Answers. Investigation 1. ACE Assignment Choices. Applications

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1 Answers Investigation ACE Assignment Choices Problem. Core,, Other Connections Problem. Core,, 5, Other Connections 7 ; Etensions 57, 5; unassigned choices from previous problems Problem. Core 5, Other Connections 5; unassigned choices from previous problems Problem. Core, 9,, 5 Other Applications 7,,, 5 7; Connections 5 5, Etensions 59; unassigned choices from previous problems Adapted For suggestions about adapting Eercise and other ACE eercises, see the CMP Special Needs Handbook. Connecting to Prior Units, 7 : Frogs, Fleas, and Painted Cubes;, 5: Covering and Surrounding; 5, 5 5: Moving Straight Ahead;, 7 5, 5: Accentuate the Negative; : Bits and Pieces II; 9: Prime Time; 55: Filling and rapping Applications. a. () + (5) + = tiles b. Possible equations: N = + + N = ( + ) + ( + ) N = ( + ) + N = + ( + ) c. See part (b) for some equations; eplanations will vary. Students might draw sketches. For eample: ( ) They might substitute values for and in the equations; for eample, when = and = : N = + + = () + () + = N = ( + ) + ( + ) = () + () = N = ( + ) + = (5) + () = N = + ( + ) = () + () =. a. (7) + (.5) = tiles b. Possible answers: N = s + N = (s +.5) N = s + (s + ) c. See equations in part (b). Students might substitute values for s [in this case values (s, N) are sufficient because these are linear relationships], generate tables for both equations, or make a geometric argument to show that the two equations are equivalent. They may also graph each equation. d. The relationship is linear; students may say that this is because the graphs are straight lines; the table increases by a constant value of for every increase of ft in the side length.. a. () + () + = tiles b. Possible answers: N = + + N = ( +.5) + ( +.5) N = ( + ) + ( ) ( ) ( ) ACE ANSERS Investigation Equivalent Epressions 7

2 c. Students might substitute values for and,create tables or graphs,or make geometric arguments to show that their two equations are equivalent.. a. First equation: ( + ) + = () + = ; 5. a. The shape is the area between the circle and the square. r Second equation: ( +.5) + ( +.5) = (.5) + (.5) = + = ; Third equation: [ ] = ( = ) b. You cannot determine whether the epressions are equivalent by checking them at one point, although students may think that they are equivalent since these epressions produced the same number of tiles for s =. c. First equation: ( + ) + = ( + ) + = ; Second equation: ( +.5) + ( +.5) = (.5) + (.5) = 5; Third equation: [ ] = = () = 5 ( ) ( ) ( ) d. Since you can determine non-equivalency of linear equations by checking one point, the first epression is not equal to the second and the third epressions because they did not produce the same number of tiles when you checked using the same side value. In general, it is not enough to show that two epressions are equivalent when they have the same value at two different points, because you need to check all points, which is impossible. However, for linear equations such as those in this problem, checking only two values would be enough because only one line can pass through the two points. So linear epressions which agree on two values (two points) contain the same two points. So, the lines that they represent must be the same. Students will either need to check all points, which is impossible, or know that two points uniquely determine a line. (This topic was addressed on the Summary Transparency for Problem..) b. The shape is all the area inside the square ecept a quarter of the area of the circle.. a. ii and iv b. i and iii 7. a. c. For part (a), ii and iv are equivalent since: (s - ) = (s - )(s - ) = s (s ) - (s - ) = s - s - s + = s - s +. For part (b), i and iii are equivalent because they both represent the same part of the pool. d. Answers will vary, but must be equivalent to A = (s - s + ) + (s - s) e. The equation in part (d) is a quadratic relationship. 5 5 y y y 5 O 5 r Say It ith Symbols

3 . a. 9. a. b. The epressions are equivalent because the table values are the same and the graph is a single line. NOTE: These are linear epressions so it is enough to show that they all pass through the same two points. c = = + (- + 5) = + b. The epressions are not equivalent because the table values are different and the graphs are separate lines; one has a negative slope and one has a positive slope. c. - 5 = y O ( ) ( ) y 5 y b. The epressions are equivalent because the table values are the same and the graph is a single line. NOTE: These are linear epressions so it is enough to show that they all pass through the same two points. c. ( + ) + ( - ) = = ( + ) + = 5 +. a. + b. 5-5 c. - d = + ( + ) + = + +. a. Possible answers: ( - 5) or ( - ) = - b. ( + ) c. 7( - ). a. equal; + 7 = ( + 7) = b. not equal; 5 - = (5 - ) = -5 5 c. equal; ( + ) - = + - = + 5 = 5 + Using the Commutative Property of Addition, 5 + = + 5. d. equal; 5 ( - ) = = - +. Step (): Distributive Property Step (): Commutative Property Step (): Distributive Property Step (): Addition. Possible answers: ( + ), + 5 +, (7 + 5)p p = p (p p) = 7 O 7. Parentheses are not needed. y y ( ) ( ) y 5 5 ACE ANSERS Investigation Equivalent Epressions 9

