11 8 = 11 (8) = = 100(2) = 10 2

Chapter 5 Radical Expressions and Equations Section 5.1 Working With Radicals Section 5.1 Page 78 Question 1 Mixed Radical Form Entire Radical Form 4 7 4 7 4 (7) 11 50 5() 5 50 11 8 11 8 11 (8) 968 00 100() 10 00 Section 5.1 Page 78 Question a) 56 4(14) 14 b) 75 5() 15 4 8() c) () d) cd c()( c d ) cd c Section 5.1 Page 78 Question a) m 4 8m 4()( m )( ) 6m, m R b) 4q ()( q )( q ) 5 q q, q R c) 5 5 6 5 5 5 5 160st (5)( s)( t )( t) 5 4st 5 t, s, t R MHR Pre-Calculus 11 Solutions Chapter 5 Page 1 of 66

Section 5.1 Page 79 Question 4 Mixed Radical Form Entire Radical Form n 5 n 5 ( n) (5) 4 ( 6) 6 1 a 7a < 45 n, n 0 or 45 n, n 0 4 1 1 7 a ( 7 a) a a 1 (7 a ) 8a 18x 4 () x ( x) 4 4x x 7, 0 8a a 4 18x Section 5.1 Page 79 Question 5 a) For 15 5 and 8 15, express the second radical in terms of 5. 8 15 8 5(5) 15 5 40 5 8 4 b) For 8 11z and 48 7z, express both radicals in terms of 7. 8 8 8 11z 8 16(7) z 4 48 7z 48z 7 4 z 7 c) For 4 5 w 4 5 w and 4 10 81w, express the second radical in terms of 4 10 4 4 8 81w ( w )( w ) 9w w 4 w. d) For 6 6 and 6 54, express the second radical in terms of. 4 6 54 6 () 18 Section 5.1 Page 79 Question 6 MHR Pre-Calculus 11 Solutions Chapter 5 Page of 66

a) 6 9(6) 10 100 7 49() 54 98 The numbers from least to greatest are 6, 7, and 10. b) 4() 1 4 16 9() The numbers from least to greatest are, 4, 18 7 7 4 14 7, and. c) 1 7() 54.8.8 1.95 The numbers from least to greatest are 1,.8, 5, and Section 5.1 Page 79 Question 7.464... 4 4.4... The numbers from least to greatest are, 4, 5 8(5). 40 7.741... 7, and. Section 5.1 Page 79 Question 8 a) 5+ 9 5 4 5 4 5 b) 1.4 + 9 7 10.4 7 c) 4 11 1 5 4 11 + 15 4 4 11 + 14 d) 9 5 1 6+ 10 10+ 6 6+ 10 Section 5.1 Page 79 Question 9 a) 75 7 5() 9() 15 1 MHR Pre-Calculus 11 Solutions Chapter 5 Page of 66

b) 18 + 9 7 6 9() + 9 7 9(7) 6 + 9 7 7 6 + 6 7 c) 8 45 + 5.1 80 + 17.4 8 9(5) 16(5) +.5 4 5 4 5 +.5 8 5 +.5 d) 75 15() 81 + 4 99 + 5 11 7() + 4 9(11) + 5 11 4 4 5 1 11 5 11 + + 4 1 7 11 4 Section 5.1 Page 79 Question 10 a) a + 6 a 8 a 8 a a, a 0 b) x + 8x x x + 4()( x) x + 6 x x 9 x x, x 0 x c) d) + + 4 4 65r 40r 4 15(5)( r) 8(5)( r )( r ) 0 5r + r 5r r ( 0) 5 r r r ( 10) 5, R 51 4 8 r w w w w 64 + 50w 4 w + 5()( w) 4 w 5 5 5 5 5 5 4w w 4 w 5 4w 6 w, w 0 5 MHR Pre-Calculus 11 Solutions Chapter 5 Page 4 of 66

Section 5.1 Page 79 Question 11 w 6. 101 p w 6. 101 965 w 6. 48 w 5. The exact wind speed of a hurricane if the air pressure is 965 mbar is 5. m/s. Section 5.1 Page 79 Question 1 c 1 + 1 c 144 + 144 c 88 c 88 c 1 The length of the hypotenuse is 1 cm. Section 5.1 Page 80 Question 1 Distance to Mars: d d d 5n 5( 704) 5[64(11)] Distance to Mercury: d d d 5n 5( 88) 5[8(11)] d 16 5(11) d 4 5(11) d 16 05 d 4 05 The difference between the distances of Mars and Mercury to the Sun is 16 05 4 05, or 1 05 million kilometres. Section 5.1 Page 80 Question 14 s s 10d 10( 1) s 10 s 0 The speed of a tsunami with depth 1 m is 0 m/s, or 11 m/s to the nearest metre per second. Section 5.1 Page 80 Question 15 MHR Pre-Calculus 11 Solutions Chapter 5 Page 5 of 66

a) Since the given area of the circle (πr ) is 8π m, the radius is 8 m. So, the diagonal of the square, which is also the diameter of the circle is 8 m. b) First, find the side length of the square. s + s ( 8 ) s 15 s 76 s 76 s 19 The perimeter of the square is 4( 19 ), or 8 19 m. Section 5.1 Page 80 Question 16 First, find the half-perimeter for a triangle with side lengths 8 mm, 10 mm, and 1 mm. 0.5(8 + 10 + 1) 15 Find the area of the triangle using Heron s formula. A s( s a)( s b)( s c) A 15( 15 8)( 15 10) ( 15 1) A 15(7)(5)() A 1575 A 15 7 The area of the triangle is 1575 mm or 15 7 mm. Section 5.1 Page 80 Question 17 Using a diagram, the points lie on the same straight line. Use the Pythagorean theorem to find the distance between the starting and ending points. c 7 + 14 c 49 + 196 c 45 c 45 c 7 5 The ant travels 7 5 units. 16 1 8 4 y 10 7 18 4 14 4 8 1 16 x MHR Pre-Calculus 11 Solutions Chapter 5 Page 6 of 66

