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2 years ago

Please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

tresset_1 [31]2 years ago

4 0

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HELP ME please!!!???

Firlakuza [10]

Answer:

sohcahtoa

Step-by-step explanation:

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3 years ago

Please help :c the first word is find

xeze [42]

Answer:

The values of $r_{2}$ and $\alpha_{2}$ are 2 and 150º.

Step-by-step explanation:

The complete statement is:

Find $\alpha_{2}$ and $r_{2}$ such that $\sin \theta - \sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos (\theta - \alpha_{2})$ .

We proceed to use the following trigonometric identity:

$\cos (\theta - \alpha_{2}) = \cos \theta \cdot \cos \alpha_{2} +\sin \theta \cdot \sin \alpha_{2}$ (1)

$\sin \theta -\sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos \theta \cdot \cos \alpha_{2}+r_{2}\cdot \sin \theta \cdot \sin \alpha_{2}$

By direct comparison we derive these expressions:

$r_{2}\cdot \sin \alpha_{2} = 1$ (2)

$r_{2}\cdot \cos \alpha_{2} = -\sqrt{3}$ (3)

By dividing (2) by (3), we have the following formula:

$\tan \alpha_{2} = -\frac{1}{\sqrt{3}}$

$\tan \alpha_{2} = -\frac{\sqrt{3}}{3}$

The tangent function is negative at second and fourth quadrants. That is:

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

There are at least two solutions:

$\alpha_{2,1} = 150^{\circ}$ , $\alpha_{2,2} = 330^{\circ}$

And the value of $r_{2}$ :

$r_{2}^{2}\cdot \sin^{2}\alpha_{2} + r_{2}^{2}\cdot \cos^{2}\alpha_{2} = 4$

$r_{2}^{2} = 4$

$r_{2} = 2$

The values of $r_{2}$ and $\alpha_{2}$ are 2 and 150º.

5 0

3 years ago

A store sells a popular toy for $29 each. the manufacturergets 25% of the total revenue. After the store sells 500 toys, how muc

ahrayia [7]

(500 x 29) x .25 = 3,625

3 0

3 years ago

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Heres a question i have 2 tries please help

KATRIN_1 [288]

Would it not just be 3.25 by five feet or am i missing something?

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3 years ago

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Please help ASAP!!!

masha68 [24]

Answer:

rotation

Step-by-step explanation:

7 0

3 years ago

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