Lecture 311 Unit 4 Lecture 31 Applications of the Quadratic Formula Applications of the Quadratic Formula.

Lecture 311 Unit 4 Lecture 31 Applications of the Quadratic Formula Applications of the Quadratic Formula

Lecture 312 Objectives Formulate a quadratic equation from a problem situationFormulate a quadratic equation from a problem situation Solve a quadratic equation by using the quadratic formulaSolve a quadratic equation by using the quadratic formula

Lecture 313 You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from the ground is given by the equation: h = -16t 2 + 1600 h is in feet and t is in seconds. How long until the rock hits the ground?

Lecture 314 You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from the ground is given by the equation: h = -16t 2 + 1600 h is in feet and t is in seconds. How long until the rock hits the ground? h = -16t 2 + 1600

Lecture 315 You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from the ground is given by the equation: h = -16t 2 + 1600 h is in feet and t is in seconds. How long until the rock hits the ground? h = -16t 2 + 1600 0 = -16t 2 + 1600

Lecture 316 You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from the ground is given by the equation: h = -16t 2 + 1600 h is in feet and t is in seconds. How long until the rock hits the ground? h = -16t 2 + 1600 0 = -16t 2 + 1600 0 = -16(t 2 – 100)

Lecture 317 You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from the ground is given by the equation: h = -16t 2 + 1600 h is in feet and t is in seconds. How long until the rock hits the ground? h = -16t 2 + 1600 0 = -16t 2 + 1600 0 = -16(t 2 – 100) 0 = -16(t – 10)(t + 10)

Lecture 318 You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from the ground is given by the equation: h = -16t 2 + 1600 h is in feet and t is in seconds. How long until the rock hits the ground? h = -16t 2 + 1600 0 = -16t 2 + 1600 0 = -16(t 2 – 100) (t – 10)=0 or (t + 10)=0 0 = -16(t – 10)(t + 10)

Lecture 319 You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from the ground is given by the equation: h = -16t 2 + 1600 h is in feet and t is in seconds. How long until the rock hits the ground? h = -16t 2 + 1600 0 = -16t 2 + 1600 0 = -16(t 2 – 100) (t – 10)=0 or (t + 10)=0 0 = -16(t – 10)(t + 10) t = 10 or t =-10

Lecture 3110 You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from the ground is given by the equation: h = -16t 2 + 1600 h is in feet and t is in seconds. How long until the rock hits the ground? h = -16t 2 + 1600 0 = -16t 2 + 1600 0 = -16(t 2 – 100) (t – 10)=0 or (t + 10)=0 0 = -16(t – 10)(t + 10) t = 10 or t =-10 t = 10 is the only reasonableanswer.

Lecture 3111 The equation for profit is, Profit = Revenue - Cost Revenue: R = 280x -.4x 2 Cost: C = 5000 +.6x 2

Lecture 3112 The equation for profit is, Profit = Revenue - Cost Revenue: R = 280x -.4x 2 P = ( 280x –.4x 2 ) – (5000 +.6x 2 ) Cost: C = 5000 +.6x 2

Lecture 3113 The equation for profit is, Profit = Revenue - Cost Revenue: R = 280x -.4x 2 P = ( 280x –.4x 2 ) – (5000 +.6x 2 ) Cost: C = 5000 +.6x 2 P = 280x –.4x 2 – 5000 -.6x 2

Lecture 3114 The equation for profit is, Profit = Revenue - Cost Revenue: R = 280x -.4x 2 P = -1x 2 P = ( 280x –.4x 2 ) – (5000 +.6x 2 ) Cost: C = 5000 +.6x 2 P = 280x –.4x 2 – 5000 -.6x 2

Lecture 3115 The equation for profit is, Profit = Revenue - Cost Revenue: R = 280x -.4x 2 P = -1x 2 + 280x P = ( 280x –.4x 2 ) – (5000 +.6x 2 ) Cost: C = 5000 +.6x 2 P = 280x –.4x 2 – 5000 -.6x 2

Lecture 3116 The equation for profit is, Profit = Revenue - Cost Revenue: R = 280x -.4x 2 P = -1x 2 + 280x - 5000 P = ( 280x –.4x 2 ) – (5000 +.6x 2 ) Cost: C = 5000 +.6x 2 P = 280x –.4x 2 – 5000 -.6x 2

Lecture 3117 How many items must be made and sold to generate a$439 profit. P = -1x 2 + 280x - 5000

Lecture 3118 How many items must be made and sold to generate a$439 profit. P = -1x 2 + 280x - 5000 439 = -1x 2 + 280x - 5000

Lecture 3119 How many items must be made and sold to generate a$439 profit. P = -1x 2 + 280x - 5000 439 = -1x 2 + 280x - 5000 Subtract 439 from both sides

Lecture 3120 How many items must be made and sold to generate a$439 profit. P = -1x 2 + 280x - 5000 439 = -1x 2 + 280x - 5000 0 = -1x 2 + 280x - 5439 Subtract 439 from both sides

