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@missradders I don't think they are terms that have a particularly formal definition, but my understanding is that it isn't incorrect to use the term variable, even when the equation is being or has been solved and the values of these variables have been found.
@mrenglishmaths @robotmaths @mrshawthorne7 NB it is interesting to see that we have to pick a convention for the square root function. If √x was allowed to be negative, we’d have something like this: 1 = √1 = -1 giving 1 = -1...
@mrenglishmaths @robotmaths @mrshawthorne7 have the space for two answers because that is not what the function is defined to do. However when just speaking about ‘square roots’, there is the option for 2 answers
@robotmaths @jessica__sands @KerryDunton@PearsonEdexcel @EmporiumMaths @MathsNot I agree with everything you say; there are fundamental problems with the question but it is good that it is encouraging debate. This is an old resource, and I don't think it is still on the website. The website is being updated and there is a better version of these on there
@robotmaths @jessica__sands @KerryDunton@PearsonEdexcel @EmporiumMaths @MathsNot ...a different point of view, eg "why is 0.025 a well chosen value"? In terms of Question 1, I may be looking at a different sheet as we have a few published, but the table is accurate I think. I do agree that 1.0 < r ≤ 2.0 is better but 1 < r ≤ 2 is still true
@robotmaths @jessica__sands @KerryDunton@PearsonEdexcel @EmporiumMaths @MathsNot If asked to use a suitable value in place of 'tr' in a data processing exercise, either the value they want you to use would be made clear or any well justified value would likely be accepted - eg 0 or 0.025 mm seem well chosen. Questions may also assess knowledge from...
@ygbjammy So essentially, the logic is 1. constrain the system so that any solution is not unique and 2. show that \vec{0} can be a solution for a suitable value of b (13). This then gives an infinite number of solutions
@ygbjammy ... if there are infinite number of solutions, then \vec{0} must be one of these. Looking at the symmetry of the inverse matrix, it is sufficient to require the third row to be 0, since the value of b we get makes all the rows 0. This is a solution, and with a=-3 it is not unique
@missradders I think they’re counting the number of terms in each case and using that as b in Q4. Similar for Q5 to get 6 (4 terms +2) and 8 (4 terms x 2)? Not sure about the 11 or 18 though...