Section 5.1 Page 80 Question 18 Since the area of the entire square backyard is 98 m, the side length is 98 m. Since the area of the green square is 8 m, the side length is 8 m. Find the perimeter of one of the rectangular flowerbeds. P 8+ ( 98 8 ) P 98 P 14 The perimeter is 14 m. 98 m 8m Section 5.1 Page 80 Question 19 Brady is correct. Kristen s final radical, 5y 4y, is not in simplest form. Express the radicand as a product of prime factors and combine identical pairs. 5y 4y 5y ()( y)( y)( y) 10y y Section 5.1 Page 80 Question 0 16 6(6) 96 16(6) 1 6 1 6 The expression 4 58 is not equivalent to 1 6. Section 5.1 Page 81 Question 1 4 58 6 4 6 4(6) 1 6 From the given information, square ABCD has perimeter 4 m, so CD 1 m CDE is an equilateral triangle, so CDE 60 CA is a diagonal, so DAC 45 In ADF, draw the perpendicular from F to AD, meeting AD at point G. Let the height of FG be x. Then, in FDG, FDG 0 and DG 1 x. 45 x G 1 x 60 1 m MHR Pre-Calculus 11 Solutions Chapter 5 Page 7 of 66

x tan 0 1 x 1 x 1 x 1 x x 1 x( + 1) 1 x + 1 1 x Next, in AFG, FG GA x. AF x + x AF 1 1 + 4 4 AF The exact length of AF is m. Section 5.1 Page 81 Question Since AB 18 cm, AC 9 cm and CD 4.5 cm. Consider right ADF, where DF 4.5 cm and AD 1.5 cm. Use the Pythagorean theorem. (AF) (AD) (DF) (AF) 1.5 4.5 (AF) 16 AF 16 AF 9 F Consider similar triangles ADF and ABE. MHR Pre-Calculus 11 Solutions Chapter 5 Page 8 of 66

AE AB AF AD AE 18 9 1.5 18 AE (9 ) 1.5 AE 1 The exact length of AE is 1 cm. Section 5.1 Page 81 Question Given t 1 7 and t 4 9. Use the formula for the general term of an arithmetic sequence with t 1 7, t 4 9, and n 4. t n t 1 + (n 1)d 9 7 +( 4 1)d 9 7 d 9 d 6 d d Find the two missing terms, t and t. t t 1 + d t t + d t 7 + t 5 + t + t 7 t 5 The common difference is and the missing terms are 5 and 7. Section 5.1 Page 81 Question 4 Examples: a) To find the greatest sum of two of the radicals, simplify the two positive radicals and then add. 1 75 + 108 10 + 6 16 b) To find the greatest difference of two of the radicals, simplify the greatest positive radical and the least negative radical and then subtract. 75 ( 1) 10 ( 6 ) 16 MHR Pre-Calculus 11 Solutions Chapter 5 Page 9 of 66

Section 5.1 Page 81 Question 5 Examples: a) Let x and x. ( x) x ( x) x ( ) ( ( )) ( ) 4 4 4 4 b) Let x and x. x ±x x ±x ± ( ) ±( ) + ± + ± Section 5. Multiplying and Dividing Radical Expressions Section 5. Page 89 Question 1 a) ( ) 5 7 (7) 5() b) ( ) 14 15 7 1(7) () c) ( ) 7 64 56 4 4 48 5 4 48(5) d) ( ) 4 40 4 4 15 4 19x x 4x 19 x() 4x 8x e) ( ) ( ) 54y 7 6y 4 y y y 6y y 1y MHR Pre-Calculus 11 Solutions Chapter 5 Page 10 of 66

f) t t 6t t 6t 4 t 6 Section 5. Page 89 Question a) 11( 4 7) 11() 11( 4 7 ) 11 4 77 b) ( 14 5 + 6 1 )) ( 14 5) ( 6 ) ( 1) c) y( y + 1) y( y) + y() 1 14 10 1 + 6 14 10 6 + 6 y+ y d) z ( z 1 5z+ ) z ( z 1) z ( 5z) + z ( ) 6 5 + z z z + 6z 5z z Section 5. Page 89 Question a) ( ) 4 + 9 + 1+ 9 b) ( ) 6 + 1 7 1 6 + 5 6+ 8 7 14 6+ 5 6+ 8 c) ( ) 1 9 6 4 5 j + 8 15j + 5 4 15j + 5 15j + 5 d) k ( ) 15 j + 5, j 0 4 1+ 8 1 4k 4 4k 16 4 k MHR Pre-Calculus 11 Solutions Chapter 5 Page 11 of 66

Section 5. Page 90 Question 4 a) ( )( ) ( ) ( ) 8 7 + 8 7 8 7 + 6 8 14 4 7 + 6 b) ( 4 9 5)( 4+ 9 5) 4(4) + 4( 9 5) 9 5( 4) 9 5( 9 5) 16 + 6 5 6 5 81(5) 89 c) ( + 15)( 15) ( ) ( 15) + 15 ( ) 15 ( 15) 45 + 45 0 7 + 45 7 + 5 d) ( 6 ) ( )( ) 4 1 6 4 1 6 4 1 6 ( 6 ) 6 ( 4 1) 4 1( 6 ) + 4 1( 4 1) 6 4 4 ( 1) 4 ( 1) + 08 6 4 48 ( 1) + 08 Section 5. Page 90 Question 5 15 c + c 6 15 c c 15 c 6 + c (6) a) ( )( ) ( ) ( ) 15c 90 c + c 1, c 0 b) ( 1 10 8 )( ) ( ) ( ) ( ) 7 5 1() 1 7 5 10 8 x + x + x x 10 8x 7 5x + 7 5x 0 8x 70 40x 4 + 7 5x 40x x 140x 10, x 0 c) ( 9 m 4 6m) ( 9 m 4 6m)( 9 m 4 6m) 9 m( 9 m) 9 m( 4 6m) 4 6m( 9 m) + 4 6m( 4 6m) 16 6 1 6 1 96 m m m + m 58m 144m, m 0 MHR Pre-Calculus 11 Solutions Chapter 5 Page 1 of 66

d) ( 10 4 )( ) ( ) ( ) ( ) ( ) 4 6 1 10 6 10 1 4 4 6 4 r r r + r r r + r r r r 4r 1r 0r 6r + 0r 1r 8 4r 1 48r 0r 6r + 0r 1r 16r 4 6r Section 5. Page 90 Question 6 a) b) 80 80 10 10 8 1 1 1 4 1 4 1 c) 11 11 d) 5 5 15m 15m 1 1m m 45m 7 5 7 9m 7 7 9m 5 7 Section 5. Page 90 Question 7 a) MHR Pre-Calculus 11 Solutions Chapter 5 Page 1 of 66

