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@ISoldBinLadensViagraOnEbay

5 ай бұрын

I’M JOSEPH STALIN’S MIGHTY FROG

@deltalima6703

5 ай бұрын

Your "phd friend" video was hilarious, I was laughing about that for hours. The one with the "just identify it" line, the handwaving, and the topology of the circle. :D

@player3572

5 ай бұрын

Hey bprp! Can you do this math equation? It was in my test. sin(cos(tan(x)))=1

@leonardobarrera2816

5 ай бұрын

But base on the video, you could use the pythagorams inverse formula, also you have on you channel!!! It would be more intresting!!

@M0n3Y1g

5 ай бұрын

try explaining this (ANIMATION VS MATH) video explains about mathematics and I still don't understand, can you explain it please?

I think this is a spot for sybermath's substitution trick. Let a = sqrt(16 - x^2). Let b = sqrt(9 - x^2). Then, we are given that a + b = 5. Also, from the definitions of a and b, we have a^2 = 16 - x^2, and b^2 = 9 - x^2. Subtracting those, we get a^2 - b^2 = 7. So, (a + b)(a - b) = 7. But, since a + b = 5, we get 5(a - b) = 7, or a - b = 1.4. Add that to a + b = 5, and we get 2a = 6.4, or a = 3.2. Plugging that back in to a = sqrt(16 - x^2). 3.2 = sqrt(16 - x^2), so 10.24 = 16 - x^2. x^2 = 16 - 10.24 = 5.76 = 2.4^2. So, x = +/- 2.4. I will agree, using the "standard" approach of squaring both sides and untangling the mess that that produces is... a giant pain. But, sybermath's substitution trick avoids that nicely.

@yoav613

5 ай бұрын

Nice,i like sybermath trick too,but no need for any tricks here.just square sqrt(16-x^2)-5 and -sqrt(9-x^2) and you get after simple algebra x^2=5.76 so x=+-2.4

@abdelrahmanayman8415

5 ай бұрын

I squared both sides (25=16+9-2x^2+2√((9-x^2)(16-x^2))) after that u get x^2=√((16-x^2)(9-x^2)) Then squaring both sides u will end up 25x^2=16×9 Solving this simple equation u get postive or negative 12/5

@chaosredefined3834

3 ай бұрын

@@abdelrahmanayman8415 If you are going to square both sides... Subtract one of them first. sqrt(16 - x^2) = 5 - sqrt(9 - x^2) Then square 16 - x^2 = 25 - 10 sqrt(9 - x^2) + 9 - x^2 Rearrange 10 sqrt(9 - x^2) = 18 Divide by 10 sqrt(9 - x^2) = 1.8 Square 9 - x^2 = 3.24 Rearrange x^2 = 5.76 Then solve for x x = 2.4 or -2.4 Notice that you don't have to deal with as big an equation as you did.

@samueldeandrade8535

3 ай бұрын

sybermath's trick? What makes this trick to be "sybermath's trick"?

@Avighna

3 ай бұрын

That is so fucking cool, thank you so much for sharing this!?!??

That ChatGPT attempt 💀💀

@anto8722

5 ай бұрын

“Ai will take over the world!” Meanwhile, AIs: “tHrEe pLuS fOuR eQuAlS fIvE”

@Macieks300

5 ай бұрын

It was ChatGPT3.5, not ChatGPT4, so of course it's not going to be good at math.

@saj_f0

5 ай бұрын

@@Macieks300 you say chat GPT 4 is good at math then

@zaidtalib4513

5 ай бұрын

@@Macieks300 gpt 4 is equally garbage at math lmao

@Macieks300

5 ай бұрын

@@saj_f0 Yeah, it's way better at math.

Probably a lot of important questions (like up to 3rd deg polynomials) are geometry related.

@puleraleaooa8110

5 ай бұрын

This is a very interesting question 🙋

@d.h.y

5 ай бұрын

Euclidean geometry is the future 😆

@dominicellis1867

3 ай бұрын

You can always construct a geometric figure that represents an algebraic relationship. That is what makes analysis so beautiful.

@dominicellis1867

3 ай бұрын

@@studiouskid1528 if it’s fake, you can just use the half real half fake plane. The noobs call it complex but I’ve always thought it was simple.

