The preview image shows a different question, so I assume you solve something entirely unretated ... zero points ... Admission failed! The 'problem' you solve in this video is childsplay ... one look and you know the first answer ... the rest is a polynominal division and a quadratic equation. Why not solve the problem you showed in the preview image?

c + c + c = c.c.c 3c = c³ c³ - 3c = 0 c.(c² - 3) = 0 First case: c = 0 Second case: (c² - 3) = 0 → c² - 3 = 0 → c² = 3 → c = ± √3

no relation with caption. irritating video. your students will get all the prizes inn the world. keep it up changing the sign in every problem.

caption is for m cube plus m square. but u r solving for m cube minus m square. you r students will acquire Phd in just a minute.!!!!!!

And you write 18 as 27-9 because you already know the answer. What a genius. That's the way to solve problems. Just have the answer before you start.

is this a joke?

Quite easy: m = ln(50)/ln(5)/2 = 1.2153

m.m.m + m.m = 18 m³ + m² = 18 m³ + m² - 18 = 0 → the aim, if we are to continue effectively, is to eliminate terms to the power 2 Let: m = x - (b/3a) → where: b is the coefficient for m², in our case: 1 a is the coefficient for m³, in our case: 1 m³ + m² - 18 = 0 → let: m = x - (1/3) [x - (1/3)]³ + [x - (1/3)]² - 18 = 0 [x - (1/3)]².[x - (1/3)] + [x² - (2/3).x + (1/9)] - 18 = 0 [x² - (2/3).x + (1/9)].[x - (1/3)] + x² - (2/3).x + (1/9) - 18 = 0 x³ - (1/3).x² - (2/3).x² + (2/9).x + (1/9).x - (1/27) + x² - (2/3).x + (1/9) - 18 = 0 x³ - (1/3).x - (484/27) = 0 → let: x = u + v (u + v)³ - (1/3).(u + v) - (484/27) = 0 (u + v)².(u + v) - (1/3).(u + v) - (484/27) = 0 (u² + 2uv + v²).(u + v) - (1/3).(u + v) - (484/27) = 0 u³ + u²v + 2u²v + 2uv² + uv² + v³ - (1/3).(u + v) - (484/27) = 0 u³ + v³ + 3u²v + 3uv² - (1/3).(u + v) - (484/27) = 0 u³ + v³ + (3u²v + 3uv²) - (1/3).(u + v) - (484/27) = 0 u³ + v³ + 3uv.(u + v) - (1/3).(u + v) - (484/27) = 0 u³ + v³ + (u + v).[3uv - (1/3)] - (484/27) = 0 → assume that: [3uv - (1/3)] = 0 ← equation (1) u³ + v³ + (u + v).[0] - (484/27) = 0 u³ + v³ - (484/27) = 0 ← equation (1) (1): [3uv - (1/3)] = 0 (1): uv = 1/9 (1): u³v³ = 1/729 ← this is the product P (2): u³ + v³ - (484/27) = 0 (2): u³ + v³ = 484/27 ← this is the sum S u³ & v³ are the solution of the following equation: z² - Sz + P = 0 z² - (484/27).z + (1/729) = 0 Δ = (- 484/27)² - [4 * (1/729)] = (484² - 4)/27² = 234252/27² = (81 * 4 * 723/)/27² = (2/3)² * 723 x = [(484/27) ± (2/3).√723]/2 x = (242/27) ± (1/3).√723 x = (242 ± 9√723)/3 ← these are u³ & v³ u³ = (242 + 9√723)/3 → u = ³√[(242 + 9√723)/3] v³ = (242 - 9√723)/3 → v = ³√[(242 - 9√723)/3] Recall: x = u + v Recall: m = x - (1/3) m = (u + v) - (1/3) m = ³√[(242 + 9√723)/3] + ³√[(242 - 9√723)/3] - (1/3) m ≈ 2.326259615

This is an utterly trivial problem. A 9th grader who just learned logarithms can solve it mentally. What "Olympiad" is it from?

Stupid solution

I almost immediate lost you. Soooo.... WHAT is m? Did you write it down or did you mention it? Am I wrong if m = 1.4142 (root of 2)? Took me only seconds to do the calculation. If it is not 1.4142, then what is it?

