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3 years ago

Pleaseeee pleaseee help mee with precal...One root of 2x2 + 8x + c = 0 is r1 = 1 + √5. Find r2.

Mathematics

2 answers:

igomit [66]3 years ago

6 0

If a root is a+√b then another is a-√b because of quadatic formula

r2=1-√5

Alenkinab [10]3 years ago

5 0

R2=1-√5 is the correct answer hope this helped

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What coordinates on the unit circle are associated with the angle measure?

Varvara68 [4.7K]

The tangent of the given angles are the ratios <em>y</em> to the <em>x</em> coordinate of

the point of the terminal side on the unit circle.

The correct options are;

$\dfrac{34 \cdot \pi}{3} \Longleftrightarrow \underline{ \left(-\dfrac{1}{2}, \ -\dfrac{\sqrt{3} }{2} \right)}$

$-\dfrac{7 \cdot \pi}{4} \Longleftrightarrow \underline{ \left(\dfrac{\sqrt{2} }{2}, \ \dfrac{\sqrt{2} }{2} \right)}$

$210^{\circ} \Longleftrightarrow \underline{ \left(-\dfrac{\sqrt{3} }{2}, \ -\dfrac{1}{2} \right)}$

<h3>How to find the points on the unit circle</h3>

The tangent of an angle is given as follows;

$tan (\theta) = \mathbf{ \dfrac{Opposite}{Adjacent}} = \dfrac{\Delta y}{\Delta x}$

First angle

An angle given is; $\mathbf{\dfrac{34 \cdot \pi}{3}}$

Therefore;

$tan \left(\dfrac{34 \cdot \pi }{3} \right) = \mathbf{ \sqrt{3}}$

The above result can be obtained as follows;

$\sqrt{3} = \mathbf{ \dfrac{-\dfrac{\sqrt{3} }{2} }{-\dfrac{1}{2} }}$

Which is obtained when we have;

$\left( \Delta x, \, \Delta y\right) = \mathbf{\left(-\dfrac{1}{2}, \, -\dfrac{\sqrt{3} }{2} \right)}$

Therefore

The required coordinates is therefore;

$\dfrac{34 \cdot \pi}{3} \Longleftrightarrow \left(-\dfrac{1}{2} , \ -\dfrac{\sqrt{3} }{2} \right)$

Second angle

The angle, $\mathbf{-\dfrac{7 \cdot \pi}{4}}$ , gives; $tan \left(-\dfrac{7 \cdot \pi}{4} \right) = 1$

The above value can be obtained as follows;

$\mathbf{\dfrac{\dfrac{\sqrt{2} }{2} }{\dfrac{\sqrt{2} }{2} }} = 1$

Which gives;

$-\dfrac{7 \cdot \pi}{4} \Longleftrightarrow \left(\dfrac{\sqrt{2} }{2}, \, \dfrac{\sqrt{2} }{2} \right)$

Third angle

The angle 210° gives; tan(210°) = $\mathbf{\frac{1}{\sqrt{3} }}$ , which can be obtained as follows;

$\sqrt{ \dfrac{1}{3} } = \mathbf{\dfrac{-\dfrac{1}{2} }{-\dfrac{\sqrt{3} }{2} }}$

Therefore;

$210^{\circ} \Longleftrightarrow \left(-\dfrac{\sqrt{3} }{2}, \ -\dfrac{1}{2} \right)$

Learn more about the unit circle here:

brainly.com/question/1673530

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Lyrx [107]

Answer:

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