Answer:
No, They will do it in more than an hour
t=1,1 h
Step-by-step explanation:
Vh=1mile/12min, Vt=1mile/15min They will meet to same time th=tt
sh=x, st=10-x, so V=s/t we have Vh=sh/t ; Vt=st/t ; t=(sh/Vh) = (st/Vt);
[tex]\frac{x}{\frac{1}{12} } = \frac{10-x}{\frac{1}{15} }\\\frac{x}{15}=\frac{10-x}{12}\\ 12x=15(10-x)[/tex]; so [tex]12x+15x=150; 27x=150; x=\frac{150}{27}=\frac{150}{9} mile[/tex]: if Vh=sh/t then [tex]t=\frac{\frac{50}{9}mile }{\frac{1mile}{12min} } =\frac{600}{9}min*\frac{h}{60min}=1,1 h[/tex]
Note: V=s/t : but in this case are 12 min (t) per mile (s), so [tex]V=\frac{mile}{12min}[/tex]
