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:
the answer is attached to the picture
Explanation:
Pair 1 is true if Jeff's monthly income is $600/20% = $3,000.
Pair 2 is true if Jeff's monthly income is $1200/10% = $12,000.
Both pairs can be true if Jeff's monthly income increased by a factor of 4 in the 20 years from 1990 to 2010.
Obviously, Jeff spent more on housing in 2010. (Fortunately for Jeff, that larger expenditure was a smaller fraction of his income.)
Answer:
a) p = 6x - 5
b) x = 6
Step-by-step explanation:
a) p = a + b + c
p = 3x - 5 + 2x - 1 + x + 1
p = 6x - 5
b) p = 6x - 5 =31
6x - 5 = 31
6x = 36
x = 6
Answer:
$8.87 per pound
Step-by-step explanation:
