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Blababa [14]
3 years ago
7

A rectangular box is constructed in 3-space with one corner at the origin and other vertices at (1, 0, 0), (0, 4, 0), (0, 0, 2).

Find the length of the diagonal of the box.
Mathematics
1 answer:
77julia77 [94]3 years ago
7 0

The space diagonal will have length ...

... d = √(1² +4² +2²) = √(1 +16 +4) = √21

_____

This can be found using the Pythagorean theorem. A drawing can help. Find the length of any face diagonal, then use that length as the leg of a right triangle whose hypotenuse is the space diagonal and whose other leg is the edge length not used in the first calculation.

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Step-by-step explanation:

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<em>Hope this helped!</em>

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Answer:

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Step-by-step explanation:

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18q^{2} - 45q + 25 = 18q^{2} - 30q - 15q + 25

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and

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So that,

\frac{18q^{2}-45q+25 }{9q^{2}-25 } = \frac{(3q-5)(6q-5)}{(3q-5)(3q+5)}

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Therefore,

\frac{18q^{2}-45q+25 }{9q^{2}-25 } = \frac{(6q-5)}{(3q+5)}

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