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Wow, Thank you for the help 😜

It was in fact not 11/3, Many Thanks!

There’s an issue with this. The question needs to be changed to him saying he TURNED 43. Humans are usually two ages during a calendar year, before and after their birthday. Unless we knew he was born on January 1st then him being 43 during 1849 means he could have been born in either 1805 or 1806.

This was a fun test. Most people don't understand this. Deducting 20% from 80 is NOT the same as adding 20% to 80. (80 x 1.2 = 96) But 80 x 1.25 = 100.

Impossible to hear what you are saying there.

1806

HypatiaMATH, the volume on this video is much too low. I tried another video of yours and that volume was OK.

THANK YOU

Thank you so much!

Good for GCSE practise.

very helpful video!

Thank you very much

thank you sm, this actually helped me!!

7=-3 mod 10, so 7^3+7= (-3)^3+(-3) mod 10=-(3^3+3)=-0=0 mod 10., Note, p divides p^3+p. so for p=5, p^3+p is a multiple of 5. also, p^3+p is even so, 10 divides p^3+p.

2³ + 2 = 10 3³ + 3 = 30 5³ + 5 = 130 7³ + 7 = 350 Done.

Thanks❤

Smart thing to do I would've never thought of that!

I saw the exact same question yesterday in a Cambridge Exam Paper. It was 4 marks. I completely agree with the suggested working steps in the video using angle properties of polygons. However my student did it in one step, which is to take angle BCD - angle ACF = 108 degrees - 90 degrees (as in, the interior angle of a pentagon - interior angle of a square). He did not even find the base angles of the isosceles triangles. His working step got the final answer in 1 step but there's it's obviously a fluke. The angles overlap with different excesses so it would not be right to subtract them. If anything, a subtraction will only result in the differences between the excesses, ie. (A+B) - (A+C) = B-C. However the interesting thing is for this regular pentagon, the base angle of the isosceles triangle of 36 degrees minus angle DCF also gives us the value of 18 degrees, which happens to be angle DCF, hence this appears to be fluke. I try this with a regular hexagon and the result is the same. I am trying to find a counter example to show why the above step is not logical but I can't seem to find the answer yet. If you could have the time to help me figure it out, it would be great. Thank you for your time in advance.

Why over complicate it so much. Use area = 1/2 ab sin C So 85 = 1/2 (AC)sq x sin 110

Nice problem and proof. Very bad delivery, however. I recommend being more enthusiastic, talking faster, align equations so that simplifications are easier to see visually, and ideally also explain where did the idea of squaring equations come from.

15

15

It’s much easier to use formula A =1/2ab sineC. So 1/2 x 5 x 5 x sine45 then multiply by 8. One calculation gets the same answer much simpler.

Nice proof!

The area of the octagon is 8 times the area of triangle AOB. The area of triangle AOB, as is the case for all triangles, is 1/2 * base * height. Let side OB be the base. We are told that it has length of 5 cm. Drop a perpendicular from A to the base. That is the height. Its length is length OA * sin of angle at O = 5 * sin(45 deg) = 5 * (sqrt(2))/2. Thus, the area of the triangle is 1/2 * (5) * (5 * (sqrt(2))/2) = 25 * (sqrt(2))/4. Now, the area of the octagon is 8 * the area of the one triangle because the octagon is made up of 8 identical triangles. Therefore, the area of the octagon is 8 * 25 * (sqrt(2))/4 = 2 * 25 * sqrt(2) = 50 * sqrt(2) = 50 * 1.4142 = 70.711 cm^2.

Perhaps a little early in using your calculator... x=5.sin(22.5), AB=10.sin(22.5)... y=5.cos(22.5) Area of Triangle AOB=½.AB.y AOB = (10.sin(22.5).5.cos(22.5))/2 AOB = 25.sin(22.5).cos(22.5) Now use the calculator... AOB = 25x0.3826x.9238 AOB = 8.839cm² (to 3 dec.places) Area of octagon = 8.AOB = 70.711cm²

Oh, I see you got the same answer.

The cost of 20º of metal equals the cost of 30º of plastic. Dial "A" cost is equal to the cost of 360º of plastic. Dial "B" cost is equal to the cost of 370º of plastic. 360 to 370 or 36 to 37

36 : 37

Nice lesson explained in gentle and easily understandable way. Thanks a lot professor. 🙏

4^x+4^x+4^x+4^x=4X4^x=4^(x+1) So x=15

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

45

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.