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
#SPJ1
V(cylinder)=πR²H
Radius of the cylinder R=x, height of the cylinder H=y.
We can write for the cylinder
V(cylinder)=πx²y
V(cone) =(1/3)πr²h
Radius of the cone r=2x.
We can write for the cone
V(cone)= (1/3)π(2x)²h=(1/3)π *4*x²h
V(cylinder) =V(cone)
πx²y=(1/3)π *4*x²h
y=(4/3)*h
h=(3/4)*y
Answer:
y-7 = 4(x+9)
Step-by-step explanation:
Point slope form is
y-y1 = m(x-x1)
where m is the slope and (x1,y1) is the point
y-7 = 4(x--9)
y-7 = 4(x+9)