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di) (x+4) > 2 / (x+3) The graph will look like this So if you can find the intersection points you can determine the values that satisfy the inequality (x+4) = 2 / (x+3) (Note x cannot equal -3) => (x+3)(x+4) = 2 => x^2 +7x +12 - 2 =0 => x^2 +7x +10 =0 => (x+5)(x+2) = 0...

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Yes, it will not matter for the most part.

Hi ROL13D, the choice does not really matter in this case. The most general approach would be x = acos(nt + alpha) + c. In this case I would use x = acosnt + c presuming you want time to be zero at 6:13AM. By making this choice a = -0.7, c = 1.2. In the case of x = c - acosnt, then a = 0.7...

//Hopefully it might be start cannot be bothered doing it all Suppose we could model the amount of the chemical by k * e ^ (-alpha*t) where alpha > 0 and t is the time in weeks. Therefore k is the initial amount of the chemical. The percentage of the chemical is given by k * e ^...

"I know it says to consider it's a Bernoulli, but would it be a binomial since there is a 3rd option on landing on the edge, and we have the probability for it landing on the edge?" No For Bernoulli and Binomial distribution there can only be two outcomes, success or failure. You cannot have a...

So you could just rewrite the expression with Tn+2 as the subject.

nC4 + nC5 = n! / [(4!) * (n-4)!] + n! / [ (5!) * (n-5)! ] = 5*n! / [(5!) * (n-4)!] + n! * (n-4) / [5! * (n-4)!] = [ 5*n! + n! * (n-4)] / [5! * (n-4)!] = [ n! (n-4 + 5)] / [5! * (n-4)!] = [ n! (n+1)] / [5! * (n-4)!] = (n+1)! / [5! * (n-4)!] Solve (n+1)! / [5! * (n-4)!] = 12C5 =>...

I would start with discrete maths and then linear algebra. Since both topics form the foundation of many other topics and can help with understanding. e.g. once you learn about logic in discrete you should understand the epsilon-delta definition, and various other theorems involving logic.

"If x is a variable then can x+1 = x be anything?" For assignments you can change the value. If x+1 = x is an equation however you will end up with 0 = 1 which is not correct. In the case of assignment you should have single term on the LHS. So instead of x+1 = x the assignment would be x =...

StuDocu, but you may not be able to find all the documents you need for a course. Having some friends can also help.

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Not really. As such you should study series before induction.

Sometimes even though you may have completed the prequisite subject might not be good to enroll as some of the material learned might be outside the syllabus so the subject might be slightly harder.

The exact order you do course does not really matter so long you meet the subjectprequisites. However it is probably best talk with someone for advice on your degree as it might be helpful to do other subjects before taxation law. Btw: you can contact UTS enquiries if need be.

What is the purpose of this question? What you need to know would be in the textbook. However, you can add inverse function, curve sketching, basic simplification/expansion, rationalisation. As function play a role in many other topics, this list is not complete but I assume you were not...

Btw if you need to find textbook solutions you can try slader.com. Access through incognitio mode if they place a limit on the number of solutions that can be viewed.

Have you done any work with the textbooks?

So I am wondering if university can give you offers base on hsc exam mark? I don't think so. It is worth it if I do Diploma at UTS or Macquarie University and transfer it to Bachelor degree afterward? If you want to to. The UTS engineering diploma will allow guaranteed transfer to a bachelor of...

You are almost there. You wrote (2(k+1) -1)^3 as (2k-1)^3 instead of (2(k+1) -1)^3 as (2k+1)^3 as mentioned by xilbur. From there it should be an easy direct proof. Note: Make sure to prove true for n=1 and then assume true for n=k. Leaving these steps out might lead to deduction of marks