Bob bikes to school every day at a steady rate of x miles per hour. On a particular day, Bob had a flat tire exactly halfway to school. He immediately started walking to school at a steady pace of y miles per hour. He arrived at school exactly t hours after leaving his home. How many miles is it from the school to Bob's home? A) (x + y) / t B) 2(x + t) / xy C) 2xyt / (x + y) D) 2(x + y + t) / xy E) x(y + t) + y(x + t)

Answer: [tex]C) \frac{2xyt}{(x+y)}[/tex]Step-by-step explanation:We start with the total time t and we define it like this [tex]t=t_{1} +t_{2}[/tex]being [tex]t_{1}[/tex] the time he was on x speed and [tex]t_{2}[/tex] the time he was on y speed Now for the distance we have the velocity equation [tex]Velocity=\frac{distance}{time}[/tex] and in this excercise we would have the two equations [tex]x=\frac{d/2}{t_{1}}[/tex][tex]y=\frac{d/2}{t_{2}}[/tex]then [tex]t_{1}=\frac{d/2}{x}[/tex][tex]t_{2}=\frac{d/2}{y}[/tex]Next[tex]t=\frac{d/2}{x}+\frac{d/2}{y}[/tex]this we simplify to get [tex]d= \frac{2xyt}{(x+y)}[/tex]