Course Review Note: Length and angle measures may vary slightly due to rounding. Course Review Question Page 8 a) Let l represent the length and w represent the width, then l + w 0. n+ q b) If n represents one number and q represents the other number, then 5. c) If q represents the number of quarters and l represents the number of loonies, then 0.5q + l 7. d) If a represents the number of adult tickets sold and s represents the number of student tickets sold, then 0a + s 950. Course Review Question Page 8 a) The point of intersection is (, ). b) The point of intersection is (, 5). c) The point of intersection is (, ). MHR Principles of Mathematics 0 Solutions
Course Review Question Page 8 a) x + y 6 x y + 6 x y Substitute y + 6 for x into equation. x y ( y + 6) y 8y + y y y Substitute y into equation. x y x () x x The solution to the linear system is x and y. b) y 6 x y x + Substitute 6 x for y into equation. y x+ 6 x x + 5x 5 x Substitute x into equation. y x+ () + The solution to the linear system is x and y. MHR Principles of Mathematics 0 Solutions
c) 5x y x + y y x Substitute x for y into equation. 5x y 5x ( x) 5x + x 8x 8 x Substitute x into equation. y x () The solution to the linear system is x and y. Course Review Question Page 8 a) x + y 55 x y x 5 + x 7 Substitute x 7 into equation. x+ y 55 ( 7) + y 55 y 8 Check by substituting x 7 and y 8 into both original equations. In x + y 55: In x y : L.S. x+ y R.S. 55 L.S. x y R.S. ( 7) + ( 8) 55 ( 7) ( 8) L.S. R.S. L.S. R.S. The solution checks in both equations. The solution to the linear system is x 7 and y 8. MHR Principles of Mathematics 0 Solutions
b) a + b 5 a b 0 a + b 0 a b 0 5a 0 + a Substitute a into equation. a+ b 5 ( ) + b 5 b Check by substituting a and b into both original equations. In a + b 5: In a b 0: L.S. a+ b R.S. 5 L.S. a b R.S. 0 ( ) ( ) + ( ) ( ) 5 0 L.S. R.S. L.S. R.S. The solution checks in both equations. The solution to the linear system is a and b. c) k + h k h h 8 h Substitute h into equation. k + h ( ) k + k 6 k.5 Check by substituting k.5 and h into both original equations. In k + h : In k h : L.S. k + h R.S. L.S. k h R.S. (.5) ( ) + ( ) (.5 L.S. R.S. L.S. R.S. The solution checks in both equations. The solution to the linear system is k.5 and h. ) MHR Principles of Mathematics 0 Solutions
d)5a b 5 a + b 9 8a a + Substitute a into equation. 5a b 5 5( ) b 5 b 0 b 5 Check by substituting a and b 5 into both original equations. In 5a b 5: In a + b 9: L.S. 5a b R.S. 5 L.S. a+ b R.S. 9 5( ) ( 5) 5 ( ) + ( 5) 9 L.S. R.S. L.S. R.S. The solution checks in both equations. The solution to the linear system is a and b 5. Course Review Question 5 Page 8 The lines have the same slope, but a different y-intercept. The lines are parallel and they have no point in common. Course Review Question 6 Page 8 a) The point of intersection is (6.7,.7). b) The point of intersection is (.,.) c) The point of intersection is ( 0., 0.9). MHR Principles of Mathematics 0 Solutions 5
Course Review Question 7 Page 8 a + b a b 80 a b 56 Multiply equation by and by. Then, subtract. 6a + 9b 7 6a b b 60 b 0 Substitute b 0 into equation. a+ b a + ( 0) a 6 a The value for a is and the value for b is 0. Course Review Question 8 Page 8 The speed up the river was 60, 5 or km/h. The speed down the river was 60, or 0 km/h. Let b represent the speed of the boat. Let r represent the speed of the river. b + r 0 b r b + b 6 Substitute b 6 into equation. b + r 0 6 + r 0 r 0 6 r The speed of the boat is 6 km/h, and the speed of the river is km/h. 6 MHR Principles of Mathematics 0 Solutions
