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Nadya [2.5K]
3 years ago
8

Write a number that has four digits with the same number in all places, such as 4,444. Circle the digit with the greatest value.

Underline the digit with the least value. Explain.

Mathematics
2 answers:
rosijanka [135]3 years ago
7 0
Use 8,888, then circle the 8 thousandth one (the first one on the left) then underline the 8 at the end (on the right)
Step2247 [10]3 years ago
6 0

Answer:

5555

Step-by-step explanation:

Here number can be many like: 1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999.

Here I take number 5555

The Number that has the greatest value is the first number from left since it is the thousandth value of a number.

The number that has the least value is the last number from left since it is the tenth value of a number.

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Find three consecutive even integers such that the product of the first and second is 8 more than 38 times the third.
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Answer:

40, 42 and 44

Step-by-step explanation:

Mark these even consecutive integers as (2n-2), 2n and (2n+2)

Note that since they are integers, n > 0 and 2n is always even

Writing the equation

(2n-2) * 2n = 38 * (2n+2) + 8

// divide both sides by 2

(2n-2) * n = 19 * (2n+2) + 4

2n^{2} - 2n = 38n + 38 + 4

2n^{2} - 2n - 38n - 38 - 4 = 0

2n^{2} - 40n - 42 = 0

// divide both sides by 2

<u />n^{2} - 20n - 21 = 0

// rewrite the equation

<u></u>n^{2} + n - 21n - 21 = 0

n * (n+1) - 21n - 21 = 0

n * (n+1) - 21 * (n+1) = 0

(n+1) * (n-21) = 0

// separate the two possible cases

n + 1 = 0 and n - 21 = 0

n = -1              n = 21

Note that here n has to be an integer, so case n = -1 can't be a solution. Therefore we use n = 21

Calculating the consecutive even integers

2n - 2 = 2 * 21 - 2 = 40

2n = 2 * 21 = 42

2n + 2 = 2 * 21 + 2 = 44

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