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Q: How to simplify n plus 1 factorial divided by n factorial?

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Factor the numerator. x+1 is one of its factors (otherwise, it wouldn't be possible to simplify it). Then cancel the identical factors in the numerator and the denominator.

9a + 1

2

3x + 8

x+5 is a factor of x2+4x-5 use synthetic division to learn that the other factor is x-1

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How about you pay attention in class so you don't have to ask the computer for the answers.be smart listen to your teacher. Just kidding ,i'm just thirteen what do i know

145 1! = 1 4! = 24 5! = 120

Factor the numerator. x+1 is one of its factors (otherwise, it wouldn't be possible to simplify it). Then cancel the identical factors in the numerator and the denominator.

unsigned __int64 factorial(unsigned __int64 number) { unsigned __int64 factorial=1; while(number) factorial*=number--; return(factorial); }

Use this function: long factorial(int N){if (N == 0){return 1;}else{return N*factorial(N-1);}}

9a + 1

2

simply, any number divided by 0 is 0.

1 factorial = 1

Actually, I don't really think that you can simplify it, because if you convert them into fractions, there all over 1 and you can't simplify it. You can simplify it if it's: 5 plus 7 DIVIDED by 3. That was when you divide the 12 by 3 in fraction form, the 3 and one switch, and u can simplify 3 into 1 and 12 into four, so there by the answer is 4. (Sorry if confusing)

Factorial in C++ is the same as factorial in mathematics. For a given integer, N, the factorial, denoted N!, is the product of all integers in the closed range 1 to N, where 0! is 1. The problem with factorials is that the largest factorial you can store in a 64-bit integer is 20!. To cater for larger factorials you need a numeric library capable of handling larger integers, such as the GMP library.

(a^2+7a-8)/(a^2+6a-7) (original equation) (a+8)(a-1)/(a+7)(a-1) (simplify top and bottom) (a+8)/(a+7) <--answer

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