<span> a) F' = 6 sin(x^2) = 0
x^2 = pi
x = sqrt(pi)
b) Fmax = F(1) + integral [1, pi] f(x) dx = 9.7743 </span>
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:
25
Step-by-step explanation:
144/60 = 2.4
60/2.4 = x
x = 25
Y > 2x - 3
3y - x ≤ 6
↓
y = 2x - 3
3y - x = 6
3(2x - 3) - x = 6
3(2x) - 3(3) - x = 6
6x - 9 - x = 6
6x - x - 9 = 6
5x - 9 = 6
<u> + 9 + 9</u>
<u>5x</u> = <u>15</u>
5 5
x = 3
y = 2x - 3
y = 2(3) - 3
y = 6 - 3
y = 3
(x, y) = (3, 3)
