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The equation of the line L1 is y = 3x – 2 The equation of the line L2 is 3y – 9x + 5 = 0 Show that these two lines are parallel.

In this question, you are being asked to show L1 and L2 are parallel. The equations of two parallel lines will have the same gradient. This is the number in front of the x term in the equation, but to compar...
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Answered by Abigail W. • Maths tutor
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log3 (9y + b) – log3 (2y – b) = 2, Find y in terms of b.

Use laws of logarithms to simplify.Log 3 ((9y+b)/(2y-b)) = 2(9y+b)/(2y-b) = 3 2 (9y+b) = 9(2y-b)9y+b = 18y-9bCollect terms.9y = 10by = 10b/9
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Answered by Roy A. • Maths tutor
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Expand and simplify (3a+b)(a-2b).

To expand, you would multiply each term in the first bracket set by each term of the second bracket set. Thus, your expansion would mean:3a^2 -6ab + ab -2b^2 To simplify, note that there are two 'ab' terms i...
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Answered by Hansika R. • Maths tutor
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The tangent to a point P (p, pi/2) on the curve x=(4y-sin2y)^2 hits the y axis at point A, find the coordinates of this point.

p=4pi 2 differentiating with respect to y we have dx/dy = 2(4y-sin2y)(4-2cos2y) substituting in the value of y =pi/2 we have dx/dy = 24pi, which means dy/dx =1/pi24using (y-y_1)=m(x-x_1) we have y-pi/2=1/24p...
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Answered by George N. • Maths tutor
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The parametric equations of a curve are: x = cos2θ y = sinθcosθ. Find the cartesian form of the equation.

x = cos2θ y = sinθcosθcos2θ = cos 2 θ - sin 2 θ cos 2 θ + sin 2 θ = 12cos 2 θ = 1 + cos2θ cos 2 θ = 1/2(1 + x)2sin 2 θ = 1 - cos2θ sin 2 θ = 1/2 (1 - x)y 2 = sin 2 θcos 2 θy 2 = ( 1/2(1 + x)) . (1/2 (1 - x))...
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Answered by Amelia N. • Maths tutor
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