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  #76    
Old February 23rd, 2011 (3:36 AM).
Dude22376 Dude22376 is offline
     
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    Uh, those questions aren't really that hard, unless you don't know algebra.

    1) Let the numbers be x, y and z.
    x + y = 11 (1)
    x + z = 17 (2)
    y + z = 22 (3)
    As there are 3 unknowns and 3 equations we should be able to solve it.
    (1) + (3): x + 2y + z = 33
    Subtract (2): 2y = 16 thus y = 8
    The rest is just basic manipulation, x = 3 and z = 14.

    2) Let the numbers be x and y, x > y.
    x + y = 8 (1)
    x^2 - y^2 = 34 (2)
    Again we have 2 equations and 2 unknowns, so it should be solvable.
    Rearranging (1): y = 8 - x
    Substituting into (2): x^2 - (8 - x)^2 = 34
    Expanding: x^2 - 64 + 16x - x^2 = 34
    Cancelling x squared and shifting -64 over: 16x = 34 + 64, thus x = 6.125
    Then y = 1.875

    If you want to get past your mental block at doing maths, just remember that if there are n equations and n unknowns, you should be able to solve for every unknown. Write down equations representing everything that is given to you in the question. Then solve those equations for the answers.
    At your level, only substitution and elimination are needed. My answers for the two questions you provided display both techniques (eliminating x + z in the first and substituting 8 - x in the second).
    If there are more unknowns than there are equations, then the question can't be solved!
    If there are 2 equations which mean the same thing (like x + y = 3 and 2x + 2y = 6), then you can totally ignore one of those equations. If this puts you in a situation where there are more unknowns than equations, then the question can't be solved.

    Oh, and if you want your equations to be displayed nicely on forums, try using Texify. It'll take some time to learn how to use, but you'll be able to display superscripts, subscripts, Greek letters, differential/integral signs, etc. Sadly I can't use it because I don't have 15 posts yet

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      #77    
    Old February 23rd, 2011 (4:29 AM).
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    ahhh, imma try all the harder ones now ,
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      #78    
    Old February 23rd, 2011 (4:46 AM).
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    This thread needs more manly trigonometry questions. I have one if anybody is bored:
    If sinθ + cosθ = √2cos(90° - θ), find cot
    θ.

    It's not really difficult, anyone wants to try?
      #79    
    Old February 23rd, 2011 (5:06 AM).
    Dude22376 Dude22376 is offline
       
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      Let theta=x, it's easier than entering Greek characters.
      cos(90-x) is sin x, substitute that in:
      sin x + cos x = sqrt(2) sin x
      Divide by sin x on both sides (assuming sin x is not 0):
      1 + cos x/sin x = sqrt(2)
      cos x/sin x is cot x, which is sqrt(2) - 1.
        #80    
      Old February 23rd, 2011 (5:36 AM).
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      Quote:
      Originally Posted by Dude22376 View Post
      Let theta=x, it's easier than entering Greek characters.
      cos(90-x) is sin x, substitute that in:
      sin x + cos x = sqrt(2) sin x
      Divide by sin x on both sides (assuming sin x is not 0):
      1 + cos x/sin x = sqrt(2)
      cos x/sin x is cot x, which is sqrt(2) - 1.
      Nice, that is the correct answer. I need to think of tougher questions now! >:D

      My final examinations are fast approaching, so gotta practise lots of math. XD
        #81    
      Old February 23rd, 2011 (11:15 AM).
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      Quote:
      Originally Posted by Pokemon Trainer Touko View Post
      Oh my god~! My classmates are doing trigonometry and they don't even know what's the sine and cosine rules~?!?! :O Unbelievable~!
      Seriousally? well heres an easy way to remember Sin,Cos,Tan
      Sin: Oh hell (O/H)
      Cos: Another Hour (A/H)
      Tan: Of Algebra (O/A)
        #82    
      Old February 23rd, 2011 (11:54 AM).
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      Quote:
      Originally Posted by NurseBarbra View Post
      Seriousally? well heres an easy way to remember Sin,Cos,Tan
      Sin: Oh hell (O/H)
      Cos: Another Hour (A/H)
      Tan: Of Algebra (O/A)
      Insulting as this may be, I find it amusing and useful. I approve of it.
      Quote:
      Originally Posted by Pokemon Trainer Touko View Post
      Oh my god~! My classmates are doing trigonometry and they don't even know what's the sine and cosine rules~?!?! :O Unbelievable~!
      Assuming you are talking about the laws of sines and cosines, and not just the definitions of the functions, I have not memorized them either. I try to avoid memorizing stuff as much as I can, since deriving them from the basics is much more rewarding. It is quite fun when my mathematician tells me to solve a trigonometry problem on the blackboard. I derive everything from the basics, which takes ages. My teacher has learned not to get me on the blackboard the hard way. That is not the case with constants, however, since I want to maximize accuracy whenever I can. I remember once when we had a problem involving circles, which brought in pi. The teacher made the foolish mistake to make me solve it, and I went and used "3,141592653589793238462643383279506884197169399375" instead of "3,14". She ended up stopping me, and making me return to my seat, when she saw me multiplying that by the square of some number.
        #83    
      Old February 23rd, 2011 (11:58 AM).
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          #84    
        Old February 23rd, 2011 (3:53 PM).
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        Quote:
        Originally Posted by NurseBarbra View Post
        Seriousally? well heres an easy way to remember Sin,Cos,Tan
        Sin: Oh hell (O/H)
        Cos: Another Hour (A/H)
        Tan: Of Algebra (O/A)
        it took me a full day to realize what that meant.

