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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(fh)(x)=2x+5 (7x/3) (fh)(4)=2*4+5 (74/3)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok. 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why are you multiplying anything against (3/3)? \[(2)(4) +5(74/3)\] \[8+5 (214)/3\] \[1317/3\] \[39/317/3=22/3\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0remember that 55 doesn't mean (5)(5).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I forgot to do the 7*3 to make it 21/3 and 4/3... so I just subtracted 74 and divided that by 3. Thanks! I get so flustered with fractions that I make stupid mistakes!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, it happens a lot. Negative signs are everyone's bane.
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