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3. Side = x + 5 P = 3(side) P = 3(x + 5) P = 3x + 15 4. Side = 2x – 1 P = 4(side) P = 4(2x – 1) P = 8x – 4 5. Side = 2x + 3 P = 5(side) P = 5(2x + 3) P.

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Presentation on theme: "3. Side = x + 5 P = 3(side) P = 3(x + 5) P = 3x + 15 4. Side = 2x – 1 P = 4(side) P = 4(2x – 1) P = 8x – 4 5. Side = 2x + 3 P = 5(side) P = 5(2x + 3) P."— Presentation transcript:

1 3. Side = x + 5 P = 3(side) P = 3(x + 5) P = 3x Side = 2x – 1 P = 4(side) P = 4(2x – 1) P = 8x – 4 5. Side = 2x + 3 P = 5(side) P = 5(2x + 3) P = 10x Side = 2x – 3 P = 5(side) P = 5(2x – 3) P = 10x – Width = x Length = 3x P = 72 P = 2(width) + 2(length) 72 = 2(x) + 2(3x) 72 = 2x + 6x 72 = 8x 9 = x Width = 9, Length = st Side = x 2 nd Side = x – 2 3 rd Side = x + 12 P = 73 P = Side1 + Side2 + Side3 73 = x + x – 2 + x = 3x = 3x 21 = x Side 1 = 21, Side 2 = 19, Side 3 = 33 Homework Answers

2 12.Width = x Length = x + 5 P = 66 P = 2(width) + 2(length) 66 = 2(x) + 2(x + 5) 66 = 2x + 2x = 4x = 4x 14 = x Width = 14, Length = Width = x – 3 Length = x P = 130 P = 2(width) + 2(length) 130 = 2(x – 3) + 2(x) 130 = 2x – 6 + 2x 130 = 4x – = 4x 34 = x Width = 31, Length = 34 Homework Answers

3 Lesson 25 Word Problems – Coin/Money

4 Coin/Money Word Problems Coin/Money problems are similar to Perimeter problems. Use information about one coin to create an expression for another coin. Multiply each coin expression by their worth.

5 Example of Coin problem In a collection of coins, there are 7 less dimes than quarters. If the total of the coin collection is $3.85, how many dimes and quarters are there? Step 1. Create expressions for each coin. Dimes = x – 7 Quarters = x

6 Step 2. Multiply each expression by their value and create and equation. Dimes =.10(x – 7) Quarters =.25x $3.85 =.10(x – 7) +.25x

7 Step 3. MOVE THE DECIMAL and solve the equation =.10(x – 7) +.25x 385 = 10(x – 7) + 25x 385 = 10x – x 385 = 35x – = 35x 35 = = x

8 Step 4. Go back to the expressions and find out how many coins there are. Dimes = x – 7 (13) – 7 = 6 Quarters = x 13 So there are 6 Dimes and 13 Quarters

9 Another Problem A collection of coins is worth $ If the number of nickels is 6 less than twice the number of dimes and the number of quarters are 6 more than twice the number of dimes, how many nickels are there?


Download ppt "3. Side = x + 5 P = 3(side) P = 3(x + 5) P = 3x + 15 4. Side = 2x – 1 P = 4(side) P = 4(2x – 1) P = 8x – 4 5. Side = 2x + 3 P = 5(side) P = 5(2x + 3) P."

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