A conjugate (in the context of math) is, simply put, two numbersseparatedby a sign, in which the sign changes.
Example:
The conjugate of a+b is a-b. Thepositivebecame negative.
For your problem (your sign is missing, so I listed two possibilities):
The conjugate of 11+17i is 11-17i
The conjugate of 11-17i is 11+17i
A conjugate (in the context of math) is, simply put, two numbersseparatedby a sign, in which the sign changes.
Example:
The conjugate of a+b is a-b. Thepositivebecame negative.
For your problem (your sign is missing, so I listed two possibilities):
The conjugate of 11+17i is 11-17i
The conjugate of 11-17i is 11+17i
11 + 17i is an IMAGINARY number. . Its conjugate is 11 - 17i NB Note the change of sign only.
To find the complex conjugate change the sign of the imaginary part: For 11 + 5i the complex conjugate is 11 - 5i.
There is not enough information. You can't calculate one root on the basis of another root. HOWEVER, if we assume that all the polynomial's coefficients are real, then if the polynomial has a complex root, then the complex conjugate of that root (in this case, 4 - 17i) must also be a root.
292
11
17 * rootover of -1
4-17i
HNO2 conjugate acid = one more hydrogen conjugate base = one less hydrogen
"Conjugate" usually means that in one of two parts, the sign is changed - as in a complex conjugate. If the second part is missing, the conjugate is the same as the original number - in this case, 100.
The conjugate base and conjugate acid for HS04 is: Conjugate acid is H2SO4 Conjugate base is SO42
The two solutions are the conjugate complex numbers, +i*sqrt(11) and -i*sqrt(11), where i is the imaginary square root of -1.
NH2- is the conjugate base of ammonia.