I’m confused about the ! being outside of the radical in the numerator but inside in the denominator. It would be better to have parenthesis around the radical 9, before the ! . That way you know to take the square root of nine, then do the factorial.
Creo que la raíz de 9 debería estar entre paréntesis para distinguir factorial de 3 de factorial de 9. En el primer caso resultará = 6. y en segundo caso = 12 por. raíz de 21.
It appears that the square root was not including the factorial sign. The denominator appeared to include the factorial sign. Since there was a difference in the length of the radical between the two I think it was intentional and so it would be correct.
this problem is disturbingly bad written .. even if u follow PEMDAS .. very ambigiuous is the Problem (√9)!/√(5!), or √(9)!/√(5!), or √(9!)/√(5!), or (√9!)/√(5!)? ... √9! = 3! or root of 9! ? (√9)! = 3! √(9)! =3! √(9!) = root of 9! or 9!=72√70 (√9!) Root of 3! or 9!=72√70 ? ..unless giving sevral different soultions seems hard to solve... √ is exponentiaton, ! (factorial) is a function() ..can be interprited as square root of a funtion of(9) OR a function() of squareroot of 9 ..here PEMDAS fails miserbly...
I’m confused about the ! being outside of the radical in the numerator but inside in the denominator. It would be better to have parenthesis around the radical 9, before the ! . That way you know to take the square root of nine, then do the factorial.
Estoy de acuerdo con Vd.
6/(120)^1/2 => 6/2(30)^1/2 = 3/(30)^1/2 ~= 3/5,447
Answer:
SQRT(30)/10≈.5477...
Creo que la raíz de 9 debería estar entre paréntesis para distinguir factorial de 3 de factorial de 9.
En el primer caso resultará = 6. y en segundo caso = 12 por. raíz de 21.
Way too many steps,
3!/sqrt(5!) × (sqrt(5!)/sqrt(5!))
=(3!×sqrt(5!))/5!
=sqrt(5!)/5×4
=sqrt (120)/20
=sqrt(30)/10
There's a mistake in the simplification of the numerator. √9! is widely different from 3!. so, the outcome is false.
It appears that the square root was not including the factorial sign. The denominator appeared to include the factorial sign. Since there was a difference in the length of the radical between the two I think it was intentional and so it would be correct.
yes
A Nice Olympiad Factorial Problem: (√9)!/√(5!) = ?
(√9)!/√(5!) = 3!/√[(5)(4)(3!)] = (√3!)/√[(5)(4)] = (√6)/(2√5) = (√30)/10
9!/5! = 3024
this problem is disturbingly bad written .. even if u follow PEMDAS .. very ambigiuous
is the Problem (√9)!/√(5!), or √(9)!/√(5!), or √(9!)/√(5!), or (√9!)/√(5!)? ...
√9! = 3! or root of 9! ?
(√9)! = 3!
√(9)! =3!
√(9!) = root of 9! or 9!=72√70
(√9!) Root of 3! or 9!=72√70 ?
..unless giving sevral different soultions seems hard to solve...
√ is exponentiaton, ! (factorial) is a function()
..can be interprited as square root of a funtion of(9) OR a function() of squareroot of 9
..here PEMDAS fails miserbly...
Well done, but not the most interesting problem
√9!/√(5!)
= 3!/√120
= 6/√2³*3*5
= 6/2√30
= 3/√30
= 3√30/30
= √30/10
Sorry thumbs down due to poor layout of problem.