Відео

Alternatively, y = x^4 - 18x^2 is an even function with a positive leading coefficient. That means it is symmetrical about x=0 and increases monotonically for sufficiently large x. Set y' = 4x^3 - 36x = 0, giving x=0, ±3. We can calculate that when x=3, y = 81 - 18 * 9 = -81 and when x=0, y=0. By symmetry, the stationary points are therefore (0, 0) and (±3, -81). Because of the monotonic increase of y for large x, the stationary point at x=3 must be a minimum, and so must the one at x=-3, by symmetry. That leaves a maximum at the origin.

You have a correct solution here, but the calculations would be made a lot easier if you solved first for "d". Assuming that no calculators are allowed, there is still a simple way to solve the equation : d^2 = 1002001 ( and I'm pretty sure that this is what the proposers of this problem intended ). Notice that the pattern in this number looks quite clearly like the binomial expansion of ( x + 1 )^2 = x^2 + 2x + 1 with "x" = 1000. Therefore, (1000 + 1) ^2 = 1,000^2 + 2(1)(1000) + 1^2 = 1,000,000 + 2000 + 1 = 1002001 . Therefore : sqrt ( 1002001 ) = 1001 = (7)(11)(13), and the problem is quickly solved from there : a = 8(1001)/ (7)(11)(13) = 8 ; therefore, a^2 = 64.

For the third case, where X is equal to 70° i wouldn't have solved this way, but rather with a similar approach to what you did in triangle 2. So for triangle 1, we have 70 and x=55 (70 + 55 + 55) for triangle 2, we have 70 and x=40 (70 + 40 + 70) and for triangle 3, we'd have 70 and x=70 (70 + 70 + 40) The question states that there are two different angles so we can't just take the one angle we know as the one we're looking for (that's what you did with you third triangle and i think it's wrong) but we can totally have two different angles that have the same measure, which is the case if we have an isosceles triangles with two 70° angles. The answer would still be 70° as you said but for a different reason.

The problem statement identifies two angles, 70 and X. In your third example, the angles are 70-55-55, so "70 and X," which by the problem statement must be two different angles, can only be 70 and 55 (same as your first example). For X to also be 70, there must be two 70s, as in your middle example 70-70-40. So in total, there are only two values for X. (70, X=55, 55 and 70, X=70, 40)

You can also solve the general case with the two angles being X and Y, and solving for Y: Case 1: X = Y. Case 2: Y is the vertex angle. Y is therefore 180 - 2X. Case 3: X is the vertex angle. Y is therefore (180 - X)/2. Add these up, and you get 270 - 3X/2. To solve this specific case, plug in 70 for X and get 165.

Poorly phrased. For the 3rd triangle, you don't have TWO angles which are 70 AND x, you have ONE angle which equals 70 AND ALSO x.

Well, it's well known that for any polynomial P with integer coeffs, a-b | P(a) - P(b)... this is nothing but a straightforward application.

this is pretty cool! you can definitely generalize this to prove (a+b)^n - a^n is always divisible by b

Use the quotient rule to expand each logarithm. Everything cancels except -log100 which equals -2.

This was super helpful!

a(n+1)=(n+2)an/(n-1)-(2n+1)/(n-1)= n(n+1)(n+2)/2!a1-(0,5/1-1/3+0,5/5 +..+0,5/(n-2)-1/n+0,5/(n+2))n(n+1)(n+2) an=(n-1)n(n+1)/2!a1-(0,5/1- 1/3+0,5/5+..+0,5/(n-3)-1/(n-1)+0,5/(n+1))(n-1)n(n+1)✓

√(-a²b²+ab²+a+a²b+b-3ab-2+2(ab+1)√((a-1)(b-1))/a/b?|√(c-1)-(√(a-1)+√ (b-1))/(ab)|,?=≤

S=9pi-18(pi/2-1)=18(cm^2)

BX->=-a->+4b-> BM->=6b->-1.5a-> BM->/BX->=1.5 ч.т.д.

¼(100π - 36 π) = 16π

All but 7 and 15

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strange voice-over. oh well. 18 - 12 - x - (28-x) 13 - 15 - (31-x) - (x-1) 20 - 10 - 11 - 17 7 - 21 - 16 - 14 Then, using the diagonal also = 58, we get 58 = 7 + 10 + (31-x) + 28-x) ... 18 = 2x ... [x = 9] Armed with that, all the rest of the values are known exactly, including the value being looked for.

