Answer: 1 is not a perfect square. 3 is the only prime number one less than a square.
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>
answer.
Answer:
x=2 and y=0 is the required result.
Step-by-step explanation:
We have been given system of equations:
5x+2y=105x+2y=10 (1)
And 3x+2y=63x+2y=6 (2)
We will use elimination method:
Multiply 1st equation by 3 and 2nd equation by 5 we get:
15x+6y=3015x+6y=30 (3)
15x+10y=3015x+10y=30 (4)
Now subtract (4) from (3) we get:
-4y=0−4y=0
y=0y=0
Now, put y=0 in (1) equation:
5x+2(0)=105x+2(0)=10
5x=105x=10
x=2x=2
Hence, x=2 and y=0
Answer:
UNIF(2.66,3.33) minutes for all customer types.
Step-by-step explanation:
In the problem above, it was stated that the office arranged its customers into different sections to ensure optimum performance and minimize workload. Furthermore, there was a service time of UNIF(8,10) minutes for everyone. Since there are only three different types of customers, the service time can be estimated as UNIF(8/3,10/3) minutes = UNIF(2.66,3.33) minutes.
