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log3 (9y + b) – log3 (2y – b) = 2, Find y in terms of b.

Use laws of logarithms to simplify.Log₃((9y+b)/(2y-b)) = 2(9y+b)/(2y-b) = 3²(9y+b) = 9(2y-b)9y+b = 18y-9bCollect terms.9y = 10by = 10b/9

Answered by Roy A. • Maths tutor

7812 Views

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=4pi² 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...

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 = cos²θ - sin²θ cos²θ + sin²θ = 12cos²θ = 1 + cos2θ cos²θ = 1/2(1 + x)2sin²θ = 1 - cos...

Answered by Amelia N. • Maths tutor

7524 Views

find dy/dx of x^1/2 + 4/(x^1/2) + 4

This can be rewritten as x^1/2 + 4x^-1/2+ 4, hence dy/dx = (1/2)x^-1/2 - 2x^-3/2

Answered by Olufayojimi O. • Maths tutor

2918 Views

(a) By using a suitable trigonometrical identity, solve the equation tan(2x-π/6)^2 =11-sec(2x-π/6)giving all values of x in radians to two decimal places in the interval 0<=x <=π .

say (2x-π/6)=qtan²q = sec²q-1so sec²q-1= 11 - sec q sec²q + sec q -12 = 0(sec q -3) (sec q + 4) = 0sec q = 3 or -4because 0<=x<= πso -π/6 <= q <= 11...

Answered by Zhaohui Z. • Maths tutor

8109 Views