        here's my way of remembering;

        S.O.H.C.A.H.T.O.A.

        ..you all know what the letters represent
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          #85    
        Old February 26th, 2011 (4:34 AM).
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          Quote:
          Originally Posted by Dude22376 View Post
          2) Let the numbers be x and y, x > y.
          x + y = 8 (1)
          x^2 - y^2 = 34 (2)
          Again we have 2 equations and 2 unknowns, so it should be solvable.
          Rearranging (1): y = 8 - x
          Substituting into (2): x^2 - (8 - x)^2 = 34
          Expanding: x^2 - 64 + 16x - x^2 = 34
          Cancelling x squared and shifting -64 over: 16x = 34 + 64, thus x = 6.125
          Then y = 1.875
          Actually, that wouldn't be necessary. Well, Unless you want to find the true value of x and y. Because

          x + y = 8 ... (1)
          x2 - y2 = 34 ... Factorize
          (x+y)(x-y) = 34

          From ... (1)
          8(x-y) = 34
          x-y = 34/8

          Since the question asked for the difference of x and y, there's no need to solve for x and y.

          Quote:
          Originally Posted by Regeneration View Post
          This thread needs more manly trigonometry questions. I have one if anybody is bored:
          If sinθ + cosθ = √2cos(90° - θ), find cot
          θ.

          It's not really difficult, anyone wants to try?
          Gah. I hate Trigonometry. idk why. >.< Couldn't be bothered to remember all the values for cos/sin/tan pi/3, pi/6 etc.

          Does someone has a formula for memorizing them? XD I know that you can basically just draw the unit circle and find the value, but I want to memorize them. It's way quicker to get, especially in exams.

          Quote:
          Originally Posted by NurseBarbra View Post
          Seriousally? well heres an easy way to remember Sin,Cos,Tan
          Sin: Oh hell (O/H)
          Cos: Another Hour (A/H)
          Tan: Of Algebra (O/A)
          XD That's awesome XP

          Quote:
          Originally Posted by ShinyMeowth View Post
          I remember once when we had a problem involving circles, which brought in pi. The teacher made the foolish mistake to make me solve it, and I went and used "3,141592653589793238462643383279506884197169399375" instead of "3,14". She ended up stopping me, and making me return to my seat, when she saw me multiplying that by the square of some number.
          WIN

          Quote:
          Originally Posted by Impo View Post
          how are you so good at maths? i'm so jealous.
          It does help that I did those during my high school years, and I'm now in college studying things like derivatives and local linear approximations and whatnot XD
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            #86    
          Old February 26th, 2011 (4:50 AM).
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          Quote:
          Originally Posted by smile! View Post
          Gah. I hate Trigonometry. idk why. >.< Couldn't be bothered to remember all the values for cos/sin/tan pi/3, pi/6 etc.

          Does someone has a formula for memorizing them? XD I know that you can basically just draw the unit circle and find the value, but I want to memorize them. It's way quicker to get, especially in exams.
          I have memorised the values of cos, sin, tan, etc. for 0, 30, 45, 60 and 90 degrees.

          I learned only the values of Sin. (In degrees)
          Sin 0 = 0 = Cos 90
          Sin 30 = 1/2 = Cos 60
          Sin 45 = 1 /√2 = Cos 45
          Sin 60 =
          √3/2 = Cos 30
          Sin 90 = 1 = Cos 0
          If you noticed, the values of Cos are just opposite of those of Sin.

