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part b: You have to solve the two equations simultaneously so that any points of intersection will be shown as the zeroes of the resultant graph \\y=x^3\\y=x^2-k\\\text{sub two equations together}\\x^3=x^2-k\\x^3-x^2+k=0 I think the hardest part of the question is finding out how many points...

for question 5 we have to find the general term of (3x-y)^{10} \text{General Term}\\\\{10 \choose k}{\cdot}(-1)^{10-k}{\cdot}y^{10-k}{\cdot}3^k{\cdot}x^k for x to be of power 7, k has to equal 7 so we sub it into the general term \\{10 \choose...

I think I might of explained it badly, the height of the triangle at the top would be the hypotenuse of the triangle at the front. Since they are both right angle triangles. \\h=\sqrt{5^2+12^2}\\\\h=13cm find the height of the top triangle...

You made a mistake in calculating the area of the top triangle. If you look at the pyramid from a birds eye view, the triangle at the very top is a right angle triangle as its vertices bend to the same degree as the triangle at the base.

you just needed a few steps more to finish the question \\\cos(x)\sin(\theta)(1+A)=\sin(x)\cos(\theta)(A-1)\\\\\frac{\sin(\theta)}{\cos(\theta)}(1+A)=\frac{\sin(x)}{\cos(x)}(A-1)\\\\\tan(\theta)(1+A)=\tan(x)(A-1)\\\\\tan(x)(A-1)=\tan(\theta)(1+A)

For part a: you basically need to know that x=-r+PA using the cosine rule we can deduce that \\PA^2=r^2+(2r)^2-2(r)(2r)\cos(\theta)\\\\PA=\sqrt{5r^2-4r^2\cos(\theta)}\\\\PA=r\sqrt{5-4\cos(\theta)} Sub PA back into the equation x=-r+r\sqrt{5-4\cos(\theta)} for part b) i): we gotta...

I think you use the area of a triangle formula (base is 4 and height is 8) and then subtract it from the area of the rectangle.

Part b: First we need to find an expression for the PB and QB that involve h \\cot(36\degree)=\frac{PB}{h}\\\\PB=h\cot(36\degree) Similarly we can deduce that QB=h\cot(27\degree) Now using the Cosine Rule 250^2=PB^2+QB^2-2(PB)(QB)\cos(74\degree) Sub in our values for PB and QB...

You gotta expand the brackets and simplify from there {10\choose k}\cdot(2x^3)^{10-k}\cdot(3x^{-2})^k ={10\choose k}\cdot(2)^{10-k}\cdot(x^3)^{10-k}\cdot3^{k}\cdot(x^{-2})^k ={10\choose k}\cdot2^{10-k}\cdot{3^{k}}\cdot{x}^{30-3k}\cdot{x}^{-2k} ={10\choose k}\cdot2^{10-k}\cdot{3^{k}}\cdot{x}^{30-5k}

let d= arc length let l= that hypotenuse created by the right angle triangle let \theta= the angle subtended the origin from point S and the final position of the man using d=r\theta radius is 1km so d=\theta using the cosine rule...

I know you

I'm pretty sure you use the equation v=f\lambda Frequency is just 200hz Wavelength is 1m (multiply 50cm by 2) sub em into the equation v=200*1=200ms^{-1}

Sorry about that, hopefully this helps you. S_{n} = \tan\theta\tan\2\theta+\tan2\theta\tan3\theta+\tan3\theta\tan4\theta+...+\tan{n}\theta\tan((n+1)\theta)\\\\\text{using the expression}\\\\tan(a\theta)\tan((a+1)\theta) = \frac{1}{\\tan\theta}(\tan((a+1)\theta) - \tan{a}\theta)-1\\\\S_{n} =...

\text{Part i)}\\\text{Formula for} \\tan(\alpha -\beta) = \frac{\\tan(\alpha) - \\tan(\beta)}{1+\\tan(\alpha)\\tan(\beta)}\\\\\text{Therefore}\\\\\\tan(3\theta - 2\theta) = \frac{\\tan(3\theta) - \\tan(2\theta)}{1+\\tan(3\theta)\\tan(2\theta)}\\\\\\tan(\theta) = \frac{\\tan(3\theta) -...

I don't think so, expanding everything out seems like the easiest way to solve it.

The particle accelerates until t = 3 so it can't be t = 1 or 2, and once it passes t = 3 it decelerates so it can't be t = 5. So I think the answer is C