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Oh no @Melody moderator now. NOW! lol what is happening

Wait, is this AoPS week 24?!!!!!!!! I'm taking the same class as you I think!

How are they repeated?

yeah!

this isn't quite right.

You've neglected to take into account the fact that some of the letters are repeated.

7) 

\(\text{In order for every pair of neighboring digits to sum to an odd number }\\ \text{the parity of the digits must strictly alternate}\\ \text{There are 4 ways to choose an even first digit}\\ \text{Then there are }5^9 \text{ ways to complete that 10 digit number alternating parity}\\ \text{There are 5 ways to choose an odd first digit}\\ \text{and then again }5^9 \text{ ways to complete the number}\\ \text{This gets us a total of }9\cdot 5^9 \text{ total numbers that strictly alternate digit parity}\\ 9 \cdot 5^9\)

\(\text{There are a total of }9\cdot 10^9 \text{ 10 digit numbers}\\ p = \dfrac{9\cdot 5^9}{9\cdot 10^9} = \left(\dfrac 1 2\right)^9 = \dfrac{1}{512}\)

I remember that problem from my math homework last week LOL!

Thanks!

Thanks Bxtterman! You really helped me understand the question and the rules for factorials!

3) In the word REARRANGE we have 9 letters.

For these letters to be distinct, we must think of each letter like a gumball in a gumball machine. Every time I use one,  there of less to be used.

We also need casework for each length of word:

CASE 1: 1 letter words

9  one letter words.

CASE 2: 2 letter words

\(9\cdot8=72\) two letter words

CASE 3: 3 letter words.

\(9\cdot8\cdot7=504\)

CASE 4: 4 letter words

\(9\cdot8\cdot7\cdot6=3024\)

CASE 5: 5 letter words

\(9\cdot8\cdot7\cdot6\cdot5=15120\)

CASE 6: 6 letter words

\(9\cdot8\cdot7\cdot6\cdot5\cdot4=60480\)

CASE 7: 7 letter words (almost done!)

\(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3=181440\)

CASE 8: 8 letter words

\(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2=362880\)

CASE 9: (DONE!)

\(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=362880\)

Now we add all the cases together for the total number of arrangments:

\(9+72+504+3024+15120+60480+181440+2(362880)=\boxed{986409}\)

SIDENOTE:

You can subsitute the decending numbers with a factorial

\(5\cdot4\cdot3\cdot2\cdot1=5!\)

\(5\cdot4\cdot3=\displaystyle\frac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1}=\displaystyle\frac{5!}{2!}\)

Just some rules for future reference, hope you enjoy and use them.

As always if you have any questions just ask.

Goodnight! 🌙

2) First we calculate how many possibilities before: 

\(9\cdot8\cdot7\cdot6\cdot5=15120\)

Now we calculate the possibilites now:

\(9\cdot9\cdot9\cdot9\cdot9=9^5=59049\)

We can get the difference:

\(59049-15120=\boxed{43929}\)

You can solve this problem easily by plotting it in on a coordinate plane and probability. 

NOTE BEFORE WE START. I MAKE MISTAKES AND I HAVE IN THE PAST PLEASE CHECK THIS ANSWER!

Thank you for reformatting.

I just graphed this. it's a diamond with equal side lengths.

We first see that there are 4 ways to get to the first letter, A

We add the first part to our equation: \(1\cdot4\)

From A, we can get to the next letter, T

We can get to there 2 ways from every A. 

Our equation is now: \(1\cdot4\cdot3\)

Finally, we can get to the letter H. 

For 4 of the 8 letter T's there are 2 H's.

Our equation is now \(1\cdot4\cdot3\cdot2=24\)

Next, we count the other 4. We have 3 H's for every T.

\(1\cdot4\cdot3\cdot3=36\)

We can add both cases:

\(24+36=\boxed{60}\)

Thank you man, I probabaly got this one wrong it's 1 AM and I'm answering math problems smh.

I wanna cuss but I can't haha but I need to vent everyday.

PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE FORMAT PLEASE PLEASE

I'm like really pissed idek why this infuriates me. I have to give you props for effort in at least putting it in latex instead of others who just paste it in.

1) Spaces are removed in LaTeX

2) LaTeX in this forum is for EQUATIONS only

3) Noone will solve your question if you make it completly illegible, is shows little effort

4) This problem is totally OK with just text, just go to Notepad and CTRL+H and find and delete the H.

5) I like your username

So yeah. Reformat and me or someone will try to solve your question.

1) The last digit must be 1 or 3.  The rest of the digits can then be arranged in 3! ways.

Thus there are 2x3! = 12 odd 4 digit numbers using digits 1-4

\(\dfrac{1}{2x} = \dfrac{r-x}{7}\\ 7 = 2rx - 2x^2\\ 2x^2 - 2rx + 7 = 0\\ \text{There will be exactly one real solution if the discriminant is equal to 0}\\ D = (-2r)^2 - 4(2)(7) = 0\\ 4r^2 = 56\\ r^2 = 14\\ r = \pm \sqrt{14}\\ \sqrt{14}(-\sqrt{14}) = -14\)

looks like it's just a matter of how old they are

Thanks so much Melody and Guests! You really helped me understnad and solve this question!

10.47 should be sufficient since you aren't finding all possible values of the sine function and there isn't a squared anywhere.

This is the answer that I got, but should there be more than one answer? Or is 10.47 sufficient

\(\frac{sin(33.2)}{x}=\frac{sin(45.6)}{13.7}\)

Alright, so the first thing you want to do here is to find the values of the sine functions.

\(sin(33.2) = 0.547563223493\)

Which rounded to 2 decimal places is 0.55.

\(sin(45.6) = 0.714472679633​\)

Rounded to 2 decimal places that's 0.72 (the 7 behind the two 4's carries down the line).

So, now we have: \(\frac{0.55}{x}=\frac{0.72}{13.7}\)

Then you have to cross multiply.

\(0.72x=(13.7)(0.55)\)

Then multiply 13.7 by 0.55.

\((13.7)(0.55) = 7.535​\)

Round to 7.54

\(0.72x=7.54\)

Then divide both sides by 0.72.

\(7.54/0.72 = 10.4722222222222222\)

Round that to 10.47

\(x=10.47\)

Does that make sense?