Review Problems for the Final

Review Problems for the Final Math 0-08 008 These problems are provided to help you study The presence of a problem on this handout does not imply that there will be a similar problem on the test And the absence of a topic does not imply that it won t appear on the test Simplify the following expressions Express all powers in terms of positive exponents (a) 4 5 5 ( 3) (b) ( x y /3 ) 6 (c) 3 8x 5 y 8 x y 4 (d) 338 + 5 98 (e) 3 50x 5 y 9 Rationalize 3 + 5 6 3 3 Calvin Butterball has 9 coins, all of which are dimes or quarters The value of the coins is $545 How many dimes does he have? 4 How many pounds of chocolate truffles worth $50 per pound must be mixed with 5 pounds of spackling compound worth $70 per pound to yield a mixture worth $00 per pound? 5 Calvin leaves the city and travels north at 3 miles per hour Phoebe starts hours later, travelling south at 7 miles per hour Some time after Phoebe starts travelling, the distance between them is 66 miles How many hours has Calvin been travelling at this instant? 6 Calvin can eat 60 double cheeseburgers in 4 hours, while Bonzo can eat 300 double cheeseburgers in 5 hours If Calvin, Bonzo, and Phoebe work together, they can eat 50 double cheeseburgers in 3 hours How many hours would Phoebe take to eat 80 double cheeseburgers if she eats alone? 7 The product of two numbers is 56 The second number is more than twice the first Both numbers are positive What are the two numbers? 8 The hypotenuse of a right triangle is more than 3 times the smallest side The third side is less than 3 times the smallest side Find the lengths of the sides 9 Solve the inequality: (a) x 7x + 0 < 0 (b) x + 4 3 x > 0 (c) x 7 > 3 0 (a) Write the expression as a single logarithm: ln + 5 lnx + lny (b) Write the expression as a single logarithm: 4 lnx lny 7 lnz 3

(a) Write the expression as a sum or difference of logarithms: log 0 x 3 y 4 z (b) Write the expression as a sum or difference of logarithms: ln x + (y ) 5 Solve for x, writing your answer in decimal form correct to 3 places: 3 x = 7 3 Solve: 5 x = 6 x 4 Solve: 5 Simplify: (a) (49 3/ ) (b) ( 49) 3/ (c) ( 64) 5/3 3 x 4 = x 3 x + (d) 360x y 3, assuming that the variables represent nonnegative quantities 6 Find values for a and b which prove that the following statement is not an algebraic identity: a + b? = a + b 7 (a) Combine the fractions over a common denominator: x 3x + x x 9 y (b) Combine the fractions over a common denominator: x xy y x y (c) Simplify: x + x x + + x x + x 8 Compute (without using a calculator): (a) log 5 5 (b) log 5 5 89 (c) log 5 5 (d) log 47 (e) log 8 4 (f) e ln 4 9 Find the inverse function of f(x) = x x + 0 Find the domain of the function f(x) = x x Solve for x: (a) x 5 = 3

(b) 3x 5 = (x + ) (c) 4(x ) < x + 7 (d) x 4x = 5 (e) x 4x = 5 Simplify the expressions, writing each result in the form a + bi: (a) i 33 (b) (3 i)(4 + 5i) (c) + 3i (d) i 6 + 8i (e) ( + 3i) (f) 3 i 4 + i 3 Find the quotient and the remainder when x 4 + 3x 3 x + is divided by x 4 Solve the following equations, giving exact answers: (a) e x + 5 = 6e x (b) 3 x+ = 7 x+ (c) 4 3x+ = 5 x (d) ln(x ) + ln(x + ) = ln 3x (e) (ln x) 3 lnx 4 = 0 5 (a) Simplify 500 (b) Simplify 300 (c) Rationalize + 7 3 7 6 Find the equation of the line: (a) Which passes through the points (, 3) and (, ) (b) Which passes through the point (3, 4) and is perpendicular to the line x 8y = 5 (c) Which is parallel to the line 3y 6x + 5 = 0 and has y-intercept 7 7 Solve the system of equations for x and y: x + 5y = 7, x + 3y = 4 8 Suppose log a x = 5 and log a y = Find: (a) log a 3 x 3

