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Hatshy [7]
3 years ago
12

Dean Baldus enjoys scuba diving. He dives to 34 ft below the surface of a lake. His partner Jeff​ Balius, dives to 39 ft below t

he​ surface, but then ascends 18 ft. Find the vertical distance between Dean and Jeff.
Mathematics
1 answer:
saw5 [17]3 years ago
5 0
The vertical distance between Dean and Jeff is 13 ft below the surface

Explanation:

Jeff dove 39 ft below the surface and then ascends 18 ft

-39 - 18 = 21

Jeff is 21 ft below the surface

------------------

Dean is 34 ft below the surface

-34 - (-21) = -13 ft

-----------------

So, Dean and Jeff are 13 feet away from each other (vertical distance).
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Step-by-step explanation:

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\mathrm{For\:convenience,\:write\:}\frac{d}{dx}\left(y\right)\mathrm{\:as\:}y^{'\:}\\6x-6\left(y+xy^{'\:}\right)+2yy^{'\:}=2-y^{'\:}\\\mathrm{Isolate}\:y^{'\:}:\quad y^{'\:}=\frac{2-6x+6y}{-6x+2y+1}\\y^{'\:}=\frac{2-6x+6y}{-6x+2y+1}\\\mathrm{Write}\:y^{'\:}\:\mathrm{as}\:\frac{d}{dx}\left(y\right)\\\frac{d}{dx}\left(y\right)=\frac{2-6x+6y}{-6x+2y+1}

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