5 5 9 4 p 7 7 p 108p p 1p p 4 87 p 61 11p 87 11p 11 b) 7 7 6 4v 4v 6 14 v 14 v 6v 7 6 6v 98 7 ( 7 ) ( ) v 7 7 Section 5. Page 90 Question 8 p p a) b) 0 0 10 10 10 10 0 10 10 10 1 1 7m 7m 7m 7m 147m 7m 7 m 7m m m MHR Pre-Calculus 11 Solutions Chapter 5 Page 14 of 66

c) 5 5 1u 1u 1u 1u 60u 1u 15u 1u 15u 9u d) 0 150 4 150 ( 5) ( ) 6t 6t 0 0 5 5 5 5 t t Section 5. Page 90 Question 9 a) The conjugate for + 1 is 1. ( )( ) ( ) + 1 1 1 1 1 11 b) The conjugate for 7 11 is 7+ 11. ( 7 11 )( 7 + 11 ) 7 ( ) 11 49 11 8 c) The conjugate for 8 z 7 is 8 z + 7. ( 8 z 7)( 8 z + 7) ( 8 z) ( 7) 64z 6 d) The conjugate for 19 h + 4 h is 19 h 4 h. MHR Pre-Calculus 11 Solutions Chapter 5 Page 15 of 66

( 19 h + 4 h)( 19 h + 4 h) ( 19 h) ( 4 h) 61h h 9h Section 5. Page 90 Question 10 a) 5 5 + + ( + ) ( ) ( + ) 5 5 4 10 + 5 b) 7 7 6 8 6+ 8 6+ 8 6 8 c) ( ) ( 6) 8 ( ) 7 6 8 7 6 8 6 64 7 1 56 58 14 56 58 7 + 8 9 MHR Pre-Calculus 11 Solutions Chapter 5 Page 16 of 66

7 7 5+ 5 5 5+ 7( 5+ ) ( 5) ( ) 7( 5+ ) 5 8 5 + 14 d) + 1 + 1 + 1 1 1 + 1 ( ) + ( 1) + ( 1) ( ) ( 1) + 9 + 1 1 16 + 9 10 8 9 5 Section 5. Page 90 Question 11 a) 4r 4r 6r 9 6r+ 9 6r+ 9 6r 9 4r( 6r 9) ( ) 6r 9 r r r 4 6 6 ±, r 6 81 6 MHR Pre-Calculus 11 Solutions Chapter 5 Page 17 of 66

b) 18 n 18 n 4n 4n 4n 4 n 18 7n 4n 108n 4n 9, n > 0 c) 8 8 4+ 6t 4 6t 4 6t 4+ 6t ( + t ) ( t ) 84 6 4 6 + 8 6t 16 6t 16 + 4 6t 8, t 0, t 8 t d) 5 y 5 y 10 10 + 10 + 10 y ( ) ( ) 10 5 10 5 0y 10 y 10 4 5 0y 10 y, y 0 6 Section 5. Page 90 Question 1 ( c+ c c)( c+ 7 c) c( c) + c( 7 c) + c c( c) + c c( 7 c) c + 7c c + c c + 7c Section 5. Page 90 Question 1 MHR Pre-Calculus 11 Solutions Chapter 5 Page 18 of 66

a) When applying the distributive property in line, Malcolm distributed the 4 to both the whole number and the root of the second term in the numerator,. The correct numerator is 1 + 8. 4 4 + + 1 + 8 9 8 1 + 8 b) Check: 4.1 and 1 + 8.1. Section 5. Page 91 Question 14 5+ 1 5 1 5 1 5+ 1 ( 5+ 1) ( 5) 1 ( 5+ 1) 5 1 ( + ) 5 1 4 5+ 1 Section 5. Page 91 Question 15 L a) T π 10 L 10 π 10 10 π 10 10 π 10 5 L L MHR Pre-Calculus 11 Solutions Chapter 5 Page 19 of 66

b) Find the period for L 7. π 10L T 5 π 10( 7) 5 π 0 5 The time to complete cycles is π 0 5, or 9π 0 5 s. Section 5. Page 91 Question 16 The triangular course is in the shape of an isosceles triangle. The length of the base is 4 units. Use the Pythagorean theorem to find the length of one of the equal sides. c 4 + c 16 + 4 c 0 c 0 c 5 Find the perimeter of the triangular course. P 4 + ( 5 ) P 4 + 4 5 Find the exact length of the track. 945 4 + 4 5 4 945 + 4 465 ( ) ( ) 4 4 5 + 4(15) 17 5 + 860 The exact length of the track is (17 5 + 860 ) m. Section 5. Page 91 Question 17 MHR Pre-Calculus 11 Solutions Chapter 5 Page 0 of 66

1 ( 5) ( ) ( ) 1+ 5 1 5 + () 1 5 4 4 + 4 7+ 4 7 4 7+ 4 ( ) ( ) 47+ 4 7 4 8 16 Section 5. Page 91 Question 18 a) 19 4 The edge length of the actual cube 4 mm. b) 19 4 48 6 c) 4 : 6 : 6 Section 5. Page 91 Question 19 a) Lev forgot to reverse the direction of the inequality sign when he divided both sides by 5. The corrected calculation follows: 5x > 0 5x > x < 5 b) Variables involved in radical expressions sometimes have restrictions on their values to ensure a non-negative value. This is the case for radicands with an index that is an even number. c) Example: x 5 14 x MHR Pre-Calculus 11 Solutions Chapter 5 Page 1 of 66

The expression does not have a variable in the denominator and radicands with an odd number index may be any real number. Section 5. Page 9 Question 0 Olivia made an error in line. She incorrectly evaluated 5 as ±5 instead of 5. The corrected solution follows: ( c c 5) c c 5 c ( c c ) 5 ( c c) 5 ( c) Section 5. Page 9 Question 1 For the right triangular prism, h 5 7, b, and l 7 14. 1 V bhl 1 V ( )( 5 7 )( 7 14 ) 105 V (7)(14) 105 V (14) V 75 The volume of the right triangular prism is 75 cm. Section 5. Page 9 Question First, determine the side length of the cube inscribed in a sphere of radius 1 m. s + s s 4 s s 1 m s MHR Pre-Calculus 11 Solutions Chapter 5 Page of 66

Determine the surface area of the cube. SA 6s SA 6( ) SA 1 The surface area of the cube is 1 m. Section 5. Page 9 Question 7 + 48 50 + 98 Midpoint AB, + 1 5 + 14, 15 9, Section 5. Page 9 Question 4 1 ( ( x ) ) 5 5 x 5 x x x 5 x x 9+ 5x+ 0 x 9 + 5x 0 x 9 5x 0 x + + 9x+ 5x + 0x x 81 450x 65x Section 5. Page 9 Question 5 a) Use the quadratic formula with a 1, b 6, and c. MHR Pre-Calculus 11 Solutions Chapter 5 Page of 66

