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Interact with answerers on the original thread!

Interact with answerers on your original thread!

Attn:  Rollingblade

I have just looked back at your more recent questions.

You have not responded to your answerers at all.

You used to respond but you seem to have stopped.

Please start again.

I and others put a lot of thought an effort into trying to help you. 

Even when our answers are not correct we still deserve a polite response.

And interaction is a key ingredient to learning.

Also
If you have any answer at all you should state what it is in the original question.

Why should we start out with less info than you have.

There are 20 chairs in a row. Find the number of ways of choosing 5 of these chairs, so that no two chosen chairs are adjacent.

Just for a starting reference.  If I hoose any 5 chairs from 20 then the answer would be 20C5= 15504 ways

so any answer must be less than that. 

Guests answer of 12376 passes that particular reasonable check but I would like him/her to expain much better how the answer was come by.

------------------------------------------------------

Here is my shot at this question.

Y stands for yes chosen

N stands for No not chosesn

*YN*YN*YN*YN*Y*

I interprete this as

Could be some chairs, yes gone, not gone, could be some chairs, etc

In order for 5 chairs to be yes chosen there must be at least  9 chairs altogether.

So it is the position of the other 11 chairs that is in question.    

The possible position of the other 11 chairs are represented by the stars.  There are 6 stars.

So my problem appears to have become.  How many ways can I put split 11 chairs into 6 groups.

11 stars and 5 bars  

11C5= 462 ways

I get 462 ways but it is quite likely wrong.   My error, if there is one, could be trivial or it could be a complete failure in logic.

Rollingblade, do you have an answer of any kind?

Since PQ  is parallel to  RS....then arc PS  = arc QR

So

y^2  + 7y  = 60 - 4y     rearrange as

y^2 - 11y  - 60  =  0      factor

(y - 15) ( y - 4)   =  0

Then y  =15   (reject)   or  y  = 4  (accept)

So  arc PS  = arc QR  =  60 - 4(4)  = 44°

Then arc   PTQ  =  

360  - 2(44)  - 30   =

242°

could have said that in a nicer way.

See answer under your first post below !!!

Well, you have 1/2 being added to smaller and smaller fractions, which means that it diverges to infinity. Or 1/2 x infinity =infinity !!.

The pattern keeps on going, so that answer is wrong

There are a total of 5 (1/2)'s. 5 * 1/2 = 5/2

1/4 + 1/8 + 1/16 + 1/32 = 8/32 + 4/32 + 2/32 + 1/32 = 15/32

5/2 = 80/32. 80/32 + 15/32 = 95/32

I hope this helped, 

Gavin. 

The  side of the equilateral triangle  =   a/3

So....the area  of the equilateral triangle  =  (1/2)(a/3)^2 sqrt (3) / 2 = a^2  [ sqrt (3)/ 36 ]

The side of the  hexagon = b / 6

So...it's area  =   6 (1/2) (b/6)^2 sqrt (3) / 2  = b^2  [ 3 sqrt (3) / 72]  =   b^2 [sqrt (3) / 24]

So

a^2 [ sqrt (3) / 36  ]  =  b^2 [ sqrt (3) / 24]

So

a^2 / b^2   =  [ sqrt (3) /24 ]  / [ sqrt (3) / 36 ]

a^2 / b^2  =  [ 36 / 24 ]

a^2/ b^2  =  3/2

a/ b  =   sqrt (3)  / sqrt (2)   =    sqrt (6) / 2

The length  of  the hypotenuse  =  50

The  inradius  can be caculated as   ( sum of leg lengths -  hypotenuse)  / 2  =  (30 + 40 - 50) / 2  = 10

And the  coordinates  of the  incenter =  (10, 10)

