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To find the equation of a line, we need; the gradient of that line and a point on that line. To find the gradient of a line, we require two points on the line which have been prov...
Find center and radiusFrom completing the quadratic:x^2 - 2x + 1 - 1 + y^2 = 3(x-1)^2 + y^2 = 4hence, center P(1, 0) and radius R = sqrt(4) = 2Find y interceptAnswered by • Maths tutor2858 Views
1/y² dy/dx = sec²(x)∫ 1/y² dy/dx dx = ∫ sec²(x) dx-1/y + C1 = tan(x) + C2y = -1/(tan(x) + A) where A = C2 - C1y(0) = -1/A so y(0) = 1 means A = -1. Finished!
To help the student understand how to expand the brackets I would start by splitting the equation to:3x2(x - 3) + 6x(x - 3) + 7(x - 3)I then would ask the student to expand these brackets to:3...
To find a gradient at a given point, first we differentiate then we sub in the x coordinate of the point. To differentiate 4(8x+2)4 we must use the chain rule. First we let u=...
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