SOLUTION:
(a,b,c)Θ(d ,e, f)=ad+ be +cf
In the expression (1,−2,3)Θ(1,−1/2,1/3) ,
a=1 d=1
b=-2 e=-1/2
c=3 f= 1/3
Using this operation, we have, ad+ be +c f
= (1.1) + (-2. -1/2) + (3.1/3)
= 1 + 1 +1
=3 (OPTION E)
For all real numbers a, b, c, d, e, and f, the operation Θ is defined by the equation (a,b,c)Θ(d,e,f)=ad+be+cf(a,b,c)Θ(d,e,f)=ad+be+cf. What is the value of (1,−2,3)Θ(1,−1/2,1/3) ? Print
Modified on: Tue, 12 Oct, 2021 at 1:04 PM
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