Відео

Yes! exactly.

Often when solving equations, you may have to square both sides of an equation. In squaring an expression, you may introduce a root that solve your new equation but doesn't work for the original. This is an extraneous root.

And 6

Excellent! Only tell me the extraneous solution.

I’m glad you find it helpful. In your opinion what is the biggest difference between BC material and Alberta’s?

​@@edyumath I've never been to high school in Alberta, so I can't comment on that. I would assume you are talking about the material of the work book? If so, it is pretty much identical, although there are some changes to where some questions and examples are in my edition. Other than that could you clarify? For example, this video for lesson 3 also has lesson 4 examples according to my workbook, and some pages are changed.

@@jasrajbrar7707 Both BC and Alberta share what is call WNCP but difference concepts are emphasized. For example BC concentrates more on rational functions in grade 11 while Alberta waits until grade 12 to study it further. Also Alberta reviews trig but BC expects mastery among others.

@@edyumath Oh interesting! I did not know that.

Thank you! He's a good boy.

They are available here. www.absolutevaluepublications.com/Products/tabid/118/c/all-workbooks-solution-manuals/Default.aspx Best of luck.

@@edyumath Thankyou. Looks like it is discontinued. 😒😒

Thanks for your comments.

Let me ask: how can a point of discontinuity Be in the domain of a function? ( Edited to humbly say I'm wrong here.)

en.m.wikipedia.org/wiki/Classification_of_discontinuities

-3 is a removable singularity and according to the wikipedia article, to label this as a point of discontinuity is abuse of terminology.

Great article. Thanks!

@@stijndevuyst8924 I understand now. Yes, I concede. A removable singularity. Thank you for the clarification. How should I have worded the question?

After studying transformations, we discover that when the x is replaced by 1/2 x then the graph appears to have been horizontal stretched by a factor of 2 (the reciprocal of 0.5)

@edyumath So if it's horizontally stretched by a factor of 2 then each part where the graph intersects the x axis is doubled in distance from the origin?

Yes. exactly. In fact, every coordinate is horizontally stretched by a factor of 2 (reciprocal of the b-value) in this case. This could include x-intercepts, asymptotes, domain restrictions etc.

@edyumath But I'm confused, if you have any function f(x), how does f(x/2) mean that the x values of the points are all doubled? If anything it would intuitively feel like it would be f(2x). I know it's a simular logic to saying f(x+h) for any positive shifts the entire graph to the left h instead of the right.

@maxhagenauer24 Think of this analogy. if you(x) look at a kitchen, let's say the stove is two steps to the right. If your little brother(1/2 x)(who is half your size and can only take steps half as big) looks at your kitchen. The stove (for them) is 4 (of their) steps to the right(Double what yours were). The stove is in the same place but the scale of the steps is transformed. You can use this analogy with all the transformations.

Ha, Ha! Yes! Train the next generation!...and have fun too!

A point of discontinuity, also called a removable discontinuity is a discontinuity that can be "fixed" by defining the function at that point. You are right there are two discontinuities but which one is fixable?

​@@edyumath I think the the more common term I've heard for this would be a 'removable singularity'. The other one at x=3 would be a pole of order 1 or a simple pole.

Yes. Thank you for the better terminology. I was limited by my resources. I appreciate the clarification.

@@PaulMurrayCanberra the graph of a function f is defined to be the set of pairs (x, f(x)) for all x in the domain of f

@@yousauce7451 @PaulMurrayCanberra has got it. The (m,n) is the point or coordinates that is the location of the "hole"

Ok, but the question is asking for the value of n at the discontinuous points. It's a trick question - there *is* no value for n at the discontinuous points by definition.

@@PaulMurrayCanberra Yes you are right but it is not intended to be a trick question. The point or coordinate in question which is undefined by the function yes, but has a location with those exact coordinates that define the only neighbourhood in that area by which it is discontinuous.

@@edyumath Huh - I just plotted it in a spreadsheet. The two expressions and the result of doing the division. I was looking at the discontinuity at 3, which is just a regular "off to infinity", but the one at -3 is the one you are talking about, and the curve does indeed look smooth. In the limit, both curves at that point are straight lines through zero and so the ratio in the limit is a constant equal to a'/b' at that point. Today I Learned.

Yes. You’re absolutely correct! I should have mentioned why I didn’t consider that discontinuity in this question.

Technically yes, but the top term factors to (x+3)(2x-3) so the discontinuity is removable

Yes. Exactly.

