6+2i
(-2 + 3i) + (-1 - 2i) = -2 + 3i - 1 - 2i = -2 - 1 + 3i - 2i = -3 + i
10 + 6i and 7 + 2i = 10 + 6i + 7 + 2i = 17 + 8i
It is imaginary; we denote sqrt of -1 = i square root (-4) = ± 2i
f(x)=x3-3x2-5x+39=(x+3)(x2-6x+13) It has three roots. One of which is x=-3. Using the quadratic equation: x = (6 +/- √(-16))/2 x = (6 +/- 4i)/2 = (3 +/- 2i) so, x=-3, x=3+2i, or x=3-2i
6+2i
the conjugate 7-2i
0.6-2i?
It is 3 minus 2i
(-2 + 3i) + (-1 - 2i) = -2 + 3i - 1 - 2i = -2 - 1 + 3i - 2i = -3 + i
Using the quadratic formula, you get the complex answers of 1 + 2i and 1 - 2i
10 + 6i and 7 + 2i = 10 + 6i + 7 + 2i = 17 + 8i
It is imaginary; we denote sqrt of -1 = i square root (-4) = ± 2i
Multiply the numerator and denominator by the complex conjugate of the denominator ... [ root(2) minus i ]. This process is called 'rationalizing the denominator'.
Yes, the reaction 2I to I2 is endothermic because it requires energy to break the bonds between the I atoms in 2I and form the I2 molecule. This process absorbs heat from the surroundings.
There are infinitely many solutions for this. For example: 6 - 0 7 - 1 8 - 2 6.5 - 0.5 5 - (-1) (8 + 2i) - (2 + 2i) etc.
The answer is 2i. When dealing with negative square roots, the expression i is used to represent the square root of -1.