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Find the value of the unknown1.)2(x^2-6)+3≡Ax^2+Bx+C2.)Ax^2-5x+44≡(Bx-C)(-x-11)3.)-A(x-5)+B(3x-1)≡5x-114.)A(2x-1)(x-3)+B(-x+4)+C≡6x^2-22+135.)prove that (3x-4)(7x-1)(x-3)=21x^3-80x^2+53x-6is an... 顯示更多 Find the value of the unknown 1.)2(x^2-6)+3≡Ax^2+Bx+C 2.)Ax^2-5x+44≡(Bx-C)(-x-11) 3.)-A(x-5)+B(3x-1)≡5x-11 4.)A(2x-1)(x-3)+B(-x+4)+C≡6x^2-22+13 5.)prove that (3x-4)(7x-1)(x-3)=21x^3-80x^2+53x-6is an identity Solve 1.)(5x-1)-6(x-3)=8x-19 2.)5x/6=(5x/3)-7 3.)-x/5=5x-1/3=6x+4/15 1.)The length of a rectangle is 11 cm longer than its width.If the perimeter of the rectangle is 88cm,find the area of the rectangle. 2.)$773is a divided among three personA,B,C.If A got $47more thanCand the moneyBgot is double than C got ,find the amount each person gets.
最佳解答:
Find the value of the unknown 1.)2(x2 – 6) + 3 ≡ Ax2 + Bx + C 2(x2 – 6) + 3 = 2x2 – 12 + 3 = 2x2 – 9 So A = 2; B = 0; C = -9 2.)Ax2 – 5x + 44 ≡ (Bx – C)(-x – 11) (Bx – C)(-x – 11) = -Bx2 + Cx – 11Bx + 11C A = -B … (1) C – 11B = -5 … (2) 11C = 44 => C = 4 Sub into (2), 4 – 11B = -5 11B = 4 + 5 B = 9/11 A = -9/11 3) -A(x – 5) + B(3x – 1) ≡ 5x – 11 -A(x – 5) + B(3x – 1) = -Ax + 5A + 3Bx – B = (3B – A)x + (5A – B) 3B – A = 5 … (1) 5A – B = -11 … (2) (A) + 3*(B) => -A + 15A = 5 – 33 14A = -28 A = -2 Sub into (2), -10 – B = -11 => B = 1 4.) A(2x – 1)(x – 3) + B(-x + 4) + C ≡ 6x2 – 22x + 13 A(2x – 1)(x – 3) + B(-x + 4) + C = A(2x2 – 7x + 3) – Bx + 4B + C = 2Ax2 – 7Ax + 3A – Bx + 4B + C = 2Ax2 – (7A + B)x + 3A + 4B + C 2A = 6 => A = 3 7A + B = 22 => B = 1 3A + 4B + C = 13 => C = 0 5.) prove that (3x – 2)(7x – 1)(x – 3) = 21x3 – 80x2 + 53x – 6 is an identity When x = 2/3, LHS = 0 RHS = 21(2/3)3 – 80(2/3)2 + 53(2/3) – 6 = 0 When x = 1/7, LHS = 0 RHS = 21(1/7)3 – 80(1/7)2 + 53(1/7) – 6 = 0 When x = 3, LHS = 0 RHS = 21(3)3 – 80(3)2 + 53(3) – 6 = 0 Coefficient of x3 terms on LHS = 21 = RHS, so the equation is an identity. Alternatively, (3x – 2)(7x – 1)(x – 3) = (21x2 – 14x – 3x +2)(x – 3) = (21x2 – 17x + 2)(x – 3) = 21x3 – 17x2 + 2x – 63x2 + 51x – 6 = 21x3 – 80x2 + 53x – 6 = RHS Solve 1.) (5x – 1) – 6(x – 3) = 8x – 19 5x – 1 – 6x + 18 = 8x – 19 -1 + 18 + 19 = 8x – 5x + 6x 36 = 9x x = 4 2.) 5x/6 = (5x/3) – 7 5x = 10x – 42 -5x = - 42 x = 8.4 3.) -x/5 = 5x-1/3 = 6x + 4/15 (Inconsistent equation, no solution) 1.)Let the length be x cm Width = x – 11 cm Perimeter = 2 * (x + x – 11) = 88 2x – 11 = 44 2x = 55 x = 27.5 cm Width = 27.5 – 11 = 16.5 cm Area = 27.5 * 16.5 = 453.75 cm2 2.)Denote the money each gets by A,B and C respectively. A + B + C = 773 … (1) A = C + 47 … (2) B = 2C … (3) Sub (2) and (3) into (1) C + 47 + 2C + C = 773 4C = 726 C = 181.5 B = 363 A = 181.5 + 47 = 228.5 A gets $228.5 B gets $363 and C gets $181.5
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Find the value of the unknown1.)2(x^2-6)+3≡Ax^2+Bx+C2.)Ax^2-5x+44≡(Bx-C)(-x-11)3.)-A(x-5)+B(3x-1)≡5x-114.)A(2x-1)(x-3)+B(-x+4)+C≡6x^2-22+135.)prove that (3x-4)(7x-1)(x-3)=21x^3-80x^2+53x-6is an... 顯示更多 Find the value of the unknown 1.)2(x^2-6)+3≡Ax^2+Bx+C 2.)Ax^2-5x+44≡(Bx-C)(-x-11) 3.)-A(x-5)+B(3x-1)≡5x-11 4.)A(2x-1)(x-3)+B(-x+4)+C≡6x^2-22+13 5.)prove that (3x-4)(7x-1)(x-3)=21x^3-80x^2+53x-6is an identity Solve 1.)(5x-1)-6(x-3)=8x-19 2.)5x/6=(5x/3)-7 3.)-x/5=5x-1/3=6x+4/15 1.)The length of a rectangle is 11 cm longer than its width.If the perimeter of the rectangle is 88cm,find the area of the rectangle. 2.)$773is a divided among three personA,B,C.If A got $47more thanCand the moneyBgot is double than C got ,find the amount each person gets.
