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Vinil7 [7]
3 years ago
15

Cube roots of negative numbers exist in the set of real numbers, but square roots of negative numbers do not. Explain why this i

s true.
Mathematics
2 answers:
raketka [301]3 years ago
3 0
When problems with negatives under a square root first happened, mathematicians thought that there is no solution for the problem.<span>

</span>In an effort to answer this problem, mathematicians made a new number, i, which was denoted to as an "imaginary number", because it was not in the set of "Real Numbers".  The imaginary number "i" is √-1.
olga2289 [7]3 years ago
3 0
<span>Cube roots of negative numbers exist in the set of real numbers, but square roots of negative numbers do not because it is not possible to get Square roots of negative real numbers because they do not exist in the real numbers. but they do exist in the complex numbers, It is not possible to square a value and arrive at a negative value.</span>
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Can anyone help me with this math problem?
Vika [28.1K]
\bold{FULL ANSWERS:}

In 1993, Moose Population: 3280

In 1999, population became: 4960

P (Population) , t (years)

t = 6 —> 4960 - 3280 = 1680

Average Change —> 1680/6 = 280 moose/year

• In terms of 1990:

t = 3 —> 3280-3 (280)

P(1990) = 2440

P(t) = 2440 + 280t

• In 2003; t = 13

P(13) = 2440 + 280 (13)
P(13) = 2440 + 3640
P(13) = 6080

• Moose population in 2003

= 6080





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2 years ago
Guys Pls help <br><br><br><br><br><br> 2+2=??????????????????/
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3 years ago
Solve the system of equations and choose the correct ordered pair.
nasty-shy [4]
The answer is D

Because if you replace the variables just by testing them they give the answer as X=-3 and Y=3.
In which replacing:
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and with the other equation we've got:
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so they comply with the required to be the correct answer, even though you can make a system of equation.  


6 0
4 years ago
Write (4x8) divided by 16 as a squared number
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Answer: =

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Step-by-step explanation:

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