Suppose the slope of the tangent line at any point on a curve is curve passes thru (1,3), approximate the tangent line at (1.2, y(1.2)). x²+4y (a) y = 0.301193x + 2.96745 (b) y = 0.03193x + 2.69745 (c) y = 0.03193x + 2.96745 (d) y = 0.30193x + 2.69745 If the

Q: y = 3x?+1 ; (0,1)

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Q: y = √1 - 8x, (-1, 3)

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Q: If the Suppose the slope of the tangent line at any point on a curve is x2+4y² curve passes thru…

A:

Q: The slope of the tangent line to the graph of the function y = 8Vx – 3 at (9, 21) is %3D 1.33 -2.33…

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Q: Find the slope of the tangent line to the graph at the given point. (x² + y²)² = 4x²y, (1,1)

A: To find slope of tangent of a curve at a given point, we need to find differential at that point

Q: Find the slope of the tangent line to the graph at the given point. 4xcosy +1 = 5 at the point (2,)

A: Solve, as follows: 4xcosy+1=541cosy+x-sinydydx=0dydx=cosyxsiny