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Find the gradient, length and midpoint of the line between (0,0) and (8,8).

let x1 = 0, y1 = 0 in (0,0) and let x2 = 8 and y2 = 8 in (8,8). To find the gradient, we would do (y2 - y1)/(x2-x1) = 1. To find the length, we would do the square root of the following: (y2-y1)^2 + (y2-y1)^...
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Answered by Jason A. • Maths tutor
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The curve C has the equation y = 1/2x^3 - 9x^3/2 + 8/x + 30, find dy/dx. Show that point P(4, -8) lies on C

y = 1/2x^3 - 9x^3/2 + 8/x + 30y = 1/2x^3 - 9x^3/2 + 8x-1 + 30dy/dx = 3/2x^2 - 27/2x^1/2 - 8x^-2 + 0dy/dx = 3/2x^2 - 27/2x^1/2 - 8/x^2substitute x=4 into equation for yy = 1/2(4)^3 - 9(4)^3 + 8/4 +30y = 32 - ...
Answered by • Maths tutor
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How do we know which formulas we need to learn for the exam?

When learning new content, I know it is really intimidating seeing how many new formulas and equations are needed to answer questions, however I also know how easy it is to get on top of it all! My main piec...
Answered by • Maths tutor
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Find all solutions of the equation in the interval [0, 2π]. 5 cos^3 x = 5 cos x

5cos 3 x = 5cosxFirstly -5cosx from both sides and divide through by 5We have:cos 3 x-cosx = 0We can factorise this:cosx(cos 2 x - 1) = 0 For this to be true either:cosx = 0 or cos 2 x = 1for cosx = 0This oc...
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Answered by Georgiana C. • Maths tutor
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Find the first three terms in the expansion of (4-x)^(-1/2) in ascending powers of x.

(4-x) -1/2 = (4 (1 - x/4)) -1/2 = 1/2 (1 - x/4) -1/2 = 1/2 ( 1 + x/8 + 3x 2 /128 + ...) = 1/2 + x/16 + 3x 2 /256 + ...Therefore the first three terms are 1/2 + x/16 + 3x 2 /256
Answered by • Maths tutor
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