What is the slope of y = –1 + 3r?

Mathematics

1 answer:

Pavlova-9 [17]3 years ago

5 0

Answer:

the slope to the problem y = –1 + 3r is M ≈ 3

Step-by-step explanation:

You might be interested in

Your classmate is unsure about how to use side lengths

DENIUS [597]

Answer:

you look at the lengths some triagles have small parallel lines anè or two of them to determine if they are equal or different from one another

3 0

3 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

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