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mikala1 Group Title

here @Arhin can some one plz help me

  • 2 years ago
  • 2 years ago

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  1. mikala1 Group Title
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    • 2 years ago
  2. mikala1 Group Title
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    @calculusfunctions could you help me i got premission to ask some one on here

    • one year ago
  3. mikala1 Group Title
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    Do you know how to do this @sue101

    • one year ago
  4. mikala1 Group Title
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    ok all i need help on is numbers 1 2 and 3 i dont even know how to make the graph and my teacher says it is ok to ask for help on making it aslong as i am doing some of it can some plz help me

    • one year ago
  5. jazy Group Title
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    1) x – 3 > –9 Add 3 to both sides. 2) –2x + 1 ≤ –11 Subtract 1 from both sides. Divide by -2 and flip the sign to ≥. 3) 10 < –3x + 1 Subtract 1 from both sides. Divide by -3 and flip the sign to > **Whenever you divide by a negative, you flip the sign. To graph: Less/Greater Than (> or <)... Open Circle Less/Greater Than AND equal to (>= or <=)... Closed Circle Greater than... Shading to the Right Less than... Shading to the left.

    • one year ago
  6. mikala1 Group Title
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    thank you so much yuo helped alot no lie thank you

    • one year ago
  7. jazy Group Title
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    No problem. I'm glad I could help! (:

    • one year ago
  8. mikala1 Group Title
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    I only ned help on number 4 but they dident know it

    • one year ago
  9. mikala1 Group Title
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    4. 2(x + 5) > 8x – 8

    • one year ago
  10. calculusfunctions Group Title
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    Solve just as you would an equation. Go ahead.

    • one year ago
  11. calculusfunctions Group Title
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    Pretend there's an equal sign. 2(x + 5) = 8x - 8 Solve for x. What would you get for x @mikala1 ?

    • one year ago
  12. mikala1 Group Title
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    5x

    • one year ago
  13. mikala1 Group Title
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    right

    • one year ago
  14. calculusfunctions Group Title
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    No. Try again. Show me your steps so that I can see where you're going wrong.

    • one year ago
  15. calculusfunctions Group Title
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    Let me get you started with the first step.

    • one year ago
  16. mikala1 Group Title
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    i added 5 + x with pemdas

    • one year ago
  17. calculusfunctions Group Title
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    2(x + 5) = 8x - 8 2x + 10 = 8x - 8 Understand so far?

    • one year ago
  18. mikala1 Group Title
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    Ihave come out with this 1. First you multiply the number outside the parentheses by both variables inside the parentheses. 2 multiplied by x=2x and 2 multiplied by 5=10 Now you have 2x + 10 > 8x - 8 2. Next you have to combine like terms and get them on separate sides. (It's easier to work with positive numbers.) You want the X's on one side and the plain numbers on the other. 2x + 10 > 8x - 8 (8 is negative so you want to add it to the 10 on the other side. But remember: what you do to one side you do to the other.) 2x + 10 > 8x -8 + 8 + 8 (When you add 8 to the negative 8, it makes it 0 so you just get rid of it.) Now you have: 2x + 18 > 8x So now you subtract 2x from both sides: 2x + 18 > 8x -2x -2x You now have: 18 > 6x which means you have combined all the like terms. 3. Now you divide the 6x by 6 to get x by itself: 18 > 6x divided by 6= x You now have: 18 > x (Remember: what you do to on side you do to the other so you have to divide 18 by 6.) 18 divided by 6= 3 4. So when it's all finished it looks like this: 3 > x which basically means 3 is more than x.

    • one year ago
  19. calculusfunctions Group Title
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    Yes so why did you ask if you already solved it?

    • one year ago
  20. mikala1 Group Title
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    no i just relised that i was asking if you could help cheak it to see if it was correct

    • one year ago
  21. calculusfunctions Group Title
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    Yup! it is correct!

    • one year ago
  22. mikala1 Group Title
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    thank you and do you have any idea how to mak a line thinkg for this on a computer because im virtual schooled and idk how

    • one year ago
  23. calculusfunctions Group Title
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    |dw:1351853689483:dw| Open circle because there is no equal sign. Arrow to the left because everything less than 3. Understand?

    • one year ago
  24. mikala1 Group Title
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    yes i do

    • one year ago
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