Hi there! The answer is B. it would be less steep.
The change of the function (in this case) influences the slope of the line. Since we've modified the slope from 5 to 3 / 4, the line will be less steep.
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
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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.
we have the function
Part a
For t=7
substitute in the given function
For t=14
For t=21
For t=28
For t=35
Observation: The values of E varies from -1 to 1, including the zero
Part B
Remember that
The Period goes from one peak to the next
so
Period=2pi/B
B=pi/14
Period=(2pi)/(pi/14)=2pi*14/pi=28
<h2>the period is 28 days</h2>
