Q: What is equal to the sqrt of negative 16?

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-sqrt(48/3) = -sqrt(16) = -4 Technically, sqrt(16) = +4 or -4 so -sqrt(16) = -4 or +4

-3*sqrt(48) = -3*sqrt(16*3) = -3*sqrt(16)*sqrt(3) = -3*4*sqrt(3) = -12*sqrt(3)

-16+(-10) = -26

There are no 'real' numbers that can do that. The numbers are 3 + j sqrt(3) and 3 - j sqrt(3). ( ' j ' is the square root of negative 1 )

sqrt(160) = sqrt(16*10) = sqrt(16)*sqrt(10) = 4*sqrt(10)

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-sqrt(48/3) = -sqrt(16) = -4 Technically, sqrt(16) = +4 or -4 so -sqrt(16) = -4 or +4

-3*sqrt(48) = -3*sqrt(16*3) = -3*sqrt(16)*sqrt(3) = -3*4*sqrt(3) = -12*sqrt(3)

sqrt(80/45) = sqrt(16/9) = sqrt(16)/sqrt(9) = 4/3 sqrt(80/45) = sqrt(16/9) = sqrt(16)/sqrt(9) = 4/3 sqrt(80/45) = sqrt(16/9) = sqrt(16)/sqrt(9) = 4/3 sqrt(80/45) = sqrt(16/9) = sqrt(16)/sqrt(9) = 4/3

A negative index is the positive index applied to the reciprocal. So, 16^(-3/2) = (1/16)^(3/2) The denominator of a fractional index represents the relevant root. Thus, (1/16)^(3/2) = {sqrt(1/16)}3 Now, sqrt(1/16) = sqrt(1)/sqrt(16) = 1/4 So the given number is {1/4}^3 = 1/64.

-16+(-10) = -26

-24 = 4 x 6 x -1 Sqrt -24 = sqrt 4 x sqrt 6 x sqrt -1 = 2 x sqrt 6 x i = 2i root6

There are no 'real' numbers that can do that. The numbers are 3 + j sqrt(3) and 3 - j sqrt(3). ( ' j ' is the square root of negative 1 )

121 < 128.3 < 144So -12 < negative sqrt(128.3) < -11and11 < sqrt(128.3) < 12121 < 128.3 < 144So -12 < negative sqrt(128.3) < -11and11 < sqrt(128.3) < 12121 < 128.3 < 144So -12 < negative sqrt(128.3) < -11and11 < sqrt(128.3) < 12121 < 128.3 < 144So -12 < negative sqrt(128.3) < -11and11 < sqrt(128.3) < 12

sqrt(160) = sqrt(16*10) = sqrt(16)*sqrt(10) = 4*sqrt(10)

The above is a statement, not a question.

Since the numbers are complex, let's re-write the question as:i*sqrt(4) * i*sqrt(9)i^2, by the good graces of algebra, is equal to -1, so your answer simplifies to:= -1 * sqrt(4) * sqrt(9)= -1 * 2 * 3= -6

112 = 16*7 So sqrt(112) = sqrt(16*7) = sqrt(16)*sqrt(7) = 4*sqrt(7)