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115 Benjamin Etgen 6 Diamonds 1 Factor the following numbers in as many ways a possible by rewriting each as a product of two integers. For example, 20=4 5, 20= 4 5, 20=2 10, 20= 2 10, 20=1 20, 20= (a) 8 (b) 16 (c) 15 (d) 21 (e) 23 (f) 36 2 Find two numbers such that their product is the number in the top diamond and their sum is the number in the bottom diamond. Example: Complete the diamond by determining two factors of 3 that have a sum of -4. Since 3=1 3 and 3= 1 3 are the ways of rewriting 3 as a product of two integers, we must chose the two whose sum is -4. Since we 1+ 3= 4 chose the pair -1 and -3. a ab a+b b (a) (b) (c) (d) (e) (f) (g) (h)

116 Benjamin Etgen Factoring 116

117 117

118 Benjamin Etgen 6 1 Factoring When we rewrite 6=3 2, we are factoring. To factor out the a common factor, we use the distributive property in reverse: Distributive Property: 3(x +7)=3 x+21 Factoring: 3 x+21=3(x+7) Factor out the Greatest Common Factor (GCF), if possible. (a) 5x + 15 (b) 8x 12 (c) 14x 2 21x (d) 9x 2 +3x (e) 45z + 3 (f) 5m + 3n (g) 8x 3 y 2 12x 2 (h) 15x 3 yz 20x 2 y 3 z 2 (i) 6s 2 t 5 4s 2 t 2 (j) 8x 3 y 2 12x 2 +14x 3 y (k) 21x 2 y 2 35xy 3 +7y 2 118

119 Benjamin Etgen (l) 2x(x+3)+5(x+3) (m) x(2x 1)+(2x 1) 6 2 Factoring By Grouping (n) x 2 x+3xy+3y (o) 3x+xy 2 x 2 y+3y (p) 5x+x 2 y+5xy 25 (q) 5x+3xy 15xy 25 (r) x y 3 y 2 +xy (s) 5x+xy 7y

120 Benjamin Etgen 6 3 Factoring One-Sum-Product Factor the following trinomials. Then, check by multiplying. (a) 1x 2 +8x+12 (b) 1x 2 +2x 15 (c) 1x 2 11x (d) x 2 2 x 15 (e) x 2 +5 x 14 (f) x 2 +10x

121 Benjamin Etgen 6 4 Factoring Trinomials Factor the following trinomials. Then, check by multiplying. (a) 3x 2 +8x+5 (b) 6 x 2 +13x+6 (3x 2 + x)+( x+5) (6 x 2 + x)+( x+6) (c) 2x 2 +2x 12 (d) 8 x 2 18x 18 2(x 2 +x 6) 4(4 x 2 9x 9) 4 x 2 x x 9 121

122 Benjamin Etgen 6 5 Factoring Special Products Difference of Squares: x 2 y 2 =(x+y)(x y) In general: (A) 2 (B) 2 =(A+B)( A B) Ex: x =( )2 ( ) 2 =( + )( ) Ex: 2x 2 50= Ex: 4x 2 49= Ex: x Perfect Square Trinomial: x 2 +2xy+y 2 =(x+y) 2 In general: (A) 2 +2(A)(B)+(B) 2 =(A+B) 2 and (A) 2 2(A)(B)+(B) 2 =(A B) 2 Ex: 16 x x+49 1) ( ) 2 +2( )( )+( ) 2 Find A and B. 2) ( ) 2 +2( )( )+( ) 2 Check that 2AB is the linear term. 3) ( + ) 2 Use the rule to factor. Ex: 9x 2 42 x

123 Difference of Cubes: x 3 y 3 =(x y)(x 2 +xy+y 2 ) In general: (A) 3 (B) 3 =(A B)((A) 2 +(A)(B)+(B) 2 ) Ex: 8x 3 1=( ) 3 ( ) 3 = Ex: 64 x 3 y 3 =( ) 3 ( ) 3 = Sum of Cubes: x 3 +y 3 =(x+y)(x 2 xy+y 2 ) In general: (A) 3 +(B) 3 =(A+B)((A) 2 (A)(B)+(B) 2 ) Ex: x 3 +1=( ) 3 +( ) 3 = Ex: 27 x =( ) 3 +( ) 3 = 123

