Melody

Thanks Jasonkiln,

You are right of course, I cannot just divide by 2 at the end

.I can tie my two Cs together though. becasue any answer with 2 Cs tied together will not be included and I will have to subtract.

So I have

aauryC    and I will add the other c  later

6!/2! = 360

Now where can the last c go

  |a|a|u|r|y|C

It can go where any of those bars are.   It can't go after the C becasue that would be the same as before the C     Cc = cC

There are 6 bars.

360*6=2160

There are 2160 permutations where 2 or more Cs are together

3360-2160 = 1200 

There are 1200 permutations where no 2 Cs are together.

This is now the same as Ginger's answer,    Builderboi agrees too.  I think his method is similar to mine.

Ginger's method is a little more simple.

Accuracy

8 letters,  3Csand 2As        8!/(3!*2!) = 3360    As the question said

aCuracy   (I tied 2 Cs together)

How many ways can this be arranged

7 letters, 2As

7!/ (2!) = 2520,  but if the little c is on either side of the C then it will be the same so I need to divide by 2

2520/2=1260

So the mumber of ways no two cs are together is     3360-1260 = 2100 ways

If you think this is incorrect then please promote a discussion.

Or at least say so and give reason  (eg answer is rejected in the app)

well

\(f(6)=-\frac{1}{6}\\~\\ f(f(6))=f(\frac{-1}{6})=6\\~\\ etc. \)

You can take if from there.

Yes good.  

And thanks for responding.

Thanks Alan  

Alice is picking up trash in the community park. She can rid the whole park of trash in 5 hours. After she works for 2 hours, her friend Brenda comes and helps her. The two of them finish in 1 more hour. How fast can Brenda clean the park, by herself?

Alice cleans 1 park in  5 hours

After 2 hours, Alice has cleanted  2/5 of the park

There is 3/5 of the park left to clean

Brenda comes and helps and the job is finished in 1 hour

In one hour Alice will clean 1/5 of the park which means that Brenda cleans 2/5 of the park in this time

So

Branda is working twice as fast as Alice.

So

Brenda can clean the whol park by herself in how much time?   

Your turn to answer.,    

everything under the square roots must be greater or equal to 0

so,    

\(1)\quad x\ge0\\ 2)\quad 5-\sqrt x \ge0\\ \qquad 5\ge \sqrt x\\ \qquad \sqrt x \le 5\\ \qquad 0\le x \le 25\\ 3)\quad \sqrt{5-\sqrt x }\le 3\\ \qquad 0 \le 5-\sqrt x \le 9\\ \qquad -5 \le -\sqrt x \le 4\\ \qquad -5 \le -\sqrt x\qquad and \qquad -\sqrt x \le 4\\ \qquad \sqrt x \le 5\qquad \qquad and \qquad \sqrt x \ge -4\\ \qquad x \le 25\qquad \qquad and \qquad x \ge -0\\ so\\ domain \;\;[0,25] \)

I have not checked my answer, you need to do that.

I would also like a response from you, last time you did not give me one.

So he investes it for   0.0375/2    for 4 time periods.

It is just compouund interestf

Future value = a(1+r)^n

If you know a question was asked before you need to give the address of the old post.

You say CPhill's answer is wrong but you have not given him a chance to comment nor have you given anyone else the chance to judge for themselves.

\(5^n + 8^n + 13^n + 17^n \mod6\\ \equiv (-1)^n + 2^n + 1^n + (-1)^n \mod6\\ \equiv 2*(-1)^n + 2^n + 1 \mod6\\\)

\(\qquad 2^1 \mod6 \equiv 2 \mod 6\\ \qquad 2^2 \mod6 \equiv 4 \mod 6\\ \qquad 2^3 \mod6 \equiv 2*2 \mod 6\equiv 4 \mod6\\ \qquad 2^4 \mod6 \equiv 4*2 \mod 6\equiv 2 \mod6\\ \qquad etc\qquad so\\ \qquad 2^n \equiv 2\mod 6\qquad \text{If n is odd}\\ \qquad 2^n \equiv 4\mod 6\qquad \text{If n is even}\\ \)

\(5^n + 8^n + 13^n + 17^n \mod6\\ \equiv (-1)^n + 2^n + 1^n + (-1)^n \mod6\\ \equiv 2*(-1)^n + 2^n + 1  \mod6\\ \text{If n is odd}\\ \qquad \equiv 2*(-1) + 2 + 1  \mod6\\ \qquad \equiv -2 + 2 + 1  \mod6\\ \qquad \equiv 1  \mod6\\ \text{If n is even}\\ \qquad \equiv 2*(1) + 4 + 1  \mod6\\ \qquad \equiv 2 + 4 + 1  \mod6\\ \qquad \equiv 1  \mod6\\ so\\ \text{For all positive integer values of n}\\ 5^n + 8^n + 13^n + 17^n \equiv 1 \mod6\\ \text{It will never have a zero remainder when divided by 6}\)

LaTex:

5^n + 8^n + 13^n + 17^n \mod6\\
\equiv (-1)^n + 2^n + 1^n + (-1)^n \mod6\\
\equiv 2*(-1)^n + 2^n + 1  \mod6\\
\text{If n is odd}\\
\qquad \equiv 2*(-1) + 2 + 1  \mod6\\
\qquad \equiv -2 + 2 + 1  \mod6\\
\qquad \equiv 1  \mod6\\
\text{If n is even}\\
\qquad \equiv 2*(1) + 4 + 1  \mod6\\
\qquad \equiv 2 + 4 + 1  \mod6\\
\qquad \equiv 1  \mod6\\
so\\
\text{For all positive integer values of n}\\
5^n + 8^n + 13^n + 17^n \equiv 1 \mod6\\
\text{It will never have a zero remainder when divided by 6}