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I think the LaTex glitched, so I can't see P. 

Also, the question isn't answerable without the diagram...

I don't understand what you're doing there. Once you get 57, you basically start over the process. You do the same thing you did to 12345 to 57.

If 57 was the FINAL result, it would make sense. But your question says: Take this last sum, double it and add it to the next number, which would be: 57 x 2 + 6=120, and double this to: 120 x 2 + 7=247......and if you keep doing this, then the 2021st term will be a HUGE number about 610 digits long!!!

The Letter P is after the word Zand and the diagram I can't upload

\(\frac{888}{1000}=\boxed{\frac{111}{125}}\)

I think it's asking you to do the same 57, which would turn into 5*2+7=17, and then do it on 17, which would turn into 1*2+7=9.

∑[  (2 n^2 + 12 n + 15),  n,  1,  5] ==365

Not sure if I understand your question!

Let us try a small number: 12345

2 x 1 + 2=4

2 x 4 + 3 = 11

2 x 11 + 4 = 26

2 x 26 + 5 = 57.  Now, what is next? Are you supposed to do the same with: 4, 11, 26, 57 ?? If so, this will take a long time without a computer code!. 19, 64, 185 and this becomes: 102, 389 and this becomes: 593 and this becomes: 79 and this becomes: 23 and this becomes: 7

Does this make sense to you? It makes no sense to me !!

This is an arithmetic sequence with common difference 7, so the next two numbers are 66 and 73.

There are: 

818 four-digit "wondeful" numbers

P.S. These numbers were found using C++ programming language.

\(x=\boxed{-\frac{2}{3},-4}\)

Seriously? Test question? Welp your test is over anyways, so the answer is \(\boxed{7}\) hours.

 (6^(273)+7^(273)+8^(273)+9^(273)) mod 49==1 - the remainder

The food cost \(\boxed{\$90}\), and the drinks cost \(\boxed{\$18}\). Each girl owes \(\boxed{\$36}\).

How many values of \(x\) satisfy both of the following conditions?

\((a)\) \(x\) is a multiple of \(7,\) not necessarily positive.

\((b)\) \(x^2\) is less than \(500\).

If \(x^2\) is \(500,\) that means \(-22\le x\le22\). Also, since \(x\) is a multiple of \(7,\) we can list all the multiples of 7 between -22 and 22 out.

-21, -14, -7, 0, 7, 14, 21

Therefore, since there are 7 terms in the list, the answer is \(\boxed{7}\).

1 ,2 ,3 ,10 ,11 ,12 ,13 ,20 ,21 ,22 ,23 ,30 ,31 ,32 ,33 ,100...........133130 ,133131 ,133132 ,133133 ,133200 ,133201 ,133202 ,133203 ,133210 ,133211)

The 2021st term ==133,211

P.S. This sequence was calculated by computer using C++ programming language.

Press the 2nd key and then the test/math key.

a(8-3x)+b=x

-3ax+8a+b=x

a would need to be -1/3 in order to keep the x term on the left isolated. 8a+b would need to be equal to 0, so be would be 8/3. 8/3+(-1/3)=7/3.

I assume \(N\) is for natural numbers and \(Z\) is for integers.

1.

A = {1, 2, 3, 4, 5, 6}

B = {4, 7, 10, 13}

Subsets of B are any combination of the elements of B (excluding some).

2. 

If A is contained within B, and B is contained with C, A is contained within C. 

This is, I suppose, one of those annoying proofs which I take for granted and see no way of prooving it other than it is logical. It is here I should also note that I have never studied set theory, and I'm probably getting everything wrong .

3.

{4(3a+b)|a,b \(\in\)Z} = {4c|c\(\in\)Z}

{3a+b|a,b\(\in\)Z}={c|c\(\in\)Z}

3a+b\(\in\)Z

c\(\in\)Z

(I don't know what to do after that....)

Again, I've never studied this, so apologies in advance for faulty logic everywhere.

Thanks CPhill!

The greatest commom divisor of the three dimensions:   

"Das also war des Pudels Kern!"

  !

thank you!

Thx for pointing out my mistake....now  corrected  !!!

wouldnt 2022/5^2 be 80 not 88?

Divide 2022  by successive positive integer powers of 5 until 2022/ 5^n  produces a division < 1

integer [ 2022 / 5^1] =  404

integer [ 2022/ 5^2 ] = 80

integer [ 2022/ 5^3 ] = 16

integer [2022/ 5^4 ]  = 3

integer [ 2022 / 5^5 ]  = 0

N =  404 + 80 +16 + 3   =  503   CORRECTED  !!!!