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anonymous

  • one year ago

Use the functions a(x) = 3x + 10 and b(x) = 2x − 8 to complete the function operations listed below. Part A: Find (a + b)(x). Show your work. (3 points) Part B: Find (a ⋅ b)(x). Show your work. (3 points) Part C: Find a[b(x)]. Show your work. (4 points)

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  1. Michele_Laino
    • one year ago
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    part A) hint: by definition, we have: \[\Large \left( {a + b} \right)\left( x \right) = a\left( x \right) + b\left( x \right)\]

  2. anonymous
    • one year ago
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    so would it be 5x+2

  3. Michele_Laino
    • one year ago
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    correct!

  4. anonymous
    • one year ago
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    how would i go about doing the other ones?

  5. Michele_Laino
    • one year ago
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    part B) again, by definition, we can write: \[\Large \left( {a \cdot b} \right)\left( x \right) = a\left( x \right) \cdot b\left( x \right)\]

  6. anonymous
    • one year ago
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    how would I do that

  7. Michele_Laino
    • one year ago
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    you have to multiply your function, like below: \[\Large \left( {a \cdot b} \right)\left( x \right) = a\left( x \right) \cdot b\left( x \right) = \left( {3x + 10} \right) \cdot \left( {2x - 8} \right) = ...\]

  8. Michele_Laino
    • one year ago
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    functions*

  9. anonymous
    • one year ago
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    so would it be 6x -80

  10. anonymous
    • one year ago
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    @Michele_Laino

  11. Michele_Laino
    • one year ago
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    I got 6x^2-4x-80

  12. anonymous
    • one year ago
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    I don't get it

  13. Michele_Laino
    • one year ago
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    you have to compute this: \[\large \left( {a \cdot b} \right)\left( x \right) = a\left( x \right) \cdot b\left( x \right) = \left( {3x + 10} \right) \cdot \left( {2x - 8} \right) = ...\] please use the "foil" method Part C) again, by definition, we get: \[\large a\left\{ {b\left( x \right)} \right\} = 3\left\{ {b\left( x \right)} \right\} + 10 = 3 \cdot \left( {2x - 8} \right) + 10=...\]

  14. anonymous
    • one year ago
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    so it would be 6x-14 right @Michele_Laino

  15. anonymous
    • one year ago
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    thank you so much for your help!!! @Michele_Laino

  16. Michele_Laino
    • one year ago
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    :)

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