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@@MariaPerez-lg4yo very welcome!

@@toffeetoe8939 that is 👍 👌

@@FoodeliciousKitchenbySabaish 👍👍👍👍

@@FoodeliciousKitchenbySabaish excellent 👌 👏

@@toffeetoe8939 thank you

@@CutePlanting you are welcome 😊

@@ishmathtestprepdouble so is this mean basically let n and m be even numbers X^n+Y^m =(X^(n/2)+Y^(m/2)+√(2X^(n/2)Y^(m/2))) (X^(n/2)+Y^(m/2)-√(2X^(n/2)Y^(m/2)) in conclusion : (a+b)(a-b) a= X^(n/2)+Y^(m/2)) b= √(2X^(n/2)Y^(m/2)) ???

@@CutePlanting it's a lot that you wrote here in terms of a template. And as I glance over it , it seems my answer to your question is yes. Obviously with actual example it's easier to not miss out on the actuality versus using letters...

@@CutePlanting czcams.com/users/shortsN-xc8mFY6Mo?si=kgQmcS3ui5tWnlE6

@@CutePlanting czcams.com/users/shorts1f29AyCPcxY?si=qmHWkTfsuLLm5gTC

@@SshopGamers_YT true, due to short time length I had to be brief with that.... I basically did the completing the square steps: For, x^2 - 2x, take -2x, divide it by 2 and then square it to obtain, x^2 -2x +1 For, y^2 - 6y, take the -6y, divide it by 2 and then square it to obtain, y^2 -6y +9 These new quadratic terns are now factored to get (x-1)^2 & (y-3)^2, the +1 and +9 go on other side of equal sign to give 10. I hope you got it?

@@ishmathtestprepdouble that helps, thank you!

@@SshopGamers_YT welcome, and yes when you "divide by two and square it", we are looking at the coefficient of the 'bx' term only, not the variable ( x or y).

FYI (this may help): czcams.com/users/shortsmcJ91CWRHAA

@@toffeetoe8939 👍👍👍👍👍

thank you, and welcome!

@@toffeetoe8939 thank you

@@benshapiro8506 makes sense. Good or bad, like Freud said, "sometimes a cigar is just a cigar".

@@toffeetoe8939 it's ok.

@@toffeetoe8939 thank you

@@neavendzonlagic2440 guess if you trace a triangle it has three vertices, like corners. Circle's just round. Or maybe don't overthink it...

You mean its not funny? or it lacks the punch line?

what? can u list every step I dont get it

@@hamayun1112 1) A^2 / x^2 = A 2) A^2 = A * x^2 3) A^2 - A * x^2 = 0 4) A * (A - x^2) = 0 5) A = 0 or A - x^2 = 0 Therefore: A = 0 or A = x^2 Looking at the infinite product in the video, anything greater than 1 to a power greater than zero is also greater than one, so each term is greater than one, so the product can never go to zero, and therefore A = x^2

Stated well!

thank you!

@@user-sr1ux8qf6r I'll look into it in the morning...

@@reduzumaki7674 thank you very much!

Only in case someone lacks essentials here and needs the more thorough approach

Because we are manipulating the definition of sinh(x) here where x = ln x. Should be no 'i' coming into the procedure, unless we actually solve for x, where based on the discriminant of the quadratic equation, imaginary solutions may come about.

Also the definition of sinh x has a 2 in denominator not 2i

You're thinking of the complex version of sin. But here is another expression for the hyperbolic sin sinh. But you are right, there is a relationshio between them

You are thinking of the complex definition of sin(x), where sin(x)=(e^(ix) - e^(-ix))/(2i)

Good post. Appreciated!

A number is what we define it to be ... 0 is defined as a number for a long way already. Your comment makes no sense.

What is a number?

@@samueldeandrade8535 yes what exactly is a number 🤔

In the academic setting (test) it would be provided, but, If you start with two equal sized circles, overlap them, connect the centers of each circle to the top and bottom points of overlap, the sectors are created, as you can picture this in your mind. To determine the angle subtending that sector requires any of several geometric procedures, one being as simple as using a protractor to determine angle and converting to radians. The radii can be extended to allow for protractor to be fitted. However, one can also connect centers of both overlapped circles with a segment, extend up or down (from midpoint of overlap) to points of top or bottom overlap and generate right triangles and then use right triangle trigonometry to determine the theta. The legs of right triangle would be measured easily, one being simply the midpoint of segment connecting both circle centers. Measuring segment from top to bottom overlap points, and the midpoint of that, will give measure of the other leg. The angle determined would be half the measure of theta, and then multiply by two to get full measure.

Thanks, really helpful!

@@wang1918 sure thing!

Excellent! :)

thank you

Thank you much!!!

😊 thank you!

Duly noted 👍👍👍

What's an "ashwag"? 🤔 or was it a typo..

@@ishmathtestprepdouble ah okay, it's another youtuber. You sound exactly like him

@@grimmmunchkin oh!! For a moment I thought it was some trendy slang or acronym and I was feeling out of touch with modern lingo! 😅

Ok! 👍

No, the definition of i is a number who's square is -1

@@hpermutation4439 👍 👍

@@hpermutation4439 yes, so by square rooting both sides i = sqrt(-1) by definition.

This radical expression can be solved in the traditional way and in this other way, an alternate way as I am demonstrating here. Which requires a substitution to drive the process and make into a Quadratic equation, which then can be factored.....do you need a video which demonstrates the two routes in a slower side by side comparison?

​@@ishmathtestprepdoubleplz

You are welcome!!

True. There are other ways as well!

The index of root must be a Natural number?

no?

Works!

You are right, the 0 is artificially brought in by pushing the -1 to other side!

Well I am happy for you, sincerely I am!

I already did in my other channel, here is the two videos links for the 2 different approaches used, in the longer video, the derivation/proof is at end of video..... czcams.com/video/v_NvjqYAl-E/video.html czcams.com/video/T4UtzP4Gerw/video.html Let me know if links did not work!

you do it with taylor series for sine, cosine and e

@@hornkneeeee very true, if you scroll through this playlist you will find videos on just that: czcams.com/play/PLNmf_rc0NI111NQMGSzv7F_-CeY5e4bsq.html

Some like em, some do not, some prefer to see limit questions solved without the hint of limits or even the sight of Limits! That is, solved in a way other than limits....I see that all the time in students uninterested in calculus.

Your point is well-noted.

Came to say the same thing, it’s kinda like saying that 2+2=4 is true because 4-2=2 It’s certainly true, but it’s not really even close to proof.

And I would agree, as I had in hindsight.....I believe with that original comment!

Thank you much.

The factoring has to occur in the manner shown as the procedure is governed by laws of exponents, main one being that, a^m * a^n = a^(m+n)

Thank you very much

:)

Great

Yes it is very easy to forget all this, I know from personal experience

Thank you for your comment!!