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The hyperoperations are basically a sequence that extends the 3 basic functions everyone is familiar with.

Addition is hyperoperation 1

Multiplication is hyperoperation 2

Exponentiation is hyperoperation 3

Tetration is hyperoperation 4

and it continues on forever (although hyperoperations greater than or equal to 4 have problems with their domains).

Let f(x) be the 'n'th hyperoperation. You can get from hyperoperation 'n' to hyperoperation 'n+1' by setting the 'x' of f(x) to a certain value and then iterating f(x) 'x' times. That takes a lot of work, but for going from hyperoperation 'n' to hyperoperation 'n-1', you just need to use the formula \(f(1+{f}^{-1}(x))\), where \(f^{-1}(x)\) is the functional inverse of \(f(x)\). Using this, you can find out how to go from hyperoperation 'n' to hyperoperation 'n-2' or other values. Basically, you can go from hyperoperation 'n' to hyperoperation 'n-v', where 'v' is an integer.

Hyperoperation 'n-2' from hyperoperation 'n': \(f(1+f^{-1}(1+f(-1+f^{-1}(b))))\)

I then remembered something I did about a year ago. I set up this equation \(f(a)=f^2(b)\) and solved for \(f(b)\), and I got \(f^{0.5}(a)=f(b)\). Then I did it for the 3rd iteration, then the 4th, and so on. In other words, I learned how to get \(f^{\frac1n}(x)\) from \(f^n(x)\). So I figured that I could apply this to hyperoperations.

Basically, if you set up this equation \(g(1+g^{-1}(a))=f(1+f^{-1}(1+f(-1+f^{-1}(b))))\) ('n-1' and 'n-2' equations equal to each other) and manage to solve the right side for \(f(1+{f}^{-1}(x))\) (the 'n-1' equation), the left side will, consequently, turn from an 'n-1' equation into an 'n-0.5' equation. This would give us the ability to define fractional hyperoperations (and not just for 'n-0.5'), and if a useful, usable pattern is found, can lead to the generalization of all hyperoperations.

Consider...

\(\bar{r} = 2-i\\ r = 2+i\\ \bar{r}z+r\bar{z}=20 \quad | \quad \bar{r}z=r\bar{z}\\ 2\bar{r}z = 20 \\ \bar{r}z = 10 \\ (2-i)z = 10\\ 2z-\underbrace{iz}_{=0} = 20 \\ 2z=10\\ \mathbf{z=5}\)

Multiply each term in the parenthesis by 3 

3x+3*5

Then multiply the numbers

3x+15

groundbreaking how?

I actually don't know the answer (or even how to go about solving it), so I asked the question to see if anyone else could figure it out. However, if it is solved, then the results could be groundbreaking.

did you get the right answer??

First, I need to find f –1(x), g –1(x), and ( f o g)–1(x):

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Inverting  f (x):

Inverting g(x):

Finding the composed function:   Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

Inverting the composed function:

Now I'll compose the inverses of  f(x) and g(x) to find the formula for (g–1 o f –1)(x):

Note that the inverse of the composition (( f o g)–1(x)) gives the same result as does the composition of the inverses ((g–1 o f –1)(x)). So I would conclude that

f (x) = 2x – 1 
    y = 2x – 1 
    y + 1 = 2x 
    (y + 1)/2 = x 
    (x + 1)/2 = y 
    (x + 1)/2 =  f –1(x)

g(x) = (1/2)x + 4 
    y = (1/2)x + 4 
    y – 4 = (1/2)x 
    2(y – 4) = x 
    2y – 8 = x 
    2x – 8 = y 
    2x – 8 = g –1(x) 

( f o g)(x) = f (g(x)) = f ((1/2)x + 4) 
    = 2((1/2)x + 4) – 1 
    = x + 8 – 1 
    = x + 7

( f o g)(x) = x + 7 
    y = x + 7 
    y – 7 = x 
    x – 7 = y 
    x – 7 = ( f o g)–1(x)

(g–1 o f –1)(x) = g–1( f –1(x)) 
    = g–1( (x + 1)/2 ) 
    = 2( (x + 1)/2 ) – 8 
    = (x + 1) – 8 
    = x – 7 = (g–1 o f –1)(x)

( f o g)–1(x) = (g–1 o f –1)(x)

While it is beyond the scope of this lesson to prove the above equality, I can tell you that this equality is indeed always true, assuming that the inverses and compositions exist — that is, assuming there aren't any problems with the domains and ranges and such.

what is your question?

i don't see it

4/sin(32)=4/0.529919246...=7.548319659...=7.55

Here is some playing with your problem!

5x^2 + 6x - 18 =0, solve for x
x =~ 1.39
x =~ -2.59
x =a + b*sqrt(c) / d
x =4 + 15*sqrt(81) / 100
x= 4 +15*9 / 100
x= (4 + 135) / 100
x = 139 / 100
x = 1.39
(acd) / b
(4*81*100) / 15 =2,160

A quadrilateral has two pairs of congruent sides and a longer diagonal of length 6, as shown. For what value of x will the area of the shaded area be 40% of the area of the unshaded region?  

