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From the reactions you have written, you can see that Ni is losing electrons, thus it is being oxidised. You can also see this from how it's oxidation state increases from 0 to 2+.

The oxidation reaction occurs at the anode, and the reduction reaction occurs at the cathode. You've already identified that though, so do you mean something else?

The table shows scores for one pair of cubes. The scores which will give a total of at least 41 are (first pair/second pair): 25/25, 20/25 and 25/16. For 25/25 you need to roll two 5s with both pairs of cubes. The probability of this is (1/6)^2 * (1/6)^2 = 1/1296. For 20/25 you need to roll a 4...

Thanks for the help!

$If $2^x=5^y=10^z$, using logarithms to base 10, or otherwise, show that $\frac{1}{z}=\frac{1}{x}+\frac{1}{y}$.

Can you give an example of when it's on the right?

The redox reaction can be broken down into a reduction and oxidation reaction. From the data sheet, you can see that there is a reaction which involves Cr2O72- which produces Cr3+ and another involving Fe2+ which produces Fe3+. These reactions are: Reduction: 2Cr2O72- +14H++6e^- <-> Cr3+ + 7H2O...

I think this is what it means: the random trial is repeated several times, so you end up with several three-digit numbers by the end. For the 5th trial to produce the second three-digit number which is a multiple of 3, there has to have been only one three-digit number which is a multiple of 3...

Re: 2015 permutation X2 marathon Okay consider if the question was instead asking for the set {1,2,3,4}. A subset from this set is a set which consists of the elements of the original set. So the subsets which have do not have any consecutive integers from this set are: With 0 elements: {} -...

Re: 2015 permutation X2 marathon The set {} has 1 subset with no consecutive integers including the empty set. The set {1} has 2, the set {1,2} has 3, the set {1,2,3} has 5. The amount of subsets with no consecutive integers increases as the amount of numbers in the set increases, in...

Re: 2015 permutation X2 marathon What's the answer?

Re: 2015 permutation X2 marathon Letters: A EEEE L N P SS T Y Case 1: 5 unique letters - ALNPT 8P5 = 6720 Case 2: 3 unique letters, 2 identical letters - EELNP (7C3*5!)/2! = 2100. Multiply by 2 as it can happen with EE or SS, so the total number of words is 4200. Case 3: 2 unique letters, 3...

Re: 2015 permutation X2 marathon Ah right thanks, I was wondering where I went wrong. a) If 0 Ps, then there is only 1 selection with the remaining 4 letters. The total selections is 11. b) If 0 Ps, then there is 4!=24 ways of arranging the letters. The total amount of words is 192.

Re: 2015 permutation X2 marathon a) If there are two Ps, we choose 2 letters from the remaining 4, so there are 4C2 selections. If there is one P, we choose 3 letters from the remaining 4, so there are 4C3 selections. 4C2 + 4C3 = 10 selections in total. b) If two Ps: From a), there are 4C2...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Okay let's focus on the part outside the brackets, that is $\frac{h}{3}$. $h$ is the height of one of the intervals. The height is also the average of the first x-coordinate of one interval, and the 2nd x-coordinate from it...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread \begin{align*}d&\approx\frac{h}{3} \left [\textup{First} + \textup{Last} + 4(\textup{Odd}) + 2(\textup{Even}) \right ] \\&=\frac{1}{3} \left [1.5+2.4+4(1.3+2)+2(1.4) \right ] \\&=\frac{1}{3} \times \frac{199}{10}...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread It isn't a normal integration like that. To do the question, you should realise that y=\sqrt{9-x^2} is the equation of a semicircle with a radius of 3 units and centre (0,0). So the area from x=-3 to x=3, is the area of the...

Re: HSC 2015 3U Marathon

There's already a thread for this: http://community.boredofstudies.org/18/physics/339609/2015-independent-paper-physics.html

Re: 2015 permutation X2 marathon It is not correct. The correct answer is in white under the question. For your calculation, what if the 6 consonants become the first 6 stars? Then we could have A C M I B N I N T O O. For this case, the last two vowels are not separated by a consonant.