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Andy Math
United States
Registrace 18. 08. 2015
Over a thousand Math Videos linked to a free math website. Andymath.com. Please check it out!
How does Tire Size change Speedometer?
Hi, this is a video I found that I made a couple years ago. I apologize for the shaky camera. And the "how excited" at the end was artificially added today.
I hope you guys have a nice week!
I hope you guys have a nice week!
zhlédnutí: 9 860
Video
Is there an easier way?
zhlédnutí 22KPřed dnem
I hope you guys like this one! Let me know if there is an easier method.
Brilliant Blue Area Question
zhlédnutí 39KPřed 14 dny
To try everything Brilliant has to offer for free for a full 30 days, visit brilliant.org/AndyMath/ . You'll also get 20% off an annual premium subscription.
4 Squares 1 Circle
zhlédnutí 201KPřed 21 dnem
I thought it would be fun to use some Coordinate Geometry on this one. I hope you guys like it! If you read this description, please comment something with the word "spicy." Have an awesome night. You are the best!
What's the angle? (With Bonus Footage)
zhlédnutí 46KPřed 28 dny
Watch to the end, the bonus footage is super cool!
What fraction is shaded?
zhlédnutí 29KPřed měsícem
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Full Video Here
zhlédnutí 220KPřed měsícem
@thefrdishow If you fold a piece of paper 42 times, its thick enough to reach the moon. How exciting!
Proving Different Pythagorean Theorems
zhlédnutí 10KPřed měsícem
Proving Different Pythagorean Theorems
Couldn't you just do x²+y²=11{y>3, y<2}
I didn't understand the joke but I understood the math. Is that weird?
Is it possible that this problem is harder than it looks? Why are we assuming we're working with rectangles? Nothing is labeled as a rectangle. There are no angles marked as right angles. Is there a way to do this without making such assumptions?
Because 7 8 9 😢
This theory assumes they aren't idiots like most of society.
Cool
More youtube bull shit
Im not buying it
Cantor's paradox
Very cool I want it
If you fold it IN HALF 42 times
But how can the vertex be at the same height when the cable is tense versus when it is slack? If you bring the ends of any cable together it'll droop, bringing the vertex of its curve closer to the ground. Am I missing something?
I got it by Pythagoras theorem ✓((80÷2)²+(50-10)²)=0
No way no time are you ever going to go from the golden gate bridge and make it to portland in 5 hrs. Its at least 10 hr trip…
You're prisoner 3. You hear "dunno" followed by "white". You realize if you were white, prisoner 1 would know they should say black. You realize you're wearing black.
Those who use radians : I still don't get it
You can’t just drop the parentheses on an infinite sum without meeting extra conditions
Nice
Bro can process information faster than flash
What grade/year would you find this level of maths in?
Number 3 just fucked huh😂😂
Can confirm, you can get to the Chicago area within 31 hours. Father and I drove from Long Beach all the way to the suburbs of Chicago in 31 hours straight lmao only pee and gas breaks.
Pythagorean thereom can be used too, divide the top into 2 right triangles where side c is half of the wire/cable. C would be 40 (just 50-10) meters, A would also be 40, so 40^2 + B^2 = 40^2. Don't actually need to simplify to solve, 40^2 - 40^2 = 0 so you're left woth B^2=0, from there I hope you can solve on your own.
I hate that the video cuts off... I hate that there's no part 2. What a waste of time. SMH
what if 1 has a white hat and 4 has a black hat? would that not make this a 50/50?
Is this the elementary school equivalent of learning limits?
well this why its impossible to fold it more then around 7 without some serious forces applied to it. it can only be folded for around as much as the area can be until it critically fails.
Look pretty easy ngl
And what does this do for you in life? Besides becoming a teacher to teach it to others
it's just neat :-)
this kind of stuff help students to be smart or the art of thinking . well my formula for this problem is ((a^2)* √3 *4) * ((3+a)^2) . which is lengthy but faster since i do not know formula but my thought process generate a formula immediately and same thought process help me to be ready when i needed to take decisions and it's nerdy fun too
you’re practicing breaking down complicated problems into simpler ones that you can solve individually, you’re using your pattern recognition skills to notice the repetition in the regularity of the shape and also to recall area formulas that you know already. you’re also working on visualization because you’re seeing the inner shape instead as a large square with four small isosceles right triangles coming off of its sides, you’re ‘seeing’ the divisions between those shapes.
This was indeed a lot of fun
I think the better explanation is that 2 has the best odds of getting their hat correctly because they know the colour in front of them is black therefore they have 2/3 chance that theirs is white, whilst everyone else has a 50/50 chance of guessing the correct colour.
Or just apply secant - tangent theorem, that's is 5² is equal to (x)(x +2r) which is the width × height And boom here's your area in one line
did this paper, was tricky under timed conditions, but got it regardless
Very good to know for next time I move
Why does one need to know this to deliver a parcel?
Why would the first man say that if the two men in front of him had black hats, he would have white? I mean, he could also have a black hat. No?
These men are logic specialists. Get rid of number 4. Tell the three guys there are three black hats and two white hats. If one can say the colorvof his hat their lives are saved. If not they all die. After a few minutes number three says his hat color correctly. What colir was it and how did he know?
