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Sagot :
a) Comenzamos con D=0.
La velocidad por los primeros 20 minutos es 13 millas por hora.
Entonces la distancia recorrida al minuto 20 (t=20/60=1/3) es:
[tex]D(20\min )=0+13\cdot\frac{20}{60}=\frac{13}{3}\approx4.3\ldots[/tex]b) la velocidad es de 15 millas por hora durante los siguientes 5 minutos. Entonces la distancia recorrida es:
[tex]D(5\min )=15\cdot\frac{5}{60}=\frac{75}{60}=1.25[/tex]Luego, la velocidad es de 12 millas por hora durante los últimos 15 minutos. Entonces la distancia recorrida total es:
[tex]D(20\min )=1.25+12\cdot\frac{15}{60}=1.25+3=4.25[/tex]c) la velocidad es M millas por hora durante los primeros 5 minutos y luego a N millas por hora durante los siguientes 15 minutos.
Podemos calcular la distancia recorrida en los 20 minutos como:
[tex]D(20\min )=M\cdot\frac{5}{60}+N\cdot\frac{15}{60}=\frac{M}{12}+\frac{N}{4}[/tex]
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