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Yr 11 MCAT Algebra Practice

Yr 11 MCAT Algebra Practice. 2. Simplify fully… a) 5a 3 b 4 x 5ab 3 = b) 8a 7 = 24a 3 c) (4a 3 ) 3 = 3. Factorise… a) a 2 – 36 = b) 2a 2 + 16a + 32 = 4. Rearrange A = π r 2 to make r the subject of the formula. 5. Simplify fully…

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Yr 11 MCAT Algebra Practice

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  1. Yr 11 MCAT Algebra Practice 2. Simplify fully… a) 5a3b4 x 5ab3 = b) 8a7 = 24a3 c) (4a3)3 = 3. Factorise… a) a2 – 36 = b) 2a2 + 16a + 32 = 4. Rearrange A = πr2 to make r the subject of the formula. 5. Simplify fully… a) 7a3 x 5ab5 = 25b3 14a2 b) a2 + 13a + 36 = a + 9 • 6. Simultaneously solve to find the solutions for both a and b • a + 3b = 45 • 2a + b = 30 • 7. Write an expression for the perimeter of the following shape… • 8. If the total perimeter of the above was 226cm what is the value of a? • 1. Solve these equations… • a) 6a + 9 = a – 5 • b) a4 = 625 • c) (3a – 1)(a + 3) = 0 • d) a2 + 9a + 14 = 0 • e) 2a(9 – a) = 0 • f) Jim buys some fish (f) and some scoops of chips (c). Jim spends a total of $23.00. Jim writes an equation for the amount he spends… • 2.6f + 2.5c = 23 • i) explain each part of this equation. • ii) Jim bought one more fish than he did scoops of chips. How many fish did he buy in total? 3a 46cm 31cm a

  2. Yr 11 MCAT Algebra Practice 2. Simplify fully… a) 5a3b4 x 5ab3 = b) 8a7 = 24a3 c) (4a3)3 = 3. Factorise… a) a2 – 36 = b) 2a2 + 16a + 32 = 4. Rearrange A = πr2 to make r the subject of the formula. 5. Simplify fully… a) 7a3 x 5ab5 = 25b3 14a2 b) a2 + 13a + 36 = a + 9 • 6. Simultaneously solve to find the solutions for both a and b • a + 3b = 45 • 2a + b = 30 • 7. Write an expression for the perimeter of the following shape… • 8. If the total perimeter of the above was 226cm what is the value of a? • 1. Solve these equations… • a) 6a + 9 = a – 5 • b) a4 = 625 • c) (3a – 1)(a + 3) = 0 • d) a2 + 9a + 14 = 0 • e) 2a(9 – a) = 0 • f) Jim buys some fish (f) and some scoops of chips (c). Jim spends a total of $23.00. Jim writes an equation for the amount he spends… • 2.6f + 2.5c = 23 • i) explain each part of this equation. • ii) Jim bought one more fish than he did scoops of chips. How many fish did he buy in total? 25 a4 b7 2a + 6b = 90 eqn 1 double eqn 1 5a = - 14 eqn 2 -24/5 or -2.8 a = -14/5 a4 . 3 subtract the two equations 5b = 60 b = 12 a = 4√625 substitute b into equation 2 2a + 12 = 30 and -5 a = 5 2a = 18 64 a9 a = 9 1/3 -3 a is ____ or ____ 3a factorise first a a ( )( ) = 0 + 2 + 7 46 – a a a ( )( ) – 6 + 6 a is ____ or ____ -2 -7 must have both the difference of two squares 46cm 31cm 9 a is ____ or ____ 0 a2 + 8a + 16 2( ) a a = 2( )( ) a + 4 + 4 take out common factor 3a + 31 can be 2(a + 4)2 perimeter = all sides added perimeter = 46+3a+46–a+31+a+3a+31 perimeter = 6a + 154 πr2 = A r2 = A/π r = ±√(A/π) 2.5 is the price of the chips. 2.6 is the price of the fish. c is thenumber of scoops. f is thenumber of the fish. 1 1 a4b5 = 10a2b3 a2b2 10 6a + 154 = 226 6a = 72 23 is the total amount he paid 2 5 a = 12 cm ( )( ) = a + 9 a + 9 a + 4 a + 4 five fish

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