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I whole-heartedly agree with sleepplease's method of stress relief :) Works great for me.

Re: 回覆: Re: How many hrs per night? The key is to take genuine interest in your subjects, then you'll actually look forward to researching and learning more about your subjects, and studying won't ever have to seem like a 'chore'. At least, that's the way I see it :wave:

But x =/= rt2, as the method used to derive an answer for x in graph (B) is incorrect. To answer your first question: the values are different because you can't use graph (B) to work out the maximum area because you have allowed the length to equal x, and you're assuming this x is equal to the x...

We're to find points of intersection that give the largest rectangle that can be inscribed in the semi-circle. So we find an expression for the area of the rectangle in terms of only one variable (x) and then solve dA/dx = 0 in order to find dimensions that will give the maximum area. This gives...

(B) doesn't give the correct answer as you're dealing with a semi-circle, so you have to take special care here. It's hard to explain but essentially, the X in the diagram doesn't correspond to the x in the equation of the semi-circle, so you're obviously not going to get the correct answer. The...

Seems as if the brochure you recieved listed the UAI's required for a commonwealth supported place (CSP), as opposed to the DFEE place, where you would have to pay more to get in, but the UAI required is lowered.

Awesome, thanks for that man.

Yeah xiao I knew that, perhaps I should of said 'it was due to a mistake'.

How can you expect us to answer a question with vital details missing? :confused: EDIT: Or perhaps that's why you got a different answer to the textbook/resource that has supplied this question? :D In that case, maybe it's the textbook's fault, but you still must review your wording, it does...

It was just a mistake.

I prefer to learn by listening and taking notes, then I can formulate my own ideas and build on my comprehension of the subject matter. Also for maths I just do questions, questions and more questions! Only way to succeed IMO. Just practice, as always.

You can't use this formula because this formula is derived as the sum of all forces acting on a rocket at the point of lift-off. There are different forces that act on a rocket once it has reached orbital velocity.

Maths ext. - 9 Maths ext.2 - 9 English adv - 7/8 Chemistry - 8 Physics - 9 German beginners - 8

Not a problem :)

Think logically; - Where f(x) has turning points, f'(x) = 0 and hence cuts the x-axis - Where f(x) cuts the x-axis, f(x) = 0 and hence f'(x) = 0 - Where f(x) is increasing, f'(x) > 0 and hence above the x-axis - Where f(x) is decreasing, f'(x) < 0 and therefore below the x-axis -...

At lift-off, this formula is applicable because we are simply summing the forces and calculating the acceleration upwards, however when we are in orbit, as suggested by the question, the two forces acting on the rocket are the thrust force, produced by the rocket engines, and the gravitational...

(b) Assuming you're asking for acceleration in the horizontal direction; f (net) = T = ma (this is the net force acting on the rocket to produce a horizontal acceleration) Hence: a = T/m where m = 15% * 32000 = 4800kg a = 83.333... = 83m/s/s g force = apparent weight/true weight = (mg +...

Yeah the weight force and the acceleration due to gravity are acting in the same direction so it's much easier to assign downwards as positive, and attain a positive answer as a result.

Because the acceleration is increasing at a rate of 9.8 m/s every second. Hence, we define it as 9.8 m/s/s. In reference to the OP's question, acceleration should be taken always as negative, as long as you're consistent with signs attributed to any values that indicate motion upwards to be...

'sif you need a song to remember y=mx+b! You deal with it so often, how could you forget? :)