Відео

This is a construction proof. We’re providing a method to find a unique quotient and remainder. The way we set this up is by taking the dividend an and subtracting multiples of the divisor b. This is a good idea, because one of these in the set is the least positive entry, and that the integer that provides q and r. Does that help?

@@PaulineFletcher-m2d that’s great to hear! I’m happy this was helpful. Good luck with the new academic venture!

@@zoiealfonsomanuelle_4917 you made my day! Thank you for the kind words. Best of luck on your upcoming exam!

@@imperator193an we can prove that if diam(Gbar) >= 3, then diam(G) <= 3 since that's the contrapositive of the given statement. Since G is arbitrary, that means that for a given graph G if diam(G) >=3, then diam(Gbar) <=3.

I'm not sure I quite follow what you're saying. Think about (a-b)(a+b) = a^2 - b^2. Since (sqrt(36-x))^2 = 36-x, (sqrt(36-x)-6)(sqrt(36-x) + 6) = (36-x) - 36 = -x. I accidentally said this equals x. Does that help?

@alirezaansari7208 understandable. I do have an Intro to Mathematica video as well if you wanted to take a look.

@@ArnoldSommerfeld thank you!

@@jackiehunke7758 awesome to hear!

Thank you for the comment! I'm glad I could help!

@@shuyu-yq5yr great!

It's worth it. Good algebra will make everything else much simpler. Good luck to you!

Honestly, I never use MakerBot anymore. Cura and Prusa are much better slicers. The Creality one though... yikes.

Because a is a divisor of b. If we use 'divides' as an action verb, then a divides b. You could think that we can split b items into a equals piles with no remainder.

@@johnestesmath3036 thank you

and for instance if the exponent (-2) wasn't a negative, and I wrote the fraction in one line, simplified it and there is a term that has a negative exponent, since moving up the terms from the denominator to the numerator will make its exponents a negative, do i need to return it into a fraction form?

@@KyLuc-v9y Absolutely, we could write the expression in (b) as (5a^(5/2)b^4a^(-3)b^(2/3))^(-2). Whether you need to write your final answer as a fraction depends on your application and what your instructor prefers. Either way, good exponent manipulation is a critical skill to be good in algebra and calculus.

You lost me. I don't understand how you teach

@@theindustryexpert I'm sorry to hear that.

OpenStax has free college algebra textbooks. That's probably the quickest way to find free algebra resources.

I'm sure a calculator could handle some of these questions, but I would advise to get familiar enough to do these by hand. Good algebraic manipulation is a valuable skill to have and will make calculus so much easier.

@@johnestesmath3036 ok sir TQ sir

Thank you for the feedback! I have several calculus videos. Let me know if you have questions.

I hope it's helpful!

If you have a coefficient matrix M, you can just apply Inverse[M] to get your inverse. Just make sure your coefficient matrix is n by n, because only square matrices have inverses.

This a Group Theory from higher algebra. Were you hoping for algebraic grouping?

​@@johnestesmath3036 I have been searching for a video about algebraic groups which are defined to be algebraic varieties endowed with group structure. But it seems that this video is only about groups, not algebraic ones. So, I wanted to imply that the usage of the term "algebraic group" was a bit misleading for me.

@@AzamatSaydaliyev ah..I see what you mean! I did not mean to mislead. I was hoping to help first semester (higher) algebra students. My mistake! Good luck with your search.

Thanks! Tell your friends 😆

3w simplifies to w after subtracting 2w from both sides.

Hey Chance, which part are you look at when you say that?

@@johnestesmath3036 sorry im in the process of going back to college so im refreshing everything i know. or dont know. you split them up like 16 * 10 an 16 * 2. my question i suppose is does anything change if you where to pull out 32 *5 instead? it was -sqrt160 - sqrt32? i think? is the question im refrencing

Practice. The thought process is that I want to two numbers a and b that multiply to -15 and add to -2. So a = -5, b = 3 works.

@@johnestesmath3036 Excellent, thank you.

Honestly, I don't know why we insist on rationalizing denominators. It's more cultural than functional.

Yes! Thanks for the feedback!

The first thing I wrote? If a term is in the denominator, you can move it to the numerator but a negative exponent. So (1/b^5) = b^(-5) for example.

That's the algorithm. Take the middle coefficient, b, divide it by 2 and then square that result. Adding and subtracting that value will allow us to ALWAYS factor the expression as a perfect square.

Many thanks!

they are in different factor groups.

Which value? That is a correct simplification.

Yeah its should be -1/sq(36-x)+6.

I do have College Algebra vids as well. You might find some of what you're looking for.

There are some really good proof resources. Hammick's Book of Proof is completely free. If you have some ideas, float them my way though!

Yes. I use an Excel grade book but also Canvas. I have videos in the description going over how I do that.

All I can say without context is that small differences over time add to big changes. Just keep at it.

We factor out x^(-1/3). Notice that if you distribute x^(-1/3) across that line, you'll get our original statement.

same issue for me, simplifying them is the real challenge.

Yeah. Algebra is what males Calculus difficult. Keep practicing it. Try to identify mistakes you may be repeating.

Check my latest video going over 25 derivatives in 30 min. That might help. Algebra definitely helps too

@@johnestesmath3036 thnx so much I'll check it out for sure

Can you clarify what you mean?

@@johnestesmath3036 like 6^3/7^3 will it be the same way(6*7 OR 6+7)^6? or in other way? sry for late reply 😅

@@thoughtsbright7928 so 6^3/7^3 is the same as (6/7)^3, but not the same as your second expression. Does that answer your q?

@@johnestesmath3036yes sir thank you. i truly do.

Don't beat yourself up too much! That's fantastic that you have a desire to learn.

thank you very much i wont fail you @@johnestesmath3036 , even tho i dont see myself