Let function be defined as follows: (x)=x!+(x+1)!/(x+2)!. What is

Question

Let function # be defined as follows:  \#(x)=\frac{x!+(x+1)!}{(x+2)!} . What is the value of #(6)?

A. 1/10
 
B. 1/9
 
C. 1/8
 
D. 1/7
 
E. 1/6

sshwetima · Answer

#(x) = x!* / 
#(x) = 1/(x+1)
#(6) = 1/7

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Charli08 · Answer

#6 = 6! + 7! / 8!

Factor out 6!:  = 6!(1+7) / 6!(8*7)

Cancel out the 6!'s  = 8/56 = 1/7

==>  D

BrushMyQuant · Answer

\#(x)=\frac{x!+(x+1)!}{(x+2)!}

To find the value of #(6) we will replace x with 6 in  \#(x)=\frac{x!+(x+1)!}{(x+2)!}

=>  \#(6)=\frac{6!+(6+1)!}{(6+2)!} = \frac{6!+7!}{8!} = \frac{6!+7*6!}{8*7*6!}
= \frac{6! (1 + 7)}{8*7*6!} = \frac{8}{8*7}  
= 1/7

So,  Answer will be D 
Hope it helps!

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