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Q: What are the answers to mathbitscom's ah-bach series questions?
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How do you find out how many possible answers there are to the following question 10 questions and the answer is either a or b for each question?

10 questions with 2 answers each equals 20 possible answers. The combination of possible answers for the series would be much higher starting with they could be all a or all b then the first one a and the rest all b then the first two a the rest b etc


What are the answers to fourth grade Houghton Miffin Harcourt chapter9 page217?

The answers are in the book, so read it and get the answers. I know the series and the answers are ALWAYS given usually in order of how it is in the book.


Why you use series expansion ie Taylor or maclaron or any other series during solving problems?

Some functions cannot be evaluated by normal means and you need an infinite series to get approximate answers.


39 clues card combinations?

You can start combining cards on the 39 clues after book four comes out. The "my cards" page will be up graded. The cards that you have already will combine automatically, just go to my clues and you will have new clues. To know if you have a card that can combine in the series two, look in the top left corner. If you see a key your card can combine with other cards with a key with the same symbol such As an alien or Africa. Ask more questions because I know the answers, but I will not give away card codes or puzzle answers.


What is complete series 2 9 1 7 9?

There are many possible answers. One possible answer is to fit the infinite series defined by the quartic equation: t(n) = (-47n4 + 586n3 - 2521n2 + 4334n -2304)/24 for n = 1, 2, 3, ... This gives the next two points as 58 and 206. Being a polynomial. there are infinitely more points so the series is never "complete".

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How do you find out how many possible answers there are to the following question 10 questions and the answer is either a or b for each question?

10 questions with 2 answers each equals 20 possible answers. The combination of possible answers for the series would be much higher starting with they could be all a or all b then the first one a and the rest all b then the first two a the rest b etc


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