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amanda320

  • 2 years ago

I don't know if linear algebra was the right sub context but here it goes... how do I compute (f-h)(4) if f(x)=2x+5 and h(x)=7-x/3?

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  1. surdawi
    • 2 years ago
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    (f-h)(x)=2x+5 -(7-x/3) (f-h)(4)=2*4+5 -(7-4/3)

  2. amanda320
    • 2 years ago
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    so it's 2/3?

  3. surdawi
    • 2 years ago
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    22/3

  4. amanda320
    • 2 years ago
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    Ok. I do the stuff in the parenthesis first and come up with 2*4+5(-3/3), 2*4 is 8, 8+5=13(-3/3) which =-39/3 and that =-13. What did I do wrong?

  5. freewilly922
    • 2 years ago
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    Why are you multiplying anything against (-3/3)? \[(2)(4) +5-(7-4/3)\] \[8+5 -(21-4)/3\] \[13-17/3\] \[39/3-17/3=22/3\]

  6. freewilly922
    • 2 years ago
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    remember that 5-5 doesn't mean (5)(-5).

  7. amanda320
    • 2 years ago
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    I forgot to do the 7*3 to make it 21/3 and -4/3... so I just subtracted -7-4 and divided that by 3. Thanks! I get so flustered with fractions that I make stupid mistakes!

  8. freewilly922
    • 2 years ago
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    Yeah, it happens a lot. Negative signs are everyone's bane.

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