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zhuklara [117]
3 years ago
9

If the same BLANK number is multiplied to each side of a true inequality, the inequality sign is reversed to make the resulting

inequality true.
Mathematics
1 answer:
Maurinko [17]3 years ago
5 0
Answer: Negative
If the same negative number is multiplied to each side of a true inequality, then the inequality sign flips to make the new inequality true as well

Example:
Take 1 < 5 and multiply both sides by -2 and we get -2 > -10. The "less than" sign flips to "greater than" since -2 < -10 is false. The value of -10 is further to the left of -2 so -10 is smaller in value. The negative basically takes the complete opposite which is why the flip must happen.

This sign flip rule does not happen if you multiply both sides by a positive number.
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C. -5 + 9 = 22 + 9 - 7 is the answer.

This is because when solved it’s : -52 = 15 which is false therefore there’s no solution.
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3 years ago
Can you solve -4a^ - (6a-2) + 3 (a^-1).the arrows that are pointing up is exponent 2.also please show work
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First step is to distribute.
-4a^2 - 6a +2 + 3a^2 -3
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3 years ago
Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

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