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Sedaia [141]
3 years ago
15

What is the solution to the inequality 5x - 11 ≥ 31

Mathematics
1 answer:
Molodets [167]3 years ago
7 0
1

 

Add 1111 to both sides

5x\ge 31+115x≥31+11


2

 

Simplify 31+1131+11 to 4242

5x\ge 425x≥42


3

 

Divide both sides by 55

x\ge \frac{42}{5}x≥​5​​42​​


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Read 2 more answers
The sum of the squares of two consecutive positive integers is 41. Find the two
Komok [63]

We have to present the number 41 as the sum of two squares of consecutive positive integers.

1² = 1

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<h3>Answer: 4 and 5</h3>

Other method:

n, n + 1 - two consecutive positive integers

The equation:

n² + (n + 1)² = 41     <em>use (a + b)² = a² + 2ab + b²</em>

n² + n² + 2(n)(1) + 1² = 41

2n² + 2n + 1 = 41     <em>subtract 41 from both sides</em>

2n² + 2n - 40 = 0     <em>divide both sides by 2</em>

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n² + 5n - 4n - 20= 0

n(n + 5) - 4(n + 5) = 0

(n + 5)(n - 4) = 0 ↔ n + 5 = 0 ∨ n - 4 =0

n = -5 < 0 ∨ n = 4 >0

n = 4

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<h3>Answer: 4 and 5.</h3>
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2 years ago
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kumpel [21]
B.) 10:18 and 25:45
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5:9.
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3 years ago
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