what is the answer to this problem: A(-4,-5), B(-2,-3), C(4,-3), D(6,-2), E(6,6), F(-1,6), G(-4,2). Find the length of BC.
nikdorinn [45]
Answer:
Length of BC is 6.
Step-by-step explanation:
B is at (-2,-3) and C is at (4,-3)
Length BC
Draw a diagram to illustrate the problem as shown in the figure below.
Let h the height of the hill. =
At position A, the angle of elevation is 40°, and the horizontal distance to the foot of the hill is x.
By definition,
tan(40°) = h/x h = x tan40 = 0.8391x
(1)
At position B, Joe is (x - 450) ft from the foot of the hill. His angle of elevation is
40 + 18 = 58°.
By definition, tan(58°) = h/(x - 450)
h = (x - 450) tan(58°) = 1.6003(x-450)
h = 1.6003x - 720.135 (2)
Equate (1) and (2).
1.6003x - 720.135 = 0.8391x 0.7612x = 720.135
x = 946.0523
From (1), obtain
h = 0.8391*946.0523 = 793.8 ft
Answer: The height of the hill is approximately 794 ft (nearest integer)
Answer:
Step-by-step explanation:
3×8-2×3=18
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.
Nah fam math hard math can solve its own problems 82773919
