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Very nice, Melody.....!!!!

a^3  + 1/a^3     if  a  + 1/a  = 6

Note that   (a + 1/a)^2  =  a^2  + 2  +  1/a^2   = 36   →  a^2 + 1/a^2  =  34

And  a^3 + 1/a^3   can be factored as a sum of cubes thusly

(a  + 1/a)  ( a^2  -  1 +  1/a^2)  =

(6)  ([ a^2  + 1/a^2] - 1 )   =

(6) ( 34 - 1 )  =

6 * 33=

198

P.S. -  thanks to hectictar for spotting my earlier error...!!!!

Is it a typing mistake?? 

Do you mean

\(\displaystyle \int^{\infty}_{-\infty}\dfrac{\sin x}{x}dx\)?

If yes, the answer is π by definition.

Thank you.

I guess it's Mode of X = 3?

Translations from Slovak:

1) The middle character of this sequence is 3, i.e., Mod (x) = 3.

2) The Middle character of this sequence has a value of 3, i.e. mod (x) = 3.

I do not understand this question,  would anyone else like to venture a guess?

I know the feeling, Sir Alan. I sometimes have to take Dramamine when doing some of the math problems on here.     

Yea they always confuse me, I can never remember the most basic conventions. :(

27=u1r^3/192=u1r^6

\(27=\frac{u_1r^3}{192}=u_1r^6\\ \frac{r^3}{192}=r^6\\ \frac{1}{192}=r^3\\ r=\frac{1}{\sqrt[3]{192}}\\ r=\frac{1}{4\sqrt[3]{3}}\\ 27=u_1 r^6\\ 27=\frac{u_1}{ 192^2}\\ u_1=27*192^2\\ u_1=995328\)

\(T_n=\dfrac{995328}{(\sqrt[3]{192})^{n-1}}\\~\\ \boxed{T_n=\dfrac{995328}{192^{((n-1)/3)}}}\\~\\ check\\ \frac{u_1r^3}{192}=\frac{995328*(192^{-1/3})^3}{192}=\frac{995328}{192^2}=27\\ u_1r^6=995328*((192)^{-1/3})^6\\ u_1r^6=995328*(192)^{-2}\\ u_1r^6=27\\ excellent \)

You are right GingerAle - these multiple exponentiations are making my brain swirl!

Actually, they are not interpreted same. Each interpretation gives a different solution for x.

As Hectictar pointed out:

Sir Alan’s interpretation is: 4^(2^x) + 2^(2^x)  =  56

And Sir  CPhill interpretation is:     (4^2)^x + (2^2)^x  =  56 

Sir Alan’s is interpretation is exactly the same as the asked question. (This follows power-tower convention, where stacked powers are multiplied from right to left or from the top down, and the resultant product becomes the exponent to the base number.)

 Sir CPhill’s solution starts with the base then increments the value by each ascending power.

The solutions for these interpretations give different results for the value of x.

-----

\(\text{For Sir Alan's solution}\\ x = log_2 (log_2(7))\\ x \approx 1.4892114692\\ \text{ }\\ \text{For Sir CPhill’s solution}\\ 2x = \dfrac {log (7) } { log (2)}\\ x= \dfrac {log (7)/ log (2)}{(2)}\\ x\approx 1.4036774610\\ \)
-----

Sir Alan notes the calculator input error of the questioner; however, the questioner may have correctly entered this into the site calculator

(4^2^log(log(7, 2), 2)) + (2^2^log(log(7 , 2), 2) )  

The calculator would interpret it as

(4^2)^(log(log(7, 2), 2)) + (2^2)^log(log(7 , 2), 2))

The site’s calculator does NOT follow hierarchal convention in interpreting stacked powers.

Just type in round(30/11) and press =.    It works!

Let V be full volume.

We are told:  V/2 -420 = V/8

Subtract V/8 from both sides: V/2 - V/8 - 420 = 0

Add 420 to both sides:  V/2 - V/8 = 420

Simplify the left hand side:  3V/8 = 420

Multiply both sides by 8/3:  V = 420*8/3

Simplify: V = 140*8  or  V = 1120 gallons

In Guest #1 #2 calculator he/she put the terms as (4²)^{log₂log₂7} + (2²)^{log₂log₂7}  but should have put 4^log₂7 + 2^log₂7 to maintain the same interpretation (noting that 2^{log₂log₂7} is the same as log₂7).

