Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify 243 = = = 9 81 • 3 81 • 3 243. 81 is a perfect square and a factor of 243. 3 Use the Multiplication Property of Square Roots. Simplify 11-1 81. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify 28x7 = = 4x6 • 7x 4x6 • = 2x3 7x 28x7. 4x6 is a perfect square and a factor of 28x7. 7x Use the Multiplication Property of Square Roots. Simplify 11-1 4x6. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. a. 12 • 12 • 32 32 = 12 • 32 Use the Multiplication Property of Square Roots. = 384 Simplify under the radical. = 64 • 6 64 is a perfect square and a factor of 384. = 64 • = 8 6 6 Use the Multiplication Property of Square Roots. Simplify 11-1 64. Simplifying Radicals ALGEBRA 1 LESSON 11-1 (continued) b. 7 5x • 3 8x 7 5x • 3 8x = 21 40x2 Multiply the whole numbers and use the Multiplication Property of Square Roots. = 21 4x2 • 10 4x2 is a perfect square and a factor of 40x2. = 21 4x2 • Use the Multiplication Property of Square Roots. = 21 • 2x = 42x 10 10 Simplify 10 Simplify. 11-1 4x2. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Suppose you are looking out a fourth floor window 54 ft above the ground. Use the formula d = 1.5h to estimate the distance you can see to the horizon. d = 1.5h = 1.5 • 54 Substitute 54 for h. = 81 Multiply. =9 Simplify 81. The distance you can see is 9 miles. 11-1 Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. a. b. 13 64 13 = 64 13 64 Use the Division Property of Square Roots. = 13 8 Simplify 49 = x4 49 x4 Use the Division Property of Square Roots. 64. 49 x4 = 7 x2 Simplify 49 and 11-1 x4. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. a. 120 10 120 = 10 12 Divide. = 4•3 4 is a perfect square and a factor of 12. = 4• =2 3 3 Use the Multiplication Property of Square Roots. Simplify 11-1 4. Simplifying Radicals ALGEBRA 1 LESSON 11-1 (continued) b. 75x5 48x 75x5 = 48x 25x4 16 Divide the numerator and denominator by 3x. = 25x4 16 Use the Division Property of Square Roots. = 25 • 16 5x2 = 4 x4 Use the Multiplication Property of Square Roots. Simplify 25, 11-1 x4, and 16. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. a. 3 7 3 = 7 = = 3 • 7 3 7 49 3 7 7 7 7 Multiply by 7 7 to make the denominator a perfect square. Use the Multiplication Property of Square Roots. Simplify 11-1 49. Simplifying Radicals ALGEBRA 1 LESSON 11-1 (continued) Simplify the radical expression. b. 11 12x3 11 = 12x3 11 • 12x3 3x 3x = 33x 36x4 Use the Multiplication Property of Square Roots. = 33x 6x2 Simplify Multiply by 3x to make the denominator a 3x perfect square. 11-1 36x4. The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 What is the length of the hypotenuse of this triangle? a2 + b2 = c2 Use the Pythagorean Theorem. 82 + 152 = c2 Substitute 8 for a and 15 for b. 64 + 225 = c2 289 = 17 = c c2 Simplify. Find the principal square root of each side. Simplify. The length of the hypotenuse is 17 m. 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 A toy fire truck is near a toy building on a table such that the base of the ladder is 13 cm from the building. The ladder is extended 28 cm to the building. How high above the table is the top of the ladder? Define: Let b = height (in cm) of the ladder from a point 9 cm above the table. Relate: The triangle formed is a right triangle. Use the Pythagorean Theorem. 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 (continued) Write: a2 + b2 = c2 132 + b2 = 282 169 + b2 = 784 Substitute. Simplify. b2 = 615 Subtract 169 from each side. b2 = Find the principal square root of each side. b 615 24.8 Use a calculator and round to the nearest tenth. The height to the top of the ladder is 9 cm higher than 24.8 cm, so it is about 33.8 cm from the table. 