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Think of the question in the context in binomial theorem. There are 8 (2*2*2) possible scenarios of which 4 of those scenarios are. 3 questions right 2 question right . . . 0 questions right

Yes that is correct. Good job

So it is important to know your definitions. Do you know how an if-then statement is written logically. A iff (if and only if) statement is just (p -> q) and (p<-p=q). You need a) know how to write using p->q using logic then use Demorgan theorem to get the negation.

Are all isoclines solutions to the DE ? Yes horizontal isoclines always horizontal asymptoetes? Horizontal lines not asymptotes. The isocline is the solution. https://en.wikipedia.org/wiki/Isocline

Thank you yeah that is probably better idea

Perform long division. You should end up with two simultaneous equtions

So x, y represent some unknown digit. There 10 possible values x could take. So if x =9, and y = 5 we would have (for the permutation xxyy) 9955. So the second character depends on the first character that is 9. Similarly the fourth character depends on the third character that is 5. That...

Altogether there are 10000 codes. let x represent digit 1 that is repeating. Let y represent digit 2 that is repeating. So the number of permutations of xxyy is: 4! / (2!*2!) = 6 # This is 4C2 or 4 choose 2. There are 10 possible digits x could take and 9 possible digits y...

Can you have go at the question first. The key is just give it go. So maybe you could try and solve some sub-problem first. And use that to get the answer.

The main math jobs I can think of are - operation research (including quantative finance analyst) - data scientist I think it might be good for you develop some of your knowledge or skills outside of uni. E.g. you perhaps you could take an online course on programming on Udemy or some other...

I presume you could do that. So you could hold off doing any science for couple years. And if something sparks your interest in science you could take it. Otherwise you could drop the science degree in the future. It best to check with your uni.

c) If he is due west his starting point. Then his bearing is 270 degrees. ii) Likewise 90 degrees since his starting point is east.

Just a minor note. Although a typical engineer will have knowledge of physic and chemistry a software engineer, IT engineer etc will not require much chemistry or physic. For the IT-related engineering degree I have taken I have only required 1 physics course and no chemistry course.

So ii) In this case there are two possibilities, x^p-1 is positive. If x^p-1 is positive then x^q-1 must be positive otherwise (x^p-1)/(x^q-1) will be negative. Similarly if x^p-1 is negative then x^q-1 is also negative. Does this this help.

Having a quick read of the handbook there is no requirement of bridging course but it may be good idea to double check this.

I think both can be both important. Let supposes you go for job in Python, it probably will not go well if the job requirement has Python. Similarly if you don't communicate respectfully it won't end well. Maybe you have taken some initiative for some project, or perhaps you have planned...

I agree they are the same. The meaning are only slightly different. As you can say the multiples of 3 are: 3, 6, 9, etc But you cannot say divisible of 3.

So this questions involves the general solution of sinusoidal equations. https://boredofstudies.org/threads/trig-help.397763/#post-7405412 z = cis(pi/4) => z^n = cis(pi/4 * n) => cos(pi/4 * n) = 0 => pi/4*n = 2*pi*k +/- pi/2 => n/4 = 2*k +/- 1/2 => n = 8*k +/- 2 => n = -6, -2, 2, 6, 10...

To be honest you have lost me with your question. So, 0 + …. + (n-2) + (n-1) and 0 + 1 + ... + (n-1) are the same. I believe that a group with one person can make zero handshakes. Of course it will be zero. It seems like the formula for is based on sum of an arithmetic sequence. (a +l)/2 *...

I think you are looking at metcalfe rule or law. initial statement is the first statement e.g. we prove true for the n = 1 inductive statement is assume true for n = k .