The answer is "i" Tada!
Answer 1:The only answer is π^-1iThe square root of -1 is i so i times the square root of pi is equal to the square root of negative piAnswer 2:More concretely, the square root of negative pi is approximately 1.772453850905i, wherein the 1.772453850905 represents an approximation of pi, and "i", which should be a lower case cursive "i", represents the square root of negative one.
Raising to the one half (1/2) power is the same as taking the square root.
"i" stands for imaginary. It represents the square root of -1.
A square root can be represented by the rational exponent of ( \frac{1}{2} ). For any non-negative number ( x ), the square root is expressed as ( x^{1/2} ). This means that taking the square root of ( x ) is equivalent to raising ( x ) to the power of ( \frac{1}{2} ).
The square root of -49 can be expressed using imaginary numbers. Specifically, it is equivalent to 7i, where "i" represents the imaginary unit defined as the square root of -1. Thus, the answer is 7i.
Which inequality represents all values of x for which the quotient below is defined?square root of 28(x-1) divided by square root of 8x^2
The square root of the expression (2x + 1) is written as (\sqrt{2x + 1}). If you meant to include the number 3 in the expression, please clarify how it relates to (2x + 1). Otherwise, (\sqrt{2x + 1}) represents the principal square root of the expression.
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The square root of 1/9 is 1/3 because the square root of 1 is 1 and the square root of 9 is 3.
Square root of 1/2 = (1)/(square root of 2) = 1/1.4142 = 0.7071 Also Square root of 1/2 = Square root of 0.5 = 0.7071
square root of (x2 + 1) = no simplification (square root of x2) + 1 = x + 1
x2+3i=0 so x2=-3i x=square root of (-3i)=square root (-3)square root (i) =i(square root(3)([1/(square root (2)](1+i) and i(square root(3)([-1/(square root (2)](1+i) You can multiply through by i if you want, but I left it since it shows you where the answer came from. Note: The square root of i is 1/square root 2(1+i) and -1/square root of 2 (1+i) to see this, try and square them!