Lines with slopes -1 and -2 are drawn through the first quadrant point (a,b) forming one triangle

with a side on the x-axis and the other with one side on the y-axis.

What is the total area of the two shaded triangles? (Write you answers in terms of a and b)

\(\mathbf{b_1=\ ?}\)

\(\begin{array}{|rcll|} \hline y&=&-x+b_1 \quad | \quad P(a,b) \text{ on line} \\ b&=&-a+b_1 \\ \mathbf{b_1} &=& \mathbf{a+b} \\\\ \mathbf{y} &=& \mathbf{-x+(a+b)} \\ \hline \end{array}\)

\(\mathbf{b_2=\ ?}\)

\(\begin{array}{|rcll|} \hline y&=&-2x+b_2 \quad | \quad P(a,b) \text{ on line} \\ b&=&-2a+b_2 \\ \mathbf{b_2} &=& \mathbf{2a+b} \\\\ \mathbf{y} &=& \mathbf{-2x+(2a+b)} \\ \hline \end{array}\)

\(\mathbf{y=0}\\ \mathbf{c_x=\ ?} \)

\(\begin{array}{|rcll|} \hline 0 &=& -x_1+(a+b) \\ \mathbf{x_1} &=& \mathbf{a+b} \\\\ 0 &=& -2x+(2a+b) \\ 2x &=& 2a+b \\ \mathbf{x_2} &=& \mathbf{a+\dfrac{b}{2}} \\\\ c_x &=& x_1-x_2 \\ &=& a+b - \left(a+\dfrac{b}{2} \right) \\ \mathbf{c_x} &=& \mathbf{\dfrac{b}{2}} \\ \hline \end{array}\)

\(\mathbf{x=0}\\ \mathbf{c_y=\ ?}\)

\(\begin{array}{|rcll|} \hline y_1 &=& -0 +(a+b) \\ \mathbf{y_1} &=& \mathbf{a+b} \\\\ y_2 &=& -2\cdot 0+(2a+b) \\ &=& 2a+b \\ \mathbf{y_2} &=& \mathbf{2a+b} \\\\ c_y &=& y_2-y_1 \\ &=& 2a+b-(a+b) \\ \mathbf{c_y} &=& \mathbf{a} \\ \hline \end{array}\)

\(\mathbf{A_x=\ ?} \)

\(\begin{array}{|rcll|} \hline A_x &=& \dfrac{c_xh_x}{2} \quad | \quad h_x = b,\qquad c_x = \dfrac{b}{2} \\\\ &=& \dfrac{\dfrac{b}{2}\cdot b}{2} \\\\ \mathbf{A_x} &=& \mathbf{\dfrac{b^2}{4}} \\ \hline \end{array}\)

\(\mathbf{A_y=\ ?}\)

\(\begin{array}{|rcll|} \hline A_y &=& \dfrac{c_yh_y}{2} \quad | \quad h_y = a,\qquad c_y = a \\\\ &=& \dfrac{a\cdot a}{2} \\\\ \mathbf{A_y} &=& \mathbf{\dfrac{a^2}{2}} \\ \hline \end{array}\)

\(\mathbf{A=\ ?}\)

\(\begin{array}{|rcll|} \hline A&=& A_x + A_y \\ &=& \dfrac{b^2}{4} + \dfrac{a^2}{2} \\ \mathbf{A} &=& \mathbf{\dfrac{1}{2}\cdot \left( a^2 + \dfrac{b^2}{2} \right) } \\ \hline \end{array} \)

Prove this identity.

\(\large \dfrac{1+\sin(\theta)}{1-\sin(\theta)}=\dfrac{\csc(\theta)+1}{\csc(\theta)-1}\)

\(\begin{array}{|rcll|} \hline && \dfrac{1+\sin(\theta)}{1-\sin(\theta)} \\\\ &=& \dfrac{1+\sin(\theta)}{1-\sin(\theta)}\cdot \dfrac{ \dfrac{1}{\sin(\theta)} } {\dfrac{1}{\sin(\theta)}} \\\\ &=& \dfrac{ \Big(1+\sin(\theta)\Big) \dfrac{1}{\sin(\theta)} } {\Big(1-\sin(\theta) \Big)\dfrac{1}{\sin(\theta)}} \\\\ &=& \dfrac{ \dfrac{1}{\sin(\theta)} + \dfrac{\sin(\theta)} {\sin(\theta)} } { \dfrac{1}{\sin(\theta)} - \dfrac{\sin(\theta)}{\sin(\theta)} } \\\\ &=& \dfrac{ \dfrac{1}{\sin(\theta)} + 1 } { \dfrac{1}{\sin(\theta)} - 1 } \quad | \quad \boxed{\dfrac{1}{\sin(\theta)} = \csc(\theta) } \\\\ &=& \dfrac{ \csc(\theta) + 1 } { \csc(\theta) - 1 } \\ \hline \end{array}\)

A four-digit hexadecimal integer is written on a napkin such that the units digit is illegible.

The first three digits are 4, A, and 7.

If the integer is a multiple of 17 base ten, 

what is the units digit?

