Let f(x)= sqrt 6x and g(x)=x+3 smallest number that is in the domain of

***The complete question is: Let f(x)= sqrt 6x and g(x)=x+3 smallest number that is in the domain of f(g(x))***

The smallest number that is in the domain is -3.

Explanation:

Given functions:

To find f(g(x)), put g(x) in f(x) as follows:

As (6x+18) is in square-root (and square-root of a negative number is an imaginary value), therefore, the domain should be the following:

Therefore, the domain is [-3, +∞).

So the smallest number that is in the domain is -3.

the answer would be d.

step-by-step explanation:

move all variables to the left side and all constants to the right side

(x−3)^2+(y−1)^2=16

this is the form of a circle. use this form to determine the center and radius of the circle.

( x − h ) 2 + ( y − k ) 2 = r 2

match the values in this circle to those of the standard form. the variable   r   represents the radius of the circle,   h   represents the x-offset from the origin, and   k   represents the y-offset from origin.

r = 4

h = 3

k = 1

the center of the circle is found at   ( h , k ) .

center:   ( 3 , 1 )

these values represent the important values for graphing and analyzing a circle.

center:   ( 3 , 1 ).

radius:   4