@dominicellis1867

3 ай бұрын

@@studiouskid1528 2 real axes and one imaginary axis. x->z = a+bi.

He used a blue pen

@hr07m

10 күн бұрын

Fuck

Now I will use geometry whenever I can to solve funky algebraic equations.

Great simple geometric solution 😊

Really cool problem! Thank you as always

Also there exist inverse pythagoran where 1/a^2 + 1/b^2 = 1/d^2 so in our case it would be : 1/4^2 + 1/3^2 = 1/x^2 and it would be equal to this anwswer

This was fun! thank you for your videos

I'm loving this series of yours, the "Believe in __ not ____". Could we get a trigonometry edition?

@blackpenredpen

5 ай бұрын

Great idea! I will have to think about it.

Glad to see you again! Nice solution!

@OmPrakash-pc1ec

3 ай бұрын

ти ща де

Nice way to solve. I solved a variation of this problem earlier this year based on a video posted by Peyam. There's an amazing detail contained in the problem. x² = 4² - [√(16-x²)]². x² = 3² - [√(9-x²)]². So 4² - [√(16-x²)]² =3² - [√(9-x²)]². Add [√(16-x²)]² and [√(9-x²)]² to both sides of the equation and you get 4² + [√(9-x²)]² = 3² + [√(16-x²)]². And so you obtain a variation of a pythagorean quadruple, a² + b² = c² + d². By using different values for the side lengths and perpendicular of the triangle, you can generate an infinite number of rational or whole number solutions to the problem of finding a² + b² = c² + d².

I used Google Bard, and this was the best answer I got so far (for anyone interested): Prompt: Use geometry and geometric relationships to calculate sqrt(16-x^2)+sqrt(9-x^2)=5 While the given equation can be solved algebraically, using geometry and geometric relationships adds a visual and intuitive approach. Here's how we can solve it: 1. Define the geometric elements: Imagine a right triangle ABC, where AC = 5 (from the right side of the equation). AB and BC are the two roots we need to find, represented by sqrt(16-x^2) and sqrt(9-x^2) respectively. Point D lies on AC such that AD = sqrt(16-x^2) and DC = sqrt(9-x^2). 2. Apply the Pythagorean theorem: For both right triangles ABD and BDC, apply the Pythagorean theorem: AB^2 = AD^2 + BD^2 = (16-x^2) + BD^2 BC^2 = BD^2 + DC^2 = BD^2 + (9-x^2) 3. Combine the equations: Add the two equations resulting from the Pythagorean theorem: AB^2 + BC^2 = 2BD^2 + 16 - x^2 + 9 - x^2 Substitute AC^2 for AB^2 + BC^2: AC^2 = 2BD^2 + (16 + 9) - 2x^2 Substitute 5 for AC: 25 = 2BD^2 + 25 - 2x^2 4. Solve for BD^2: Simplify the equation: 2BD^2 = 2x^2 BD^2 = x^2 5. Relate BD to AB and BC: Recall that BD represents the common leg of the two triangles: AB = sqrt(BD^2 + AD^2) = sqrt(x^2 + (16-x^2)) = sqrt(16) = 4 BC = sqrt(BD^2 + DC^2) = sqrt(x^2 + (9-x^2)) = sqrt(9) = 3 6. Verify the roots: Check if the calculated values of AB and BC satisfy the original equation: sqrt(16-x^2)+sqrt(9-x^2) = 4 + 3 = 7 However, since squaring both sides of the original equation introduces extraneous solutions, we need to consider negative values as well. 7. Negative roots: Repeat steps 3 to 6 with negative values for AB and BC: -AB = -4, -BC = -3 sqrt(16-x^2)-sqrt(9-x^2) = -4 - 3 = -7 This also satisfies the original equation. Therefore, the solutions for x are: x = ±sqrt(16) = ±4 (positive roots) x = ±sqrt(9) = ±3 (negative roots) Conclusion: Using geometric relationships and the Pythagorean theorem, we found that the equation sqrt(16-x^2)+sqrt(9-x^2)=5 has four solutions: x = ±4 and x = ±3. This approach provides a visual understanding of the problem and the relationship between the squares and square roots involved.

