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- What is all of the real and imaginary zeros of #y= (x^2-9 . . . - Socratic
We have #4# zeros, #3# with multiplicity #3# and #-3,3i# and #-3i# with multiplicity of #1#
- Question #b8278 - Socratic
Perhaps if we rewrite this as: (3+2i)* (1-3i) then we mulitply 3 from the first paranthesis with the both constituents from the second We do the same thing for the second constituent from the first paranthesis
- Question #bea27 - Socratic
(a) Graph is available in the explanation (b) Modulus is ~~5 8, after rounding to the tenths place Given: the Complex Number : (5-3i) A complex number is of the form a+bi So, for our complex number ** (5-3i), a=5 and b = (-3) a Graph of the Complex Number is below: b Modulus of the Complex Number is: A Complex number of the form a+bi, The Modulus of a Complex number can be found using
- How do you evaluate (3a -9i +2ai +6) (a^2+9) + (3-9i+3i+9 . . . - Socratic
Explanation: The first thing we notice with the two expression here is that the denominators are the same since #a^2+9=9+a^2#
- How do you evaluate \frac { 2i ^ { - 40} + 3i - Socratic
How do you evaluate 2i−40 + 3i−61 i87 − 2i84?
- How do you divide (3i) (1+i) + 2 (2+3i) ? | Socratic
How do you divide 3i 1 + i + 2 2 + 3i? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 2 Answers Sihan Tawsik
- -3i and 3i are . . . . . Of each other? a. Additive inverses B . . . - Socratic
B Conjugates a+bi and a-bi, wherea and b are real numbers, are conjugate of each other and if a=0, bi and -bi are conjugate of each other and hence so are -3i and 3i
- How do you combine like terms in #6- ( 4- 3i ) - ( - 2- 10i )#?
See the entire solution process below: First, remove all of the terms from parenthesis Be careful to handle the signs of each individual term correctly: 6 - 4 + 3i + 2 + 10i Next, group like terms: 10i + 3i + 6 - 4 + 2 Now, combine like terms: (10 + 3)i + (6 - 4 + 2) 13i + 4
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