anonymous
  • anonymous
A rectangle for a pet is 4 ft longer than it is wide. Give the possible values for the width W of the pen if its area must be between 165 and 480 sq ft, inclusively . so what would the widths of the smaller and larger values be?
Mathematics
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anonymous
  • anonymous
A rectangle for a pet is 4 ft longer than it is wide. Give the possible values for the width W of the pen if its area must be between 165 and 480 sq ft, inclusively . so what would the widths of the smaller and larger values be?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
A rectangle for a pet is 4 ft longer than it is wide It means that if the width is X the length is X+4
radar
  • radar
Solve for area. A= lw. Solve sides for 165. Solve for 480. This will give you the range of the sides. x(x+4)=165, or x^2 +4x=165 x^2+4x-165=0 factor (x+15)(x-11)=0. roots are x=-15, x=11 choose x=11, then x+4 =15 the sides are 11 and 15 for the 165 area
radar
  • radar
Now solve for the larger area of 480 sq ft x(x+4)=480 x^2 + 4x=480 x^2+4x-480=0 (x+24)(x-20)=0 x=-24, x=20 choose x=20 x+4=24. sides are 20 and 24 ft. To express x for the range of area would be \[11\le x \le 20\]

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anonymous
  • anonymous
Thank you for explaining that unlike the person whom posted before you!

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