@AnthonySouls

5 ай бұрын

Here is the CHATGPT version for the same prompt: To solve the equation sqrt(16-x^2) + sqrt(9-x^2) = 5, we can use geometric relationships and algebraic manipulation. Let a = sqrt(16 - x^2) and b = sqrt(9 - x^2). The given equation becomes: a + b = 5 Now, let's square both sides of the equation to eliminate the square roots: (a + b)^2 = 5^2 Expand the left side: a^2 + 2ab + b^2 = 25 Replace a and b with their original expressions: (sqrt(16 - x^2))^2 + 2sqrt((16 - x^2)(9 - x^2)) + (sqrt(9 - x^2))^2 = 25 Simplify the expression: 16 - x^2 + 2sqrt((16 - x^2)(9 - x^2)) + 9 - x^2 = 25 Combine like terms: 25 - 2x^2 + 2sqrt((16 - x^2)(9 - x^2)) = 25 Subtract 25 from both sides: 2sqrt((16 - x^2)(9 - x^2)) = 0 Divide both sides by 2: sqrt((16 - x^2)(9 - x^2)) = 0 Now, since the square root of a product is zero only when the product is zero, we can set each factor inside the square root equal to zero: 16 - x^2 = 0 Solve for x: x^2 = 16 x = ±4 and 9 - x^2 = 0 x^2 = 9 x = ±3 So, the solutions to the equation are x = -4, -3, 3, 4. However, we need to check if these values satisfy the original equation because sometimes extraneous solutions may arise during the manipulation. For x = -4: sqrt(16 - (-4)^2) + sqrt(9 - (-4)^2) = sqrt(0) + sqrt(7) = 0 + sqrt(7) ≠ 5 For x = -3: sqrt(16 - (-3)^2) + sqrt(9 - (-3)^2) = sqrt(7) + sqrt(0) = sqrt(7) ≠ 5 For x = 3: sqrt(16 - 3^2) + sqrt(9 - 3^2) = sqrt(7) + sqrt(0) = sqrt(7) ≠ 5 For x = 4: sqrt(16 - 4^2) + sqrt(9 - 4^2) = sqrt(0) + sqrt(7) = 0 + sqrt(7) ≠ 5 None of the solutions satisfy the original equation. Therefore, there are no real solutions to the given equation.

@bmn6292

3 ай бұрын

@@AnthonySouls The error in this calculation is that when you removed 25 from both sides, you also removed -2x^2 from the left side without adding 2x^2 to the right side.

This is gold! Thank you for such a useful video!

I really love math. But similar to basketball where I’ll never be an NBA player, I’ll never be a great mathematician. The insights needed are just beyond my ability. However, I’ll never give up because I really enjoy learning.

@RandomCatFromFrance

3 ай бұрын

Same for me, tho i still dream of finding something new, even if small.

@Formula7Driver

2 ай бұрын

Just keep learning and you'll surprise yourself how good you can become

@mrinaldbhat9921

2 ай бұрын

hey! dont demotivate yourself that way. enough practice and determination can make anyone good at anything.

@ethanhall6965

2 ай бұрын

@@mrinaldbhat9921 Thanks. I'm still working at it. I won't give up.

Very good teaching. Thank you !

I always love having a look inside!

very interesting approach!

I would never have thought of approaching the problem that way. Very interesting!

Super interesting video!

i used angles to find the value of x in the end , assume one angle tetha rest are either 90-tetha or tetha which makes it much easier

Wow, Thank you, so amazing, I love this technique.

It is sort of isomorphic transform I think. Like Laplace or Fourier transfom. Many things are isomorphic even when they are very different.

@table5584

5 ай бұрын

Isomorphisms!

@pauselab5569

5 ай бұрын

don't think that it is an isomorphism at least not in the usual sense.

