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Find the coordinate of the stationary point on the curve y = 2x^2 + 4x - 5.

The important point in the question is the term 'stationary point'. This is where the graph of y will 'flattern out'. If we look at this graph, we can say that the gradient is equal to 0 at this point. Th...

SM
Answered by Serkan M. Maths tutor
4602 Views

write (x+2)(x+3)(x+5) in the form ax^3+bx^2+cx+d

First take (x+2)(x+3) and expand to get x2 +3x+2x+6, which can be simplified to equal x2+5x+6.Then mutiple x2+5x+6 by (x+5). This will give you x3+5x2

CH
Answered by Cora H. Maths tutor
3082 Views

Using the identity cos(A+B)= cosAcosB-sinAsinB, prove that cos2A=1-2sin^2A.

A curve has the equation: x^3 - x - y^3 - 20 = 0. Find dy/dx in terms of x and y.

x3 - x - y3 - 20 = 0 Find dy/dx. Differentiate with respect to x.
3x2 - 1 - 3y2(dy/dx) = 0Therefore: dy/dx = (3x

KP
Answered by Karishma P. Maths tutor
3229 Views

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