1. ENM Review 503
Midterm

2. Running the Bases
1. Find a 3-digit number in base 7 that is equivalent to 147 in base 10 representation. Q R 7 | ACDC16 = ________ Write the first 10 primes in base 7.

3. Functions
4. For a and b > 0, prove that geometric mean , < arithmetic mean (a+b)/2. Assume ab < [(a + b)/2]2 Then 4ab < a2 + 2ab + b2 => a2 -2ab + b2 = (a-b)2 > Determine conditions on a, b, and c so that the function f(x) = ax2 + bx + c satisfies f(x) = x2f(1/x) for all x  0. ax2+ bx + c = x2(a/x2 + b/x + c) = a + bx + cx2 => a = c; b arbitrary 6. Suppose a function f satisfies f(1) = 1 and f(x) = f(x - 1) + 2 for all x. Find f(3) and f(4). f(2) = f(1) + 2 = 3; f(3) = f(2) + 2 = 5; f(4) = f(3) + 2 = If f(x) = x2 and x = g(t) = t + 2, find f[g(3)]. f(g(t) = (t + 2)2 => f(g(3) = 25
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4. Graphs
8. Sketch Domain, range, symmetry, intercepts, and asymptotes. 9. Sketch |x| + |y| = Describe location in xy-plane of points P(x, y) for which a. x > 0 b. x > 0 and y < 0 c. x = 0 d. y = 0 e. x and y have same sign. f. y = -|x| g. xy = 0 a. rt hand plane b. R-Bot plane c. y-axis d. x-axis e. quads 1 & 3 f. Inverted v g. xy-axes
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5. Lines
11. Find a) inclination and slope of the line y = x; b) equation of line through (-2, 0) with inclination 45°. Inclination is 45° and slope is tan (45°) = 1; y = x Points (4, -2) and (8, 6) are endpoints of a diameter. Write the equation of the circle. d = => d2 = 80 => 4r2 = 80 => r2 = 20. midpoint = (6, 2) => circle equation is: (x - 6)2 + (y - 2)2 = Show that the points (-3, 2), (4, 1), and (5, 8) are vertices of a right triangle. Find the center and radius of the circumscribed circle. D2: (-3, 2) & (4, 1) = 50; D2: (-3, 2) & (5, 8) = 100; D2: (4, 1) & (5, 8) = 50 => isosceles right triangle (x - 1)2 + (y - 5)2 = 50 with center at (1, 5) with radius = 50½
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6. Lines
14. Show that the line through points (4, -1) and (5, 1) is perpendicular to the line through points (11, 1) and (9, 2). slope m1 = 2 and slope m2 = -1/2 and product = Find the slopes of the lines y = ax + b and ax + by + c = 0. a and -a/b 16. a. Show that the line 6x - y - 9 = 0 is tangent to y = x2. b. The slope of the curve y = |x| at x = p is 1. 6x - 9 = x2 => (x - 3)2 = 0 => line intersects parabola at x = 3 and y = 9 => the slope is y' = 2x = 2(3) = 6.
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7. 17. Find a real root solution to the equation
e5x + e4x - 6e3x = e3x(e2x + ex - 6) = 0
Let y = ex; Then
y5 + y4 - 6y3 = 0
y3(y2 + y – 6) = 0 =>
(y – 2)(y + 3) = 0
y = 2 = ex => x = ln 2; ex > 0

8. Solving logarithmic Equations
18. log 3x + log 3 = 2 => log 9x = 2 => 9x = 102 => x = 100/9 19. e3x = 14 => 3x = Ln 14 = => x = Ln (logx3) = 2 => logx3 = e2 => x7.389 = 3 => Ln x = Ln 3 => x = 1.160
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9. Solving logarithmic Equations
x = 15 => (3 – 4x) Ln 6 = Ln 15 => 3 – 4x = => x = Ln (x + 2) = x2 – 7 x + 2 = e(x^2 - 7) x = 2.932
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10. Inverses
23. Find the inverse of y = x2 – 16. x =  (y – 16)½ and the function y is not 1-1 and the inverse does not exist. 24. Find and plot the inverse of y = 3x Use truth tables to prove De-Morgan's Law. Include a Venn diagram to clarify. of ~A~B = ~(A + B)

11. 26. Solve for x.
Log = logx 4532
27. A B C D
D = abc + abC + Abc + AbC
= ab + Ab
= b

12. Tableaus
Max 3x1 + 4x2 s.t. 2x1 + 3x2 <= 12 AX = B 5x1 + 3x2 <= 15 Initial Tableau at corner point (0, 0) Tableau at corner point (0, 4) -1/3 0 4/ /3 1 1/ Final Tableau Tableau at corner point (3, 0) /9 1/9 49/3 0 -7/5 0 ¾ /9 -2/9 10/3 0 7/5 1 -2/ /3 1/ /5 0 1/5 3 X2 5 4 X1
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13. Dual Solutions
Max Z = 10X1 + 24X2 Min W = 120Y Y2 s.t. 1X1 + 2X2 <= 120 s.t 1Y1 + 1Y2 >= 10 1X1 + 4X2 <= 180 2Y1 + 4Y2 >= 24 (LP '(( )( )( )))  /2 1/2 30 W = 120(8) + 180(2) = 1320
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14. Converting (<= >=) Constraints
Min W = 4X1 + 3X2 s.t. 3X1 - 1X2 >= 2  3X1 - 1X2 >= 2 1X1 + 1X2 <= 1  -1X1 - 1X2 >= -1 -4X1 + 1X2 <= 3  4X1 - 1X2 >= -3 LP '(( )( )( )))  0 1/3 17/3 2/3 0 8/3 1 -1/3 4/3 1/3 0 4/3 0 -4/3 1/3 1/3 1 13/3 W = 4(2/3) + 3(0) = 8/3
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15. Converting Equality Constraints
2X1 + 3X2 = 6 becomes 2X1 + 3X2 <= 6 2X1 + 3X2 >= 6
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16. Inverse Matrix
28. Find the inverse and determinant of matrix … and write its characteristic equation. Show that the matrix satisfies its characteristic equation and again find its inverse The characteristic equation is 1t2 -10t + 25 = 0 7/25 -4/25 1/25 3/25 A2 – 10A + 25I = = -