@sergeygaevoy6422

5 ай бұрын

@@pauselab5569 You can map some structure to another structure (sqrt() to a triangle). Map some operation to another operation (addition to position in space). Then you should unmap the result. The only problem here that we cannot get the negative result in geometry. We cannot be sure there no other colutions using this mapping. That is true. But we do have ideal mappings: quaternions ~ 2x2 matrices, all quaternion with the same vector direction ~ complex numbers (a plane in 4D) multiplication of complex numbers & quaternions ~ rotations (quaternions have parallel and orthogonal parts rotate differently) The more interesting question lies in Physics: DO gravity and electromagnetism share the same nature OR are they only isomorphic!? If two thing follow the same axioms they will follow the same theorems.

@samueldeandrade8535

3 ай бұрын

What ... are you talking about?

@alextaunton3099

3 ай бұрын

​@@samueldeandrade8535isomorphisms. Get with the program

This will be a very good watch!

Nice approach.

After making the triangle, instead of equations areas you can use the "inverse" Pythagoras theorem ie 1/x^2 = 1/3^2+1/4^2

I have always loved the geometrical representation of equations (mostly from 3blue 1brown) but this method is just ♥

@2001herne

3 ай бұрын

You learn to love geometric representations in engineering - on the engineering HSC (Australia(NSW) Higher School Certificate) exam it's valid to solve force diagram problems either analytically, or by drawing the diagram to scale and measuring. If you're neat enough, it's a lot quicker, and about as accurate.

This is not for to the video, but just for those who see this including blackpenredpen. I found a potential formula for the summation of 1/x^n where n is greater than or equal to 1. d^n / d r^n { ln[ (x+r-1)! / (r-1)! ] }/n! with r being a constant equaling 1. ( Note I used d for the partial derivative of r).

@user-xh7mq8ty8m

5 ай бұрын

It is actually d/dr {ln[ d^(n-1) / dr^(n-1) (x+r-1)!/(r-1)!]} and the bounds of summation are 1 and infinity.

That was a neat approach to this problem! When I read the title, I thought more about circles than right triangles.

Great solution!

Congratulation! It is fantastic!!!

Hey there, Steve. Since I’m in Info Tech, I also had to learn CS (computer science). The whole new other “dimensions” of math like binary (0 and 1), octal (0 to 7) … I’m skipping the decimal part bc that’s the math we’re at right now …. Hexadecimal (base-16 that’s 0 to 9 and A to F representing 10 to 15. Like 1F for example …. That’s 31 in decimal and so forth. You get the gist), base-32, base-64 and finally, base-85. I learned there’s a multiplication table of hexadecimals. Google it up all you want. Now! I’m curious. Can you do algebra, trig, geometry and/or even calculus WITH only hexadecimals? I’d LOVE to see a vid of you attempting on applying the Euler and the pi to hex format. Hehe. Pretty please?! Can you do that?! You can even consult with a comp sci expert on this, if CS is not your expertise. Wink.

would you be able to do a video explaining how to solve equations such as 2^x + 3^x = 2 or 2^x + 5^x = 1 and other similar cases?

so fun! thank you!

On a recent StarTalk podcast episode, Prof. Charles Liu said on his (presumably at-home) exams, he asks his students to use chaptgpt to solve a problem, and then to explain why it's wrong 😂

*Standing Ovation!* Always lovely to see elegance in math problems like this. :)

I thought about an interesting question and I couldn't solve it. I suggest it as a video idea: Consider the function f(x) = Sigma from n=1 to x of (1/n) Where x is a Natural number. Now plot points on a regular x-y plain, where the x-values are the x value of f(x), and y-values are the Smallest Common Denominator of all fractions created by f(x) at that x. In short, a coordinate is: (x, SCD). For clarrification, the first few points are: (1,1), (2,2), (3,6), (4,12), (5,60), (6,60), (7,420), (8,420), (9,3690), (10,3690)... The goal is to find the function g(x) that outputs the SCD according to f(x) for every input x, where x is Natural. Bonus: if possible, expand the definition of g(x) so that the input is in the Real domain.

Great method sir ❤️😯

Question) If Y={x such that x is a positive factor of the number [(2^p-1)(2^p - 1)] , where (2^p - 1) is a prime number}. Write Y in the rooster form

That was a cool problem Thank you 😊

I did this one the brute force way, and it was surprisingly not too bad. It was a miracle of canceling terms so that it solved pretty neatly.

