WYSE Academic Challenge Sectional Mathematics 2006 Solution Set

Transcription

1 WYSE Academic Challenge Sectinal 006 Slutin Set. Cect answe: e. mph is 76 feet pe minute, and 4 mph is 35 feet pe minute. The tip up the hill takes 600/76, 3.4 minutes, and the tip dwn takes 600/35,.70 minutes. The ttal tip takes = 5. minutes, which is appximately 5 minutes, 7 secnds. Cect answe: e. Angle 4 and the angle which measues 30 ae supplementay angles. S, 80 = 30 + m 4 and m 4 = 50. Since the sum f the angles f a tiangle is 80, m 3 + m = 80. Thus, m = 80 and m 3 = Cect answe: b. The length f the ladde need t be at least 0 inches. 0, which is.03 feet, feet, sin Cect answe: c. The median is 7, and the mean is 9, s the diffeence is. 5. Cect answe: a. t / 3.8 If S is the safe level, then slving f t in the equatin S = 50S(0.5) wuld give us u t / 3.8 t / 3.8 t 3.8ln(0.0) answe. S = 50S(0.5), 0.0 = (0.5), ln(0.0) = ln(0.5), t = = 3.8 ln(0.5).44, but need t wait until nd day. 6. Cect answe: c. The vlume f the cylinde is Ah=π Ah = π h 6000 π. The vlume f the cne tp is Ah = π h π. Cmbined, the vlume is π = 9, cubic feet Divide this by 50 since gain is pued at a ate f 50 cubic feet pe minute. This gives minutes, 6 hus 38 minutes. 7. Cect answe: b *0 + 6* 6 = 50, ,50 = Cect answe: b. Dup = Ddwn uptup = dwntdwn ( 4 x)3 = 4 + x and 3x = 4 + x and x = 006 Sectinal Slutin Set

2 9. Cect answe: a. The vlume in the tank when the wate is at height h can be given by the fmula h V ( h) = 60 h = h 0 3. The deivative f this with espect t time is 3 dv dh = h 0 3 = 00. Slve f dh dh when h is. 0 3 = 00, dh 00 = = Cect answe: b. 3*4C3*48C/5C5=.05690=.3%. We culd subtact ff the fu f a kind (appx 0.04%) and full-huse cmbinatins (appx (0.4%), but u answe des nt change.. Cect answe: b. x + x y + y The midpint f a line segment with endpints ( x, y ), ( x, y ) is,. S, the midpint f A and B is, = ( 6,).. Cect answe: d. Each pint n the tie mves a ttal f 7.5*580 feet = 39,600 feet. Evey time the axle spins, the pint mves *pi*0/=6.857/=5.38 feet. Theefe 39,600/5.38 = 7,560 tatins. 3. Cect answe: c = 3(7) + 6(7) + (7) = = 9 4. Cect answe: c. 3 4 The sequence is fllwing the fm a, a, a, a, a, whee a = 64, and =.5. Althugh we culd wite ut the next five by hand and sum them, we can als d the n fllwing: The sum f the fist n numbes is S( n) = a. We essentially want S(0) S(5) = = Cect answe: b. x x+ Taking the natual lg f bth sides yields ln0 = ln 6, ln 6 x ln 0 = ( x + ) ln 6, x ln 0 = xln6 + ln 6, x = = ln0 ln 6 6. Cect answe: e. Facting x x 3 esults in (x-3)(x+). Multiplying each side f the equatin by this esults in x(x+)+(x-3)=8x, x + x + x 6 = 8x, x + 3x 6 = 8x, x 5x 6 = 0, ( x 6)( x + ) = 0. Since (x+) was in the denminat f the iginal equatin, x -. S, x = Sectinal Slutin Set

3 7. Cect answe: c. Fist, daw the aea the dg is able t cve. Yu will ntice that it is a ectangle with tw semicicles n the ends. The aea f the ectangle is 5*8=0 ft². The aea f the semicicles is 4²π=6π. Cmbined, the aea is 0+6π=70.3 ft². 8. Cect answe: d. Simila tiangles ae fmed by the figue. Theefe, x 0 = x +, which yields x + 30 = 8x, x = 6. Cmpaing the smallest tiangle with the lagest yields x x + 5 =, whee d is the diamete f the lagest ball. Substituting x = 6 gives 3 d d = Cect answe: c. If x = the time pump A wks, then we get: x x = 0 6x = x 44 6x = 44 x = 7.333hus 6 0. Cect answe: c. We fist need t detemine hw many a s and l s we have in the wd. We can have 3 a s and l s; 3 a s and diffeent lettes (ut f 6); a s, l s, and ne the lette (ut f 5); l s and 3 diffeent lettes (ut f 6), 5 diffeent lettes (ut f the 7). This gives us a ttal f 5C3 + 5C3*6P + 5C*3C*5 + 5C*6P3 + 7P5 = Cect answe: e. The ellipse has a cente at (3,0). The fcal length wuld be 3, the semi-maj axis wuld be 5 (a + c + a - c = 0), and the semi-min axis wuld be 5 3 = 4. This geneal ( x 3) ( y 0) fm f the ellipse wuld be + = = 3 3 d 006 Sectinal Slutin Set