Course Review Question 9 Page 8 Let s represent the required volume of 60% hydrochloric acid. Let t represent the required volume of 0% hydrochloric acid. s+ t 5 0.6s + 0.t 0.6 5 0.6s+ 0.t 5 Multiply equation by 0.6. Then, subtract equation. 0.6s + 0.6t 75 0.6 0.6s + 0.t 5 0.t 0 t 00 Substitute t 00 into equation. s+ t 5 s + 00 5 s 5 The required volume of 60% acid is 5 ml, and the required volume of 0% acid is 00 ml. MHR Principles of Mathematics 0 Solutions 7
Course Review Question 0 Page 8 x y+ + 5 x y+ 5 + 5 5 5 5x 0+ y+ 0 5x+ y 7 x+ y+ 5 7 x+ y+ 5 ( ) 7 x+ 6 7y 5 x 7y Multiply equation by, and equation by 5. Then, subtract. 5x + 9y 5x 5 y 65 5 y 76 y Substitute y into equation. 5x+ y 7 5x + ( ) 7 5x 5 x 5 The solution to the linear system is x 5 and y. 8 MHR Principles of Mathematics 0 Solutions
Course Review Question Page 9 Line segment AB: x+ x y+ y ( xy, ), + 6 + 5,, ( ) ( x x ) ( y y ) ( ) ( 5 ) ( 8) ( ) AB + 6 + + 80 Line segment CD: x+ x y+ y ( xy, ), ( 5, 0) ( ) ( ) 7 + +, Line segment EF: x+ x y+ y ( xy, ), (,.5) ( ) + 6 +, ( x x ) ( y y ) ( ) ( ) ( ) ( 8) CD + 80 7 + + ( x x ) ( y y ) ( ) ( ) ( 8) ( ) EF + 6 + + 65 MHR Principles of Mathematics 0 Solutions 9
Course Review Question Page 9 a) Let the midpoint of KL be M. x + x y + y,, ( xy) m JM ( 0,. 5) ( ) + +, y x y x ( 5). 0 65. The y-intercept is.5. The equation of the line through JM is y 6.5x.5. b) Let the midpoint of JL be N. x + x y + y,, ( xy) m KN (,0) ( ) + +, y x y x ( ) ( ) 0 0. y 0.x+ b 0.( ) + b 0.6+ b 0. b The equation of the median from vertex K is y 0.x 0.. 0 MHR Principles of Mathematics 0 Solutions
c) From part b), the midpoint of JL is N(, 0). m JL y x y x The slope of the right bisector of JL is. y x+ b 0 ( ) + b 0 + b b The equation of the right bisector of JL is y x. Course Review Question Page 9 a) AC ( x x ) + ( y y ) ( 0) ( 8 6) ( ) ( 5) 5.6 + + ( x x ) ( y y ) BC + ( 87) ( 8 0) + ( 6) ( ) 7.5 + Fire station B is closer. b) Answers will vary. For Example: Plot points A, B, and C. Use the Measure menu to calculate the distances. MHR Principles of Mathematics 0 Solutions
Course Review Question Page 9 ( x x ) ( y y ) ( 0) ( 0) ( 8) ( ) DE + 80 + + ( x x ) ( y y ) ( 8) + 8 ( 6) ( ) ( ) FG + 6 00 + ( x x ) ( y y ) EF + ( 8 ) ( 6 ) + ( 0) ( 0) 00 + ( ) ( ( 6 0) ( 8 0) ( ) ( 8) ) DG x x + y y 80 + + Adjacent sides are equal in length. DEFG is a kite. Course Review Question 5 Page 9 a) MHR Principles of Mathematics 0 Solutions
b) The coordinates of M are The coordinates of N are x+ x y+ y ( xy, ), ( ) x+ x y+ y xy,, + 6 0 + ( 6) + 0+ 6,,, ( 8,8) ( ) c) MN ( x x ) + ( y y ) ( 8 ) ( 8 ) ( ) ( 6) 5 + + The length of MN is half the length of KL. y y mmn x x 8 8 6 d) ( x x ) ( y y ) KL + m KL ( 6) 6 ( 6) + ( 8) ( ) + 08 y y x x 6 6 ( ) 6 8 The slopes are the same. MN is parallel to KL. MHR Principles of Mathematics 0 Solutions
Course Review Question 6 Page 9 Let M be the midpoint of QR. The coordinates of M are x+ x y+ y y y ( xy, ), mqr x x + 5+ 5, (,) 6 ( ) The coordinates of M are (, ). The slope of the right bisector of QR is. y x+ b () + b b y x+ ( ) + The coordinates of point P do not satisfy the equation of the right bisector of the line segment QR. The point P(, ) does not lie on the right bisector of line segment QR. MHR Principles of Mathematics 0 Solutions
