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2010 March Regional Algebra 1 Individual Test Solutions: 7 1. 13 + 3 7 13 −3 4 – − 2 7– 3 2( 7+ 3) 4 7 13−3 13 + 3 13−3 – 2( 7+ 3) 7 – 3 ( 7+ 3) D. 2. 5x + 8y =3, m = -5/8. Perpendicular line will have opposite reciprocal of the slope new slope = 8/5. Using m = 8/5 and (1, 4), we find y = (8/5)(x) + (12/5). Writing the equation in standard form, we obtain 5y – 8x = 12 C. 3. (1/20) + (1/15) + (1/x) = 1/5 15 + 20 + (300/x) = 60 25 x = 300 x = 12 minutes D. 4. By definition, it is the Commutative Property B. 5. 1569978 is divisible by 1, 2, 3, 6, and 9. The sum of these divisors is 21 B. 6. 119 is divisible by 7 and by 17 C. 7. (x3 + 15x + 2)2 + (–x6 – 20x5 –4x3 - 225x2 – 60x4 – 4) x6 + 30x4 + 225x2 + 4x3 + 60x + 4 – 4 – x6 – 20x5 – 4x3 - 225x2 – 60x4 -30x4 – 20x5 + 60x 10x(-3x3 – 2x4 + 6) E. 8. (x + y)2 = 225 x + y = 15, x + y = -15 (second possibility is ruled out, though, because both integers must be positive). (x – y)2 = 49 x – y = 7, x - y = -7 (x + y) + (x – y) = 2x = 22 x = 11. 11 + y = 15, y 4. OR (x+y) + (x – y) = 15 – 7 = 8 = 2x x = 4 x + y = 15, 4 + y = 15 y = 11. In either case, the difference between x and y is 7 units. However, the magnitude refers to the value of the number, and not to its sign answer is E. 9. 36e2g2 – 36e2 – 4g2 + 4 36e2(g2 – 1) – 4(g2 – 1) (36e2 – 4) (g2 – 1) 4(9e2 – 1)(g+1)(g-1) 4(3e +1)(3e – 1)(g+1)(g-1) C. 10. 243𝑦 − 147𝑦 + 75𝑦 9 3𝑦 − 7 3𝑦 + 5 3𝑦 = 7 3𝑦 C = 7, D = 3. C2 - D2 = 49 – 9 = 40 A. 11. C = (7/10)(W), 2(C – 11) = W – 11 + 9. Substituting, and simplifying: 2C – 22 = W – 2 2C = W + 20 (7/5)(W) = W + 20 (2/5)(W) = 20 W = 50. If W = 50, then C = (7/10)(50) = 35. Wenxuan is currently 50 years old, while Clara is 35 C. 12. y = k/x2. y = 4, x = 100. k =(y)(x2) = 4(100)2 = 40,000. y = 40,000/x2; x = 160, y = (40,000)/(160)2 = 1.5625 C. 13. I – False. There are 8 primes between 1 and 20 (2,3,5,7,11,13,17,19) II - False. There are 5 triangular numbers between 1 and 20 (1,3,6,10,15). III - False – 3/2 is both rational and an integer. IV – True. Answer – C. 14. x – 5 = 3x + 3 x2 – 10x +25 = 3x + 3 x2 – 13x + 22 = 0 (x – 2)(x – 11) = 0 x = 2, 11. However, x =2 is an extraneous solution since it yields an impossible solution ( 3x + 3 = −3). Hence the answer is B. 15. f(x) = 5x2 – 23x + 17. The discriminant of a function is equal to B2 – 4AC (-23)2 – 4(5)(17) = 529 – 340 = 189 C. 16. f(x) = 2x2 – 21x – 65. To find the x-intercepts, we set the function equal to zero. 2x2 – 21x – 65 = 0 2x2 – 26x + 5x – 65 = 0 2x(x -13) + 5(x – 13) = 0 x = -2.5, x = 13. X-intercepts are then (-2.5, 0) & (13,0) E. 17. If ΔABC is a right triangle, then it is true that a2 + b2 = c2. Therefore, (2)2 + (x)2 = (2x2 – 9x + 12) 4 + x2 = 2x2 – 9x + 12 x2 – 9x + 8 = 0 x =1, 8. However, we cannot evaluate the value of M(x) for a prime solution to the triangle since neither 1 nor 8 is prime answer is E. 18. 1 2187 = 35y-13 3-7 = 35y-13 5y – 13 = –7 5y = 6 y = 6/5 or B) 1.2. 19. 0= h +vt – 16t2. In this case, –16t2 + 32t + 40 = 0. This parabola achieves its maximum value at t = -b/2a or -32/-32 = 1. At t = 1, the book is at a height of -16(1)2 + 32 + 40 = 56 feet above the ground. However, Bo is 40 feet above the ground. Therefore the distance between him and the book is only 16 feet B. 20. 3*y = (9 – 3y +y)(3y) = 30 y(9 – 2y) = 10 2y2 – 9y + 10 = 0 (2y – 5)(y-2) = 0 y = 2.5, 2. Since 2.5 is not an integer, the only solution is y = 2. 2*4 = 0 C. 21) f(x) = 2x2 + 5x + 3, then f(x+2) = 2(x+2)2 + 5(x+2) + 3 = 2x2 + 13x + 21= 0 (2x + 7)(x +3) = 0 x = –3.5, –3. A2 + B2 = (-3.5)2 + (-3)2 = (49/4) + (36/4) = 85/4 = 21.25 C. 22) g(x) = 8719x2 + 1085x – 18. Sum of the roots = -b/a -1085/8719 B. 23) y = 4x – 5, 3x = 2y – 4 -3y = -12x +15 8y – 16 = 12x 5y – 16 = 15 5y = 31, y = 31/5. If y = 31/5, then x = 14/5 -D/C = -31/14 B. 24) Distance between (a,b) and (c,d) = [(c-a)2 + (d-b)2] Plugging in for (1,14) and (-4,2), we get a distance of 13 units C. 25) C – (a/r) = L +(o/s) L + (o/s) – C = - a/r r = -a/[L + (o/s) – C] r = -as/(Ls + o – Cs) as/(Cs – Ls – o) C. 26) By definition, the domain is {1,2,3} B 27) 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32….the pattern for the last digit repeats every fourth power. 2010/4 = 502 with remainder 2. The remainder of two indicates that the last digit will be equal to 4 B. 28) Answer is E. 8[2(62 – 52) + 11(–2)] simplifies out to zero, and multiplication by zero always implies a product of zero. 29) (a + 3)(a – 1)(a + 4)(a + 5) = (a + 5)(a + 7)(a – 1) a + 5 = a2 + 7a + 12 a2 + 6a + 5 = 0 a = -1, -5. However, if a = -5, the denominator of the original expression is equal to zero a = -1 is the only solution C. 30) The term with the highest sum of exponents determines the degree of the polynomial. In this case, 8xy5z9 has the greatest sum of 15 C.