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 Determine whether the given lengths are sides of a right triangle. a. 5 in., 5 in., and 7 in. 52 + 52 72 25 + 25 49 50 =/ 49 Determine whether a2 + b2 = c2, where c is the longest side. Simplify. This triangle is not a right triangle. b. 10 cm, 24 cm, and 26 cm 102 + 242 262 Determine whether a2 + b2 = c2, where c is the longest side. 100 + 576 676 Simplify. 676 = 676 This triangle is a right triangle. 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 If two forces pull at right angles to each other, the resultant force is represented as the diagonal of a rectangle, as shown in the diagram. The diagonal forms a right triangle with two of the perpendicular sides of the rectangle. For a 50–lb force and a 120–lb force, the resultant force is 130 lb. Are the forces pulling at right angles to each other? 502 + 1202 2500 + 14,400 1302 Determine whether a2 + b2 = c2 where c is the greatest force. 16,900 16,900 = 16,900 The forces of 50 lb and 120 lb are pulling at right angles to each other. 11-2 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 Find the distance between F(6, –9) and G(9, –4). d= ( x2 – x1)2 + (y2 – y1)2 Use the distance formula. d= (9 – 6)2 + [–4 – (–9)]2 Substitute (9, –4) for (x2, y2) and (6, –9) for (x1, y1). d= 32 + 52 Simplify within parentheses. d= 34 Simplify to find the exact distance. d 5.8 Use a calculator. Round to the nearest tenth. The distance between F and G is about 5.8 units. 11-3 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 Find the exact lengths of each side of quadrilateral EFGH. Then find the perimeter to the nearest tenth. The perimeter = EF = = = = [4 – (–1)]2 + (3 + 5)2 52 + (–2)2 25 + 4 29 FG = = = = (3 – 4)2 + (–2 – 3)2 (–1)2 + (–5)2 1 + 25 26 GH = |–2 – 3| = 5 EH = = = = [–2 – (–1)]2 + (–2 – 5)2 (–1)2 + (–7)2 1 + 49 50 29 + 26 + 5 + 50 11-3 22.6 units. The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 Find the midpoint of CD. x1+ x2 y1+ y2 , 2 2 (–3) + 5 7 + 2 = , 2 2 2 9 = 2,2 Substitute (–3, 7) for (x1, y1) and (5, 2) for (x2, y2). Simplify each numerator. 1 Write 9 as a mixed = 1, 42 2 number. 1 The midpoint of CD is M 1, 4 2 . 11-3 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 A circle is drawn on a coordinate plane. The endpoints of the diameter are (–3, 5) and (4, –3). What are the coordinates of the center of the circle? x1+ x2 y1+ y2 , 2 2 (–3) + 4 5 + (–3) = , 2 2 1 2 = 2,2 1 = 2, 1 1 The center of the circle is at 2 , 1 . 11-3 Substitute (–3, 5) for (x1, y1) and (4, –3) for (x2, y2). Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 Simplify 4 4 3+ 3=4 3+1 = (4 + 1) =5 3 3+ 3 3 3. Both terms contain 3. Use the Distributive Property to combine like radicals. Simplify. 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 Simplify 8 8 5– 45 = 8 5+ =8 5– =8 5–3 = (8 – 3) =5 5 5– 9•5 9• 5 5 5 45. 9 is a perfect square and a factor of 45. Use the Multiplication Property of Square Roots. Simplify 9. Use the Distributive Property to combine like terms. Simplify. 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 Simplify 5( 8 + 9) = = =2 5( 40 + 9 4• 8 + 9). 5 10 + 9 10 + 9 Use the Distributive Property. 5 Use the Multiplication Property of Square Roots. 5 Simplify. 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 Simplify ( ( 6–3 6 – 3 21)( 6 + 21) = 36 + 126 – 3 126 – 3 21)( 441 6+ 21). Use FOIL. =6–2 126 – 3(21) Combine like radicals and simplify 36 and 441. =6–2 9 • 14 – 63 9 is a perfect square factor of 126. =6–2 9• Use the Multiplication Property of Square Roots. =6–6 14 – 63 = –57 – 6 14 – 63 Simplify 14 Simplify. 