Answer:
We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.
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Explanation:
It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.
I'm going to use these three log rules, which apply to any base.
- log(A) + log(B) = log(A*B)
- log(A) - log(B) = log(A/B)
- B*log(A) = log(A^B)
From there, we can then say the following:
Circumference is equal to diameter multiplied by pi. C=d X pi. To find the radius we need to rearrange this equation. We know what pi and the circumference are, so to find diameter we do circumference divided by pi. D=c/pi. 942/pi = 299.85. This gives us the diameter, for the radius simply divide by 2 which is 149.92 :)
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Step-by-step explanation:
Answer:
The water temperature that produces the maximum number of salmon swimming upstream is approximately 12.305 degrees Celsius.
Step-by-step explanation:
Let , for . represents the temperature of the water, measured in degrees Celsius, and is the number of salmon swimming upstream to spawn, dimensionless.
We compute the first and second derivatives of the function:
(Eq. 1)
(Eq. 2)
Then we equalize (Eq. 1) to zero and solve for :
And all roots are found by Quadratic Formula:
,
Only the first root is inside the given interval of the function. Hence, the correct answer is:
Now we evaluate the second derivative at given result. That is:
According to the Second Derivative Test, a negative value means that critical value leads to a maximum. In consequence, the water temperature that produces the maximum number of salmon swimming upstream is approximately 12.305 degrees Celsius.
Answer:
See below
Step-by-step explanation:
The number of loaves of bread remaining can only go as low as ZERO (you can't have negative loaves) ...and can only go as high as when h = 0
so 0 = 42-3.5 h
3.5 h = 42
h = 12 and min is h = 0 ( when there are 42 loaves to start)
So:
<u>0≤ h ≤ 12 </u>