WYSE Academic Challenge 2004 State Finals Mathematics Solution Set
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1 WYSE Academic Callenge 00 State Finals Matematics Solution Set. Answer: c. We ave a sstem of tree equations and tree unknowns. We ave te equations: x + + z 0, x z 9600, and 7x + + z 90. Wen we solve, we end up wit x 560, 0, and z 60. x 56 6 x. Answer: x x 5 x x + x 5 ( f ( x)) ( g( )). Answer:. We ll put te equation into te form: +. a 6x 60x ( x 0x + 5) 6 * 5 + 5( 6( x 5) + 5( 5) 0 + 5) 5* ( x 5) ( 5) + 5 Tis tells us tat te semi-maor axis is along te x-direction, and te center of te ellipse is at te point (5, 5). Te focal lengt is equals 5, so te two foci are at (5, 5) and (5 +, 5), or (, 5) and (, 5).. Answer: * Answer: e. We ave een given two vectors in polar form. We will need to find te sum of te vectors also in polar form. If we treat due east as 0 degrees, ten te plane s vector in Cartesian coordinates would e (0cos75)i + (0sin75), and te wind s vector would e (60cos5)i + (60sin5). Wen we add tese two togeter and simplif, te resultant vector is 0.07i We find te speed evaluating te expression wic equals 75, and we find te eading evaluating 7.5 tan wic equals 6 degrees nort of east, wic we will ten switc to degrees east of nort. 6. Answer: c. A s( s a)( s )( s c) A.5(.5 95)(.5 0)(.5 0),707.97, and, * were s ( ). 5, giving us 00 State Finals Matematics Solution Set
2 7. Answer:. Te sum of te previous ten is 0*, or 0. Te new mean would e (0+5)/. We use ϑ ( x) E( x ) ( E( x)) to find standard deviation. Our old standard deviation was 6 and our old mean was, so E ( x ) was Our new ( x ) E will e ( 00*0 5 )/ E( x ) ( E( x)), we get 9.7 (), or , or 00. +, or 9.7. If we ten evaluate our new. Answer: e. 5 + x +x + x + 0, x, x 6, A degrees 9. Answer: d. Te multiplier is, and if we consider te numer 0 as te start of our gives us te sum to e 5,65,000. numers, ten evaluating ( ) 0. Answer:. 7 x ( ) x(7 ) 0 x x 6. Answer: d. Te cross product A x B can e found evaluating te determinant: i k i k wic equals -7i + 5k. If C ai + + ck, ten would give us te value of A x C. Tis simplifies to (c )i + (a c) + ( a)k. Tis must also equal 7i + - 5k if it is to e equal lengt ut opposite direction of A x B. We terefore ave te equations c 7, a c, and a -5. Wen we tr to solve for a,, and c, we find tat tere are infinitel man solutions.. Answer: a. Te two data points are (, 500) and (7, 9000) giving a slope of -75. Tis gives a linear equation of (x - ) and -75x + 5,5. Finall, - 75(.5) + 5,5,.5. Answer: c. x researc udget, advertising udget, z qualit control udget, and A equals total udget for te tree departments. We ave te equations x + + z A, and 0.5x z (left over amounts are equal). We find tat.5x, and z.75x. Tis means tat x +.5x +.75x A, or x A. Tis means x 0.5A, 0.5A, and z 0.75A. Te spent portion is tus 0.65A for researc,.5a for advertising, and 0.5A for qualit control. Totaling tis up, we get.775a, or rougl 7% of te original udget. a c. Answer: c. d r d.5 r t r t (.5)(.5) r (.5) 5. Answer: c. If te radius of A is x, ten te area of A is π x, te area of B is π x, and te area of C is 9π x. Ring I as an area of 9π x πx 5πx, and Ring II as an area of π x πx πx. Te ratio of Ring I to Ring II would tus e 5 to. 00 State Finals Matematics Solution Set