124 Benjamin Etgen 6 7 Solve by Factoring Zero Factor Property a b=0 a=0 b=0 Ex: Solve (x 3)(x+2)=0 x 3=0 x=3 x+2=0 x= 2 Ex: (9x+7)(3 x 8)(2x+1)=0 Zero: Factor: Property: Ex: x 2 +2x 8=0 Zero: Factor: Property: 124

125 Ex: 49x 2 =9 Zero: Factor: Property: Ex: 49x 2 =3(10 x 3) Zero: Factor: Property: Ex: x(6 x 1)=2 Zero: Factor: Property: 125

126 Benjamin Etgen 7 Simplifying Rational Expressions Compare the way we simplify a fraction with the way we simplify a rational expression: Ex. 1: Simplify 2x 2 +7 x+3 4 x 2 1 Same as fractions! Step 1: Factor the numerator and the denominator. (x+3)(2x+1) (2 x+1)(2 x 1) Step 2: Cancel the common factors. x+3 2x Ex. 2: Determine where the rational expression is undefined. x 2 2x 15 x 4 Step 1: Set the denominator equal to zero. x 4=0 Step 2: Solve the resulting equation. x=4 Same as fractions! When x=4, x 2 2x 15 = 7 x 4 0 Recall that you can not divide by zero. 126

127 Ex. 3: Multiply the rational expressions. ( 9 x 2 1 x x 2x 15)( 2 5x 2 3x +10x+3) 2 Same as fractions! Step 1: Factor all numerators and denominators. (3 x+1)(3 x 1) (x+3)(x 5) x(x 5) (3 x+1)(x+3) Step 2: Cancel any common factors. x(3 x 1) (x+3)(x+3) Step 3: Perform any remaining multiplication. 3x 2 x x 2 +6x Ex. 4: Divide the rational expression. x+3 x 2 +5x x2 9 x+5 Same as fractions! Step 1: Multiply by the reciprocal. x+3 x 2 +5x x+5 x Step 2: Multiply the rational equation. x+3 x(x+5) x+5 (x+3)(x 3) = 1 x(x 3) = 1 x 2 3x = =

128 Ex. 5: Combine the rational expressions. x+9 x 2 3x 10 x+5 x 2 5x Same as fractions! Step 1: Factor denominators. x+9 (x 5)(x+2) x+5 x( x 5) Step 2: Multiply numerator and denominator by the missing factors of each term separately. x x x+9 (x 5)(x+2) x+5 x(x 5) (x+2) (x+2) Step 3: Combine. x(x +9) (x +5)( x+2) x(x 5)(x+2) Step 4: Simplify the numerator. (multiply, then combine like terms) (x 2 +9x) (x 2 +7x+10) = 2x 10 x(x 5)(x+2) x(x 5)(x+2) = Step 5: Factor the numerator. 2( x 5) x(x 5)(x+2) Step 6: Cancel any common factors. 2 x(x+2) = 2 x 2 +2x =

129 Benjamin Etgen 7 1 Simplify Rational Expressions Evaluate 4 x+7 3x+9 at x=2 x= 4 x= 3 What values make the rational expression undefined? Ex: 3x x 2 +6x+8 Ex: 2x+6 x

130 Simplify: Ex: x x Ex: 12 x 2 3 8x 4 Ex: x 2 +3x x 2 9 Ex: x 2 4 x 5 x 2 +x 130

131 Benjamin Etgen 7 2 Multiply Or Divide Rational Expressions Ex: 2x 2 +7 x+3 x2 3x x 2 9 2x 2 +11x+5 Ex: x 2 +4 x 5 x 2 +7x+10 x 1 x+5 131

132 Benjamin Etgen 7 3 Least Common Denominator Ex: LCD is: Ex: x 4+ 12x 6 LCD is: Ex: 6 x 2 4 x + 3x 1 x 2 16 LCD is: Ex: 12 x x LCD is: 132

