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anonymous
 4 years ago
Could someone help me with the following question?
anonymous
 4 years ago
Could someone help me with the following question?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here's the question.. I have no idea on what to do!

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353507196703:dw i might be thinking this to hard, but id say we would need to define the volume of water that can be collected in a year

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353507466106:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't get what youre doing :/

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0im trying to determine the volume of water that the roof collects by finding the area along the edge of the roof that would collect 50cm of water a year

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0then using similar triangles i can add up all the areas along the roof line to cover 7 meters and then multiply it by 12 to get the total volume on one side of the roof

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353507706506:dw but it loks like the area im looking for is 1/4 the area of the bigger triangle

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0and 4 of those rake down the roof, which means the area of the big triangle equals the area of the rake .... area of a triangle is base times height divided by 2

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0Area = sqrt(48)/2 times 12 is: 6 sqrt(48) 6 sqrt(4.4.3) 24 sqrt(3) is the total volume of water that one side of the roof collects during the year

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0how many cubed meters are in a liter?

phi
 4 years ago
Best ResponseYou've already chosen the best response.0Imagine all the rain fell at once dw:1353508016540:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0the rest of it seems pretty basic, except for the last one

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0phi, should we assume its asking for total roof, or just the side thats showing?

phi
 4 years ago
Best ResponseYou've already chosen the best response.0good question. I initially focused on 1/2 the roof (due to the picture), but the roof is both sides... I hate trying to be a mindreader for people that make up these questions

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0indeed, but then i got thinking (since a gable end) that the only rain that they barrel catches is the part that falls down the one side ....

phi
 4 years ago
Best ResponseYou've already chosen the best response.0sounds reasonable. 1/2 the roof it is!

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353508980599:dw \[V=12\int_{0}^{\sqrt{48}}{(1.5\frac{x}{\sqrt{48}})(1\frac{x}{\sqrt{48}})}~dx\] \[12\int_{0}^{\sqrt{48}}{.5}~dx=12(.5\sqrt{48})\] \[V=6\sqrt{48}\] lol :)
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