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find points of intersection. x^2 + y^2 = 25, 2x + y =10
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x=3 y=4
dont care. i dont know how
\[x^2+y^2=25, y=10-2x\] \[x^2+(10-2x)^2=25\]
and it = (5,0) too
Well its basically from the equation, a^2+b^2=c^2. I remembered my special 3,4,5 Triangle and got the answer.
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\[x^2+10^2-2(10)(2x)+(2x)^2=25\] \[x^2+100-40x+4x^2-25=0\] \[5x^2-40x+75=0\]
\[x^2-8x+15=0\]
\[(x-5)(x-3)=0=>x=5 or x=3\]
\[y=10-2x=> y=10-2(5)=10-10=0 or y=10-2(3)=10-6=4\]
so we have the ordered pairs where the curves/lines intersect are (5,0) or (3,4)
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abe gj on that thinking
i like that
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