Midterm Review Answers - Lexington Public Schools /...

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Name Honors Math 2 review answers February, 2015 page 1 4 3 2 1 l k j Honors Math 2 – Answers to Midterm Review Unit 1: Deductive Geometry The following are outlines of proofs. 1. a. ΔPAB ΔPCB by SSS, then APB = CPB by CPCTC. b. ΔPAD ΔPCD by SSS, then APD = CPD by CPCTC. c. Parts a and b show that PB and PD are angle bisectors of the same angle, but an angle has only one angle bisector, so PB and PD must be the same line, so P, B, and D are collinear. 2. See diagram. 1 = 2 by PAI using j || k. 2 = 3 because they are vertical angles. 3 = 4 by PAI using j || l. So 1 = 2 = 3 = 4. Finally j || l by AIP using 1 = 4. 3. a. This is false. Any non-rhombus kite is a counterexample. b. Suppose diagonals of rhombus ABCD intersect at E. ΔABD ΔCBD by SSS. ABD = CBD by CPCTC. ΔABE ΔCBE by SAS. AEB = CEB by CPCTC. AEB and CEB are both equal and supplementary, so they must be right angles. 4. ABC = ACB by the Isosceles Triangle Theorem. because they are both right angles. ΔBDC ΔCEB by AAS, then BE = CD by CPCTC. 5. Let ABCD be a quadrilateral with AB = DC and AB || DC. Draw diagonal AC. BAC = ACD by PAI. ΔABC ΔCDA by SAS. DAC = ACB by CPCTC. AD || BC by AIP. 6. Rewrite the given equation as mP + mS = mQ + mR. Since the four angles of the quadrilateral add to 360°, it follows that mP + mS = mQ + mR = 180. Since consecutive angles are supplementary, PQ || SR, so the PQRS must be a trapezoid or a parallelogram. Unit 2: Exponents and Radicals 7. a. 2 b. 4 c. 2 d. 2 e. 6 8. a. 2 0 by definition and 3 3 > 0 because 3 > 0 , so 2 3 3 0 . 2 3 3 ( ) 6 = 2 ( ) 6 3 3 ( ) 6 = 2 ( ) 2 # $ % & ' ( 3 3 3 ( ) 3 # $ % & ' ( 2 = 2 3 3 2 = 89 = 72 . b. 2 3 2 " # $ % & ' 3 = 2 3 ( ) 3 2 3 = 2 8 = 1 4 . 9. a. 6 b. 1 c. 4 d. 2 10. 5 times greater ( 75 = 5 3 )

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  • Name Honors Math 2 review answers February, 2015 page 1

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    Honors Math 2 Answers to Midterm Review

    Unit 1: Deductive Geometry The following are outlines of proofs.

    1. a. PAB PCB by SSS, then APB = CPB by CPCTC.

    b. PAD PCD by SSS, then APD = CPD by CPCTC. c. Parts a and b show that PB and PD are angle bisectors of the same angle, but an angle has

    only one angle bisector, so PB and PD must be the same line, so P, B, and D are collinear.

    2. See diagram. 1 = 2 by PAI using j || k. 2 = 3 because they are vertical angles. 3 = 4 by PAI using j || l. So 1 = 2 = 3 = 4. Finally j || l by AIP using 1 = 4.

    3. a. This is false. Any non-rhombus kite is a counterexample. b. Suppose diagonals of rhombus ABCD intersect at E.

    ABD CBD by SSS. ABD = CBD by CPCTC. ABE CBE by SAS. AEB = CEB by CPCTC. AEB and CEB are both equal and supplementary, so they must be right angles.

    4. ABC = ACB by the Isosceles Triangle Theorem. because they are both right angles. BDC CEB by AAS, then BE = CD by CPCTC.

    5. Let ABCD be a quadrilateral with AB = DC and AB || DC. Draw diagonal AC. BAC = ACD by PAI. ABC CDA by SAS. DAC = ACB by CPCTC. AD || BC by AIP.

