Answer:
D. 10
E. 23
F. -7
G. -2x+18
Explanation:
It doesn't matter how we name that operation. I don't remember what that particular symbol is called so I'm just going to use the @ symbol.
The operation, @, on two real numbers is defined as:
a@b=a-b(a-2b)
That is how the operation is defined.
Now can start the problems.
D.
We are asked to evaluate 4@3 given:
a@b=a-b(a-2b)
So we need to replace a with 4 and b with 3:
4@3=4-3(4-2*3)
Now we just compute the right hand side as we normally would.
You can use PEMDAS to help if you like.
4@3=4-3(4-6)
4@3=4-3(-2)
4@3=4+6
4@3=10
So the answer is 10 for D.
E.
We are asked to evaluate 3@4 given:
a@b=a-b(a-2b)
So we need to replace a with 3 and b with 4:
3@4=3-4(3-2*4)
3@4=3-4(3-8)
3@4=3-4(-5)
3@4=3+20
3@4=23
So the answer is 23 for E.
F.
We are asked to evaluate -5@-2 given:
a@b=a-b(a-2b)
So we need to replace a with -5 and b with -2:
-5@-2=-5-(-2)(-5-2(-2))
-5@-2=-5+2(-5+4)
-5@-2=-5+2(-1)
-5@-2=-5+-2
-5@-2=-7
The answer for F is -7.
G.
We are asked to evaluate x@3 given:
a@b=a-b(a-2b)
So we need to replace a with x and b with 3:
x@3=x-3(x-2*3)
x@3=x-3(x-6)
x@3=x-3x+18
x@3=-2x+18
The answer for G is -2x+18.
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We could have simplified the operation, @, defined on the two real numbers a and b, as such:
a@b=a-b(a-2b)
Distributive property:
a@b=a-ba+2b^2
There are no like terms here so nothing else could be done.
So we could have used a@b=a-ba+2b^2 and gotten the same answers per question.