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Varvara68 [4.7K]
3 years ago
15

What’s 33345 times 234654?

Mathematics
1 answer:
Rudiy273 years ago
5 0

Answer: 7824537630

Step-by-step explanation:

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Answer:

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

Step-by-step explanation:

The complete statement is:

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

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.

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Step-by-step explanation:

In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree, a generalised version of the familiar arithmetic technique called long division. ... The result R = 0 occurs if and only if the polynomial A has B as a factor.

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