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a) radius = 12/2 = 6cm. Area of a circle = pi x r^2 = 36pi. Area of semi-circle = 36pi / 2 = 18pi cm^2b) Area of full circle = 120. pi x r^2 = 120. r^2 = 120/pi. r = sqr root (120/pi)= 6.18cm

We are going to have to use the cosine rule for this type of question. The cosine rule is a2=b2+c2 - 2bcCosATherefore a2= (6.46.4) + (5.65.6) - (26.45.6)*Cos107.9 If we plug this into ou...

= (2x^2-3)(2x^2-3) -{(2x^2+2)(2x^2+2)}= 4x^4-6x^2-6x^2+9 -(4x^4+4x^2+4x^2+4)= 4x^4-12x^2+9-4x^4-8x^2-4= -20x^2+5= -5(4x^2-1) or 5(-4x^2+1)

For both the two main subjects I intend on teaching in this platform, I highly recommend to my students to focus on the theory as a first step (this could be a definition, some notes, etc). So, by trying ...

This is a inequality question. There are two separate inequalities and the values n can take are the solutions of n that overlap between the 2 inequalities. First inequality: 3n< 12 therefore n <4 ....