133 Benjamin Etgen 7 4 Combining Rational Expressions Ex: 5x x 1 x+5 x 1 Ex: 2 x x x 2 1 Ex: x x x 2 4 Ex: 1+ 2 x+2 133

134 Ex: x+47 x 2 +4 x 5 + x+12 x 2 +11x

135 2 x 7 x 2 2 x 8 + x 3 x 2 6 x+8 135

136 Benjamin Etgen 7 7 Solving Rational Equations Worked Example 1: Solve. Check your answer. 1 x 1 +x+4 x+3 = 1 x+3 Step 1: When solving we need the least common denominator. First we factor the denominators. 1 x 1 +x+4 x+3 = 1 x+3 Step 2: Then, multiply each term by the LCD. In this case LCD: (x 1)(x+3) 1 x+4 (x 1)(x+3)+ x 1 x+3 (x 1)(x+3)= 1 x+3 (x 1)(x+3) (x+3)+(x+ 4)(x 1)= x 1 Step 3: Solve the resulting equation. Ask yourself, if the equation is linear or quadratic. (x+3)+(x 2 +3x 4)=x 1Since this equation is quadratic, we will use the Zero Factor Property (ZFP). x 2 +3 x=0 becomes x(x+3)=0 so x=0 or x= 3. Step 4: Check the proposed solutions in the original equation. When x=0 the equation is solved, 1 3= 1 3. When x= 3 the equation is undefined. Thus, x=0 is the only solution. 136

137 Ex 1: Solve. Check your answer. 10 x 2 x+3 x+2 = 2 x+2 Step 1: When solving we need the least common denominator. First we factor the denominators. 10 x 2 x+3 x+2 = 2 x+2 Step 2: Then, multiply each term by the LCD. In this case LCD: ( x 2 )x+2 ( ) Step 3: Solve the resulting equation. Ask yourself, if the equation is linear or quadratic. Step 4: Check the proposed solutions in the original equation. 137

138 Worked Example 2: Solve. Check your answer. 1 x+2 + x+1 x 2 +x 2 = 1 x 1 Step 1: When solving we need the least common denominator. First we factor the denominators. 1 x+2 + x+1 (x+2)(x 1) = 1 x 1 Step 2: Then, multiply each term by the LCD. In this case LCD: (x+2)(x 1) 1 x+2 (x+2)(x 1)+ x+1 (x+2)(x 1) (x+2)(x 1)= 1 x 1 (x+2)(x 1) (x 1)+(x+1)=x+2 Step 3: Solve the resulting equation. Ask yourself, if the equation is linear or quadratic. 2 x=x+2 Since this equation is linear, we will isolate the variable. x 2=0 becomes x=2. Step 4: Check the proposed solutions in the original equation. When x=2 the equation is solved, 1=1. 138

139 Ex 2: Solve. Check your answer. 4 x+2 + x 4 x 2 +7 x+10 = 3 x+5 Step 1: When solving we need the least common denominator. First we factor the denominators. 4 x+2 + x 4 (x+5)(x+2) = 3 x+5 Step 2: Then, multiply each term by the LCD. In this case LCD: (x+5)(x+2) Step 3: Solve the resulting equation. Ask yourself, if the equation is linear or quadratic. Step 4: Check the proposed solutions in the original equation. 139

140 Benjamin Etgen Solve. Check your answer. 1 x+2 + x+1 x 2 +x 2 = 1 x 1 Step 1: When solving we need the least common denominator. First we factor the denominators. 1 x+2 + x+1 (x+2)(x 1) = 1 x 1 Step 2: Then, multiply each term by the LCD. In this case the LCD is: (x+2)(x 1) (x 1)+(x+1)=x+2 Step 3: Solve the resulting equation. Ask yourself, if the equation is linear or quadratic. 2x=x+2 Since this equation is linear, we will isolate the variable. x 2=0 becomes x=2. Step 4: Check the proposed solutions in the original equation. When x=2 the equation is solved, 1=1. Solving Rational Equations Combine the rational expressions. 1 x+2 + x+1 x 2 +x 2 Step 1: Factor denominators. 1 x+2 + x+1 (x+2)(x 1) Step 2: Multiply numerator and denominator by the missing factors of each term separately. x 1 x 1 1 x+2 + x+1 (x+2)(x 1) Step 3: Combine. (x+1)+(x 1) (x+2)(x 1) Step 4: Simplify the numerator. (multiply, then combine like terms) 2x (x+2)(x 1) Step 5: Factor the numerator. Step 6: Cancel any common factors. 2x (x+2)(x 1) 140

