Practice Problems for MTH 11 Exam Prof. Townsend Fall 013 The problem list is similar to problems found on the indicated pages. means I checked my work on my TI-Nspire software Pages 04-05 Combine the following terms into a single fraction. 1) 3 4x + 7a 4 + a) Factor all denominators first term: x second term: third term: 1 b) Form the Lowest Common Denominator (LCD) such that it includes all the factors found from the denominators. LCD = x 1 = 4x c) Find the denominators in the LCD. first term: x 1 second term: x 1 third term: x 1 d) Multiply all numerators and denominators by the factors missing in the original denominator. You can ignore the 1 as it does not change the answer. 3 4x + 7a x 4 x + 4x 4x e) Combine factors in numerators and denominators for each term. 3 4x + 7ax 4x + 8x 4x f) Combine the terms into a single term. 3+ 7ax + 8x 4x Page 1 of 19
) 6 5x + a 3 5x a) Factor all denominators first term: 5 x x x second term: 5 5 x b) Form the LCD such that it includes all the factors found from the denominators. LCD = 5 5 x x x = 5x 3 c) Find the denominators in the LCD. first term: 5 5 x x x second term: 5 5 x x x d) Multiply all numerators and denominators by the factors missing in the original denominator. 6 5 5x 3 5 + a x 5x x e) Combine factors in numerators and denominators for each term. 6 5 5x + ax 3 5x 3 f) Combine the terms into a single term. 30 + ax 5x 3 Page of 19
3) x x 6 + 1 4 3x 4x 1 a) Factor all denominators first term: (x 3) second term: third term: (x 3) b) Form the LCD such that it includes all the factors found from the denominators. LCD = x 3 ( ) = 4( x 3) <- - - - - - - - - - - - - Corrected c) Find the denominators in the LCD. first term: (x 3) second term: (x 3) third term: (x 3) d) Multiply all numerators and denominators by the factors missing in the original denominator. x x 6 + 1 x 3 3x 4 x 3 4x 1 e) Combine factors in numerators and denominators for each term. x 4 x 3 4 x 3 4 x 3 ( ) + x 3 ( ) 3x ( ) f) Combine the terms into a single term. x + x 3 3x 4 x 3 ( ) = 3 4 x 3 ( ) Page 3 of 19
4) x 1 3x +1 3x 13x + 4 4 x a) Factor all denominators first term: (x 4) (3x 1) second term: 4 x= (x 4) b) Form the LCD such that it includes all the factors found from the denominators. LCD = (x 4) 3x 1 ( ) c) Find the denominators in the LCD. first term: (x 4) (3x 1) second term: (x 4) (3x 1) d) Multiply all numerators and denominators by the factors missing in the original denominator. x 1 1 3x +1 3x 1 x 4 1 4 x 3x 1 ( )( 3x 1) ( ) e) Combine factors in numerators and denominators for each term. x 1 ( 3x +1) ( 3x 1 ) x 4 x 4 ( )( 3x 1) ( )( 3x 1) f) Combine the terms into a single term. Note that the minus signs cancel in each term. ( x 1) + ( 3x +1) ( 3x 1) x 4 ( )( 3x 1) x 1+ 9x 1 ( x 4) ( 3x 1) = 9x + x 3x 13x + 4 Page 4 of 19