Lecture 3121 Use the Quadratic Formula to solve: 0 = -1x 2 + 280x - 5439

Lecture 3122 Use the Quadratic Formula to solve: 0 = -1x 2 + 280x - 5439 0 = ax 2 + bx + c

Lecture 3123 Use the Quadratic Formula to solve: 0 = -1x 2 + 280x - 5439 a = -1 0 = ax 2 + bx + c

Lecture 3124 Use the Quadratic Formula to solve: 0 = -1x 2 + 280x - 5439 a = -1 b = 280 0 = ax 2 + bx + c

Lecture 3125 Use the Quadratic Formula to solve: 0 = -1x 2 + 280x - 5439 a = -1 b = 280 c = -5439 0 = ax 2 + bx + c

Lecture 3129 To find the square root of a number, use key in row 6 column 1. Use of the calculator to evaluate a square root. is keyed in as 2nd,, 56644, ENTER

Lecture 3133 Use the Quadratic Formula to solve: 0 = -1x 2 + 280x - 5439 Two solutions to make P = $439: x = 21 and x =259 items x = 21 and x =259 items

Lecture 3134 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea

Lecture 3135 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea 5

Lecture 3136 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea 5 40-2(5)=30

Lecture 3137 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea 5 40-2(5)=30 5(30)=150

Lecture 3138 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea 5 40-2(5)=30 5(30)=150 10

Lecture 3139 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea 5 40-2(5)=30 5(30)=150 10 40-2(10)=20

Lecture 3140 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea 5 40-2(5)=30 5(30)=150 10 40-2(10)=20 10(20)=200

Lecture 3141 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea 5 40-2(5)=30 5(30)=150 10 40-2(10)=20 10(20)=200 15

Lecture 3142 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea 5 40-2(5)=30 5(30)=150 10 40-2(10)=20 10(20)=200 15 40-2(15)=10

Lecture 3143 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea 5 40-2(5)=30 5(30)=150 10 40-2(10)=20 10(20)=200 15 40-2(15)=10 15(10)=150

Lecture 3144 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea 5 40-2(5)=30 5(30)=150 10 40-2(10)=20 10(20)=200 15 40-2(15)=10 15(10)=150 X

Lecture 3145 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea 5 40-2(5)=30 5(30)=150 10 40-2(10)=20 10(20)=200 15 40-2(15)=10 15(10)=150 X 40-2(X)

Lecture 3146 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Make a table and find the formula for the area of the rectangle. Wall L WW WidthLengthArea 5 40-2(5)=30 5(30)=150 10 40-2(10)=20 10(20)=200 15 40-2(15)=10 15(10)=150 X 40-2(X) X(40-2X)= 40X-2X 2

Lecture 3147 A = 40X-2X 2 65 = -2x 2 + 40x Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Find the dimensions of the rectangle if the area is 65 square inches. Wall L WW Subtract 65 from both sides

Lecture 3148 A = 40X-2X 2 65 = -2x 2 + 40x 0 = -2x 2 + 40x - 65 Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the rectangle. Find the dimensions of the rectangle if the area is 65 square inches. Wall L WW Subtract 65 from both sides

Lecture 3150 Use the Quadratic Formula to solve: 0 = ax 2 + bx + c 0 = -2x 2 + 40x - 65

Lecture 3151 Use the Quadratic Formula to solve: a = -2 0 = ax 2 + bx + c 0 = -2x 2 + 40x - 65

Lecture 3152 Use the Quadratic Formula to solve: a = -2 b = 40 0 = ax 2 + bx + c 0 = -2x 2 + 40x - 65

Lecture 3153 Use the Quadratic Formula to solve: a = -2 b = 40 c = -65 0 = ax 2 + bx + c 0 = -2x 2 + 40x - 65

Lecture 3160 Use the Quadratic Formula to solve: 0 = -2x 2 + 40x - 65 Two solutions to make A = 65 sq in: x = 1.8 and x =18.2 inches x = 1.8 and x =18.2 inches

Lecture 3161 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H

Lecture 3162 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 2 4 6 8 10 b

Lecture 3163 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 2 4 6 8 10 b

Lecture 3164 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 4 6 8 10 b

Lecture 3165 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 10 -(2)=8 4 6 8 10 b

Lecture 3166 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 10 -(2)=8 (.5)(2)(8)= 8 sq cm 4 6 8 10 b

Lecture 3167 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 10 -(2)=8 (.5)(2)(8)= 8 sq cm 4 10-(4)=6 6 8 10 b

Lecture 3168 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 10 -(2)=8 (.5)(2)(8)= 8 sq cm 4 10-(4)=6 (.5)(4)(6)=12 sq cm 6 8 10 b