          Then, Tan = Sin/ Cos
          Cosec = 1/Sin
          Sec = 1/Cos
          Cot = 1/Tan = Cos/ Sin

          That way, you can remember the values with the help of just memorising the values of just Sin. We get to use a log table for all other values. XD

            #87    
          Old February 26th, 2011 (5:34 AM).
          Dude22376 Dude22376 is offline
             
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            Yeah, that is a faster way of doing it. I guess it depends on whether you can spot such a shortcut right away or not.

            I memorize all trigonometric values by imprinting a picture of their graphs in my mind (i.e. y = sin x, y = cos x, y = tan x). So I know right away that, for example, sin x starts at 0, goes up to 1, back down to 0, then -1, then 0 again.

            One more thing when converting between radians and degrees: pi/6 is 30 and pi/3 is 60.
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              #88    
            Old February 26th, 2011 (7:22 AM).
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              Quote:
              Originally Posted by Regeneration View Post

              I learned only the values of Sin. (In degrees)
              Sin 0 = 0 = Cos 90
              Sin 30 = 1/2 = Cos 60
              Sin 45 = 1 /√2 = Cos 45
              Sin 60 =
              √3/2 = Cos 30
              Sin 90 = 1 = Cos 0
              If you noticed, the values of Cos are just opposite of those of Sin.

              Then, Tan = Sin/ Cos
              Cosec = 1/Sin
              Sec = 1/Cos
              Cot = 1/Tan = Cos/ Sin

              That way, you can remember the values with the help of just memorising the values of just Sin. We get to use a log table for all other values. XD

              Ooh. I remember trying the same approach last year. XD But when we moved into Calculus this year, I have forgotten all those Precalc methods I've devised. XD THANKS!

              Quote:
              Originally Posted by Dude22376 View Post
              Yeah, that is a faster way of doing it. I guess it depends on whether you can spot such a shortcut right away or not.

              I memorize all trigonometric values by imprinting a picture of their graphs in my mind (i.e. y = sin x, y = cos x, y = tan x). So I know right away that, for example, sin x starts at 0, goes up to 1, back down to 0, then -1, then 0 again.

              One more thing when converting between radians and degrees: pi/6 is 30 and pi/3 is 60.
              Mhm. I used to get stumped on such questions as well. Like, this can't be factorized. Heck, how am I going to get the value? Then my friend showed me that method, and I realized that you don't really need the value. It's more of how creative you are.

              Memorizing graph pattern, cool. That could help. Thanks
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                #89    
              Old February 26th, 2011 (7:23 AM).
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                Quote:
                Originally Posted by ShinyMeowth View Post
                Assuming you are talking about the laws of sines and cosines, and not just the definitions of the functions, I have not memorized them either. I try to avoid memorizing stuff as much as I can, since deriving them from the basics is much more rewarding. It is quite fun when my mathematician tells me to solve a trigonometry problem on the blackboard. I derive everything from the basics, which takes ages. My teacher has learned not to get me on the blackboard the hard way. That is not the case with constants, however, since I want to maximize accuracy whenever I can. I remember once when we had a problem involving circles, which brought in pi. The teacher made the foolish mistake to make me solve it, and I went and used "3,141592653589793238462643383279506884197169399375" instead of "3,14". She ended up stopping me, and making me return to my seat, when she saw me multiplying that by the square of some number.
                Wow, that's just awesome.
                And I agree with you on deriving any formulae, mainly because I don't trust my brain to get ALL the numbers correct when memorising, plus it's fun, and as you said it feels rewarding, and I can trust that I've gotten it right at the end.
                As for memorising constants, I would like to memorise up to about 6 decimals (an accuracy to a millionth) but I know that Pi is 3.1415926 (7 decimals), e begins with 2, and phi begins with 1.618.

                Quote:
                Originally Posted by Regeneration View Post

                I have memorised the values of cos, sin, tan, etc. for 0, 30, 45, 60 and 90 degrees.

                I learned only the values of Sin. (In degrees)
                Sin 0 = 0 = Cos 90
                Sin 30 = 1/2 = Cos 60
                Sin 45 = 1 /√2 = Cos 45
                Sin 60 =
                √3/2 = Cos 30
                Sin 90 = 1 = Cos 0
                If you noticed, the values of Cos are just opposite of those of Sin.

                Then, Tan = Sin/ Cos
                Cosec = 1/Sin
                Sec = 1/Cos
                Cot = 1/Tan = Cos/ Sin
                That's actually very useful, perhaps I'll try and memorise that. It's a shame that, in school, I've only done SohCahToa, Law of Sine and Law of Cosine. I haven't been taught how the rules are related to each other (like Cosine and Sine are "reversed" and Tan is Sin/Cos), or anything like that.
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                  #90    
                Old February 26th, 2011 (7:24 AM).
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                Quote:
                Originally Posted by Regeneration View Post

                Sin 0 = 0 = Cos 90
                Sin 30 = 1/2 = Cos 60
                Sin 45 = 1 /√2 = Cos 45
                Sin 60 =
                √3/2 = Cos 30
                Sin 90 = 1 = Cos 0
                If you noticed, the values of Cos are just opposite of those of Sin.

