Se desarrollan funciones teniendo en cuenta metodología TDD y se da solución al reto #7
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… solución al reto
hyfi06
reviewed
May 15, 2020
| # Use math.pi for π value | ||
| pass | ||
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| area = round((math.pi * pow(radius, 2)),2) |
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Por qué reondeas a 2? esto depende del caso de uso. Si tu instrumento de medición, normalmente se toda la mitad de la mínima escala. Por ejemplo una regla que mide milímetros, tendrías que tomar 0.0005 m como incertidumbre. Manejar más decimales no tiene sentido.
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Usando TDD? ver para creer, hubieras hecho dos commits ;) |
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