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  • one year ago

Gaming systems are on sale for 20% off the original price (g), which can be expressed with the function p(g) = 0.8g. Local taxes are an additional 12% of the discounted price (p), which can be expressed with the function c(p) = 1.12p. Using this information, which of the following represents the final price of a gaming system with the discount and taxes applied? c(p) + p(g) = 1.92g c[p(g)] = 0.896g g[c(p)] = 1.92p c(p) ⋅ p(g) = 0.896pg

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  1. help_people
    • one year ago
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    @triciaal @mathstudent55 @welshfella

  2. mathstudent55
    • one year ago
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    p(g) gives you the discounted price. You need to apply function p first. Then c(p) gives you the final price after the tax is added to the discounted price. You apply function c after function p.

  3. triciaal
    • one year ago
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    start with the original price and take the 20% off . this is now the "x" to use to figure the taxes

  4. mathstudent55
    • one year ago
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    That means you need: c(p(g))

  5. mathstudent55
    • one year ago
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    First, let's write c(p) |dw:1437507064770:dw|

  6. help_people
    • one year ago
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    ok

  7. mathstudent55
    • one year ago
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    Now to find c(p(g)), you replace p(g) by what the function p(g) is equal to. |dw:1437507122077:dw|

  8. mathstudent55
    • one year ago
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    Now just multiply out 1.12 * 0.8 to simplify c(p(g))

  9. help_people
    • one year ago
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    0.896 @mathstudent55

  10. mathstudent55
    • one year ago
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    |dw:1437507412489:dw|

  11. mathstudent55
    • one year ago
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    Correct.

  12. help_people
    • one year ago
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    thank you do you mind helping me w. one more

  13. mathstudent55
    • one year ago
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    That means if you take each original price and just multiply it by 0.896, you are discounting the price by 20% and adding a 12% tax. With one simple multiplication, you are discounting and adding tax. For example if a system had an original price of $100, the final price after the discount and tax is 0.896 * $100 = $89.60

  14. help_people
    • one year ago
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    OK :)

  15. mathstudent55
    • one year ago
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    yes, but pls start a new post

  16. mathstudent55
    • one year ago
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    You're welcome.

  17. help_people
    • one year ago
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    OK (again)

  18. help_people
    • one year ago
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    thank you :) if i did not say it srry

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