                The observation is correct, but the opposite of x is -x. What you meant is sin(45°±a)=cos(45°∓a).

                I never memorize trigonometric values, since those most commonly used ones can be easily found by constructing triangles, and whenever you need something like sin(17), you are definitely going to have access to a table anyway.
                  #91    
                Old February 26th, 2011 (7:42 AM).
                Dude22376 Dude22376 is offline
                   
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                  I, on the other hand, always memorize formulas. I may have a go at deriving them when I have time, but I never derive formulas during tests/exams. Try deriving the ideal gas equation and tell me that memorizing PV = nRT is harder.
                  It actually isn't that hard, if you've done enough practice problems you can always fill in the correct numbers in the correct places. It's a bit like muscle memory.
                  Writing down every necessary formula on a piece of paper also helps.

                  I don't memorize constants like pi either - your answer is always either computed using a calculator, or given in terms of pi.
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                    #92    
                  Old February 26th, 2011 (8:22 AM).
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                    Quote:
                    Originally Posted by ShinyMeowth View Post
                    I never memorize trigonometric values, since those most commonly used ones can be easily found by constructing triangles, and whenever you need something like sin(17), you are definitely going to have access to a table anyway.
                    Yeah, but constructing triangles take time, and time is essential in a test. >.< Or in class, when your lecturer asks something and wants the answer fast. That is why I prefer memorizing. But you have a point there. If you suddenly get blank in an exam, you can start from scratch if you understand the basics and know how to construct the table.

                    Quote:
                    Originally Posted by Dude22376 View Post
                    I, on the other hand, always memorize formulas. I may have a go at deriving them when I have time, but I never derive formulas during tests/exams.

                    Writing down every necessary formula on a piece of paper also helps.

                    I don't memorize constants like pi either - your answer is always either computed using a calculator, or given in terms of pi.
                    I concur.

                    THAT. I seriously need to do that. First Principles, Newton's Method, Rolle's Theorem, Local Linear Approximation... They're all scattered throughout my notes. XD

                    We used to substitute pi for 22/7 or 3.142 in the high school, but now we give the answers in terms of pi instead. I love that better, because substituting pi for 22/7 or 3.142 doesn't really give a correct answer.


                    Also, I don't know whether you guys already know about this, but.
                    Speaking about pi, if you use 22/7 as pi, the equivalent value is 3.142857142857142857....

                    If you notice, we have a recurring sequence there, which is 142857. It's a roundabout number, i.e whenever you multiply that number by e.g 2, you get the same numbers, only in different sequence.

                    142857 x 2 = 285, 714
                    142857 x 3 = 428, 571
                    142857 x 4 = 571, 428
                    .
                    .
                    .
                    142857 x 7 = 999, 999
                    142857 x 8 = 1, 142, 856 (notice that 6+1 = 7 so essentially it's still 142, 857)

                    I don't know how far this number will stay the same though. XD
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                    Old February 26th, 2011 (9:38 AM).
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                    Derive a formula during tests? No way. Unless that is what the question asks. XD

                    Memorising formulae is not difficult at all if you're practising maths a lot. You'll keep using them and hence easily memorise them all.
                      #94    
                    Old February 28th, 2011 (7:42 AM).
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                      If there's one thing I hate more than trigonometry, it's modulus, I swear. They drive me nuts on limit questions. x.x I wonder if someone can help me with this?

                      Determine whether all the hypotheses of the Mean-Value Theorem are satisfied on the stated interval. If they are not satisfied, find all values of c guaranteed in the conclusion of the theorem.

                      f(x) = |x-1| on [-2,2]

                      ...I didn't even understand the last part of the question. o.o Anyone? :3
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                        #95    
                      Old February 28th, 2011 (8:02 AM). Edited February 28th, 2011 by Dude22376.
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                        The hypotheses required are:
                        (1) f(x) is continuous on [-2,2], and
                        (2) f(x) is differentiable on (-2,2)

                        The first statement is easy to prove: |x-1| can be piecewise defined, as 1-x if x < 1, 0 if x=1, and x-1 if x > 1. Since x-1 is a polynomial, and all polynomials are continuous over R, f(x) is continuous over [-2,1) and (1,2].
                        Next, we must check for continuity at x=1.


