Answer:
13.8 days
Step-by-step explanation:
In this case we can solve it by means of a rule of three, knowing the amount of life for 1000 grams, it must be proportional for 600 grams, like this:
grams days
1000 23
600 x
x = 600 * 23/1000
x = 13.8
which means that for the 600 grams it would be an average life of 13.8 days.
the derivitive is just the slope
minimum happens when the derivitive goes from negative to positive, imagine a slope of the function, the minimum is where the slope goes from neative to positive, and to get there, it has to pass through 0
max happens when the derivitive goes from positive to negative
increaseing is when the derivitive is positive
so, based on what you said, the slope of f(x) is 0 at x=-3, x=1 and x=2 since those are where the derivitive is 0 (derivitive is just the slope)
A and B are wrong because the derivitive isn't 0 at those points
C is correct because increasing means that the derivitive is positive, and so therefo since the only hoirontal place in between 1 and 2 is 1.5, it must remain positive throughout and not dip down, C is right
D is wrong then
answer is C
Answer:
3.45
Step-by-step explanation:
you have to do tihis then that
Answer:
Step-by-step explanation: the first time you get 0.75 and the second time you get 0.125 hope this helped
Answer:
- number of multiplies is n!
- n=10, 3.6 ms
- n=15, 21.8 min
- n=20, 77.09 yr
- n=25, 4.9×10^8 yr
Step-by-step explanation:
Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...
mpy[n] = n·mp[n-1]
mpy[2] = 2
So, ...
mpy[n] = n! . . . n ≥ 2
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If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...
10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10
Then the larger matrices take ...
n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min
n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years
n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years
_____
For the shorter time periods (less than 100 years), we use 365.25 days per year.
For the longer time periods (more than 400 years), we use 365.2425 days per year.
