Question 818363: The depth of a lake can be calculated by the function y=.2x^2-10x+6 . Where x represents the distance from the Western edge.
b) What is the diameter of the lake? Round to the nearest thousandth.
Found 2 solutions by stanbon, TimothyLamb:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The depth of a lake can be calculated by the function y=.2x^2-10x+6 .
Where x represents the distance from the Western edge.
b) What is the diameter of the lake? Round to the nearest thousandth.
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Comment:: That quadratic has minimum depth value but no maximum depth.
The maximum depth should be at the middle of the lake and the radius
of the lake could be determined. Did you post the quadratic correctly?
Cheers,
Stan H.
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Answer by TimothyLamb(4379)  (Show Source):
You can put this solution on YOUR website! y = 0.2x^2 - 10x + 6
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the above quadratic equation is in standard form, with a=0.2, b=-10, and c=6
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to solve the quadratic equation, by using the quadratic formula, plug this:
0.2 -10 6
into this: https://sooeet.com/math/quadratic-equation-solver.php
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the two real roots (i.e. the two x-intercepts), of the quadratic are:
x = 49.3926218
x = 0.607378165
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What is the diameter of the lake? Round to the nearest thousandth:
49.3926218 - 0.607378165 = 48.785243
answer:
48.785
NOTE: the length units of variable x were not given in the problem statement, could be foot, meter, mile, km, etc.)
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