<h2>
Explanation:</h2><h2>
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The diagram is missing but I'll assume that the arc BDC is:

And another arc, let's call it FGH. measures:

If those arc are equal, then this equation is true:

Substituting k into the first equation:

<span>erry starts rowing at 2 pm from the west end of the lake, and Tao starts rowing from the east end of the lake at 2:05 pm.
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Terry DATA:
rate = 60 meter/min ; time = x min ; distance = r*t = 60x meters
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Tao DATA:
rate = 40 meter/min ; time = x-5 min ; distance = r*t = 40(x-5) meters
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If they always row directly towards each other at what time will the two meet?
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Equation:
dist + dist = 2000 meters
60x + 40(x-5) = 2000
20x = 1800
x = 90 minutes
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To find the area of a building.
and to make a small farm