CPhill

Sum of the last three terms is       ar^(n - 1) + ar^(n - 2) + ar^(n - 3)  ⇒

ar^(n - 3)  [ r^2 + r  + 1 ]  = ar^(n - 3) [ 1 + r + r^2]

Sum of the first three terms  is     a + ar + ar^2   =  a [ 1 + r + r^2 ]

And the 3rd term is  5  ⇒  ar^2  = 5   ⇒  a = 5 / r^2 

And we're told that  the sum of the lst 3 terms  =  1024 times the sum of the 1st three

So we have that

1024a  [ 1 + r + r^2 ]  = ar^( n-3)  [ 1 + r + r^2]

1024  =  r^(n - 3)

Note that  1024  = 2 ^10....which implies that

2^10  = r^(n - 3)

So    r  = 2      and  n = 13

And the first term is 5 / 2^2  =  5/4

And the third term  is    ar^2  =   (5/4)*2^2  =  (5 /4) * 4  =   5

So.... the last term is

ar ^ (n -1)  = (5/4) (2)^(13 - 1)  =  (5/4) *2^12  =  5120

Yep......bedtime calls...!!!!!

Well...let's see

-10 (t - 1) - (t - 1)(t - 3) + 5

Let's break this up like this

-10 (t - 1) -  [(t - 1)(t - 3)] + 5

-10t  + [-10*-1] - [ t^2 - 3t - 1t + 3 ]  +   5

-10t  + [-10*-1] - [ t^2 - 4t + 3 ]  +   5

-10t  +  10  - t^2 + 4t - 3 +  5

-10t + 4t    + 10 -  3  + 5   - t^2 

-6t   +   12   -  t^2

-t^2 - 6t  + 12

You teacher made a mistake

-10t + 10 - t^2 + 3t - t - 3 + 5  should be  -10t + 10 - t^2 + 3t + t - 3 + 5

Which leads to my result  of   -10t  +  10  - t^2 + 4t - 3 +  5

The thing about this, juriemagic, is there isn't really one way to solve these things.....if we asked ten different people, we might get ten different approaches....heck....I'm not even sure mine is necessarily the best....LOL!!!!

Just reverse my steps......  

There are 15 Blue Jays and 14 Orioles that wish to rest on the branches of three older trees. Each of the trees will have at least 4 Blue Jays and 2 Orioles. However, no tree may have more Orioles than Blue Jays in its branches. Determine the largest number of birds that can be in one tree.

Put   4 Jays  and 4 Orioles in each tree

So......we have   used 24  and there are 5 left  -  3 Jays and 2 Orioles

Put the 3 Jays in each of the trees   and 2 Orioles in any of  two trees

So....we will  have

5 Jays - 5 Orioles      5 Jays - 5 Orioles         5 Jays - 4 Orioles

Take   1 Jay and 1 Oriole from each of the last two trees and put them in the first tree

So we have this  arrangement

7Jays - 7 Orioles       4 Jays - 4 Orioles           4 Jays - 3 Orioles

Each tree has at least 4 Jays and  2 Orioles....and no tree contains more Orioles than Jays

And I agree with hectictar.....14 is the greatest number in any tree  !!!

[ t - 2 ] / [ t - 1 ]   =  10 / [ 3-t ]   - 1   + 5 / [ t^2  - 4t + 3]

Subtract  10 / [3 - t ]  form both sides   and factor the last denominator

[ t - 2 ] / [ t - 1]   - 10  / [ 3 - t ]   =   -1  +  5 / [ (t - 1) (t - 3) ]

And we can factor a negative out of   - 10 / [ 3 - t]     so it becomes  + 10 / [ t - 3]

[ t - 2 ] / [ t - 1]   + 10  / [ t - 3 ]   =   -1  +  5 / [ (t - 1) (t - 3) ]

Subtract   5 / [ (t - 1) (t - 3) ]  from both sides

[ t - 2 ] / [ t - 1]   + 10  / [ t - 3] ]   - 5 / [ (t - 1) ( t - 3) ]   = -1

Get a common denominator on the left  = [  [t - 1 ] [ t - 3]...so we have

 ( [ ( t - 2) ( t - 3) ] + 10 [ t - 1 ]  -  5) / [ (t - 3) ( t - 1) ]    = - 1

Simplify the left side

(  t^2 - 5t + 6  + 10t - 10  - 5 )  / [ ( t - 1) (t - 3) ]  =  - 1 

( t^2 + 5t  - 9 )  /  [ ( t - 3) ( t - 1) ]  =  -1

Multiply both sides by  [ ( t - 3) ( t - 1) ] 

( t^2 + 5t  - 9 )   =   -1 [ (t - 3) ( t - 1) ]

(t^2 + 5t - 9)  = - [ t^2 - 4t +  3 ]

(t^2 + 5t - 9 ] =  -t^2 + 4t - 3      rearrange as

2t^2 + t  - 6   = 0      factor as

(2t  - 3) ( t + 2)  =  0

Set each factor to 0 and solve

2t - 3  = 0                                                 t   +   2   = 0

Add 3 to both sides                                  Subtract 2 from both sides

2t  = 3                                                        t  = -2

Divide both sides by 2

t  = 3 / 2

So    t  =  -2  , 3/2

Here you are, Manuel :

http://www.dictionary.com/browse/adjective

L =  A  =  180 - 35 - 95  =  180 - 130  =  50°

Mmmmmm...that answer looks a little "fishy,"  AT   .....!!!!