Pages 09-10 Solve for x using the method of multiplication by LCD. 5) 1 x 5 = 3 6 4 a) Find the LCD. LCD= 3 b) Multiply all terms by it. LCD 1 x 5 = 3 6 4 = 3 1 x 5 = 3 6 4 x 5 3 1 3 6 = 3 3 4 c) Cancel the appropriate terms in the LCD with the denominators 3 x 5 ( ) = 3 3 d) Do the algebra to solve for x. 1 x +10 = 9 Add and subtract terms from both sides x = 9 = 13 Divide by. x = 13 = 6.5 6) 3x 7 5 1 = x 14 a) Find the LCD. LCD=7 3 b) Multiply all terms by it. 3x LCD 7 5 1 = x 14 = 3 7 3x 7 5 1 = x 14 3x 5 x 3 7 7 3 7 1 = 3 7 14 c) Cancel the appropriate terms in the LCD with the denominators 3 3x ( ) ( 5) = 3 ( x) d) Do the algebra to solve for x. 18x 10 = 6 3x Add and subtract terms from both sides 1x = 16 Divide by 1. x = 16 1 Page 5 of 19
7) 1 4x + 3 x = x +1 a) Find the LCD. LCD= x (x+1) b) Multiply all terms by it. 1 LCD 4x + 3 x = 1 x +1 = x ( x +1) 4x + 3 x = x +1 1 3 x ( x +1) 4x + x ( x +1) x = x ( x +1) x +1 c) Cancel the appropriate terms in the LCD with the denominators ( x +1) ( 1) + ( x +1) ( 3) = x ( ) d) Do the algebra to solve for x. 7 x +1 ( ) = 8x 7x + 7 = 8x Add and subtract terms from both sides x = 7 Page 6 of 19
8) S = P A + Mc I Solve for A a) Find the LCD. LCD=A I b) Multiply all terms by it. LCD S = P A + Mc I = A I S = P A + Mc I P Mc A I S = A I A + A I I c) Cancel the appropriate terms in the LCD with the denominators A I S = I P ( ) + A ( Mc) d) Do the algebra to solve for x. AIS = IP + AMc Add and subtract terms from both sides AIS AMc = IP Factor out the A. A IS Mc ( ) = IP ( ) Divide by IS Mc A = IP IS Mc Page 5 and 9-30 Quadratic Equation and Parabolas Page 5 - Analytically find the values of x Quadratic Equation: ax + bx + c = 0 Quadratic Formula: x = b ± b 4ac a Which means there are two solutions for x: x = b + b 4ac a and x = b b 4ac Use the discriminant to determine the type of solution. D = b 4ac D>0 Two solutions D=0 One Solution D<0 No real solution a Page 7 of 19
9a) x + x + = 0 ax + bx + c = 0 so a=1, b=, c= Find the discriminant, D = b 4ac = 4( 1) ( ) = 4 D<0 so no real solution 10a) x 6x + 8 = 0 ax + bx + c = 0 so a=, b= 6, c=8 Find the discriminant, D = b 4ac = 6 D>0 Two solutions ( ) 4( ) ( 8) = 100 ( ) ± 100 ( ) x = 6 = 6 ±10 4 x = 1, 4 11a) x( x 1) = 18 Rearrange the terms so the equation looks like ax + bx + c = 0. x x 1 ( ) = 18 x 1x = 18 x 1x +18 = 0 ax + bx + c = 0 so a=, b= 1, c=18 Find the discriminant, D = b 4ac = 1 D=0 One solution ( ) ( ) x = 1 = 3 ( ) 4( ) ( 18) = 0 Page 8 of 19
Page 9-30 Graphically find y- intercept, zeros(roots), vertex. Locate them on the graph y = ax + bx + c a) y- intercept x=0 so yintercept=c b) zeros(roots) y=0 so ax + bx + c = 0 (You should get the same answers as you did above using the quadratic formula) c) vertex x vertex = b a y = ax vertex vertex + bx vertex + c means the graph corroborates the math. 9b) y = x + x + a=1, b=, c= yintercept= x vertex = (1) = 1 y = ( 1 vertex ) + 1 No real solution from 9a ( ) + = 1 Page 9 of 19
