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3^352^log(10^(sqrt(sqrt3(

$$\\y=3^{{352}^{log\left(10^\sqrt{\sqrt{3}}}\right)}}\\\\
y=3^{{352}^{log\left(10^{3^{0.25}}}}\right)}}\\\\
y=3^{{352}^{\left(3^{0.25}log10}\right)}}\\\\
y=3^{{352}^{\left(3^{0.25}}\right)}}\\\\
log(y)=log\left(3^{{352}^{\left(3^{0.25}}\right)}}\right)\\\\
log(y)={352}^{(3^{0.25})}log(3)\\\\
log(y)=1071.685578\\\\
y\approx 10^{1071.7}\\\\$$

This answer is different from Chris's and these large powers really confuse me.

Could other mathematicians (including Chris) take a look and comment please.  

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Wow, Melody....you made that one look easy....very nice........!!!!!!

1

0.0512820512820513

0.0057971014492754

The thousands column is the middle column!

    1       2       3                4                  5                          6                                7

units  tens  hundreds  thousands  ten-thousands  hundred-thousands   millions

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Thanks, Melody......I "stole" that pic from Wikipedia.....LOL!!!

I already answered this today !

DELETED - I'm an idiot.   Only kidding

Thanks Alan :))

Hi Chris,   

That makes sense :)  I like your pic too :)

You are right about the probability of the thousands digit equalling 5 being 1/3 Melody.  I just checked my program and realised I'd put the counter in the wrong place!  

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This is a very good question and really confuses a lot of students.

The compound interest formula is     

$$FV=P(1+r)^n$$

FV stands for future value and P stands for principal - that is how much you start with.

When you are introduced to this in year nine you are told that r is your interest rate and n is the time.

This is true BUT  the interest rate is stated as an annual interest rate and the time is stated in years so kids often get it cemented into their brains that  r is the (annual) rate and n is the time (years).  This is WRONG!

r is the interest rate for the compounding period of time.

and

n is the number of compounding time periods.

SO     if the interest is 6% per annum compounding quarterly and the money is invested for 2 years then

the interest is compunded 4 times every year so each time it must be a quarter of 6%      

$$r=\frac{6\%}{4}=\frac{0.06}{4}=0.015$$

and

the interest is compounded 4 times a year for 2 years so that is 

$$n=4*2 = 8$$     compounding periods

sorry I just read your question better. If it were compounding monthly then r=6%/12  and n=2*12

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Sometimes different letters are used - but this is always the idea behind it.

sometimes the formual is written like this

$$PV=P(1+\frac{r}{t})^{n*t}$$

Where r is the annual rate, n is the number of years and t is the number of compounding periods each year.

but it really is the same.   (although n and r stand for the yearly figures not the compounding period figures)

I never thought that you did post it twice.

I was just inviting you to repost (like you have done here) if you still needed the answer - that is all.  :)

I have now answered you properly on your original post.  

Can I make this a shorter answer?    YES I CAN     

Pat writes all the 7-digit numbers in which all the digits are different and each digit is greater than the one to its right (so the tens digit is greater than the units, the hundreds greater than the tens, and so on). For example, 9,865,320 is one of the numbers that Pat writes down.

(a) How many numbers does Pat write down?

9,8,7,6,5,4,3,2,1,0     3 of the 10 must be removed    10C3 = 120

(b) One of Pat's numbers is chosen at random. What is the probability that the tens digit is a 1?

(9,8,7,6,5,4,3,2) 1,0       3 of  the 8 must be left out     8C3= 56

P(1 in the 10s column) = $${\frac{{\mathtt{56}}}{{\mathtt{120}}}} = {\frac{{\mathtt{7}}}{{\mathtt{15}}}} = {\mathtt{0.466\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

(c) One of Pat's numbers is chosen at random. What is the probability that the middle (thousands) digit is a 5?

(9,8,7,6 one of these four must be left out) 5 ( 4,3,2,1,0 two iof thse 5 must be left out) 4C1*5C2=4*10=40

P(5 is in the middle) =  $${\frac{{\mathtt{40}}}{{\mathtt{120}}}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}} = {\mathtt{0.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

Thanks Anon,  

My instructions for reposting were followed to the T so I am going to give some time to your question.

Thank you!

You can easily solve this with a 3 way rule (regla de 3).   $${\frac{{\mathtt{30}}}{{\mathtt{45}}}} = {\frac{{\mathtt{x}}}{{\mathtt{100}}}} \Rightarrow {\mathtt{x}} = {\frac{{\mathtt{200}}}{{\mathtt{3}}}} \Rightarrow {\mathtt{x}} = {\mathtt{66.666\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

I'll answer you with another question. 

NOTE you are not being asked for the temperature- you are being asked for the overall change in temperature!

If you climb three rungs of a ladder and then go down 3 rungs again where are you

(in relation to where you started?)

Here is a pic that might help you to think about the temperature

Thanks for showing us this Nauseated  

The power of the Web2 calc continues to amaze me  !!      

Hi anon,

You cannot give thumbs up unless you are a member. 

If you were a member you could have give CPhill 5 points:)

Why don't you join up - there are a number of plusses to be had and absolutely no negatives :)

If log10 = log(2x-3) then 10 = 2x-3 so 2x = 13 and x = 13/2

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Press the 2nd button and select asin, which is sin^-1

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$$x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)$$

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1 nm = 1000 pm so 95 pm = 0.095 nm 

You can do the addition.

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BLESS YOU CPHILL ALL MY LOVE TO YOU