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zahabaha

  • 2 years ago

Consider the functions f(x) = 8x + 2 and g(x) = 5x − 3. What is the value of ( )(4)? Please show all of your work for full credit.

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  1. zahabaha
    • 2 years ago
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    URRI AND ANNAS DO U NO THIS STUFF??

  2. zahabaha
    • 2 years ago
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    IM BAD AT IT

  3. uri
    • 2 years ago
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    Yes,let me try.

  4. uri
    • 2 years ago
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    fog or gof

  5. zahabaha
    • 2 years ago
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    Consider the functions f(x) = 8x + 2 and g(x) = 5x − 3. What is the value of (F/G )(4)? Please show all of your work for full credit.

  6. zahabaha
    • 2 years ago
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    THERE THIS IS THE RIGHT VERSION OF THE QUESTION :)

  7. uri
    • 2 years ago
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    |dw:1342122906892:dw|

  8. uri
    • 2 years ago
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    this is f/g as u asked

  9. zahabaha
    • 2 years ago
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    THANK U

  10. uri
    • 2 years ago
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    welcome

  11. uri
    • 2 years ago
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    @annas its already done

  12. zahabaha
    • 2 years ago
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    WAIT THOUGH U DO 34/7 RIGHT

  13. uri
    • 2 years ago
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    Sorry? @zahabaha

  14. zahabaha
    • 2 years ago
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    UM U NO HOW AT THE END U GOT 34/7 DO U DIVIDE 34 BY 7

  15. uri
    • 2 years ago
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    YES.@zahabaha

  16. Krabsen
    • 2 years ago
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    i get 2 when i make it

  17. zahabaha
    • 2 years ago
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    OK

  18. zahabaha
    • 2 years ago
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    IS TWO RIGHT

  19. Krabsen
    • 2 years ago
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    f(x) = 8x + 2=34 g(x) = 5x − 3=17 34/17=2

  20. uri
    • 2 years ago
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    @Krabsen How are you getting 2?

  21. zahabaha
    • 2 years ago
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    I GOT 4.85 URI ARE U SURE U DIVIDE 34 BY 7

  22. uri
    • 2 years ago
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    @Krabsen Can you show me how 17 is coming? :/

  23. Krabsen
    • 2 years ago
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    yes g(4)=5x4 -3

  24. zahabaha
    • 2 years ago
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    f(x) = 8x + 2=34 g(x) = 5x − 3=17 34/17=2

  25. uri
    • 2 years ago
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    @Krabsen Oh my mistake! sorry i though it was 2 xd.

  26. Krabsen
    • 2 years ago
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    :)

  27. zahabaha
    • 2 years ago
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    KRABSEN CAN U HELP ON THIS ONE TEW Consider the functions f(x) = 3x^2 and g(x) = 2(x + 9). Find f[g(−7)]. Please show all of your work for full credit.

  28. uri
    • 2 years ago
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    Ask in question box @zahabaha

  29. zahabaha
    • 2 years ago
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    OK

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