All Answers 356736 Answers

its not so hard if you try

Rotating Q 180° clock-wise about the origin produces (-3, 2)

Reflecting this about the y axis  produces Q' = ( 3, 2)

6 × 6! + 7 × 7! + 8 × 8! + 9 × 9! + 10 × 10! + 11 × 11! + 12 × 12! + 13 × 13! + 14 × 14! = a! - b!

a + b= ?

\(\begin{array}{|rcll|} \hline && n\times n! \\ &=& (n+1-1)\times n! \\ &=& [(n+1)-1]\times n! \\ &=& (n+1)\times n! -1\times n! \\ &=& n!\times (n+1) -n! \\ &=& (n+1)! -n! \\\\ &&\mathbf{ n\times n! = (n+1)! -n! } \\ \hline \end{array}\)

\(\begin{array}{|rcrcr|} \hline 6\times 6! &=& 7!- 6! \\ 7\times 7! &=& 8!- 7! \\ 8\times 8! &=& 9!- 8! \\ 9\times 9! &=& 10!- 9! \\ 10\times 10! &=& 11!-10! \\ 11\times 11! &=& 12!-11! \\ 12\times 12! &=& 13!-12! \\ 13\times 13! &=& 14!-13! \\ 14\times 14! &=& 15!-14! \\ \hline \text{sum} &=& 7! &\mathbf{-}& \mathbf{6!} \\ &+& 8! &-& 7! \\ &+& 9! &-& 8! \\ &+& 10! &-& 9! \\ &+& 11! &-& 10! \\ &+& 12! &-& 11! \\ &+& 13! &-& 12! \\ &+& 14! &-& 13! \\ &+& 15! &-& 14! \\\\ &=& \not{ {\color{green}7!}} && -6! \\ && {\color{red}{-}}\not{ {\color{red}7!}} && {\color{green}{+}} \not{ {\color{green}8!}} \\ && {\color{green}{+}}\not{ {\color{green}9!}} && {\color{red}{-}}\not{ {\color{red}8!}} \\ && {\color{red}{-}}\not{ {\color{red}9!}} && {\color{green}{+}} \not{ {\color{green}10!}} \\ && {\color{green}{+}} \not{ {\color{green}11!}} && {\color{red}{-}}\not{ {\color{red}10!}} \\ && {\color{red}{-}}\not{ {\color{red}11!}} && {\color{green}{+}} \not{ {\color{green}12!}} \\ && {\color{green}{+}} \not{ {\color{green}13!}} && {\color{red}{-}}\not{ {\color{red}12!}} \\ && {\color{red}{-}}\not{ {\color{red}13!}} && {\color{green}{+}} \not{ {\color{green}14!}} \\ && \mathbf{+15!} && {\color{red}{-}}\not{ {\color{red}14!}} \\\\ &\mathbf{=}& \mathbf{15! - 6!} \\ \hline \end{array} \)

so a = 15 and b = 6

a+b = 15+6 = 21

No, I don't believe so, but @Melody or @CPhill might now better.

only post topics about mathematics and thereabouts,

you shouldnt advertise

thanks

what is pi to the power of pi

\(\pi^{\pi} = 36.46215960720791177099082602269212366636550840222881873870...\)

Anything to the power of 0 is 1, so 0⁰=1

108251299468315690105155229 how do i prime factorize this?

factor  \(108\ 251\ 299\ 468\ 315\ 690\ 105\ 155\ 229\)

Result: \(4466771201309 \times 24234798378881\) (2 distinct prime factors)

i think it is like this but i am not too sure.

6(1/2x+3)                          =7(5)-2

6*1/2(x)=3x                      7*5=35

6*3=18                             35-2=33

so:

6(1/2x+3)=3x+18             7(5)-2=33

                            3x+18=33

take 18 from both sides:

                             3x=15

divide both sides by 3

                              x=5

Thanx a lot Melody,

Like you said...this does make sense!. Thank you for your input!!

Thanx guest,

I agree 100%..Ta.

Actually this makes sense.

This is the question....if 7 dogs consume 3 bags of food in 12 weeks, how many weeks will it take 10 dogs to consume 3 bags?

When you are working our problems, usually rates problems, you can use the units to help you work out what you need to do.

here     7dogs*3bags*12weeks = 7*3*12  dogs bags weeks

The units are  dogs bags weeks

So yes, this must equal

          10dogs*3bags*y weeks = 7*3*y  dogs bags weeks

7*3*12 = 10*3*y

etc :)

I have not used this units method for a question quite like this before. ://

Try it on a few different questions, I mean just change the numbers around and check it works Juriemagic. :))  I think you will find that it does.

Using the units in questions  is often extremely helpful.

Yea, it appears to me to be a simple proportion problem (I'm not he/she that came up with it)!. 

7 x 3 x 12 =10 x 3 x W(weeks), solve for W.

Divide both sides by 10 x 3

W =[7 x 3 x 12] / [10 x 3]

W = 252 / 30

W = 8.4 weeks!.

Note: If you had twice as many dogs or 14 dogs, you would expect the time to be halved from 12 weeks to 6 weeks, and that is exactly what you get when you substitute 14 dogs instead of 10!.

How old are the children ??

Solved by the distinguished super sleuth Ginger !!

Very impressive Ginger   

This needs to be added to our Puzzle thread!!      

This is not likely to be a very good 'proof' but here goes.

since f(x) and g(x) are odd funtion,

\(f(-x)=-f(x)\)            and      \(g(-x)=-g(x)\)

Let  \(x_1\)  be a point such that    

\(a=g(x_1)   \qquad     -a=g(-x_1)\)

\(f(g(x_1))=f(a) \qquad f(g(-x_1))=f(-g(x_1))=f(-a)=-f(a)\\ so\\ f(-g(x_1))=-f(g(x_1)) \\ \text{by definition this means that }f(g(x)) \text{ is an odd function.}\)

oops!!..then I guess that says a lot about me too!!!...lol.

Melody, yes, I certainly do understand what you are saying. Well, looking at this it certainly calculates to the correct answer. I kind of understand the principle behind this which was to take the first scenario and devide that by the 2nd scenario, making sure the unknown is at the bottom, meaning this:

\({7 \over10}:{3 \over3}:{12 \over x}\)

Then multiply the X out and put that equal to the rest. Multiply the top row and devide by the product of the bottom row.

anycase, I know you know what I'm saying, so it's not like I have to explain this to you..(sorry)..uhm, I think to really test this out would be to create a few different scenarious and try to calculate the unknown using this approach.

Thanx for being out there!! 

Wow Ginger! That took all kinds of thinkin' !!  

Yes we do.... need another question that is :)

I put this thread in sticky topics.

If it is used it can stay there but otherwise I will eventually take it down :)

If you are an honours student and you do not already know how to find the percentage of a number then I think you need to drop down several levels and study introductory arithmetic instead.  

15/75  *  100% = 20%

I want to know how to do this problem too.....

Thanks Melody!! Now, we need a new question!!