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I assume that "I" is a variable

2+5i+6+3i
7i+6+3i
10i+6
16i is the answer

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15y ago

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8 plus 6i-2 plus 3i?

(8+6i)-(2+3i)=6+3i 8+6i-2+3i=6+9i


P and q are complex number's where p equals 2 plus 5i and q equals 6 minus i - find pq?

p = (2 + 5i)q = (6 - i)pq = (2 + 5i) (6 - i)pq = 12 + 30i - 2i - 5i2pq = 12 + 28i + 5pq = 17 + 28iIts the same as multiplying polynomials. Just multiply all the combinations of terms, group, and simplify.


What is the complex conjugate -6-5i?

6+5i


Where do the parentheses go in this problem 12 plus negative 6 divided by negative five plus 2?

it would go inbetwwen 12 and negative 5i hoped this helped u /\__/\( + + )| |-----------\| \( )---------( )\ \| | | | \ \___/__| /__| \--------/\|Kitty


What plus what is -6?

Choose ANY number, for example, 5, -100, square root of 2, pi, or (5 + 3i). Subtract -6 minus your number, to get the other number.


What is 2i times 3i?

To multiply (2i) by (3i), you can use the property of multiplication for complex numbers. Multiplying the coefficients gives (2 \times 3 = 6), and multiplying the imaginary units results in (i \times i = i^2), which is equal to (-1). Therefore, (2i \times 3i = 6i^2 = 6(-1) = -6). Thus, the result is (-6).


What two numbers multiply to 45 and add to 12?

57


How do you factor and equation 2X2-6X plus 9?

The best way to factor this is to use the quadratic equation. Using an equation of the form: ax2 + bx + c = 0 You use the quadratic formula like this: x = [-b +(or -) sqrt(b2-4ac)]/2a For your problem, a = 2, b = -6, and c = 9. Therefore: x = [6 +(or -) sqrt(36 - 4(2)(9))]/4 = [6 +(or -) sqrt(36-72)]/4 = [6 +(or -) sqrt(-36)]/4 = (6 +(or -)6i)/4 = 3/2 +(or -)3i/2 Therefore the factorization is (x - 3/2 - 3i/2)(x - 3/2 + 3i/2).


What is the magnitude of -1 - 5i?

-6


What is the conjugate of 6-5i?

The idea here is to change the sign before the imaginary term. In this case, since there is a minus, you change it to a plus.


What is 6 plus 6 plus 6 plus 2 plus 6 plus 3 plus 6 plus 2 plus 8 plus 6 plus 6 plus 3 plus 6 plus 2 plus 3 plus 6 plus 6 plus 2 plus 2 plus 2 plus 6 plus 6 plus 6 plus 5 plus 6 plus 6 plus 6 plus 2 p?

132


How do you do the math problem (4-3i)(5 2i)?

To multiply complex numbers you can use the same FOIL rule that you use for multiplying binomials (First, Inside, Outside, Last).(4 - 3i)(5 + 2i) = (4)(5) +(4)(2i) - (3i)(5) - (3i)(2i) = 20 + 8i-15i - 6(i)^2= 20 -7i - 6(-1) = 20 + 6 -7i = 26 -7i.