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Imagine a sector of a circle called AOB. With center O and radius rcm. The angle AOB is R in radians. The area of the sector is 11cm². Given the perimeter of the sector is 4 time the length of the arc AB. Find r.

11 = 1/2 r 2 RAB = x = rRr + r + x = 4x2r = 3rRR = 2/3r²R = 22r² = 33r = √33
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Answered by Alice S. • Maths tutor
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x^3 + 3x^2 + 2x + 12

3x 2 + 6x + 2
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Answered by Jiawang Q. • Maths tutor
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How to perform integration by substitution. (e.g. Find the integral of (2x)/((4+(3(x^2)))^2)) (10 marks)

We begin by looking at the integral by itself. The first thing we must do is evaluate what type of integration we should perform. It's always important to make sure that it's not a standard integral such as ...
KB
Answered by Kenneth B. • Maths tutor
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A function is defined as f(x) = x / sqrt(2x-2). Use the quotient rule to show that f'(x) = (x-2)/(2x-2)^(3/2)

u = x v = (2x-2)^(0.5)u' = 1 v' = (2x-2)^(-0.5)f'(x) = (vu' - uv') / v^2Therefore, f'(x) = (((2x-2)^(0.5) * 1) - (x * (2x-2)^(-0.5))) / ((2x-2)^(0.5))^2f'(x) = (2x - 2 - x) / (2x-2)^(3/2) = (x-2) / (2x-2)^(3...
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Answered by Isaac F. • Maths tutor
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The equation of a circle is x^2+y^2-6x-4y+4=0. i) Find the radius and centre of the circle. ii) Find the coordinates of the points of intersection with the line y=x+2

i) x 2 + y 2 - 6x - 4y + 4 = 0 x 2 - 6x + y 2 - 4y + 4 = 0 group the terms together x 2 - 6x = (x - 3) 2 - 9 factorise the x terms y 2 - 4y = (y - 2) 2 - 4 factorise the y terms x 2 - 6x + y 2 - 4y + 4 = (x ...
RL
Answered by Ralf L. • Maths tutor
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