Question

??????? please I have no idea what to do.

Answer 1

this is a ridiculous question, but the answer is B

Answer 2

why is it ridiculous?

Answer 3

because on the right side you are defining the domain, but the domain is definded incorrectly

Answer 4

- the second answer is right sure

Answer 5

when you define a piecewise function like this
\[f(x) = \left\{\begin{array}{rcc}
\frac{1}{x}& \text{if} & x \leq 0 \\
2x& \text{if} & x >0
\end{array}
\right.
\]
a an example, the part after the "if" is defining the domain. but you cannot in my example make x = 0, so what i wrote doesn't make any sense. it has to be
\[f(x) = \left\{\begin{array}{rcc}
\frac{1}{x}& \text{if} & x < 0 \\
2x& \text{if} & x >0
\end{array}
\right.
\]

Answer 6

when you define a function, you define the domain along with it. it is part of the definition

so you cannot define a function, define the demain incorrectly, and then ask "what is the domain"
the question makes no sense and is completely idiotic.

tell those fools at Khan that they do not know what they are doing

Answer 7

so how did you know how to solve it?

Answer 8

because the denominator cannot be 0, so x cannot be 9

Answer 9

would you mind showing me now you solved it?

Answer 10

i am afraid there is nothing really to show

Answer 11

you have a denominator, so you know that it cannot be zero
i guess you could write
\[\sqrt{x-9}\neq 0\]
\[x-9\neq 0\]
\[x\neq 9\]

Answer 12

I usually use inequalities when dealing with numbers under the radical sign

Answer 13

i see,yes, but look at how they defined the domain of the function

Answer 14

\[f(x) = \left\{\begin{array}{rcc}
\frac{1}{\sqrt{x-9}}& \text{if} & x >9 \\
\frac{1}{\sqrt{9-x}}& \text{if} & x <9
\end{array}
\right.\]
they have already taken care of where the expression inside the radical should be positive or negative

Answer 15

to be honest with you satellite, i have no idea what this expression is telling me.