Course Review Question 7 Page 9 y y a) mab x x 6 8 8 m CD y y x x 5 ( ) 9 9 m m BC AD y x 6 y x ( ) ( ) 5 y x y x 6 7 One pair of opposite slopes are equal, but the others are not. ABCD is a trapezoid. b) Answers will vary. For example: Plot the given points and construct line segments joining them. Use the Measure menu to measure the slopes of the line segments. MHR Principles of Mathematics 0 Solutions 5
Course Review Question 8 Page 9 a) Let I be the point at which the connector should be located. The shortest distance is the perpendicular distance HI to WM, so point I should lie on the perpendicular to line segment WM that passes through point H(, ). m WM y y x x 0 8 y x+ b ( ) + b 6 b y x 6 The slope of HI is y x+ b ( ) + b 8 b. The equation of the line passing through points M(, ) and W(0, ) is Substitute x 6 for y in the equation y x+ 8. x 6 x+ 8 y x 6 6x x+ ( 8) 6 7x 6 6 x 8 y x+ 8. The water main should be located at point I(8, 6). 6 MHR Principles of Mathematics 0 Solutions
b) d ( x x ) + ( y y ) HI ( 8) ( 6) ( 6) ( ) 7 6.5 + + Since each grid interval represents 0.5 m, the connection requires about 0.5 6.5, or 8.5 m of pipe. Course Review Question 9 Page 9 a) The radius of the circle is 7. x + y 9 b) ( ) 5 + 6 6 ) x ( ) + y 6 c) ( ( ) 8 + 67 x + y 67 Course Review Question 0 Page 9 x + y 6 The radius of the circle is 8 units. The diameter is 6 units. Aπr π( 8) 0 The area of the circle is about 0 square units. Course Review Question Page 0 s + s 60 s 600 s The side length of the chute is about cm. MHR Principles of Mathematics 0 Solutions 7
Course Review Question Page 0 a) The centroid is the point where the three medians of a triangle intersect. b) Determine the equation of two of the medians of the triangle and then find the point of intersection of these two lines. c) Answers will vary. For example: Construct the triangle. Construct the midpoints of the sides. Join the midpoints to the vertices to form the medians. Measure the coordinates of the point of intersection of the medians. Course Review Question Page 0 a) Answers will vary. Draw any triangle. Draw a median. Find the length of the base and height of each of the two triangles formed. Calculate the area of each triangle. b) Answers will vary. For example: Construct a triangle. Construct a median. Construct triangle interiors in the two triangles formed. Measure the area of each triangle using the Measure menu. Course Review Question Page 0 ( x x ) ( y y ) AC + ( 6) ( 0) ( ) ( ) 6 + + 60 ( x x) ( y y) ( 6 ) ( ) ( ) ( ) BC + + + 60 The length of AB equals the length of BC. ΔABC is isosceles. 8 MHR Principles of Mathematics 0 Solutions
Course Review Question 5 Page 0 a) Answers will vary. For example: y y y y mde mef x x x x 0 8 8 8 0 6 6 0 5 5 The slopes of DE and EF are negative reciprocals. ΔDEF is a right triangle. b) Answers will vary. For example: Another way to show that ΔDEF is a right triangle is to show that the sides satisfy the Pythagorean theorem. Course Review Question 6 Page 0 a) MHR Principles of Mathematics 0 Solutions 9
