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MolloyMaths
Ireland
เข้าร่วมเมื่อ 22 ธ.ค. 2013
This channel includes videos on various Mathematics topics on the Irish Junior and Leaving second level syllabus (as well as topics not on either syllabus)
See www.molloymaths.com for more information.
See www.molloymaths.com for more information.
วีดีโอ
The Factor Theorem An Example
มุมมอง 5014 วันที่ผ่านมา
This video shows one example of how the factor theorem can be used to solve a quintic equation.
Theorem 13
มุมมอง 60หลายเดือนก่อน
This is a proof that given two similar triangles, sides are in proportion, in order.
I thought I understood it until we added a 2Npi why do these appear isn’t it just cos pi raised to 1/3
Like why did it get used in this question and not all the other ones what indicator was there
@@danielcobban8755 When converting a Complex Number that's in rectangular form to Polar form, there is just one answer. You do not include 2nPi. If on the other hand you have to SOLVE a Cubic equation (as in this case), you have to allow for all possible solutions - in this case three solutions. So you have to write out the general form i.e. include the 2nPi.
@ if I’m understanding correctly if it asks for roots add the 2npi if it just wants say to the power of 9 don’t add the 2npi
@@danielcobban8755 Yes. If it's a cubic equation for example there are three roots. You use 2nPi - the general form to find all the roots starting at n = 0 etc. If you are raising a complex number to the power of three for example you don't.
If you multiplied (3+2i) by (1-i)^8 and got (5-i)^8, almost working backwards to your method, and then doing De Moivre's Theorem would you get the same answer? And if not, how many marks would you get?
Should work. How did you do the multiplication? Seems like a very long procedure.
@@molloymaths1092 Sub in values for initial geometric sequence and find an overall complex number. (5-i)^8 i think
@@Shauna_OConnor If you multiplied (3+2i) by (1-i)^8 you should get (48+32i). If you do it using a Geometric sequence when do you use De Moivre's Theorem? I think that they require you to do it the way that I've done it but I'd be interested to see your method in full.
Would theta not be 11 pie /6 because of its position in the 4th quadrant?
11 Pi/6 is the same angle as - pi/6
@@molloymaths1092 thanks
Why is x=p
You're given it as x in the question.
Thanks for the videos they are making a big difference in my grades.
You're welcome. Glad they help.
Thank you so much!
You're welcome.
On the graph, your x axis shows the time of day starting at 00:00. But t is the time in hours starting from the first high tide. So shouldn’t P be 0? And the start of the graph (at midnight) be minus 2?
You are told in the question that p high tide. If you take the final formula 3.6 + 1.9Cos(0.5t) and put in for example 2 hours (i.e. at 4 am) you get 4.6 which is correct from the graph. The same if you put in -2 hrs (Midnight). You also get 4.6. t is the time from the first high tide which is at p.
Excellent job. Thank you, easy to follow. 🎉🎉🎉🎉
You're welcome.
how do you know what to do with each row?
You need to end up with 1, 1, 1, in the diagonal and the rest zeros. How you get there may be different to the way I got there.
for the Θ part why is did you put -π/3? is the answer not 2π - π3 which is 5π/3?
The marking scheme allows for either one. They both get you to the correct answer. Conventionally, when converting from rectangular form to polar form, theta is between zero and Pi.
@@molloymaths1092 I know that the marking scheme accepts both answers, I want to understand why both values are acceptable.
@@Reddglawer They're the same angle.
@@molloymaths1092 thank you.
So I’ve seen two comments wondering about why g(x) is set to K*exp(kx^2). So I’ll explain why. If you use the substitutions: h(z) = log(g(sqrt(z))) u = x^2 v= y^2 You get h(u+v) = h(u) + h(v) Where h(w) = aw+b are a class of functions that satisfy the property. If I remember correctly, there are other classes of functions that satisfy this property, but they are weird, I think pathological if I recall correctly, which I guess aren’t as “nice” as linear functions.
For b (ii), it says "find as a percentage, correct to 3 significant figures". So why take the significant figure approximation from 'i' rather than the percentage value. As in, use 1.654%. Maybe I'm being a bit pedantic!
I presume you mean parts (ii) and (iii) from b. Normally in these questions you would take the answer from the previous part of the question unless otherwise told. It would be harsh to penalise you for using the unrounded number but they may do so. I'm not sure but thanks for your comment.
sortest path is SCGIT -->>34 Is it correct or not ply confirm me.
helpful
The square root of w^2 is plus or minus w, so you only need to find one value of w then just get the negative of that. That means you don’t need to write it in the generalised format and the question is much easier. Excellent set of videos, though!
