challenge me to a hard math problem?

Answer 1

Roman Numeral Problem:
CDXXXVI+CCXL=

Try This!It may not be that hard for you!

Answer 2

enable a be the sum of x,y and z (x + y)a = one hundred twenty (y + z)a = ninety six (x + z)a = seventy 2 xa + ya = one hundred twenty ya + za = ninety six xa + za = seventy 2 upload them up 2xa + 2ya + 2za = 288 xa + ya + za = one hundred and forty four a (x + y + z) = one hundred and forty four all of us recognize that a = x + y + z (x + y + z) (x + y + z) = one hundred and forty four (x + y + z)^2 = one hundred and forty four x + y + z = 12 or -12 so the sum of the x,y,z is 12 or -12 plug 12 decrease back in (x + y) * 12 = one hundred twenty (y + z) * 12 = ninety six (x + z) * 12 = seventy 2 simplify x + y = 10 y + z = 8 x + z = 6 upload them up 2x + 2y + 2z = 24 x + y + z = 12 all of us recognize that x + y = 10 10 + z = 12 z = 2 all of us recognize that y + z = 8 8 + x = 12 x = 4 all of us recognize that x + z = 6 6 + y = 12 y = 6 so the answe is: (4, 6, 2) use the an same procedure for x + y + z = -12 to get the different answer wish this permits!

Answer 3

What is the equation of the parabola having points (10,1),(-2,1),(12,5) & axis of symmetry line has slope=1?

Whoever voted thumbs down, lets see you do it chump.
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Here's what I get for the others' questions. How about you?
Schuetzy- (1000-369)/2 = 315.5 or 316 stupid answers at this time.
Christine...-Not possible
Gilbert M-72136637394
Me-?
birgit - not possible
get- no variable z showing - however ~ -6.029021932408822E-6
ashley h - (cosx + isinx)^3
= cos^3(x) + 3icos^2(x)sin(x) - 3cos(x)sin^2(x) - isin^3(x)
= cos^3(x) - 3cos(x)sin^2(x) + i [3cos^2(x)sin(x) - sin^3(x)]
= cos(3x) + isin(3x) by deMoivre
→ cos(3x) = cos^3(x) - 3cos(x)sin^2(x)
= cos^3(x) - 3cos(x)[1-cos^2(x)] = 4cos^3(x) - 3cos(x)
&
sin(3x) = 3cos^2(x)sin(x) - sin^3(x)
= 3[1-sin^2(x)]sin(x) - sin^3(x) = 3sin(x) - 4sin^3(x)

honeypot56000 - not original - not her challenge
MIRAGE_KID2 - not a problem - not true either.
heyhelpme41 - 1000! C(1000000,1000) x^999000
..where C(a,b) is the combination function
rjr - 12345678987654321
weyboom - x=-3, y=4, z=2
MantisDream - http://mathworld.wolfram.com/HeronsFormula.html
rena - DCLXXVI
captn_sal - (√2/2 ± √2i/2)c
stephen m - no one on the planet has a proof of these. Not to mention that his statement is condescending: how could he possibly know what anyone might or might not understand? Does he think he's the resident math wiz here? stephen m - if you're so good let's see you solve my problem: I could have done it in 9th grade.

I'm still waiting for you 4 chumps who voted thumbs down to come up with a solution to mine - what? no answer? - I thought so... It's really not that hard, but i guess you prefer to criticize others rather than do any actual work...it draws attention from the fact that you can't.

Answer 4

Decide whether P=NP, prove the Riemann hypothesis, the Poincare conjecture, or any one of the other millenium prize problems, and you'll get $1,000,000.
I'll be surprised if you can even understand the problems (the three I mentioned I do understand, but thats because I've done a lot of mathematics and computer science :))

Answer 5

111111111x111111111 = ??? Maybe not hard , but portrays the beauty of maths!

Answer 6

Use De Moivere's Theorem with n = 3 to express cos3(-) and sin3(-) in terms of cos(-) and sin(-).

Answer 7

Given a triangle with sides a,b,c, semiperimeter s, and area A,
show that A^2 = s(s-a)(s-b)(s-c)

Answer 8

45454258963214741563325822147258522582(z) divided by 52584216941369 +52116341879845198451-958746*987456321
EQUALS 56566566566


WHAT DOES Z EQUAL

Answer 9

if I answer stupid questions like this and not receive any best answers, how many answers would it take to get my current point value to 1000?

Answer 10

3*5(5*5/8)>18/67(9*9/12)