I wish I had the time to do so but it's not going to happen this year. Sorry.

@@malloryplug ah yes the kids!! Love them. ..still

If there is a negative inside a square root, it forces us to ensure that the x values must be negative in order for us to be able to take a square root. Because as your intuition tells you in the real numbers, we cannot take a square root of a negative. So it's not so much as having a negative sign in the square root but the most important thing is that the result "of the inside" the radicand must end up being positive.

Glad you found it helpful!

it may be that it can't be converted into a fraction, or it could be as simple as you are in degree mode by accident, and should change to radian mode? Can you give me a specific example?

Yes!

yes. Thanks. Watch it at 1.25 x too. :)

You’re welcome nephew. 🙂

the domain does have an important effect and you must always consider it but in this case, we were given information to confirm though since the angle is second quadrant and tan is negative there. If you check the reference angle, it will show the reference angle is about 63 degrees make the angl about 117 or something. then the double angle will be in quadrant III (about 234 degrees)

​@@edyumathohhh right thanks

This is only true if the terms behave "nicely" . Sometimes one of the terms in the binomial have either higher degrees or the negative exponent like (-3). It is always worth taking time puzzling it out(if you have the time) rather than relying on a shortcut.

Hint: Does a row in Pascal's triangle correspond with these numbers? Anything familiar about the term 2^n?

@edyumath thanks for the hint it seems i need to revisit pascal's triangle because this doesn't ring any bells for me... Do you have any long form videos on the topic?

no video yet. Think of this: 4 activities. These are the possibilities 4C0+4C1+4C2+4C3+4C4 Recognize this pattern?

@@edyumath Ohh

@@edyumath so its like out of 4 activities how many you could choose to do & that gives you the fifth row of pascal's triangle

I'll do my best to add more log videos.

Glad you find it helpful! Best of luck on your diploma. It’s soon🙂

This is not a mistake but I didn't mention it. 2 factorial is 2 x 1. so 2 factorial is just 2.

Sorry, I don't have resources to share publicly at this time.

@@HarryGuit the x intercept always means y (another term for f(x) ) is zero

You have it!

Yes awesome. I guess I thought I would use all the digits. I guess both of us are missing nothing. (0^3). 🙂

If you press 2nd TBL , it will take you to the table settings. Set the independent variable as "Ask" Then you can enter whatever values you need.

Remember that 2pi can be thought of as 12pi over 6.

It may be that you don’t have enough digits. Put some more sixes on the end.

Great that helped!

Hi Hassan, I teach at a school East of Calgary in Canada.

@ oh that’s cool I go to western Canada high school in Calgary

Hey that’s great. I think there are various strategies that work. Glad you tried it. Find a method that works well for you.

This is only easier in the specific case of multiplying by 25 which isn't really the point of the video

@@kennyduncan7 Nah, since multiplying any number with 20 is just multiplying with 2 and adding a 0 to the result. Generally you can always split multiplication problems into smaller easier chunks which often result in an easy solution, even if the numbers are not "close to 20".

Of course use what strategy can work for you!

​@@doodlePimp In the case of something like 24 × 27, most people would probably find the method highlighted in the video to be easier and quicker.

Have a look at the radicals playlist. ua-cam.com/play/PLH2aYkBnhvWUjvq3kxMeSgZSuL6xlQ24b.html

@edyumath thank you

You are welcome. Browse the other playlists to find something appropriate to your course. And of course you can subscribe to see the updated content. :)

Excellent explanation! Will it work for numbers that are under 40?

@@edyumath Yes, it will work, but working with negative numbers is more error prone when doing it in your head. You could also change the base number from 40 to any other number, but then change the multiplication should change to the new base number. Expl: 50 is 100/2 48 x 44 = (50 - 2) (50 - 6) 50 - 2 - 6 = 42 42 / 2 = 21 21 * 100 = 2100 -2 x -6 = 12 2100 + 12 = 2112 ✅

Excellent observation! And great insight on thinking of 50 as 100/2!

@@MrMousley you are right! Just thinking about how to use base 20 in multiplication lends well to how polynomials can be utilized. But you are correct. Many strategies can be used and some are just as fast or faster. Thanks for your comment.

You are most welcome! Best of luck to you!

Great idea! I'll think about that.

Yes. exactly. How does it compare to the combination formula.

Absolutely! Very interesting how it relates to combinations. I was stuck in the combinations mindset. :)

Thanks!

I will see what I can do! I am glad to hear you found the videos useful! Good luck in your math journey.

:)