He can use:
a quarter and a dime (difference: 15)
a quarter and two nickels (difference: 20)
a dime and three nickels (difference: 5)
two dimes and a nickel (difference: 5)
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:
The ratio of red to green crayons is 18 to 24 which can be reduced to 3 to 4.
Step-by-step explanation:
Answer:
Emilio will have both activities again on the same day after 90 days from this Saturday.
Step-by-step explanation:
<em>To solve this question list the multiples of 30 and 9 to find the first common multiple because the two activities will happen again in </em><em>a number divisible by both 30 and 9.</em>
∵ The multiples of 30 are: 30, 60, 90, 120, ............
∵ The multiples of 9 are: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, .....
→ The common multiple is 90
∴ The common multiple of 30 and 9 is 90
→ That means the activities will meet again after 90 days from
this Saturday
Emilio will have both activities again on the same day after 90 days from this Saturday.