10b) y = x 6x + 8 a=, b= 6, c=8 yintercept=8 x vertex = 6 ( ) = 3 y vertex = 3 6 3 + 8 = 5 = 1.5 <- - - - - - - - - - - - - Corrected x = 1, 4 from 10a Page 10 of 19
11b) x( x 1) = 18 0 = x 1x +18 a=, b= 1, c=18 Generalize to y = x 1x +18 yintercept=18 ( ) x vertex = 1 () = 3 y = ( 3 vertex ) 1 3 ( ) +18 = 0 x = 3 from 11a where y = 0 Page 11 of 19
Page 111 # Angles 1) Draw the given angles inside a circle. θ = 0, θ = 100, θ = 00, θ = 300 0 100 00 300 13) Determine one positive and one negative coterminal angle for the given angle. θ = 0 θ = 100 θ = 00 θ = 300 θ = 0 + 360 = 380 θ = 100 + 360 = 460 θ = 00 + 360 = 560 θ = 300 + 360 = 660 θ = 0 360 = 340 θ = 100 360 = 60 θ = 00 360 = 160 θ = 300 360 = 60 14) Convert the following angles from radians to degrees θ = 0 θ = 100 θ = 00 θ = 300 θ = 0deg π rad 360 deg θ = 100deg π rad 360 deg θ = 00deg π rad 360 deg θ = 300deg π rad 360 deg = π 9 = 0.3491rad = 5π 9 = 1.745rad = 10π 9 = 3.491rad = 5π 3 = 5.36rad 15) Convert the following degrees from radians to angles θ = 1 rad., θ = rad., θ = 4 rad., θ = 6 rad., θ = 1rad 360 deg π rad θ = rad 360 deg π rad θ = 4rad 360 deg π rad θ = 6rad 360 deg π rad = 180 π = 57.3 = 360 π = 114.6 = 70 π = 9. = 1080 π = 343.8 Page 1 of 19
16) The given angles are in standard position. Identify the quadrant in which the terminal side of the angle lies. Note: if θ = 0,90,180,70,360 etc., the angle is called a quadrantal angle. θ = 0 θ = 100 θ = 00 θ = 300 0 < 0 < 90 90 < 100 < 180 180 < 00 < 70 70 < 300 < 360 I II III IV 17) Draw angles in standard position such that the terminal side passes through the given point. Identify the quadrant in which the terminal side of the angle lies. ( 1, ) ( 1, ) ( 1, ) ( 1, ) (1,) (-1,) 1 1 1 1 (1,) (-1,) (-1,-) (1,-) (-1,-) (1,-) I II III IV Page 13 of 19
Page 115 Defining the trig functions 18) Find the values of the six trig functions of the angle (in standard position) whose terminal side passes through the given points. ( 1, ) ( 0,) (,0) ( 3,4 ) x = 1 y = r = 1 + = 5 x = 0 y = r = 0 + = x = y = 0 r = + 0 = x = 3 y = 4 r = 3 + 4 = 5 cosθ = x r = 1 5 cosθ = x r = 0 cosθ = x r = = 1 cosθ = x r = 3 5 sinθ = y r = 5 sinθ = y r = 1 sinθ = y r = 0 = 0 sinθ = y r = 4 5 secθ = 1 cosθ = r x = 5 1 = 5 cscθ = 1 sinθ = r y secθ = 1 cosθ = r x = 0 = cscθ = 1 sinθ = r y secθ = 1 cosθ = r x = = 1 cscθ = 1 sinθ = r y tanθ = y x = 1 = tanθ = y x = 0 = tanθ = y x = 0 = 0 tanθ = y x = 4 3 = 11 3 secθ = 1 cosθ = r x = 5 3 = 1 3 cscθ = 1 sinθ = r y = 5 cotθ = 1 tanθ = x y = = 1 cotθ = 1 tanθ = x y = 0 = cotθ = 1 tanθ = x y = 5 4 = 1 1 4 cotθ = 1 tanθ = x y = 1 = 0 = 0 = 0 = = 3 4 Page 14 of 19
