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.
Answer:
x = 13
Step-by-step explanation:
First start with the blank equation
Plug in the legs and hypotenuse. Remember c squared is always the hypotenuse.
Then solve
25 + 144 = 
169 =
c = 13
Step-by-step explanation:
See attached picture.
First, compare the highest term of the dividend (x²) to the highest term of the divisor (x). We need to multiply the divisor by x.
When we do that, we get x² + 5x. Subtracting this from the dividend, we get -9x + 11.
Now repeat the process. Compare the highest term of the new dividend (-9x) to the highest term of the divisor (x). We need to multiply by -9.
When we do that, we get -9x − 45. When we subtract from the new dividend, we get 56.
So the quotient is x − 9, and the remainder is 56.
The slope would be B) 0.05
Answer:
-0.96774193548
Step-by-step explanation:
kind a mouth full
