SOLUTION: im having trouble with two questios, i just dont understand them. i know i have to use the formula for distance or radius (r = xsqaured + y sqaured) but i dont know how to apply it

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Question 162570This question is from textbook
: im having trouble with two questios, i just dont understand them. i know i have to use the formula for distance or radius (r = xsqaured + y sqaured) but i dont know how to apply it. the questions are
A rock dropped into a pond sends out a circular ripple whose radius increases steadily at 6cm/s. A toy boat is floating on the pond 2 m east and 1 m north of the spot where the rock is dropped. How long does it take for the rippe to reach the boat
the second question is
A triangle has vertices at s (6,6), t(-6,12), and u(0,-12). SM is the median from the vertex S. Find the coordinates of the point that is two-thirds of the way frm S to M.
I dont know how to apply toe distance formuls, radius or mid point formula. I really need your help, i have a test tomorrow.
Thank you
- Neha =)
This question is from textbook

Answer by Fombitz(13823) About Me  (Show Source):
You can put this solution on YOUR website!
2 meters east, 1 meter north.
The distance (radius) from the center of rock drop to toy boat is,
D=sqrt%28x%5E2%2By%5E2%29
D=sqrt%282%5E2%2B1%5E2%29=sqrt%284%2B1%29=sqrt%285%29
D=2.23606
So the center of the rock drop is 2.23606 meters away from the toy boat.
The ripple is moving towards the boat at 6 cm/s.
This is now a rate x time=distance problem.
You know the rate, the speed of the ripple.
You calculated the distance.
Now calculate the time.
First, the units have to match (can't mix cm and m).
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Change the distance to cm.
2.23606 meters= 223.6cm
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R%2AT=D
T=D%2FR
T=223.6%2F6=37.27
It will take the ripple 37.27 seconds to reach the boat.
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drawing%28+300%2C+300%2C+-12%2C+12%2C+-15%2C+15%2Cgrid%28+1+%29%2Ccircle%28+0%2C+-12%2C+.2+%29%2C%0D%0Alocate%286%2C6%2CS%29%2C%0D%0Alocate%28-7%2C12%2CT%29%2C%0D%0Alocate%281%2C-10%2CU%29%2C%0D%0Acircle%28+6%2C+6%2C+.2+%29%2C%0D%0Acircle%28+-6%2C+12%2C+.2+%29%2C%0D%0Agreen%28line%280%2C-12%2C6%2C6%29%29%2C%0D%0Agreen%28line%286%2C6%2C-6%2C12%29%29%2C%0D%0Agreen%28line%28+-6%2C+12%2C+0%2C+-12%29%29%29
M is the midpoint of the line from T to U.
x%5BM%5D=%28-6%2B0%29%2F2=-3
y%5BM%5D=%2812%2B%28-12%29%29%2F2=0
The midpoint M has coordinates (-3,0).
The change in x from S to M is
DELTA%2Ax=%28-3-6%29=-9
The change in y from S to M is
DELTA%2Ay=%280-6%29=-6
Using only 2/3 of the change in x, you would have,
%282%2F3%29%28-9%29=-6
Using only 2/3 of the change in y, you would have,
%282%2F3%29%28-6%29=-4
My starting point is S at (6,6).
The point P is 2/3 of the way from S to M.
The coordinates of P are found by starting at S and adding the 2/3 change in x and y.
(6,6)+(-6,-4)=(0,2)
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P is located at (0,2)
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drawing%28+300%2C+300%2C+-12%2C+12%2C+-15%2C+15%2Cgrid%28+1+%29%2Ccircle%28+0%2C+-12%2C+.2+%29%2C%0D%0Alocate%286%2C6%2CS%29%2C%0D%0Alocate%281%2C2.5%2CP%29%2C%0D%0Alocate%28-7%2C12%2CT%29%2C%0D%0Alocate%28-3%2C2%2CM%29%2C%0D%0Alocate%281%2C-10%2CU%29%2C%0D%0Acircle%28+6%2C+6%2C+.2+%29%2C%0D%0Acircle%28+-6%2C+12%2C+.2+%29%2C%0D%0Acircle%28-3%2C0%2C.4%29%2C%0D%0Acircle%280%2C2%2C.4%29%2C%0D%0Acircle%28-3%2C0%2C.2%29%2C%0D%0Acircle%280%2C2%2C.2%29%2C%0D%0Ablue%28line%286%2C6%2C-3%2C0%29%29%2C%0D%0Agreen%28line%280%2C-12%2C6%2C6%29%29%2C%0D%0Agreen%28line%286%2C6%2C-6%2C12%29%29%2C%0D%0Agreen%28line%28+-6%2C+12%2C+0%2C+-12%29%29%29