Lösung 4.1:6a
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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If we write the equation as | If we write the equation as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>(x-0)^2 + (y-0)^2 = 9</math>}} |
we can interpret the left-hand side as the square of the distance between the points (''x'',''y'') and (0,0). The whole equation says that the distance from a point (''x'',''y'') to the origin should be constant and equal to <math>\sqrt{9}=3\,</math>, which describes a circle with its centre at the origin and radius 3. | we can interpret the left-hand side as the square of the distance between the points (''x'',''y'') and (0,0). The whole equation says that the distance from a point (''x'',''y'') to the origin should be constant and equal to <math>\sqrt{9}=3\,</math>, which describes a circle with its centre at the origin and radius 3. | ||
Version vom 08:48, 22. Okt. 2008
If we write the equation as
| \displaystyle (x-0)^2 + (y-0)^2 = 9 |
we can interpret the left-hand side as the square of the distance between the points (x,y) and (0,0). The whole equation says that the distance from a point (x,y) to the origin should be constant and equal to \displaystyle \sqrt{9}=3\,, which describes a circle with its centre at the origin and radius 3.
