Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

really comfused with the acts. Check it out please. (conjugate of z)*w=2+3x+(2x-3)i and (conjugate of w)*z=2+3x-(2x-3)i Find x when you know this :[(conjugate of z)*w-(conjugate of w)*z]/(2*i)=5*[(conjugate of z)*w+(conjugate of w)*z]/2

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

now when you say (conjugate of z)*w you mean 2+3x+(2x-3)i correct? so in the actual question you just replace the word with the equation am i getting that correct?
yes
okay give me a sec and ill work through it for you

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

thanks,I am waiting,step by step please cause I lose it for some reason. :/
Start with the right side \[5(2+3x+(2x-3)i)+(2+3x-(2x-3)i)\]/2 \[10+15x+(10x-15)i+10+15x-(10x-15)i\] \[(20+30x)/2\] left side \[2+3x+(2x-3)i-(2+3x-(2x-3)i)\] \[[(2x-3)i+(2x-3)i]/[2i]\] in total we then have \[[(2x-3)i+(2x-3)i]/[2i] = (20+30x)/2\]
now are you trying to solve for x? or just simplyfy
I found it,thanks a lot.

Not the answer you are looking for?

Search for more explanations.

Ask your own question