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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 :( |
ahhh, imma try all the harder ones now :) ,
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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? |
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. |
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My final examinations are fast approaching, so gotta practise lots of math. XD |
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Sin: Oh hell (O/H) Cos: Another Hour (A/H) Tan: Of Algebra (O/A) |
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Username: Drekin
Overall Education Level: Highschool math plus 13 credit hours after this semester is over Mathematics Education Level (Or most recent/advanced math subject): College Calculus I Do you think you can be asked for help in your level or lower?: Definitely Life = The Universe = Everything = : 42 (shouldn't it be Life + The Universe + Everything = ? ) |
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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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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:
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:
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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 |
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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Memorizing graph pattern, cool. That could help. Thanks :) |
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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:
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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. |
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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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 |
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. |
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 |
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. http://www.texify.com/img/%5CLARGE%5C%21%5Cnormalsize%20%5Clim_%7Bx%5Cto1%5E-%7Df%28x%29%20%3D%201-1%20%3D0.gif http://www.texify.com/img/%5CLARGE%5C%21%5Cnormalsize%20%5Clim_%7Bx%5Cto1%5E%2B%7Df%28x%29%20%3D%201-1%20%3D0.gif 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: http://www.texify.com/img/%5CLARGE%5C%21%5Cnormalsize%20%5Clim_%7Bx%5Cto1%5E-%7D%5Cfrac%7Bf%28x%29-f%281%29%7D%7Bx-1%7D%3D%20%5Clim_%7Bx%5Cto1%5E-%7D%5Cfrac%7Bf%28x%29%7D%7Bx-1%7D%3D%20%5Clim_%7Bx%5Cto1%5E-%7D%5Cfrac%7B1-x%7D%7Bx-1%7D%3D-1.gif Noting that f(1) = 0, and using the piecewise definition of f(x) as described earlier. Then calculate the right-hand derivative: http://www.texify.com/img/%5CLARGE%5C%21%5Cnormalsize%20%5Clim_%7Bx%5Cto1%5E%2B%7D%5Cfrac%7Bf%28x%29-f%281%29%7D%7Bx-1%7D%3D%20%5Clim_%7Bx%5Cto1%5E%2B%7D%5Cfrac%7Bf%28x%29%7D%7Bx-1%7D%3D%20%5Clim_%7Bx%5Cto1%5E%2B%7D%5Cfrac%7Bx-1%7D%7Bx-1%7D%3D1.gif 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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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. |
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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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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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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