(b) log a (a 6 y ) (c) log a x 3 y 9 (a) Simplify and write the result using positive exponents: ( 3x ) 5 y 7 9x 3 y 6 (b) Simplify and write the result using positive exponents: 4(x /3 y /5 ) ( 3y 3/5 ) x /6 30 (a) Simplify, cancelling any common factors: x x x 4 x 3 3x x x 6 (b) Simplify, cancelling any common factors: a 3 a b a 3 4ab a 4 + 3a 3 b a + ab b (c) Simplify, cancelling any common factors: x 5x x 3 4x x 0x + 5 x 6 3 (a) Find the equation of the parabola which passes through the point (3, 7) and has vertex (, 3) (b) Write the quadratic function f(x) = x 6x + 34 in standard (vertex) form 3 Find the center and the radius of the circle whose equation is x 4x + y + 6y = 33 Suppose that $000 is invested at 6% annual interest, compounded monthly How many years must pass before the account is worth at least $000? (Round up to the nearest year) Solutions to the Review Problems for the Final Simplify the following expressions Express all powers in terms of positive exponents (a) 4 5 5 ( 3) 4 5 5 ( 3) = 4 5 5 + 3 = 9 = 9 (b) ( x y /3 ) 6 (c) 3 8x 5 y 8 x y 4 ( x y /3 ) 6 = ( ) 6 (x ) 6 (y /3 ) 6 = 64x y 3 8x 5 y 8 x y 4 = ( 8x 5 y 8 x y 4 ) /3 = ( 8x 6 y ) /3 = ( 8) /3 (x 6 ) /3 (y ) /3 = x y 4 = x y 4 4

(d) 338 + 5 98 338 + 5 98 = 69 + 5 49 = 3 + 5 7 = 3 + 35 = 48 (e) 3 50x 5 y 9 3 50x5 y 9 = 3 50 3 x 5 3 y 9 = 3 5 3 x 3 x 3 y 9 = 3 5 3 3 x 3 3 x 3 y 9 = 5xy 3 3 x Rationalize 3 + 5 6 3 3 + 5 6 3 = 3 + 5 6 3 6 + 3 6 + 3 = 8 + 9 + 30 + 5( ) 36 9( = ) 48 + 39 8 = 6 + 3 6 3 Calvin Butterball has 9 coins, all of which are dimes or quarters The value of the coins is $545 How many dimes does he have? Let d be the number of dimes and let q be the number of quarters number value = total value dimes d 0 = 0d quarters q 5 = 5q coins 9 545 The first column says d + q = 9 The last column says 0d + 5q = 545 The first of these equations gives d = 9 q Substitute this into 0d + 5q = 545 to get 0(9 q) + 5q = 545 Now solve for q: 0(9 q) + 5q = 545, 90 0q + 5q = 545, 5q = 55, q = 7 There are 7 quarters and d = 9 7 = dimes 4 How many pounds of chocolate truffles worth $50 per pound must be mixed with 5 pounds of spackling compound worth $70 per pound to yield a mixture worth $00 per pound? Let x be the number of pounds of truffles pounds price per pound = value spackling compound 5 70 = 850 truffles x 50 = 50x mixture (x + 5) 00 = 850 + 50x The last line of the table gives Solve for x: 00(x + 5) = 850 + 50x 00x + 000 = 850 + 50x, 00x + 50 = 50x, 50 = 50x, x = 3 5

The mixture needs 3 pounds of truffles 5 Calvin leaves the city and travels north at 3 miles per hour Phoebe starts hours later, travelling south at 7 miles per hour Some time after Phoebe starts travelling, the distance between them is 66 miles How many hours has Calvin been travelling at this instant? Let t be the time Calvin has travelled Then Phoebe has travelled t hours Let x be the distance Calvin has travelled in this time Then Phoebe has travelled 66 x miles speed time distance Calvin 3 t x Phoebe 7 t 66 x The first line says 3t = x The second line says 7(t ) = 66 x Plug x = 3t into 7(t ) = 66 x and solve for t: 7(t ) = 66 3t Calvin has been travelling for 5 hours 7t 34 = 66 3t 40t = 00 t = 5 6 Calvin can eat 60 double cheeseburgers in 4 hours, while Bonzo can eat 300 double cheeseburgers in 5 hours If Calvin, Bonzo, and Phoebe work together, they can eat 50 double cheeseburgers in 3 hours How many hours would Phoebe take to eat 80 double cheeseburgers if she eats alone? Let x be the number of burgers Calvin can eat per hour Let y be the number of burgers Bonzo can eat per hour Let z be the number of burgers Phoebe can eat per hour burgers per hour hours = burgers Calvin x 4 = 60 Bonzo y 5 = 300 Phoebe z t = 80 together (x + y + z) 3 = 50 The first equation says 4x = 60, so x = 40 The second equation says 5y = 300, so y = 60 The last equation says 3(x + y + z) = 50 Divide by 3: x + y + z = 70 Substitute x = 40 and y = 60: 40 + 60 + z = 70, z = 70 The third equation says zt = 80 Substitute z = 70: 70t = 80, so t = 4 hours 7 The product of two numbers is 56 The second number is more than twice the first Both numbers are positive What are the two numbers? Let x and y be the two numbers The product of two numbers is 56, so xy = 56 6

