Let Fn be the number of ways of arranging such flagpole with the given conditions.
When arranging a flagpole of n feet high, consider the following cases
If the last flag used is a red flag, then the other flags are n-1 foot high, so they can be seen as arranged on a smaller flagpole of n-1 feet high, which can be done in Fn-1 ways.
Similarly, If the last flag used is a gold flag, then the other flags can be seen as arranged on a smaller flagpole of n-1 feet high. This can be done in Fn-1 ways.
If the last flag used is green, the other flags are n-2 feet high, so the flagpole can be arranged in Fn-2 ways.
Using the sum rule, we obtain that Fn = 2Fn-1 + Fn-2 for all n≥3. Listing all the combinations of flags, the initial conditions are F1 = 2 and F2 = 3.
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Answer:
Let 3rd digit be x
then 2nd digit = 4x & 1st digit = 4x-2.
It is three digit number. So, there are two case either x=1 or x=2.
Case 1,
3rd digit = 1, 2nd digit = 4 & 1st digit = 4–2=2
Hence number = 241 (ans)
Case 2,
3rd digit = 2 ,2nd digit 8 & 1st digit = 8–2=6.
Hence number = 682 (ans).
Step-by-step explanation: hope this helps :P
Answer:
87%
Step-by-step explanation:
In order to get 87% you divide 16000 by 13920
16000/13920= 0.87
To convert this into a percentage you can multiply by 100
0.87 * 100 =87
Answer: 87%
I hope this helps :)
Textbook is y
Notebook is x
Formula
800 = 80y + 4x
Answer:
9x6=54
2x2=4
6+9=15
1+2+3=6
Step-by-step explanation:
