it is either -2, 1, 4, 5 or if it is absolute value then it is 1, -2, 4, 5 . let me know if that helps.
For this case, we have to:
By definition, we know:
The domain of ![f (x) = \sqrt [3] {x}](https://tex.z-dn.net/?f=f%20%28x%29%20%3D%20%5Csqrt%20%5B3%5D%20%7Bx%7D)
is given by all real numbers.
Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.
So, we have:
![y = \sqrt [3] {x-2}](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%20%5B3%5D%20%7Bx-2%7D)
with
: ![y = \sqrt [3] {- 2}](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%20%5B3%5D%20%7B-%202%7D)
is defined.
![y = \sqrt [3] {x+2}](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%20%5B3%5D%20%7Bx%2B2%7D)
with ![x = 0:\ y = \sqrt [3] {2}](https://tex.z-dn.net/?f=x%20%3D%200%3A%5C%20y%20%3D%20%5Csqrt%20%5B3%5D%20%7B2%7D)
is also defined.
has a domain from 0 to ∞.
Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.
Thus, we observe that:
is not defined, the term inside the root is negative when.
While 
if it is defined for .
Answer:
Option b
Answer:
Approximately 51 Yards
Step-by-step explanation:
You can solve this by cutting the square in half to make a right triangle. Than, you can use the Pythagorean theorem to solve for the hypotenuse.
a^2+b^2=c^2
36^2+36^2=c^2
1296+1296=c^2
2592=c^2
50.9116
Round to the nearest whole number since the question says approximately: 51
<span>7√7 - 2√28
2 sqrt(28) = 4 * sqrt(7)
7*sqrt(7) - 4*sqrt(7) =
3 * sqrt(7)
</span>
A. C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.
B. P(13,10)= 13! =13! =13·12·11·10·9·8·7·6·5·4.
(13−10)! 3!
C. f there is exactly one woman chosen, this is possible in C(10, 9)C(3, 1) =
10! 3!
9!1! 1!2!
10! 3!
8!2! 2!1!
10! 3!
7!3! 3!0!
= 10 · 3 = 30 ways; two women chosen — in C(10,8)C(3,2) =
= 45·3 = 135 ways; three women chosen — in C(10, 7)C(3, 3) =
= 10·9·8 ·1 = 120 ways. Altogether there are 30+135+120 = 285
1·2·3
<span>possible choices.</span><span>
</span>
