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Never mind, this is a really simple question.

The denominator simplifies to

2y -10

Since  this cannot = 0     then  y = 5   is NOT in the domain

So

domain  =   (-inf , 5) U ( 5 ,inf )

We need y not equal to 2, and y not equal to 8, so the domain is (-inf,2) U (2,8) U (8,inf).

X

                 28

Z                            Y

cos Y =   ZY / XY

7/29  = ZY / 28

(28 * 7) / 29 = ZY  =  196/ 29

XZ =  sqrt (28^2  -  (196/29)^2)   =  (168)sqrt (22) / 29

Tan X =    ZY / XZ =      (196/29) / [ 168 sqrt (22) / 29] =    196 /  [168 sqrt (22)]  =

7/ [6sqrt (22) ] =

7 sqrt 22  / ( 6 * 22)  =

7sqrt (22)  / 138

@CPhill

Im not sure if thats how it works

The perimeter of the trapezoid is 56.

No denominator can =  0      .... so.....

x - 64 =   0

x =64  is not in the domain

x^2  - 64  =0

x^2 =64

x = -8  or x =8     are not in the  domain

x^3 - 64  = 0

x^3 =64

x =4 is not in the  domain

umm.....

(x + 4) / (x + 5)  =  (x - 5 + 7)  / (2x)

(x + 4) / ( x + 5)  = ( x +2) / (2x)      cross-multiply

(2x) ( x + 4) =  (x + 5) ( x + 2)

2x^2 + 8x =   x^2  + 7x + 10        rearrange as

x^2 + x   =  10        complete the square on x

x^2 + x + 1/4 =  10 + 1/4

(x + 1/2)^2 =  41/4       take both roots

x + 1/2  =  ±  √41 / 2

x =  -1/2 ±  √41 / 2

I'm getting m = 7/2.

(1 - 2) + (3 - 4)  + .... + (2005- 2006) + (2007 -2008) + ( 2009 -2010)

We have    (2010 / 2)  =     1005  pairs each summing to -1

So

1005 (-1) =  - 1005

It might help if you put your user name for the mods..

Well....just ask.....somebody might know the answer

Draw QC and PC

Since angle QRC  =35

Then angle CRP  also = 35 

Then angle QRP  = 70

And angle   QPR =   180  - 68 - 70 =  180  - 138  =   42°

And you are EXACTLY correct !!!!

 In triangle ABC let the angle bisectors be BY and CZ. Given AB=8, AY=6, and CY=3, find BZ.

In the other problem we found that BC = 8

So, again, by the angle bisector theorem

AC / BC = AZ / BZ

9 / 8  =  AZ / BZ

So....there  are 17 parts to AB

And BZ  =  8 of them

So

BZ =  (8/17)(AB)  =  (8/17)(8) =   64 /17

3) 

 In triangle ABC let the angle bisectors be BY and CZ. Given AB=8, AY=6, and CY=3, find BZ.

In the other problem we found that BC = 8

So, again, by the angle bisector theorem

AC / BC = AZ / BZ

9 / 8  =  AZ / BZ

So....there  are 17 parts to AB

And BZ  =  8 of them

So

BZ =  (8/17)(AB)  =  (8/17)(8) =   64 /17

 In triangle ABC let the angle bisectors be BY and CZ. Given AB=8, AY=6, and CY=3, find BZ.

In the other problem we found that BC = 4

So, again, by the angle bisector theorem

AC / BC = AZ / BZ

9 / 4  =  AZ / BZ

So....there  are 13 parts to AB

And BZ  =  4 of them

So

BZ =  (4/13)(AB)  =  (4/13)(8) =   32/ 13

CORRECTED  !!!

 In triangle ABC let the angle bisectors be BY and CZ. Given AB=8, AY=6, and CY=3, find BC.

Angle  Bisector Theorem

BC / CY = BA / AY

BC / 3 = 8 / 6

BC  =  3 * 8 /  6  =    24 / 6   =   4

Since angle AIB  =  106

Then    angle (A  / 2) +   angle( B /2)  + 106  =180

A/2 + B/2 =  180-106

A/2 + B/2  =  74

A + B =  148

So

C =  180   -  148  =  32°

sorry, thats not correct.

thanks guest, my answer is very far off lol

By the angle bisector theorem, BC = 13/3.

The way I think about it, the answer is 42. I may be wrong though, since my logic isn't very solid.

The common ratio is  r  / 2

And       l common ratio  l   <  1

So

-1  < r/2 <  1

-2 < r < 2