141 Benjamin Etgen 8 Roots And Radicals 1 Complete the following pairs of squares and square roots. (a) 2 2 = and =2 (c) 4 2 = and =4 (e) 6 2 = and =6 (g) 8 2 = and =8 (i) 10 2 = and =10 (k) 12 2 = and =12 (m) 14 2 = and =14 (b) 3 2 = and =3 (d) 5 2 = and =5 (f) 7 2 = and =7 (h) 9 2 = and =9 (j) 11 2 = and =11 (l) 13 2 = and =13 (n) 15 2 = and =15 2 Complete the following pairs of cubes and cube roots. (a) 2 3 = and =2 3 (c) 3 3 = and =3 3 (e) 4 3 = and =4 3 (g) 5 3 = and =5 3 (i) 6 3 = and =6 3 (b) ( 2) 3 = and = 2 3 (d) ( 3) 3 = and = 3 3 (f) ( 4) 3 = and = 4 3 (h) ( 5) 3 = and = 5 3 (j) ( 6) 3 = and =

142 Benjamin Etgen 8 1 Square Roots Operations come in pairs with their inverses: 3+5=8 is undone with 8 5=3 7 6=42 is undone with 42 6=7 4 2 =16 is undone with 16=4 16 is the square root of 16. The square (like all other even powers) looses the sign. The square of 5 and -5 are both ( 5) 2 ± Since there is no information about the sign remaining after squaring, we can not know if a positive or a negative was squared. This means that there are two square roots of 25, both 5 and and 5 square 25 square root 5 and 5 We use the sign to tell us which square roots we want: Ex: 16= Ex: 81= Ex: ± 4= and Ex: 49= 142

143 Since the product of fractions is simply the product of the numerators and the product of the denominators, taking the square roots of fractions is the same as taking the square root of the numerator and the denominator separately. Ex: = since ( ) 2 =. Ex: = Because the square of all (real) numbers are positive (or zero), negative numbers do not have square roots. Taking the square root of a negative requires finding a number that is negative when you square it. Suppose that there was a number x that was the square root of -9, 9=x. Then, that would mean that x 2 = 9. Since there is no (real) number that makes this true, then 9 is not a real number. (In intermediate algebra 9=3 i.) Ex: 9 Ex: 9 Ex: 9 Ex: 9 Product Rule: (No Sum Rule! Both x+y and x+ y can not be simplified.) Simplify: 28 Simplify: 7 75 Simplify:

144 If we know that x is positive, then x 2 =x. (In intermediate algebra x 2 = x.) All variables are positive: Simplify: 3 72x y 2 z 3 Simplify: x 4 y 5 z 6 144

145 Benjamin Etgen 8 2 Higher Roots 3 3 = ( 3) 3 = Since the cube of a positive is positive and the cube of a negative is negative, the cube keeps its sign. Thus, the cube root of a negative is a real number. 3 cube 27 cube root 3 3 cube 27 cube root 3 Ex: 3 27= since 3 =. Ex: 3 27= since 3 =. Ex: 3 8= Ex: 3 375= Ex: 3 72x 2 y 3 z 4 = Ex: 3 189x 6 y 7 z 10 = 145

146 Benjamin Etgen 8 3 Combining Radicals Ex: 2x 5x= Ex: 2x 5 y= Ex: = Ex: = Recall that x+ y can not be simplified. Ex: x 2 x= Ex: = Ex: 6+ 7= Ex: = Ex: = Ex: = 146

147 Benjamin Etgen 8 4 Multiply and Divide Radicals...just like multiplying polynomials. Because of the product rule, we can distribute, FOIL and use the multiplication algorithm. Product Rule: x y= x y Ex: 2 5= Ex: = Ex: x 2 3 y= Ex: 6 15= When we are squaring, it is easier to not use the product rule: Ex: 7 2 = Ex: 3x+7 2 = Ex: 2 2= Ex: x+5 x+5= When we have a special product, it is easier to not use the product rule: Rule for a sum and difference: (A+B)(A B)=A 2 B 2 ( 5+3)( 5 3)= =5 9= 4 Ex: ( 7+ 2)( 7 2) Ex: (2 3 1)(2 3+1) Rule for squaring binomials: (A+B) 2 =A 2 +2 AB+B 2 ( 5+2) 2 =( 5) 2 +2( 5)(2)+(2) 2 = =9+4 5 Ex: ( 10 5) 2 Ex: ( 15 6) 2 147