The second number is more than twice the first, so y = + x Substitute y = + x into xy = 56: x( + x) = 56, x + x = 56 Then x + x = 56 56 56 x + x 56 = 0 Factor and solve: x x 56 = 0 (x 3)(x + ) = 0 ւ ց x 3 = 0 x + = 0 x = 3 x = x = is ruled out, because x is supposed to be positive Therefore, x = 3, and y = + x = 4 8 The hypotenuse of a right triangle is more than 3 times the smallest side The third side is less than 3 times the smallest side Find the lengths of the sides Let s be the length of the smallest side, let t be the length of the third side, and let h be the length of the hypotenuse By Pythagoras theorem, h = s + t The hypotenuse of a right triangle is more than 3 times the smallest side, so h = 3s + The third side is less than 3 times the smallest side, so t = 3s Plug h = 3s + and t = 3s into h = s + t and solve for s: h = s + t (3s + ) = s + (3s ) 9s + 6s + = s + 9s 6s + 0 = s s 0 = s(s ) The possible solutions are s = 0 and s = Now s = 0 is ruled out, since a triangle can t have a side of length 0 Therefore, s = is the only solution The other sides are h = 3 + = 37 and t = 3 = 35 9 Solve the inequality: (a) x 7x + 0 < 0 7

The graph of y = x 7x + 0 is a parabola opening upward x 7x + 0 = 0 gives (x )(x 5) = 0, so the roots are x = and x = 5 5 The solution is < x < 5 (b) x + 4 3 x > 0 y = x + 4 equals 0 when x = 4 and is undefined when x = 3 Set up a sign chart with these break 3 x points: - + - 5 y(-5)=-/8-4 y(0)=4/3 3 y(4)=-8 The solution is 4 < x < 3 (c) x 7 > 3 Translate the inequality: Draw the picture: x 7 > 3 the distance from x to 7 is greater than 3 3 3 4 7 0 Since x is greater than 3 units from 7, I must have x < 4 or x > 0 These inequalities give x < or x > 5 0 (a) Write the expression as a single logarithm: ln + 5 lnx + lny ln + 5 lnx + lny = ln + ln(x ) 5 + lny / = ln + lnx 0 + ln y = ln x 0 y (b) Write the expression as a single logarithm: 4 lnx lny 7 lnz 3 4 lnx ( ) x 3 lny 7 lnz = lnx4 lny /3 lnz 7 4 = ln y /3 z 7 8

(a) Write the expression as a sum or difference of logarithms: log 0 x 3 y 4 z log 0 x 3 y 4 z = log 0 (x 3 y 4 ) log 0 z = log0 x 3 + log 0 y 4 log 0 z / = 3 log 0 x + 4 log 0 y log 0 z (b) Write the expression as a sum or difference of logarithms: ln ln x + (y ) 5 x ( ) / + x + (y ) 5 = ln (y ) 5 = x + ln (y ) 5 = ( ln x + ln(y ) 5 ) = ( ln(x + ) / ln(y ) 5) = ( ) ln(x + ) 5 ln(y ) = 4 ln(x + ) 5 ln(y ) Solve for x, writing your answer in decimal form correct to 3 places: 3 x = 7 3 x = 7 ln 3 x = ln 7 (x) ln3 = ln 7 x = ln 7 ln 3 x = ln 7 ln3 08856 3 Solve: 5 x = 6 x Square both sides: ( 5 x ) = ( 6 x), (5 x) 5 x + = 6 x, 6 x 5 x = 6 x Then Square both sides: 6 x 5 x = 6 x + 6 x 6 x 5 x = x ( 5 x) = x, 4(5 x) = x, 0 8x = x Then Factor and solve: 0 8x = x + 0 8x 8x 0 0 = x + 8x 0 x + 8x 0 = 0 (x + 0)(x ) = 0 ւ ց x + 0 = 0 x = 0 x = 0 x = 9

