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Mathematics
OpenStudy (anonymous):

Solve: (n+2)!/(n-1)! = 210

OpenStudy (anonymous):

1/(n-1)(n)(n+1)=210

OpenStudy (anonymous):

??

OpenStudy (agreene):

\[\frac{(n+2)!}{(n+1)!}=n(n+1)(n+2)\] it is a bit more straight forward when you realise that.

OpenStudy (anonymous):

Mr.klinker is 35 and his daughter is 10 , in how many years will mr.klinker be twice as old as his daughter? ~~PLZ HELP

OpenStudy (anonymous):

\[\frac{(n+2)!}{(n-1)!}=(n+2)(n+1)(n)\]

OpenStudy (anonymous):

can someone wplease explain how to do a problem like this

OpenStudy (anonymous):

i i get it i was doing it as plus one but its minus one

OpenStudy (anonymous):

so basically you have 3 consecutive integers whose product is 210, which gives n = 5

OpenStudy (anonymous):

Mr.klinker is 35 and his daughter is 10 , in how many years will mr.klinker be twice as old as his daughter? ~~PLZ HELP sorry i posted here but wont let me post

OpenStudy (anonymous):

Thank you guys! :) I got it now.

OpenStudy (anonymous):

thank you satelite73 ! but how would you find the answer

OpenStudy (anonymous):

15

OpenStudy (anonymous):

years

OpenStudy (anonymous):

well for a small number like 210 i would guess and check

OpenStudy (anonymous):

35+x=2(10+x) 35+x=20+2x 15=x

OpenStudy (anonymous):

thanks!

OpenStudy (anonymous):

?? what is this x = 15?

OpenStudy (anonymous):

zalxo asked this: Mr.klinker is 35 and his daughter is 10 , in how many years will mr.klinker be twice as old as his daughter? ~~PLZ HELP sorry i posted here but wont let me post

OpenStudy (anonymous):

so for my second question George is 7 and his mother is 37. In how manyyears will his mother be 3 times as old as he is. would the answer be 37+x=3(7+x) 37+x=21+2x

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