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Lunna [17]
3 years ago
14

mula1" title="\sqrt{ x^{2}-10x+25}+25+12\sqrt{x} =15\sqrt{x}" alt="\sqrt{ x^{2}-10x+25}+25+12\sqrt{x} =15\sqrt{x}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
vodomira [7]3 years ago
8 0

Step-by-step explanation:

\sqrt{ x^{2}-10x+25}+25+12\sqrt{x} =15\sqrt{x} \\  \\  \therefore \: \sqrt{ x^{2}-10x+ {5}^{2} }+25 =15\sqrt{x} - 12\sqrt{x}\\  \\  \therefore \: \sqrt{( x - 5)^{2}}+25 =3\sqrt{x}  \\  \\ \therefore \:  x - 5+ 25 = 3 \sqrt{x}  \\  \\ \therefore \: x + 20 = 3 \sqrt{x}  \\  \\ squaring \: both \: sides \\ (x + 20)^{2}  = ( {3 \sqrt{x} })^{2}  \\ \therefore \:  {x}^{2}  + 40x + 400 = 9x \\ \therefore \:  {x}^{2}  + 40x + 400  - 9x = 0 \\  \therefore \:  {x}^{2}  + 31x + 400  = 0 \\

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cos 2Ф = - 161/289 , tan 2Ф = - 240/161

Step-by-step explanation:

* Lets explain how to solve the problem

∵ cos Ф = - 8/17

∵ Ф lies in the 3rd quadrant

- In the 3rd quadrant sin and cos are negative values, but tan is

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