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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:
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Answer:
A
Step-by-step explanation:
The velocity of a moving body is given by the equation:
Is the velocity is <em>positive </em>(v>0), then our object will be moving <em>forwards</em>.
And if the velocity is negative (v<0), then our object will be moving <em>backwards</em>.
We want to find between which interval(s) is the object moving backwards. Hence, the second condition. Therefore:
By substitution:
Solve. To do so, we can first solve for <em>t</em> and then test values. By factoring:
Zero Product Property:
Now, by testing values for t<1, 1<t<4, and t>4, we see that:
So, the (only) interval for which <em>v</em> is <0 is the second interval: 1<t<4.
Hence, our answer is A.