This is conerning the normal curve, properties and how you dertermine the shape
121239
The areas bounded by any s.d. can be looked up ona table, but be aware that the tables can be set to give out the answer in several different ways. One such table is on the link below. To answer your question you would have to multiply each number by 2, eg, for 3 s.d. the table gives out 0.4987 (the area between 0 and 3 s.d.) so what you are looking for is 0.9974. You don´t do "calculations" to find these numbers. That would require solving an integral that so far no one has been able to do so analitically. What you do is work a numeric solution like simpsons integration or a polinomial approximation like Abramowitz´s
200343
You motivate differentiation with something students are familiar with; speed, rates of change, etc.  You do first have to give them a comfortable feel with limits, since the definition of the derivative involves limits and differentiation is a limiting process.  You then show the difference between average rate of change, and instantaneous rate of change (anyone taking a trip from point A to B in a car can get this one).  You introduce the slope of the line as the rate of change and then work to the idea that the slope of the secant line to a curve is the average rate, and by taking smaller differences in the independent variable you approach the instantaneous rate of change(this is where the whole limit thing is important).  Once the students feel comfortable with this explanation of what we are trying to do, then you move into the technical and rigorous side.
287529
a nasty curve and a changeup
516732
Plot 3 points, connect them with lines. you get a triangle\n\nList most acute (sharp) angle 1st, most obtuse (blunt) angle last, and remaining one in the middle.\n\nIt looks like C will be the most obtuse
201089
Aha, geometry!\n\nDraw a diagram and follow the steps below.\n\nThe trick is to show that the triangles DBE and \ntriangle DAF are similar.\n\nProof That triangles DBE and triangle DAF are similar.\n\nLet <ABC> donote angle ABC.\nWe need to show any two corresponding angles of these traingles are equal.\n\nFirst angle,\n<FDA> = <BDE>  because D,B,F collinear and D,E,A collinear.\n\nSecond angle,\n<DAF> = <DAC> because A,F,C collinear\n      = (pi) - <ACD> - <ADC> angles in triangle ACD\n      = (pi) - <ACD> - <ADC>  ...(1)\n\nNow, separately  manipulate,<ACD> and <ADC>.\n\n<ACD> = <ACB> + <BCD> splitting <ACD> with line CB\n      = <ACB> + <CAB>, tangent property <BCD> = <CAB> \n      = (pi) - <CBA>  angles in triangle ABC\n      = <ABE>  because C,B,E are all colinear\n\nNow, similarly\n\n<ADC> = <ADB> + <BDC> splitting <ADC> with line DB\n      = <ADB> + <BAD>, tangent property <BDC> = <BAD> \n      = (pi) - <ABD>  angles in triangle ABD\n         \nSubstituiting these two info into (1), we get,  \n\n<DAF> = (pi) - <ACD> - <ADC>\n      = (pi) - <ABE> - [ (pi) - <ABD> ]\n      = <ABD> - <ABE>\n      = <BDE> because BE splits <ABD> \n\nDone. In showing the triangle DAF and triangle DBE are similar.\n\nHence, by ratio for similar triangles,\n\n        DA/DF = DB/DE\n\nOr Rewriting we get\n \n      DA * DE = DB * DF\n\nProved.\n\nNot so difficult, was it. hehehe.\n\nRegards
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The length of the curve between the two points according to my calculation was:\n\nd = (3/2a) + (a*ln(2)/4)\n\nThe length is minimum under the condition that\n\na = √(6/ln(2))\n\n\n\nI don't have the time to show the full calculations, but I'll give you some key points:\n\n1. Find the line element 'ds' using the formula:\n   ds = √(dx^2 + dy^2)\n   -or-\n   ds = dx * √(1 + (dy/dx)^2)\n\n2. Evaluate (dy/dx) and substitute in above equation. \n\n3. Integrate the line element 'ds' determined above from x=1 to x=2 - this should give you the length of the curve in terms of 'a'\n\n4. To find the minimum length, differentiate the equation for length with respect to 'a', make the equation equal zero & solve for 'a'. This should give you the values for 'a' that is either for minimum or maximum. Now, differentiate the new equation once more & substitute the different values of 'a'. For each case, check if the equation yields a positive value - this means that it is a minimum case.
