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tangare [24]
3 years ago
14

find three consecutive odd integers such that the sum of the middle and largest integer is 21 more than smallest integer. ​

Mathematics
1 answer:
mina [271]3 years ago
8 0

Answer:

15, 17, 19

Step-by-step explanation:

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What’s the square root of negitive 64
vladimir2022 [97]

Answer:

± 8i

Step-by-step explanation:

note that \sqrt{-1} = i

using the rule of radicals

• \sqrt{a} × \sqrt{b} ⇔ \sqrt{ab}, hence

\sqrt{-64}

= ±  \sqrt{64(-1)}

= ± \sqrt{64} × \sqrt{-1} = ± 8i


6 0
3 years ago
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Option B

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Step-by-step explanation:

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If a number is a nagative, then it's absolute value is nagative?
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8 0
3 years ago
I need help with this question.
Paul [167]
All we need to do is plug in the values and check.
15*60 + 10 * 20 ≥ 1100.
Is this statement true?
15 * 60 = 900
20 * 10 = 200
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Yes! So, (60, 20) is a solution because 1100 is greater than or equal to 1100.
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Anyone know the answer? Pls help
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