b) From the diagram, the midpoint of JL is X(, ), the midpoint of LK is Y(5, ), and the midpoint of KJ is Z(, 5). ( x x ) ( y y ) ( 5 ) ( ) ( ) ( ) XY + 0 + + ( ) ( ( 5) + ( 5 ) ( 6) ( 8) ) YZ x x + y y 00 0 + ( ) ( ( ) ( 5 ) ( ) + ( 7) ) XZ x x + y y + 58 ( x x ) ( y y ) ( ) + ( 6) ( 6) ( ) JK + + 0 0 ( x x ) ( y y ) ( ) 0 ( 6) ( ) ( 6) JL + 8 + + 00 0 ( ) ( ( 8 ) 0 ( ) ( 6) ( ) ) KL x x + y y 58 + + Corresponding sides are in the same proportion of :. ΔXYZ is similar to ΔJKL. 0 MHR Principles of Mathematics 0 Solutions
Course Review Question 7 Page 0 a) From the diagram, the midpoint of JL is X(, ), the midpoint of LK is Y(7, ), and the midpoint of KJ is Z(5, 5). m JL y y x x 0 5 The slope of the right bisector of JL is. y x+ b ( ) + b 5 b The equation of the right bisector of JL is y x 5. m LK y x y x 8 0 9 5 8 The slope of the right bisector of LK is. y x+ b ( 7) + b 5 b 5 The equation of the right bisector of LK is y x+. MHR Principles of Mathematics 0 Solutions
m JK y x y x 8 9 6 8 The slope of the right bisector of JK is. y x+ b 5 ( 5) + b 5 b The equation of the right bisector of JK is y x+ 5. b) Use the method of substitution to find the circumcentre. y x 5 5 y x+ Substitute x 5 for y in equation. 5 x 5 x+ x 0 x+ 5 5x 5 x 5 y x 5 ( 5) 5 5 Two of the right bisectors intersect at (5, 5). Check to see that (5, 5) also lies on the third bisector. 5 ( ) 5 x + 5 + 5 The coordinate (5, 5) satisfies the equation of the third right bisector. All three intersect at the circumcentre C(5, 5). MHR Principles of Mathematics 0 Solutions
c) JC ( x x ) + ( y y ) 5 5 ( 5 ) ( 5 ) ( ) ( ) + + ( x x ) ( y y ) ( 9 5) ( 8 5) ( ) ( ) CK + 5 5 + + ( x x ) ( y y ) CL + 5 5 ( 5 5) ( 0 5) ( 0) ( 5) + + The circumcentre of ΔJKL is equidistant from its vertices. Course Review Question 8 Page a) Squares and rectangles have diagonals that are equal in length. b) Squares, rhombi, and parallelograms have diagonals that bisect each other. c) Squares, rhombi, and kites have diagonals that meet at right angles. Course Review Question 9 Page a) Answers will vary. For example: Draw a parallelogram. Find the midpoints of two opposite sides. Join the midpoints. Find the length of the line segment created, and the lengths of the other two sides of the parallelogram. b) Answers will vary. For example: Construct a parallelogram. Construct the midpoints of two opposite sides. Join the midpoints. Measure the length of the line segment created, and the lengths of the other two sides of the parallelogram. MHR Principles of Mathematics 0 Solutions
Course Review Question 0 Page m y y a) AB x x 7 5 5 m BC y y x x ( ) 6 m CD y y x x ( ) 8 5 m AD y y x x 7 8 6 Opposite sides are parallel. Adjacent sides are not perpendicular. ABCD is a parallelogram. b) AC ( ) ( x x + y y 85 ( ) + ( 7) ( ) ( 9) + ) ( ) ( ( ) ( ) ( 0) + ( ) ) BD x x + y y 0 8 + The diagonals are not equal in length. c) Find the midpoint of each side. For AC: x+ x y+ y ( x, y), (,.5) ( ) + 7+, For BD: x+ x y+ y ( x, y), + 8 +,,.5 ( ) The midpoints of the diagonals are the same. The diagonals bisect each other. y y d) mac x x 7 9 8 ( ) m BD y x 0 y x The slopes are not negative reciprocals. The diagonals are not perpendicular. MHR Principles of Mathematics 0 Solutions
Course Review Question Page x y+ 0 y x+ y x y x+ 5 x+ y+ 0 y x 6 There are two pairs of equal slopes, of and. They are not negative reciprocals. No sides are perpendicular. The quadrilateral is a parallelogram. Course Review Question Page a) PC ( x x ) + ( y y ) ( ) ( 5) ( ) ( ) + + ( x ) ( y y ) x QC + ( 6) ( ) + ( ) ( ) + ( ) ( ( 7) ( ) ( ) ( ) ) RC x x + y y + + All three points are equidistant from C. They lie on a circle with centre at C. b) Calculate the midpoint for PQ. x+ x y+ y ( x, y), (,) ( ) + 6 5+, This is the centre of the circle. The centre of the circle lies on the right bisector of chord PQ. MHR Principles of Mathematics 0 Solutions 5