“To my regular followers, they’re used to seeing lots of As” clever, and kind!
If number 4 says he has a white hat then number 1 would solve it.
Took me about 5 minutes with a few of those drawing triangles with c2 ect... Lol good one for sure
Come on man....
26+112+x=180
Why does the keyboard sound reversed?
Scarlet and Gray
I got there by adding a square in the top left of the circle and making a fourth point. After that I could make multiple lines and calculate the radius that way. But your solution is very nice too, even nicer I'd say.
TF2 MENTIONED AT 0:01 RAHHHHH 🗣️🔥🔥🔥🔥🤽♂️🧚♂️🤑🪂⚡🚵🏿♀️🥵🐡😂🚵🏿♀️💯🧡💔❣️🫀👨🚒👳♂️👨🚀🤴👲🤴🦄🐁🐂🐀🐸🐘🐁🐍🦘🦄🦎🐃🛢️🛵🛢️🚧🏎️🚅🛢️🏎️⛱️🏰🏤🏗️🏩🏥🏩🏙️
Who's gonna tell him what 2³ is?
If you just mulitple the powers by rule it should be 0.5^infinity which should be 0. isnt that the case?
this couldve been done way faster, since we know theres a formula for the radius of an inscribed circle (2 x area / perimeter) and the area was easy to solve, using herons formula (Area= sqrt(s x (s-a)x(s-b)x(s-c), where s is semi perimeter) and the area, in this case would be 84, and the perimeter 42. 168 / 42 = 4
and then you find the area of the circle whatever
I learnt in the sec school that 0.9™ equals to 9/9 or 1 so 1 - 1 = 0 Im aware about this being 1-0.999999.... but 0.9 periodic is infinitely close to 1 so who cares let it be 1 LOL!
'but 0.9 periodic is infinitely close to 1' It's not 'infinitely close to 1' (whatever that means). It is exactly 1.
@@thetaomegatheta prove it ez
Here are a couple of proofs of the fact that 0.999... = 1: 1) Consider relation R between Cauchy sequences of rational numbers: for any two Cauchy sequences of rational numbers a=(a_1, a_2, a_3,...) and b=(b_1, b_2, b_3,...) the relation aRb holds iff lim(a_n-b_n)=0. Any given real number is an equivalence class of such sequences with respect to R. Any given digital representation corresponds to a Cauchy sequence of rational numbers, for example, 0.999... corresponds to (0.9, 0.99, 0.999,...), and 1 corresponds to (1, 1, 1,...). Let's check if (0.9, 0.99, 0.999,...)R(1, 1, 1,...): lim(1-sum(9/10^k) for k from 1 to n) as n->inf = lim(1/10^n) as n->inf = 0. That means that (0.9, 0.99, 0.999,...)R(1, 1, 1,...) and 0.999... = 1. 2) If x is some real number, |x|<=1, and r = p/10^n, where p is integer, and n is natural, we have: If x = r+x*1/10^n, then x*(1-1/10^n) = r x = r/(1-1/10^n) As said previously, r is rational, 1-1/10^n is a sum of two rational numbers, meaning that it's rational (because p_1/q_1+p_2/q_2 = (p_1*q_2+p_2*q_1)/(q_1*q_2), and a product of two integers is an integer, and a sum of two integers is an integer, meaning that p_1*q_2+p_2*q_1 = p_3a+p_3b = p_3, q_1*q_2 = q_3), and the division of a rational number by a rational number is also rational (because p_1/q_1/(p_2/q_2) = p_1*q_2/(q_1*p_2)), i.e. r/(1-10^n) is rational, and, because x = r/(1-10^n), so is x. If |x| was greater than 1, then we could do either of the substitutions x_1 = x/(floor(|x|)+1) or x_2 = x-floor(|x|) and prove that x_1 and/or x_2 is a rational number, from which it would follow that x is rational. That means that every real number x that can be represented with repeating decimals, i.e. x = p/10^n+x*1/10^n, where p is integer, and n is natural, is a rational number. And, of course, 0.999... = 9*1/10+0.999...*1/10, so 0.999... has to be rational, and 0.999... = 9*1/10/(1-1/10) = 9/10/(9/10) = 90/90 = 9/9 = 1. 3) 0.999... is the sum of the series 9/10+9/100+9/1000+..., i.e. it is the limit of the sequence of partial sums of that series. The sequence of partial sums of that series is s = (0.9, 0.99, 0.999,...). Let's denote the nth element of that sequence as s_n. Let's see if 1 is the limit of that sequence: for every neighbourhood U(1) there exists a metric ball B = B(1, 1/10^N), centered at 1 and of radius 1/10^N, where N is natural, such that B is a subset of U(1). However, d(1, s_n) = |1-s_n| = |1-(9/10+9/100+9/1000+...+9/10^n| = 1/10^n, but for all n > N it is true that 1/10^n < 1/10^N, meaning that for all n > N every s_n is in B, and, thus, is in U(1). That means that 1 is the sum of the series 9/10+9/100+9/1000+... by definition, i.e. this means that 1 is 0.999... by definition.