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Great stuff you are nearly there, using the fact that \sin{2x}=2\sin{x}\cos{x} what should \sin{x}\cos{x} give us? Work this out then someone will give you the next clue.

Okay, du question ay. Well, start with making y the subject and this will allow you to have 3y=2x+6 giving us y=\frac{2x}{3}+2 point P can be anywhere on the line. Now Q is where you want the line OP to be divided 3 to 2 internally so what happens is that we will pretend that OP=\left(x...

Someone decided to respond to a 20-year-old question, mate they're like 40 years old now and this response is 20 years late.

By letting \tan{\theta}=\frac{1}{3} you are stating that \theta=\tan^{-1}\frac{1}{3} which is exactly what you want considering where you finished from part ii. Considering the question is stating that \tan^{-1}\frac{24}{7}=4\tan^{-1}\frac{1}{3} by using \theta=\tan^{-1}\frac{1}{3} we then have...

Now one-tailed test. That is where you need to know what the alternative hypothesis is pointing at and then you will have the P-value pointing in the same direction towards the test statistic which would be found using a modified z-score technique.

I believe for accuracy you should put a tilde above H_{0} because H_{0}:\mu=70 is the boundary condition for the question. Like this \widetilde{H_{0}}:\mu=70

IIRC to determine null and alternative hypothesis the alternative hypothesis is when this statement is true null hypothesis is when it is not.

\left(\cos{\theta}+i\sin{\theta}\right)^{6}=\cos^{6}\theta+6i\cos^{5}\theta\sin{\theta}-15\cos^{4}\theta\sin^{2}\theta-20i\cos^{3}\theta\sin^{3}\theta+15\cos^{2}\theta\sin^{4}\theta+6i\cos{\theta}\sin^{5}\theta-\sin^{6}\theta...

This is just from what I know the function of a function, replace g\left(x\right) with the equation given and then let f(...) where ... is the equation and then sub that new function into what f\left(x\right) is.

For the question \frac{\sin{x}+\sin{3x}+\sin{5x}}{\cos{x}+\cos{3x}+\cos{5x}}=\tan{3x}, we can start like this \frac{\sin{x}+\sin{3x}+\sin{5x}}{\cos{x}+\cos{3x}+\cos{5x}}=\frac{\sin\left(3x-2x\right)+\sin{3x}+\sin\left(3x+2x\right)}{\cos\left(3x-2x\right)+\cos{3x}+\cos\left(3x+2x\right)}...

Okay, in a race if someone finished ahead of a certain amount of competition then they must have finished the race with less time than their competitors. In that case, what should happen is that suppose Ozzie finished ahead of 16% of the competitors then he would have 65 minutes because he is...

Have a look at this solution.

Find a reduction formula question and then in the question it should have find I_{5}[\TEX] or something similar. What you do is find I_{0}[\TEX] and then work your way up to the number desired.

@Drongoski’s advice is just in general but specifically for what you specified there is the fact that when you are solving a reduction formula one can start at I_{0} to minimise all forms of difficulty because the reduction formula you found on the previous part will help you if you follow this...

Computer science and nursing are two completely different things. In nursing, you are interacting with humans and listening to the doctors and what medication to take and with computer science you are working with programming and stuff, which one do you want?

Steve Howard has a very interesting way of interpreting this idea so what happens is that if a = b = 0 then the numerators will be 0 and therefore there will be no square roots.

Is that your question Find the minimum value of x^{2}+y^{2} given that xy\left(x^{2}-y^{2}\right)=x^{2}+y^{2}, x\neq{0}

Which Dr Du class are you in? a) Okay with this question we are required to find the auxiliary angle which is in the form R\sin\left(x+\alpha\right) for \sqrt{2}\sin{x}+\sqrt{2}\cos{x} Well, R\sin\left(x+\alpha\right) = R\sin{x}\cos{\alpha}+R\cos{x}\sin{\alpha} where R > 0 \&...

@xibu34 In this question, we are told that the side length is increasing at 0.12mm/s. and that the side is 150mm. To start off we can see that \frac{ds}{dt}=0.12 noting that s represents the side and t represents the time. Now, what is the volume of a cube? V=s^{3}, and \frac{dV}{ds}=3s^{2}...