So......the length of the segment is 

sqrt  (10^2 + 10^2) + 10  =

10sqrt (2)  + 10  =

10  [1 + sqrt (2)  ]  units

a=0;b=0;p=0; cycle:d=a*10+b;if(a^2+b^2 - a*b==49, goto loop, goto next); loop:printd,", ",;p=p+1; next:b++;if(b<10, goto cycle, 0);b=0;a++;if(a<10, goto cycle, 0);print"Total = ",p

x                y

7                7

3                8

5                8

7                0

8               3

8               5

Acute isosceles triangle.
Sides: a = 28.302 b = 28.302 c = 30

Area: T = 360
Perimeter: p = 86.604
Semiperimeter: s = 43.302

Angle ∠ A = α = 57.995° = 57°59'41″ = 1.012 rad
Angle ∠ B = β = 57.995° = 57°59'41″ = 1.012 rad
Angle ∠ C = γ = 64.011° = 64°39″ = 1.117 rad

Height: ha = 25.44
Height: hb = 25.44
Height: hc = 24

 From height h we calculate side a - Pythagorean theorem:
a^2 = h^2 + (c/2)^2, a = sqrt{ h^2 + (c/2)^2 } = sqrt{ 24^2 + (30/2)^2 } = 28.302 

EP: He/she are asking:" Formula for the sum of the squares of the first n positive odd integers"

OddSum  = (Sum of Squares of all 2n numbers) - 
          (Sum of squares of first n even numbers)
        = 2n*(2n+1)*(2*2n + 1)/6 - 2n(n+1)(2n+1)/3
        = 2n(2n+1)/6 [4n+1 - 2(n+1)] 
        = n(2n+1)/3 * (2n-1)
        = n(2n+1)(2n-1)/3 - This is the formula for the sum of consecutive ODD squares.

First Problem: We can convert this to a problem where x1 + x2 + x3 + x4 + x5 + x6 = 18, where the xi are positive integers.  By stars and bars, the number of solutions is C(17,6) = 12376.

Second Problem: Let a, b, c, d be the number of cookies that the children get.  Then a + b + c + d = 6.  By stars and bars, the number of solutions is C(9,3) = 84.  But we must subtract the solutions where any of a, b, c, d is at least 4.  The number of such solutions is 12, so the answer is 84 - 12 = 72.

-----------  Misread the question

See below: 

(x^2-16)(x^2-1)

x^4 -16x^2 - x^2 + 16

x^4 -17x^2+16     take derivative and set = 0

4x^3-34x = 0

2x(2x^2-17)

x = 0       2x^2 = 17     x^2 = 8.5   x = +-sqrt8.5    

SUb these values in to the original equation to see which produces the smalles result  

    you will find at x = sqrt 8.5    or   sqrt -8.5    the result will be the minimum (-56.25) 

       at x = 0 the function = 16

Here is a graph:

https://www.desmos.com/calculator/l4gbkfzoa7

Maybe try Yahoo answers they have answers for all subjects. I am in grade 9 and have science next semester so I am not much help.

To clarify the above answer, consider this:

Sum of squares of consecutive numbers =1/6[2n +1) * n * (n +1)..........(1)

Sum of consecutive numbers                     =1/2[n * (n +1)].......................(2)

Divide (1) by (2) and you get:                     = 1 /3 *(2n +1)

1/3[2n +1] =43/3           Cross multiply

2n +1 = 43

Therefore: n = 21

Second one.....let  A  be the apex of the triangle....Let AB  be the tangent drawn to the top circle tangent to this circle at  B....and let  C be the center of the top circle

So  triangle  ABC  is a 30- 60 - 90  right triangle  with angle CBA = 90°

CB  = the radius of the  top circle  = 3

Angle CAB  = (1/2)60°  = 30°.....so  angle ACB  = 60°

So   AB  =  CB*sqrt (3)  =  3sqrt(3) 

So....by symmetry.....the perimeter  of the equilateral triangle   =  3 times twice the radius of each circle  +  6 * 3sqrt(3) 

=     3 * 2 (3)  +  18sqrt(3)  = 

18 + 18sqrt(3)  =

18 [ 1 + sqrt(3) ]   units