Course Review Question Page Answers will vary. For example: Plot the three given points. Join the points with line segments to form two chords. Find the midpoints of the chords. Construct perpendicular lines through the midpoints. The perpendicular lines meet at the centre of the circle. Course Review Question Page a) The relation is neither linear nor quadratic. x y First Differences Second Differences 9 0 0 6 6 9 6 6 5 b) The first differences are constant. The relation is linear. c) The first differences are constant. The relation is linear. x y First Differences 6 0 0 6 9 x y First Differences 7 5 0 5 5 8 5 5 Course Review Question 5 Page a) The ball was released from a height of. m. b) Use graph paper or a graphing calculator to graph the relation. The maximum height reached by the ball was about. m. c) h 7.t + 8.5t+. () () 7. + 8.5 +..05 The height of the hoop was.05 m. 6 MHR Principles of Mathematics 0 Solutions
Course Review Question 6 Page a) b) c) d) MHR Principles of Mathematics 0 Solutions 7
Course Review Question 7 Page a) b) The height is 0 m at t 0 s and t s. The maximum height occurs at t 5.5 s. t( t ) ( 5.5) ( 5.5) h 5 5 5 The maximum height is approximately 5 m. c) The maximum height occurs after 5.5 s. d) Answers will vary. For example: The lava will be ejected away from the crater and so it will probably fall on land that is below the crater. The length of time in the air will probably be more than s. Course Review Question 8 Page a) The vertex is (0, ). b) y ax + 5 a ( ) + y a 7 x ( x+ 5 + 5 5 a + a 7 7a a ( )( ) a )( ) y x + ( )( 5) y x x+ x x 5 + Course Review Question 9 Page The horizontal distance to the vertex from the given x-intercept is 9. The other x-intercept is the same distance past the vertex, at. 8 MHR Principles of Mathematics 0 Solutions
Course Review Question 0 Page 0 a) 8 b) c) ( ) 5 d) 5 5 e) ( 8) 0 f) Course Review Question Page a) At 500ºC, there are 0 doublings. 0 t 0 It would take the wood 0 s to burn. b) At 650ºC, there are 5 halvings. t 5 It would take the wood s to burn. Course Review Question Page a) ( ) ( ) x + 5 x+ 6 x + 5x+ 0 8x + 8 b) ( ) ( ) 6 a+ a 5 6a+ 8 a+ 0 a + 8 c) ( ) ( ) k k k + e) ( ) ( ) 8k 6 + p 8 0p +.5p 6 + 7.5p + 9p f) ( ) ( ) y t t + t t+ 5 6t 8t+ t + 5t k d) ( ) ( ) y y y y y y+ y y y y + y y 7 y t t 8 MHR Principles of Mathematics 0 Solutions 9
Course Review Question Page a) SA ( x + )( x ) + ( x )( x + ) + ( x + )( x + ) ( x x x x x x x x x ) ( x x ) + 6+ + + + 6 + + 5 + 5 5 + 0x 0x 0 b) SA x + x 0 0 0 ( ) ( ) 0 5 + 0 5 0 90 The surface area of the box is 90 cm. Course Review Question Page ( ) + + + ( )( ) a) x x 8x 6 b) y y+ y 6 c) ( 5) 0 +5 d) ( )( ) a a a t+ t 9t e) ( )( ) 5a+ b 5a b 5a 9b f) ( m+ ) 9 ( m + 6m+ ) + + 8m m 0 MHR Principles of Mathematics 0 Solutions
Course Review Question 5 Page m m+ + m m 9+ m 8m+ 6 a) ( )( ) ( ) b) ( ) ( )( ) + m 8m 7 t+ + t t+ t + t+ + 8t + + 0t t x+ y x y x y 8x 8y 7x + 8xy y c) ( )( ) ( ) 9x + 8xy y y + y+ y + y y y+ + y + y+ + y d) ( ) ( ) ( )( ) e) ( ) ( ) + 9y m+ n + m n 6m + 8mn+ n + m 8mn+ 8n 8m + 9n f) ( ) ( )( ) 5 t 5z + t z t+ z 0t 00tz+ 5z + 8t 7z 68t 00tz+ 98z Course Review Question 6 Page a) 5k 5 5( k 7) b) h 0h h( h 5) c) xy 8xy xy( y) d) x 5 ( x+ 5)( x 5) + m f) a 6b ( a b ) e) 9m ( 7m)( 7 ) ( a b)( a+ b) MHR Principles of Mathematics 0 Solutions