Yes but it's part of a general procedure for higher powers. Thanks for your comment.
thank you
You're welcome.
these solutions are wrong. q 7 - look at the marking scheme. your answer is not correct.
Other than the fact that I didn't convert to minutes and seconds in part (e), I can't see what you are talking about.
amazing
Legend is learning 1 day before exam
You can do a general proof by contradiction to show that the square roots of all positive integers (that are not perfect squares) are irrational.
Thank you very much
You're welcome.
Thank you for this! Very clear and straightforward and actually makes Gauss-Jordan elimination fun!
Thanks. Glad it helps.
Wow! Thank you so much for making me great
No problem. Thanks.
Not a gauss-jordan!
Yes it is! When using RREF it's Gauss Jordan Elimination. When using REF it's Gaussian Elimination. I have used RREF in this video. I think I mention Gauss Jordan at the end of the video. Thanks for the comment.
Thank you.
You're welcome.
Vos explications sont tellement simples, j'ai le bac à passer et j'me fais une remise à niveau force à vous
Merci. Bonne chance.
Can equation #7 be factorized any further into (x - 1)(2x + 5)?
Yes it can. The idea is to simplify the fraction so either way is fine. Thanks for the comment.
Safe to say I couldnt understand anything past 9:00. Could you leave a trail of references and math topics that a student could look into to better understand?
You'll have to be more specific.
Where are you from brother
Ireland
The work is moving on fantastically😅😅
Great
Will watch this later, need it for my engineering foundation maths. Eeep
Good luck
I am grateful that teachers like you exist
Thank you.
Thanks ❤
You're welcome!
Lovely video, great intro to the formula used for parametric diff.
Thanks.
Can you do the physics paper by any chance?
Sorry. I'm afraid I will not be doing the Physics paper.
@@molloymaths1092 by any chance could you just do the circular motion question on its q7.
For the force diagram, could you just symbolically put R as normal force, Ff as friction and such? Like I didn't specifically include the actual value of those forces.
Yes. I see in previous marking schemes that they didn't include all the values. Have a look at last year's marking scheme for the connected particles question. I probably over did it a bit.
Thanks for clarifying
Thanks for reply U would be forgiven to think it wasn’t a right angle triangle I would suggest the examiners made it look anything but a right angle triangle in their sketch which I feel is sloppy for a state exam Would you agree ? Really enjoy your channel
There is a little square at O indicating a right angle. With 3-D questions, right angles very often don't appear to be right angles in the diagram. You've got to imagine the real-life situation as described.
Could you please do the 2024 applied maths paper solutions🙏🏿
I'll try and get them done this week.
@@molloymaths1092 thank you so much
Is the triangle RHO supposed to be or assumed to be a right angles triangle ?
Yes. It is given as a right angled triangle in the question.
why is RVA = v1 - V2 and not V2 - V1. if its final vel - initial vel. same with u
V1 and V2 are final velocities of the two particles. The difference is the relative velocity (After)
was the best long question on the paper honestly
Would u lose many marks if u switched the null hypothesis and alternative hypothesis but still got the right answer and right conclusion
Not many I would imagine. I would think HPC at least.
Well explained
Thanks
man you are the king you have helped me so much i used to be failing hl maths 🙏
Thanks. Glad to have helped.
have paper 2 tmr cheers for the help
Good luck.
For simultaneous equations can't you minus the two equations instead of adding them
I think I did after multiplying by minus 3.
Any predictions for tomorrow?
Difficult to predict. Hopefully it will be at the same level as paper 1.
for part c, how do u know when the cosine / sine value is the reference angle, and not just the angle?
In part c, you have to solve two equations. Sin t = 0 and Cos t = 1/2. In all these cases there will be two solutions. Sin and Cos (and indeed Tan) are positive in two quadrants and negative in two quadrants so you will get two solutions between 0 to 360 degrees. Any angle greater than 90 degrees will require you to look for the reference angle, find the Sin, Cos or Tan of the reference angle, and then decide what the sign will be.
@@molloymaths1092 ok thank you so much!
For part C why did you do cos 135 could it not have been cos 45???
45 is in the 1st quadrant. 135 (and Q) are in the second quadrant. 45 is therefore only the reference angle. If you did the cos of 45 you would get (1/root2, 1/root2) so the sign would be incorrect.
@@molloymaths1092 thank you so much you're a life saver! Maths paper 2 is tomorrow 🤞
@@Shawnmusic1224 Best of luck.