Problem 1
Draw a straight line and plot P anywhere on it. Use the compass to trace out a faint circle of radius 8 cm with center P. This circle crosses the previous line at point Q.
Repeat these steps to set up another circle centered at Q and keep the radius the same. The two circles cross at two locations. Let's mark one of those locations point X. From here, we could connect points X, P, Q to form an equilateral triangle. However, we only want the 60 degree angle from it.
With P as the center, draw another circle with radius 7.5 cm. This circle will cross the ray PX at location R.
Refer to the diagram below.
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Problem 2
I'm not sure why your teacher wants you to use a compass and straightedge to construct an 80 degree angle. Such a task is not possible. The proof is lengthy but look up the term "constructible angles" and you'll find that only angles of the form 3n are possible to make with compass/straight edge.
In other words, you can only do multiples of 3. Unfortunately 80 is not a multiple of 3. I used GeoGebra to create the image below, as well as problem 1.
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
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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:
3 - (-7)
= 3 + 7.
Step-by-step explanation:
A double negative is equivalent to a plus.
The sum will be x+y because their ages are variables.
In parallelogram LMNO, NO = 10.2, and LO = 14.7. The basic property of a parallelogram that the opposite sides in a parallelogram are equal; so, two sides are 10.2 (NO=LM=10.2), and two sizes are 14.7, = (LO=MN=14.7).
The perimeter of the parallelogram LMNO is:
P= 2NO + 2LO.
P= (2 x 10.2)+(2 x 14.7).
P= 20.4 + 29.4.
So, P= 49.8.