4 . Cect answe: c. If x is five feet less than the height f the mnument, and y is the distance fm the pint n the gund belw the tp f the mnument t the gils stating pint, then we have the pictue as shwn, and we ae tying t slve f x x, then add five. We end up with tan 30 = y x x and tan 0 =. Afte we eaange y these, we get x = y tan 30 = ( y + 0 ) tan 0. y 0 ft 0 tan 0 Slving f y gives us y =, tan 30 tan s x = y tan 30 = If we add five t that, we end up with Cect answe: d. Afte pefming lng divisin we get 4x R 4. Cect answe: c. tan(4) = y/x and y = 5*tan(4 ) =.5 5. Cect answe: a. Let the adius f the lage cicle be and the smalle be. Then, by the Pythagean theem, aeas is π ²- π ²= π ²- π( =. The aea f the lage cicle is π ² and the aea f the smalle cicle is π ². The diffeence in these tw π )²=π ²- π ²= ², the aea f the shaded egin. S, the shaded egin is (π the lage cicle. 6. Cect answe: a. This functin ceates a cicle with adius 0.5, s the aea is squae units, und t. 0.5 π ²)/( π ²)=/ f squae units, which is 7. Cect answe: d. lg ( ),, 6 6 x x lg 6 x + lg 6 ( x ) = lg 6 x ( x ) = = 6, x( x ) = 6, x x 6 = 0, ( x 3)( x + ) = 0, x = 3 x = -. The value f - makes bth (x-3) and (x+) negative. The dmain f the lgaithmic functin des nt include negative numbes. S, x = Cect answe: b. Use He s fmula, A ea = s( s a)( s b)( s c) whee s = ( a + b + c) /. S = 88, s Aea = 88 (6)(34)(8) = Sectinal Slutin Set

5 9. Cect answe: d. B A is i 4 j + 3k, and the length is + ( 4) + 3 = Cect answe: d. Let x epesent the numbe f hus pe week Billy wked at the pl, and y epesent the numbe f hus pe week Billy mwed lawns. Then, x + y = and x = 3y. Substitutin esults in 3y + y =, y = 5.5. Placing that value back int the fist equatin esults in x =, x = 6.5. Thus, Billy wked 5.5 hus a week mwing lawns and 6.5 hus pe week life guading at the pl. Multiplying by thei espective wages gives 5.5($8.30) + 6.5($6.50) = $5.90, the amunt eaned pe week. 3. Cect answe: e. d 5 4 dy dy 6x ( 3x 4y = 7) 6x 0y = 0 = at (4,-) is. 4 dx dx dx 0y 3. Cect answe: d x 75 Z M = Z J = x = Cect answe: a. ± 44 4* 4*( 7) ± 56 x = =, s x = 7/ and -/ * Cect answe: e. The best way t slve this pblem is gaphically. If we gaph the functin when a = 0, we have thee eal ts, a maximum at (0.5,.75), and a minimum at (, -4). We have thee eal ts as lng as -.75 < a < 4. When a = -.75 and a = 4, we have nly tw eal ts. If a < -.75 a > 4, we have nly ne eal t, s the the tw wuld be 3 imaginay. An algebaic slutin exists, but it s nt petty. We take 4x 5x + x + a and fact it int ( bx c)( dx e) t find the values f a that fm the bundaies f the tw eal ts slutin. We then slve f b, c, d, and e. but this is a vey difficult ad t take. 35. Cect answe: a. x + x + y 4y = 4 x + x + + y 4x + 4 = x ( y ) =, s the cente is at (-,), and the pduct is -. ( ) ( ) ( ) ( ) 4 ( +) Cect answe: c. T get i t wk, the fu vets teat,,, 3, 4 animals. By ii and iii., Bill teats animal (because f ii and iii), Chalie teats (because f iv), Anne teats 4 (iv again), leaving Dave teating 3. Bill teats just the hse (v). Each f the the vets teat ne cat (iv.) plus enugh dgs t bing thei ttal up t the cect numbe. This means Chalie teats a cat and a dg, Anne teats a cat and thee dgs, and Dave teats a cat and tw dgs. 006 Sectinal Slutin Set

6 37. Cect answe: b. x 3xR9 3 x + x + x 6x x 4x 3 0x 3x 6x + 9 3x + 6x 0x + 0x + 9 Remainde f Cect answe: a. The matix that wuld slve this fllws the fm X = [X ] = A ( A B), which wuld give us 39. Cect answe: d. The distance fm the igin at time t is dwn t t. Desied distance is 0, s time = 0. ( t sin t 0) + ( t cst 0), which simplifies 40. Cect answe: e. With the dtted lines, as shwn, the length f the metal sheet must be 35 + x. The measue f x can be fund by using the Pythagean theem with 35 f the hyptenuse and x as leg lengths. x² = 35², 35 s x =. The length f the side is then 35 + x = Sectinal Slutin Set