Check: x = 0 gives 5 x = 5 = 4, 6 x = 6 = 4 (Checks) x = gives 5 x = = 0, 6 x = 4 = (Doesn t work) The solution is x = 0 4 Solve: 3 x 4 = x 3 x + Factor the denominator on the left: 3 (x )(x + ) = x 3 x + Multiply both sides by (x )(x + ) to clear denominators: (x )(x + ) ( 3 = (x )(x + ) (x )(x + ) x 3 ), x + 3 = (x + ) 3(x ), 3 = x + 4 3x + 6, 3 = 0 x Then Check: x = 3 gives 3 = 0 x 0 0 3 = x 3 = x 3 x 4 = 3 5, x 3 x + = 5 ( 3) = 3 5 (Checks) The solution is x = 3 5 Simplify: (a) (49 3/ ) (49 3/ ) = ( 49) 3 = (7 3 ) = 343 (b) ( 49) 3/ ( 49) 3/ is undefined (c) ( 64) 5/3 ( 64) 5/3 = ( 64) = 5/3 ( 3 64) = 5 ( 4) 5 = 04 (d) 360x y 3, assuming that the variables represent nonnegative quantities 360x y 3 = 360 x y 3 = 36 0 x y y = 6xy 0y 0

6 Find values for a and b which prove that the following statement is not an algebraic identity: a + b? = a + b If a = and b =, then a + b = while a + b = Since, a + b a + b for a = and b = 7 (a) Combine the fractions over a common denominator: x 3x + x x 9 = x(x 3) + x 3x + x x 9 x (x 3)(x + 3) = x + 3 x + 3 (x + 3) + x x(x 3)(x + 3) = x + x + 3 x(x 3)(x + 3) (b) Combine the fractions over a common denominator: y x xy y x y = y x(x y) y x xy y x y x(x 3) + x x y (x y)(x + y) = y x(x y) x + y x + y x x 9 = y (x y)(x + y) x x = (c) Simplify: y(x + y) x(x y)(x + y) xy y(x + y) xy = x(x y)(x + y) x(x y)(x + y) = y x(x y)(x + y) x + x x + + x x + x x + x x + + x x + x = x + x x + + x x + x x x = x(x + ) x(x + ) + x x x + = x(x + ) x(x ( (x + ) ) + ) ( (x x + ) + ) = x x x x + x = x x x x x + x x x = x x x x + = x x x + x + x 3 + x x 3 + x (x )(x + ) x x + x x = x (x + ) x (x ) = (x )(x + ) = x (x )(x + ) 8 Compute (without using a calculator): (a) log 5 5 log 5 5 = 3, because 5 3 = 5

(b) log 5 5 89 log 5 5 89 = 89 (c) log 5 5 log 5 5 =, because 5 = 5 (d) log 47 log 47 = 0 (e) log 8 4 log 8 4 = 3, because 8/3 = 4 (f) e ln 4 e ln 4 = 4 9 Find the inverse function of f(x) = x x + Write y = x + Swap x s and y s: x = y + Solve for y: x y x = y y +, x = y, (x )(y ) = y, (x )y (x ) = y, y (x ) = y (x )y, (x ) = ( (x ))y, (x ) = ( x)y, y = The inverse function is f (x) = (x ) x (x ) x 0 Find the domain of the function f(x) = f(x) = x x (x )(x + ) ; x = and x = are not in the domain, because those values cause division by zero I must throw out values of x for which (x )(x+) < 0, since I can t take the square root of a negative number To find out which values I need to throw out, I solve the inequality The graph of y = (x )(x + ) is a parabola opening upward The roots are x = and x = - -

Thus, (x )(x + ) < 0 for < x < If I throw out these points, together with x = and x =, I m left with the domain: x < or x > Solve for x: (a) x 5 = 3 x 5 = 3 the distance from x to 5 is 3 3 3 Thus, x = or x = 8, so x = or x = 4 5 8 (b) 3x 5 = (x + ) Then (c) 4(x ) < x + 7 Then (d) x 4x = 5 Factor and solve: (e) x 4x = 5 3x 5 = (x + ), 3x 5 = x + 3x 5 = x + x x x 5 = + 5 5 x = 7 4(x ) < x + 7, 4(x ) < x + 7, 4x 8 < x + 7 4x 8 < x + 7 x x 3x 8 < 7 + 8 8 3x < 5 / 3 3 x < 5 x 4x = 5 5 5 x 4x 5 = 0 x 4x 5 = 0 (x 5)(x + ) = 0 ւ ց x 5 = 0 x + = 0 x = 5 x = x 4x = 5 + 5 5 x 4x + 5 = 0 3

