Answer:
The value of x that gives the maximum transmission is 1/√e ≅0.607
Step-by-step explanation:
Lets call f the rate function f. Note that f(x) = k * x^2ln(1/x), where k is a positive constant (this is because f is proportional to the other expression). In order to compute the maximum of f in (0,1), we derivate f, using the product rule.

We need to equalize f' to 0
- k*(2x ln(1/x) - x) = 0 -------- We send k dividing to the other side
- 2x ln(1/x) - x = 0 -------- Now we take the x and move it to the other side
- 2x ln(1/x) = x -- Now, we send 2x dividing (note that x>0, so we can divide)
- ln(1/x) = x/2x = 1/2 ------- we send the natural logarithm as exp
- 1/x = e^(1/2)
- x = 1/e^(1/2) = 1/√e ≅ 0.607
Thus, the value of x that gives the maximum transmission is 1/√e.
Answer:
4/5
Step-by-step explanation:
Square of 1: 1*1=1
Square of 2: 2*2=4
Square of 3: 3*3=9
<u>Square of 4: 4*4=16</u>
<u>Square of 5: 5*5=25</u>
Square of 6: 6*6=36
Square of 7: 7*7=49
Square of 8: 8*8=64
Square of 9: 9*9=81
Square of 10: 10*10=100
Those are the basics.
I hope this helps!
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