57!
arg(-2-i) = sqrt[22 + 12] = sqrt(5)
This is a complex number, not an algebraic expression. The letter i represents the imaginary unit (which is equal to sqrt(-1)). Graphiclly, with real numbers on a horizontal axis, and imaginary numbers on a vertical axis, this means starting at the origin, go to the left 5 units, and then go down 12 units.
9 to 12 is the ratio. BUT, if you simplify it (divide them both by the same number which makes both of them come out to a whole number) it's 3 to 4
To simplify 158/12, divide 158 by 12. The answer is 13 remainder 2. 13 is a whole number. Now, pit the 2 over 12. It is 2/12, but it needs to be simplified, so divide the denominator and numerator by 2. The answer is 1/6. Put it with the 13 and the final answer is 13 1/6.
4/7
-12 = 4 x 3 x -1 so sqrt = 2i root 3
sqrt(-12/4) = sqrt(-3) = +/- i*sqrt(3) where i is the imaginary sqrt of -1.
6
- 24 sqrt(6) / 6 sqrt(12) = (-24/6) sqrt(6/12) = (-4) sqrt(1/2)
Factorise 12.Replace each pair appearing in this factorisation as by the same number outside the radical and then put everything under the radical sign. sqrt(12) = sqrt(2*2*3) = 2*sqrt(3)
26
sqrt(12) = sqrt(4*3) = sqrt(4)*sqrt(3) = 2*sqrt(3)
To find the square root of 12, you can simplify it by breaking it down into its prime factors. Since 12 can be expressed as (4 \times 3) (where 4 is a perfect square), the square root can be calculated as (\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}). If you need a decimal approximation, (\sqrt{12} \approx 3.464).
9
You can't simplify a number into a bigger number, stupid
-6 +/- i*sqrt(139) where i is the imaginary square root of -1.
To simplify ( 3 \sqrt{432} ), first factor ( 432 ) into its prime factors: ( 432 = 2^4 \times 3^3 ). Then, rewrite the square root: ( \sqrt{432} = \sqrt{2^4 \times 3^3} = \sqrt{(2^2)^2 \times (3^1)^2 \times 3} = 2^2 \times 3 \times \sqrt{3} = 12\sqrt{3} ). Finally, multiply by ( 3 ): ( 3 \sqrt{432} = 3 \times 12\sqrt{3} = 36\sqrt{3} ).