11-4 9. Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 Simplify 8 7– = 3 7+ 7+ • 3 3 8 7– 3 . Multiply the numerator and denominator by the conjugate of the denominator. = 8( 7 + 3) 7–3 Multiply in the denominator. = 8( 7 + 4 Simplify the denominator. = 2( 7+ =2 7+2 3) 3) 3 Divide 8 and 4 by the common factor 4. Simplify the expression. 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 A painting has a length : width ratio approximately equal to the golden ratio (1 + 5 ) : 2. The length of the painting is 51 in. Find the exact width of the painting in simplest radical form. Then find the approximate width to the nearest inch. Define: 51 = length of painting x = width of painting Relate: (1 + Write: 5) : 2 = length : width 51 (1 + 5) = x 2 x (1 + 5) = 102 x(1 + 5) = 102 (1 + 5) (1 + 5) Cross multiply. Solve for x. 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 (continued) x= (1 – 102 • (1 – (1 + 5) x= 102(1 – 5) 1–5 x= 102(1 – –4 5) 5) 5) Multiply the numerator and the denominator by the conjugate of the denominator. Multiply in the denominator. Simplify the denominator. x = – 51(1 – 5) Divide 102 and –4 by the common factor –2. x = 31.51973343 Use a calculator. 2 x 32 The exact width of the painting is – 51(1 – 5) inches. 2 The approximate width of the painting is 32 inches. 11-4 Solving Radical Equations ALGEBRA 1 LESSON 11-5 Solve each equation. Check your answers. a. x–5 =4 x=9 ( x)2 = 92 Isolate the radical on the left side of the equation. Square each side. x = 81 Check: x–5 –5 9–5 = 4 4 4 Substitute 81 for x. 4=4 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 (continued) b. x–5 =4 ( x – 5)2 = 42 x–5=9 Square each side. Solve for x. x = 21 Check: x–5 21– 5 16 = 4 = 4 = 4 Substitute 21 for x. 4=4 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 On a roller coaster ride, your speed in a loop depends on the height of the hill you have just come down and the radius of the loop in feet. The equation v = 8 h – 2r gives the velocity v in feet per second of a car at the top of the loop. 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 (continued) The loop on a roller coaster ride has a radius of 18 ft. Your car has a velocity of 120 ft/s at the top of the loop. How high is the hill of the loop you have just come down before going into the loop? Solve v = 8 120 = 8 120 = 8 8 15 = h – 2r for h when v = 120 and r = 18. h – 2(18) Substitute 120 for v and 18 for r. h – 2(18) 8 h – 36 (15)2 = ( h – 36)2 225 = h – 36 261 = h The hill is 261 ft high. Divide each side by 8 to isolate the radical. Simplify. Square both sides. 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 Solve ( 3x – 4 = 3x – 4)2 = ( 2x + 3)2 3x – 4 = 2x + 3 3x = 2x + 7 x=7 Check: 3x – 4 = 3(7) – 4 17 = 2x + 3. Square both sides. Simplify. Add 4 to each side. Subtract 2x from each side. 2x + 3 2(7) + 3 17 Substitute 7 for x. The solution is 7. 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 Solve x = x + 12. (x)2 = ( x + 12)2 Square both sides. x2 = x + 12 x2 – x – 12 = 0 Simplify. (x – 4)(x + 3) = 0 Solve the quadratic equation by factoring. (x – 4) = 0 or (x + 3) = 0 x = 4 or x = –3 Use the Zero–Product Property. Solve for x. Check: x = x + 12 4 4 + 12 4 = 4 –3 –3 =/ –3 + 12 3 The solution to the original equation is 4. The value –3 is an extraneous solution. 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 Solve 3x + 8 = 2. 3x = –6 ( 3x)2 = (–6)2 Square both sides. 3x = 36 x = 12 Check: 3x 3(12) 36 6 +8=2 +8 2 +8 2 + 8 =/ 2 Substitute 12 for x. x = 12 does not solve the original equation. 3x + 8 = 2 has no solution. 