The illustration shows how a geometric wall hanging can be created by stretching yarn from peg to peg across a wooden ring. The relationship between the number of pegs p placed evenly around the ring and the number of yarn segments s that crisscross the ring is given by the formula p(p – 3) S = How many pegs are needed if the designer wants 44 segments to crisscross the ring? (Hint: Multiply both sides of the equation by 2.) See Example 1. p = pegs

College Algebra
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ISBN:9781938168383
Author:Jay Abramson
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Chapter2: Equations And Inequalities
Section2.3: Models And Applications
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The illustration shows how a geometric wall hanging can be created by stretching yarn from peg to peg across a wooden ring. The relationship between the number of pegs p placed evenly around the
ring and the number of yarn segments s that crisscross the ring is given by the formula
Р(р — 3)
S =
2
How many pegs are needed if the designer wants 44 segments to crisscross the ring? (Hint: Multiply both sides of the equation by 2.) See Example 1.
pegs
Transcribed Image Text:The illustration shows how a geometric wall hanging can be created by stretching yarn from peg to peg across a wooden ring. The relationship between the number of pegs p placed evenly around the ring and the number of yarn segments s that crisscross the ring is given by the formula Р(р — 3) S = 2 How many pegs are needed if the designer wants 44 segments to crisscross the ring? (Hint: Multiply both sides of the equation by 2.) See Example 1. pegs
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