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Hellas
Username
Hellas
Score
14
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7
0 Questions
7 Answers
#1
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0
3^(x-1)-3^(x)=-6 => -(3^χ)/3-3^χ=-6 =>3^-3*3^χ=-18 => -2*3^χ=-18 => 3^χ=9
=> 3^χ=3^2 => χ=2
Hellas
Oct 2, 2012
#1
+14
0
3x-2(5x-3)=7 => 3χ-2*5χ+2*3=7 => 3χ-10χ+6=7 => 7χ=7-6 => 7χ/7=1/7 =>χ=1/7
Hellas
Oct 2, 2012
#3
+14
0
If you can tell me exactly what you want to factorization may be able to help
Hellas
Sep 30, 2012
#1
+14
0
what you think you can do is (4ψ^2-3ψ-5)/2ψ^2+2ψ-2)
=[2ψ(2ψ-1)-ψ-5]/[2ψ(ψ+1)-2]
Hellas
Sep 30, 2012
#1
+14
0
8-4x<3-5x => 5χ-4χ<3-8 => χ<-5
Hellas
Sep 30, 2012
#3
+14
0
The correct answer is χ^3-χ^2χ+1 => χ^2*(χ-1)+χ-1 => (χ^2+1)*(χ-1)
Hellas
Sep 30, 2012
#3
+14
0
if you mean i^17
17/4=16 with remainder 1 so i^17=i^1=i
Hellas
Sep 30, 2012
