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Here's the question.. I have no idea on what to do!
Do you need help?
I kind of do..

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Other answers:

|dw: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
I don't get what youre doing :/
im 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
then 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
|dw:1353507706506:dw| but it loks like the area im looking for is 1/4 the area of the bigger triangle
and 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
Area = 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
how many cubed meters are in a liter?
  • phi
Imagine all the rain fell at once |dw:1353508016540:dw|
the rest of it seems pretty basic, except for the last one
phi, should we assume its asking for total roof, or just the side thats showing?
  • phi
good 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 mind-reader for people that make up these questions
indeed, 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
sounds reasonable. 1/2 the roof it is!
Im confused..
|dw: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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