148 Distribute: Ex: 2 5(3+ 2)= Ex: 6 15( )= FOIL: Ex: ( 2+1)( 5+3)= Ex: (2 6+3)(3 6+7)= Quotient Rule: x y = x y Recall that the same number can be written in different ways: 2 3 =4 6 = 6 = and 75= 25 3=5 3=5 3 9 When we simplify radicals involving fractions, we want: 1) no radicals in the denominator, and 2) no fractions in radicals. We don t actually divide! 148

149 Ex: 5 6 = 5 6 = = = = Ex: 12 3 = Ex: 5 45 = Ex: 1 3 = Ex: = Ex: 7 11 = Ex: = 149

150 Benjamin Etgen Ex: 5 =5 8 5 Rationalize Radicals Ex: 3 2 =2 Ex: 18 =6 Ex: 3 4 =2 Ex: ( 3 5)( )=3 25= 22 Ex: (2 5+3)( )= What is the conjugate of 3 5+3? Ex: = Ex: = 150

151 Ex: = Ex: = Ex: = Ex: 1 3 4x 2 y 3 z 4= 151

152 Benjamin Etgen 9 1 Solving with Radicals Looking at the equation x+5=3 it is natural to square both sides in order to undo what was done to x. However, squaring loses the sign. 5 and 5 square 25 square root 5 and 5 Look at what happens to this false statement when we square both sides: 5= 5 (5) 2 =( 5) 2 25=25 Squaring made this false statement look like a true statement because it lost the sign. For this reason we MUST CHECK when we square both side of an equation. Ex: x+5=3 1) Isolate a radical: Ex: x+5=3 2) Square both sides: 3) Simplify and solve: 4) Check: 152

153 Ex: 10 x 8=3 x Ex: 3 x 5= 2 x+1 Ex: 7 x= 49x 2 +2x 10 Ex: 2x+4=x 153

154 Benjamin Etgen 9 1 Solving Equations With Two Radical Terms Ex. 1: Solve. Check your answer. x+ 4 x 1=1 Step 1: Isolate a radical term and then square both sides. (Repeat as needed.) isolate: x+4=1+ x 1 square: ( x+4) 2 =(1+ x 1) 2 The resulting equation, x+4=2 x 1+x, still has a radical we perform step 1 again: Step 1: Isolate a radical term and then square both sides. isolate: 2 x 1=4 square: (2 x 1) 2 =(4) 2 Step 2: The resulting equation, 4(x 1)=16, no longer has a radical term. We solve the equation. In this case the equation is linear, so we isolate the variable. x 1=4 becomes x=5. Step 3: Check the proposed solutions in the original equation. When x=5 the equation is solved, 1=1. 154

155 Ex. 2: Solve. Check your answer. 2 x+3= x+13+1 Step 1: Isolate a radical term and then square both sides. (Repeat as needed.) isolate: A radical is already isolated. square: ( 2 x+3) 2 =( x+13+1) 2 The resulting equation, 2x+3=2 x+13+x+14, still has a radical we perform step 1 again: Step 1: Isolate a radical term and then square both sides. isolate: 2 x+13=x 11 square: (2 x+13) 2 =(x 11) 2 Step 2: The resulting equation, 4(x+13)=x 2 22x+121, no longer has a radical term. We solve the equation. In this case the equation is quadratic, so we use the ZFP. 0=x 2 26x+69 becomes 0=(x 23)(x 3) so x = 23 or x = 3. Step 3: Check the proposed solutions in the original equation. When x = 23 the equation is solved, 1 = 1. When x = 3 the equation is not solved, -1 = 1. Thus, x = 23 is the solution to the radical equation. 155

156 Solve the radical equation. Ex.1: x+3 x 1=2 156

157 Benjamin Etgen 9 2 Solving with Exponents Looking at the equation (x+5) 2 =25 it is natural to take the square root of both sides in order to undo what was done to x. However, the there are two square roots. 5 and 5 square 25 square root 5 and 5 When we take the square root, we must remember that there is both a positive and negative root: (x+5) 2 =25 (x+5) 2 =± 25 x+5=±5 x+5=5 or x+5= 5 x=0 or x= 10 Remembering that we MUST HAVE BOTH ROOTS is called the Square Root Property. We should not be surprised that there are two solutions, since we found two solutions when we used the Zero Factor Property. Zero Factor Property x 2 =81 x 2 81=0 (x +9)(x 9)=0 x+9=0 or x 9=0 x= 9 or x=9 Square Root Property x 2 =81 x 2 =± 81 x=±9 157

