Vastor

wow, never see that coming! so, basically, this would answer my wonder about this shape also that: - area of triangle AED = area of triangle BEC; because triangle EDC is the "intersection" between the triangle (and obviously both has same amount of area) - 'every' kind of this shape have same amount of area, right?; I mean: at first, when I thought if we move the "B" only, the shape will distort thus make both triangle have different area, etc.. but then, "E"(the intersection line) will changed also for keeping the shape. Thus make both triangle have equal area whatsoever! thank you very much. P.S. my ambition is to become a theoretical physicist/mathematician. Overlooking the answer of such simple "puzzle" give me a bad feeling! D; *kicking myself*

what I know about DAC? diagram below and the stuff that I post before! try my best(thinking for hours), and the best I got is [math] 12 = \frac{1}{2}*(AC)*(H) = \frac{1}{2}*(BD)*(BC)*(sin DBC) [/math] I don't know what kind of 'logical approach' you are trying to use, but the sure thing is I can't grasp it at all. D:

Hey guys, There are 2 school activities that I'm getting into: 1. build Paper Plane (only A4 paper and stapler) 2. make fancy stuff from a program called Geometer's Sketchpad. for (1), I found these site: http://howto.wired.com/wiki/Fold_Your_Own_Sky_King_Paper_Airplane http://www.paperairplanes.co.uk/planes.php hope anyone can contribute any better site! for (2), can anyone give some tip / idea on what should I do?!

correct me if I'm wrong but doesn't the (2) method is just derived from (1) method [math] \Delta ABC = \frac{1}{2} * BC * AB * sin B [/math] [math] \Delta ABC = \frac{1}{2} * BC * AB * \frac{h}{AB} [/math] where h = height [math] \Delta ABC = \frac{1}{2} * BC * h [/math] I don't get this (as a) hint. I mean the "h" is unknown too. there too much unknown already (look below picture) what I know about triangle DAC? not much, maybe you can know it by look at my messy perspective. well let's see: 1. The angle of A = angle of B = k1 2. (assumption) that AD = BC and AC = BD (and always?) uh, forgot to say, I got the answer already (30 degree). From my assumption, but I doesn't know it will always happen like that because I can't find logical reasoning behind it. I mean you can move B further, and the only thing that change is angle, thus making it length equal no more! btw, for (1), from this point, if BD and BC are known, we can get answer by using your (2) method.

ahah, when register with fb account, the 'username' were taken from my fb account name..

that's what I'm confuse about... I don't have any idea about (on how to obtain) the ratio of AP and PB so that I can use cos rule. Thus leave me with only length of 1 of the side(BC).

Hey guys, In the figure above, ABC and APQ are two triangles such that PQ is parallel to BC. Given that BP = 2cm, CQ = 4cm, BC = 6cm and PQ:BC = 1:3. Find (a) angle of BAC (b) the ratio of the area of [math] \Delta APQ [/math] to the area of trapezium BCQP. for (a). Well, I'm assumed that AP:AB = 1:3 = AQ:QC beforehand, but the calculation doesn't work out. (Answer = 75"31') and what I learn on this chapter is sine rule, cosine rule, and area of triangle based on these rule. so, can anyone give some hint about the trick behind this question because non of the method given can be used! thnx.

Well, I'm just insecure with my time management because I spend too much time just to understand everything in math's subject ONLY, there 3 more science subject to study for the exam incoming on November. Thus, I ask every question that I found "time-consuming" to figure out by myself or the info not really available on net. Anyway, because of your respond to some of my question by not replying anything a.k.a. "you should think for yourself" respond (but it worked ), I think I just can't help myself but force my memory to work and remember the raw calculation. I still get some intuitive by using the method though I still unable to "read" calculus as proficient as algebra or simpler topic. when you said so, does Khan Academy video that cover differentiation topic do well?

