This is the equation which you can add:
![[c + (3(c + 2))] + [5c]](https://tex.z-dn.net/?f=%5Bc%20%2B%20%283%28c%20%2B%202%29%29%5D%20%2B%20%5B5c%5D)
Simplify:
![[c + 3c + 6] + [5c] = 9c + 6](https://tex.z-dn.net/?f=%5Bc%20%2B%203c%20%2B%206%5D%20%2B%20%5B5c%5D%20%3D%209c%20%2B%206)
And, that's your answer!
Simplify:
And, that's your answer!
[67-(9+11)÷10]-[32÷(19-17]
2 answers:
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Hello,
if the question is to solve the equation:
1) log must exist:
a)==>2r²-27>0==>(r-√(27/2))(r+√(27/2))==> outside the roots
r<-√(27/2) or r>√(27/2)
b)==>3r²+12r>0 ==>3r(r+4)>0 ==>outside the roots
r>0 or r<-4
==>r<-4 or r>√(27/2)=3.6742...
If this is realized:
2r-27=3r²+12r
==>r²+12r+27=0
Δ=12²-4*27=36=6²
r=(-12+6)/2=-3 or r=(-12-6)/2=-9
One solution r=-9
Proof
if the question is to solve the equation:
1) log must exist:
a)==>2r²-27>0==>(r-√(27/2))(r+√(27/2))==> outside the roots
r<-√(27/2) or r>√(27/2)
b)==>3r²+12r>0 ==>3r(r+4)>0 ==>outside the roots
r>0 or r<-4
==>r<-4 or r>√(27/2)=3.6742...
If this is realized:
2r-27=3r²+12r
==>r²+12r+27=0
Δ=12²-4*27=36=6²
r=(-12+6)/2=-3 or r=(-12-6)/2=-9
One solution r=-9
Proof