The conjugate of 6 + i is 6 - i.
the conjugate 7-2i
How about "-i" since "i" is just "0 + i" so the conjugate should be "0 - i" or "-i"
semi conjugate diameter of ellipse
If you are referring to conjugate acids and bases, a conjugate acid is an acid that can donate a H+ in order to form a conjugate base. For example, HCl can donate it's H+ and create the conjugate base Cl-. On the other hand, a conjugate base would just be the opposite where chloride could add a hydrogen in order to create the conjugate acid.
The conjugate acid of ClO- is HClO. The conjugate acid of HClO is ClO2. The conjugate acid of HCI is H2Cl. The conjugate acid of Cl- is HCl. The conjugate acid of ClO is HClO2.
The conjugate base and conjugate acid for HS04 is: Conjugate acid is H2SO4 Conjugate base is SO42
"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 acid of NO2- is HNO2 (nitrous acid).
The conjugate acid of LiOH is considered Li+.
The conjugate acid of H2O is H3O+ (hydronium ion). When an acid donates a proton, it forms its conjugate base, and when a base accepts a proton, it forms its conjugate acid.
The conjugate of 6 + i is 6 - i.
It depends on what the denominator was to start with: a surd or irrational or a complex number. You need to find the conjugate and multiply the numerator by this conjugate as well as the denominator by the conjugate. Since multiplication is by [conjugate over conjugate], which equals 1, the value is not affected. If a and b are rational numbers, then conjugate of sqrt(b) = sqrt(b) conjugate of a + sqrt(b) = a - sqrt(b), and conjugate of a + ib = a - ib where i is the imaginary square root of -1.
The conjugate base of NH3 is NH2-, formed by removing a proton (H+) from NH3.
The conjugate base of HSO3- is SO32-.
the conjugate 7-2i
The conjugate base for CH3CH2COOH is CH3CH2COO-.