19) Given one trig function, find some of the others by extracting the implied triangle. Given cosθ = 3 cosθ = x r = 3 x = 3 r =, find sinθ and cotθ. Since similar triangles have the same trig functions, assume x = 3 and r = y = r x = ( 3) = 4 3 = 1 y 3 sinθ = y r = 1 cotθ = x y = 3 1 = 3 Given sinθ = 5, find secθ and cscθ. 13 Since similar sinθ = y r = triangles have 5 the same trig functions, assume r = 5 and y = 13 x 5 x = r x = ( 5 ) = 5 4 = 1 r = 5 y = secθ = r x = 5 1 = 5 cscθ = 1 sinθ = 5 Given tanθ = 1.0, find sinθ and cosθ. tanθ = y x = 1 = 1 Since similar triangles have the 1 same trig functions, assume x = 1 y = 1 and y = 1 r 1 1 x = 1 y = x + y = 1 +1 = sinθ = y r = 1 cosθ = x r = 1 Page 15 of 19
Given cotθ = 1 5, find sinθ and cosθ. cotθ = x y = 1 5 Since similar triangles have the same trig functions, assume x = 1 and y = 5 r 1 5 x = 1 r = x + y = 1 + 5 = 144 + 5 = 13 y = 5 sinθ = y r = 5 13 cosθ = x r = 1 13 Page 118-119 Values of trig functions and inverse trig functions. 0) Using inverse trig functions, findθ for each given trig function. Find θ for cosθ = 3 Find θ for sinθ = 5 13 Find θ for tanθ = 1.0 Find θ for cotθ = 1 5 θ = cos 1 3 = 30 5 θ = sin 1 13 =.6 θ = tan 1 1 = 45 θ = cot 1 1 5 =.6 1) Using inverse trig functions, find cosθ, given cosθ Note the angles were found in problem 0. Given cosθ = 3, find sinθ and cotθ. θ = 30 sin 30 = 1 cot 30 = 3. Given sinθ = 5, find secθ and cscθ. 13 θ =.6 sec.6=1.083 csc.6=.6 Given tanθ = 1.0, find sinθ and cosθ. θ = 45 sin 45 = 1 cos 45 = 1. Given cotθ = 1 5, find sinθ and cosθ. θ =.6 sin.6 = 0.3846 cos.6=0.931 Page 16 of 19
Page 13 Solving right triangles ) Solve the right triangle for the missing angle(s) and side(s). Given one angle and one side. First find the other angle since A + B = 90 Then use a trig function using the given side and angle to find another side. Use the Pythagorean Theorem to find the third side. Check using trig function and the two numbers you just found. The given numbers are in black. The found numbers are in red A = 53.13 a = 8 A = 53.13 a = 8 B = 36.37 b = 6 C = 90 c = 10 B = 90 53.13 = 36.87 sin A = a sin53.13 = 8 c c b = c a = 10 8 = 6 Check: sin B = b sin 36.87 = 0.6 c c = 8 sin53.13 = 8 0.8 = 10 b c = 6 10 = 0.6 B = 36.87 c = 10 A = 53.13 a = 8 B = 36.87 b = 6 C = 90 c = 10 A = 90 36.87 = 53.13 sin B = b c a = c b = 10 6 = 8 Check: sin A = a c sin 36.87 = b 10 sin53.13 = 0.8 b = 10sin 36.87 = 10 0.6 = 6 a c = 8 10 = 0.8 Page 17 of 19
Given two sides. Find an angle using the two sides and an appropriate inverse trig function. Then find the other angle since A + B = 90 Then use a trig function using the given side and angle to find the other side. Check using trig function and the two numbers you just found. The given numbers are in black. The found numbers are in red. a = 10 c = 6 A =.6 a = 10 B = 67.38 b = 4 C = 90 c = 6 a c = sin A sin A = a c = 10 6 = 5 13 A = sin 1 5 13 =.6 B = 90.6 = 67.38 sin B = b c b = csin B = 6sin67.38=4. Check: cos A = b c cos.6 = 0.931 b c = 4 6 = 0.931 Page 18 of 19
a = 10 b = 4 A =.6 a = 10 B = 67.38 b = 4 C = 90 c = 6 a b = tan A tan A = a b = 10 4 = 5 1 A = tan 1 5 1 =.6 B = 90.6 = 67.38 sin B = b c Check: c = b sin B = 4 0.931 = 6. cos A = b c cos.6 = 0.931 b c = 4 6 = 0.931 Page 19 of 19