The _____is the set of dependent or y-values of a function. A. domain B. range

Mathematics

1 answer:

irinina [24]3 years ago

4 0

Answer: OPTION B

Step-by-step explanation:

First it is importanty definition, a relation is a function if and only if each input value has one and only one output value.

The input values are the x-values and the output values are the y-values.

Therefore, "y" is the Dependend variable and "x" is the Independent variable.

Knowing the above, you can understand the definition of "Domain" and "Range".

The Domain is defined as the set of all the possible values of "x" or the Independent variable

The Range is defined as the set of all the possible values of "y" or the Dependent variable.

Therefore, you can conclude that the correct answer is the given in the Option B.

You might be interested in

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

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