470707
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
13129
F'(x) is the derivitive of F(x), which means it describes the change in F with respect to x.  Another way to say this is that F'(x) is the slope of F(x).  \n\nTo sketch the graph of F'(x), look at your graph of F(x).  Pick a point on that graph (maybe the first point). See what the slope (remember rise over run?) of F(x) is at that point.  That slope is the derivitive F' for that value of x.  Plot this on your F'(x) skecth.  Go the the next point and repeat.
58519
1-The points to be brought out on the drawing of this aspect are the relative positions of the contact areas , the relationship of the cusps to each other, the design of the grooves on teh occlusal surfaces , the width of the crown buccolingually as compared to mesiodistally , and the rhomboidal outline of teh crown generally.\n2-A good way to study the design of the occlusal aspect is to drawa rectangle on a piece of tracing paper with the same length and breadth as the crown measurements buccolingually and mesiodistally , place this over the illustration and check the salient points with geometric figure as constrast (check a view of the occlusal surface of the upper 6).\n3-This procedure may be adapted to any of the illlustrations of crown form or root form wherever desired . Some students are able to get a clean mental picture this way than studying graphs
621156
For a fixed a and n, and allowing b and c to vary, the curve would be a hyperbolar centered at the origin. You will only be taking the rational points on that curve, though.
529849
The curved surface area is 2*pi*r*h\n\nor 2*3.142*27*12 sq cms\n\nand 2*3.142*17*35 sq mms\n\nyou can do the rest
290410
i think it's a parabola.. im not sure. u should try plotting it on a graph.. substitute  the x and y values and see what form comes out..
296376
u can even use calculus to find this out
155314
KNOW YOUR RELATED RATES FOR CALC!! thats one thing that killed me and know your 2-D rotation around a line intergrals.
495883
If you mean an isometric drawing, it will appear as an ellipse.
124292
1 (a):  Bit of partial differentiation.  The slope is dy/dx; however, to get this we need to take partial derivatives of the function so we can calculate it.\n\ndy / dx = (Df / Dx) * (Dy / Df) [where the D denotes partial derivative].\n\nDf / Dx = 2x + y;\nDf / Dy = 2y + x.\n\nHence dy / dx = (2x + y) / (2y + x).\n\n1 (b):  x-intercepts of the curve have x = 0, and hence gradient is given by:\n\ndy / dx = (2*0 + y) / (2*y + 0) = y / 2y = 1/2.\n\nThis is the same for both intercepts, and so they are parallel.\n\n1 (c):  For vertical tangents, dy/dx is infinite; hence the denominator must be zero.  \n\n2y + x = 0   =>  x = -2y.\n\nSubstitute in:\n\n4y^2 - 2y^2 + y^2 = 27\n3y^2 = 27\ny^2 = 9\ny = +/- 3 and x = -/+ 6.\n\n----------------------------------------------------------\n\n2 (a):  Circumference = 2pi*r, and is changing by 6" / sec.\n\nHence d/dt [2pi*r] = 6 and dr / dt = 3/pi.\n\nSquare perimeter P is 8r, and so dP / dt is 8*3/pi = 24/pi inches / second.\n\n2 (b):  The area between the two is 4*r^2 - pi*r^2;\n\nA = (4 - pi)r^2.\ndA / dt = (4 - pi) * 2r * (dr / dt).\n\ndr / dt we have, from (a), and we know that the circle's area is 25pi:\n\n25pi = pi*r^2\nr^2 = 25\nr = 5\n\nSo:  dA / dt = 2*(4 - pi)* 5 * (3 / pi) inches squared / second.  And that might need a calculator :)
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Draw a straight line.\nDraw a good cuve (mirror), using a compass or a protractor(for later mark a point as centre and place the protractor on this point coinciding with the axis.)\nThe curve should be of equal length on each side of principle axis.\nNow mark the pole, radius and focal point (twice the radius).\nIn case u use protractor, take focal lenght as twice the distance between the marked pont and pole.\n\nRemember how the rays reflect/refract, and accordingly draw two incident and two reflected/refracted rays.\nGive the standard sign convention...I hope plane mirror and prism are easier.