3 cos q sin q sin q 6. Answer: c. tan q sin q cos q sin q cos q sin q cos q sin q cos q cos q 7. Answer: a. Te tird root is at x - i, so f ( x) ( x )( x + i)( x + + i), wic simplifies down to f ( x) x + x + x 6.. Answer: c. k x + k +, k 9 + k, 7, Answer: e. ( x) sin as a period of π cos x as a period of π. Te period of te comined function is te LCM of te periods of te parts, wic is π., ( ) 0. Answer: Answer: d. We can reak te triangle ABC up into six congruent triangles, all toucing te center of te center triangle, one edge along alf an edge of ABC, and a point on one of te tree points A, B, or C. Te potenuse of eac triangle is 0 inces, so te sort ase is 0 inces and te long ase is 7. inces. Eac triangle as an area of 0.5*0*7.6.6, so te area of ABC is 6* Answer:. Graping te aove information ields a trapezoid wit eigt of, a ottom ase of.5 and an upper ase of.5 giving us (/)()(.5+.5) square units.. Answer: d. Te oter transformations are a) clockwise rotation 90 degrees, ) clockwise rotation of 5 degrees wit expansion of, c) counterclockwise rotation of 5 degrees wit expansion of, and e) counterclockwise rotation of 5 degrees.. Answer: e. x so x x + x + 5. Answer: d. A Bae s teorem example: 0.0* * * * Answer: d. Rater eas, can simpl use a graping calc to find te intersection. 7. Answer: e. Let te measure of angle ADE equal x. Because te lines are parallel, ever oter angle will end up eing eiter x or 0 x. We ten find tat W 0sinx, W + Z X 0sinx, Y sinx, and Z 6sinx. If we evaluate te, we end up wit X + Y... Answer: a. 50(00 + R ) 00R 50(00 + R) 5, R 00R 95, R, giving us 5, R 00R 5, R R 5., or 00R 95, R 50R R State Finals Matematics Solution Set
4 9. Answer: c. We can eiter consider te possiilities for te first word and ten te possiilities for te second word, or we could consider were te space goes, ten were te ten letters would fit. Eiter wa, we get 9*P(0,0),659,00 0. Answer:. F < 500 cos 0, 500sin 0 >< 65.06, 50 >, F < 00,0 > F < 65.06, 50 F Answer:. Because x ( + z) and z, we can find tat /x and z /6x. In x ears, we ave one alf life of A, tree alf lives of B, and 6 alf lives of C. Tis means we ave /A left, /B left, and /6C left. Since eac sustance was a tird of our total amount, we are left wit , or %. 6. Answer: a. x + 60,.5x (60 ) x 90. Answer: c. Tis equation can e solved grapicall, or: ( x + x ) ( x + x ) x + x x + x x x x + 0 x ( x + )(x )( x ) 0, so x -0.5, 0.5,, ut onl 0.5 and are allowed as solutions for rational exponents. ( x + ) ( x + ) + 00 [x x + 00]. Answer:. ( x + x + ) x + 00 x + x 00 x + and 0,x x + 5. Answer: d. Since: X AX + B X AX B X ( I A) B X ( I A) B, we ave ( I A), ( I A) 0. 0., and ( I A) B Answer: c. P(get in) P(unlocked) or P(locked) and P(ave correct ke).75 So,,, * p and p.5 so n 7. Answer: d. If we consider te distance from te position of te particle at t0 to te position of te particle at time t, we end up wit a ver complicated expression to maximize. It s muc easier to realize tat we ave a parametric expression for an ellipse of te general form x +. Wen t 0, we are at te point (0, ). Since tis is at one end of te maor axis, te fartest awa te particle will e is wen it is at te oter end of te maor axis. Tis would e units awa. x x / x 5/ ( x) + 5( x / ) + 0( x 5/ ). Answer: a x + 5x 5 / + 0x 00 / 60 7x 5/ 0 / 7 x 5/. 0 x, and State Finals Matematics Solution Set
5 9. Answer: a. Te current speed of starsip A is k 0, and te speed of starsip B 6 is k 0. In te seven das of catcing up, if we let te new speed of starsip B 6 equal k 0, ten we get 7 k 0 ( k 0 k 0 ) + 7 k 0. After simplifing, we 799 end up wit 6 + log Answer: d. p at (. +.)/.75 so. 005 n and n 9, 00 State Finals Matematics Solution Set
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