Originally posted by: Toastedlightly
math != hard.
Yeah, that! This is an application of abstract algebra. You use the definition given and replace the letters.Originally posted by: Anonemous
? just plug in the numbers and chug.
b#a = (-b+a)(a+b)
if b =5 and a =2
(-5 +2)(2+5) = -3(7) = -21
"#" is not just a +,-,/,* it's a combination of those operations.
Originally posted by: Gobadgrs
For all integers a and b, the operation # is defined as a # b = (?a + b)(b + a). If a = 2 and b = 5, then b # a =
?25
?21 <--answer
?7
21
25
I guessed 21. I dont understand how to interpret what b#a means. It could be + - / or *.
Someone clue me in please.
Originally posted by: iversonyin
haha...I wonder which B School is going to accept this guy
Originally posted by: mitaiwan82
damn trick question...the definition is a # b, but they ask for b # a :frown:
Originally posted by: Gobadgrs
For all integers a and b, the operation # is defined as a # b = (?a + b)(b + a). If x= 5 and y =2, then x # y = ?
You seem to be confused by that fact that they are using the same variable names to mean different things.Originally posted by: Gobadgrs
im quite capable of calculating (-2+5)(5+2) = 21. I was confused if they wanted us to flip the whole equation around or just change what was being substituted. All I thought the equation would change to would be (5+2)(-2+5)
Think of it this way, if they told you:Originally posted by: Gobadgrs
im quite capable of calculating (-2+5)(5+2) = 21. I was confused if they wanted us to flip the whole equation around or just change what was being substituted. All I thought the equation would change to would be (5+2)(-2+5)