LoJao(37-3八):log 20 (3rniay)
1 answer:
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
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Answer:
Step-by-step explanation:
A) From the order of the exercise we already know that the intersection points lies on the Y-axis, so its coordinates are P(0;y;0). In order to find it, we only need to substitute the equation 4x+4z=0 into the equation 4x+3y+4z=1. Then,
1=4x+3y+4z = 3y + (4x+4z)= 3y+0.
From the expression above it is easy to obtain that y=1/3, and the intersection point is P(0;1/3;0).
B) To obtain the parallel vector to both planes we use the cross product of the normal vector of the planes.
As we want a unit vector, we must calculate the modulus of u:
.
Thus, the wanted vector is . Therefore,
.
C) In order to obtain the vector equation of the intersection line of both planes, we just need to put together the above results.
where is a real number.