SOLUTION: What is n in the equation?
n!/84 = (n-2)!/[(n-2)-(n-4)]
do I cancel n-2 and n-4 on the right side?
Algebra.Com
Question 1139281: What is n in the equation?
n!/84 = (n-2)!/[(n-2)-(n-4)]
do I cancel n-2 and n-4 on the right side?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
You can't cancel the (n-2) and (n-4) in the denominator on the right; they are not factors.
The denominator on the right is simply (n-2)-(n-4) = 2.
Then write the n! on the left as n(n-1)(n-2)!; do some canceling of common factors; cross-multiply; and solve.
RELATED QUESTIONS
nP4=84nC2 Solve for n using factorial notation
I'm really struggling with this... (answered by ikleyn)
if n = -5, what numbers do n, n + 2, n + 4, and n + 6... (answered by richard1234)
I've been working through factoring polynomials and have solved the following as such;... (answered by stanbon)
-4 n( n+ 2) =... (answered by stanbon)
4*n-2=34 what is n?
(answered by josgarithmetic)
I have tried this problem for rational equations and I am unable to get the "work shown"... (answered by scott8148)
n!
_______
2!(n-2)!
I understand that 5! is 5*4*3*2*1
I do not understand how to... (answered by Earlsdon,stanbon)
Dear math teacher,
I am having difficulties solving for n in the following problem:
(answered by Theo)
Solve via direct proof:
n^5 - 64n^3 - n^2 ∈ Θ (n^5)
I'm not exactly sure... (answered by ikleyn)