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nevermind, i solved it!

I found similar   BUT

x+y = 4  is given    ....   then if

x-y = sqrt28       then  if you add two equations together

2x = 4 + sqrt28 = 4 + 2 sqrt 7

then

x = 2  + sqrt 7

Now    x+y = 4     means      2+ sqrt7    + y = 4     then y =  4-(2 + sqrt7)  = 2 - sqrt7

Then does x^2 + y^2 = 22 ???   Yes it does

Does   x^4 = y^4 - 176 sqrt7  ?    No it does not,,,,,   I do not think the question is written correctly !   Did you find this too?

Sorry, I have fixed my problem

THAT is illegible to me...Yikes !   

If you draw a Venn diagram.....you will have    28 who speak Greek       43 who speak Port   and 16 who speak BOTH and 13 who speak NEITHER .

By the angle bisector theorem, XY = 7/2.

Hey there guest! I'm doing this at 12:30 rn so I didn't get to finish the whole question. I will share what I've gotten thru so far:

Let's first start off with some basic logarithm properties: first,

\(c\log_{a}b = \log_{a}b^c\)

Next:

\(\log_{a}b + \log_{a}c = \log_{a}b+c\)

With these 2 properties, we have enough to finish our question!

For the first term, a * log 10 of 2, we can rewrite that as:

\(\log_{10}2^a\)

Repeating this for all the terms in the given equation, we eventually reach:

\(\log_{10}2^a + \log_{10}3^b + \log_{10}5^c + \log_{10}7^d\)= 2005

This simplifies to:

\(\log_{10}(2^a*3^b*5^c*7^d) = 2005\)

Using the basic definition of a log, we can rewrite this as:

\(10^{2005} = (2^a*3^b*5^c*7^d)\)

That's about as far as I got! :/ sorry for the inconvenience!

Now adding on to what I did yesterday:

As far as we're concerned, this problem is now just asking us for which of the prime factor "four-tuples"(in 2, 3, 5, and 7) multiply to equal \(10^{2005}\). We clearly see that there is only 1 distinct tuple that gives us \(10^{2005}\), because 10 factors out into \((2*5)^{2005}\), meaning that a must be 2005, c must be 2005, and b and d are 0.

Yup, and exponential graphs are NEVER linear! Keep this in mind :D

Combinations!! That's it!!! Nice, jfan17.

Just in case you don't know what a combination is, the formula is n!/r!(n-r)!

Hope this helped!

Thanks, even though it was mostly jfan17 :D

Nice, jfan17!!!

Hey nice to see you again Magnus! This is another case of stars and bars like I was explaining to you earlier. Because we have 8 cherry candies, let's first look at how to distribute that to 4 children. That would be:

\({8+4-1 \choose 8}\)= \({11 \choose 8}\)= 165

Now that we have the number of ways to distribute the cherry candies, we can multiply that with the number of ways to distribute the melon candies(correct me if I'm wrong with this reasoning).

Distributing 7 things to 4 people, we get:

\({7+4-1 \choose 7}\)= \({10 \choose 7}\)= 120

165 * 120 = 19800 ways. Now that I think about it, I feel like my number is a bit high, but feel free to double check this reasoning!

Wow! Thank you guys so much for helping me! I learned so much from you guys! Hope you guys stay safe and have a good day!

yep!

Start with a "reference bead" that we can use to indicate the position of each bead, with the reference bead being one of the distinct colored beads. Next, realize that of the 5 remaining beads left, the possible ways to select the same-colored beads is \({5 \choose 2}\)

= 10 ways to pick the pair. After you pick where the pair goes, the remaining beads have 4! possibilities to go to(do you see why? The first bead has 4 places, the second has 3, etc.) The total number of possibilities is then 10 * 24 = 240 arrangements. If you see something wrong with my solution, plz lmk. I haven't been the best historically with counting  ;).

The minimum value is 2, which occurs for z = i.

1998 - 1950 =48 / 14 =3 3/7 - common difference

[1950 + 1998] / 2 x 15 =29,610

Would n = 50 and k = 4?

ok I'm not really sure why, but web2.0 calc keeps constantly flagging my responses for this question for moderation. I'll just explain the basic jist of it. This problem employs a common counting method known as "stars and bars." with stars and bars, you can find out how many ways there are to distribute n things into k groups. For stars and bars, there are two forms of it: non negative and positive stars and bars(in other words, non negative is where a group is allowed to have 0 things in it, and positive at least 1 thing). This problem is asking for the non negative form of stars and bars, which is :

\({n+k-1 \choose n}\) With this formula, the problem becomes a lot easier!

This is a really common method of counting known as "stars and bars". If you want more reference information, I'd highly recommend looking to wikipedia or other sources online. The basic idea is that we can think of 50 votes as "50 balls" or "stars" laid out in a row.

These 50 stars are divided by 3 "bars", which form 3 groups. You can picture it like this:

if the O's are balls, and the |'s are dividers, we have:

O O O O O O | O O O O O | O O O O | O O .. ..... x 50.

These bars divide how many votes each of the four candidate gets(can you see why there are 3 bars? because if n is the number of bars there are, n+1 regions are formed from n bars). 

Because we want to see how many votes each candidate could get, the answer is stars and bars' non negative form(there are two forms of stars and bars, watch out!)

With stars and bars, we get the equation:

(50+4-1) c (50),  or (53c50). This gives us 53*52*51/3!

This simplifies to:

53*26*17 = 23426 totals

1.  This is a textbook case of a method of counting known colloquially as "stars and bars" or "balls and urns"(I'd recommend you to search it up online). We can think of the 50 votes as 50 balls laid out in a row, divided by 3 "dividers" which separate the balls into 4 separate regions for each candidate, with each region being how many votes a candidate gets. Using stars and bars, we get the equation:

(50+4-1) c (50),  or (53c50). This gives us 53*52*51/3!

This simplifies to:

53*26*17 = 23426 totals

For stars and bars:

like the above example, just think of the votes as 50 "balls" that are divided by 3 dividers, which create 4 regions for vote allocation(for 4 candidates). the formula for stars and bars(non negative, meaning 0 is allowed) is:

(n + k -1) choose (n) 

where n is the number of things you want to distribute or "stars", and k is the number of groups you want to distribute those stars into.

I tried your way, but it wasn't correct. Thanks for trying though.

I used a method of counting known as "stars and bars". I'd recommend you search it up! It's really interesting imo

I'm confused on what equation is needed to solve this problem. I don't know which formula or something that relates to this problem. Also, I have seen the second part, but all I need to solve is the first part.