± x b b ac a 6± 4 4 6± 6 4( 1)( ) x ( 1) x 6± 6 x x ± 6 The exact roots are x + 6 and x 6. b) + 6+ ( 6) 6 c) ( )( ) + 6 6 9 6 d) Example: The answer to part b) is equal to Section 5. Page 9 Question 6 b. The answer to part c) is equal to c a a. r r r c c n 1 a a r n n n 1 c a r n 1 r Section 5. Page 9 Question 7 First, find the length of the hypotenuse of the triangular faces. c c c ( 5 7) ( ) + 175 + 18 19 c 19 MHR Pre-Calculus 11 Solutions Chapter 5 Page 4 of 66

( ) ( ) ( ) ( ) SA 5 7 + 5 7 7 14 + 7 14 + 19 7 14 15 14 + 5 98 + 1 8 + 7 70 15 14 + 45 + 4 7 + 7 70 The exact surface area of the right triangular prism is (15 14 + 45 + 4 7 + 7 70 ) cm. Section 5. Page 9 Question 8 Example: You can multiply and divide polynomial expressions with the same variables, and you can multiply and divide radicals with the same index. Section 5. Page 9 Question 9 To rationalize a square-root binomial denominator, multiply the numerator and denominator by the conjugate of the denominator. The product of a pair of conjugates is a difference of squares. 5 5 + + ( + ) ( ) ( + ) 5 5 4 10 + 5 Section 5. Page 9 Question 0 a) Substitute t 0. h(t) 5t + 10t + h(0) 5(0) + 10(0) + h(0) The snowboarder s height above the landing area at the beginning of the jump is m. b) h(t) 5t + 10t + h(t) 5(t t) + h(t) 5(t t + 1 1) + h(t) 5(t 1) + 5 + h(t) 5(t 1) + 8 Isolate t. h(t) 5(t 1) + 8 h(t) 8 5(t 1) MHR Pre-Calculus 11 Solutions Chapter 5 Page 5 of 66

ht () 8 ( t 1) 5 8 ht ( ) t 1 5 t 1+ 8 ht ( ) 5 c) First, determine the time to complete the jump by using the result from part b) with h(t) 0. 8 ht ( ) t 1+ 5 t 1+ 8 0 5 8 t 1+ 5 5+ 8 5 t 5 5 t 5+ 40 5 5+ 10 t 5 So, the time halfway through the jump is 5 + 10 10 Substitute t 5 + 10 into h(t) 5t + 10t +. 10 s. MHR Pre-Calculus 11 Solutions Chapter 5 Page 6 of 66

5+ 10 5+ 10 5+ 10 h 5 + 10 + 10 10 10 5 + 0 10 + 40 5 + 5+ 10+ 100 65 0 10 + 8 + 10 0 1 4 10 + 8 + 10 4 1 4 10 + + 8 10 4 19 + 4 10 4 The exact height of the snowboarder halfway through the jump is Section 5. Page 9 Question 1 Use the quadratic formula with a, b 5, and c 1. ± m b b 4ac a 5± 5 4( )( 1) m ( ) 5 ± 1 m 6 The solutions m 5+ 1 6 and m 5 1 6 Section 5. Page 9 Question are correct. 19 + 4 10 4 m. MHR Pre-Calculus 11 Solutions Chapter 5 Page 7 of 66

a) V π V 4π V 1 π V 1 4π 6V ( V 1) ( V ) V 1 1 6 VV ( 1) V 1 b) For volumes greater than 1 the ratio is a real number. Section 5. Page 9 Question Step 1 Step Example: The values of x and y have been interchanged. Step Example: The restrictions on the radical function produce the right half of the parabola. MHR Pre-Calculus 11 Solutions Chapter 5 Page 8 of 66

Section 5. Radical Equations Section 5. Page 00 Question 1 a) ( ) z z x 4 x 4 b) ( ) c) ( ) x+ 7 4( x+ 7) d) ( ) 4x + 8 4 9 y 16(9 y) 144 y Section 5. Page 00 Question Isolate the radical expression. Square both sides of the equation. x + 5 11, x 0 x 6 ( ) x 6 x 6 Section 5. Page 00 Question a) x, x 0 ( ) x x 9 9 x Check for x 9. Left Side Right Side x MHR Pre-Calculus 11 Solutions Chapter 5 Page 9 of 66

9 9 Left Side Right Side The solution is x. b) 8x 4, x 0 ( ) 8x 4 8x 16 x Check for x. Left Side Right Side 8x 4 8( ) 16 4 Left Side Right Side The solution is x. c) 5 7 5 x, x ( x ) 7 5 49 5 x x 44 x Check for x. Left Side Right Side 7 5 x 5 49 7 Left Side Right Side The solution is x. ( ) MHR Pre-Calculus 11 Solutions Chapter 5 Page 0 of 66

Section 5. Page 00 Question 4 a) z + 8 1, z 0 z 5 ( ) z 5 z 5 Check for z 5. Left Side Right Side z + 8 1 5 + 8 5+ 8 1 Left Side Right Side The solution is z 5. b) y 4, y 0 y 6 y 6 ( ) y 6 y 6 Check for y 6. Left Side Right Side y 4 6 6 4 Left Side Right Side The solution is y 6. c) x 8 6, x 0 x ( ) x x 4 4 x MHR Pre-Calculus 11 Solutions Chapter 5 Page 1 of 66

Check for x 4. Left Side Right Side x 8 6 4 8 4 8 6 Left Side Right Side The solution is x 4. d) 5 6 m, m 0 6m 7 ( ) 6m 7 6m 49 49 m 6 49 Check for m. 6 Left Side Right Side 5 6m 49 6 6 49 5 Left Side Right Side 49 The solution is m. 6 Section 5. Page 00 Question 5 k+ 4 k, k 0 ( k) ( k+ 4) + 8 + 16 k + 10k+ 16 0 ( k+ 8)( k+ ) 0 k k k MHR Pre-Calculus 11 Solutions Chapter 5 Page of 66

k + 8 0 or k + 0 k 8 k Check for k 8. Check for k. Left Side Right Side Left Side Right Side k + 4 k k + 4 k 8 + 4 ( 8) + 4 ( ) 4 16 4 4 Left Side Right Side Left Side Right Side k 8 is an extraneous root. Section 5. Page 00 Question 6 a) n 1+ 7 14, n 1 n 1 1 n 1 7 ( ) n 1 7 n 1 49 n 50 1 b) 7 4 x 1 17, x 4 x 1 4 x 1 6 There is no solution. c) 1 + 5 8 x, x 8 15 5 8 x 8 x ( x ) 8 9 8 x x 1 MHR Pre-Calculus 11 Solutions Chapter 5 Page of 66

