You are very welcome, friend. Thank you for leaving your message. Kind regards from Australia.

That is definitely one way to perform the calculation, Antonio! Another would be to calculate 10% of £8 (which equals 160 shillings) ... i.e. 16 shillings. Then one could calculate 10% of 5s 4d (which equals 64 pence) ... i.e. 6.4 pence. The total would be 16s 4d, and one would proceed in the same way that you described. Which method would be more appropriate would depend largely on the figures involved (and how one felt at the time). Thank you very much for your helpful contribution!

@@CrystalClearMaths Your method sounds a lot better, in separating pounds from shillings and pence, and I will try to memorise it. I didn't realise how straightforward it was, and thought that's why the £sd system was abandoned. I can calculate percentages mentally using your method. I bought an old pre 1971 arithmetic book, via ebay, to practise the maths. I've improved my arithmetic skills since doing this: I've memorised the 12 times table, and all the times-tables as we did when children at school. The half-crown was 2s/6d, (12.5 new pence) and there were 8 to a pound. A guinea is 21 shillings. There are four farthings to a penny.

@@atonio2476 I am glad that you have been discovering all this, Atonio. In some ways, it is quite an enjoyable exercise (changing bases between measurements). Thank you for your reflections.

@@CrystalClearMaths We weren't told about the older system at school, and would have benefited from one lesson, to familiarise us. The 'shilling' is a feature in older literature such as Charles Dickens, and this cultural aspect has been lost with decimalisation. I find that with decimal addition, the loss of the shilling creates a gap in the psychological processes.

@@atonio2476 I agree. It is mathematically (arithmetically) beneficial to learn other base systems and culturally and historically important to be aware of these measurements, too. I enjoyvreading historical books and it helps hugely when we understand references like 'shillings.' Thank you for your important observation.

Hehehehe ... thank you :-).

Thank you, kindly, John. I greatly appreciate your encouragement, and am glad that you are finding worthwhile things on this channel. Kind regards, Graeme

You are welcome, Komitet Gosudarstvennoy Bezopasnosti, and thank you for your observation. Kind regards to you from Australia.

@@CrystalClearMaths sir yuh still active here,how’s life going for ya

@@komitetgosudarstvennoybezo5216 A few ongoing challenges, K. Thank you for asking. I am still responding to comments, but have some significant matters in real life to sort out before I can see my way clear to resume posting videos. I am sorry. Kind regards to you, and thank you very much for your support and concern. Both are appreciated!

The second graph certainly LOOKS like a negative quartic graph, but it is not necessarily one. I randomly sketched it and no formula was provided. The object of the video was to explain and teach a principle or method for sketching gradient functions when the formula for the original function is unknown. It is quite possible for a graph of that shape (not a negative quartic) to have a gradient function as I depicted. You are, however, quite correct in this ... IF it was a negative quartic graph, the derivative function would most certainly be a cubic function.

Because the equation for the x-axis is y = 0. When we set y = 0, we are solving two equations simultaneously ... that of the polynomial (or any function for that matter), and the line y = 0. The results are all the points where the polynomial crosses or touches the x-axis (i.e. the points of intersection). Algebraically, we refer to these values as the 'roots' or 'zeros' of the polynomial. Graphically, they are simply referred to as x-intercepts. Thank you for asking, Tran.

Thank you very much, Jim. I have been unwell and have not been monitoring my website for some time. I appreciate your drawing my attention to it and will try to rectify the matter asap. Kind regards to you. Graeme

Greetings, Arshan. Thank you very much for your questions. Unfortunately, I have been dealing with a long list of family matters (currently I am dealing with cancer). This has kept me from making videos for years. I still hope to resume working on them, but need to get my health re-established first. I am sorry. I have seen some quite good KZfaq videos in the past about Richard Feynman's integral technique. It is quite ingenious and, if you search KZfaq for them, you will find good explanations in that way. Sadly, I do not have the resources at the moment to help you with this matter. I greatly appreciate your concern and interest and hope to be able to produce videos again 'soon.' Kind regards to you!

I would love to hear how you go, Bubblepop, thank you. I am glad that my suggestions have made sense to you and very much appreciate your making contact and telling me about it! Warm regards to you. Graeme

@@CrystalClearMaths Dear sir, these suggestions magically worked for me I got 29/30 in my latest math test... I hope to continue this marks for the upcoming tests also. Thank you again 😊

@@Jess.is.a.mess_08 That is brilliant and exciting news, Bubblepop! Congratulations! I very much appreciate your letting me know of your results. I am glad that my suggestions helped you, and wish you every success for the future.

@@CrystalClearMaths thank you sir, I'll work very hard from now on!!

Thank you, carvelbell181. I am glad that my videos help you 'see and learn concepts clearly.' That is a great encouragement to me. May you always continue to enjoy learning. Warm wishes to you. Graeme

Occasionally, that happens. It is simply an estimate. In that case, choose 9 and see if it 'works.' This is not an exact predictive method. It is an iterative process using approximations. Thank you for seeking clarity, WoC. Sometimes, the first approximation needs to be adjusted.

You are welcome, CMP.

It very likely would be ... a good iterative technique. Thank you for your astute observation, Silly Me Silly.

I agree, Ivan. When I started producing this series, I was intent on demonstrating and teaching particular substitution techniques and realise that, for a number of the integrals, I therefore adopted a less efficient method of arriving at a satisfactory answer. I left the videos because the primary objective was still achieved ... that of showing how to use certain trigonometric substitutions. Students facing this question during an examination would be well advised to adopt the method that you describe so well! Thank you.