Course Review Question 7 Page a) The length is n and the width is n. b) P ( n ) + ( n ) A n n ( 8 ) ( 8 ) + 5 + 6 ( 8) + 8 5 6 0 The perimeter is cm and the area is 0 cm. Course Review Question 8 Page a) x x ( x )( x+ ) b) y + y 8 ( y+ 6)( y ) c) m + m+ ( m+ 8)( m+ ) d) t 8t+ 5 ( t 5)( t ) e) x x cannot be factored. f) + + n n+ 0 ( n 8)( n 5) g) w w 0 ( w 6)( w+ 5) h) + 5m m ( 7 m)( + m) Course Review Question 9 Page + + ( + ) y y+ 6 ( y 6) a) x 0x 5 x 5 b) c) m 6m 6 cannot be factored. d) x + x+ 9 x+ + + ( ) + ( ) ( ) e) 5r 0rs s 5r s f) 5x 0xy + 0y 5 x xy + y Course Review Question 50 Page Answers will vary. ( x y) 5 MHR Principles of Mathematics 0 Solutions
Course Review Question 5 Page + + ( + ) x x+ 9 ( x ) 8 + 6 ( ) a) x 8x 6 x b) x x x Possible values for p are 8 and 8. c) ( ) 5x + 0x+ 6 5x+ The only possible value for p is 5. Course Review Question 5 Page m n + ( m n)( m n) The only possible value for p is 9. The only pairs of integers whose product is are (, ), (, 7), (, ), and (, 7). Possible values for (m, n) are thus (, 0), (5, ), (, 0), and ( 5, ). Course Review Question 5 Page a) y x x + + ( x x ) + + + ( x ) + The vertex is (, ), and the axis of symmetry is x. b) y x x 6 5 ( ) x + 6x+ 9 9 5 ( x ) + + The vertex is (, ), and the axis of symmetry is x. c) y x x ( ) x + x+ + ( x ) + + 7 The vertex is (, 7), and the axis of symmetry is x. MHR Principles of Mathematics 0 Solutions
Course Review Question 5 Page a) The minimum is (.5, 0.5). b) The maximum is (0.,.). c) The minimum is ( 0.,.7). Course Review Question 55 Page a) y x x + ( x+ )( x ) The x-intercepts are and. b) y x x + 6 + 5 ( x )( + 5 x+ ) The x-intercepts are 5 and. c) y x x + ( x )( x ) The x-intercept is. d) y x x+ 9 ( x )( x ) x 0 x.5 The x-intercept is.5. MHR Principles of Mathematics 0 Solutions
Course Review Question 56 Page a) x + x 8 0 ( x )( x ) + 7 0 x 7 or x L.S. x + x 8 ( ) ( 7) R.S. 0 L.S. x + x 8 ( ) ( ) 7 + 8 + 8 0 0 L.S. R. S. L.S. R. S. R.S. 0 The roots are 7 and. b) m + 7m+ 0 0 ( m )( m ) + 5 + 0 m 5 or m L.S. m + 7m+ 0 ( ) ( ) R.S. 0 L.S. m + m+ 7 0 ( ) ( ) 5 + 7 5 + 0 + 7 + 0 0 0 L.S. R. S. L.S. R. S. R.S. 0 The roots are 5 and. c) n 7 5n ( n n) ( n ) ( + ) ( n+ ) ( n+ 9)( n ) 0 ( n ) ( n ) L.S. n n + 5n 7 0 n + 8n n 7 0 + 8 + 7 0 n n 9 9 0 ( ) + 9 0 or 0 n 9 or n.5 R.S. 7 5n ( ) L.S. n R.S. 7 5n ( ) ( ) 9 7 5 9.5 7 5.5 6 6.5.5 L.S. R. S. L.S. R. S. The roots are 9 and.5. MHR Principles of Mathematics 0 Solutions 5
d) k k + k + k + 0 ( ) ( ) + + + 0 k k k k 7 + 0 k + 0 ( k k) ( k ) 0 ( ) ( k ) ( k )( k ) 0 ( k ) ( k ) k k 0 k k k k 0 or 0 k or k k( k ) k ( k ) ()() () () 0 L.S. + + + R.S. + + + 0 L.S. R. S. ( ) ( ) L.S. k k + k + k + R.S. 0 + + + 8 7 + + 8 + + 0 L.S. R. S. The roots are and. 6 MHR Principles of Mathematics 0 Solutions
Course Review Question 57 Page a) Answers will vary. For example: y ( x 5)( x+ ) x x 0 b) Answers will vary. For example: y ( x+ )( x+ ) x + x+ 7 6 Course Review Question 58 Page a) b) c) b ( ) ( k )( ) ac 0 5 0 5 8k 0 k 5 8 b ac 0 ( k ) ( )( 9) 0 k 6 0 k ± 6 b ac 0 ( 0) ( 5)( k ) 0 00 00k 0 k Course Review Question 59 Page x ( ) x x 5 6 5x 6 0 ( x )( x ) 9 + 0 The negative root is inadmissible. The dimensions are 9 cm by 9 5, or cm. MHR Principles of Mathematics 0 Solutions 7