Section 5. Page 01 Question 7 a) m m m ( m ) 5, or 5 m 5 m 8 m ± 8 m ± 7 b) x x x x ( x x) + 1 8, 1 or 0 + 1 8 x + 1x 64 x + 1x 64 0 ( x+ 16)( x 4) 0 x + 16 0 or x 4 0 x 16 x 4 c) q + 11 q 1 q + 11 ( q 1) q + 11 q q+ 1 q + q 4q+ 0 q 4q 0 Use the quadratic formula with a 1, b 4, and c 0. MHR Pre-Calculus 11 Solutions Chapter 5 Page 4 of 66

± q b b 4ac a ( 4 4 1)( 0 q ( 1) 4± 96 q 4± 4 6 q q ± 6 ) ± ( ) 4( ) d) n n 7 14, n 7 or x 7 + n 7 14 n ( n ) 7 7 n 7 (7 n) n 7 49 14n+ n 14n 56 n n 56 14 4 Section 5. Page 01 Question 8 a) 5 5+ x 5 x, x x 5 x 5 ( ) x 5 ( x 5) n x 5 x 10x+ 5 0 x 1x+ 0 0 ( x 10)( x ) x 10 0 or x 0 x 10 x Since x is an extraneous root, the solution is x 10. MHR Pre-Calculus 11 Solutions Chapter 5 Page 5 of 66

b) x x x x ( x x) x + 0 8, 0 or 0 + 0 8 x + 0x 64 + 0x 64 0 ( x+ )( x ) 0 x + 0 or x 0 x x c) d + 5 d 1, d 5 ( ) d + 5 ( d 1) d + 5 d d + 1 0 d d 4 0 ( d 4)( d + 1) d 4 0 or d + 1 0 d 4 d 1 Since d 1 is an extraneous root, the solution is d 4. d) j + 1 + 5 j j 1, j 1 j + 1 j 1 j + 1 ( 1) j j + 1 4 j + 4 j + 1 j+ 1 1j + 1j+ 0 1j + 11j+ 0 (j+ )(4j+ 1) j + 0 or 4j + 1 0 j j 1 4 Since j 1 4 is an extraneous root, the solution is j. MHR Pre-Calculus 11 Solutions Chapter 5 Page 6 of 66

Section 5. Page 01 Question 9 a) b) c) k 8, k 0 ( k ) ( 8) k 8 k 4 m 7 m, m 0 ( m) ( 7m) m 7m 4m 0 m 0 j 5 00, j 0 j 5 00 5 j 00 5 j 400 j 16 ( ) MHR Pre-Calculus 11 Solutions Chapter 5 Page 7 of 66

d) 5+ n n, n 0 5 n ( n n) 5 n ( ) n n n 5 + 5 4n n 5 n 4 5 n 4 5 4 + n 4 4 + 100 + 50 n 4 50 + 5 n Section 5. Page 01 Question 10 a) 1 z+ 5 z 1, z ( z+ 5) ( z 1) z+ 5 z 1 z 6 z 6 b) y + y y 6 1 17, 17 ( 6y 1) ( 17+ y ) 6y 1 17+ y 0 y 6y 16 0 ( y 8)( y+ ) y 8 0 or y + 0 y 8 y Since y is an extraneous root, the solution is y 8. MHR Pre-Calculus 11 Solutions Chapter 5 Page 8 of 66

c) 9 5r 9 r+ 4, r 5 5r 9 1 r+ 4 ( 5r 9 1) ( r+ 4) ( ) 4r 1 ( 5r 9) ( r 6) ( 5r 9) 5r 9 5r 9 + 1 r+ 4 r r+ r 4 4 6 5 9 r r+ 4 9 45 0 (4r 9)( r 5) 0 4r 9 0 or r 5 0 r 9 4 r 5 Since r 9 4 is an extraneous root, the solution is r 5. d) x+ 19 + x 7, x x+ 19 7 x ( x+ 19 7) ( x ) ( ) 14( x + 19 ) 70 ( x ) x+ 19 14 x+ 19 + 49 x x + 19 5 + 19 5 x + 19 5 x 6 Section 5. Page 01 Question 11 Isolating the radical in the equations y 1 5 and 4 m + 6 9 results in it being equated to a positive value, which has a solution. The equation x + 8+ 9 will have an extraneous root. Isolating the radical results in it being equated to a negative value, which has no solution. MHR Pre-Calculus 11 Solutions Chapter 5 Page 9 of 66

Section 5. Page 01 Question 1 Jerry forgot to note the restriction in the first step, and he made a mistake in the third line when he incorrectly squared the expression x. The corrected solution is shown: + x+ 17 x, x 17 x+ 17 x ( x+ 17 ) ( x ) 17 6 9 x+ x x+ 0 x 7x 8 0 ( x 8)( x+ 1) x 8 0 or x + 1 0 x 8 x 1 Since x 1 is an extraneous root, the solution is x 8. Section 5. Page 01 Question 1 Substitute v 50 and solve for l. v 1.6 l + 8 50 1.6 l + 8 4 l 1.6 4 ( l ) 1.6 11.111... l The length of skid mark expected is 11.1 m, to the nearest tenth of a metre. Section 5. Page 01 Question 14 a) Substitute v 40. B 1. v+ 10.0.49 B 1. 40 + 10.0.49 B 1. 50.0.49 B 5.914... The wind scale is 6. b) Substitute B. MHR Pre-Calculus 11 Solutions Chapter 5 Page 40 of 66

B 1. v+ 10.0.49 1. v + 10.0.49 6.49 1. v + 10.0 6.49 1. v + 10.0 ( v 10.0 ) 6.49 + 1. 6.49 v + 10.0 1. 6.49 10.0 v 1. 1.811... v The wind velocity is approximately 1.8 km/h. Section 5. Page 0 Question 15 Substitute t 4. 1 m t 5 1 m 4 5 m 0 m 0 m 400 100 m It can support a mass of 100 kg. MHR Pre-Calculus 11 Solutions Chapter 5 Page 41 of 66