Apply the quadratic formula: x = 4 ± 6 0 = 4 ± i = ± i Simplify the expressions, writing each result in the form a + bi: (a) i 33 (b) (3 i)(4 + 5i) i 33 = i i 3 = i (i ) 6 = i ( ) 6 = i = i (3 i)(4 + 5i) = 4i + 5i 5i = + i 5( ) = 7 + i (c) + 3i + 3i = + 3i 3i 3i = 3i ( + 3i)( 3i) = 3i 4 + 9 = 3 3 3 i (d) i 6 + 8i i 6 + 8i = i 6 + 8i 6 8i 0 0i = = 6 8i 00 0 5 i (e) ( + 3i) ( + 3i) = + ()(3i) + (3i) = 4 + i + 9i = 4 + i 9 = 5 + (f) 3 i 4 + i 3 i 4 + i = 3 i 4 + i 4 i 3i 8i + i = 4 i 6 i = 0 i 7 3 Find the quotient and the remainder when x 4 + 3x 3 x + is divided by x x + 3x - x - x x 4 4 + 3 x 3x 3x 3 3 3 - x + - x - x - 3x -x + 3x + -x + 3x x 4 + 3x 3 x + = (x )(x + 3x ) + 3x 4 Solve the following equations, giving exact answers: (a) e x + 5 = 6e x e x + 5 = 6e x 6e x 6e x e x 6e x + 5 = 0 (e x ) 6e x + 5 = 0 4

Let y = e x The equation becomes y 6y + 5 = 0 Factor and solve: y = gives e x = Then x = lne x = ln = 0 y = 5 gives e x = 5 Then x = lne x = ln 5 The solutions are x = 0 and x = ln 5 (b) 3 x+ = 7 x+ The solution is x = (c) 4 3x+ = 5 x y 6y + 5 = 0 (y )(y 5) = 0 ւ ց y = 0 y 5 = 0 y = y = 5 3 x+ = 7 x+ ln 3 x+ = ln 7 x+ (x + ) ln 3 = (x + ) ln 7 x ln3 + ln 3 = x ln 7 + ln 7 x ln 7 ln 3 x ln 7 ln3 x ln 3 x ln7 = ln 7 ln3 x(ln3 ln 7) = ln 7 ln3 / ln 3 ln 7 ln 3 ln7 x = ln 7 ln3 ln 3 ln7 ln 7 ln 3 ln 3 ln 7 (d) ln(x ) + ln(x + ) = ln 3x Factor and solve: 4 3x+ = 5 x ln 4 3x+ = ln 5 x (3x + ) ln4 = x ln5 (3 ln 4)x + ln 4 = (ln5)x ln 4 = (ln5)x (3 ln4)x ln 4 = (ln 5 3 ln 4)x ln 4 ln 5 3 ln4 = x ln(x )(x + ) = ln 3x e (ln(x )(x+)) = e (x )(x + ) = 3x x 4 = 3x 3x 3x x 3x 4 = 0 (ln 3x) x 3x 4 = 0 (x 4)(x + ) = 0 ւ ց x 4 = 0 x + = 0 x = 4 x = x = can t be plugged into the original equation, because you can t take the log of a negative number 5

x = 4 gives ln(x ) + ln(x + ) = ln + ln 6 = ln = ln 3x The only solution is x = 4 (e) (ln x) 3 lnx 4 = 0 Let y = ln x The equation becomes y 3y 4 = 0, or (y 4)(y + ) = 0 The solutions are y = 4 and y = y = 4 gives ln x = 4, so e ln x = e 4, and x = e 4 y = gives ln x =, so e ln x = e, and x = e The solutions are x = e 4 and x = e 5 (a) Simplify 500 500 = 00 5 = 0 5 (b) Simplify 300 300 = 00 3 = 0i 3 (c) Rationalize + 7 3 7 + 7 3 7 = + 7 3 7 + 3 7 + 3 7 = ( + 7)( + 3 7) ( 3 7)( + 3 7) = + 6 7 + 7 + 3( 7) 9( = 7) + 7 7 + 3(7) 9(7) = 3 + 7 7 6 = 3 + 7 7 6 6 Find the equation of the line: (a) Which passes through the points (, 3) and (, ) The slope is 3 = The point-slope form for the equation of the line is 3 (x ) = y 3 3 (b) Which passes through the point (3, 4) and is perpendicular to the line x 8y = 5 x 8y = 5 x x 8y = x + 5 / 8 8 8 y = 4 x 5 8 6

