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Thank you, I sort of get it, so that means if cosx is a negative, the root of cos^2(x) is still a positive cosx but the numerator is a negative cosx so when times with -1 = +1 right? Can you show me how to write the answer in exam form

well its from excel 3 u , it says that +-1, as cosx -+, i am not sure whether is it to do with the absolute value thing or not. Not sure whether the answer is correct or not...

differentiate cos^-1(sinx) I got -cos/ square root (cos^2(x)) so cos cancel out and is -1. But the answer is +- 1. I don't really understand it... can any one explain? Thanks

o right thanks guys

Just wondering, for 4 unit, do u need to remember Product-to-Sum Formulas and Sum-to-Product Formulas in trig identities. Thanks

your right, i missed about bunch of other negative independent terms... thanks I got it now

ok, so this is the question: to find the independent term of x where (x+(1/x))^40)(x-(1/x))^40) when I didn't factorise this and working with the first bracket and then times the second bracket , it's this 40C20 * 40C20 yet if i factorise it into (x^2- x^(-2)) the answer is 40C20. Wth... can...

Can any one explain this to me, i am lost

make the volume spin at y axis x^2=a^2 - y^2 where limits are y= a and y= a/2 integrate that: a^2y- y^3/3 then substitute a and a/2 subtract them and you get 5a^3pi/24

Sorry, the y is a+a(p^2) But yea,I got it, thank you deswa and others!

The first one, a+ p^2, not (a+p)^2

thanks guys but still struggling with the focal length... i don't even get wat is the focal point in here...

Show that the locus of M is a parabola and find its vertex and focal length, where M is (ap, a+ap^2) How do you the focal length and vertex after I worked out to be this: x^2=ay-a^2? Can anyone help me out... thx

Its 5C2(0.5)^2(0.5)^３＝　５／１６ If p is the probability of success and q is the probability of failure for an event, then the probability of r successes in n independent events is given by P r successes ｎＣｒ　Ｐ＾ｒ　ｑ＾（ｎ－ｒ）

In part 3, should it the 2 equations be 2-2a-b=0 and 6-4a+b=0 , therefore a=2 and b=2?

well, last digit with 3 or 9 may not be divisible by 3, for example, 13 is not, nor 19... the only method i can find is what i did, for example the sum of digit 138 is 12 which is divisible by 3, there for, all arrangement of 138 is divisible by 3... like 381,813 etc. But this method is really...

The answer is 50.

Oops forgot to mention , no repetition

Yea

number less than 4000 using 1,3,5,8,9 how many of them are divisible by 3 I did this question in a long way, using that the sum of digit equals is divisible by 3 like 1359, there is 2*3! number which is divisible by 3. can someone show me a short method for this? Thanks