Section 5. Page 0 Question 16 n+ n, n 0 n n ( n) ( n ) n n 4n+ 4 0 n 5n+ 4 0 ( n 4)( n 1) n 4 0 or n 1 0 n 4 n 1 Since n 1 is an extraneous root, the solution is n 4. Section 5. Page 0 Question 17 a) v h(9.8), h 0 v 19.6h b) Substitute v 0 and solve for h. v 19.6h 0 19.6h ( h ) 0 19.6 900 19.6h 900 h 19.6 45.918... h The spray is expected to reach a height of approximately 45.9 m. c) Substitute v 5 and solve for h. v 19.6h 5 19.6h ( h ) 5 19.6 900 19. 6h 15 h 19.6 6.5 h The pump can reach a maximum height of 6.5 m, which meets the 60-m minimum requirement if only just. MHR Pre-Calculus 11 Solutions Chapter 5 Page 4 of 66

Section 5. Page 0 Question 18 Substitute h 00 and d 1609 and solve for r. d rh+ h 1609 ( r 00) + 00 1609 400r + 40 000 1609 400( r + 100) 1609 0 r + 100 1609 0 r + 100 + ( r ) 80.45 100 647.05 r + 100 67.05 r The radius of Earth is approximately 67. km. Section 5. Page 0 Question 19 x ax + x ax ( x ) ( ax ) x 4 x + 4 ax x 4 x + 4 a x Section 5. Page 0 Question 0 a) Example: A radical equated to a negative value will result in no solution. 1 x + 0 1 x ( ) 1 x ( ) 1 x 4 x The result x is an extraneous solution and no solutions are valid. b) Example: One extraneous root can occur when the squaring of both sides results in a quadratic expression. MHR Pre-Calculus 11 Solutions Chapter 5 Page 4 of 66

( x ) ( x+ 10) x x x x x x+ 10 4 + 4 + 10 5x 6 0 ( x 6)( x+ 1) 0 x 6 0 or x + 1 0 x 6 x 1 Since x 1 is an extraneous root, the solution is x 6. Section 5. Page 0 Question 1 tm te 0.5 h h 0.5 1.8 4.9 h h 0.5 1.8 4.9 h h h h 0.5 1.8 1.8 4.9 + 4.9 6.7h h 0.5 8.8 8.8 6.7h 1 h 0.5 8.8 8.8 6.7h 8.8 h 0.5 8.8 8.8 6.7 8.8 h 0.5 8.8 8.8 h 0.5 6.7 8.8 h.900... The object was dropped from a height of.9 m. Section 5. Page 0 Question Substitute r 1740 and d 610 and solve for h. MHR Pre-Calculus 11 Solutions Chapter 5 Page 44 of 66

d rh+ h 610 ( 1740) h+ h 610 480h+ h ( h h ) 610 480 + 7100 480h+ h 0 h + 480h 7100 Use the quadratic formula with a 1, b 480, and c 7 100. b± b 4ac h a 480 ± 480 4( 1)( 7 100) h ( 1) 480 ± 1 598 800 h 480 ± 0 997 h h 1740 ± 10 997 h 1740 + 10 997 or h 1740 10 997 h 10.87... h 58.857... Since h must be positive, the distance to the horizon is 104 km, to the nearest kilometre. Section 5. Page 0 Question a) Complete the square to find the vertex. P n + 00n P (n 00n) P (n 00n + 10 000 10 000) P (n 100) + 10 000 The maximum profit is $10 000 with 100 employees. b) P (n 100) + P 10 000 (n 100) P + 10 000 (n 100) P + 10 000 n 100 P + 10 000 + 100 n c) P 10 000 MHR Pre-Calculus 11 Solutions Chapter 5 Page 45 of 66

d) The original function has domain {n 0 n 00, n W} and range {P 0 P 10 000, n W}. The restriction in part c) is similar to the range of this function. Section 5. Page 0 Question 4 Example: Both types of equations may involve rearranging and factoring. Solving a radical involves squaring both sides while solving a quadratic equation by completing the square involves taking a square root. Section 5. Page 0 Question 5 Example: Extraneous roots may occur because squaring both sides and solving the quadratic equation may result in roots that do not satisfy the original equation. For example, x x+ 10 ( x ) ( x+ 10) x 4x+ 4 x+ 10 x 5x 6 0 ( x 6)( x+ 1) 0 x 6 0 or x + 1 0 x 6 x 1 Since x 1 is an extraneous root, the solution is x 6. The extraneous root is introduced in line. The expression x could have a positive or negative value, but when squared the result is the same positive number. However, x + 10 only represents a positive value. Section 5. Page 0 Question 6 a) Substitute P i 0 and P f 90. Pf r 1+ P i r 1+ 90 0 r 0.068... The annual growth rate is 6.8% to the nearest tenth of a percent. MHR Pre-Calculus 11 Solutions Chapter 5 Page 46 of 66

b) i ( r 1) f + P i ( 1) r 1+ r + 1 Pf ( r + 1) Pi P r+ P Pf P f i P Pf P i c) The initial population is 0 moose. After 1 year, the population is 0(1.068), or about 4 moose. After years, the population is 0(1.068), or about 65 moose. After years, the final population is 90 moose. d) The set of populations in part c) represent a geometric sequence. Section 5. Page 0 Question 7 Step 1 The table shows the number of nested radicals, the expression, and the decimal approximation. Step The predicted value for 6+ 6+ 6+ 6 +... is. MHR Pre-Calculus 11 Solutions Chapter 5 Page 47 of 66

Step x 6 + x, x 6 ( 6 x) x 6 + x x x 6 0 ( x )( x+ ) 0 x 0 or x + 0 x x x + Step 4 Since x is an extraneous root, the solution is x. Step 5 Example: 1 + 1 + 1 + 1 +... results in a rational root of 4. Step 6 Example: x 1 + x, x 1 x ( 1 x) + x 1 + x x x 1 0 ( x 4)( x+ ) 0 x 4 0 or x + 0 x 4 x Since x is an extraneous root, the solution is x 4. Chapter 5 Review Chapter 5 Review Page 04 Question 1 a) b) c) 8 5 8 (5) 0 5 5 5 ( ) () 5 96 y 7 ( y ) (7 6y 6 ) MHR Pre-Calculus 11 Solutions Chapter 5 Page 48 of 66

