2006-09-01 02:17:02 · 5 answers · asked by ›tªmmy‹ 3 in Education & Reference ➔ Homework Help
what do you want for an answer? the value of b? whether the equation is consistent? whether i am smart enough to solve it?
whatever the case, you got to be more specific.
2006-09-01 02:26:40 · answer #1 · answered by J S 3 · 0⤊ 0⤋
265+b^2=b^2-32b+256 ->b=0
2006-09-01 09:22:41 · answer #2 · answered by amin s 2 · 0⤊ 0⤋
256 + (b^2) = (16-b)^2
(16 - b)^2 = (16 - b)(16 - b) = 256 - 32b + b^2
256 + (b^2) = 256 - 32b + b^2
256 + (b^2) - 256 + 32b - b^2 = 0
32b = 0
b = 0
2006-09-01 09:23:16 · answer #3 · answered by ³√carthagebrujah 6 · 0⤊ 0⤋
256+b^2 = (16-b)^2
256+b^2=(16-b)(16-b)
256+b^2=b^2-32b+256
b^2 = b^2-32b
if b is 0
256+b^2=0^2-32(0)+256
b^2 =0-0+256-256
b = 0
so..b is equal to 0
2006-09-01 09:56:49 · answer #4 · answered by rayna 3 · 0⤊ 0⤋
Nope.YOU answer it,it's YOUR homework!
2006-09-01 09:18:34 · answer #5 · answered by Anonymous · 0⤊ 0⤋