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ollegr [7]
2 years ago
11

What are the last two digits of 7^1867? ​

Mathematics
1 answer:
Kipish [7]2 years ago
3 0

Considering that the powers of 7 follow a pattern, it is found that the last two digits of 7^{1867} are 43.

<h3>What is the powers of 7 pattern?</h3>

The last two digits of a power of 7 will always follow the following pattern: {07, 49, 43, 01}, which means that, for 7^n, we have to look at the remainder of the division by 4:

  • If the remainder is of 1, the last two digits are 07.
  • If the remainder is of 2, the last two digits are 49.
  • If the remainder is of 3, the last two digits are 43.
  • If the remainder is of 0, the last two digits are 01.

In this problem, we have that n = 1867, and the remainder of the division of 1867 by 4 is of 3, hence the last two digits of 7^{1867} are 43.

More can be learned about the powers of 7 pattern at brainly.com/question/10598663

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Answer:

The cost C as a function of t is C(t) = 30,000 + 6,400,000 t - 40,000 t²

Step-by-step explanation:

The function N(t) = 800 t -  5t², represents the number of cars produced at a time t hours in a day, where 0 ≤ t ≤ 10

The function C(N) = 30,000 + 8,000 N, represents the cost C​ (in dollars) of producing N cars

We need to find The cost C as a function of the time t

That means Substitute N in C by its function by other word find the composite function (C о N)(t)

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∴ C(N(t)) = 30,000 + 8000(800 t - 5 t²)

- Multiply the bracket by 8000

∴ C(N(t)) = 30,000 + 8000(800 t) - 8000(5 t²)

∴ C(N(t)) = 30,000 + 6,400,000 t - 40,000 t²

- C(N(t) = C(t)

∴ C(t) = 30,000 + 6,400,000 t - 40,000 t²

The cost C as a function of t is C(t) = 30,000 + 6,400,000 t - 40,000 t²

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