11-5 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 Find the domain of each function. a. y = x+5 x+5> –0 Make the radicand > – 0. x> – –5 The domain is the set of all numbers greater than or equal to –5. b. y = 6 4x – 12 4x – 12 > –0 Make the radicand > – 0. 4x > – 12 x> – 3 The domain is the set of all numbers greater than or equal to 3. 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 The size of a television screen is the length of the screen’s diagonal d in inches.The equation d = 2A estimates the length of a diagonal of a television with screen area A. Graph the function. Domain 2A > –0 A> –0 Screen Area (sq. in.) 0 50 100 200 300 400 Length of Diagonal (in.) 0 10 14.1 20 24.5 28.3 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 Graph y = y= x + 4 by translating the graph of x. For the graph y = the graph of y = 11-6 x + 4, x is shifted 4 units up. Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 Graph ƒ(x) = y= x + 3 by translating the graph of x. For the graph ƒ(x) = the graph of y = 11-6 x + 3, x is shifted to the left 3 units. Trigonometric Ratios ALGEBRA 1 LESSON 11-7 Use the triangle. Find sin A, cos A, and tan A. opposite leg 6 3 sin A = hypotenuse = = 10 5 adjacent leg 8 4 cos A = hypotenuse = = 10 5 opposite leg tan A = adjacent leg = 6 = 3 8 4 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 Find sin 40° by using a calculator. To find sin 40°, press Use degree mode when finding trigonometric ratios. Rounded to the nearest ten-thousandth, the sin 40° is 0.6428. 11-7 40 . Trigonometric Ratios ALGEBRA 1 LESSON 11-7 Find the value of x in the triangle. Step 1: Decide which trigonometric ratio to use. You know the angle and the length of the hypotenuse. You are trying to find the adjacent side. Use the cosine. Step 2: Write an equation and solve. adjacent leg cos 30° = hypotenuse x cos 30° = 15 x = 15(cos 30°) 15 30 12.99038106 x 13.0 The value of x is about 13.0. Substitute x for adjacent leg and 15 for hypotenuse. Solve for x. Use a calculator. Round to the nearest tenth. 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 Suppose the angle of elevation from a rowboat to the top of a lighthouse is 708. You know that the lighthouse is 70 ft tall. How far from the lighthouse is the rowboat? Round your answer to the nearest foot. Draw a diagram. Define: Let x = the distance from the boat to the lighthouse. Relate: You know the angle of elevation and the opposite side. You are trying to find the adjacent side. Use the tangent. 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 (continued) Write: opposite leg tan A = adjacent leg 70 tan 70° = X x(tan 70°) = 70 Substitute for the angle and the sides. Multiply each side by x. x= 70 tan 70 Divide each side by tan 70°. x 25.4779164 Use a calculator. x 25 Round to the nearest unit. The rowboat is about 25 feet from the lighthouse. 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 A pilot is flying a plane 15,000 ft above the ground. The pilot begins a 3° descent to an airport runway. How far is the airplane from the start of the runway (in ground distance)? Draw a diagram. Define: Let x = the ground distance from the start of the runway. Relate: You know the angle of depression and the opposite side. You are trying to find the adjacent side. Use the tangent. 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 (continued) Write: opposite leg tan A = adjacent leg tan 3° = 15,000 x x(tan 3°) = 15,000 Substitute for the angle and the sides. Multiply each side by x. x= 15,000 tan 3 Divide each side by tan 3°. x 286217.05 Use a calculator. x 290,000 Round to the nearest 10,000 feet. The airplane is about 290,000 feet (or about 55 miles) from the start of the runway. 11-7