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    S - Alg 2 Team Sol

    Uploaded by

    lusienopop
    The document contains solutions to 15 algebra 2 practice problems from a regional competition. Some key details include: 1) Problem 1 calculates the total distance a ball bounced, traveling 45 feet. 2) Problem 2 finds the vertex and equation of a parabola by solving a system of equations. 3) Problem 3 determines the interval that a fraction is within using interval notation. 4) Problem 4 calculates the distance and rate of two cousins traveling using a system of equations. 5) Problem 15 calculates the determinant and inverse of a matrix.

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd

    S - Alg 2 Team Sol

    Uploaded by

    lusienopop
    101 views4 pages
    The document contains solutions to 15 algebra 2 practice problems from a regional competition. Some key details include: 1) Problem 1 calculates the total distance a ball bounced, traveling 45 feet. 2) Problem 2 finds the vertex and equation of a parabola by solving a system of equations. 3) Problem 3 determines the interval that a fraction is within using interval notation. 4) Problem 4 calculates the distance and rate of two cousins traveling using a system of equations. 5) Problem 15 calculates the determinant and inverse of a matrix.

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    S_Alg 2 Team Sol

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    S_Alg 2 Team Sol

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    The document contains solutions to 15 algebra 2 practice problems from a regional competition. Some key details include: 1) Problem 1 calculates the total distance a ball bounced, traveling 45 feet. 2) Problem 2 finds the vertex and equation of a parabola by solving a system of equations. 3) Problem 3 determines the interval that a fraction is within using interval notation. 4) Problem 4 calculates the distance and rate of two cousins traveling using a system of equations. 5) Problem 15 calculates the determinant and inverse of a matrix.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    101 views4 pages

    S - Alg 2 Team Sol

    Uploaded by

    lusienopop
    The document contains solutions to 15 algebra 2 practice problems from a regional competition. Some key details include: 1) Problem 1 calculates the total distance a ball bounced, traveling 45 feet. 2) Problem 2 finds the vertex and equation of a parabola by solving a system of equations. 3) Problem 3 determines the interval that a fraction is within using interval notation. 4) Problem 4 calculates the distance and rate of two cousins traveling using a system of equations. 5) Problem 15 calculates the determinant and inverse of a matrix.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 4

    Capital City Regional

    3/8/2008

    Algebra 2 Bowl Solutions


    Algebra Two Team Question # 1

    Infinite geometric series:


    9
    6
    Bounces Down a =
    + Bounces Up a =
    2
    2
    1
    1
    3
    3
    Ball traveled 27 + 18 = ;45 feet.

    Algebra Two Team Question # 2


    3+7
    x- coordinate of Vertex:
    = 5.
    2
    Points: (3,0)-> 0 = a (3) 2 + b(3) + c
    (7,0)->

    0 = a (7 ) 2 + b (7 ) + c

    (5,8)->

    8 = a (5) 2 + b(5) + c

    Solve the system:


    9a + 3b + c = 0
    49a + 7b + c = 0

    f ( x) = 2 x 2 + 20 x 42

    25a + 5b + c = 8

    -2 + 20 + -42 = -24
    Algebra Two Team Question # 3
    3x 5
    3x + 5
    5 or 3
    5
    2
    2
    6 3 x 5 10 or 6 3 x + 5 10
    1 3 x 15 or 11 3 x 5
    1
    11
    5
    x 5 or
    x
    interval notation:
    3
    3
    3
    3

    Cousin 1
    Cousin 2

    Rate
    r
    r+8

    time
    2
    2

    5
    3 ,5

    2r + (2r + 16) = 224; 4r + 16 = 224; 4r = 208; r=52, r+8=60


    A= 52mph
    B= 60mph
    C= 3(52)+3(60) = 336 miles apart.
    336-2(52+60) = 112

    Algebra Two Team Question # 4


    distance
    2r
    2r + 16

    Capital City Regional

    3/8/2008

    Algebra 2 Bowl Solutions


    Algebra Two Team Question # 5

    p
    = (1,2,3,6) use synthetic division to try possible roots:
    q
    -1 | 1 -2 -5 4 6
    -1 3 2 -6
    1 -3 -2 6 | 0  x 3 3 x 2 2 x + 6 , factor x 2 ( x 3) 2( x 3) = ( x 2 2)( x 3)

