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40 ozkoreapost.com FRI. 3. JUNE. 975 가로열쇠 2.수익과배당이높은일류회사의주식 4.닥치는대로마구하는짓 5.돈을모아둠 7.지독한감기 8.말이나행동으로더럽혀욕되게함 10.강의를하지않고쉼 11.세차게타오르는불꽃 13.어떤세력의관할아래 15.이름을지정하여가리킴 16.군대의기강 18.천가지모습과만가지형상 20.전지를두번접어자른크기의종이 세로열쇠 1.집안살림에쓰는기구 2.쇠귀에경읽기 3.머뭇거리며망설임 4.네변의길이가같은사각형 6.비단에수를놓은것처럼아름다운산천 9.자기생각대로혼자서처리하는사람 12.아주환하게밝은세상 14.낮은사람이규율을무시하고윗사람을꺾고오름 17.말을탄무사 19.마음껏먹고마심 낱말퍼즐 *정답은아래에있습니다. 1 2 3 4 5 6 7 9 8 9 10 11 12 13 14 15 16 17 18 19 20 김선생 수학 이번호 정답 스도쿠 이번호 정답 낱말퀴즈 이번호 정답 *문제는40p에있습니다. Solve the followings (1-4) 1 − 2 − 3 − (5 − 3) = 3 + 8 ( 6 , 7 학년 ) 2 3 −2 + 4 − −2 + = 4 − 10 − 3 + 4 ( 7, 8 학년 ) 3 3 − 4 − 2 = 0 ( 8, 9 학년 ) 4 3 2 − 7 + 2 = 0 ( 9, 10 학년 ) Factorise the followings (5-8) 5 5 − 10 − 20 ( 6, 7 학년 ) 6 − + 6 − 3 + 2 ( 8 학년 ) 7 2 − 9 − 10 ( 9 학년 ) 8 4 2 − 12 + 9 − 25 + 40 − 16 2 ( 10 학년 ) (9) The area of a rectangle is 200 2 and its perimeter is 60 m. Find its dimensions of the rectangle. 김선생수학알제브라연습문제 (268) ( Exercise of Algebra ) ( Answer ) 1 − 2 − 3 − (5 − 3) = 3 + 8 − ( − 1) − 2 = 3 + 8 − 2 − 3 = 8 − 4 = 8 ∴ = − 2 2 3 − 2 + 4 − − 2 + = 4 − 10 − 3 + 4 3 2 − − = 4 − 40 − 3 + 4 6 + = − 36 6 + − = − 36 6 = − 36 ∴ = − 6 3 3 − 4 − 2 = 0 3 − = 0 4 − 2 = 0 = 3 2 = 4 ∴ = 2 , 3 4 3 2 − 7 + 2 = 0 − 2 3 − 1 = 0 − 2 = 0 3 − 1 = 0 = 2 3 = 1 ∴ = 2 , 1/3 5 5 − 10 − 20 = 5 × − 5 × 2 − 5 × 4 = 5 ( − 2 − 4 ) 6 − + 6 − 3 + 2 = − + 2 + 6 − 3 = − + 2 + 3 (2 − ) = 2 − + 3 (2 − ) = 2 − ( + 3 ) 7 2 − 9 − 10 = ( 2 − 9 − 10) = − 10 + 1 8 4 2 − 12 + 9 − 25 + 40 − 16 2 = (2 ) 2 − 2 2 3 + 3 2 − (25 − 40 + 16 2 ) = (2 − 3) 2 − 5 2 − 2 5 (4 ) + (4 ) 2 = (2 − 3) 2 − (5 − 4 ) 2 = (2 − 3) − 5 − 4 {(2 − 3) + 5 − 4 } = 2 − 3 − 5 + 4 (2 − 3 + 5 − 4 ) = 2 − 8 + 4 (2 + 2 − 4 ) = 2 + 2 − 4 × 2( − 2 + 1) = 4 + 2 − 4 ( − 2 + 1) (9) Let x = width of the rectangle y = length of the rectangle area of the rectangle = width × length 200 = xy xy = 200 ① Solve the followings (1-4) 1 − 2 − 3 − (5 − 3) = 3 + 8 ( 6 , 7 학년 ) 2 3 −2 + 4 − −2 + = 4 − 10 − 3 + 4 ( 7, 8 학년 ) 3 3 − 4 − 2 = 0 ( 8, 9 학년 ) 4 3 2 − 7 + 2 = 0 ( 9, 10 학년 ) Factorise the followings (5-8) 5 5 − 10 − 20 ( 6, 7 학년 ) 6 − + 6 − 3 + 2 ( 8 학년 ) 7 2 − 9 − 10 ( 9 학년 ) 8 4 2 − 12 + 9 − 25 + 40 − 16 2 ( 10 학년 ) (9) The area of a rectangle is 200 2 and its perimeter is 60 m. Find its dimensions of the rectangle. 