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Find the greatest common factor of all nubmers of the form 2^n*3^(2n) - 1, where n is an integer greater than 1.

\(2^n*3^{2n} - 1=2^n*9^n-1=18^n-1\) 

I think the greatest common factor is probably 1.  

I don't know how to show or prove it though. 

I tried that but got it wrong

Because  angle  ADE  = angle B  and   angle EAD  = angle BAC....then

Triangle ADE  is similar to triangle ABC

So

DE / AD  = BC / AB

DE / 6.8  =  5.5 / (8.6 + 2.4)

DE / 6.8  =  5.5 / 11 

DE / 6.8  = 1 / 2       multiply both sides by 6.8

DE =  6.8  / 2   =   3.4

Thanks!

f^-1 (-3)     means that  f ( x)  =  - 3

This happens when x  =  3

So  f^-1 ( -3)   = 3

f^-1 (0)  means that   f(x)  =  0

This happens when x = 2

So  f^-1(0)  =  2

f^-1(3)   means  that  f(x)  = 3

This happens when  x = -1

So   f^-1(3)  = -1

So

f^-1 (-3)   + f^-1(0) +  f^-1(3)  =

     3       +     2       +  (-1)  =

4

We have that

2x + 7   =  -5      subtract  7 from both sides     or       -x^2  - x  +  1    = -5

2x  = -12            divide both sides by  2                        rearrange as

x  = - 6                                                                            x^2 + x - 6    =  0

                                                                                       factor

                                                                                       (x + 3) ( x  - 2)  =  0

                                                                                       Set each factor to 0 and solve for  x

                                                                                        x  = -3          x   = 2

                                                                                       Only  x = 2  is defined for the  function

The sum of  the x values  =

-6 + 2   

-4

This is just tedious

f(1) = -1

f(-1) = 7

f(7) = -49

f(-49) = -41

f(-41) = -34

23:59:59    12/31/19    ....and then it's

2020 !   Happy New Year !!

     I do not see any zeros at the end of 2019!           

(but ...there is ONE at the end of 2020   !  )

jk, it was -4

3+4+5 =12    180/12=15  so the angles are

3(15)    4(15)   5(15)

45         60        75

SMALLEST ANGLE   45    SIN = SQRT2 / 2        NEXT LARGER ANGLE  SIN60 = SQRT3/2

RATIO =  SQRT2 /2   :  SQRT 3/2

           = SQRT 2  : SQRT 3  

2019 / 5  +   2019 / 25  +  2019 / 125  +  2019 / 625 = 502 zeros.

The cross product of A + B = (1,2,1) and B + C = (2,1,3) is (5,-1,-3).  We can then scale this to (25,-5,-15), to get the sum of the entries to be 5.

332 is not correct

You can construct right triangles by dropping a perpendicular from D to the lines.  This gives you an answer of 332.

Factor out the expression.

\(2^{28}(2^{6}-1)\)

\(2^{28}(64-1)\)

\(2^{28}*63\)

Since all the factors of \(2^{28}\) are 2's, for now, the smallest prime factor is 2.

Next factor out 63 into \(3^{2}*7\). 

The prime factors in this expression are 2,3, and 7 now.

7, being the largest, is the greatest prime factor of this expression.

The first condition should be followed for numbers that are lower than or equal to 1.

The second condition should be used for numbers that are higher than 1.

Since f^-1(-3) follows the first condition, do 2-(-3)=5.

Similarly, since f^-1(0) also follows the first condition, do 2-0=2.

Since f^-1(3) follows the second condition, as 3 is bigger than 1, do 2*(3)-3^2 = 6-9 = -3

5 + 2 - 3 = 4

The answer is 4. 

We have that

9x +  16  =  -2        or              2x   - 14   =  -2

9x  =  -18                                 2x  =  12

x = -2                                         x  = 6

So....the sum of the possible values of  x=    6 - 2   = 4

Geometric mean of  7 and 10  =  sqrt  (  7 * 10)  = sqrt (70)  = median length

See the  image below

Let AD  = BD   = x

By the Law of Cosines   we have these equations

[ Note angle BDC  is supplemental to angle  ADC...so  cos BDC  = -cos (CDA)   ]

7^2  =  70 + x^2  - 2(√70) (x) cos (ADC)

10^2  = 70 + x^2 - 2(√70)(x) cos (BDC)             so.....         

7^2  =  70 + x^2  - 2(√70) (x) cos (ADC)

10^2  = 70 + x^2 - 2(√70)(x) (-cos (ADC))         so......

7^2  =  70 + x^2  - 2(√70) (x) cos (ADC)

10^2  = 70 + x^2  + 2(√70)(x) cos (ADC)                add these    

7^2 + 10^2 =  140 +  2x^2      simplify

149 =  140 +  2x^2

9 = 2x^2     divide both sides  by 2

x^2=  9/2

x =  3/ √2 =   3√2 / 2

So....the length of the remaining side  =  3√2  units

Is it possible to inscribe the ellipse in that shaded rhombus, and what would be the area of it?

We can find the radius of the  circumcircle with the Law of Cosines

AB^2  = 2R^2  - 2R^2  cos  (2 * angle BAC)

6^2 =  2R^2 - 2R^2 cos (120°)

36  =  2R^2  + R^2

36 = 3R^2

R^2  = 12

So....the area of the  circumcircle =   pi * R^2   =  pi * 12  =   12 pi   units^2 

Second one

By the intersecting chord theorem

PV * VR  = QV * SV

3 * VR   =  4 * 9

3 * VR   =36

VR  = 12

So....by the Pythagorean Theorem

sqrt [ VS^2  + VR^2 ] = SR =  sqrt { 9^2 + 12^2)   = sqrt (225) =   15

First one...... by....the secant- tangent theorem  .... we have

XY^2  = XA * XB

9^2  = 15 * XB

81  = 15 * XB

81/15 =  XB  =  27/5   =  5.4

So  AB  = AX  - XB   =   15  - 5.4   =   9.6

ab^4  =  384    

a^2b^5 =  4608

Divide the second  equation by the  first

ab  = 12    ⇒    b = 12/a

Sub this into the  first equation

a [12/a]^4  = 384

12^4 / 384  = a^3

a = ∛54

a = 3∛2

b  =  12 /[3∛2]=    4 /∛2  =  4 * ∛4 / 2 =  2∛4

cos 30 = sqrt3/2   = adj/hyp = diameter/hyp

                sqrt 3  /2   =   dia/(1/pi)  =   dia (pi)

               (sqrt 3 /2)/pi = dia

circumference = pi x dia   =    sqrt3 /2  = ~ .866