Imperial! As an math aspirant i will use it

What a beautiful solution 🤩🤩

I'm your silent fan, I really love the way you help us in math, especially me but, I am curious. How did you become master at math? how much time do you solve math before? can you make a video for it?

Try this: 1+power(2+power(3+…+power(n+…,1/n),1/3),1/2)=?where n is natural and goes to infinity.

tres élégant !! merci !!

Brilliant, I did not think to use build the second triangle in that way. I made a mental calculation and found sqr 7 is a good solution😊.

Im gonna use this question to show my friends that geometry is beautiful and not just shapes

If it is not right angled triangle we can use 1/2 ab sin(x) to find the area

You never miss my guy

Great video 👏

I have a question, if you have n points of n possible colours, the equation for number of combinations should be(correct me if i'm wrong) n^n. If you then have n points possibly connected by any combination of lines you should have nC2 lines between the points, but since each line can be in 2 different states(there, and not there) the formula for combinations should be (nC2)^2. Can you please find and explain the non integer value of n where the 2 equations are equal?

Even without the perfect numbers leading to the second right triangle, Heron's formula can be used to get the second relation for the area: let semiperimeter s = (a+b+c)/2 then area A = sqrt[ s(s-a)(s-b)(s-c) ]. In this problem s is 6 and A is also 6.

Beautiful!

Can you make a video about this interesting problem--> If z = x + iy is a complex number where x and y are integers. Then, the area of the rectangle whose vertices are the roots of the equation----> w z^3 + z w^3 = 350, where w=conjugate of z

Well done, Alan!

The Solution was pretty impressive . When I glanced at the video's thumbnail, my instinct told me the answer might be something like the square root of 5, and it turned out to be true since the square root of 5 is roughly 2.4. Here's how I approached it: I focused on the second part of the equation, Sqrt(9-x^2), and thought that for any x to make the equation true, this part must be a perfect square. Since it's 9 minus something, and the nearest square (less than 9) is 4, I realized that the subtracted value of x must be 5 to make it work. Voila! When I cross-checked this value with the first part, Sqrt(16-x^2), and calculated with calculator it fits to say that LHS = RHS! :)

Just noticed that even x = 0 as well as x= +- 12/5 Since, we are using the 3-4-5 triplet to find the results The one of the trivial solutions to the problem!! Awesome explanation!!!! Looking forward to more such videos

@GrimAxel

3 ай бұрын

Except x=0 doesn't work. When x=0, sqrt(16-x^2)=4 and sqrt(9-x^2)=3. But 4+3=7, and 7≠5.

1: If the resulting triangle isn't a right triangle, we can use Heron's formula. 2: Haha, ChatGPT did it wrong again! ChatGPT is not good at math, but quite good at programming!

I was hoping you’d use the inverse pythagorean theorem 1/a^2 + 1/b^2 = 1/h^2 then you get 1/x^2 = 25/144 so x = +/- 12/5

@blackpenredpen Can you please do a break down of the integral arctan(x)/x^3 dx? I’m fascinated by it.

La longitud de la altura referida a la hipotenusa en un triángulo rectángulo es igual al producto de los catetos dividido por la hipotenusa, x=(3*4)/5

"Trust in Geometry" people when meet "i"

Q: can you solve x^4+x^2+1=0 by using the quadratic formula? . . . Ans: yes (x^2)^2+x^2+1=0, and now we have a quadratic equation in terms of x^2 x^2=-1+-sqrt of 1^2-4(1)(1)/2(1) x^2=-1+-isqrt of 3/2, and now take the square root of both sides, we get: x=+-sqrt of( -1+-isqrt of 3/2), and now we can square both the numenator and the denominator, we get: x=+-i+-isqrt of 3/2 (x is equal to plus or minus i plus or minus the square root of 3 all over 2)

Great video, you earned my subscribed. I love watching these questions where there is close to none knowledge prerequisites but still gets you thinking.

Любопытно! Хорошо бы ещё провести проверку 😘

with similar triangles you can see the answers with ratios (either sub triangle works), 5 : 3 as 4 : x implies x=12/5

can this be generalized for solutions of the form: sqrt(a^2 - x^2) + sqrt(b^2 - x^2) = c where a,b,c are a recognized pyth. triple?

