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I am Lyosha [343]
2 years ago
10

*math please help* Simplify.

Mathematics
2 answers:
Ghella [55]2 years ago
3 0

Given : \frac{\frac{4x}{5 + x}}{\frac{6x}{x + 2}}

\implies \frac{4x}{5 + x}\times\frac{x + 2}{6x}

\implies \frac{2}{x + 5}\times\frac{x + 2}{3}

\implies \frac{2(x + 2)}{3(x + 5)}

\implies \frac{2x + 4}{3x + 15}

amm18122 years ago
3 0
\frac{ \frac{4x}{5 + x} }{ \frac{6x}{x + 2} } \\ = \frac{4x}{5 + x } ( \frac{x + 2}{6x} )
= \frac{4x^{2} + 8x}{30x + {6x}^{2} } \\ = \frac{2x(2 {x} + 4) }{2x(3 {x}+ 15)} \\ = \frac{2 {x} + 4 }{3 {x} + 15}
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There is a typo error, the perimeter of equilateral triangle ABC is 81/√3 centimeters.

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A diagram is attached for reference.

Step-by-step explanation:

Given,

The perimeter of equilateral triangle ABC is 81/√3 centimeters.

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tan\theta =\frac{perpendicular}{base} \\tan\theta=\frac{OE}{BE} \\tan\theta=\frac{OE}{\frac{27\sqrt{3}}{2}  } \\tan30\times {\frac{27\sqrt{3} }{2} }= OE\\\frac{1}{\sqrt{3} } \times\frac{27\sqrt{3} }{2} =OE\\

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Now for radius,

We consider ΔOBE

cos\theta=\frac{base}{hypotenuse} \\cos30= \frac{BE}{OB} \\Cos30 = \frac{\frac{27\sqrt{3} }{2}}{OB}  \\OB= \frac{\frac{27\sqrt{3} }{2}}{cos30} \\OB= \frac{\frac{27\sqrt{3} }{2}}{\frac{\sqrt{3} }{2} } \\OB =27 \ cm

Thus for

Perimeter of equilateral triangle ABC is 81/√3 centimeters,

The radius of equilateral triangle ABC is 27 cm

The apothem of equilateral triangle ABC is 13.5 cm

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