202 High School Math Contest 6 Algebra II Eam 2 Lenoir-Rhyne University Donald and Helen Schort School of Mathematics and Computing Sciences Algebra II Solutions This eam has been prepared by the folloing faculty from Western Carolina University: Mark Budden Andre Chockla Aelle Faughn Risto Atanasov Nathan Borchelt Geooff Goehle John Wagaman
ALGEBRA II - Solutions, March, 202. Solution Consider the folloing: g(2) = 5 and f(5) = 0 Thus f(g(2)) = 0 2. Solution (A) The dimension-equation could be ritten as ( 2)*(a b) = ( 2). To be able to multiply at all, a must be 2 and to obtain the matri ith to columns b must be 2 thus (2,2).. Solution The equivalence equations for taking eponential form to logarithmic form are as follos: Thus, y = 2 / can be reritten as log 2 y =. b n = c log b c = n. Solution Using the properties of natural logarithms, observe that e 2+ = = 2+ = ln() = 0 = = 0. Note that solutions (a), (b), (c) all yield = 0, so the anser is. 5. Solution Observe that = = = ( ) 2, so the anser is. 6. Solution If varies jointly as y and z then = kyz, so e have that = k(2 y )(z 2 ) and using the given information = k(2 )(2 2 ) = 2k = k = 2, so = 2 (2 )( 2 )2 =, so the anser is. 7. Solution -f() - 2 consists of an inversion of the original equation over the -ais and a translation of 2 don the y-ais. Since all are graphically represented, there is little question as to hat the anser is. 8. Solution (C) g(2) = 2 (2) 2 + 2 = 0; f(0) = 2 0+ = 2 = 7 9. Solution (A)
log ( 9 ) = 2 ( 2/) = 9 = 2/ 9 = / = 27 = 27 0. Solution ( ( + i))( ( i)) = ( i)( + i) = 2 + i + 9 i i + i i 2 = 2 6 + 9 + = 2 6 + 0. Solution By multiplying the denominator by the comple conjugate of ( 0i), ( + 0i), e have +5i (+5i)(+0i) ( 0i)(+0i) = +5i+50i2 00i 2 2. Solution = 5i 9 0, so the anser is. We obtain the equation for a circle as follos: 0i =. Solution ( + translation) 2 + (y + ytranslation) 2 = radius 2 2 + y 2 = 25 2 + () 2 = 25 = ± Solving the equations for y, the first three, and, the final, e obtain the folloing: y 2 + 2 y 2 2
y 2 2 7 Which produces:(mathematica). Solution 6 + 8 y = 6y + 8 = = 6y + 8 6 2 = 6y + 8 6 2 8 = 6y 6 2 8 = y 6 8 2 = y 5. Solution Using the properties ln(a b ) = b ln(a) and ln(ab) = ln(a) + ln(b), observe that 6 = 0 = 6 ln() = ln(0) = = ln(0) 6 ln() = ln()+ln(0) ( ) 6 ln() = 6 + ln 0 ln, so the anser is.
6. Solution Observe that 2 2 + + 2 ) 2 7 2 2 5 2 + 8 2 2 2 2 + 2 5 2 + 8 7 so the anser is 2 2 + + 2 7, hich corresponds to. 7. Solution (A) First observe that g() = + = g () = = g ( ) = = then use in place of in f() and evaluate, here f(g ( )) = f( ) = ( ) + 2 +2 2 = +52 2 = 2 + 5 2, so the anser is (A).. We ( )2 2 = 8. Solution Since f() is undefined at = there is a vertical asymptote at =. There is an -intercept at = since f( ) = 0 but not at = due to the vertical asymptote. The y intercept is since f(0) =. The anser is. 9. Solution Both f() and g() have a horizontal asymptote at y = 0. The graph of f() is undefined at = 2 so a one-unit shift of f() to the right yields the graph of + g(). Hoever, a one-unit shift of the graph of g(), hich is undefined at = ill transform it to +2 = f(). The anser is. 20. Solution To more easily solve this problem, one should first translate any hours into minutes. It, then, takes Amanda 2 minutes and both still minutes. One may notice that 2 =. Thus, in minutes Amanda can only paint /rd of the room. This means that her roommate must complete 2/rds of the job in minutes and so / = 6.5 minutes = hour and &/2 minutes. 2. Solution The volume of the cylinder is V = πr 2 h and r = d/2 = h 2 2 = h so r2 = h 6. Since V = π = πr2 h = π h2 6 = h2 = 6 = h = 8. The anser is.
