The image is decomposed as follows: H1 and H2. Where original graph is Hx.
<h3>Are the images (attached) valid decompositions of the original graph?</h3>
- Yes, they are because, H1 and H1 are both sub-graphs of Hx; also
- H1 ∪ H2 = Hx
- They have no edges in common.
Hence, {H1 , H2} are valid decomposition of G.
<h3>What is a Graph Decomposition?</h3>
A decomposition of a graph Hx is a set of edge-disjoints sub graphs of H, H1, H2, ......Hn, such that UHi = Hx
See the attached for the Image Hx - Pre decomposed and the image after the graph decomposition.
Learn more about decomposition:
brainly.com/question/27883280
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Answer:
b^2-4b+3=0
b²-3x-b+3=0
b(b-3)-1(b-3)=0
(b-3)(b-1)=0
either
b=3 or b=1
.
2n^2 + 7 = -4n + 5
2n²+4n+7-5=0
2n²+4n+2=0
2(n²+2n+1)=0
(n+1)²=0/2
:.n=-1
.
x - 3x^2 = 5+ 2x - x^2
0=5+ 2x - x^2-x +3x^2
0=5+x+2x²
2x²+x+5=0
comparing above equation with ax²+bx +c we get
a=2
b=1
c=5
x={-b±√(b²-4ac)}/2a ={-1±√(1²-4×2×5)}/2×1
={-1±√-39}/2
Answer:
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Step-by-step explanation:
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Answer:
here:
1. -9
2. -28
3. 54
4. -100
5. -60
6. -0
7. 49
8. -135
9. -96
10. 300
11. -153
12. -192
THIS TOOK LONG BECAUSE I DID A LONG ONE BUT THEN I MADE IT SHORTER. :D
Let the require value be x, then
P(z > x) = 0.025
1 - P(z < x) = 0.025
P(z < x) = 1 - 0.025 = 0.975
P(z < x) = P(z < 1.96)
x = 1.96
Therefore, <span>the positive critical value that corresponds to a right tail area of 0.025</span> is 1.96