Respuesta :
recall your d =rt, or distance = rate * time
so... when the boat is going upstream, is going against the current, hmmmm let's say the boat has a speed rate of "r", we know the current has a rate of 2, so... .if the boat is going against the current, its actual speed is not really "r", but "r - 2", since the current is taking away speed from it
now...going downstream, is going with the current, is not really going "r" fast either, is going "r + 2"
let us notice, the time for upstream travel, is really the same time for the downtream travel, let's say hmm it took "t" long
thus [tex]\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ upstream&15&r-2&t\\ downstream&35&r+2&t \end{array}\\\\ -----------------------------\\\\ \begin{cases} 15=(r-2)t\implies \cfrac{15}{r-2}=\boxed{t}\\\\ 35=(r+2)t\\ ----------\\ 35=(r+2)\left( \boxed{\frac{15}{r-2}} \right) \end{cases}[/tex]
solve for "r"
so... when the boat is going upstream, is going against the current, hmmmm let's say the boat has a speed rate of "r", we know the current has a rate of 2, so... .if the boat is going against the current, its actual speed is not really "r", but "r - 2", since the current is taking away speed from it
now...going downstream, is going with the current, is not really going "r" fast either, is going "r + 2"
let us notice, the time for upstream travel, is really the same time for the downtream travel, let's say hmm it took "t" long
thus [tex]\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ upstream&15&r-2&t\\ downstream&35&r+2&t \end{array}\\\\ -----------------------------\\\\ \begin{cases} 15=(r-2)t\implies \cfrac{15}{r-2}=\boxed{t}\\\\ 35=(r+2)t\\ ----------\\ 35=(r+2)\left( \boxed{\frac{15}{r-2}} \right) \end{cases}[/tex]
solve for "r"
