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It's that time again! (Homework help)
Gigazombie Cybermage
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So I'm taking Math again (Didn't make it through the first time) and I need some help double checking my work. It's pretty simple, but he didn't really explain stuff very well so I'm a little unsure. I get how Pascal's Triangle works, like I said, easy stuff, but can't you use it to solve map questions like
Get from point A to B without backtracking:
B
|_|_|_|
|_|_|_|
A
If I'm not mistaken, aren't there ten ways to get from point a to b without backtracking?
He also didn't explain how to do those kinds of questions when the blocks on the map are weirdly shaped, unless I just do the same thing?
(edit: No idea what's up with point B up there. It's supposed to be on the other side)
Get from point A to B without backtracking:
B
|_|_|_|
|_|_|_|
A
If I'm not mistaken, aren't there ten ways to get from point a to b without backtracking?
He also didn't explain how to do those kinds of questions when the blocks on the map are weirdly shaped, unless I just do the same thing?
(edit: No idea what's up with point B up there. It's supposed to be on the other side)
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I'm thinking you are too low with only 10 paths, but if you could clarify what you mean with the drawing I'd be able to help further.
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TylerJ on League of Legends (it's free and fun!)
I labeled the edges we could traverse. We go from the 8-12 corner to the 3-7 corner.
So, we can go along the left and top (8-4-1-2-3). Or, we can shave a bit off the corner, cut through the middle a bit, which will give us a few more:
You should see the pattern there, yes? Notice also that if we reflect our path we can get several more. So again, start along the outside edge and go 12-13-14-11-7. Then, we can start cutting through the middle from that corner, like we did on the other side. But make sure you don't trace a path that you did already, which means you should get only two more paths by cutting through the middle.
So, so far we've got 7.
We can get a few more pretty easily, too. Take, for example, 12-9-5-2-3. See how that's a bit different? There's also the other very similar path, which you should hopefully see.
That's 9 then. Those were the easiest to find and not repeat or anything. That's because they were basically the most direct. We were always going the way we needed to go (right and up), just in different orders.
The problem doesn't tell you to find the shortest paths, though, but all of them. So what if we go in a sort of zig-zag pattern? Try... 8-4-1-5-9-13-10-6-3. We could also do another zig-zag, like 8-4-1-5-9-13-14-11-7.
Now I've got 11, which is more than you found. There are quite a few other paths still to find that require you to, at some point, go down or left.
I don't know a very "foolproof" way of doing it, I don't think there is, like, a method or anything. You just want to be very methodical when you do it. Start with the easy paths, vary things in a pattern to get some more paths, keep in mind if the shape is somehow symmetric so your paths be reflected and double your count with little effort on your part (just keep in mind to avoid overlaps!).
If I didn't understand the problem fully, just let me know. If you want me to walk through some more of them, just let me know. If you have any other questions, also just let me know :P.
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Here's how I think of it. To start, you have an n x m grid. You have two options, you can go up or you can go right. If you go up, you now have an n x (m-1) grid. And if you go right, you now have an (n-1) x m grid. So the solution for an n x m grid is just the sum of the solutions for the n x (m-1) and (n-1) x m grids.
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That's how I would have interpreted it, also. But that project Euler problem also says "without backtracking", and they interpreted it the way I described above. That's why I think the problem description is ambiguous.
Okay, I'm feeling ovewhelmed here, so I'll guess I'll just start with the easy stuff.
234 in scientific notation is "2.34x10^-2" right? I got being able to go back and forth, but the trouble is big problems with it.
Say: (2.5x10^3)(6.6x10^3)/8.25x10^3
How would you go about solving it without a calculator? Don't you start by getting it out of scientific notation?
Oh yeah, the other thing is about the Nature of Sets. I've seen a symbol that I can't identify no matter where I look. It's a "w" but it's similar to ℕ. I have no clue.
Can well-defined sets be something specific, yet kinda ridiculous? Like {All the grains of sand in the world} is -technically- well-defined isn't it despite being a mind-boggling huge set? I know that "subjective" things are not well-defined.
*deep breath*
Here's a problem directly from the book.
Write out in words the descriptions for the sets given and then list each in roster form.
{x|x∈ℕ, x≠8}
My answer was: {x is an element of the set of natural numbers and is not the number 8}
x = {1, 2, 3, 4, 5, 6, 7, 9, ...}
Sorry if that's completely idiotic, I'm out of my element.
(2.5x10^3)(6.6x10^3)/8.25x10^3 = 2.5*6.6*10^3/8.25 = 16.5*10^3/8.25=2*10^3
(If you need to exit scientific notation, it should solely be because you don't like looking at 2.5 * 6.6 but think 25 * 66 is nicer, or whichever other combination you feel to. Personally, I tend to split them into chunks 2.5*6.6=(2+0.5)*6.6 which reduces annoying multiplication into two simple multiplications and some addition. But this depends on what you personally think is easiest)
I'd say you've got the sets under control. Yes, all the grains of sand in the world is well-defined (the set of prettiest piecest of sand not so much), and the answer to the set question is correct - all the natural numbers except 8.
Calixtus basically nailed it, I just wanted to point out that 234 is 2.34*10^2, not 10^-2. The way I think of it is that positive powers are the number of zeroes after the one (so 10^2 is 100, two zeroes), and the opposite for negative powers, counting the zero before the point (so 10^-2 is 0.01 or 1/100, as mentioned).
No, that should be ten to the positive 2 power. 2.34*10^-2 is 0.0234 in standard.
Definitely not! The chief purpose of scientific notation is that is dramatically simplifies calculations.
