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A medical battery loses 5% of its power each week. How long will it take until only 20% of
its power remains.

Answer if you only know

Respuesta :

[tex]\star[/tex] [tex]20=100(0.95)^t[/tex] 

[tex]\star[/tex] [tex]0.2=0.95^t[/tex] 

[tex]\star[/tex] [tex]Log \ 0.2 = t \ log.95[/tex] 

[tex]\star[/tex] [tex] \dfrac{log0.2}{log.95} = t[/tex] 

[tex]\star[/tex] [tex]31.4 \ weeks = t[/tex] 

[tex]\star[/tex] [tex]Hence \ solved [/tex]

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