The slope of the given line is The line I want is perpendicular to the given line, so the line I want 4 has slope 4 (the negative reciprocal of ) The point-slope form for the equation of the line is 4 4(x 3) = y 4, or y = 4x + 6 (c) Which is parallel to the line 3y 6x + 5 = 0 and has y-intercept 7 3y 6x + 5 = 0 3y = 6x 5 y = x 5 3 The given line has slope The line I want is parallel to the given line, so it also has slope Since it has y-intercept 7, the equation is y = x 7 7 Solve the system of equations for x and y: x + 5y = 7, x + 3y = 4 Multiply the second equation by, then subtract it from the first: x + 5y = 7 x + 6y = 8 y = 5 y = 5 Plug this into the first equation: x 75 = 7 Then The solution is x = 4, y = 5 x 75 = 7 + 75 75 x = 8 / x = 4 8 Suppose log a x = 5 and log a y = Find: (a) log a 3 x = 3 log a x = 5 3 (b) log a (a 6 y ) = log a a 6 + log a y = 6 + = 0 (c) log a x 3 y = ( 3) log a x log a y = 5 4 = 9 (a) Simplify and write the result using positive exponents: ( 3x ) 5 y 7 9x 3 y 6 ( 3x ) 5 y 7 9x 3 y 6 = 43x0 y 7 9x 3 y 6 7 = 7x7 y 3

(b) Simplify and write the result using positive exponents: 4(x /3 y /5 ) ( 3y 3/5 ) x /6 4(x /3 y /5 ) ( 3y 3/5 ) x /6 = 4x /3 y 4/5 9y 6/5 x /6 = 36x / y /5 = 36x/ y /5 30 (a) Simplify, cancelling any common factors: x x x 4 x 3 3x x x 6 x x x 4 x 3 3x x x 6 = x x x 4 x x 6 x(x ) (x 3)(x + ) x 3 3x = (x )(x + ) x = (x 3) x (b) Simplify, cancelling any common factors: a 3 a b a 3 4ab a 4 + 3a 3 b a + ab b a 3 a b a 3 4ab a 4 + 3a 3 b a + ab b = a3 a b a 3 4ab a + ab b a (a b) (a b)(a + b) a 4 + 3a 3 = b a(a b)(a + b) a 3 = a b (a + 3) a (a + 3b) (c) Simplify, cancelling any common factors: x 5x x 3 4x x 0x + 5 x 6 x 5x x 3 4x x 0x + 5 x 6 = x 5x x 3 4x x 6 x(x 5) (x 4)(x + 4) x = 0x + 5 x (x 4) (x 5) = x + 4 x(x 5) 3 (a) Find the equation of the parabola which passes through the point (3, 7) and has vertex (, 3) The vertex is (, 3), so the parabola s equation has the form y = a(x ) + 3 Plug in x = 3, y = 7: 7 = a(3 ) + 3, 7 = a + 3, 7 3 = a + 3 3, 4 = a The equation is y = 4(x ) + 3 (b) Write the quadratic function f(x) = x 6x + 34 in standard (vertex) form Complete the square: ( 6 ) = ( 3) = 9, so f(x) = x 6x + 34 = (x 6x + 9) + (34 9) = (x 3) + 5 8

3 Find the center and the radius of the circle whose equation is x 4x + y + 6y = Half of 4 is, and ( ) = 4 I need to add 4 to the x-stuff to complete the square Half of 6 is 3, and 3 = 9 I need to add 9 to the y-stuff to complete the square x 4x + 4 + y + 6y + 9 = + 4 + 9, (x ) + (y + 3) = 5 The center is (, 3) and the radius is 5 33 Suppose that $000 is invested at 6% annual interest, compounded monthly How many years must pass before the account is worth at least $000? (Round up to the nearest year) The compound interest formula is A = P ( + r n) nt P is the amount invested, t is the number of years, n is the number of compoundings per year, r is the interest rate, and A is the value of the account In this case, P = 000, n = (monthly compounding, with months in a year), r = 006, and A = 000 I want to find t, where Simplify, and solve for t: Thus, t = ( 000 = 000 + 006 ) t 000 = 000 005 t = 005 t ln = ln 005 t ln = t ln 005 ln ln 005 = t ln 583 Rounding up, I find that the account must mature for years ln 005 The best thing for being sad is to learn something - Merlyn, in T H White s The Once and Future King c 008 by Bruce Ikenaga 9