Evelyn and Meredith were paddling at a rate of [tex]\frac{3+\sqrt{493}}{11}mph[/tex] which is equivalent to 2.29mph approximately.
Step 2: See attached picture
Step 3: In this case we only have different variables, there is the rate at which Evelyn and Meredith were paddling, the rate at which they moved when rowing upstream, the rate at which they moved when rowing downstream, the time it took them to paddle upstream and the time it took them to paddle downstream.
v = rate at which Evelyn and Meredith were paddling.
[tex]v_{up}[/tex]= velocity at which they were moving when paddling upstream.
[tex]v_{down}[/tex]= velocity at which they were moving when paddling downstreamstream.
[tex]t_{up}[/tex]= time it took them paddling upstream.
[tex]t_{down}[/tex]= time it took them paddling downstream
Se attached picture for the table.
Step 4: Building this equation will require us to combine different equations into a single one. Let's start with the equation for the final rate at which they paddled when rowing upstream.
[tex]v-2=v_{up}[/tex]
When rowing upstream, the current will drag the kayak, so we subtract it from the rate at which they were rowing.
Let's find the final rat at which they moved when rowing downstream.
[tex]v+2=v{down}[/tex]
next, the problem tells us it took them 3 hours and 40 minutes to row up and down the channel so we can convert it into just hours like this:
[tex]40min*\frac{1hr}{60min}=\frac{2}{3}hr[/tex]
[tex]3hr+\frac{2}{3}hr=\frac{11}{3}hr[/tex]
so now we can build our equation for time.
[tex]t_{up}+t_{down}=\frac{11}{3}[/tex]
We also know that the rate is built by dividing the distance over the time it took them to travel the distance, so:
[tex]v_{up}=\frac{1}{t_{up}}[/tex]
[tex]v_{down}=\frac{1}{t_{down}}[/tex]
If we solved each of those equations for their respective times, we would end up with the following:
[tex]t_{up}=\frac{1}{v_{up}}[/tex]
[tex]t_{down}=\frac{1}{v_{down}}[/tex]
so we can now combine all the equations together so we get:
[tex]t_{up}+t_{down}=\frac{11}{3}[/tex]
[tex]\frac{1}{v_{up}}+\frac{1}{v_{down}}=\frac{11}{3}[/tex]
[tex]\frac{1}{v-2}+\frac{1}{v+2}=\frac{11}{3}[/tex]
So this equation models the problem.
Step 5: We can solve this equation by multiplying everything by the LCD
In this case the LCD is:
3(v-2)(v+2)
so, when doing the respective multiplications we end up with:
[tex]\frac{3(v-2)(v+2)}{v-2}+\frac{3(v-2)(v+2)}{v+2}=\frac{11(3)(v-2)(v+2)}{3}[/tex]
We can now simplify to get:
3(v+2)+3(v-2)=11(v+2)(v-2)
We can now do the respective multiplications to get:
[tex]3v+6+3v-6=11(v^{2}-4)[/tex]
and we can further simplify:
[tex]6v=11v^{2}-44[/tex]
Step 5: So we can now solve it by using the quadratic formula, first, we need to rewrite the equation in standard form:
[tex]11v^{2}-6v-44=0[/tex]
So we can now use the quadratic formula:
[tex]v=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex]
we substitute:
[tex]v=\frac{-(-6) \pm \sqrt{(-6)^{2}-4(11)(-44)}}{2(11)}[/tex]
and simplify:
[tex]v=\frac{6 \pm \sqrt{36+1936}}{22}[/tex]
[tex]v=\frac{6 \pm \sqrt{1972}}{22}[/tex]
[tex]v=\frac{6 \pm \sqrt{4(493)}}{22}[/tex]
[tex]v=\frac{6 \pm 2\sqrt{493}}{22}[/tex]
[tex]v=\frac{3 \pm \sqrt{493}}{11}[/tex]
this gives us two possible results:
[tex]v=\frac{3 + \sqrt{493}}{11}[/tex] and [tex]v=\frac{3 - \sqrt{493}}{11}[/tex]
Step 6: We only pick the first result since it's the positive result. We don't take the second one because a negative result represents the kayak moving in the opposite direction which is not how the situation was modeled.
For further information, take a look at the following link:
https://brainly.com/question/12919292?referrer=searchResults
