<p>((x-5)/4)<sup>2</sup>- ((y-3)/9)<sup>2</sup>=1      How does one solve this equation?</p>
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<p>"A matrix A is said to be skew symmetric if A^T = - A.Show that if a matrix is skew symmetric, then its diagonal must allbe 0."</p><p>Where A^T works as such: say you have a 2x3 matrix suchthat row one is [ 1 2 3 ] and row two is [ 4 5 6 ]. Then the resultwould be a 3x2 matrix such that the first Column is 1, 2, 3 and thesecond Column is 4,5,6 {Sorry, can't seem to put matrices in here.}</p><p>I roughly understand how A^T=-A but I have no idea how toprove it and have been stuck on it for a couple days. Any helpwould be very much appreciated. </p>
For all the problems, all of the greater thanand less than signs (>,<) should be a greater/less than equalto signs.
-Solve using the simplexmethod:
1) Maximizesubject: p= x + 2y +3z
3x + y+ 3z<1
2x + y + z<2
x> 0, y> 0, z >0
2) Minimize subject to: c=10s + 3t,
5s + 4t > 4
2s + 3t > 2
s> 0, t> 0
-Solve:
3)Maximizesubject to P = 3x+ 2y + z
2x + 3y + z <30
x - 2y + z > 20
2x + 5y + 2z >0
x > 0, y > 0, z >0
I know The answer for these question , but the graph missed ,,,So >> The question here is to Graph the answers
Here is the answer
a) |2y+5|>1
so -1 <2y+5< +1
-6<2y<-4
so -3<y<2 answer
b) 7+ |3y-2| <= 10
|3y-2| <= 3
-3<3y-2<3
-1<3y<5
so -1/3 < y < 5/3
c) |6x+5| >= -5
5 > 6x+5 > -5
-10 < 6x < 0
-10/6 < x
d) |3x| < -1
not possible a positive quantity can not be negative
NOW JUST GRAPH .