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Harvest Sprite
post Sep 11 2009, 02:10 AM
Post #41


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(IMG:http://www.dailycognition.com/content/image/15/99692_10.jpg)

Vulcanized rubber is a very menacing thing. D:

My math teacher thought it would be funny to give us a bunch of logic problems for the first week.... >_> Luckily its the same sheet as 3 years ago :3
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Lindstrom
post Sep 11 2009, 02:11 AM
Post #42


irgendwie, irgendwo, irgendwann
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Vulcanized rubber = tires, bottoms of shoes, and hockey pucks. I can find you a picture of a tire if you need one.
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Katie
post Sep 11 2009, 02:15 AM
Post #43


--valiant||
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Ahhh, okay. That makes a lot more sense. It sounds a lot more complicated by its name than it is, then. Suuupear. Thanks, guys. And nah, I can draw one. I just expected it to be some crazy weird thing that is hard to explain. BD
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Lindstrom
post Sep 11 2009, 02:16 AM
Post #44


irgendwie, irgendwo, irgendwann
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My offer still stands.
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Saffy
post Sep 11 2009, 02:21 AM
Post #45


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There's two solutions for the problem (at least). It all depends on if you're asking the user for a number and then retrieving it through cin, gets, getline, or anything similar to those. Or are you passing an int into a function?

Assuming you've just learned cin -- such a downer, I wanted to teach you math -- you'd just traverse each character of the array and check against the value at index 0. If they all match then you're good to go.
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Rui L'cie
post Sep 11 2009, 02:28 AM
Post #46


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hm, our lecturer hasnt taught us any of those functions(cin, gets, getline, ...) before.
So I guess we have to keep the things simple and basic, like comparing each of the digits
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Saffy
post Sep 11 2009, 02:29 AM
Post #47


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So you haven't learned how to retrieve user input then? Ah, okay. So are you expected to just create a function which takes in an int? Then we can do some math :]
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Rui L'cie
post Sep 11 2009, 02:49 AM
Post #48


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Yep, we havent learned that..
Please guide me Saffy-sensei :3
I have to pass this subject no matter what it takes
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Saffy
post Sep 11 2009, 02:51 AM
Post #49


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Well, what do you know so far? I'm not entirely sure how I can help you quite yet.
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Jason
post Sep 11 2009, 03:16 AM
Post #50


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Rui L'cie, is this what you're trying to make?

Attached File  ex1.zip ( 23.95K ) Number of downloads: 69
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Rui L'cie
post Sep 11 2009, 11:18 AM
Post #51


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@Jason
Just call me Rui :D
Yea, I think the result should be something like that..! You're awesome..!
Thank you for the help..could you show me the codings? :3
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Jason
post Sep 11 2009, 12:59 PM
Post #52


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I can't just give you the code, but I will try to help you out.

Look into the "cin >>" method for assigning each character entered to a separate variable. Then you check each character variable to see if they all match each other. That's basically it, along with "cout <<" to make it look pretty.
MOD:: Fixed the arrows
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Chicken
post Sep 11 2009, 11:55 PM
Post #53


Because I said so.
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Just so you know, I'm going to post a practice ACT essay up here, and I want you guys to see what you think it rate, assuming I give you the rubric. But, first I have to complete the essay.
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Saffy
post Sep 11 2009, 11:58 PM
Post #54


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Jason, you're failing with those arrows. Other way around.
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Chicken
post Sep 12 2009, 06:39 PM
Post #55


Because I said so.
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l1/4-5l =8

I get this part:

1/4x-5= 8 1/4x-5 = -8
+5 +5 +5 +5
1/4x=13 1/4 = =-3

But...then what? Fractions are baffling to me. :/
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Saffy
post Sep 12 2009, 06:43 PM
Post #56


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You're going to have to redo and clarify your math. I have no idea what you're trying to do.
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Chicken
post Sep 12 2009, 06:46 PM
Post #57


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It's solving an absolute value equation. I get most of it, but not all. Now do you understand?

l1/4x-5l= 8
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Jason
post Sep 12 2009, 06:50 PM
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You can't bring over the abs, but you can with the 5...I think...so it's 1/4x = 13. But I'm most likely wrong, so don't take my math for it.
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Chicken
post Sep 12 2009, 06:53 PM
Post #59


Because I said so.
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It's a dual equation problem. :/

I guess I should just ask, how do you divide 13 over 1/4x and -3 over 1/4x?

It's not the problem itself. I just don't speak fraction.
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Saffy
post Sep 12 2009, 06:59 PM
Post #60


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Well, for one I don't believe you can just remove the absolute value indicators as you did. You seemed to have to just used it to invert 8 into -8 which from what i recall is a no no. You need to keep those bars in place until there's a solvable solution.

Now since you have the absolute value bars you need to realize you're diverging onto two separate answers which will require two different equations.

Equation the First:

(1/4x - 5) = 8

Equation the Second:

-(1/4x - 5) = 8

Note the additional minus (-) sign on the second equation. The absolute value bars indicate that we'll take positive value of the solution. Essentially if the value is negative we'll flip it =\ I'm not too good at explaining why there's two equations, but hopefully you can follow along anyway.

Continuing forward we just solve both equations.

(1/4x - 5) = 8
1/4x - 5 = 8
1/4x = 8 + 5
1/4x = 13
x = 4*13
x = 52

-(1/4x - 5) = 8
-1/4x + 5 = 8
-1/4x = 3
x = -4*3
x = -12

So x = -12, 52. And we'll prove this to be correct now!

|-12/4 - 5| = 8
|-3 - 5| = 8
|-8| = 8
8 = 8

|52/4 - 5| = 8
|13 - 5| = 8
|8| = 8
8 = 8

If necessary I'll explain better :)
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