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Solve the simultaneous equations y = x^2 - 6x and 2y + x - 6 = 0

Rearrange the second equation in terms of y: meaning that the equation is of the form y = ....-this will give y = 3 - x/2You may now substitute the y in the left hand equation with what y in the right hand e...
TD
Answered by Tabi D. • Maths tutor
7734 Views

Express (4x)/(x^2-9) - (2)/(x+3) as a single fraction in its simplest form.

First we must expand the demoninator to; (x+3)(x-3)Then we can multiply the left hand fraction on top and bottom by (x-3) to get a common demoninatorthis gives us; (4x)/((x+3)(x-3)) - ((2)(x-3))/((x+3)(x-3))...
EE
Answered by Eddie E. • Maths tutor
4519 Views

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. There...
SM
Answered by Serkan M. • Maths tutor
5429 Views

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

Use cos(A+B)=cosAcosB-sinAsinB and let A=B so cos(A+A)=cosAcosA-sinAsinA this means cos(2A)=cos 2 A-sin 2 A and since cos 2 A+sin 2 A=1, cos 2 A=1-sin 2 A. Therefore, by subbing cos 2 A=1-sin 2 A into cos(2A...
RF
Answered by Rebecca F. • Maths tutor
22827 Views

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

x 3 - x - y 3 - 20 = 0 Find dy/dx. Differentiate with respect to x. 3x 2 - 1 - 3y 2 (dy/dx) = 0Therefore: dy/dx = (3x 2 - 1)/3y 2
KP
Answered by Karishma P. • Maths tutor
3925 Views