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m_a_m_a [10]
3 years ago
9

Is the inequality sometimes,always, or never true ?

Mathematics
1 answer:
Paha777 [63]3 years ago
3 0
A. 3(x + 3) > -3(2 + x)  SOMETIMES TRUE

3(x + 3) > -3(2 + x)      |use distributive property a(b + c) = ab + ac

3x + 9 > -6 - 3x      |add 3x to both sides

6x + 9 > -6     |subtract 9 from both sides

6x > -15    |divide both sides by 6

x > -15/6

x > -2.5

B. 9 - x - 5 < - x + 4      NEVER TRUE

9 - x - 5 < - x + 4

4 - x < - x + 4    |add x to both sides

4 < 4 FALSE
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Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.

So, if we sum the first N odd numbers, we have

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)

The second sum is simply the sum of N ones:

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\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2

which ends the proof.

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