Given the product of two numbers 7π²√π - 33, find the difference of the square of their sum and the sum of their squares.
easy...
^_^
2006-08-24 02:28:15 · 7 answers · asked by kevin! 5 in Science & Mathematics ➔ Mathematics
(x+y)^2-(x^2+y^2) = x^2+2x*y+y^2-x^2-y^2 = 2x*y = 2(7π²√π - 33) = 14π²√π - 66
2006-08-24 02:50:20 · answer #1 · answered by Anonymous · 0⤊ 0⤋
2
2006-08-24 02:46:26 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Let the two numbers be a and b respectively.
So, ab=7π²√π - 33.
What you require is (a²+b²)-(a+b)²=a²+b²-(a²+b²-2ab)
=2ab
=14π²√π - 66.
2006-08-24 02:47:32 · answer #3 · answered by Veefessional 2 · 0⤊ 0⤋
If the pages are numbered n and n+a million, then: n x (n+a million) = 1190 => n^2 + n - 1190 = 0. =>(n-34)(n+35) = 0. in view that n won't have the capacity to be damaging, which skill n = 34. subsequently the pages are numbered 34 and 35.
2016-12-11 14:27:04 · answer #4 · answered by spadafora 4 · 0⤊ 0⤋
14 π² √π - 66 . that's right its easy.
2006-08-24 03:05:10 · answer #5 · answered by sweetie 5 · 0⤊ 0⤋
Do your own homework.
2006-08-24 02:46:22 · answer #6 · answered by Anonymous · 0⤊ 0⤋
2.7(pi²)(pi)^1/2-33
tht was simple
2006-08-24 02:33:13 · answer #7 · answered by priya 2 · 0⤊ 0⤋