30 points which pair of complex numbers has a real-number product? (1 + 2i)(8i) (1 + 2i)(2 – 5i) (1 + 2i)(1 – 2i) (1 + 2i)(4i)

calculating each of the products

noting that i² = (√- 1 )² = - 1

(1 + 2i)(8i) ( distribute by 8i )

= 8i + 16i² = - 16 + 8i ← complex number

(1 + 2i)(2 - 5i) ( expand using FOIL )

= 2 - 5i + 4i - 10i²

= 2 - i + 10 = 12 - i ← complex number

(1 + 2i)(1 - 2i) ( expand using FOIL )

= 1 - 2i + 2i - 4i²

= 1 + 4 = 5 ← real number

(1 + 2i)(4i) ( distribute by 4i )

= 4i +8i² = - 8 + 4i ← complex number

(1 + 2i)(1 - 2i) is the only product which results in a real number

yeet

step-by-step explanation:

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!