Answer:
Domain: (
−
∞
,
∞
)
,
{
x
|
x
∈
R
}
Range: (
−
∞
,
∞
)
,
{
y
|
y
∈
R
}
3,5,7,2 is the correc tnaswer
Answer:
-10x^2 +9+8x^2-5x+7=-2x^2-5x+16
Answer:
b
Step-by-step explanation:
3/2
There are two of them.
I don't know a mechanical way to 'solve' for them.
One can be found by trial and error:
x=0 . . . . . 2^0 = 1 . . . . . 4(0) = 0 . . . . . no, that doesn't work
x=1 . . . . . 2^1 = 2 . . . . . 4(1) = 4 . . . . . no, that doesn't work
x=2 . . . . . 2^2 = 4 . . . . . 4(2) = 8 . . . . . no, that doesn't work
x=3 . . . . . 2^3 = 8 . . . . . 4(3) = 12 . . . . no, that doesn't work
<em>x=4</em> . . . . . 2^4 = <em><u>16</u></em> . . . . 4(4) = <em><u>16</u></em> . . . . Yes ! That works ! yay !
For the other one, I constructed tables of values for 2^x and (4x)
in a spread sheet, then graphed them, and looked for the point
where the graphs of the two expressions cross.
The point is near, but not exactly, <em>x = 0.30990693...
</em>If there's a way to find an analytical expression for the value, it must involve
some esoteric kind of math operations that I didn't learn in high school or
engineering school, and which has thus far eluded me during my lengthy
residency in the college of hard knocks.<em> </em> If anybody out there has it, I'm
waiting with all ears.<em>
</em>