Greetings, Vidhya. You have asked a very interesting question. Thank you. Not all children will respond to the "uncomfortable feeling of solving challenging problems." I think of it like asking children to run or hike up a tall hill or mountain, or to engage in a multi-day hike. Some children respond positively. Some do not. I have taken reluctant children on a multi-day hike or had them camp alone in the 'bush' for a night and had some of them respond extremely positively. I believe it is because they realise that they have succeeded at something big and momentous ... something that they did not believe that they could do (or wanted to do). In all cases, I believe it was MY PARTICIPATION and ENCOURAGEMENT at all steps that made the difference. Their sense of achievement is what may motivate them to engage in future challenges. In the case of mathematical problems, I do not leave them alone (at first). If they struggle and are on the verge of 'giving up,' I ask a QUESTION. Good leading questions lead to discussions that allow them to make discoveries, so that they can take another step. Finally, when they reach the goal and solve the difficult problem, I am full of PRAISE for their achievement. I will tell others and tell their parents (and, sometimes, though rarely, provide a extra reward). You will not win with everyone (as not everyone will develop a passion for mathematics), but your successes will sustain you ... and you will profoundly help and motivate some students.

Thank you, Rajesh. I greatly appreciate your feedback, too. Kind regards, Graeme

I feel for you, TK, and wish you the best for this exam.

Thank you!

You are very kind, branflakes. Thank you. I appreciate your support. Kind wishes to you from "Down Under!"

❤🎉😢

@@frankadinortey8518 😃

Thank you, cellotownn. Being from Australia, I am not aware of what 'classes in 901' are, but your comment is very encouraging. Kind regards and best wishes from 'Down Under.'

oh hi ethan

@@theguywhoasked7962 bro shut up Thomas

@@CrystalClearMaths 901 is a classroom in sydney boys high school its where we get taught pe theory and we watch your videos from there

@@liwendeng6923 Thank you very much for the 'heads up.' That is helpful to know. I taught in a few schools in Sydney before moving south ... now in the Shoalhaven. I hope some of the videos are of use to you. Kind regards, CCM

You are most welcome, Stefan. I am glad that I was able to provide what you needed. Kind regards to you (and thank you for your feedback!).

I also hope so

I am glad that this video helped you, petznassloukas, and hope that you do well in your exam! Kind regards to you, and thank you for providing your feedback.

All the best for your finals, petznassloukas ... and thank you for both your comments!

Thank you, Abhishek. Kind regards from CCM.

Thank you, friend. I am glad that you liked the limerick.

I hope it helped, friend. I notice that you have an excellent name ... "the guy who asked." A Nobel prize winner was asked who was the greatest influence on him and his success. He replied that it was his mother, and explained that every afternoon when he arrived home from school, she never asked him if he had a good day or how he felt. She always asked, "Did you ask any good questions today?" Keep asking questions, and you will live a rich and fulfilling life. Best wishes to you from Australia (and pass my greetings on to your teacher).

The ln(5) is part of the pattern. If you differentiate y = e^(ln5)x, you obtain dy/dx = ln5xe^(ln5)x. This means that integrating (finding the antiderivative of) ln5xe^(ln5)x results in e^(ln5)x + C. Thank you for asking, pied6. I hope my explanation helped.

@@CrystalClearMaths thank you for your explanation

@@sometimes-xd2fl You are welcome 🙂.

Cheers to you, too, rev0cs ... 👍.

Thank you, Math-Skill. You are very kind.

Another good pun! Thank you, davea4245.

You are welcome, crystallising7068.

Henry, your message to me is most encouraging. Thank you. I am glad that you have been able to make such massive gains in your mathematics in such a short amount of time. Congratulations! I am also delighted to learn that my videos have been a help to you. I hope to resume posting videos again soon ... but have been overcoming a number of significant setbacks. Much will be revealed when I am able to resume such work. In the meantime, very best wishes to you for your studies and kind regards from the Shoalhaven (just south of Sydney).

You are welcome, user-vm5yf4ul7b. Thank you for watching my video (and for taking the time to leave your encouraging comment).

:-) Thank you, Castillomn971.

Thank you, Rifat. Sorry for my late reply, but I greatly appreciate your taking the time to leave your (encouraging) comment!

Unless I have seriously misunderstood you, user-zo3wd8ui7m, I don't think I agree. Perhaps you mean that the exponent could have been written x·ln(5) istead of ln(5)·x. Either would be quite satsifactory, although we do tend to write the constant before the pronumeral. Kind regards to you, and thank you for commenting.

You are correct, friend. I should have pointed out that the method is quite useless if we are integrating 1^x ... but that simply reduces to 1 and the integral proceeds just as you indicated. Thank you.

Thank you so much for taking the time to let me know, toxicdubz. I really appreciate your feedback and am quite encouraged that my video has helped you along your journey! Very kind regards to you, and I hope that your (mathematical) future is bright 🙂.

@majikazem6926

Thank you very much for your encouragement, maryamdeen. God bless you, too!

You are so welcome, pompompurin. I am glad that I have been able to make a difference for you (and really appreciate your support and your taking the time to let me know!). Warm regards to you from Australia (Sorry for the delay in replying ... I am a bit preoccupied with life in general atm.)

Thank you, Lee William. That is very kind of you.

Thank you very much, Lee William.

I am glad that you are, wanda. I hope you found it helpful. Thank you very much for taking time to leave your comment. Kind regards from Australia 🙂.

✋

@@parokyatv2256🙂Thank you.

@karoldrawings5135Thank you, karold.

Me

For simple problems, you could try this, but for the accuracy you may want, you would need a calculator to save time. If you have a calculator to find the value of 2.318⁵, you may as well use it to find the fifth root anyway.