<em>So</em><em> </em><em>the</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>of</em><em> </em><em>option</em><em> </em><em>B</em><em>.</em>
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>
<em>hope </em><em>it</em><em> </em><em>will</em><em> </em><em>be</em><em> </em><em>helpful</em><em> </em><em>to</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em>
Answer:
Step-by-step explanation:
Rotations are easy when its multiples of 90 degrees; 90,180,270,360. Treat them like they’re a complex number like [x+yi]*i=[-y,xi] so rotating by 90 degrees and 180 is i squared [-1]! So [-3,2] rot90 is [-2,-3].
Reflection about the y=x line is change places. [x.y]=[y,x].
So [-2.-3] reflected about the y=x line is [-2.-3]=[-3.-2],
Compile a list of these transform is best practice technique in this area.
The statement "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g" is FALSE.
Domain is the values of x in the function represented by y=f(x), for which y exists.
THe given statement is "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g".
Now we assume the
and 
So here since g(x) is a polynomial function so it exists for all real x.
<em> </em>does not exists when
, so the domain of f(x) is given by all real x except 6.
Now,

So now (fg)(x) does not exists when x=4, the domain of (fg)(x) consists of all real value of x except 4.
But domain of both f(x) and g(x) consists of the value x=4.
Hence the statement is not TRUE universarily.
Thus the given statement about the composition of function is FALSE.
Learn more about Domain here -
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Can't help without a pic or a problem to solve. sorry.