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Theta Equations and Inequalities Solutions FAMAT State Convention 2015 π 1. B- (8 + 2π)(4 β 3π) = 38 β 16π β π = 38, π΅ = β16 β (2 β ββπ΅) = (19 β β16) = (19 β 4) = 15 (Square root denotes a positive number only, in this case, 4.) 2. B- To find the average speed, take the harmonic mean of the speeds each way to and from 2ππ 2(60)(80) 9600 480 the competition. π+π β 60+80 = 140 = 7 3. A- (π β π)(π₯) = π(π(π₯)) Therefore β π (π((π β π)(π₯))) = π (π (π(π(π₯)))) . π(β1) = β1 β π(β1) = 2 β π(2) = 2 β π(2) = 5 4. D- π₯ + 2βπ₯ β 7 = 10 β 2βπ₯ β 7 = 10 β π₯ β 4(π₯ β 7) = 100 β 20π₯ + π₯ 2 β π₯ 2 β 24π₯ + 128 = 0 β (π₯ β 8)(π₯ β 16) β π₯ = 8, 16. Substituting these values into the original equation gives "8 + 2β8 β 7 = 10 β― 8 + 2 = 10" and "16 + 2β16 β 7 = 10 β― 16 + 6 β 10". Therefore, π₯ = 8 is the only solution. 5. C- π = π0 β π‘ 9.6 1 β (2) 1 1.2 (2) β π = 320 β 1 8 320 5 β π = 320 β (2) β π = 256 = 4 = 1.25 6. B- We are given the following important equations, with Q, D, and N representing the number of quarters, dimes, and nickels, respectively: π + π· + π=105, 5Q=N, and D+5=N. Substituting the last π 11π two equations into the first in terms of N gives: 5 β 5 + π + π = 105 β 5 = 110 β π = 50. If there are 5 fewer dimes than nickels, there are 45 dimes. π₯β7 7. D- To find π β1 (π₯), solve for π₯ in π(π₯) and then flip the variables. π(π₯) = π¦ = π₯+15 β π₯π¦ + 15π¦ = π₯ β 7 β π₯π¦ β π₯ = β15π¦ β 7 β π₯(π¦ β 1) = β15π¦ β 7 β π₯ = β15π¦β7 π¦β1 . Then, exchange, giving π β1 (π₯) = β15π₯β7 π₯β1 8. E- π₯ = ππ¦ π§ β 12 = 3π 1 8 1 β π = 2. Using the new values for π¦ and π§, π₯ = 1 (9) 2 1 4 = 18 9. A- First, we can multiply the entire equation by the quantity 6(π₯ β 4)2 in order to clear all denominators. From this, we are given 3π₯(π₯ β 4) β 48 = 2(π₯ β 2)(π₯ β 4). Simplifying gives π₯ 2 = 64, so π₯ = ±8. Thus, the largest solution is 8. 10. B- If the second term is 9 and the eighth term is 33, the common difference is 4, and the first term is π 5. The sum of an arithmetic sequence can be given by 2 (π1 + ππ ), where N is the number of terms and ππ is the n-th term. Thus, it is 101 2 (5 + (5 + 4 β 100)) = 101 2 (410) = 20705 11. A- To find the quadratic equation, you must multiply the roots. The proper form is: (π₯ β 6 β πβ3)(π₯ β 6 + πβ3) β π₯ 2 β 6π₯ + π₯πβ3 β 6π₯ + 36 β 6πβ3 β π₯πβ3 + 6πβ3 β 3π 2 β π₯ 2 β 12π₯ = β39 Theta Equations and Inequalities Solutions FAMAT State Convention 2015 π ππ‘ 12. B- The formula for compound interest is π = π0 (1 + ) . Thus, from Wayne Enterprises, Matt makes π = 1000 (1 + π .04 2 .03 2 2 ) = 1000(1.02)2 = 1040.40. Dave, with Stark Industries, makes π = 1000 (1 + 1 ) = 1000(1.03)2 = 1060.90. Therefore, Stark Industries was the better investment with 20.50 more profit. 13. C- This could occur when (π₯ 2 β 9π₯ + 19) = 1 β π₯ 2 β 9π₯ + 18 = 0 β (π₯ β 3)(π₯ β 6) = 0 β π₯ = 3, 6. β3 This could also occur when (2π₯ 2 β π₯ β 6) = 0 β (2π₯ + 3)(π₯ β 2) = 0 β π₯ = 2 , 2. Finally, it could occur when (π₯ 2 β 9π₯ + 19) = β1 and the exponent is even. So, (π₯ 2 β 9π₯ + 19) = β1 β π₯ 2 β 9π₯ + 20 = 0 β (π₯ β 4)(π₯ β 5) = 0 β π₯ = 4, 5. However, the exponent is only even when π₯ = 4. Therefore, β3 27 you must take the arithmetic mean of 2 , 2, 3, 4, and 6, which is 10. β9 14. A- Both lines have a slope of and a y-intercept of β2, so the system is consistent and dependent, 2 meaning they are the same line. 15. A- When you graph this system of inequalities, you are given an irregular pentagon. By breaking the 31 pentagon into smaller rectangles and triangles, you can find that the area is equal to 2 . 16. D- First, you must put the ellipse into the form (π₯ββ)2 π2 + (π¦βπ)2 π2 = 1. 