Question

If R={(x, y ):where x and y are whole numbers,x2+y2=169}then domain of R ?

Answer 1

X+y=17. X2+Y2=169. (17)2= 169+2XY. 289-169=2XY. 120=2XY. XY=60. then x+y=17. XY=60then X=12. Y=5. 3.4. 5 votes. 5 votes ... To find the value of x & y if,. x + y = 17

Answer 2

Answer:

Required domain of R is {0,5,12,13}.

Step-by-step explanation:

Given,R={(x, y ):where x and y are whole numbers,x2+y2=169}

${x}^{2} + {y}^{2} = 169$

${y = 169 - {x}^{2} }$

$y = \sqrt{169 - {x}^{2} }$

For $x = 0$,

$y = \sqrt{169 - 0} = 13$

For $x = 1$,

$y = \sqrt{169 - {1}^{2} } = \sqrt{168}$

For $x = 2$,

$y = \sqrt{169 - {2}^{2} } = \sqrt{165}$

For $x = 5$,

$y = \sqrt{169 - {5}^{2} } = \sqrt{169 - 25} = \sqrt{144} = 12$

For $x = 12$,

$y = \sqrt{169 - {12}^{2} } = \sqrt{169 - 144} = \sqrt{25} = 5$

For $x = 13$,

$y = \sqrt{169 - {13}^{2} } = \sqrt{169 - 169} = 0$

For$x = 14$,

$y = \sqrt{169 - {14}^{2} } = \sqrt{169 - 196} = \sqrt{ - 27}$

(not define)

So, domain of R is {0,5,12,13}