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Paul [167]
3 years ago
11

Substitute the values of x and y into the expression -3x∧2 + 2y∧2 + 5xy - 2y + 5x∧2 - 3y∧2. Match that value to one of the numbe

rs.
1. x = 2, y = -1
__14
2. x = 0, y = 2.5
__0.44
3. x = -1, y = -3
__ -1
4. x = 0.5, y =
__1.665
5. x = , y =
__9.17
6. x = √2, y = √2
__-11.25

Mathematics
1 answer:
Tems11 [23]3 years ago
5 0
A couple of comments:
1) The character used to indicate exponentiation in plain text is the caret (^), above the 6 on most keyboards. This character (∧) signifies a logical AND, which is not appropriate for your use here.

2) Variable values are missing for problems 4 and 5. We can guess that the missing value in problem 4 is either -0.1 or 0.6.

_____
For f(x, y) = 2x^2 -y^2 +5xy -2y
1. f(2, -1) = -1 . . . . . . . . . . 3rd selection
2. f(0, 2.5) = -11.25 . . . . 6th selection
3. f(-1, -3) = 14 . . . . . . . . 1st selection
4. f(0.5, 0.1) = 0.44 . . . . 2nd selection
5. f( __, __) = 1.665 . . . 4th selection
6. f(√2, √2) ≈ 9.17 . . . . 5th selection

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Remark

Watch out for the -4<=x<=6 thing. It causes lot's of problems if you are not careful.


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The second best answer is A and if that restriction, -4<=x<=6, wasn't there, the answer would be A.


When y = - 4, the x value falls outside the restricted domain ( -4<=x<=6) we have been given. so B is not the answer.


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So the answer is C <<<<< Answer.


When y = 1 , x = - 2 1/2. That is within the domain, but it is not the minimum value so D is wrong.

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