김선생수학알제브라연습문제 (268) ( Exercise of Algebra ) ( Answer ) 1 − 2 − 3 − (5 − 3) = 3 + 8 − ( − 1) − 2 = 3 + 8 − 2 − 3 = 8 − 4 = 8 ∴ = − 2 2 3 − 2 + 4 − − 2 + = 4 − 10 − 3 + 4 3 2 − − = 4 − 40 − 3 + 4 6 + = − 36 6 + − = − 36 6 = − 36 ∴ = − 6 3 3 − 4 − 2 = 0 3 − = 0 4 − 2 = 0 = 3 2 = 4 ∴ = 2 , 3 4 3 2 − 7 + 2 = 0 − 2 3 − 1 = 0 − 2 = 0 3 − 1 = 0 = 2 3 = 1 ∴ = 2 , 1/3 5 5 − 10 − 20 = 5 × − 5 × 2 − 5 × 4 = 5 ( − 2 − 4 ) 6 − + 6 − 3 + 2 = − + 2 + 6 − 3 = − + 2 + 3 (2 − ) = 2 − + 3 (2 − ) = 2 − ( + 3 ) 7 2 − 9 − 10 = ( 2 − 9 − 10) = − 10 + 1 8 4 2 − 12 + 9 − 25 + 40 − 16 2 = (2 ) 2 − 2 2 3 + 3 2 − (25 − 40 + 16 2 ) = (2 − 3) 2 − 5 2 − 2 5 (4 ) + (4 ) 2 = (2 − 3) 2 − (5 − 4 ) 2 = (2 − 3) − 5 − 4 {(2 − 3) + 5 − 4 } = 2 − 3 − 5 + 4 (2 − 3 + 5 − 4 ) = 2 − 8 + 4 (2 + 2 − 4 ) = 2 + 2 − 4 × 2( − 2 + 1) = 4 + 2 − 4 ( − 2 + 1) (9) Let x = width of the rectangle y = length of the rectangle area of the rectangle = width × length 200 = xy  xy = 200 ① perimeter of the rectangle = 2 ( width + length ) 60 = 2 ( x + y )  x + y = 30 ② From ① xy = 200 = 1 × 200 = 2 × 100 = 4 × 50 = 8 × 25 = 10 × 20 If x = 10 and y = 20 , the equation of ② is satisfied. The dimensions of the rectangle is 10 × 20. ( width = 10 and length = 20 ) Solve the followings (1-4) 1 − 2 − 3 − (5 − 3) = 3 + 8 2 3 −2 + 4 − −2 + = 4 − 10 − 3 + 4 3 3 − 4 − 2 = 0 ( 8 4 3 2 − 7 + 2 = 0 ( 9, Factorise the followings (5-8) 5 5 − 10 − 20 ( 6 − + 6 − 3 + 2 7 2 − 9 − 10 ( 9 학 8 4 2 − 12 + 9 − 25 + 40 − 16 2 (9) The area of a rectangle is 200 2 and its perimet Find its dimensions of the rectangle. 김선생수학알제브라연습문제 (268) ( E ( Answer ) 1 − 2 − 3 − (5 − 3) = 3 + 8 − ( − 1) − 2 = 3 + 8 − 2 − 3 = 8 − 4 = 8 ∴ = − 2 2 3 − 2 + 4 − − 2 + = 4 − 10 − 3 + 3 2 − − = 4 − 40 − 3 + 4 6 + = − 36 6 + − = − 36 6 = − 36 ∴ = − 6 3 3 − 4 − 2 = 0 3 − = 0 4 − 2 = 0 = 3 2 = 4 ∴ = 2 , 3 4 3 2 − 7 + 2 = 0 − 2 3 − 1 = 0 − 2 = 0 3 − 1 = 0 = 2 3 = 1 ∴ = 2 , 1/3 5 5 − 10 − 20 = 5 × − 5 × 2 − 5 × 4 = 5 ( − 2 − 4 ) 6 − + 6 − 3 + 2 = − + 2 + 6 − 3 = − + 2 + 3 (2 − ) = 2 − + 3 (2 − ) = 2 − ( + 3 ) 7 2 − 9 − 10 = ( 2 − 9 − 10) = − 10 + 1 8 4 2 − 12 + 9 − 25 + 40 − 16 2 = (2 ) 2 − 2 2 3 + 3 2 − (25 − 40 + 16 2 ) = (2 − 3) 2 − 5 2 − 2 5 (4 ) + (4 ) 2 = (2 − 3) 2 − (5 − 4 ) 2 = (2 − 3) − 5 − 4 {(2 − 3) + 5 − 4 } = 2 − 3 − 5 + 4 (2 − 3 + 5 − 4 ) = 2 − 8 + 4 (2 + 2 − 4 ) = 2 + 2 − 4 × 2( − 2 + 1) = 4 + 2 − 4 ( − 2 + 1) (9) Let x = width of the rectangle y = length of the rectangle area of the rectangle = width × length 200 = xy  xy = 200 ① perimeter of the rectangle = 2 ( width + length ) 60 = 2 ( x + y )  x + y = 30 ② From ① xy = 200 = 1 × 200 = 2 × 100 = 4 × 50 = If x = 10 and y = 20 , the equation of ② is satisfi The dimensions of the rectangle is 10 × 20. ( width