Answer:
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we have
y > -2
x + y < 4
using a graph tool
see the attached figure
The shaded area is the solution of the system
<u>Part 1) </u>Name an ordered pair that is a solution to this system and explain how you know that this is a solution point
Let
A ( -40,20)
The point A is solution of the system because the point lie on the shaded area
<u>Check</u>
If the point A is solution of the system must satisfy both system inequalities
point A
x=-40
y=20
substitute
y > -2-------> 20 > -2-------> is ok
x + y < 4----> -40+20 < 4-----> -20 < 4-----> is ok
therefore
<u>the answer Part 1) is</u>
The point A is a solution of the system
Part 2) Name an ordered pair that is not a solution to the system and explain how you know that it is not a solution
Let
B(20,20)
The point B is not solution of the system because the point not lie on the shaded area
<u>Check</u>
If the point B is not solution of the system must not satisfy both system inequalities
point B
x=20
y=20
substitute
y > -2 -------> 20 > -2-------> is ok
x + y < 4---->20+20 < 4-----> 40 < 4------> is not ok
therefore
<u>the answer part 2) is</u>
The point B is not a solution to the system
Answer:
- 41
- -5
- 42
- 21
- 4
- 7
Step-by-step explanation:
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.