anonymous
  • anonymous
Charlie's Computer Company charges $0.65 per pound to ship computers. Part A: Write an equation to determine the total cost, c, to ship p pounds of computers. Use your equation to determine the cost of shipping 2 pounds of computers. (6 points) Part B: If the company reduces the cost to ship computers by 0.05 per pound, write an equation to determine the total cost, c, to ship p pounds of computers with the reduced cost. (4 points)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
hint: if one pound costs 0.65 dollars, then p pounds cost: 0.65*p
anonymous
  • anonymous
are we doing part a?

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Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
okay.. 0.65p ?
Michele_Laino
  • Michele_Laino
correct! \[c = 0.65 \cdot p\]
anonymous
  • anonymous
so what do we say for part a?
Michele_Laino
  • Michele_Laino
we can say this: "the requested equation, is: \(c = 0.65 \cdot p\)"
anonymous
  • anonymous
ok and part b?
Michele_Laino
  • Michele_Laino
now, we have to replace p with 2, so we get: \[c = 0.65 \cdot p = 0.65 \cdot 2 = ...?\]
anonymous
  • anonymous
i dont know that one :(
Michele_Laino
  • Michele_Laino
it is simple: c=0.65*2=1.3 dollars
anonymous
  • anonymous
oh okay.. :)
Michele_Laino
  • Michele_Laino
for part B) if one pound is shipped for 0.05 dollars, then p pounds are shipped for: 0.05*p
anonymous
  • anonymous
what do we say for part b?
Michele_Laino
  • Michele_Laino
we can write this statement: "the cost to ship \(p\) pounds, with the reduced unitary cost is \(c=0.05 \cdot p\)"
anonymous
  • anonymous
so whats part A and Part B... please include all of the work :)
Michele_Laino
  • Michele_Laino
summarizing, we can write this: part A) requested formula is \(c=0.65 \cdot p \) so, specializing to p=2, we have: \(c=0.65 \cdot 2=1.30$\) part B) the new formula, using the reduced unitary cost, is: \(c=0.05 \cdot p \)
anonymous
  • anonymous
ok thanks :) i have more questions :)
Michele_Laino
  • Michele_Laino
please wait, since someone has tagged me
anonymous
  • anonymous
okay ;)

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