    6. Rewrite the given equation as mP + mS = mQ + mR. Since the four angles of the quadrilateral add to 360, it follows that mP + mS = mQ + mR = 180. Since consecutive angles are supplementary, PQ || SR, so the PQRS must be a trapezoid or a parallelogram.

    Unit 2: Exponents and Radicals 7. a. 2 b. 4 c. 2 d. 2 e. 6

    8. a.

    2 0 by definition and

    33 > 0 because

    3 > 0, so

    2 33 0 .

    2 33( )6

    = 2( )6 33( )

    6= 2( )

    2# $ % &

    ' ( 3

    33( )3#

    $ % &

    ' ( 2

    = 23 32 = 8 9 = 72 .

    b.

    23

    2

    "

    # $

    %

    & '

    3

    =23( )

    3

    23=28

    =14

    .

    9. a. 6 b. 1 c. 4 d. 2

    10. 5 times greater ( 75 = 5 3)

  • Name Honors Math 2 review answers February, 2015 page 2 11. a.

    x (24 )12 = (26)

    43 82 22 =x 64=x

    b.

    32 = (33)x

    32 = 33x

    x =23

    12. . !!"

    ! = !!

    b. !!"#

    !!! = !"#!

    !! = !

    !

    != !"

    !

    c. 9!!! = !

    !

    !! = !

    !

    != !

    ! d. No real solution

    12A. No solution ( !! is extraneous)

    Unit 3 and 4: Quadratics and Polynomials 13.

    x = 2, 3 5 14. Put q(x) in normal (standard) form 15. a.

    x 2 + 3x 28 = 0 b.

    x 2 10x + 23 = 0 16. a.

    2x 2(5x 2 9) b.

    10x(x 3 + 4) c.

    mn(m n) d.

    a2(ab3 b2 c 2) 17. a.

    k(x) = 2x 2 + 8x + 2 b.

    m(x) = x 2 + 3x 11 c.

    n(x) = x or any linear function 18. b = 5 19. 16 20. a.

    p = 0 since

    (15 + p)(15 p) =152 p2 b.

    p = q =15 21. a.

    k = 3 b.

    k = 112 22. a.

    m = 48 b.

    n = 25 23. a.

    p < 8 b.

    p = 8 c.

    p > 8 24. a.

    2 3 b.

    8 3 c.

    13 25. a. -7 b. -9 c. -56.25 26. a. 52 ft by 52 ft b. 52 ft by 104 ft 27. a. (8, 0) b.

    y = (x 4)2 +16 28. a. (5, 10) b.

    y = 3(x 5)2 +10 c. 85 29. (3,

    443)

    30. a.

    x = 0, 4 5 b.

    x = 0, 2, 2 c.

    x = 6 5, 14 d.

    x = 3, 8 e.

    x = 1, 6 f.

    x = 1, 2 3 g.

    x = 2 11 h.

    x = 6, 6 i.

    x = 3, 12 j. no solution k.

    x = 14 , 9 4

    Unit 5: Functions 31. a. c.

  • Name Honors Math 2 review answers February, 2015 page 3

    b. D: 1 ,0 > xx R: 2y

    d. f(x) is not a one-to-one function because: (any of these reasons is sufficient)

    Does not pass the horizontal line test

    More than one input for each output

    Inverse is not a function (fails vertical line test)

    32.

    g(x) = f 1(x) = 5x 31 x

    = !!!!!!!

    To confirm, show

    f (g(x)) = x or

    g( f (x)) = x .

    33. a.

    f ! g (x) = 3x 2 6x + 4

    b.

    g ! f (x) = 9x 2 36x + 38

    c. g(a + 1) g(a) = 2a 1

    d. f 1(25) = 10 34. a. See table on right

    b.

    f (x) =5 x = 0

    f (x 1) + 6x +1 x > 0# $ %

    c.

    f (x) = 3x 2 + 4x 5

    35. a. f(4) = 32 c. f 1(32) = 4 b.

    x f(x)

    2 0 5 7 6 1 2 13 6 2 15 19 6 3 34 25 6 4 59 31 5 90