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dedylja [7]
3 years ago
13

Show that 3 · 4^n + 51 is divisible by 3 and 9 for all positive integers n.

Mathematics
1 answer:
-BARSIC- [3]3 years ago
4 0

Answer:

Step-by-step explanation:

Hello, please consider the following.

3\cdot 4^n+51=3\cdot 4^n+3\cdot 17=3(4^n+17)

So this is divisible by 3.

Now, to prove that this is divisible by 9 = 3*3 we need to prove that

4^n+17 is divisible by 3. We will prove it by induction.

Step 1 - for n = 1

4+17=21= 3*7 this is true

Step 2 - we assume this is true for k so 4^k+17 is divisible by 3

and we check what happens for k+1

4^{k+1}+17=4\cdot 4^k+17=3\cdot 4^k + 4^k+17

3\cdot 4^k is divisible by 3 and

4^k+17 is divisible by 3, by induction hypothesis

So, the sum is divisible by 3.

Step 3 - Conclusion

We just prove that 4^n+17 is divisible by 3 for all positive integers n.

Thanks

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If you have a cubic polynomial of the form y = ax^3 + bx^2 + cx + d and lets say it passes through the points (2,28), (-1, -5),
nikklg [1K]

Step-by-step explanation:

<u>Step 1:  Solve using the first point</u>

<em>(2, 28)</em>

28 = a(2)^3 + b(2)^2 + c(2) + d

28 = 8a + 4b + 2c + d

<u>Step 2:  Solve using the second point</u>

<em>(-1, -5)</em>

-5 = a(-1)^3 + b(-1)^2 + c(-1) + d

-5 = -a + b - c + d

<u>Step 3:  Solve using the third point</u>

<em>(4, 220)</em>

220 = a(4)^3 + b(4)^2 + c(4) + d

220 = 64a + 16b + 4c + d

<u>Step 4:  Solve using the fourth point</u>

<em>(-2, -20)</em>

-20 = a(-2)^3 + b(-2)^2 + c(-2) + d

-20 = -8a + 4b - 2c + d

<u>Step 5:  Combine the first and fourth equations</u>

<u />28 - 20 = 8a - 8a + 4b + 4b + 2c - 2c + d + d

8 = 8b + 2d

8 - 8b = 8b - 8b + 2d

(8 -8b)/2 = 2d/2

4 - 4b = d

<u>Step 6:  Solve for c in the second equation</u>

-5 + 5 = -a + b - c + d + 5

0 + c = -a + b - c + c + d + 5

c = -a + b + d + 5

<u>Step 7:  Substitute d with the stuff we got in step 5</u>

c = -a + b + (4 - 4b) + 5

c = -a + b + 4 - 4b + 5

c = -a - 3b + 9

<u>Step 8:  Substitute d and c into the first equation</u>

<u />28 = 8a + 4b + 2(-a - 3b + 9) + (4 - 4b)

28 = 8a + 4b - 2a - 6b + 18 + 4 - 4b

28 - 22 = 6a - 6b + 22 - 22

6 / 6 = (6a - 6b) / 6

1 + b = a - b + b

1 + b = a

<u>Step 9:  Substitute a, b, and c into the third equation</u>

220 = 64(1 + b) + 16b + 4(-(1 + b) - 3b + 9) + (4 - 4b)

220 = 64 + 64b + 16b + 4(-1 - b - 3b + 9) + 4 - 4b

220 - 100 = 60b + 100 - 100

120 / 60 = 60b / 60

2 = b

<u>Step 10:  Find a using b = 2</u>

a = b + 1

a = (2) + 1

a = 3

<u>Step 11:  Find c using a = 3 and b = 2</u>

c = -a - 3b + 9

c = -(3) - 3(2) + 9

c = -3 - 6 + 9

c = 0

<u>Step 12:  Find d using b = 2</u>

d = 4 - 4b

d = 4 - 4(2)

d = 4 - 8

d = -4

Answer:  a = 3, b = 2, c = 0,d = -4

6 0
3 years ago
Read 2 more answers
The 8th grade is selling tickets to the 8th grade formal.The cost per ticket is $15.The decorations and food cost $600,and each
MArishka [77]

Answer:

Step-by-step explanation:

Each ticket is $15.  The number of tickets is what we are trying to solve for.  The class spends a certain amount of money to prepare for the formal.  They hope that the money they make in ticket sales is MORE than what they spend.  The expression that represents the number of tickets at $15 each is 15x, where x is the number of tickets.  They hope that the sales are greater than what they spend, so what we have so far is

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Solve this inequality for x.  Begin by subtracting .5 from both sides to get

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Because we are not selling (or printing) .3 of a ticket, it's safe to say (and also correct!) that they need to sell (and print) 41 tickets.  If they sell 41 tickets, the profit is found by

15(41) > 600 + .5(41)

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This means that at 41 tickets, they make a profit.  At 40 tickets, the inequality looks like this:

15(40) > 600 + .5(40) and

600 > 620.  This is not true, so 40 tickets isn't enough.

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Brut [27]

Answer:

d. A five-hour bus trip is divided into six equal legs.

Step-by-step explanation:

hope it helps

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