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gregori [183]
4 years ago
8

What is the minimum value of the expression $x^2+y^2-6x+4y+18$ for real $x$ and $y$?

Mathematics
1 answer:
Aleonysh [2.5K]4 years ago
3 0
X^2+y^2-6x+4y+18=0
x^2-6x+y^2+4y+18=0
x^2-6x+9+y^2+4y+4 =-18+9+4
(x-3)^2+(y+2)^2=-5
-5 is the minimum value of the given circle
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DiKsa [7]

4.a) 2x=7*3

2x=21

x=21/2 OR

x=10.5

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5x=150

x=150/5

x=30

4.c)4x=5*14

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X=0.15

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X=1.75

4.g)x=0.4*1.5

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X=1.54

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X/7=4

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B)2x/5=3+8

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C)2x/3=74-26

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(I can’t finish all but it is the same method )

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