Group your decimals and your powers of ten together in your mind and do each separately. (2.5*6.6/8.25 ) * (10^3 * 10^3/10^3). For the powers of ten remember your laws of exponents. When multiplying powers of the same base add exponents, when dividing powers of the same base, subtract exponents:
10^3 * 10^3 / 10^3 = 10^6 / 10^3 = 10^3
Once you're done your decimal answer may have more than one digit in front of the decimal, don't forget to adjust your decimal and increase or decrease your exponent as necessary.
That sounds like the symbol for the whole numbers which are simply the natural numbers plus zero.
I would say yes but for your purposes that depends on precisely how your book defines a well-defined set.
That translation appears to be fine. Your teacher may be happier seeing you change the words somewhat rather than simply stating in english how you would read the symbols because that suggests an understanding of the meaning rather than simple memorization of the words.
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Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{2, 6, 8} u {6, 8, 10}
For this first problem, is the answer {2, 6, 8, 9}?
{2, 6, 8} n {6, 8, 9}
And I got {6, 8} for this one.
I have to find the compliment of {1, 2, 5, 7, 9}, is the answer {3, 4, 6, 8}?
Second looks good.
Third looks good. Make sure you always keep track of what the Universe is, like you did.
I assume you understand why for each of these, and you are just checking to make sure you did it correctly?
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{x | x is a positive integer} n {x | x is a negative integer}, This one confuses, because they don't give a U set for these next set of problems. Is the answer x = {0}?
Also I have some odd ones like
natural numbers union whole numbers (Except the book uses the actual symbols of course) would I just write {0, 1, 2, 3...}?
and one problems asked for the compliment of an italicized letter "u", whatever the fuck that means, and another wants the compliment of null.
If they don't give a Universe and they don't ask for the compliment, then just take the Universe to be all numbers (or whatever makes sense in the context).
So with the Universe being everything, your answer looks to be incorrect. Unless you didn't give the whole problem? How did you get x = {0}?
I.... think so? I don't remember 100% what the difference is, but I think I remember and that seems right...
Ummm.... is the italicized "u" the Universe, maybe? And on that note, null is nothing, so what would be the opposite of nothing?
Set theory is kinda weird sometimes :P. Just imagine having a set with sets in it! And even more fun stuff!
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Would the compliment of the Universal set be nullset (and vice versa)? This stuff makes my head hurt. And what the hell would universal set union nullset look like? *headdesk*
Let U = {1, 2, 3, 4, 5, 6, 7}, A = {1, 2, 3, 4}, B = {1, 2, 5, 6}, C = {3, 5, 7}
I already did some problems.
A u C = {1, 2, 3, 4, 5, 7}
A n C = {3}
Complement of C = {1, 2, 4, 6}
Complement of U = { }
Then I get to this one.
{ } u A = ???
0 is considered neither a positive integer nor a negative integer, so what would be the union, then? What do they have in common?
The compliments look right, yes. As for the union, what would you get if you put together all the elements of the Universe and the null set?
(As a note, when doing some googling, I learned that "null set" actually means something different than "empty set," but that most textbooks that are not super-high-level use null set to mean empty set.
That goes with any union with the null set. Union means you basically put allll the elements of both sets into a set, and remove the repeats (that's the straightforward way to do it). So what happens when you put alll the elements of a set A and of { } into a set? What would your answer look like?
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So {1, 2, 3, 4} u { } = {1, 2, 3, 4}
The second set contains nothing, so when you put them together, you get the first set + nothing. So yeah, it is that simple
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Let U = {1, 2, 3, 4, 5, 6, 7}, A = {1, 2, 3, 4}, B = {1, 2, 5, 6}, C = {3, 5, 7}
List all the members in each of the sets in problems 3-10
3. (A u
{1, 2, 3, 4, 5, 6} n C = {3, 5}
Does that look right?
7. "Complement of A" u (B n C)
"Complement of A" u {5} = {5, 6, 7}
I'm drawing Venn diagrams now. In "A Union B Complement" do you shade in the intersection? Does B complement mean that nothing should be in B? (Cuz I have vertical lines where they intersect)
Both of those look good, yeah. It's pretty straightforward if you just remember what everything means :P
What does the "complement" mean? That should answer your question.
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I whipped this up real quick. Is it right?
Yeah, that's a good strategy, if you understand what it means
So you would vertically draw lines in A, and horizontally in B, THEN it matters what the question is. If it's union, then wherever you have ANY lines you would shade. If it's intersection, then where there are BOTH you would "shade" as your answer.
Or if you have A u ~B (that's "not B"), then you would put horizontal lines in A, then IGNORE A and just look at B, and put vertical lines everywhere that is "not B". Then, where they cross, you have your answer (for intersection). For union, it would be anywhere at all you have drawn lines, horizontal or vertical or both.
So.... your answer is right. (Edit: I'm not sure if my following concern is actually well-founded. It's late and now I see that I misunderstood a bit. Just be careful and you'll be fine :P) I'm a bit concerned by the lines I see, though. It looks like you did vertical lines for "not (A u
In the future, I would probably ignore the "not," do the rest of it, and then just take the complement of what you get as the answer to get the final answer. Make sense? It's late, not sure how rambling I'm getting :P
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I'm out of time at the moment, but I can answer your other question later if its still unanswered
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My final diagram shaded in the intersect of a and c, and everything inside b. Does that sound right?
Dividing is subtraction wrt exponents. ie (x^6)/(x^2) = (x*x*x*x*x*x)/(x*x), which (just be cancelling) clearly is equal to x*x*x*x. So x^4. Which is also x^(6-2)
Sounds good, yeah.
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