d) z 4 z ( z) (4 z) 108z 4 Chapter 5 Review Page 04 Question a) 7 6() 6 b) 40 4(10) 6 10 c) d) m 7 9()( ) m m 80x y 8(10)( x )( x )( y ) 5 6 6 xy 10x Chapter 5 Review Page 04 Question a) 1 + 1 1 b) 4 7 11 4 7 16(7) 4 7 8 7 4 7 c) + 4 + 8() + Chapter 5 Review Page 04 Question 4 a) 4 45x 7x + 17 x 9 15 x, x 0 1x 5x x + 17 x 45x 5x x 5x + 14 x MHR Pre-Calculus 11 Solutions Chapter 5 Page 49 of 66

b) 11a a + a a 5 44 144, 0 4 11a 11a + 1a a 5 11 a + 1 a a 10 Chapter 5 Review Page 04 Question 5 11 8 7 448 8 7 4 4 8 8 7 The expression 4 is not equivalent to 8 7. Chapter 5 Review Page 04 Question 6 7 6 65 17 68 8 64 The numbers from least to greatest are 7, 8, 65, and 17. Chapter 5 Review Page 04 Question 7 a) v 169d v 1 d b) Substitute d 1.4. v 1 d v 1 1.4 v 47.587... The speed of the car is 48 km/h, to the nearest kilometre per hour. Chapter 5 Review Page 04 Question 8 Determine the perimeter of a square with side length of s 4.0. P 4s P 4 4.0 P 8 6 The perimeter of the city is 8 6 km. Chapter 5 Review Page 04 Question 9 a) The equation ±9 is not true. The left side is equivalent to 9 only. b) The equation ( ) 9 is true. The left side is equivalent to 9 only. MHR Pre-Calculus 11 Solutions Chapter 5 Page 50 of 66

c) The equation 9 ± is not true. The left side is equivalent to only. Chapter 5 Review Page 04 Question 10 a) ( ) 6 1 4 f 15 f 5 6f 75 b) ( )( ) 4 0 f 4 4 4 c) ( )( ) 8 18 144 4 6 9 Chapter 5 Review Page 04 Question 11 a) ( )( ) ( ) 5 + 5 5 4 5 1 b) ( 5 8) ( 5 8)( 5 8) 75 10( )( 8) + 8 8 10 4 8 0 6 c) ( )( ) a+ a a+ 7 4a a + 7a 4a + a a + 1 4 a + 14 + + 4 a a a a a a 17 4, 0 a + a a + a a Chapter 5 Review Page 04 Question 1 ( ( ) ) 5+ 17 5 17 1 5 17 4 MHR Pre-Calculus 11 Solutions Chapter 5 Page 51 of 66

5+ 17 5 17 Since the product is the difference of two squares, x and x are a conjugate pair. Solving x 5± 17 5x + 0 using the quadratic formula yields x. So these are solutions to the equation. Chapter 5 Review Page 05 Question 1 a) b) c) 6 6 1 1 1 1 7 1 6 1 5 5 5 1 1 5 65 5 5 5 5 5 5 a a 4 4 9 9 4a MHR Pre-Calculus 11 Solutions Chapter 5 Page 5 of 66

Chapter 5 Review Page 05 Question 14 a) b) c) d) 4+ 4 4 4+ ( ) 4+ 16 8 1 7 7 5+ 7 5 7 5 7 5+ 7 5+ 7 0 7 5+ 7 1 18 18 6 7m 6+ 7m 6+ 7m 6 7m 108 18 7m 6 7m 1 7m 4 m 1 6 m 4, m 0, m 4 m a+ b a+ b a+ b a b a b a+ b + + a b a a b b, 0, b a ± b MHR Pre-Calculus 11 Solutions Chapter 5 Page 5 of 66

Chapter 5 Review Page 05 Question 15 Use the Pythagorean theorem to find the lengths of sides of the triangle. For side from ( 4, 0) to (0, 4): c 4 + 4 c c For side from (0, 4) to (4, 4): c 8 + 4 c 80 c 80 The third side is the same length as the one above. Find the perimeter. P + 80 P 4 + 8 5 Chapter 5 Review Page 05 Question 16 a) 5 7 5 1 6 1 16 5 1 1(6) 5 6 6 6 5 6 18 b) MHR Pre-Calculus 11 Solutions Chapter 5 Page 54 of 66

a a 1 4a a 9 8a 9 8a 4a a 18 a 4a a 6 4 a Chapter 5 Review Page 05 Question 17 Substitute A 1 and w 4 and solve for l. A lw ( ) 1 l 4 1 l 4 1 4 + l 4 4 + 48 + 1 l 16 48 + 1 l 14 4 + 6 l 7 An expression for the length is 4 + 6 units. 7 Chapter 5 Review Page 05 Question 18 a) x 7, x 0 x 7 ( ) x 7 x 49 Check for x 49. MHR Pre-Calculus 11 Solutions Chapter 5 Page 55 of 66

Left Side Right Side x 7 49 7 Left Side Right Side The solution is x 49. b) 4 x, x 4 ( ) 4 x ( ) 4 x 4 x 0 Check for x 0. Left Side Right Side 4 x 4 0 Left Side Right Side There is no solution. c) 5 x 1, x 0 6 x ( x ) 6 6 x 18 x Check for x 18. Left Side Right Side 5 x 1 5 (18) 1 Left Side Right Side The solution is x 18. MHR Pre-Calculus 11 Solutions Chapter 5 Page 56 of 66

d) 7x 1+ 8, x 0 7x 7 7x 7 7x 49 7x 147 x 1 Check for x 1. Left Side Right Side 1+ 7x 8 1+ 7( 1) 1+ 49 8 Left Side Right Side The solution is x 1. Chapter 5 Review Page 05 Question 19 a) b) 1 5x 7x 1, x 7 ( 5x ) ( 7x 1) 5x 7x 1 x 9 9 x y y, y ( y ) ( y ) y y 6y+ 9 0 y 7y+ 1 0 ( y 4)( y ) MHR Pre-Calculus 11 Solutions Chapter 5 Page 57 of 66

y 4 0 or y 0 y 4 y c) 5 7n+ 5 n 1, n 7 7n+ 5 1+ n ( 7n+ 5) ( 1+ n) 7n+ 5 n + n+ 1 0 n 5n 4 0 ( n 8)( n+ ) n 8 0 or n + 0 n 8 n Since n is an extraneous root, the solution is n 8. d) m 8 m 4,0 m 4 m 8 m 4 m 8 ( 4) m m 8 m 8 m+ 16 10m 8 8 m 5m 1+ m 1 5m 1+ ( m ) 1 10m 5m 1+ + m 1 144 144 + 10m+ 5m 4m 5m 1m+ 144 0 ( m 1)(5m 1) 0 m 1 0 or 5m 1 0 m 1 m 1 5 MHR Pre-Calculus 11 Solutions Chapter 5 Page 58 of 66

