Answer :
To determine the domain of the function
[tex]$$
h(x) = \sqrt{x-7} + 5,
$$[/tex]
we first need to ensure that the expression under the square root is non-negative. This is because the square root of a negative number is not defined in the set of real numbers.
The expression under the square root is [tex]$x-7$[/tex], so we need to have:
[tex]$$
x - 7 \geq 0.
$$[/tex]
Now, solve this inequality:
[tex]\[
x \geq 7.
\][/tex]
This means that the function [tex]$h(x)$[/tex] is defined for all real numbers [tex]$x$[/tex] that are greater than or equal to [tex]$7$[/tex].
Thus, the domain of the function is:
[tex]$$
\{ x \in \mathbb{R} : x \geq 7 \}.
$$[/tex]
So, the correct choice is:
B. [tex]$x \geq 7$[/tex].
[tex]$$
h(x) = \sqrt{x-7} + 5,
$$[/tex]
we first need to ensure that the expression under the square root is non-negative. This is because the square root of a negative number is not defined in the set of real numbers.
The expression under the square root is [tex]$x-7$[/tex], so we need to have:
[tex]$$
x - 7 \geq 0.
$$[/tex]
Now, solve this inequality:
[tex]\[
x \geq 7.
\][/tex]
This means that the function [tex]$h(x)$[/tex] is defined for all real numbers [tex]$x$[/tex] that are greater than or equal to [tex]$7$[/tex].
Thus, the domain of the function is:
[tex]$$
\{ x \in \mathbb{R} : x \geq 7 \}.
$$[/tex]
So, the correct choice is:
B. [tex]$x \geq 7$[/tex].