22. Solution 2 + + 7 2 28 = 2( + 7) ( 2 )( + 7) 2 = ( + 2)( 2) Vertical asymptotes at = 2 and = 2 2. Solution log ( 2 + 7 + 6 2 2 ) log ( 2 + 9 + 78 2 + 7 + 2 ) ( + 6)( + ) log ( ( 6)( + ) ) log ( + 6)( + ) ( ( + )( + ) ) ( + 6)( + ) log ( ( 6)( + ) ( + )( + ) ( + 6)( + ) ) ( + ) ( + ) log ( ( 6) ( + ) ) ( + )( + ) log ( ( 6)( + ) ) log ( 2 + + 2 + 7 78 ) 2. Solution (A) Note: = 5 and = cannot be part of the solution as the cause the denominator of the fraction to equal zero! =, = and = 0 all cause the numerator to equal zero hich causes the entire fraction to equal zero and are thus part of the solution. 25. Solution (C) (, 5) [, ] 0 (, ) () 8H + 6F + 6C = 26.0 (2) 0H + 6F + 8C =.60 () H + 2F + C = 0.95 Combine equations -() and (2) to get equation (): 2H + 2C = 5.5 Combine equations () and -() to get equation (5): H 6C = 6.75 Combine Equations () and 2(5) to get equation (6): 0C = 8 5
Solve for C: C =.80 Plug.80 in for C in equation () and solve for H, hich leads to H =.95 2 delue hamburgers plus large cola is $.70 26. Solution 2 8 + 20 has a verte at (,-7) and 2 2 + 2 + 77 has a verte at (-6,5): d = ( 6 ) 2 + (5 ( 7)) 2 = (9) 2 + (2) 2 = 8 + = 225 = 5 27. Solution Solution: Here is a picture. I don t kno ho big the cardboard ill be yet, so I ll label the sides as having length. Since I kno I ll be cutting out three-by-three squares to get sides that are three inches high, I can mark that on my draing. The dashed lines sho here I ll be scoring the cardboard and folding up the sides. Since I ll be losing three inches on either end of the cardboard hen I fold up the sides, the final idth of the bottom ill be the original inches, less three on the one side and another three on the other 6
side. That is, the idth of the bottom ill be = 6. Then the volume of the bo, from the draing, is: ( 6)( 6)() = 8 ( 6)( 6) = 6 ( 6) 2 = 6 6 6 This is the quadratic I need to solve. I can take the square root of either side, and then add the to the right-hand side: = 6 ± 6 = 2 or 0 in Ho do I kno hich solution value for the idth is right? By checking each value in the original ord problem. If the cardboard is only 2 inches ide, then ho on earth ould I be able to fold up three-inch-deep sides? But if the cardboard is 0 inches, then I can fold up three inches of cardboard on either side, and still be left ith inches in the middle. Checking: ()()() = 8 28. Solution We first should determine the equation for the parabola. Setting the minimum of the parabola at the origin, e find that the horizontal line y = ill represent the average height of the river graphically. The normal parabola y = 2 is only 0 meters ide (20 + 20) at y = so e need to reduce its y-values by a tenth, creating y = (/0 ) 2 = /00 2. We ant to find at hat distance from the bank to indicate a depth of three meters. Therefore = /00 2 00 = 2 0 () /2 = The distance form the bank to the point here the depth of the river is m is 20 0 m. 7
(Generated in Mathematica) 29. Solution 0. Solution r = ( soldprice ) n 0.065 = 9 2000 = ( 0.065)9 2000 = 2000 ( 0.065) 9 i 00 Let d denote the rate of depreciation here d =, thus the house depreciates in value by a factor of d during the first year and by a factor of 2d during the second year. No observe that 00, 000( d)( 2d) = 7, 500 = ( d)( 2d) =.75 = d + 2d 2 =.75 = 2d 2 d +.625 = 0 = d = ± ( ) 2 (2)(.625) 2(2) = ±2 =.25 so i = (.25)(00) = 25, so the anser is. 8