最佳解答:
Find the value of the unknown 1.)2(x2 – 6) + 3 ≡ Ax2 + Bx + C 2(x2 – 6) + 3 = 2x2 – 12 + 3 = 2x2 – 9 So A = 2; B = 0; C = -9 2.)Ax2 – 5x + 44 ≡ (Bx – C)(-x – 11) (Bx – C)(-x – 11) = -Bx2 + Cx – 11Bx + 11C A = -B … (1) C – 11B = -5 … (2) 11C = 44 => C = 4 Sub into (2), 4 – 11B = -5 11B = 4 + 5 B = 9/11 A = -9/11 3) -A(x – 5) + B(3x – 1) ≡ 5x – 11 -A(x – 5) + B(3x – 1) = -Ax + 5A + 3Bx – B = (3B – A)x + (5A – B) 3B – A = 5 … (1) 5A – B = -11 … (2) (A) + 3*(B) => -A + 15A = 5 – 33 14A = -28 A = -2 Sub into (2), -10 – B = -11 => B = 1 4.) A(2x – 1)(x – 3) + B(-x + 4) + C ≡ 6x2 – 22x + 13 A(2x – 1)(x – 3) + B(-x + 4) + C = A(2x2 – 7x + 3) – Bx + 4B + C = 2Ax2 – 7Ax + 3A – Bx + 4B + C = 2Ax2 – (7A + B)x + 3A + 4B + C 2A = 6 => A = 3 7A + B = 22 => B = 1 3A + 4B + C = 13 => C = 0 5.) prove that (3x – 2)(7x – 1)(x – 3) = 21x3 – 80x2 + 53x – 6 is an identity When x = 2/3, LHS = 0 RHS = 21(2/3)3 – 80(2/3)2 + 53(2/3) – 6 = 0 When x = 1/7, LHS = 0 RHS = 21(1/7)3 – 80(1/7)2 + 53(1/7) – 6 = 0 When x = 3, LHS = 0 RHS = 21(3)3 – 80(3)2 + 53(3) – 6 = 0 Coefficient of x3 terms on LHS = 21 = RHS, so the equation is an identity. Alternatively, (3x – 2)(7x – 1)(x – 3) = (21x2 – 14x – 3x +2)(x – 3) = (21x2 – 17x + 2)(x – 3) = 21x3 – 17x2 + 2x – 63x2 + 51x – 6 = 21x3 – 80x2 + 53x – 6 = RHS Solve 1.) (5x – 1) – 6(x – 3) = 8x – 19 5x – 1 – 6x + 18 = 8x – 19 -1 + 18 + 19 = 8x – 5x + 6x 36 = 9x x = 4 2.) 5x/6 = (5x/3) – 7 5x = 10x – 42 -5x = - 42 x = 8.4 3.) -x/5 = 5x-1/3 = 6x + 4/15 (Inconsistent equation, no solution) 1.)Let the length be x cm Width = x – 11 cm Perimeter = 2 * (x + x – 11) = 88 2x – 11 = 44 2x = 55 x = 27.5 cm Width = 27.5 – 11 = 16.5 cm Area = 27.5 * 16.5 = 453.75 cm2 2.)Denote the money each gets by A,B and C respectively. A + B + C = 773 … (1) A = C + 47 … (2) B = 2C … (3) Sub (2) and (3) into (1) C + 47 + 2C + C = 773 4C = 726 C = 181.5 B = 363 A = 181.5 + 47 = 228.5 A gets $228.5 B gets $363 and C gets $181.5
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