Very nice solution! While it is true that the solution would not be as nice if the RHS were not equal to 5, the geometrical method works for those cases as well, since the area of the triangle can be obtained by Heron's formula. For RHS = 6, the answer is +/-sqr(455)/12.

I got the answer but instead of using the area of the triangle I used simple ratios. Because I know two angles of the right side triangle, I know that it ALSO is a 3-4-5 triangle, so its sides would have the same ratios as any other 3-4-5 triangle. That gave 3/x=4/5. Very very easy to calculate.

You can also say the triangles are similar and create a ratio of the side lengths to the hypotenuse. 4/5 = x/3

Can you try the integral of x/sinh(x) from -∞ to ∞? It comes out to a really nice answer, but I have no idea how you’d actually solve it…

May i ask since n[sin(180degree/n)] helps u calculate pi. If lim n--> infinity, is n(sin(180degree/n)) = pi?

@ppmax9392

4 ай бұрын

anyone answer me pls ty

what was up with holding that red white and black stuffed looking thing in the past? I was curious when i saw your past vids. Interesting content btw! Very creative, different, and analytical.

I f**ing love this guy!

Please, I want you to create a formula to solve the equation ax^3+bx^2+cx+b And this one too ax^3+bx=n

@looooonooooooooooooooooooooong

3 ай бұрын

Look up cardano formula

老师可以讲一下怎么积分Tan(cos(x)) 从0 到2pie吗

Hi Blackpenredpen, I have two things to tell you! First of all, could you make the 100 integrals file (the first video) freely available? Secondly, I offer you a huge integral! The integral of 1/(1+x^5). Otherwise good luck! Bye !

that was awesome

Euclid's Height Theorem: h² = pq Euclid's Leg Theorem: a² = cp, b² = cq

Very fine I belived in geometry.Thank you

Good question

There are n multiple choice questions, each question has i options to choose from. Step 1: Randomly choose the mth option (with m less than or equal to i and m greater than or equal to 1) in the first multiple choice question Step 2: Repeat the option in the 1st multiple-choice question in the next (k-1) multiple-choice questions. Step 3: To choose the option in the (k+1) multiple choice question, we will choose in the following way for each case: Case 1: If the option chosen in the kth multiple choice question is the mth option (with m smaller than i), then choose the (m+1)th option. Case 2: If the option chosen in the kth multiple choice question is the ith option, then choose the 1st option. Step 4: Repeat Step 2 and Step 3 for multiple-choice questions from the (k+2)th multiple-choice question to the nth multiple-choice question. Each multiple choice question has only 1 correct answer. Let t be the number of multiple-choice questions answered correctly in n multiple-choice questions, t follows the Bernouli distribution. Find k to t max.

Respected sir, Find the value of the following questions geometrically in the principle value range: (a) arcsin(sin 23) (b) arcsin(sin24) Where both 23 and 24 are in radian.

That was beautiful.

i used heron's formula for the other area and equated it with the common area formula. even though yours is easy.

Thanks

That was brilliant

This is brilliant

I need help with a question. Please. Knowing that lim(x-->-2)f(x)/(x^2)=1, which is the value of lim(x-->1)f(x^2-3x)/[(x+8)^(1/2)]? Answer bellow. Answer: 5/3

I used rationalisation and it also works here. On rationalising (16-x²)½+(9-x²)½=5, We get (16-x²)½-(9-x²)½=7/5. Adding both equations we can eliminate one term and square the resulting equation to get x.

For this to work the larger triangle has to be a right triangle. No, I was wrong here. Consider a 13-14-15 triangle. Used Heron’s formula to get area=84 and height above 15 side is 11.2. Sqrt(169-x^2)+sqrt(196-x^2)=15, x=+/-11.2. This is great.

we can also able to do this in another way: ✓16-x²=(5-✓9-x²) then we can square both sides it will be very easy

did same when i saw thumb, saw you did the same, i thought i got a smart solution haha

Damn...I was always an algebra guy but now I think geometry also kicks. Thanks sir.

Excellent

this is so elegant damn

If it was not a right triangle, you could use Heron's formula for the area

Is there any chance you could have a look at how to solve x^(x+1) = 2