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anonymous
 5 years ago
Set both equations to y=mx+b
x^2+y^2=17
x^2+y^22x=13
(sqrtx^2 + 17 ) (sqrtx^2+17)
(sqrtx^2+2x+13) (sqrtx^2+2x+13)
Is this right?
anonymous
 5 years ago
Set both equations to y=mx+b x^2+y^2=17 x^2+y^22x=13 (sqrtx^2 + 17 ) (sqrtx^2+17) (sqrtx^2+2x+13) (sqrtx^2+2x+13) Is this right?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0circles arent normally linear...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0those may be more ovalish if anything, but same principle :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cant tell, im still confused by the question :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im supposed to put the equations in mx+b so I can put it in my graphing calculator. Then im supposed to use the intersect function on my calculator to find the points. there should be 4 points.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0there are up to 4 possible points. that is right... but: y=mx+b is the equation for a line. What you have here are 2 ovals that might be intersecting adn there is no way to make any part of them a straight line that I am aware of

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0your best bet would be to "eliminate" a variable first

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(x^2+y^20x=17) (1) x^2+y^22x =13 x^2 y^2 +0x =17 x^2 +y^2 2x = 13  0x^2 +0y^2 2x = 4 2x = 4 x = 2 plug in x=2 for your x values and solve for y :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04 + y^2 = 17 y^2 = 13 y = sqrt(13) or y=sqrt(13) 4 +y^2 4 = 13 y^2 = 13 y= sqrt(13) or y=sqrt(13) your points of intersection are: (2,sqrt(13)) and (2,sqrt(13))
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