-8i
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
To simplify the expression (2i \cdot 6u), you multiply the coefficients and combine the variables. This gives you (12iu). If there are no further simplifications or like terms to combine, the final simplified form is (12iu).
(-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.