks!

Is it negative sqrt 196?

(t^2 - 82)^2  =  324     expand

t^4  - 164t^2  + 6724  = 324    subtract 324 from borh sides

t^4 - 164t^2  +  6400   =   0     we can factor this as

( t^2  - 100) ( t^2 - 64)  =  0    factor these terms  as differences of squares

(t-10)(t+10)(t-8)(t+8)  =0

Setting each factor to 0  and solving for t  produces  the solutions

t  =10   t  = -10    t  =  8    t  = -8

No problem!

hugo, you are also missing the negative solution... (-10)^2 = 100 and (10)^2 = 100. So if c - 3 = -10, then c can also equal -7.

Thanks

It said it was wrong...