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Anna35 [415]
1 year ago
10

find and solve a recurrence relation for the number of ways to arrange flags on an n-foot flagpole using three types of flags: r

ed flags 2 feet high, yellow flags 1 foot high, and blue flags 1 foot high. page 4 of 8
Mathematics
1 answer:
DerKrebs [107]1 year ago
3 0

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.

To Learn more about similar flagpole questions:

brainly.com/question/14277535

#SPJ4

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Answer:

Result:

  • The value of x = 12
  • The value of y = 21

Step-by-step explanation:

Given

The parallelogram DEFG

DE = 6x-12

FG = 2x+36

EF = 4y

DG = 6y-42

We know that the opposite sides of a parallelogram are equal.  

As DE and FG are opposite sides, so

DE = FG

substituting DE = 6x-12 and FG = 2x+36 in the equation

6x-12 = 2x+36

6x-2x = 36+12

simplifying

4x = 48

dividing both sides by 4

4x/4 = 48/4

x = 12

Therefore,

The value of x = 12

Also, EF and DG are opposite sides, so

EF = DG

substituting EF = 4y and DG = 6y-42 in the equation

4y = 6y-42

switching sides

6y-42 = 4y

6y-4y = 42

2y = 42

dividing both sides by 2

2y/2 = 42/2

y = 21

Therefore,

The value of y = 21

Result:

  • The value of x = 12
  • The value of y = 21
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