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).
I'll just take the points thanks XD
Answer:
64 carnations and 16 roses.
Step-by-step explanation:
The events committee buys 80 flowers
The flowers are a combination of carnations and roses.
Each carnation costs $0.50 and each rose costs $2.50
The committee spends a total of $72
<u>Solution:</u>
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Let there be (c) carnations and (r) roses.
So,
c + r = 80 flowers
and $0.50c + $2.50 = $72
This forms a simultaneous equation:
c + r = 80 ... (i)
0.5c + 2.5r = 72 ... (ii)
Multiplying equation (i) by 0.5 and equation (ii) by 1 gives;
0.5c + 0.5r = 40 ... (i)
0.5c + 2.5r = 72 ... (ii)
Subtracting (i) from (ii) gives;
0c + 2r = 32
2r = 32
r = 32 ÷ 2 = 16
Therefore, there are 16 roses
and 80 - 16 = 64 carnations
14^15/14^5=14^(15-5)=14^10
Hope this kinda helps !!
The area would be 36 each side is 6 inches therefore you would multiply 6 by 6 to find the area
