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OK folks. I figured it out and the result is: Euler's constant "e". But, I was only able to get 160 accurate digits!.Let us see if anybody can improve on that!. 

You have taken the entertainment of the world very seriously I see :))

Melody: I thought I should keep everybody entertained over the Christmas holidays!!.

Thanks Alan,

More brain food for me  :)))

Here's my take on 1):

For 2) I would start by writing  I₂ = x*I₁   and then for 3) by writing I₃ = x*I₂

[1 + 9^-4^(6*7)]^3^2^85=

Ah ha.    Very interesting     

what is the value of x if it is a right triangle. The hypotenuse is 2*sqrt(3). One side is (x*sqrt(3))/3, the other is x

\(x^2+(\frac{x\sqrt3}{3})^2=(2\sqrt3)^2\\ x^2+(\frac{3x^2}{9})=(4*3)\\ x^2+(\frac{x^2}{3})=12\\ x^2(1+\frac{1}{3})=12\\ x^2=12\div \frac{4}{3}\\ x^2=12\times \frac{3}{4}\\ x^2=9\\ x=3 \qquad \text{x is a distance so it cannot be negative}\)

Much appreciated  Alan,    

It is very interesting how calculus works and if amnesia can sort out these concepts she will have a good head start on her class mates I think. :))

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So amnesia, be a sponge and soak up as many concepts as you possibly can :)

I have another demonstration graph in mind but I might not get up to constructing it.   ://

Thanks Ginger, you have made some very good points.  

I did not realize that the correct (by convention) answer for 48/(2)(9+3) is 2 but I can see why this is the case.

However, I do not accept that tthe following does not have brackets.  It is like saying that 6x does not have a multiply sign. The multiply sign is implied just as the brackets are implied in your example.

\(\dfrac {5+20}{4+1} = 5\)

this simpy means

(5+20)/(4+1)=5

The same as square root is done before addition but

\(\sqrt{21+4}=\sqrt{25}=5\)

the brackets are implied.

Let's take Melody's picture idea a little further.  Imagine you have the following function, f(x), which is comprised of a sequence of straight lines (this is an arbitrary function I made up off the top of my head - it isn't specially chosen):

The gradient, df(x)/dx, of this function is easily plotted:

The integral of df(x)/dx with respect to x from, say x = 1 to x = 7, is just the area under the curve. It is easy to calculate because df(x)/dx is just constructed from a number of constants (i.e. horizontal straight lines).  We can see that the area under df(x)/dx from x = 1 to x = 7 is 9.  

Using the plot of f(x) we can also see this is the same as the difference between f(7) and f(1). i.e. f(7) - f(1) = 9

You can change the values of the gradients of  the various sections of f(x) as much as you like and you will always find the area under the corresponding graph of df(x)/dx will equal f(x2) - f(x1).  Try it!

Now a general curve can be thought of as an infinite number of (infinitely small) straight line segments, so the same result applies to "curvy" curves as well!

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Hi Amnesia,

I felt like drawing a picture.  I might do another if I get up to it.   I have a second slightly different pic already in my mind  :)

Take a look, see if you can see the relevance :))   Ask questions if you want to.

I am just looking at the the function   y=0.75x+3   and its derivative   y'=0.75  I have graphed both thes lines.  See if you can work out the rest of what I have done.

Yes i want some !    

Here is something for you to think about amnesia :))

It is not very technical but

You are graphing  dy/dx against  x from x1 to x2

The integral

\(\int_{x_1}^{x_2}\;\frac{dy}{dx}\;dx\)

is the sum of an infinite number of rectangles from x1 to x2

remember that d stands for difference and The integral sign is a stylized S which stands for sum (that is my interpretation anyway and it makes total sense)

So with the integral we have (from x1 to x2)

\(\frac{\text{difference in ys}}{\text{difference in xs}}*\text{difference in xs}\\ =\text{difference in ys because the x'es cancel out}\\ =y_2-y_1\)

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This is why definite integration gives an area.

\(\int_{x_1}^{x_2}\;y\;dx\\ \text{This is all the infinitely thin y heights multiplied by all the infinitely thin x widths }\\\text{over the domain }x_1 \;to \;x_1\\ and\\ \text{y *x is the area of a rectangle. Add all these skinny areas together and you have the area under the curve.}\)

In all this discussion i have used y instead of f(x) because it is usually easier to interprete, but they are the same thing.

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Anyone is welcome to comment of course but do not criticize my lack of technical talk unless you truly believe it to be totally incorrect.

This is how I see it in a fairly non-mathematically precise way.  :)

Hit the  "2nd" button.......you will see asin, acos, etc......

28  =    2 x 2 x 7

66 =     2            x 3   x 11

GCF  =  2

See if this makes sense, amnesia  :

With the wind.....the plane covers  (D/60)  every minute  and against the wind, the plane covers (D/90)  every minute.....where D is the total disance beween the cities

So......the  average rate per minute  [ assuming no wind ] is given by  :

[D / 60  + D / 90]  / 2   =

[90D + 60D] / [ 60 * 90 *2]  =

[150D]  / [ 10800]  =

[15D ] / 1080]  =

D/72  ...... means that the plane would make the flight in  a normal time of 72 minutes witn no wind

Thus....the plane is 90 − 72   = 18 min late

Happy B'Day, HSC.......that cake does look good.....!!!!

Can we all share this cake with you?  PLEASE :)

 55371057962*x=99

\( 55371057962*x=99 \) 

\(\large x=\frac{ 99 ​}{55371057962}\)

\(x=1.78793766353\times 10^{-9}\)

   !

True, I did not include WEB2.0 calculator in my list...I just kinda forgot it, but it delivers 288 if you enter this question as written

48/2(9+3)

48/(2)(9+3)     Note the EXTRA parentheses     delivers   ' 2 '     ...with which I am OK.....kinda

By the way:

this is just putting a^2 + b^2 = c^2

Just the sentence of pythagoras saying that a squared + b squared equals the area of c squared.

[2*sqrt(3)]^2 =[ (x*sqrt(3))/3]^2 + x^2, solve for x
Solve for x:
12 = (4 x^2)/3

12 = (4 x^2)/3 is equivalent to (4 x^2)/3 = 12:
(4 x^2)/3 = 12

Multiply both sides by 3/4:
x^2 = 9

Take the square root of both sides:
Answer: |x = 3       or         x = -3

Jea, but why is the reduction in time 50% greater in the 2nd flight?