Course Review Question 60 Page a) ± x b b ac a ( ) ± ( ) ( )( ) () b) ± k b b ac a ( ) ± ( ) ( 7) ( ) ( 7) ± 7 ± 60 5 ± 7 b± b ac b± b ac c) x d) h a a 8± ( 8) ( )( ) ± ( 5 ( ) ( ) 8± 88 ± 96 8 ± ± 6 ( ) )( ) e) ± a b b ac ± a ± 8 6 ± 7 ( ) ( )( ) ( ) 8 MHR Principles of Mathematics 0 Solutions
Course Review Question 6 Page a) Let R represent the revenue. Let x represent the number of $0 decreases in price. ( 00 0 )( 90 5 ) R x + x 7 600 8 000 + 00x 50x 0 00 + 00x 50x 0 50 8+ ( x x ) ( x)( x) 0 50 + The negative root gives a price increase. The value for x is. The number of jackets sold is 90 + 5(), or 0. The selling price for a revenue of $7 600 is 00 0(), or $60. b) R ( 00 0x)( 90 + 5x) 5 600 8 000 + 00x 50x 0 00+ 00x 50x 0 50 8+ ( x x ) ( x)( x) 0 50 8 6 + The negative root yields a higher selling price. The value for x is 8. The number of jackets sold is 90 + 5(8), or 0. The lowest selling price that gives a revenue of $5 600 is 00 0(8), or $0. Course Review Question 6 Page Let x represent the length of one leg of the triangle. x ( x) + 8 0 x + 78 56x+ x 00 x x+ 56 8 0 ( x x ) 8 + 9 0 ( x )( x ) 6 0 The sides of the triangle measure cm and 6 cm. MHR Principles of Mathematics 0 Solutions 9
Course Review Question 6 Page Let x represent the width of the deck. ( x)( x) 0 + + 5 + x+ x 0 8 5 + x+ x 95 8 0 ± x a b b ac ( ) ( )( 95) ( ) 8 ± 8 8 ± 0 8 8 ± 8 8 The negative root is inadmissible. The width of the deck is.5 m. Course Review Question 6 Page 5 ΔADE ~ ΔACB A is common ADC ACB (corresponding angles). Course Review Question 65 Page 5 h. 0. 0.h. h.5 The height of the tree is.5 m. 0 MHR Principles of Mathematics 0 Solutions
Course Review Question 66 Page 5 a) 8.5 tan A. A b). tan A 0.7 A 60 Course Review Question 67 Page 5 tanθ 0.08 θ.6 The angle of inclination of the hill is approximately.6. Course Review Question 68 Page 5 x a) cos5.6 x.6cos5 x 7.6 The length of side x is about 7.6 m. x b) sin 8.5 x.5sin 8 x 5.9 The length of side x is about 5.9 cm. MHR Principles of Mathematics 0 Solutions
Course Review Question 69 Page 5 b a) sin 56 b 56sin b 0 a cos 56 a 56cos a 7 A 90 57 In ΔABC, A 57, a 7 cm, and b 0 cm. e b) tan 60 e 60tan e 5 60 cos d 60 d cos d 80 In ΔDEF, F 9, d 79 m, ande 5 m. 0 c) sint 5 F 90 9 T U 90 5 0 + u 8 u In ΔUST, T, U 8,andu m. d) 8 tan P P R 90 58 q + 8 q 5 In ΔQRP, P, R 58, and q 5 cm. MHR Principles of Mathematics 0 Solutions
Course Review Question 70 Page 6.5 sin Z.8 Z 7 Y 90 7 +.8.5 XZ XZ. In ΔXYZ, Z 7, Y, andxz. cm. Course Review Question 7 Page 6 sin A 0.5 A 0 B 90 0 60 Course Review Question 7 Page 6 a) l 7.5 +.8 l 6.6 The coast guard boat is approximately 6.6 km from the yacht. b) 7.5 tan B.8 B 6.9 The coast guard boat must travel at an angle of approximately 6.9 south of west to reach the yacht. MHR Principles of Mathematics 0 Solutions
Course Review Question 7 Page 6 h tan 6.6 00 + x 00 tan 6.6 + x tan 6.6 h h 00tan6.6 x tan 6.6 h tan 7.7 x x tan 7.7 h h x tan 7.7 h 00tan 6.6 h tan 6.6 tan7.7 htan7.7 00tan7.7 tan 6.6 htan 6.6 htan7.7 htan 6.6 00tan7.7 tan 6.6 00tan7.7 tan 6.6 h tan7.7 tan 6.6 h 60.0 The height of the bridge is approximately 60.0 m. Course Review Question 7 Page 6 v 5 + 8 v 6 x 8 65 6 6x 80 x 0 The length of side x is about 0 cm. MHR Principles of Mathematics 0 Solutions
