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3 years ago

6

Use the quadratic formula to solve the equation.

1 answer:

allochka39001 [22]3 years ago

7 0

=1.3+0.1sqrt29, 1.3-0.1sqrt29

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When dividing two different Integers what would their sign be?

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Answer:

Negative

Step-by-step explanation:

When you divide two integers with the same sign, the result is always positive. Just divide the absolute values and make the answer positive. When you divide two integers with different signs, the result is always negative. Just divide the absolute values and make the answer negative.

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The function f(x) = 800(1.09)3x shows the growth due to interest that Jim earns on an investment.

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3 years ago

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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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3 years ago

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