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AlekseyPX
3 years ago
7

Which algebraic expression has a term with a coefficient of 3?​

Mathematics
1 answer:
kirill115 [55]3 years ago
8 0

Answer:

180

Step-by-step explanation:

in Geometry if you want a farewell you need to know the parallel lines they never touch they're always the same distance apart it's as if they don't like each other much and there's only one thing that comes between the line that passes right through each of them called the transversal line it has several angle properties just like these: vertically opposite angles are equal corresponding angles are equal alternate interior angles are equal co-interior angles equal 180* vertically opposite angles are equal corresponding angles are equal alternate interior angles are equal co-interior angles equal 180*

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Katyanochek1 [597]

Write tan in terms of sin and cos.

\displaystyle \lim_{t\to0}\frac{\tan(6t)}{\sin(2t)} = \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)}

Recall that

\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1

Rewrite and expand the given limand as the product

\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)

Then using the known limit above, it follows that

\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}

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