Hey guys, this rule doesn't seems clear to me, If [math] \delta x [/math] and [math] \delta y [/math] represent small changes in x and y respectively, then [math] \frac{\delta y}{\delta x} \approx \frac{dy}{dx} [/math] [math] \delta y \approx \frac{dy}{dx}* \delta x [/math] and one of the example given its answer that its 'equal' rather than 'approximation', what varies between the two?

heh, that's answered all of the question. and the picture is the "question picture" itself that I draw on paint. (yes, I sure the picture show straight line => constant ac/deceleration)

The question is, "Calculate the value of t, if the average speed for the t seconds is 22ms" for imatfaal, I don't know how this calculation become wrong...

well, most of the question still can be answered by using simultaneous equation and some pure logic. like this question:- A piece of wire of length 120 cm is bent to form a shape as shown in the figure above. Show that the area enclosed by the wire, in cm^2, is given by [math] 480x - 60x^2 [/math]. Hence, find the maximum area of the figure and state the corresponding value of x. let [math] area-of-diagram = A = 480x - 60x^2 [/math] because it's said "maximum area of the figure", so it should be [math] \frac{d^2A}{dx^2} < 0 [/math] [math] \frac{dA}{dx} = 0 [/math] [math] \frac{dA}{dx} = 480 - 120x = 0 [/math] [math] 480 - 120x = 0 [/math] [math] 120x = 480 [/math] [math] x = 4 [/math] "because the above *x* is the maximum value for the function, so, the maximum area should be" [math] 480x - 60x^2 = 480(4) - 60(4)^2 [/math] [math] = 1920 - 960 = 960cm^2 [/math] So, I thought I'm on the right road...

22 is given. for 285:- [math] \frac{40 * 9}{2} + \frac{40 + 30}{2} * (12 - 9) = 285[/math]

the title... =|

Hey guys, based on the picture above, the answer given by my teacher :- [math] 22 = \frac{285 + \frac{t-12}{2} * 30}{t} [/math] [math] 22t = 285 + 15t - 180 [/math] [math] 7t = 105 [/math] [math] t = 15 [/math] but for my answer:- [math] 22 = \frac{180 + \frac{t-9}{2} * 40}{t} [/math] [math] 22 = 180 + 20t - 180 [/math] [math] 2t = 0 [/math] ??!

already "found" the answer for the question before. for (b):- let [math] \Delta AEF = y [/math] [math] k^2 - kx + x^2 = y [/math] "*Delta*AEF is minimum" [math] \frac{d^2y}{dx^2} > 0 [/math] (I do it right?) [math] \frac{dy}{dx} = 0[/math] [math] \frac{dy}{dx} = 2x - k = 0[/math] (I don't know why I take this step *finding derivative*, but it works, whatsoever, I really need to understand how that happen!) [math] 2x = k [/math] [math] x = \frac{k}{2} [/math] Unlike I thought before, from this sentence "Express x in terms of k" something like [math] \frac{dx}{dk} [/math] for ©:- the question sound like "find y in terms of k", so... [math] k^2 - kx + x^2 = y [/math] [math] k^2 - k(\frac{k}{2}) + (\frac{k}{2})^2 = y [/math] [math] \frac{4k^2 - 2k^2 + k^2}{4} = y [/math] [math] \frac{3k^2}{4} = y [/math] hope someone bother to answer my confusion now.

oh, forgot to clarify about my "real" confusion. how does you actually apply differentiation for geometry? I'm hardly see any relationship there =S

