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Mathematics

1 answer:

rosijanka [135]3 years ago

3 0

The two equations are equal, so "The graphs of the equations are parallel lines." is false.

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A function g is defined by g:x→3−2sinx,for 0◦ ≤x≤A◦,where A is a constant.

Crank

<span>hmmm: g maps x onto 3-2sin(x) for all x from 0 to A degrees g(x) = 3-2sin(x) the inverse would have to be arcsin (3-x)/2, which only has a radian output between -pi and pi i believe. but this is just from memory</span>

6 0

4 years ago

39-50 find the limit.<br> 41. <img src="https://tex.z-dn.net/?f=%5Clim%20_%7Bt%20%5Crightarrow%200%7D%20%5Cfrac%7B%5Ctan%206%20t

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

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