9π₯ 2 + 25π¦ 2 β 36π₯ = 189 β 9(π₯ 2 β 4π₯) + 25π¦ 2 = 189. Complete the square, giving 9(π₯ β 2)2 + 25π¦ 2 = 225 β 1. The latus rectum is given by 2π 2 π β 2(32 ) 5 = (π₯β2)2 25 + (π¦βπ)2 9 = 18 5 17. E- A conic can be classified by the product of its C and A values in the general equation π΄π₯ 2 + πΆπ¦ 2 + π·π₯ + πΈπ¦ + πΉ = 0. In this case, the product of C and A is a negative value (2 β β5) = β10. When AC< 0, the conic is a hyperbola. 18. E- Using synthetic division, you can determine that the roots to this equation are β2, 1, and 3. Using a number line, you can determine that the only solutions to this inequality occur between β2 and 1, as well as 3 to infinity β (β2, 1) βͺ (3, β). We cannot include β 2, 1, and 3 in the solution set because the inequality is not greater than or equal to -6. π¦ π 19. C- The fourth term of the expanded form of π(π₯) = (3π₯ β ) 6 can be found by (πβ1 ) β (π)πβπ+1 β 2 (π)πβ1 , following basic binomial expansion. n is the exponent, and p is the p-th term you are βπ¦ 3 attempting to find, while a is the first term in the binomial and b is the second. (63) β (3π₯)3 β ( 2 ) = βπ¦ 3 20(27π₯ 3 ) ( 8 )=β 135 2 π₯3π¦3 20. B- The solution we are looking for, π½2 πΈπ + π½πΈ 2 π + π½πΈπ 2 , can be factored to (π½ + πΈ + π)(π½πΈπ). This βπ΅ is the sum of the roots multiplied by the product of the roots. The sum of roots is π΄ and the product of roots is βπΎ π΄ for any odd function in the form π΄π₯ π + π΅π₯ πβ1 + πΆπ₯ πβ2 + β― + π½π₯ + πΎ = 0. Thus, the sum of roots for this polynomial is βπ΅ π΄ and the product of roots is βπ· π΄ . So, π½2 πΈπ + π½πΈ 2 π + π½πΈπ 2 = π΅π· π΄2 Theta Equations and Inequalities Solutions 21. D- Factoring the expression π₯β4 π₯ 2 β4 β€ 0 gives FAMAT State Convention 2015 (π₯β4) (π₯β2)(π₯+2) β€ 0. β2, 2, and 4 are the critical numbers that we must test on a number line. It can be found that the inequality works for numbers from negative infinity to -2, and from 2 to 4. However, only 4 is included in the solution set, because the others (-2, 2, and negative infinity) are undefined. Therefore, the solution is (ββ,β2)βͺ(2,4]. 22. A- 6π₯ β |4π₯ + 2| = 24 β |4π₯ + 2| = 6π₯ β 24. First, we will solve 4π₯ + 2 = 6π₯ β 24. With this, π₯ = 13. Plugging this value back into the original equation, we find that it is a solution. Next, we must 11 solve 4π₯ + 2 = β6π₯ + 24. With this, π₯ = 5 . Plugging this value back into the original equation, we find that it is not a solution. Therefore, the sum of solutions is 13, as it is the only solution. 23. C- π₯ = β10 + 3β10 + 3β10 β¦ β π₯ = β10 + 3π₯ β π₯ 2 β 3π₯ β 10 = 0 β (π₯ β 5)(π₯ + 2) = 0 β π₯ = 5 24. C- When computing the determinant, we must set it equal to 36. This gives us: 2π₯ β 48 = 0 after combining like terms and simplifying. Therefore, π₯ = 24. 25. B- Using the remainder theorem, the remainder is given by π(β2) = 2π₯ 5 β 3π₯ 4 + 4π₯ 3 β 5π₯ 2 + 6π₯ β 7 β β64 β 48 β 32 β 20 β 19 = β183. 26. C- Using Descartesβ Rule of Signs, first compute the equation π(βπ₯). π(π₯) = 4π₯ 5 β π₯ 4 β 10π₯ 3 + 4π₯ 2 β ππ₯ β 24 β π(βπ₯) = β4π₯ 5 β π₯ 4 + 10π₯ 3 + 4π₯ 2 + ππ₯ β 24. Because there are two sign changes in π(βπ₯), there are a maximum of 2 possible negative zeroes. 27. D- The sum of the reciprocals of the roots is given by βπ½ πΎ for any function in the form π΄π₯ π + π΅π₯ πβ1 + β0 πΆπ₯ πβ2 + β― + π½π₯ + πΎ = 0. Thus, for this equation, it is equal to 36 , or 0. 28. D- Because the volume and surface area are numerically equal given π(π) = π(π), we must start by π 4 π π setting the volume and surface area formulas equal to each other. 4π( 2 )2 = 3 π( 2 )3 β 1 = 6 β π = 6 29. B- 42x β 4x+1 = β3 β 42π₯ β 4π₯+1 + 3 = 0 β (4π₯ β 3)(4π₯ β 1) = 0 β Because we know that π₯ β 0, 1 we can say that 4π₯ β 3 = 0. β 4π₯ = 3 β 4βπ₯ = 3 β 2βπ₯ = 1 β3 = β3 3 30. A- Each week, Cody loses a net amount of $20. We must set up the following inequality: 500 β 20π₯ < 150 β 350 < 20π₯ β 17.5 < π₯. Rounding to the fewest integer number of weeks, Cody will have less than $150 after 18 weeks.