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\(\text{The Fibonacci Sequence is a sequence of positive integers }\\ \text{that is defined by the recursive rule}\\ a_0=1,~a_1=1,~a_{n+1}=a_n + a_{n-1}\\ a_2 = a_{1+1} = a_1 + a_0 = 2\\ a_3 = a_{2+1}=a_2 + a_1 = 3 \text{ and so on to obtain }\\ 1, 1,2,3,5,8,13,21, 34 \dots\)

In English each number in the sequence is the sum of the two numbers immediately behind it.

Since  M is a median of AB and  N is the median of AC, their intersection occurs (2/3) of the way from vertex B to N

So... TN     is (1/3) of the distance from B to N =  (1/3)(12) = 4

And  TNC  is a right triangle  with legs NC, TN  and hypotenuse CT

So

CT  =  sqrt ( NC^2 + TN^2 )  =  sqrt ( 7^2 + 4^2) = sqrt ( 65)

No problem 

I can see a graphical solution shows x= 0...and I can see it by LOOKING at it....but I am kinda stuck as to how to solve it algebraically...   Hmmmmm....

Second one

Because VU   splits QS and RS proportionally, then VU is  ll QR is  ll TR

And because VT splits QR and QS proportionally, then VT is  ll RS ll RU

So...TVUR is a parallelogram because opposite sides are parallel

Then   VU = TR  = 6

So....QR  = 2*TR = 2*6  =   12

the domain restrictions in this case will arise from the denominator being zero for a couple values of x.

\(\dfrac{x^2+5x+6}{x^2-9} = \dfrac{x^2+5x+6}{(x-3)(x+3)} \\ \text{and it is seen that the denominator will be zero at }x=\pm 3\\ \text{thus }x \neq 3,~x \neq -3\)

NOW I see what you are saying......

   working on yet ANOTHER (in?)correct solution...

       Thanx  for keeping me mathematically honest !  

First one

DE/ DB = AC /AB

DE / DB =  AC / [ 2DB]         multiply through by DB

DE / 1 = AC / 2

DE = AC /2

DE = 12.6 / 2 =   6.3

6 and 9 have are common factors of  18  so re-write

2 + 8/18 x = 15/18      Now multiply through by 18

36 + 8x = 15

Inverting the fraction in the denominator and multiplying, we have

(4x^2 + 2x) /  ( x^2 + x - 2)   *    (3x^2 + 10x + 8) / (8x^2 + x)   =

( 4x^2 + 2x) (3x^2 + 10x + 8)

________________________            factor each term

(x^2 + x - 2) (8x^2 + 4x)

(2x) ( 2x + 1) (3x + 4) (x + 2)

_______________________         "cancel" where possible

(x +2) (x - 1) (4x) (2x + 1) 

(3x + 4)

__________   =    the first answer

2(x - 1)

the equation is 3^2x+2+27^x+1=36, not 3*(2x+2)+27*(x+1)=36. 

you got from 3^2(x+1) + 3^3(x+1) = 3² * 4 to 1^x+1 + 3^x+1 = 4. Check your answer, you divided the right side by 9 and the left side by (3²)^x+1=9^x+1.

The answer is x=(16/5).

3/4   and  2/5 have a common denominator of 20, so re-write

15/20x - 2 = 8/20       Now multiply both sides by 20

15x-40 = 8

No.....I divided by 9     then I simplified  1^(x+1)   to ' 1 '   since 1 to any power is still just 1 

I think this can be further simplified:

(2x+x²) /(24x+8x²)=(2+x)/(24+8x) (when x!=0)

Simplify    (2+x)/8x  /  (3+x)/x     ?

Flip the lower fraction and multiply

(2+x)/8x  *  x/(3+x)   = (2x+x^2) / (24x+8x^2)

EP on the right side you divide by 9 and on the left side you divide by 9^x+1, you can't do that

Good, busy week of school and work. Tiring, but worth it, lol.

Yeah...I see my error....!!!

You are correct....!!!

Sure you can.....let's call it "divide both sides of the equation by 9 '

You can't divide by 3² like that

Here is a better solution:

3^2(x+1) + 3^(3(x+1) = 3^2 * 4    Divide by 3^2

1^(x+1) + 3^(x+1) = 4

1 + 3^(x+1) = 4

3^(x+1) = 3

(x+1) LOG 3 = LOG 3

x+1 = 1

x=0

The diagonals of a rhombus intersect at right angles

So....the tangent ratio of c1 is

OD / OC   =  sqrt ( 5^2 - 3.9^2) / 3.9   =  sqrt (9.79) / 3.9

d1 can be found thusly

tan d1   = 3.9 / sqrt (9.79)        take the arctan

artan [ 3.9 / sqrt (9.79) ] = d1 ≈ 51.26°

I don't think your factoring is correct, the equation

(3²)^x+1 + (3³)^x+1 =(3²)^x+1*(1+3)

Is not always true (although the only solution is x=0)