Lecture 3169 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 10 -(2)=8 (.5)(2)(8)= 8 sq cm 4 10-(4)=6 (.5)(4)(6)=12 sq cm 6 10-(6)=4 8 10 b

Lecture 3170 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 10 -(2)=8 (.5)(2)(8)= 8 sq cm 4 10-(4)=6 (.5)(4)(6)=12 sq cm 6 10-(6)=4 (.5)(6)(4)=12 sq cm 8 10 b

Lecture 3171 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 10 -(2)=8 (.5)(2)(8)= 8 sq cm 4 10-(4)=6 (.5)(4)(6)=12 sq cm 6 10-(6)=4 (.5)(6)(4)=12 sq cm 8 10-(8)=2 10 b

Lecture 3172 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 10 -(2)=8 (.5)(2)(8)= 8 sq cm 4 10-(4)=6 (.5)(4)(6)=12 sq cm 6 10-(6)=4 (.5)(6)(4)=12 sq cm 8 10-(8)=2 (.5)(8)(2)=8 sq cm 10 b

Lecture 3173 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 10 -(2)=8 (.5)(2)(8)= 8 sq cm 4 10-(4)=6 (.5)(4)(6)=12 sq cm 6 10-(6)=4 (.5)(6)(4)=12 sq cm 8 10-(8)=2 (.5)(8)(2)=8 sq cm 10 10-(10)=0 b

Lecture 3174 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 10 -(2)=8 (.5)(2)(8)= 8 sq cm 4 10-(4)=6 (.5)(4)(6)=12 sq cm 6 10-(6)=4 (.5)(6)(4)=12 sq cm 8 10-(8)=2 (.5)(8)(2)=8 sq cm 10 10-(10)=0 (.5)(10)(0)=0 sq cm b

Lecture 3175 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 10 -(2)=8 (.5)(2)(8)= 8 sq cm 4 10-(4)=6 (.5)(4)(6)=12 sq cm 6 10-(6)=4 (.5)(6)(4)=12 sq cm 8 10-(8)=2 (.5)(8)(2)=8 sq cm 10 10-(10)=0 (.5)(10)(0)=0 sq cm b 10-(b)

Lecture 3176 A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed. Base H BaseHeight Area (A=.5bh) 0 10 -(0)=10 (.5)(0)(10)=0 sq cm 2 10 -(2)=8 (.5)(2)(8)= 8 sq cm 4 10-(4)=6 (.5)(4)(6)=12 sq cm 6 10-(6)=4 (.5)(6)(4)=12 sq cm 8 10-(8)=2 (.5)(8)(2)=8 sq cm 10 10-(10)=0 (.5)(10)(0)=0 sq cm b 10-(b) (.5)(b)(10-b)= 5b-.5b 2

Lecture 3177 What should the base of the sail be if the area must be 10 sq cm? Base H A = 5b-.5b 2

Lecture 3178 What should the base of the sail be if the area must be 10 sq cm? Base H A = 5b-.5b 2 10 = 5b-.5b 2

Lecture 3179 What should the base of the sail be if the area must be 10 sq cm? Base H A = 5b-.5b 2 10 = 5b-.5b 2 Subtract 10 from both sides

Lecture 3180 What should the base of the sail be if the area must be 10 sq cm? Base H A = 5b -.5b 2 10 = 5b -.5b 2 0 = -.5b 2 + 5b -10 Subtract 10 from both sides

Lecture 3181 Use the Quadratic Formula to solve: 0 = -.5b 2 + 5b - 10

Lecture 3182 Use the Quadratic Formula to solve: 0 = ax 2 + bx + c 0 = -.5b 2 + 5b - 10

Lecture 3183 Use the Quadratic Formula to solve: a = -.5 0 = ax 2 + bx + c 0 = -.5b 2 + 5b - 10

Lecture 3184 Use the Quadratic Formula to solve: a = -.5 b = 5 0 = ax 2 + bx + c 0 = -.5b 2 + 5b - 10

Lecture 3185 Use the Quadratic Formula to solve: a = -.5 b = 5 c = -10 0 = ax 2 + bx + c 0 = -.5b 2 + 5b - 10

Lecture 3191 Use the Quadratic Formula to solve: 0 = -.5x 2 + 5x - 10

Lecture 3194 Use the Quadratic Formula to solve: 0 = -.5x 2 + 5x - 10 Two solutions to make A = 10 sq in: x = 2.8 and x =7.2 inches x = 2.8 and x =7.2 inches

Lecture 3195

Lecture 3196

Lecture 3197

Lecture 311 Unit 4 Lecture 31 Applications of the Quadratic Formula Applications of the Quadratic Formula.

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Lecture 311 Unit 4 Lecture 31 Applications of the Quadratic Formula Applications of the Quadratic Formula.

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Presentation on theme: "Lecture 311 Unit 4 Lecture 31 Applications of the Quadratic Formula Applications of the Quadratic Formula."— Presentation transcript:

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