I hope you are not a teacher!

4^11-4^1=4194300 It’s in my head.

by inspection!

-1/3

You gave me a headache! 🤕😳

This problem, with the identical solution strategy, was posted back in November by "Math Beast". kzbin.info/www/bejne/m5rXk4GGj9WLhZIsi=L9mVmlGSPeOLfPN5

For any question of this type. Find natural numbers a,b such that 1/a+1/b=1/c The solutions are a=b=2c and a=(c+1) . b=c(c+1) If sometimes helpful to look at an easier problem. c=2 1/2= 1/4+1/4 1/2=1/3+1/6

I wish you learn speak Oxford english, ?

m = 2

Wrong at 7.06

m=1+log_2(81)

We can do it by using division method for square root. It only needs a few lines to get the answer and we can also use similar methods to find cube root or fourth root.

When m^4 = 16, m may be 2, -2 or 2i, as (-4)^2 as well equals 16. I think, when you want to study in Stanford you should see that immediately, because that's school mathematics.

Whoa! Hold the red pen! you started out by dividing the values by 13 - then worked out the area of 13th the area, so 144 is not the answer. Think about it for just a few plank moments a smaller subset of the quad is 65*156 = 13*13 *5*12 = 169 * 60 >>>>>> 144. you have to multiply the 144 by 13 = 1872

Realschulen Niveau

If you have a set of equations with same base and exponent is a variable, it's typically better to subtract the equations so you can eliminate the need to use logarithms. 3^m + m^2 = 13 and you subtract 3^m - m^2 = 5 yields 2(m^2) = 8 yields m^2 = 4 yields m = 2. 3^m + m^2 = 65 and you subtract 3^m - m^2 = 1 yields 2(m^2) = 64 yields m^2 = 32 yields m = 4*sqrt(2). 4*sqrt(2) is not an integer, so reject. If you were interested in all real solutions though, much simpler to solve m^2 = 32 than 3^m = 33.

Why was 500,000 written as 50,0000?

2^19-32 = (2^20-64)/2 = (1048576-64)/2 = 1048512/2 = 524256. Knowing the 2 powers table up to 2^16 and then significant higher powers is assumed for any competent student, in the same way, knowing the 12 times table is assumed for any school student.

How about two to the fourteenth.

Nonsense question ... anyone with a decent knowledge of maths and IT knows 2"20 is 1048576, and then you simply compute off the top of your head (took me 2 seconds).

I feel it's much more straightforward to go with 2^19 = 1024*512 and calculate 512000+10240+2048-32.

32

Ans:option B 4^13

Harvard University Admission Interview, Simply: (2/3)²⸍³ =? (2/3)²⸍³ = (2²⸍³)/(3²⸍³) = [(2²⸍³)(3)]/[(3²⸍³)(3¹⸍³)] = {[(2²)(3)]¹⸍³}/3 = (³√12)/3 Answer check: (2/3)²⸍³ = 0.667²⸍³ = 0.763, (³√12)/3 = 2.289/3 = 0.763; Confirmed The calculation was achieved on a smartphone with a standard calculator app Final answer: (2/3)²⸍³ = 0.763 = (³√12)/3

0

9

its 9 D is wrong.

18

D

9

9

D

D

With the fact that x>y and the assumption that x = y+1 or y=x-1 the task becomes solvable. Basically you can't do it like that. It doesn't work like that with a triangle with different lengths. Still... a nice task.

Brilliantly solved

Im confused.....to my mind there is not sufficient info to determine this. I say this because i could draw a similar triangle where one side is still 11 but the x and y sides are either increased or decreased in length, which will in turn change their actual value as well as change the area of the triangle! Is this fancy maths not just finding only one possible value of x and y. If so, i could do that by guessing a value for one of the variables and using pythagerous to calculate the other sides and the area! WTF????? Can someone please explain why i am wrong?

Muffled and inconsistent sound

99% failed? Really? Come on, this is simple math, nothing special. m1 = 3; m2, m3 = -2(1 ± i√2)

At 1:12 the answer is on the board. No need to go further. EDIT: You are right. I am part of 100% that did not apply. I will de better next time! 😂