The Solution:
Given:
Center = (0,0)
Point A = (-5,2) being a point on the circle.
We are required to check if point P = (2,-5) is on the circle.
Solving the given problem graphically, we have:
From the above graph, it is clear that point P(2,-5) is a point on the circle.
Answer:
f) a[n] = -(-2)^n +2^n
g) a[n] = (1/2)((-2)^-n +2^-n)
Step-by-step explanation:
Both of these problems are solved in the same way. The characteristic equation comes from ...
a[n] -k²·a[n-2] = 0
Using a[n] = r^n, we have ...
r^n -k²r^(n-2) = 0
r^(n-2)(r² -k²) = 0
r² -k² = 0
r = ±k
a[n] = p·(-k)^n +q·k^n . . . . . . for some constants p and q
We find p and q from the initial conditions.
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f) k² = 4, so k = 2.
a[0] = 0 = p + q
a[1] = 4 = -2p +2q
Dividing the second equation by 2 and adding the first, we have ...
2 = 2q
q = 1
p = -1
The solution is a[n] = -(-2)^n +2^n.
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g) k² = 1/4, so k = 1/2.
a[0] = 1 = p + q
a[1] = 0 = -p/2 +q/2
Multiplying the first equation by 1/2 and adding the second, we get ...
1/2 = q
p = 1 -q = 1/2
Using k = 2^-1, we can write the solution as follows.
The solution is a[n] = (1/2)((-2)^-n +2^-n).
Answer:
Evelyn's dance class on Saturday starts at 11:05 A.M.
Step-by-step explanation:
I know this because if her class takes an hour and 40 minutes you would subtract the hour and 40 minutes from 12:45 so you could do 12 - 1 which equals 11. Then you would do 45 - 40 which equals 5. And last you would put the new times together so it would be, 11:05. Her dance class starts at 11:05 A.M.
Answer:
l think the answer is B
Step-by-step explanation:
The eighth patern should give the result 81
21x-33°=180°-135°
21x=12°
x=12/21
