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Anit [1.1K]
3 years ago
7

Nadiya determines that there are no fractions equivalent to 2/10 with a denominator greater than 10, but less than 20, that have

a whole number numerator and denominator. Is Nadiya correct?
Yes, Nadiya is correct. By multiplying the numerator and denominator by 2, the first fraction equivalent to 2/10 is 4/20.

Yes, Nadiya is correct. By multiplying the whole fraction by any whole number, each equivalent fraction will continue to have a denominator of 10.

No, Nadiya is incorrect. She could multiply both the numerator and denominator of the unit fraction equivalent to 2/10 by 3 and arrive at the equivalent fraction 3/15.

No, Nadiya is incorrect. She could add any number between 1 and 8 to the numerator and denominator and arrive at an equivalent fraction with a denominator greater than 10 but less than 20.

Mathematics
2 answers:
frosja888 [35]3 years ago
8 0

Answer:

Yes, Nadiya is correct. By multiplying the numerator and denominator by 2, the first fraction equivalent to 2/10 is 4/20.

Step-by-step explanation:

To find a fraction equivalent to 2/10, we need to multiply (or divide) the numerator and denominator by the same nonzero whole number.

As we need an equivalent fraction with a denominator greater than 10, we will need to multiply and not divide.

The first nonzero whole number we have is 1. If we multiply the numerator and denominator by 1, we get 2/10. Obviously, 2/10 is equivalent to 2/10 but the denominator is not greater than 10, so it doesn't help us.

The next whole number is 2. When we multiply the numerator and denominator by 2, we get 4/20. The denominator is not less than 20.

If we keep going, we will get 6/30, 8/40 and so on.  

Therefore, Nadiya is correct.

The correct answer is the first one.

vampirchik [111]3 years ago
3 0

Answer: It is C

Step-by-step explanation: I literally just took the test bruhh

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Solve the following system of equations:
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Step-by-step explanation:

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--------------------

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=================

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1 year ago
How much p² – q² + 2pq is less than -2p² -2q² + 7pq +10 ?
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Answer:   3p² + q² - 5pq - 10 < 0

<u>Step-by-step explanation:</u>

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since you didn't specify what to solve for, move everything from the right side to the left side of the inequality symbol using inverse operations.

(p² + 2p²) + (- q² + 2q²) + (2pq - 7pq) + (- 10) < 0

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3p² + q² - 5pq - 10  < 0

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