                        f(1) = 0
                        All 3 numbers match, so f(x) is continuous at x=1.

                        The second statement can be seen as false graphically: there is a corner point at x=1. However, we must prove this rigorously (no graphical methods are acceptable).

                        Start with the definition of derivative, calculate the left-hand derivative:

                        Noting that f(1) = 0, and using the piecewise definition of f(x) as described earlier.
                        Then calculate the right-hand derivative:

                        Since they are not the same, f(x) is not differentiable at x=1, so (2) is false.

                        One of the conditions was not satisfied, so no values of f'(c) are guaranteed. (1 - 3)/(2 - (-2)) = -0.5.
                        The only values of f' at the points where it is differentiable are 1 or -1.

                        p.s. I'm pretty sure the last part of the question is worded wrongly, because:
                        1) "not satisfied" = theorem doesn't count for anything
                        2) MVT involves f'(c), and does not state exactly which, or how many, values c could take.
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                          #96    
                        Old February 28th, 2011 (8:32 AM).
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                          Quote:
                          Originally Posted by Dude22376 View Post
                          The hypotheses required are:
                          (1) f(x) is continuous on [-2,2], and
                          (2) f(x) is differentiable on (-2,2)

                          The first statement is easy to prove: |x-1| can be piecewise defined, as 1-x if x < 1, 0 if x=1, and x-1 if x > 1. Since x-1 is a polynomial, and all polynomials are continuous over R, f(x) is continuous over [-2,1) and (1,2].
                          Next, we must check for continuity at x=1.


                          f(1) = 0
                          All 3 numbers match, so f(x) is continuous at x=1.

                          The second statement can be seen as false graphically: there is a corner point at x=1. However, we must prove this rigorously (no graphical methods are acceptable).

                          Start with the definition of derivative, calculate the left-hand derivative:

                          Noting that f(1) = 0, and using the piecewise definition of f(x) as described earlier.
                          Then calculate the right-hand derivative:

                          Since they are not the same, f(x) is not differentiable at x=1, so (2) is false.

                          One of the conditions was not satisfied, so no values of f'(c) are guaranteed. (1 - 3)/(2 - (-2)) = -0.5.
                          The only values of f' at the points where it is differentiable are 1 or -1.

                          p.s. I'm pretty sure the last part of the question is worded wrongly, because:
                          1) "not satisfied" = theorem doesn't count for anything
                          2) MVT involves f'(c), and does not state exactly which, or how many, values c could take.
                          Thank you! That was a great help
                          And yeah, I don't know whether the last part of question is worded correctly. I'll ask my lecturer again to clarify. Anyway, thanks again! 8D

                          I didn't quite get that last part, though. The last 2 lines before the postnote.
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                            #97    
                          Old February 28th, 2011 (8:45 AM).
                          Dude22376 Dude22376 is offline
                             
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                            This should be clearer.

                            f(2) = 1
                            f(-2) = 3.
                            (f(2) - f(-2))/(2 - (-2)) = -0.5.

                            d/dx (x-1) = 1 and d/dx (1-x) = -1.
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                              #98    
                            Old February 28th, 2011 (8:50 AM).
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                              Quote:
                              Originally Posted by Dude22376 View Post
                              This should be clearer.

                              f(2) = 1
                              f(-2) = 3.
                              (f(2) - f(-2))/(2 - (-2)) = -0.5.

                              d/dx (x-1) = 1 and d/dx (1-x) = -1.
                              Ah, of course. It's to clarify f(b)-f(a)/b-a = f'(c) right? I forgot the Theorem itself, lol. Thanks very much
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                                #99    
                              Old March 1st, 2011 (3:01 AM).
                              Impo's Avatar
                              Impo Impo is offline
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                              Join Date: Feb 2010
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                              i had my maths test today.
                              i feel pretty confident .

                              i know i got one question wrong though (i accidentally added the dividends instead of leaving them, so the answer was too big) .

                              but i managed to conquer the speed questions :D
                              they were actually fun

                              thanks for all the help on here, too
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                                #100    
                              Old March 2nd, 2011 (3:01 PM).
                              smile!'s Avatar
                              smile! smile! is offline
                                 
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                                Quote:
                                Originally Posted by Impo View Post
                                i had my maths test today.
                                i feel pretty confident .
                                I had my Calculus test yesterday as well And I felt it was better than the first one. At least I could answer all questions. XD

                                Although, I did one silly mistake by separating integral cos x sin x into integral cos x times sin x. =\ I thought you could do so, and only after the test when I checked I found out you cannot. >.
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