Hey guys, 62. The figure above shows a rectangle ABCD. Given that AB = 2k cm and AD = k cm. E and F are points on BC and DC respectively where BE = x cm and FC = 2x cm. (a) Show that the area of [math] \Delta AEF [/math] is [math] (k^2 - kx + x^2) cm^2 [/math] [math] 2k * k - [\frac{2k*x}{2} + \frac{k*(2k - 2x)}{2} + \frac{2x * (k - x)}{2}] = \Delta AEF [/math] [math] 2k * k - [ k*x + k*(k - x) + x *(k - x) ] = \Delta AEF [/math] [math] 2k * k - [ kx + k^2 - kx + kx - x^2 ] = \Delta AEF [/math] [math] 2k^2 - kx - k^2 + x^2 = \Delta AEF [/math] [math] k^2 - kx + x^2 = \Delta AEF [/math] perfect, but wthell it has anything to do with differentiation?! are there any alternative way to calculate this? (b) Express x in terms of k such that the area of [math] \Delta AEF [/math] is minimum. similar question to the post before, still vague on how to pick which one should be "the first derivative of *unknown1* with respect to *unknown2* which written as [math] \frac{d*unknown1*}{d*unknown2*} [/math] or [math] f'(*unknown2*) [/math] © Find the minimum area of [math] \Delta AEF [/math] in terms of k. and the difference with "(b)" question is?!!! thank you for the one who keep helping me, here some cookie.

Hey guys, 57. Variables x and y are related by the equation [math] x + y = 10 [/math]. A third variable p is defined by [math] p = x^2y [/math]. Find the values of x and y such that p is maximum. what a question, I'm not even understand what p is, firstly, it said [math] p = x^2y [/math] then, "p is maximum", which from what I understand [math] \frac{dp}{dx} < 0 [/math] or [math] p = 0 [/math]

anyone?

Hey guys, got 3 more question here, 1. if the charge model below is the right one, how does 2 negative charge repel if both are pulling? 2. From the textbook: "Household electrical appliances that work on the heating effect of current are usually marked with voltage and power ratings. A typical filament bulb shown in Photograph 2.9 is marked 240V 60W. This means that the bulb will consume 60J of electrical energy every second if it is connected to 240V supply." what happen if it's supplied with more or less than 240V? as far as I thought. [math] E = VIt [/math] and [math] P = VI [/math] thus, with constant Power, Voltage varies inversely to Current. so, even though how different voltage is, it's only affect the current but not amount of energy, doesn't it?! [math] E = Pt [/math] 3. Actually, I'm not quite understand of resistance yet, so hope anyone can help me on this question and give me a little insight about resistance itself. based on the circuit below: (i) What will happen to the bulb if the resistance of the rheostat is reduced? (ii) Explain your answer in (a) (i). for first question, as far as I learned about resistance, I just understand it as an invariant for Voltage that directly varies to Current (in Ohmic conductor) [math] V \propto I [/math] or [math] \frac{V}{I} = R [/math] thus, if R is reduced, the Voltage decreased OR the Current increased. thank you for everyone's help!

Hey guys, How do you find the derivative from first principles for: [math] y = \frac{4}{x} + 5x[/math], and [math] y = \frac{3}{x^2} [/math] anyone have any tip on how to solve the denominator that causing me to get indeterminant? such as for second question: problem with indeterminant and to find the derivative at the same time. [math] y = \frac{3}{x^2} [/math] [math] y + \delta y = \frac{3}{(\delta x + x)^2} [/math] [math] (y + \delta y)(\delta x + x)^2 = 3 [/math] [math] yx^2 + 2yx\delta x + y\delta x^2 + x^2\delta y + 2x\delta x\delta y + x^2\delta y = 3 [/math]' then...?! =S thank you.

This explain what you said? if my thought about voltage as "potential energy" is wrong, then it should be the "potential energy difference", based on your analogy and my current understanding of the voltage now. when you said "there must be a difference in the amount of potential given to the system", so basically voltage is like a resultant differences that cause something to happen? I mean, like resultant force, if it's not 0 N, then the object will accelerate, or something like that?! Except for the fact that I learned electromotive force as : e.m.f. = Potential Difference(Volt) + Drop in potential difference due to internal resistance(Resistance in the battery) so basically, even though both things are the same, the textbook claim that "voltage is e.m.f., but e.m.f. is not voltage"

thnx, and about the L'Hopital's rule, should have re-read or something before asking that, nvm...