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change dim uche to (2,0)

why did you do it so hardly , you can simply use the formula AP=a+b/2

The mean difference is deducted from the value obtained in case of cosine while calculating the value using the log table

thank you very much for shearing your knowledge I have been benefited from this video and I thank you for your efforts. have a nice one :)

Reference books pls?

How do I make backspace?

Promo_SM 😋

2, by five seconds of inspection. Another couple seconds show that 3 also works.

Uncle you are doing it more complex. Let, perpendicular of the triangle be "a" Base of triangle be "b" Given, Perimeter=120 and Hypotenuse=51 ATQ a+b+51=120 Or, a+b=120-51 Or, a+b=69 According to Pythagoras Theorem, a²+b²=51² Or, (a+b)²-2ab=51² Or, 69²-2ab=51² Or,69²-51²=2ab Or,(4761-2601)÷2= ab Or,ab= 1080 We know that area of triangle 1/2 × perpendicular × Base Or 1/2ab Or 1080/2 Or 540 So area =540

How come you assumed that 2a+1 and 2b+1 are integers (as factors of 13)? Was that given? For example, a=5/3, b=1 is a solution for a+2ab+b=6, yet 2a+1 = 13/3. It's not necessarily 13, 1, -13, -1 as you concluded in your solution.

We can follow and read the solution

dude... at 1:16, you were done if you really understood what logarithms mean. you had 3^x = 6. By definition of logarithm, x = log_3(6). you waste like 4 minutes using log base 10 unnecessarily.

I got there in the end with minimal help from you but it took me a long time. I don't think I would have thought to make triangle ABE on my own. I enjoyed the challenge. Thanks.

I'm sure it would be much more clear if you upgraded your microphone and tried to avoid background noise. Still, following along with the notation doesn't seem too difficult.

More than a minute in before you bother to mention that it's a right triangle.

Given that an integer solution is stipulated, it takes five seconds to work this out mentally. Applying pythagoras you need two squares whose difference is 25, and a quick estimate shows 12 and 13 are the values - even if you don't remember the first few standard integer pythag triangles in the first place.

damn bro

If ab =y , if ac = 5 then area = 5y/2 finding bc would bebirrelevant

@lustepfordresourcecenter6176 Actually, "0" is a *VALID* answer. Simply because the DEFINITION of a triangle does NOT exclude sides of length 0. It ONLY excludes NEGATIVE line-lengths. *AND* using the formula of Pythagoras, "0^2 + 0^4 = 0^6" is a True statement. Way too many people think that ALL sides of a shape MUST be >0.🤔 Granted that in the PRACTICAL world of engineering, a value of "0" is more-often-than-not a useless answer.

Vc enrolado demais.......complica as coisas fáceis....ok

Thank you man for the help, Im in 7th grade, and we have this problems in my mathbook and i didn't know how to do it. Thank you so much, because of you I legit got so much better.

(2;1), (1;2).

Awesome, thank you

6^m + 6^n = 42 6^n [6^(m-n) + 1 ] = 6.7 6^(n-1) . [6^(m-n) + 1] = 7. The left side of the last equation must be factors of 7, which is a prime, so they must be 1 & 7, i.e. 6^(n-1) = 6^0 and 6^(m-n) +1 = 7 n-1 = 0 and m - n = 1 n = 1 and m = 2. HOWEVER, this solution only works if we knew in advance that m and n are integers OR if there were some theorem that tells us that the left side of 6^x . (6^y + 1) = 7 must be factors of 7. It feels instinctive, but I couldn't think of a theorem to justify it.

Nice problem and a nice solution. Thank you! My only comment is that you didn't need to prove that x > y because the hypotenuse of a right angle triangle is always greater than any of its two sides.

sin2a= 7/16

Thank you...

like 49 super ayudemonos a crecer en youtube

That took me back 40 years.

very nice

fist

Direct apply a+b whole square.

(1,2) & (2,1)

16 and 9 by inspection. Not really an Olympiad question.

Simply splendid

I'm sorry, but I might be missing something. 5 x (5/24) = (5x5)/24 = 25/24 , not 125/24. If it were 5 + 5/24, then it would be 120/24 + 5/24 = 125/24. I don't think that the two sides in each of these two equations are equal. There must be something wrong.

Thank you!

There is a condition here, though. Because a and b have to be whole numbers, a has be to greater than 0.

81 - 16 = 3^4 - 2^4 . m = 4

I take it you are not reading your comments you respond. Everyone is pointing out your mistake.

(X,Y):(6,9),(-10,-7)

You picked about the longest hardest way to do this. Solve the second equation for a or b and substitute into first and solve. Then solve for other variable. Then take the time you spent to take nap.

It's mindblowing to see so many people not having heard of mixed fractions. I would guess the USA started that, since they created another worldwide issue with juxtaposition ('1÷2(1+1)=1'). Just of curiosity have you heard of mixed fracture in school and where and when have you finsihed school? I finished school in 1996 in Germany (NRW) and we were taught mixed fractures (and that '1÷2(1+1)=0.25').

3*3/а, or 3+3/а?and 3rootor cubic rooy

Indian students who know this by heart seeing the video like

This is confusing because you left out the + for sqrt ( 3 + 3/a) in the thumbnail

3*(3/a)=9/a, not 3+3/a Squaring both sides yields 9/a=27/a, which has no solutions.

So plug your solution a = 8 into the original equation. Hint: it isn't a solution.

There must be 12 pairs solutions of (x,y), not just 3 as in the video, i.e. 99=1×99, 3×33, 9×11, giving 3 solutions only. But 99=-1×(-99), -3×(-33), -9×(-11), another 6 solutions, and their inversion as the question is cyclical.

Excellent!