Its the differences in the travel times and lengths which matter. When travelling an additional 75m, the train took an additional 3 seconds. The train is travelling at 25 m/s.

Intermediate Maths Olympiad question. good answer

Lots of videos upload train based sum please

Thanks mam❤🎉

There is a slight error in your reasoning here. After noticing that AD must be 14, you assume that AB must be 2. But that can't be known yet; it could be that the BC, the distance of the two interior points, could be 2. (There's no need to consider CD = 2 because of symmetry with AB = 2.) The proper way, after noticing that AD = 14, is to say that either AC or BD must be 12, the next longest length. Due to symmetry, it doesn't matter which you choose, so set BD = 12, and then AB = 2, and the rest of your solution follows.

The ending still needs proper reasoning. First, the only two possible values for y are 0 and 5 because 165 is divisible by 5. Second, the option Y=0,X=4 is impossible because 165 is also divisible by 3. This means that XX4XY is divisible by 3, and thus that X+X+4+X+Y is divisible by 3. X=4,Y=0 doesn't work for divisibility by 3, but X=9,Y=5 does

good answer

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HypatiaMATH, Your videos always brighten my day, so I subscribed!

y = 5 and x = 9. There are only 9 digits to chose from to find x, so it was easy to try them one by one. We know the value of y right away as either 5 or 0 to be an end number on a number evenly divisible by five.

165 = 11*5*3 so the entire string must be divisible by 11, divisible by 5, and divisible by 3 Y *must* equal 5 because 0 won’t work and any other number fails the ‘divisible by 5’ test 0 won’t work because Y+4 must be divisible by three (see the Mod 3 cheat code) any value for X satisfies the ‘divisible by 3’ condition, so check elevenses decollate the first two X’s*** (that is, chop off the 10,000s digit and the 1000s digit) and just consider “4X5”, which needs to be divisible by 11 only 495 fits the bill, so 99495 ***XX000 is always divisible by 11, so the difference, namely 4X5, will be also

11418 = 1028*11, but X+4+Y = 1+4+8 = 13 != 2 = 1+1 = X+X

You said X cannot be 4 without explaining why. You needed to say why that was the case.

To determine the values of 𝑋 X and 𝑌 Y in the 5-digit number "XX4XY" so that it is divisible by 5, we start with the key property of numbers divisible by 5: the number must end in either 0 or 5. Given the number is "XX4XY", the last digit 𝑌 Y must be either 0 or 5. Case 1: 𝑌 = 0 Y=0 The number becomes "XX4X0". Now we need to find out if this form satisfies any constraints for 𝑋 X. Since the divisibility rule for 5 is satisfied (the number ends in 0), we move to check if there are any other constraints. There are no further conditions given for 𝑋 X when 𝑌 = 0 Y=0, hence 𝑋 X can be any digit from 0 to 9. Case 2: 𝑌 = 5 Y=5 The number becomes "XX4X5". Similarly, this number satisfies the divisibility rule for 5. There are no further constraints for 𝑋 X given, so 𝑋 X can also be any digit from 0 to 9 when 𝑌 = 5 Y=5. Determining 𝑋 + 𝑌 X+Y From the two valid cases, we have: If 𝑌 = 0 Y=0, then 𝑋 + 𝑌 = 𝑋 + 0 = 𝑋 X+Y=X+0=X. If 𝑌 = 5 Y=5, then 𝑋 + 𝑌 = 𝑋 + 5 = 𝑋 + 5 X+Y=X+5=X+5. Since 𝑋 X can range from 0 to 9 in both cases, let's examine the outcomes: When 𝑌 = 0 Y=0, 𝑋 + 𝑌 = 𝑋 X+Y=X ranges from 0 to 9. When 𝑌 = 5 Y=5, 𝑋 + 𝑌 = 𝑋 + 5 X+Y=X+5 ranges from 5 to 14. Since we do not have restrictions on specific values of 𝑋 X other than being digits (0 through 9), we summarize the possible sums 𝑋 + 𝑌 X+Y: For 𝑌 = 0 Y=0, 𝑋 + 0 X+0 can range from 0 to 9. For 𝑌 = 5 Y=5, 𝑋 + 5 X+5 can range from 5 to 14. However, the problem requires a single result for 𝑋 + 𝑌 X+Y. By standard practice in such puzzles, the smallest valid digit sums that appear consistently in both scenarios (to ensure general correctness) are of interest: If 𝑋 = 2 X=2 and 𝑌 = 5 Y=5, then 𝑋 + 𝑌 = 7 X+Y=7. So, the value of 𝑋 + 𝑌 X+Y is calculated as: 7 7 ​

2^64 is a perfect square, a^2, with a= 2^32 . So write 2^64 -1 as a^2 -1 = ((a-1)(a+1)=(2^32-1)(2^32+1). In fact, you can show that 2^32-1 has two factors (2^16-1)(2^16)+1), and so on. So 2^64-1 = (2^32+1)(2^16+1)(2^8+1)(2^4+1)(2^2+1)(2^2-1). And these are almost all prime numbers. You will recognize primes 2,3,5,17,257. 2^16+1 is also a prime. But 2^32+1 can be factored.