Since m 1 5 is an extraneous root, the solution is m 1. e) x 1+ 7 x 1 4 ( x 1) ( 4) x 1 64 x 6 x 1 Chapter 5 Review Page 05 Question 0 Example: Isolate the radical. Next, square both sides. Then, expand and simplify. Factor the quadratic equation. Possible solutions are n and n 8. The root n is an extraneous root because when it is substituted into the original equation a false statement is reached. Chapter 5 Review Page 05 Question 1 Substitute d 7.1 and solve for h. d h 7.1 h h 7.1 h 50.41 100.8 h h.606... The height of the crew is about.6 m. MHR Pre-Calculus 11 Solutions Chapter 5 Page 59 of 66

Chapter 5 Practice Test Chapter 5 Practice Test Page 06 Question 1 ( ) ( ) Choice B. 54 Chapter 5 Practice Test Page 06 Question The condition on the variable in 7n is 7n 0, or n 0. Choice D. Chapter 5 Practice Test Page 06 Question x 6x + 5x 6x x 6x Choice C. Chapter 5 Practice Test Page 06 Question 4 ( y) ( y) 540 6 6 15 6 Choice D. 6 90y 18 10y Chapter 5 Practice Test Page 06 Question 5 x + 7 x ( x + 7) ( x ) x + 14x+ 49 x x + 15x+ 6 0 ( x+ 1)( x+ ) 0 x + 1 0 or x + 0 x 1 x Since x is an extraneous root, the solution is x. Choice B. Chapter 5 Practice Test Page 06 Question 6 MHR Pre-Calculus 11 Solutions Chapter 5 Page 60 of 66

5 5 7 7 5 6 7 5 6 14 Choice C. Chapter 5 Practice Test Page 06 Question 7 11 99 5 6 150 9 16 160 The numbers from least to greatest are 11, 5 6, 160, and 9. Chapter 5 Practice Test Page 06 Question 8 ( 5n )( 8n) 6 40n 1 1 1 1 1n 10 1+ 1 1 1 1 1 + 1n 10 + 144n 0 1 88 1n 10 + 88n 5 87 Chapter 5 Practice Test Page 06 Question 9 x x 5, x 5 or x 5 ( x) ( x 5) x x x 9 6 + 5 6x 14 x x 14 6 7 MHR Pre-Calculus 11 Solutions Chapter 5 Page 61 of 66

Chapter 5 Practice Test Page 06 Question 10 9y+ 1 + 4y ( 9y+ 1) ( + 4y ) 9y+ 1 9+ 6 4y + 4y 5y 6 6 4y ( 5y 6) ( 6 4y ) y y+ y 5 60 6 6(4 ) 5y 04y+ 108 0 Use the quadratic formula with a 5, b 04, and c 108. b± b 4ac y a ( 04) ± ( 04) 4( 5)( 108) y ( 5) 04 ± 0 816 y 50 04 ± 1 14 y 50 10 ± 6 14 y 5 10 + 6 14 Check for y. 5 Left Side Right Side 9y + 1 + 4y 10 + 6 14 10 + 6 14 9 + 1 5 + 4 5 8.57... 8.57... Left side Right Side 10 6 14 Check for y. 5 Left Side Right Side 9y + 1 + 4y MHR Pre-Calculus 11 Solutions Chapter 5 Page 6 of 66

10 6 14 10 6 14 9 + 1 5 + 4 5.474....557... Left side Right Side 10 6 14 Since y is an extraneous root, the solution is y 5 10 + 6 14 5. Chapter 5 Practice Test Page 06 Question 11 Combine pairs of identical factors. 450 ()()(5)(5) (5) 15 Chapter 5 Practice Test Page 06 Question 1 Use the Pythagorean theorem to find the lengths of sides of the two right isosceles triangle. For triangle with side lengths of 4 km: c 4 + 4 c c For triangle with side lengths of 5 km: c 5 + 5 c 50 c 50 Find the sum of the hypotenuses. d + 50 d 4 + 5 d 9 The boat is 9 km from its starting point. Chapter 5 Practice Test Page 06 Question 1 a) To rationalize the denominator of by 6 because 6( 6) 6. 4, multiply the numerator and denominator 6 MHR Pre-Calculus 11 Solutions Chapter 5 Page 6 of 66

b) To rationalize the denominator of by y because ( ) y y y y., multiply the numerator and denominator c) To rationalize the denominator of 7, multiply the numerator and denominator by ( ) 7 because ( ) 7 7 7. Chapter 5 Practice Test Page 07 Question 14 Substitute m and solve for C. C m 700 C 700 C 700 C 9 700 C 600 The cost of a -carat diamond in $600. Chapter 5 Practice Test Page 07 Question 15 Teya s solution is correct. Chapter 5 Practice Test Page 07 Question 16 a) Let x represent the length of the other leg. Then, an expression for the hypotenuse, c, is c 1 + x c 1+ x c 1+ x b) Substitute c 11 and solve for x. MHR Pre-Calculus 11 Solutions Chapter 5 Page 64 of 66

c 1+ x 11 1+ x ( x ) 11 1+ 11 1+ x 10 x x 10 x 0 The length of the unknown leg is 0 units. Chapter 5 Practice Test Page 07 Question 17 a) I P R P I R P I R P R I b) Substitute I 0.5 and P 100. P R I 100 R 0.5 R 400 The resistance in the bulb is 400 Ω. Chapter 5 Practice Test Page 07 Question 18 a) Let s represent the edge length of the cube. SA 6s SA s 6 SA s 6 b) Substitute SA. MHR Pre-Calculus 11 Solutions Chapter 5 Page 65 of 66

s SA 6 s 6 s 11 s The edge length is cm. c) Let s new represent the edge length of the new cube. SA snew 6 SA snew 4 6 snew 4s The edge length will change by a scale factor of 4. Chapter 5 Practice Test Page 07 Question 19 a) Substitute P 500, n, and A 71.15. A P(1 + i) n 71.15 500(1 + i) b) 71.15 500(1 + i) 71.15 (1 + i) 500 1.0609 (1 + i) 1.0609 1+ i i 1.0609 1 i 0.0 The interest rate is %. MHR Pre-Calculus 11 Solutions Chapter 5 Page 66 of 66