Also see GingerAle's post below.

how many dogs can fit in heaven?

In people-heaven, none. They voted a long time ago to kick all dogs out of people-heaven, because they were tired of stepping in dog dirt. After they were kicked out, the dogs went to dog-heaven.    In dog-heaven, bunches can fit.  This is where I want to go. Dogs don’t mind people coming there –they even let cats in. I have my own pooper-scooper.   

who started time?

I did. As soon as I start answering these questions, I started time. Wasting time, but I started it.  

when do people start to remember?

I used to know the answer to that but I can’t remember.

are we too evolved?

After reading and answering these questions, I am certain we are not.  

will we ever find something in space worth leaving our home for?

Yes. Just ask Neil Armstrong.

is there anything out there?

Yes. There are lots of rocks. Lots and lots of them.  There’s also some space junk.

are we insignificant?

For sure, I am not. You, however, are decidedly insignificant. 

why is space so big?

For the same reason that your brain is so small.

is there a higher power?

Yes, there is always a higher power. Take X to the power of Y, for any value of Y you choose, I can choose a higher one.  

what is at the core of the earth?

Evil people and a few really bad dogs. That is where heII is. Evil people are mostly composed of ion and nickel, though.   I’m not sure what the bad dogs are composed of --dehydrated vomit or cat-food, maybe.

if someone got down there, do you think the could control the earths movement through space?

A long time ago, they could control the Earth’s movement through space, but the panel was deactivated and control transferred to heaven. I’m not sure if it was people-heaven or dog-heaven. We should know in a few more years, though. If the Earth goes to the dogs then it was dog- heaven, if it’s just generally screwed up, then it was people-heaven. If Earth goes to heII then obviously they regained control.   

If you have any more questions, just ask. I usually make some time to waste.  

Do you want the equation of a line that passes through the given points? If so.....

slope  =  (change in y's) / (change in x's)

slope  =  (2 - 5) / (3 - 9)

slope  =  (-3) / (-6)

slope  =  1/2

Now we know a point,  (3,2)  , and the slope,  1/2  , so we can write the equation of this line in point-slope form.

y - 2  =  (1/2)(x - 3)            And we can get this in slope-intercept form. Distribute the  1/2  .

y - 2  =  (1/2)x - 3/2           Add  2  to both sides.

y  =  (1/2)x + 1/2

I could be wrong but it looks like Alan interpreted the question this way:     4^(2^x) + 2^(2^x)  =  56

And CPhill did this:     (4^2)^x + (2^2)^x  =  56        

\(2010/4=502R2 \\ 1+i+i^2+i^3+i^4+\cdots+i^{2009} \\=1+i+-1+-i\\ +1+i+-1+-i+.......\\~\\ =0+1+i\\ =1+i \)

'Best wishes'    perhaps..... I think.

That was one of Googles althernate translations :)

Probability of 0 heads =  C(10, 0) / 2^10

Probabilty of 1 head = C(10,1) / 2^10

Probability of 2 heads  = C(10,2) / 2^10

Probability of 3 heads = C(10,3) / 2^10

Probabilty of 4 heads  = C(10, 4)  / 2^10

So....the probabilty of getting fewer heads than tails  =

[ C(10,0)  + C(10,1) + C(10,2)  + C(10,3) + C(10,4) ]  / 2^10   = 193/512 ≈ .377  = about 37.7%

Answered here :

https://web2.0calc.com/questions/i-have-2-questions#r1

The "formula" for the total  charge,  C(h)  =

 $80  + 45 (h)      where h is the number of hours worked after the first hour

If the domain of f(x)  =  (-1, 1)...then the domain of h(x)  must be :

f(x) + 1  =

( -1  + 1,  1 + 1 )  =

(0, 2)

I will assume for this problem that \(x2\hspace{1mm}\text{means}\hspace{1mm}{x^2}\). In this problem, there is only one like term in the expression \( x^2+2x-1+x^2+6y^9\); the 2 x^2-terms. Let's combine them:
 

\( x^2+2x-1+x^2+6y^9=2x^2+2x-1+6y^9\)

Normally, I would stop here, but sometimes you should order your terms in term of its individual degree. If this is the case, your final answer is:
 

\(6y^9+2x^2+2x-1\)

According to Google Translate it's "Kumkum Bhagya"