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OpenStudy (anonymous):

find points of intersection. x^2 + y^2 = 25, 2x + y =10

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OpenStudy (anonymous):

x=3 y=4

OpenStudy (anonymous):

dont care. i dont know how

myininaya (myininaya):

\[x^2+y^2=25, y=10-2x\] \[x^2+(10-2x)^2=25\]

OpenStudy (anonymous):

and it = (5,0) too

OpenStudy (anonymous):

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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myininaya (myininaya):

\[x^2+10^2-2(10)(2x)+(2x)^2=25\] \[x^2+100-40x+4x^2-25=0\] \[5x^2-40x+75=0\]

myininaya (myininaya):

\[x^2-8x+15=0\]

myininaya (myininaya):

\[(x-5)(x-3)=0=>x=5 or x=3\]

myininaya (myininaya):

\[y=10-2x=> y=10-2(5)=10-10=0 or y=10-2(3)=10-6=4\]

myininaya (myininaya):

so we have the ordered pairs where the curves/lines intersect are (5,0) or (3,4)

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myininaya (myininaya):

abe gj on that thinking

myininaya (myininaya):

i like that

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