n^8
basically it's like this n^12/n^4 = n^(12-4) = n^8
The ball was initially thrown from a height of 5.5 feet and 5.5 is y intercept that it throm from 5.5 feet.
In the given you wrote -2, whereas it should be -2i
z = a + bi and in trigo form | z |.(cos Ф + i.sin Ф).
z = 0 -2i → | z |.(cos Ф + i.sin Ф)
Now let's calculate z:
z² = a²+b² → z² = 0² (-2.i)² → z = -2(i)² →z= -2(-1) → z = |2|
tan Ф = b/a = -2/0 → tan Ф → - ∞ ↔ Ф = -90° or Ф = 270°
then the - 2 I ↔ |2|(cos 270° + i.sin 270°)
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).
