anonymous
  • anonymous
4/5(2x-1)>12
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Are you solving for x?
anonymous
  • anonymous
Excuse me, is it (4/5)*(2x-1)>12 or 4/[5(2x-1)]>12? could you make that clearer
anonymous
  • anonymous
if it is (4/5)*(2x-1)>12 then x>8; if it is 4/[5(2x-1)]>12 then x>(4/5)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Yes, solving for x, and the first of the options is what it is supposed to be. :Thank you. :)
anonymous
  • anonymous
It told me the answer was incorrect. :( Here is a similar equation, with as far as I can get on my own: 2/3(3x-6)>12 6/3x-12/3>12 12(6/3x-12/3)>12(12) (72/3)x-(144/3)>144 I can not seem to get to the answer from here. I'm sure it's probably easy, but I need it explained in simple terms.
anonymous
  • anonymous
2/3(3x-6)>12 3x-6>18 3x>24 x>8
anonymous
  • anonymous
Also needing assistance with compound inequalities. Will be posting one soon. I don't just want an answer. I want someone to teach me (simply) how to solve them, step-by-step. Note: The compound inequalities have some odd fractions, so I may have to explain them differently.
anonymous
  • anonymous
Dhat: Where did the 2/3 go?
anonymous
  • anonymous
I multiplied by 3/2 on both sides.
anonymous
  • anonymous
OK, I see. Thank you. :)
anonymous
  • anonymous
you are welcome
anonymous
  • anonymous
Would you like me to post the compund inequality here, or as a new question?
anonymous
  • anonymous
new thread please

Looking for something else?

Not the answer you are looking for? Search for more explanations.