2005 Palm Harbor February Invitational Algebra II Answer Key

Transcription

1 005 Palm Harbor February Invitational Algebra II Answer Key Individual. D. C. B 4. A 5. E 6. B 7. E 8. B 9. D 0. A. B. D. C 4. C 5. A 6. C 7. B 8. C 9. E 0. D. A. B. B 4. C 5. E 6. B 7. A 8. D 9. A 0. C Team (seconds) e 4. 4x x y = or x y 8 0

2 February Palm Harbor Invitational. D log w+ log x log( wx) = = log yz ( wx) log y+ log z log( yz) Combine logarithms and apply change of base rule.. C Using the given formula, (0)(0) 6(7)() x = x x = = 48, x= 7 x=. B ( ) = = = ( ) 4. A Use the polynomial remainder theorem, with 4 x = = E Use the Binomial Theorem. The fourth term corresponds to: 7 ( ) 4 ( 5 ) x y = x y B i to any exponent which is a multiple of 4 is equal to. The rule for divisibility by 4 is if the number formed with last two digits of some number is divisible by 4, then the original number is as well. Thus,,45 i,45 (,45) 4,690 4,688 = i = i = ( i )( i ) = ()( ) =.,45 i 7. E Even: f ( x) = f( x) Odd: f ( x) = f( x) Choices A, B, and D are all even (substituting x = x has no effect). Choice C is odd. Therefore, none of the functions are neither even nor odd.

3 February Palm Harbor Invitational 8. B Let u equal the strength of uber-fire, and s equal the strength of super-duper fire. k Then, u = kv and s =. Using the given data, we find k = and k = 96. v 96 5 Setting the two equations for strength equal to each other, v =, v =, and v v =. 9. D Only matrices of the form a b and b c can be multiplied. The order here matters. Thus, no answer matrix exists. 0. A Adding the two matrices yields matrix A, k+ 5 8 k k 6 k+ 5 A k + 4 k = = k+ 4 k+. Thus, the determinant of A equals, k 6 k+ 5 det ( k 6)( k ) ( k 5)( k 4) k 4 k = = ( k k 8) ( k + 9k+ 0) = k 8 = 0 Finally, k = 48 and k = 4.. B Consider the U and C as just one letter then. Since there are then 9 letters in the word, and there are two T s, so the total number of permutations is = 8, 440. However, we can have either UC or CU, so the final answer is (8,440) = 6,880.. D This means that exactly one of the remaining students got an A. This will 7 7 happen with probability, since one must an A, the other two must not get an A, and there are three different students, each of whom could get the A.. C a b i 5 i 8 i 40 i 9 i = = + = + = c i+ 8 i

4 February Palm Harbor Invitational 4. C Applying the Distance Formula twice, equating the two sides, and squaring, we have, ( x 7) + ( y 6) = ( x+ 4) + ( y ). Expanding both sides and combining like terms, 4x y = = 65, and simplifying, 4x 4y = A Putting this function into vertex form, we have y + = ( x + x+ ). 9 4 y+ = ( x+ ). Thus the minimum value of the function is. 6. C 9x Using substitution would be most appropriate. y = 8x 9, z =. 9x Substituting into the first equation, x+ 4(8x 9) 9 = 6. Solving, we get x =, and plugging back into the other equations, y = 7 and z =. 7. B Factoring the top and bottom, we have ( x 5)(x+ 4). Thus, there is a ( x 5)(9x+ 4) removable discontinuity at x = 5, and the only vertical asymptote is at 8. C Remembering to use change of base, and bringing the exponents out to the log 6 log8 x log 6 coefficient, we have = 0. Canceling and xlog 5 log 6 xlog8 log6 rearranging, x = log5 9. E f (9) can be expressed as 9 = 9x 6. Solving for x, x = 9. g (9) = 9 4 = 5. Thus the answer is 9+ 5= 4, or E. 4 x = D The units digit of any number ending in a six raised to a positive integral power is always 6. The units digit of number ending in 7 cycles every 4, so 7 4, 7 8, 7, etc. all end in. Thus, 7 04 ends in a, and so we know 7 06 ends in a 9. The units digit of a number ending in 9 cycles every, so all odd exponents of 9 end in 9, and all even exponents end in. Thus, the units digit of this number seems to be

5 February Palm Harbor Invitational 6 9+ =. However, we merely need to carry a ten over from the tens digit, and 8(mod0), so the final answer is 8.. A ( i) = i and 4 ( i) 4 =. Thus, ( ) ( i) ( i) ( 4) 4096 = = =.. B The arithmetic mean is equal to + 68,99 = 4,507. The geometric mean is equal to ()(68,99) =,. Thus the answer is 4,507 +, = 5, 79.. B The Quadratic Formula gives the two roots as 6 ± ( 6) 4()(8) 6 ± 08 6 ± 6 i ± i = = =. The product of these () 4 4 two roots is 9, the sum of the squares of the roots is 9. Thus, the final answer is 9+ 8= C The fines each time make up a geometric progression. The sum of the first n terms of this geometric progression with a common ratio of is n+ n+ r 5 n+ Sn = 47,60 = t0 = 5 = ( ). r Solving for n, we have 59,048 + = n+, and log59,049 n = = 0. log 5. E We have two equations in two variables, one for number of flowers and one for cost. t is for tulips, l is for lotuses. t+ l = 50 0t+ 5l = 000 Solving, we get l = 00 and t = B Since we are only concerned with the center, we shall ignore the right side of the equation. Distributing and completing the square on the left side, we have 6( x 0x+ 5) + 9( y + 4y+ 4). Thus, the center is (5, -). 7. A Again applying the polynomial remainder theorem, plug in x = 5. We have 4(5) + (5) + = 5.

6 8. D February Palm Harbor Invitational The length of each jump forms a geometric progression, with common ration 5. Thus the limit of the sum of the distances he jumps as the number of jumps t0 97 approaches infinity is given by the formula.5 r = 4 = A This conic section is an ellipse. Rearranging the equation and completing the squares, we have 6( x 4x+ 4) + 4( y + 6y+ 9) = 44, and the more useful form 0. C x y+ + =. In this case, the focal distance is given by 6 c b a = = 6 4 = (log x ) x x + log 40 = + xlog 40 = x + x= log Solving for x, x + x+ = 0 = ( x+ ), and x =.