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October 24, 2025, 08:54:41 pm

Author Topic: My Solutions (Core; Graphs; Networks; Matrices)  (Read 29980 times)  Share 

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djsandals

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #30 on: November 02, 2012, 03:40:33 pm »
Is question 8 in graphs and relations not B? Because it says at least 5 loaves of white bread are made for each loaf of brown bread...so wouldn't that mean that one white = five or more brown? 

aka w =< 5b?
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depressedchild

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #31 on: November 02, 2012, 03:41:07 pm »
can someone clarify graphs and relations q7? I got E
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Will T

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #32 on: November 02, 2012, 03:47:29 pm »
How can you transform a complete graph into a planar graph? I believe it is impossible when the number of vertices is greater than 4. I can see how you can transform the complete graph with No. of vertices = 4. But the Pentagon, Heptagon and Nonagon shapes are not able to be transformed into a planar graph?
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Varunchka

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #33 on: November 02, 2012, 03:52:40 pm »
can someone clarify graphs and relations q7? I got E

Me too, but turns out it's wrong, the k value is 5/2
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Yendall

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #34 on: November 02, 2012, 03:54:08 pm »
How can you transform a complete graph into a planar graph? I believe it is impossible when the number of vertices is greater than 4. I can see how you can transform the complete graph with No. of vertices = 4. But the Pentagon, Heptagon and Nonagon shapes are not able to be transformed into a planar graph?
Euler circuits can exist on non-planar graphs. I don't know why I said planar transformation, must've been thinking of something else. The fact that three of the graphs have non-odd vertices shows that an Eulerian circuit is possible.
« Last Edit: November 02, 2012, 03:57:25 pm by Yendall »
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julie9300

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #35 on: November 02, 2012, 03:54:54 pm »
Euler circuits exist when all vertices have an even degree.
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wallaced93

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #36 on: November 02, 2012, 03:56:57 pm »
Guys i'm pretty sure networks question 7 is A.
Line 1 was 250+60=310. 300 and 120 don't count towards the cut.

Yendall

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #37 on: November 02, 2012, 03:58:02 pm »
Guys i'm pretty sure networks question 7 is A.
Line 1 was 250+60=310. 300 and 120 don't count towards the cut.
The source is down the bottom, not to the left. 120 is flowing out of the cut.
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julie9300

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #38 on: November 02, 2012, 03:59:34 pm »
Guys i'm pretty sure networks question 7 is A.
Line 1 was 250+60=310. 300 and 120 don't count towards the cut.

The cut is suppose to be drawn in a way that if you remove those lines, you can't get from one end to the next. If you cut out the lines in line 1, you could still get from the town to the freeway.
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Will T

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #39 on: November 02, 2012, 04:00:11 pm »
Guys i'm pretty sure networks question 7 is A.
Line 1 was 250+60=310. 300 and 120 don't count towards the cut.

You made the same mistake as me! I feel slightly better now knowing I wasn't the only one.
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wallaced93

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #40 on: November 02, 2012, 04:05:28 pm »
Guys i'm pretty sure networks question 7 is A.
Line 1 was 250+60=310. 300 and 120 don't count towards the cut.

The cut is suppose to be drawn in a way that if you remove those lines, you can't get from one end to the next. If you cut out the lines in line 1, you could still get from the town to the freeway.
Oh crap you're right! Damnnn  :-[
What's the general consensus on Networks Q9? I've seen some places say D and some E. I have a feeling it's E though.

Yendall

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #41 on: November 02, 2012, 04:06:51 pm »
Guys i'm pretty sure networks question 7 is A.
Line 1 was 250+60=310. 300 and 120 don't count towards the cut.

The cut is suppose to be drawn in a way that if you remove those lines, you can't get from one end to the next. If you cut out the lines in line 1, you could still get from the town to the freeway.
Oh crap you're right! Damnnn  :-[
What's the general consensus on Networks Q9? I've seen some places say D and some E. I have a feeling it's E though.
I didn't even know what to do with that question haha I said 17 :p
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oneoneoneone

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #42 on: November 02, 2012, 04:24:49 pm »
I drew a really detailed diagram and found that D worked, so it can't be E. I'll redraw it and upload it soonish.

Edit: Attached an image of Q9 Networks
« Last Edit: November 02, 2012, 04:36:05 pm by oneoneoneone »

mooimachicken

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #43 on: November 02, 2012, 04:29:07 pm »
The last question in business maths is C, not A

michelleeee

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #44 on: November 02, 2012, 04:31:05 pm »
nooo whaat D:
for graphs and relations - question 8
isnt it A??? i have a formula that i got from my teacher that says
"for each x there is at least 5y" and then its 2y<(including arrow)x

:(