conic sections, Mathematics

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The locus of the midpoint of the chords of an ellipse which are drawn through an end of minor axis is called

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Let u = sin(x). Then du = cos(x) dx. So you can now antidifferentiate e^u du. This is e^u + C = e^sin(x) + C.  Then substitute your range 0 to pi. e^sin (pi)-e^sin(0) =0-0 =0

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y 2 = t 2 - 3 is the actual implicit solution to y'= t/y, y(2) = -1. At such point I will ask that you trust me that it is actually a solution to the differential equation. You w

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How can I solve x in a circle? For example.. m

Linear algebra, solve for k such that the system 4x+ky=6 kx+y=-3

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