4 Connections.. 9. ( + ) = + ( ) + = ( + ) or 7. a. Area of water = p() = p < 5 ft b. Area of border = p(5 ) - p( ) = 5p - p = 9p < ft c. Area of water = pr d. Area of border = p(r + ) - pr,or pr + p 5. B. J 7. ( - ) = -. 5 ( + )( + ) = or ( + 5) = = ( + ) ( + 5)( + ) = or = ( - ) 5 5 (5 + ) = 5 + or Say It ith Symbols

5 = + + = ( + )( + ) = ( + 5)( + ) = ( + 7)( + 7) a. Possible answers: ()() + p() or + + p() p() b. The fencing needed for the rectangular region is + = since you don t count the two shorter sides. The two half circles each have a perimeter of, p() which is half of the circumference p(). So the perimeter is + f or p +. p()g c. Possible answers: p + p + + or (p + ). 5. a. Yes. + (s ) = + s - = - + s = + s = s + b. Hank 5. a. Since the epression represents her money after one year, she would have the money she put in, which is D, plus the interest the account accrues in that year, which is. times D,so the epression D +.D is correct. b. D( +.) c. $,5(.) = $,5 5. a. Corey s estimate is correct: C = + (5) = + 5 = $7. b. Duncan performed the operations incorrectly by doing the addition first: C = ( + )5 = $,5. () 5. a. S = = = = $ b. S = (N) N () c. S = = = = $5 55. a. The volume of the prism is cubic units = cubic units. So the volume of the pyramid is = cubic units. b. A cube with edges units would have volume cubic units. A pyramid that fits inside of this cube would have the given volume. c. A cube with edge units would have volume 7 cubic units, so a pyramid that fits inside this cube would have the given volume. d. A cube with edge units would have volume 7 cubic units. So a pyramid with base by and height units would have a volume of = 9. (7 ) ACE ANSERS Investigation Equivalent Epressions

6 5. a. Sarah performed the calculations correctly. b. Emily did not use the order of operations correctly. In the second line, she added and before the multiplication of and. In the third line, she added 5 and to get 5 instead of multiplying the 5 by. Etensions 57. a. For s =, tiles are needed. For s =, + tiles are needed. For s =, + + tiles are needed. Thus, for any s,the number of tiles needed is equal to plus (s - ) fours, or N = + (s - ). b. Percy s equation is equivalent to Stella s equation, (s + ). Eplanations will vary; they may be based on tables, graphs, the substitution of specific values of s,or the sameness of the epressions. 5. For (s +.5) + (s +.5), the picture should look like: s s s (s ) For [ ],the picture should look like: s (s ) where [ ] is the area of half the shaded region multiplied by. Half the shaded region can be represented by one of the four rectangles to the right of the equal sign. 59. Puzzle : a. (n - ) + n + n + = n - 5 b. (n - ) + n + n + = n - + n + n + Distributive Prop. = n - + ( + )n + Distributive Prop. = n - + n + Addition = n + n - + Comm.Property = ( + )n - 5 Distributive Prop. = n - 5 Addition Puzzle : s s s s s s a. n - + n + (n + ) = n + b. n - + n + (n + ) = n - + n + n + Distributive Prop. = n - + ( + )n + Distributive Prop. = n - + n + Addition = n + n - + Comm.Property = ( + )n + Distributive Prop. = n + Addition Say It ith Symbols

7 Puzzle : a. n - + n + n + = n - ; no need for parentheses b. n - + n + n + = n - + ( + )n + Distributive Prop. = n - + n + Addition = n + n - + Comm.Property = ( + )n - Distributive Prop. = n - Addition Puzzle : a. n - ( + )n + n + = n + b. n - ( + )n + n + = n - 7n + n + Addition = ( )n + Distributive Prop. = n + Add. and Subtr. Possible Answers to the Mathematical Reflections. Two epressions are equivalent when they are symbolic representations for the same situation. For all values of n,they should give the same result. The same table and the same graph can represent the epressions.. The Distributive Property can be used to rewrite epressions as the product of two or more factors (factored form) or as the sum of two or more terms (epanded form). For eample, the epression ( + 5) can be written as the sum of two terms using the Distributive Property: +.The epression 9 can be written in factored form using the Distributive Property: ( ). The Commutative Property states that we can change the order of addition or multiplication and still have equivalent epressions. For eample, + = + and ( + ) = ( + ).. To show that two epressions are equivalent, apply the Distributive and Commutative properties to one of the epressions until the original epression is identical to the second epression. If the two epressions are not equivalent, then this procedure will result in a contradiction. For eample, the epressions ( + ) and + 5 are not equivalent. If we apply the Distributive Property to the first epression, we get: ( + ) = + and ACE ANSERS Investigation Equivalent Epressions