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kondor19780726 [428]
2 years ago
10

11 points

Mathematics
1 answer:
anyanavicka [17]2 years ago
5 0

Answer:

9.100

9.10,00

1. 1,098,000

1.109,800,000

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Write the explicit formula
Margaret [11]

Answer:

Step-by-step explanation:

Since there is no single number we can add to one number in the sequence to get to the next one, this is not arithmetic. That means it's geometric, if it's a sequence at all. To get from 2 to 6 we multiply by 3. To get from 6 to 18 we multiply by 3. To get from 18 to 54 we multiply by 3. Therefore, this is in fact geometric and the common ratio is 3. The standard form of a geometric explicit formula is

A_n=a_1(r)^{n-1 where n is the position of a number in the sequence, a1 is the first number in the sequence, and r is the common ratio. We have then

a1 = 2 and r = 3. Therefore, the explicit formula is

A_n=2(3)^{n-1 and you can use this to find the value of any number in the sequence. Very handy; much more so than the recursive formula, which requires that all the numbers in a sequence be found in order in order to get to a desired value.

6 0
3 years ago
WILL MARK BRAINLIEST!!! Can someone help me wit dis question?
Advocard [28]

Answer:

B.) -4

Step-by-step explanation:

y=mx+b

b is equal to the y-intercept.

7 0
3 years ago
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What did the doctor say after examining yunn yunsberger?
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4 0
3 years ago
This is another one helpzzz mee
Varvara68 [4.7K]
I don't know about the first one but the second one is 34.9°. It would be that because are right angles add up to 90° and if you subract 55.1 and 90 it would equal 34.9. But this rule only works on right angles.
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3 years ago
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Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
1 year ago
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