11 + 17i is an IMAGINARY number. .
Its conjugate is 11 - 17i
NB Note the change of sign only.
11 minus 17i
The conjugate of 6 + i is 6 - i.
8 - 8i
The product is a^2 + b^2.
3 + 11 + 11 + 11 + 12 + 13 + 15 + 15 + 16 + 17 + 19 = 143
[ 2 minus square root of 5 ] is the only one.
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 11-17iThe conjugate of 11-17i is 11+17i
To find the complex conjugate change the sign of the imaginary part: For 11 + 5i the complex conjugate is 11 - 5i.
The conjugate of 6 + i is 6 - i.
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.
4-17i
NH3 is the conjugate base of NH4+
IN this reaction the conjugate base is the nitrate ion, NO3-
The two solutions are the conjugate complex numbers, +i*sqrt(11) and -i*sqrt(11), where i is the imaginary square root of -1.
The conjugate is 7-5i
When finding the conjugate of a binomial, you just reverse the sign. So the conjugate of 3+4i is 3-4i.
The concept of conjugate is usually used in complex numbers. If your complex number is a + bi, then its conjugate is a - bi.
The conjugate acid of any substance is given by removing an acidic hydrogen. In the case of ammonium ion, the conjugate base is ammonia.