eric9723 eric9723

31-08-2019
Mathematics

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Since the expression (x-4)^2+(y+1)^2 evaluated at (3,6) is

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jdoe0001 jdoe0001

31-08-2019

[tex]\bf (x-4)^2+(y+1)^2\qquad (3,6)~~ \begin{cases} x = 3\\ y = 6 \end{cases}\implies (3-4)^2+(6+1)^2 \\\\\\ (-1)^2+7^2\implies 1+49\implies 50[/tex]

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