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2 years ago

5

Which equations represent linear functions? Select all that apply.

1 answer:

liraira [26]2 years ago

4 0

Answer: x-3y=3x

8y=6x8

y=6x

Step-by-step explanation:

<h2>Meaning of linear equation</h2><h3>A linear equation is a equation that goes at a constant pace and won't change overtime, so functions like x squared and functions that multiply x and y together doesn't go at a constant pace, there not linear equations, but functions like y=mx, M representing any constant, that is a linear equation, therefore x-3y=3x</h3>

8y=6x8

y=6x

<h3>Are the answers to your Question,</h3>

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Rectangle GHIJ inscribed in a circle, GK⊥JH, GK = 6cm, m∠GHJ = 15°. Find the radius of the circle. pls help !!

Mariulka [41]

Answer:

6(2 + √3)

Step-by-step explanation:

Given : Rectangle GHIJ inscribed in a circle. <em>GK⊥JH</em>, <em>GK</em> = 6 cm and <em>m∠GHJ</em>=15°.

To find: Radius<em>(KH)</em> =?

Sol: As given in figure 1, Since <em>GK⊥JH ∴ m∠GKH = 90° . Let GK = x cm.</em>

Now, In ΔGKH,

$tan\Theta =\frac{perpendicular}{base}$

$tan\Theta =\frac{GK}{KH}$

$tan 15^{\circ} =\frac{6}{x}$

$2-\sqrt{3} =\frac{6}{x}$ (∵<em> tan 15°</em> = 2 - √3)

$x = \frac{6}{2-\sqrt{3} }$

On rationalizing the above expression,

$x = \frac{6}{2-\sqrt{3} } \times \frac{2+\sqrt{3} }{2+\sqrt{3} } =\frac{6(2+\sqrt{3} )}{4-3}$

Therefore, radius of the circle <em>(KH)</em> = 6 (2+√3)

This is how to find the value of <em>tan 15°</em>

<em>tan</em> 15° = <em>tan </em>(45° -30°)

Now using,

$tan(A-B) = \frac{tanA-tanB}{1+tanA tanB}$

$tan(45^{\circ} - 30^{\circ}) = \frac{tan 45^{\circ}-tan30^{\circ}}{1+tan 45 tan30}$

$tan(45^{\circ} - 30^{\circ}) = \frac{1-\frac{1}{\sqrt{3}}}{1+1\times \frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

On rationalizing,

$tan 15^{\circ} = \frac{\left (\sqrt{3}-1 \right )^{2}}{3-1} =\frac{4-2\sqrt{3}}{2}$

Taking 2 common from numerator,

<em>tan</em> 15° = 2 - √3

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