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victus00 [196]
3 years ago
6

Find two consecutive odd integers whose sum is 111

Mathematics
2 answers:
notka56 [123]3 years ago
8 0
The two odd numbers who’s sum is 111 are 53.5 and 57.5.
VARVARA [1.3K]3 years ago
7 0

Answer:

54.5 and 56.5

Step-by-step explanation:

To solve this, we need to know the meaning of the term 'consecutive odd integers'

Consecutive odd integers are odd odd integers which follow each other in the pattern such as; 1, 3, 5, 7,9,11,13,15------etc

This implies they take the form n, n+2, n+4, n+6, n+8, n+10--------etc where n is an odd number.

Haven gotten the above definition, we will now proceed to solve the question.

we are ask to find the two consecutive odd numbers whose sum is 111

let the two numbers be n and n+2

n + n+2 =111

2n + 2 = 111

subtract 2 from both-side of the equation

2n + 2 - 2 = 111 -2

2n = 109

Divide both-side of the equation by 2

\frac{2n}{2}   =  \frac{109}{2}

(at the left-hand side of the equation 2 will cancel-out 2, leaving us with just n and on the right-hand side of the equation 109 will be divided by 2)

n = 54.5

The first number is n=54.5

The second number is n+2 = 54.5 + 2 =56.5

Therefore, the two consecutive numbers whose sum is 111 is 54.5 and 56.5

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I am a number greater than 40,000 and less than 60,000. My ones digit and tens digit are the same. My ten-thousands digit is 1 l
jarptica [38.1K]
I am a number greater than 40,000 and less than 60,000:

40,000 < n < 60,000

This means that:

n = 10,000n₁ + 1,000n₂ + 100n₃ + 11n₄

And also:

4 ≤ n₁ < 6

0 ≤ n₂ ≤ 9

0 ≤ n₃ ≤ 9

0 ≤ n₄ ≤ 9

My ten thousands digit is 1 less than 3 times the sum of my ones digit and tens digit:

n₁ = 3*2n₄ - 1

n₁ = 6n₄ - 1

This means that:

n = 10,000*(6n₄-1) + 1,000n₂ + 100n₃ + 11n₄

n = 60,000n₄ - 10,000 + 1,000n₂ + 100n₃ + 11n₄

n = 60,011n₄ - 10,000 + 1,000n₂ + 100n₃

<span>My thousands digit is half my hundreds digit, and the sum of those two digits is 9:

n</span>₂ = 1/2 * n₃
<span>
n</span>₂ + n₃ = 9
<span>
Therefore:

n</span>₂ = 9 - n₃
<span>
Therefore:

9 - n</span>₃ = 1/2 * n₃
<span>
9 = 1/2 * n</span>₃ + n₃
<span>
9 = 1.5 * n</span>₃
<span>
Therefore:

n</span>₃ = 6
<span>
If n</span>₃=6, n₂=3.
<span>
This means that:

</span>n = 60,011n₄ - 10,000 + 1,000*3 + 100*6

n = 60,011n₄ - 10,000 + 3,000 + 600

n = 60,011n₄ - 6,400

Therefore:

0<n₄<2, so n₄=1.

If n₄=1:

n = 60,011 - 6,400

n = 53,611

Answer:

53,611
3 0
3 years ago
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x⁴ + 10x² + 25 = (x² + 5)²

Next, we have

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So, the complete factorization over the complexes is

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