    So, ( x + 1)( x +

    2 )( x 2 )( x 3).
    Algebra Two Team Question # 6

    1 2
    10 9 3
    5 3 5
    C=

    3 2 3
    B=

    3 2 3
    =
    5 3 5
    3 34 21
    =
    2 21 13
    AA 1 = I

    2 3 34 21 5 3
    B X C = BXC =

    =
    ( hint: and
    3 5 21 13 3 2
    IA = A

    5 3

    Answer:
    3 2
    Algebra Two Team Question # 7
    9

    Let A = (2i ) 13i

    61

    = (512i ) 13i = 525i

    18

    B = (1 + i ) = ((1 + i ) 2 ) 9 = (2i ) 9 = 512i


    C = (2 + 3i )(1 2i ) = 8 i
    -525i + 512i - (8 - i) =

    8 12i
    Algebra Two Team Question # 8

    4 9
    20 45
    5( = 48)

    = 240
    x y
    x
    y
    97
    97
    1
    algebraically:
    +
    
    = 388 ;
    = y; y =
    5 13
    20 52
    y
    388
    4
    4( + = 37)

    = 148
    x y
    x
    y
    4 9
    13( = 48)
    x y
    +
    5 13
    9( + = 37)
    x y

    52 117

    = 624
    x
    y
    
    45 117
    +
    = 333
    x
    y

    97
    97
    1
    = 291 ;
    = x; x =
    x
    291
    3

    1 1
    ,
    3 4
    Algebra Two Team Question # 9
    3
    1
    5 3
    1
    5
    3
    15
    1
    A= , B= , C = ;
    + 2( ) 3( ) = + 1 + 5 = or 7 .
    2
    2
    3 2
    2
    3
    2
    2
    2
    2

    Capital City Regional

    3/8/2008

    Algebra 2 Bowl Solutions

    Algebra Two Team Question # 10


    2

    y
    x

    =1
    12 4
    A  y 2 = 3(0) 2 + 12 = 12
    y 2 3 x 2 = 12 

    B  y = 3x; m 2 = 3
    C  center (0,0).x-coordinate for foci will be 0.
    D  y 2 = 3(1) 2 + 12 = 15

    (12)(3)-(0)(15) = 36
    Algebra Two Team Question # 11
    105
    Geometric progression: a 3 = 44000(1.05) 31 ; a 3 = 44000(
    )2
    100

    a3 = 4.4(11025) = $48510.00
    Algebra Two Team Question # 12
    n = 4d
    System of equations:
    0.05n + 0.10d = 45
    multiply the bottom by 100 to simplify and use substitution. 5(4d ) + 10d = 4500 
    30d = 4500  150 dimes, so 150 X 4 = 600 nickels

    (600,150)
    Algebra Two Team Question # 13
    by elimination, +

    x + y = 25
    x y =7

     2 x 2 = 32  x 2 = 16 ; x = 4

    By substitution: (4) 2 + y 2 = 25  y 2 = 9 ; y = 3
    (4,3),(-4,3),(4,-3),(-4,-3).
    A=4+4=8
    B = -3 + -3 = -6

    A-B = 8 - -6 = 14
    Algebra Two Team Question # 14
    c = a + b 12
    1
    1
    1
    b + 10 by substitution you get: ( b + 10) + b + (( b + 10) + b 12) = 188
    2
    2
    2
    a + b + c = 188
    a=

     b = 60; so a = 40 and c = 88
    40, 60, 88
    3

    Capital City Regional

    3/8/2008

    Algebra 2 Bowl Solutions

    Algebra Two Team Question # 15


    det A = 1(2 11 6 2) 5(2 1 7 2) = 1(10) 5(16) = 70

    8
    8 5
    1
    3 = 51
    B 1 =
    6
    1
    (48 3)
    5

    85
    350
    560

    51
    51
    answer :
    14
    140

    17
    17

    5
    51
    6

    51

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