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professor190 [17]
4 years ago
5

The lock is numbered from 0 to 49. Each combination uses three numbers in a right, left, right pattern. Find the total number of

possible combinations for the lock.
Mathematics
1 answer:
oksano4ka [1.4K]4 years ago
5 0
The <u>correct answer</u> is:

125,000.

Explanation:

For the first turn, there are 50 numbers to choose from.

On each turn of a lock, the numbers can be repeated; this means on the second turn, there are 50 numbers to choose from, and on the third turn, there are 50 numbers to choose from.  This gives us:

50³ = 50×50×50 = 125,000.
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Lesechka [4]
For this case, the first thing we must do is define a variable.
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 We now write the equation that models the problem.
 We have then:
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 Answer:
 a function, p (x), which can be used to determine his pay for the week is:
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5 0
3 years ago
Prove that given any three consecutive integers, one of them is divisible by 3. Hint: What are the possible remainders when we d
juin [17]

Answer:

it is proved that from the any three consecutive integers one of them is divisible by 3.

Step-by-step explanation:

Let the first integer = x

The second  consecutive integer = x + 1

The third consecutive integer = x + 2

Case 1. take x = 1

The value of first integer = 1

The value of second integer = 1 +1 = 2

The value of third integer = 1 + 2 = 3

Here the third integer is divisible by 3.

Case 2. take x = 2

The value of first integer = 2

The value of second integer = 2 +1 = 3

The value of third integer = 2 + 2 = 4

Here the second  integer is divisible by 3.

Case 3. take x = 3

The value of first integer = 3

The value of second integer = 3 +1 = 4

The value of third integer = 3 + 2 = 5

Here the first  integer is divisible by 3.

Thus it is proved that from the any three consecutive integers one of them is divisible by 3.

8 0
3 years ago
114 seniors out of a HS of size 430
Goshia [24]

Answer:

114/430 not that hard

Step-by-step explanation:

7 0
3 years ago
Find the fifth roots of 32(cos 280° + i sin 280°)
shusha [124]

Answer:

The fifth root is 2[cos(56°) + i sin(56°)]

Step-by-step explanation:

* To solve this problem we must revise De Moiver's rule

- In the complex number with polar form

∵ z = r(cosФ + i sinФ)

∴ z^n = r^n(cos(nФ) + i sin(nФ))

* In the problem

- The fifth root means z^(1/5)

- We can put 32 as a form a^n

∵ 32 = 2 × 2 × 2 × 2 × 2 = 2^5

∴ z = 2^5[cos(280°) + i sin(280°)]

* Lets find z^(1/5)

*z^{\frac{1}{5}}=[2^{5}]^{\frac{1}{5} } (cos(\frac{1}{5})(280)+isin(\frac{1}{5})(280)

*(2^{5})^{\frac{1}{5}}=2^{5.\frac{1}{5}}=2

∴ z^(1/5) = 2[cos(56) + i sin(56)]

* The fifth root of 32[cos(280°) + i sin(280°)] is 2[cos(56°) + i sin(56°)]

 

4 0
3 years ago
1. You have two boxes of colored pens. The first box contains a red pen, a blue pen, and a green pen. The second box contains a
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