Answers Investigation 1

Transcription

1 Applications. a. () + () + = tiles b. Possible epressions: + + ( + ) + ( + ) ( + ) + + ( + ) c. See part (b) for some epressions; eplanations will vary. Students might draw sketches. For eample: + + ( + ) + ( + ) + ( + ) + ( + ) They might substitute values for and in the epressions; for eample, when = and = : + + = () + () + = ( + ) + ( + ) = () + () = ( + ) + = () + () = + ( + ) = () + () =. a. (7) + (.) = tiles b. Possible answers: s + (s +.) s + (s + ) c. See epressions in part (b). Students might substitute values for s [in this case two 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: + + ( +.) + ( +.) ( + ) + c. Students might substitute values for and, make tables or graphs, or make geometric arguments to show that their two epressions are equivalent.. a. First equation: ( + ) = () + = ; Second equation: ( +.) + ( +.) = (.) + (.) = + = ; 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: ( +.) + ( +.) = (.) + (.) = ; + ( + ) Third equation: [ ] = ( ) = () = d. Since you can determine nonequivalency of linear equations by checking one point, the first epression is not equal to the second and 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 Eercise, checking only two values would be enough because only one line can pass through the two points. So linear epressions that agree on two values (two points) contain the same two points. So, the lines that they represent must be the same. Say It ith Symbols

2 Students will either need to check all points, which is impossible, or know that two points uniquely determine a line. y. a. The shape is the area between the circle and the square. r O y = + y = + + b. The shape is all the area inside the square ecept a quarter of the area of the circle. r 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 pass through the same two points. c = = + ( - + ) = +. a. ii and iv. a. 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 since s(s - ) = s - s and s + s(s - ) = s - s. d. Answers will vary, but must be equivalent to A = (s - s + ) + (s - s). e. The equation in part (d) is a quadratic relationship Say It ith Symbols y = y y = O 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. - = - + -

3 9. a. ( + ) + ( ) y y = ( + ) + ( ) y = + O 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. ( + ) + ( - ) = = ( + ) + = +. a. + b. - c. - d = + ( + ) + = + +. a. ( + ) b. 7( - ) c. Possible answers: ( - ) or ( - ) = - d. ( + ). a. equal; + 7 = ( + 7) = b. not equal; - = ( - ) = - c. equal; ( + ) - = + - = + = + Using the Commutative Property of addition, + = +. d. equal; - ( - ) = - + = - +. Step (): Distributive Property Step (): Commutative Property Step (): Distributive Property Step (): Addition. Possible answers: ( + ), + +, (7 + )p - p = p. 7 + (p - p) = 7 7. Parentheses are not needed. Connections. 9. ( + ) = + ( ) ( - ) = - Say It ith Symbols

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

5 . 7 s + (s + ) For [ ], the picture should look like: s s = = ( + 7)( + 7). For (s +.) + (s +.); the picture should look like: s s s s s s =. 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 a. Possible answers: ()() + p() or p() + p() + b. The fencing needed for the rectangular region is + = since you do not 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 + [ p() ], or p +. c. Possible answers: p + p + + or (p + ) Say It ith Symbols

6 . a. Yes. + (s - ) = + s - = - + s = + s = s + b. Hank. 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. +,(.) = +,. a. Corey s estimate is correct: C = + () = + = +7. b. Duncan performed the operations incorrectly by doing the addition first: C = ( + ) = +,. + (). a. S = = = + + (N) b. S = N + () c. S = = = = = 7. a. Sarah performed the calculations correctly. b. Emily did not use the Order of Operations correctly. In the second line, she added and instead of multiplying and. In the third line, she added and to get instead of multiplying the by (her incorrect calculation). Etensions. The number of tiles in the first pool = ( * ) + () The number of tiles in the second pool = ( * ) + ( ) The number of tiles in the third pool = ( * ) + ( ) Therefore, the equation is N = () +, or N = +. (See Figure.) Another solution path is N = ( + ), which simplifies to N = +. (See Figure.) Figure Figure Say It ith Symbols

7 9. The number of tiles in the first pool = ( * ) + ( * ) + ( ) The number of tiles in the second pool = ( * ) + ( * ) + ( ) The number of tiles in the third pool = ( * ) + ( * ) + ( ) Therefore, the equation is N = ( * ) + ( * ) + ( ), which simplifies to N = ( + ) + = +. (See Figure.). Puzzle: a. (n - ) + n + n + = n - b. (n - ) + n + n + = n - + n + n + Distributive = n - + ( + )n + Distributive = n - + n + = n + n - + Comm. Property = ( + )n - Distributive = n - Puzzle : a. n - + n + (n + ) = n + b. n - + n + (n + ) = n - + n + n + Distributive = n - + ( + )n + Distributive = n - + n + = n + n - + Comm. Property = ( + )n + Distributive = n + Puzzle : a. n - + n + n + = n - ; no need for parentheses b. n - + n + n + = n - + ( + )n + Distributive = n - + n + = n + n - + Comm. Property = ( + )n - Distributive = n - Puzzle : a. n - ( + )n + n + = n + b. n - ( + )n + n + = n - 7n + n + = ( )n + Distributive = n + Figure Say It ith Symbols 7