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
__
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:
This is a guess!
When looking at the given equation I can not help but think of compound interest. So I am going to convert this into that format.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Within the context of financial interest:
Looking for:
P
(
1
−
x
)
n
Where P is the principle sum,
x
is the interest and n is the number of interest cycles (annual)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:
y
=
5100
(
0.95
)
x
But
0.95
=
1
−
0.05
so we have
y
=
5100
(
1
−
0.05
)
x
But
.
0.05
=
5
100
So we have
y
=
5100
(
1
−
5
100
)
x
Thus the percentage change each year is
−
5
%
Step-by-step explanation:
Answer:
bhdwkvdveqvhjvhjawvhjcddvscg
wvvdgvewagvdgwvgdvwgevdegw
Step-by-step explanation:
Answer:
x < 33.84
Step-by-step explanation:
we have
13.48x-200 < 256.12
Solve for x
Adds 200 both sides
13.48x-200 +200 < 256.12+200
13.48x < 456.12
Divide by 13.48 both sides
13.48x/13.48 < 456.12/13.48
x < 33.84
The solution is the interval ----> (-∞, 33.84)
All real numbers less than 33.84
