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a_sh-v [17]
3 years ago
6

A proportion question ​

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0

Answer:

because x,y,z are in continuous proportion

=> x/y = y/z

<=> xz = y² =>

\frac{xz}{y} =y\\\\=>\frac{x^{3}z^{3}  }{y^{3} }=y^{3}(1)

with (1), we have:

x^{2}y^{2}z^{2}(\frac{1}{x^{3} }+\frac{1}{y^{3} }+\frac{1}{z^{3} })\\\\=  (xyz)^{2}(\frac{1}{x^{3}  }  +\frac{y^{3} }{x^{3}z^{3} }+\frac{1}{z^{3} })\\\\=(xyz)^{2} (\frac{x^{3}+y^{3}+z^{3}   }{x^{3}z^{3}}  )\\\\=y^{2}.\frac{x^{3}+y^{3}+z^{3}  }{xz}  \\\\= xz.\frac{x^{3}+y^{3}+z^{3} }{xz} \\\\=x^{3}+y^{3}+z^{3}

Step-by-step explanation:

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Answer:

Step-by-step explanation:

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alexandr402 [8]

Answer:

when (-1, -4) is substituted into the first equation it's true

when (-1, -4) is substituted into the second equation it's true

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Step-by-step explanation:

original equation

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Answer:

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