\(\begin{array}{|rcll|} \hline \mathbf{4A7x_{16}} &=& \mathbf{ ( ( 4\cdot 16+A )\cdot 16+7)\cdot16+x } \quad | \quad A = 10 \\ &=& ((4\cdot 16+10)\cdot 16+7)\cdot16+x \\ &=& (74\cdot 16+7)\cdot16+x \\ &=& 1191\cdot16+x \\ &=& \mathbf{19056 +x},\qquad x=0,1,\ldots, 15 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{19056 +x} &=& \mathbf{n\cdot 17},\ n \in \mathbf{Z} \\\\ 19056 +x &\equiv& 0 \pmod {17} \quad | \quad 19056 \pmod {17} = 16 \\ 16 +x &\equiv& 0 \pmod {17} \\ 16 +x &=& n\cdot 17 \\ x &=& n\cdot 17 -16 \quad | \quad n = 1 \\ x &=& 1\cdot 17 -16 \\ \mathbf{x} &=& \mathbf{1} \\ \hline \end{array}\)

What is the least positive integer value of \(x\) such that \((2x)^2 + 2\cdot 37\cdot 2x + 37^2\) is a multiple of \(47\)?

\(\begin{array}{|rcll|} \hline \mathbf{(2x)^2 + 2\cdot 37\cdot 2x + 37^2} &=& \mathbf{\left(2x+37\right)^2} \\\\ \left(2x+37\right)^2 &=& \left(47n\right)^2 \\ 2x+37 &=& 47n,\ \qquad n \in \mathbf{Z} \\ 2x &=& 47n - 37 \\\\ \mathbf{ x } &=& \mathbf{\dfrac{ 47n - 37 } {2}} \quad | \quad 47n - 37 \text{ is even, so } n \text{ is odd} \\\\ \mathbf{ x_{\text{least positive integer}} } &=& \mathbf{\dfrac{ 47n - 37 } {2}},\ \text{if } n = 1 \\\\ &=& \dfrac{ 47 - 37 } {2} \\\\ &=& \dfrac{ 10 } {2} \\\\ \mathbf{ x_{\text{least positive integer}} } &=& \mathbf{5} \\ \hline \end{array}\)

\(3^3\sqrt{2a}-6^3\sqrt{2a}=\\ \sqrt{2a}\left(3^3-(3\cdot 2)^3\right)=\\ \sqrt{2a}3^3(1-2^3) = \\ 27\sqrt{2a}(-7) = \\ -189\sqrt{2a}\)

\(3^5(x+2)^3+3=27\\ 3^5(x+2)^3=24\\ (x+2)^3 = \dfrac{24}{3^5} = \dfrac{8}{81}\\ x+2 = \dfrac{2}{3\sqrt[3]{3}}\\ x = \dfrac{2}{3\sqrt[3]{3}} -2 = \\2\left( \dfrac{1}{3\sqrt[3]{3}} -1\right)\)

technically it rounds to $2.38 but yeah

see edited version

\(\int 4 \cos(\pi x) ~dx = 4 \cdot \dfrac 1 \pi \sin(\pi x) = \dfrac{4}{\pi}\sin(\pi x)\)

\((2-i)z + (2+i)\bar{z} = 20\\ 2(z+\bar{z})-i(z-\bar{z}) = 20\\ 4Re(z) - 2iIm(z)= 20\\ \text{On the real axis }Im(z)=0\\ 4Re(z) = 20\\ Re(z) = 5\\ z=Re(z)+i Im(z)= 5 + i0 = 5\)

\(\dfrac{1+\sin(\theta)}{1-\sin(\theta)} \cdot \dfrac{1+\sin(\theta)}{1+\sin(\theta)} = \\ \dfrac{(1+\sin(\theta))^2}{1-\sin^2(\theta)} = \\ \left(\dfrac{1+\sin(\theta)}{\cos(\theta)}\right)^2 = \\ (\sec(\theta)+\tan(\theta))^2\)

Re-arrange to find the radius of the circle:

2x^2 - 12x + 2y^2 +4y = 20     divide through by 2

  x^2-6x + y^2 + 2y = 10      complete the squares for x and y

(x -3)^2   + (y+1)^2  = 10 +9 +1

    radius = sqrt 20

2 x radius =diameter AND diagonal of square

2 sqrt20 = diagonal of square

s^2 + s^2 = (2 sqrt20)^2   (pythagorean theorem)

2 s^2 = 80

s^2 =40

s = sqrt 40        area = sqrt40 x sqrt 40 = 40 sq units     

This is a short computer code to find the above numbers:

n=1; a=((541 - n*17) / 9);printn,a; n++; if(n<=50, goto1, discard=0;

Adults            Children

8                      45

17                    28

26                    11

This is how I would solve it:

17 - 9 = 8 -number of adults the first night

8 x $17 =$136   and 45 children x $9 =$405 for a total of $541.

Then the adults number goes up by 9 and childrens' number comes down by 17.

8+9 =17 x $17 =$289 and 28 children x $9 =$252 for a total of $541.

17+6 =26 x 17 =$442 and 11 children x $9 =$99 for a total of $541.