Algebra II - Rational Applications?

Question

Can someone help me understand these problems? I don't get how to do them at all.

Taylor and Michael are in their boat headed upstream. They travel a distance of 20 miles to the nearest beach. The current is going 4 mph. The trip to the beach and back downstream takes 8 hours. What is the speed of the boat?

In order to feed her baby, a bird needs to get a worm that is fifteen meters away. The bird canfly at a rate of 3 meters per second through stll air. However, there is a wind blowing. So, the bird must fly into the wind to get the worm, and with the wind to return to her nest. The entire trip takes 18 seconds. What is the speed of the wind?

I understand that I'm supposed to use the formula D=RT for the first one, and work rate x time = work done for the second, but I don't understand how to apply them. Can someone help me? You don't have to give me the answer, but explaining how would be greatly appreciated. My algebra teacher is horrible...

Answer 1

Let's call the speed of the boat v
On the way up the stream the speed of the travelers is
v-4. the current works against the boat. It takes 20/(v-4) hrs

on the way downstream the speed is v+4. The current helps the boat It takes 20/(v+4) hrs

the total time is 20/(v+4)+20/(v-4)=8
solve for v

The second is the same as the first.

The resultant speed as the bird goes against
the wind is 3-v m/s
The resultant speed as the bird goes with the wind
is 3+v m/s

we have distance is 15 m and round-trip time is 18 s

15/(3+v)+15/(3-v)=18 solve for v

Answer 2

you ought to harm up the 80^(a million/4) into sixteen^(a million/4) and 5^(a million/4). This will become sixteen^(a million/4)x5^(a million/4) over 5^(-a million/4). this is important understand that sixteen^(a million/4)=2 and 5^(a million/4) over 5^(-a million/4) will become 5^((a million/4)+(a million/4)) it particularly is 5^(a million/2) and for this reason you get 2 x 5^(a million/2).