158 Ex: x 2 =80 Ex: (3x+2) 2 =49 The Square Root Property is how we started the semester, using the inverse operation in the opposite order from the order of operations. Solve 5x 2 +4=8 Solving for x Arithmetic: Algebra: x 2 5x 2 5x 2 +4 Since algebra undoes arithmetic, we do the inverse of each step and in the opposite order. 158

159 Ex: x 2 +5=0 Ex: 3x 2 +1=73 Ex: 5x 2 =80 Ex: (2x+1) 2 +3=21 159

160 Benjamin Etgen 9 3 Solving Equations By Completing The Square Solve x 2 10x 11=0 by completing the square. We need (1) the variables on one side and (2) the leading coefficient to be 1. x 2 10x+ =11 b: -10 half: -5 square: 25 We add the square of the half, 25, to both sides. x 2 10x+25=11+25 To factor the left side, a perfect square trinomial, we use the half. x 2 10x+25=11+25 (x 5) 2 36 (x 5) 2 =36 The square root property will allow us to solve for x. (x 5) 2 =± 36 x 5=±6 x 5=6 x=6+5 x=11 x 5= 6 x= 6+5 x= 1 Check: 160

161 Ex. 1 x 2 4 x=60 b: half: square: x 2 4 x+ =60+ (x ) 2 = (x ) 2 =± x =± Solve the equation by Ex. 2 x 2 +4 x=117 b: half: square: completing the square. Make notes to yourself in the margin. Be sure to explain to yourself x 2 +4 x+ =117+ when to use the half and the (x ) 2 = square. (x ) 2 =± x =± Ex. 3 x 2 2 x=8 b: half: square: x 2 2 x+ =8+ (x ) 2 = (x ) 2 =± x =± 161

162 Benjamin Etgen 9 4 Quadratic Formula Completing the square on the general quadratic equation, ax 2 +bx+c=0, we derive the quadratic formula: Ex: x x 13=0 x= b± b2 4ac 2a Ex: 2x x= 5 162

163 Ex: 2x 2 +x+5=0 Ex: 4x 2 x+4=x+7 163

164 Benjamin Etgen 9 7 Rectangles Ex: In a landscape plan, a rectangular flowerbed is designed to be 4 meters longer than it is wide. If 60 square meters are needed for the plants in the bed, what should the dimensions of the rectangular bed be? Ex: A rectangular lot is 20 yards longer than it is wide and its area is 2400 square yards. Find the dimensions of the lot. 164

165 Ex: The area of a rectangle is 48 sq-ft and its perimeter is 32 ft. Find its length and width. 165

166 Benjamin Etgen 9 8 Teamwork Ex. 1: Lea can pack for a summer vacation in Monterey in 3 hours while Chris takes 6 hours. Complete the table to approximate how long it will take them working together. Lea takes 3 hours to pack: time (hours) fraction of the job Chris takes 6 hours to pack: time (hours) 1 1 fraction of the job Working together: fraction of the job t t Now, write the equation for the time it will take them to pack when working together. Solve: 166

167 Ex. 2: Jill can paint a room in 12 hours while Carl takes only 8 hours. How long it will take them working together? Ex. 3: Working together, Ellen and Dianna can plant the garden in 3 hours. Working alone would take Ellen 12 hours. How long would it take Dianna working alone? Ex. 4: Working together, Steve and Peter can trim the bushes in 4 hours. Working alone would take Steve 15 hours. How long would it take Peter working alone? 167

168 Benjamin Etgen 9 11 Graphs of Quadratics Ex: Graph y=x 2 x y Ex: Graph y= x 2 +4 x y

169 Ex: Graph y=(x+3) 2 x y

170 Ex: Graph y=x 2 +2x 3 x-intercept: y-intercept: x-vertex: y-vertex: Vertex: 170

171 Ex: Graph y=x 2 +2x x-intercept: y-intercept: x-vertex: y-vertex: Vertex: 171

172

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