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Unit 5. Day 15.
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Modeling Proportional Relationships
Traditional Alternate
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Official Math Standard 7.RP.2.c
Represent proportional relationships by equations.Β For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn. $ πππππ $ πππππ 0.30 1.20 Γ·4 Three apples cost $1.20 4 Γ·4 π π‘ = 0.3 π β π π₯:πππππ π¦:πππ π‘ π¦ =0.3π₯
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Official Math Standard 7.RP.2.c
Represent proportional relationships by equations.Β For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn. 4 shirts cost $50. 25 2 50 4 $ π βπππ‘ 50 $ π βπππ‘ $ π βπππ‘ 50 $ π βπππ‘ 12.50 Γ·4 Γ·4 4 4 Γ·4 Γ·4 π₯:π βπππ‘π π¦:πππ π‘ 25 2 π‘ π π¦ = π₯ π β π π π‘ π¦ = 12.5 β π₯ π π
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Official Math Standard 7.RP.2.c
Represent proportional relationships by equations.Β For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn. 20 kilometers in 6 hours 20 6 10 3 20 ππ 1 βπ 3. 3 ππ βπ ππ βπ 20 Γ·6 ππ 1 βπ Γ·6 6 6 Γ·6 Γ·6 π₯:π‘πππ (βππ ) π¦:πππππ 10 3 π¦ π = β π₯ π‘ π π¦ = 3. 3 β π₯ π‘
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Official Math Standard 7.RP.2.c
Represent proportional relationships by equations.Β For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn. 4 pizzas were ordered to feed 10 teens 4 10 2 5 πππ§π§π π‘πππ 4 πππ§π§π 1 π‘πππ πππ§π§π π‘πππ 4 0.4 πππ§π§π 1 π‘πππ Γ·10 Γ·10 10 10 Γ·10 Γ·10 π₯:π‘ππππ π¦:πππ§π§π 2 5 π = β π‘ π π¦ = 0.4 β π¦ π₯ π₯ π‘
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Modeling Proportional Relationships
Traditional Alternate
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If we have a proportional relationship
And we want to represent it with an equation (model) We have two options to create our equation. Find Unit Rate Cross Multiply
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π¦= π₯ π Three apples cost $1.20 1.2 3 = π¦ π₯ π₯:ππππππ π¦:πππ π‘ π=0.4π 1.20
Example A: Three apples cost $1.20 Find Unit Rate Cross Multiply $ πππππ = $ πππππ 1.20 π¦ 3 π₯ 1.20 0.40 $ πππππ Γ·3 $ πππππ = π¦ π₯ 3 Γ·3 3 π¦ = 1.2 π₯ π¦= π₯ π 0.4 3π¦ = 1.2π₯ 3 3 π₯:ππππππ π¦:πππ π‘ π=0.4π π¦ = 0.4π₯
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π¦= π₯ π We walked 10 miles in 4 hours. 10 4 = π¦ π₯ π₯:π‘πππ π= 5 2 π‘
Example B: We walked 10 miles in 4 hours. Find Unit Rate Cross Multiply ππ βπ = ππ βπ 10 π¦ 4 π₯ 10 4 5 2 10 ππ βπ Γ·4 ππ 1 βπ = π¦ π₯ 4 Γ·4 4 π¦ = 10 π₯ 5 2 π¦= π₯ π 4π¦ = 10π₯ 4 4 π₯:π‘πππ π¦:πππ π‘ππππ π= 5 2 π‘ 10 4 π₯ 5 2 π₯ π¦ =
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Example C*: Mr. Jordan paid $6 for eight donuts.
Model each situation with an equation Example C*: Mr. Jordan paid $6 for eight donuts. Example D*: We rode a bike miles in hour. Example E*: A truck uses gallons to travel 45 miles.
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π¦= π₯ π Mr. Jordan paid $6 for eight donuts. 6 8 = π¦ π₯ π₯:ππππ’π‘ π= 3 4 π
Example C*: Mr. Jordan paid $6 for eight donuts. Find Unit Rate Cross Multiply 6 πππ π‘ ππππ’π‘ = πππ π‘ ππππ’π‘ π¦ 8 π₯ 6 8 3 4 6 $ ππππ’π‘ Γ·8 $ ππππ’π‘ 6 8 = π¦ π₯ 8 Γ·8 8 π¦ = 6 π₯ 3 4 π¦= π₯ π 8π¦ = 6π₯ 8 8 π₯:ππππ’π‘ π¦:πππ π‘ π= 3 4 π 3 4 π₯ 6 8 π₯ π¦ =
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π¦= π₯ π We rode a bike 3 1 4 miles in 1 2 hour. π= 13 2 π‘ π₯:π‘πππ
Example D*: Find Unit Rate Cross Multiply 13 4 π¦ ππ βπ = ππ βπ π₯ 1 2 3 1 4 13 4 Γ· 1 2 26 4 13 2 ππ βπ ππ 1 βπ = π¦ π₯ 1 2 Γ· 1 2 1 2 π¦ π₯ = 1 2 π¦ 13 4 π₯ 13 2 π¦= π₯ π = 1 2 1 2 π= 13 2 π‘ π₯:π‘πππ π¦:πππ π‘ππππ (ππ) 13 2 π₯ 26 4 π₯ π¦ =
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π¦= π₯ π A truck uses 2 1 4 gallons to travel 45 miles. π=20π π₯:πππππππ
Example E*: Find Unit Rate Cross Multiply 45 ππ πππ = ππ πππ π¦ 9 4 π₯ Γ· 9 4 180 9 20 45 ππ πππ ππ 1 πππ = π¦ π₯ 2 1 4 9 4 Γ· 9 4 9 4 π¦ = 45 π₯ 9 4 π¦ π¦= π₯ π = 45π₯ 20 9 4 9 4 π=20π π₯:πππππππ π¦:πππ π‘ππππ (ππ) 180 9 π₯ π¦ = 20π₯
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