This is nice 💯

Wrong explanation. Divisible by 11 only means that the alternate sum of digits is divisible by 11, which would also allow for x=4, y=0. E.g., 44440 is divisible by 11. But since the number has to be divisible by 3 also, the sum of the digits must be as well, which implies that 4+y is. So y=0 can be excluded, leaving us only with the other solution you gave, 99495

Let's use 165 = 3×5×11 The last digit Y can only be 0 or 5. Using mod 3, we have 4+Y = 0 This implies Y=5. Now, using mod 11, -X+9 = 0 X = 9 Done. Checking: 99495 + 165 = 99660 = 330×302 = 165×604 Which means 99495 = 165×603

You could just check the most obvious solutions of 165 x N = 4XY since it is very easy to do in head. The first possible one is 165 x 3 = 495 to get X= 9, Y=5 and then check that 99495 / 165 = 603 and see that it works to solve it with. So you can ignore all but the 4XY part at first and solve it in a couple of seconds, before even using a more arduous solution edit: reposted my comment to make sense after being called out, since I left out the most important points. Sorry I ended up deleting your great comment as well Samuel de Andrare

Just n! - (n-1)! = 16(n-2)! n(n-1) - (n-1) = 16 (n-1)² = 16 n-1 = ±4 n = 3 or -5 ...

... the reasoning om the video is NOT good. Look, the middle b+c tell us b+c=4 or b+c=13 then the right b+c (equals a or 1a) combined with the previous options tell us a=4 or a=3 If a=4, the a+a=c implies c=8. Then b+c=4 or 14 gives us b=6, which doesn't work. So, we just have the option a=3, which implies c=6, then b=7, which works.

y=-4 is not a solution because taking the logarithm of a negative number yields a complex number. Then the proper constraint is y>-1/2. When y=-1/2, you're taking the logarithm of zero, which is undefined. It's OK to say, "Taking a logarithm of zero is undefined and taking the logarithm of a negative number give you complex results. We're only dealing with real numbers in this class." It takes the mysticism out of just setting y>0 with no explanation. Also, who is going to come up with that factorization for the quadratic 2y^2 + 7y - 4 = 0? You put that out and act like it's easy. NO! Any normal person is going to use the quadratic equation there. Factoring is the hard way to do it.

4:52 why X can't be 4?

X = 0

Trivially false for x = 1.

I am a bit confused as to why you chose to develop instead of factorize. keep the left side intact, factorize by 2 on the right, and move it to the left with a simple addition which does NOT take the risk of changing the sign of the inequation: (x+2)(x-1)-2(x-1)>0. Now you factorize once more, by (x-1) this time and have a simple x(x-1)>0. Now the solution is obvious as you need either both factors to be positive (x>1) or both factors to be negative (x<0) for this inequation to be true. Anything in ]0;1[ is not true. check for x=1/2: 5/2.(-1/2)>2(-1/2) or -1.25>-1 which is false, you get your counter example

Clearly not correct as it is no true for x=0 or x=1 (or, for that matter, it is not true and value of x between 0 and 1 inclusive). Simply just look for intersections of the expressions on the right and left. If there is one then it cannot be true (in this case there are 2 intersections as the line y=2x-2 cuts the parabola (x+2)(x-1) at two points). Not exactly elegant, but the question only requires one counter example.

Put in x=1, get 0 on both sides, done ✅

x = 0

Inequality can be simplified to x^2 > x. That's clearly not true and x = 0 and x = 1, where they are the equal, and between those two values x^2 < x.

Just reading the thumbnail, didn't watch the video... Tried x=0 in my head and turns out -2>-2 Tried x=1 and turns out 0>0 Also tried -1 and 2 it works for those

The line must lie below the curve for the inequality to be true, but in fact, the line crosses the parabola in two places so the line must be above and below the curve at different points.