Course Review Question 75 Page 6 sin A sin B a b sin A sin 8 5..6.6sin A 5.sin 8 5.sin 8 sin A.6 sin A 0.860... A 57 Course Review Question 76 Page 6 c b sinc sin B c 5 sin 65. 6 sin 8. csin8. 5sin65.6 5sin65.6 c sin8. c 79.5 The length of side c is about 79.5 cm. MHR Principles of Mathematics 0 Solutions 5
Course Review Question 77 Page 6 a) sin R sin P r p sin R sin5 8 sin R 8sin5 8sin5 sin R sin R 0.6867... R. Q 80. 5 85.6 q p sinq sin P q sin85.6 sin5 q sin 5 sin85. 6 sin85.6 q sin5 q 55. In ΔQRP, R., Q 85.6, and q 55. cm. b) J 80 75 7 k j sin K sin J k 0.5 sin 75 sin7 k sin 7 0.5sin 75 0.5sin75 k sin7 k 0.7 l j sin L sin J l 0.5 sin sin 7 l sin7 0.5sin 0.5sin l sin7 l. In ΔJKL, J 7, k 0.7 cm, and l. cm. 6 MHR Principles of Mathematics 0 Solutions
Course Review Question 78 Page 6 ABC + 75 C 80 75 05 A 05 x 8.5 sin sin05 x sin05 8.5sin 8.5sin x sin05 x 6.0 y 8.5 sin sin05 y sin05 8.5sin 8.5sin y sin05 y.7 The sides of the parallelogram measure approximately.7 cm and 6.0 cm. Course Review Question 79 Page 6 s r + t rt s ( coss) ( 0)( 0)( 0 ) 0 + 0 cos s.9 The length of side s is about.9 m. MHR Principles of Mathematics 0 Solutions 7
Course Review Question 80 Page 7 l m k cos L mk.5..8 cos L cos L 0.660... L 8.7 (.)(.8) k m l cos K ml.8..5 cos K (.)(.5) cos K 0.506... K 57. M 80 57. 8.7 7.0 In ΔKLM, L 8.7, K 57., and M 7.0. Course Review Question 8 Page 7.+ 7. + 9.8 s.65 ( )( )( ) A s s a s b s c (.65.)(.65 7.)(.65 9. 8).65 0 The area of the triangle is approximately 0 m. 8 MHR Principles of Mathematics 0 Solutions
Course Review Question 8 Page 7 80 8 B 7 b a sin B sin A b 0 sin7 sin8 b sin 8 0sin 7 0sin 7 b sin8 b 5. The perimeter of the triangle is about 5. + 5. + 0, or 0.8 cm. Course Review Question 8 Page 7 h a) tan 7 50 + x 50tan 7 + xtan 7 h h 50tan 7 x tan 7 h tan5 x xtan5 h h x tan5 h 50 tan 7 h tan 7 tan 5 htan5 50tan5 tan 7 htan 7 htan5 htan 7 50tan5 tan7 50 tan5 tan 7 h tan5 tan 7 h 9.6 The length of side h is about 9.6 m. MHR Principles of Mathematics 0 Solutions 9
h b) tan 7 00 x 00tan 7 xtan 7 h h tan x xtan h h x tan h + 00tan 7 x tan 7 h+ 00tan 7 h tan 7 tan htan + 00tan tan 7 htan 7 htan htan 7 00tan tan7 00tan tan 7 h tan + tan 7 h 97.9 The length of side h is about 97.9 m. Course Review Question 8 Page 7 After 5 min, the first boat travelled 0.75 0, or 7.5 km, and the second boat travelled 0.75 8, or 6.0 km. The angle between the bearings is 79 7, or. ( )( )( ) d 7.5 + 6.0 7. 5 6.0 cos d.0 The distance between the boats after 5 min was about.0 km. 50 MHR Principles of Mathematics 0 Solutions
Course Review Question 85 Page 7 a) F 80 58 5 69 f d sin F sin D f 8 sin69 sin58 f sin 58 8sin 69 8sin 69 f sin58 f 8.8 e d sin E sin D e 8 sin5 sin58 e sin 58 8sin 5 8sin 5 e sin58 e 7.5 In ΔDEF, F 69, f 8.8 cm, and e 7.5 cm. b) S 80 6 7 6 s r sins sin R s 8 sin 6 sin 7 s sin 7 8sin 6 8sin6 s sin7 s 7. sint sin R t r sin T sin7 6 8 8sinT 6sin7 6sin7 sint 8 sint 0.77... T 6 In ΔRST, T 6, S 6, ands 7. m. MHR Principles of Mathematics 0 Solutions 5
c) a b + c ab( cosa) a 5 + 7 ( 5)( 7)( cos68 ) a 6. 9 b a c cos B ac 5 6.9 7 cos B ( 6.9)( 7) cos B 0.7... B C 80 68 70 d) In ΔABC, B, C 70, and a 6.9 cm. w x y y x w cosw cos Y xy xw 0 0 cosw cos Y ( 0)( ) ( 0)( ) cosw 0.65 cos Y 0.6... W 5 Y 8 X 80 5 8 6 In ΔWXY, W 5, Y 8, and X 6. Course Review Question 86 Page 7 The largest angle is opposite the longest side, 8 cm. Let A be the largest angle. 8 7 5 cosa ( 7)( 5) cosa 0.75... A 68 The largest angle is 68. 5 MHR Principles of Mathematics 0 Solutions