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Erzeon
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How does ((10-n)!)/((8-n)!) = (10-n)(9-n)? I take my 2nd exam tomorrow and I came across this question
Erzeon said:How does ((10-n)!)/((8-n)!) = (10-n)(9-n)? I take my 2nd exam tomorrow and I came across this question
Erzeon said:I think I found out, is it because if you expand the factorials out all the way to 0, the (8-n),(7-n),(6-n),(5-n),(4-n),(3-n),(2-n),(1-n) all cancel out in the numerator and denominator? So (10-n) and (9-n) is left?
Erzeon said:Thanks, I was blind to not see it.
Erzeon said:What do you mean by definition?
six789 said:no problem... we are all here to benefit frm each others knowledge...
Tom Mattson said:Yes, but we do have rules here, which you all agreed to. We don't give assistance until the person asking the question shows an attempt at the problem. Guiding questions are OK, but complete solutions are not.
The equation (10-n)/(8-n) is used to find the value of a variable when given a certain numerical relationship between two other variables. It is commonly used in algebra and can be used to solve for n in various equations.
The parentheses in (10-n)/(8-n) indicate that the operations inside should be performed first. In this case, the numerator should be evaluated before dividing by the denominator. Without the parentheses, the equation would have a different meaning and the answer would be incorrect.
Yes, the equation (10-n)/(8-n) can be simplified to 1+(2/(8-n)). This can be done by using the distributive property to expand the numerator and then simplifying the resulting terms.
The domain of the equation (10-n)/(8-n) is all real numbers except for n=8, as this would result in division by zero. In other words, the value of n can be any real number except 8.
Yes, the equation (10-n)/(8-n) can be used to solve real-world problems that involve numerical relationships between variables. For example, it can be used to calculate the speed of an object when given the distance and time it traveled.