The standard form for the equation of a circle is :
<span><span> (x−h)^2</span>+<span>(y−k)^2</span>=r2</span><span> ----------- EQ(1)
</span> where handk are the x and y coordinates of the center of the circle and r<span> is the radius.
</span> The center of the circle is the midpoint of the diameter.
So the midpoint of the diameter with endpoints at (2,-5)and(8,-9) is :
((2+(8))/2,(-5+(-9))/2)=(5,-7)
So the point (5,-7) is the center of the circle.
Now, use the distance formula to find the radius of the circle:
r^2=(2−(5))^2+(-5−(-7))^2=9+4=13
⇒r=√13
Subtituting h=5, k=-7 and r=√13 into EQ(1) gives :
(x-5)^2+(y+7)^2=13
LCM of 15 and 60 is 60.
GCF of 15 and 60 is 15.
le tienes que sumar 5,400 para que te de el resultado
Answer:
ON ATTACHMENT
Step-by-step explanation:
Plugin values of x:
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