Solved Practice Final Exam - Intermediate Algebra | MATH 105, Exams of Algebra

Download Solved Practice Final Exam - Intermediate Algebra | MATH 105 and more Exams Algebra in PDF only on Docsity! MATH 105 Practice Final 1) The function A described by A(r) = r2 √ 3 4 gives the area of an equilateral triangle with side r. Find the area when the side is 6 in. Find A(6). A(6) = (6)2 √ 3 4 = 36 √ 3 4 = 9 √ 3  2) Write an equation of the line containing the point (3,−2) and (a) Parallel to the line 3x + 4y = 5 (b) Perpendicular to the line 3x + 4y = 5 First find the slope of 3x + 4y = 5 by putting it into slope-intercept form. 3x + 4y = 5 4y = −3x + 5 y = −3 4 x + 5 4 So the slope m = −34 . (a) Parallel lines have the same slope so use slope-point form to get the equation. y− (−2) = −3 4 (x− 3) y = −3 4 x + 9 4 − 2 y = −3 4 x + 1 4 1 (b) Perpendicular lines have slopes which are negative reciprocals of one another. So the slope of a perpendicular line is 43 . y− (−2) = 4 3 (x− 3) y = 4 3 x− 4− 2 y = 4 3 x− 6  3) For f (x) = 2xx+1 , g(x) = 2x + 5, determine the domain of f g . f (x) is undefined when x + 1 = 0 so x 6= −1. g(x) is defined for all real numbers. fg is undefined when either f or g is undefined or when g(x) = 2x + 5 = 0 so x 6= −52 and x 6= −1. In set-builder notation: {x|x is real and x 6= −52 and x 6= −1} In interval notation: (−∞,−52) ∪ (− 5 2 ,−1) ∪ (−1, ∞)  4) Solve 2x 3 + 3y 4 = 11 12 x 3 + 7y 18 = 1 2 Multiply the second equation by −2, add and then solve. 2x 3 + 3y 4 = 11 12 −2x3 − 7y 9 = −1 0+ 3y4 − 7y 9 = − 1 12 27y− 28y = −3 y = 3 2 10) Factor completely: p6 − q6 Solve using the identities a2 − b2 = (a + b)(a− b), a3 − b3 = (a− b)(a2 + ab + b2), and a3 + b3 = (a + b)(a2 − ab + b2). p6 − q6 (p3)2 − (q3)2 (p3 + q3)(p3 − q3) (p + q)(p2 − pq + q2)(p− q)(p2 + pq + q2)  11) Solve (x + 2)(x− 5) = 8 (x + 2)(x− 5) = 8 x2 + 7x− 10 = 8 x2 + 7x− 18 = 0 (x− 6)(x + 3) = 0 x = 6 or x = −3  12) Divide and simplify: a3 + 4a a2 − 16 ÷ a2 + 8a + 15 a2 + a− 20 Flip and multiply, then simplify. a3 + 4a a2 − 16 · a2 + a− 20 a2 + 8a + 15 a(a2 + 4)(a− 4)(a + 5) (a− 4)(a + 4)(a + 3)(a + 5) a(a2 + 4) (a + 3)(a + 4)  5 13) Combine and simplify if possible: 2 + t t− 3 − 18 t2 − 9 2 + t t− 3 − 18 (t− 3)(t + 3) 2 · (t− 3)(t + 3) (t− 3)(t + 3) + t t− 3 · t + 3 t + 3 − 18 (t− 3)(t + 3) 2(t− 3)(t + 3) + t(t + 3)− 18 (t− 3)(t + 3) 2(t2 − 9) + t2 + 3t− 18 (t− 3)(t + 3) 2t2 − 18 + t2 + 3t− 18 (t− 3)(t + 3) 3t2 + 3t− 36 (t− 3)(t + 3) 3(t2 + t− 12) (t− 3)(t + 3) 3(t + 4)(t− 3) (t− 3)(t + 3) 3(t + 4) t + 3  6 14) Simplify x y − y x x2 y − y x y − y x x2 y − y · xy xy x2y y − xy 2 x x3y y − xy2 x2 − y2 x3 − xy2 x2 − y2 x(x2 − y2) 1 x  15) Solve y + 5 y + 1 − y y + 2 = 4y + 15 y2 + 3y + 2 y + 5 y + 1 − y y + 2 = 4y + 15 (y + 2)(y + 1) (y + 2)(y + 1) ( y + 5 y + 1 − y y + 2 ) = 4y + 15 (y + 2)(y + 1) (y + 2)(y + 1) (y + 2)(y + 5)− y(y + 1) = 4y + 15 y2 + 7y + 10− y2 − y = 4y + 15 2y = 5 y = 52  7 Checking: √ 20− 80 9 + 8 = √ 9− 80 9 + 11√ 180− 80 9 + 8 = √ 81− 80 9 + 11√ 100 9 + 8 = √ 1 9 + 11 10 3 + 8 = 1 3 + 11 34 3 = 34 3 True So x = 809 is a solution.  22) Divide and Simplify. Write in the form a + bi: 5 + 3i 7− 4i 5 + 3i 7− 4i · 7 + 4i 7 + 4i 35 + 20i + 21i− 12 49 + 16 23 + 41i 65 23 65 + 41 65 i  23) Completely solve x2 + 9 = 4x x2 − 4x = −9 x2 − 4x + 4 = −9 + 4 (x− 2)2 = −5 x− 2 = ±i √ 5 x = 2± i √ 5  10 24) Two hoses are connected to a swimming pool. Working together, they can fill the pool in 4 hr. The larger hose, working alone, can fill the pool in 6 hr less time than the smaller one. How long would the smaller take, working alone, to fill the pool? Set the time the smaller hose would take to be s, then the time the larger hose takes is s− 6. 1 s− 6 + 1 s = 1 4 4s + 4(s− 6) = s(s− 6) 4s + 4s− 24 = s2 − 6s 0 = s2 − 14s + 24 0 = (s− 2)(s− 12) s = 2 or s = 12 But if s = 2 then the time the larger hose takes would be -4, so the only solution is s = 12. Thus the smaller hose takes 12 hrs to fill the pool.  25) Solve y 1 3 − y 1 6 − 6 = 0 Substitute u = y 1 6 into the equation and solve. u2 − u− 6 = 0 (u− 3)(u + 2) = 0 u = 3 or u = −2 Substitute y 1 6 = u back in. y 1 6 = 3 or y 1 6 = −2 y = 36 or y=(−2)6 y = 729 or y = 64 Checking y = 729 729 1 3 − 729 1 6 − 6 = 0 9− 3− 6 = 0 0 = 0 True 11 Checking y = 64 64 1 3 − 64 1 6 − 6 = 0 4− 2− 6 = 0 −4 6= 0 False So y = 729 is the only solution.  26) Find the vertex, axis of symmetry, maximum or minimum value, x- and y-intercepts and graph: f (x) = 2(x− 5)2 − 3 The function is in a(x − h)2 + k form so its vertex is (5,−3) and its axis of symmetry is x = 5. The leading coefficient is positive so it has a minimum value which is f (x) = −3. To find the y-intercept, find f (0) f (0) = 2(0− 5)2 − 3 = 2 · 25− 3 = 47 So the y-intercept is (0, 47). To find the x-intercepts, set f (x) = 0 0 = 2(x− 5)2 − 3 (x− 5)2 = ±3 2 x− 5 = ± √ 3√ 2 x = 5± √ 6 2 So the x-intercepts are (5 + √ 6 2 , 0) and (5− √ 6 2 , 0). 12