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En

  • one year ago

WILL MEDAL AND FAN!!! Express the surface area of a cube as a function of the length of a diagonal of one face.

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  1. anonymous
    • one year ago
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    |dw:1444471441259:dw|

  2. anonymous
    • one year ago
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    now we can use Pythagorus theorem one one face (Shown above)

  3. anonymous
    • one year ago
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    |dw:1444471558016:dw|

  4. anonymous
    • one year ago
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    i have to leave sorry, perhaps @imqwerty can help?

  5. imqwerty
    • one year ago
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    ok :)

  6. En
    • one year ago
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    My book says that the answer is 3d^2. I dont get it :/

  7. En
    • one year ago
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    I know that the SA=6s^2 and the diagonal as √3 s.. idk how to go from there :/

  8. imqwerty
    • one year ago
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    hah XD this is not correct wait a sec i read it wrong

  9. En
    • one year ago
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    I keep solving it i only get 2d^2. The answer from my book is 3d^2. This is heartbreaking

  10. imqwerty
    • one year ago
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    |dw:1444473276715:dw|by pythagorus theorem \[a^2+a^2=d^2\]\[2a^2=d^2\]\[a=\frac{ d }{ \sqrt{2} }\]surface area is given by-\[Area=6a^2\]putting the value of a we got-\[Area=6 \times \left( \frac{ d }{ \sqrt{2} } \right)^2\]\[Area=6 \times \frac{ d^2 }{ 2}\] \[Area=3d^2\]

  11. anonymous
    • one year ago
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    sorry, i'm back. ye well done @imqwerty

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