Algebra & Geometry Semester 1 & 2 Final Exams Solved Problems

This document presents detailed solutions to Algebra and Geometry final exams for semesters 1 and 2. The Algebra 1 solutions cover topics like number properties, algebraic expressions, and equation solving. Algebra 2 solutions include solving quadratic equations, inequalities, and graphing linea...Read More

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Algebra 1 Semester 1 Final Exam
1) (…, -3, -2, -1, 0, 1, 2, 3,)
2) (a× b) ×c=a× (b×c) is an associative property of multiplication
3) 4x+20-3x-6=14
4) x(5−3) y(2−6) z
= x2 z
y4
5) (x2y3)2*(x3y) 3= ( x2.2 y3.2) × (x3.3 y3
)
=x4y6*x9y3=x13y9
1.256097 rounded off to the nearest thousandth is 1.256
7) 36/48 into percentage =36/48*100=75%
8) First step in evaluating {6 [2 ( 6−4 ) ] } ÷ 4 is 6-4
9) 1 oz=28.3495 g
22 oz=22*28.3495=623.69 g=624 g
10) (x-4) (x+7) =0
x-4=0; x=4 or x+7=0; x=-7
11) 3(x-2) =3x(x-2); (x-2) cancels out on both sides hence we are left with 3=3x. dividing
through by 3; x=1
12) 5x+y=-23; when x is 0, y=-23 and when y=0, x=-23/5

13) Cannot be determined
14) Graph D
15) Find (f⁰g) (4) when f(x) =4x+5 and g(x) =4x2-5x-3
Evaluate (f (g)) by substituting in the value of g into f
4(4x2-5x-3) +5=16x2-20x-12+5
=16x2-20x-7
16) A
17) B
18) Graph C
19) Graph of g(x) =|x|+3 is C
20) Graph A
21) Inverse of the function -8-5x= −x
5 + 8
5 (B)
Algebra 2 Semester 1 Final Exam
1) A: 14x-4
2) B. transitive axiom
3) A. x=-2

4) y= m+d
x
5) A. Always
6) 1 ≤d ←3
7) x>3
8) |x-8|≥ 4; |x| ≥ 4 +8 ≥12; Number line D
9) A. I
10) C; y=3/2x-8
11) A. −1
5
12) Graph of the equation 5x+2.5y=20 is (A), the coefficients of x and y are 4 and 8 respectively
13) Equation of the line
x/1.5+y/-3=1; -2x+y=-3
y=2x-3 (B)
14) Parallel Lines
15) Fundamental Theorem of Arithmetic
16) C. 6y+3+2y=5
17) A. Add the 2 equations together to eliminate y
18) B. 2x=8

19) D. (3, 1)
Geometry Semester 1 Final Exam
1.) DB
↔
2.) GFC
3.) AC
4.) 6 planes
5.) Postulate
6.) b.
7.) <MRQ
8.) <PRN
9.) d. 154⁰
10.) c. congruent angles
11.) b. Perpendicular
12.) Lines l and m must be parallel
13.) a. All quadrilaterals are squares
Algebra 2 Semester 2 Final Exam
Q1: x2=10x-24

x2-10x+24=0
x2-4x-6x+24=0
x(x-4)-6(x-4) =0
x-6=0 or x-4=0
x=6 x=4
The solution is thus (4, 6)
Q2: x2=49
In this case it problem is solved using the first identity equation
x2-49=0
x2-7x+7x-49=0
x(x-7) +7(x-7) =0
X-7=0; x=7
Q3: 2n2=-10n+7
2n2+10n+7=0
The quadratic formula is n=−b ± √ b2 −4 ac
2 a
=−10 ± √102−4∗2∗−7
4 =−10 ± 44
4 ={−10+ √ 11
2 , −10−√ 11
2 }

Q4: x2+x+4=0
x=−b ± √ b2−4 ac
2 a =−1± √12−4∗1∗4
2 ={−1+i √ 15
2 , −1−i√ 15
2 }
Q5: Completing square method
z2+16z+44=0
z2+16z=-44
z2+ 16z+ (16*1/2)2=-44+ (16*1/2)2
z2+82=-44+64
2
√ ( z+ 8 ) 2=√ 20
Z+8= ± √ 20; z=± √20−8={−8+ 2 √5 ,−8−2 √5 }
Q14: SAS similarity: The ratio between two sides is the same as the ratio between another two
sides and the included angle angles are equal
Q15: Reflection in the line y=x since there is change in the places of the x-coordinate and y-
coordinate
Q16: The length WZ is determined from the ratio of any known two sides of the two
parallelograms
XW /EF =6 in/2 in=3
The length of WZ=length EH*3 (the ratio)
=3*3 in=9 in

Q17: Inductive and deductive arguments make different claims about their conclusion: In a
deductive argument, the premises are a guarantee to the truth of the conclusion while in an
inductive argument; the premises are anticipated only to be very strong that should they be true
then the conclusion is true
Q18: Certain
Q20: If Marie does not have soccer practice, and then it is not Tuesday
Q21: If we cannot go hiking, then there is lighting
Algebra 2 Semester 2
(6) s2+3s-4=0
D=b2-4ac; 9-(-4) =13; Two rational solutions
(7) t2+8t+16=0
D=b2-4ac; 64-64=0: One rational solution
(8) 4y2=6y-7; 4y2-6y+7=0
D=b2-4ac; 36-112=-76: Two no rational complex solutions
(9)log b ( q2 + y3 ): C
(10) log 4 3
log4 x
(11) log [ 1
1000 ]=log10 0.001=−3

(12) ln e=1 since e1=e
(13) Express in terms of natural log e5 x=2
{ ln2
5 }
(14) Conic section represented by the equation x2+6x+4y2=9; Ellipse
(15) Conic section represented by the equation x2+6x+4y2+12y=9; Circle
(16) Conic section represented by the equation x2+6x+4y2-18y=9; Circle
(17) Conic section represented by the equation x2+6x-4y=9; Hyperbola
(18) Center of a circle x2+y2+4x-8y+10=0; (-2, 4)
(19) Foci of the hyperbola 16x2-9y2=144
(16x2/144)- ( 9y2/144)=1; (x2/9)-( y2/16)=1
a=3, b=4 and since c2=a2+b2, c=5; the foci is thus (5, 0) and (-5, 0)
Algebra 2
(1) -4x+1;
(2) 14x-13
(3) e4 f 6
(4) v6
2

(5) 9 z8
y8
(6) x2 y +2 x y2 +6 x +4
(7) x2 y −2 x y2
(8) n2 −5 n+5
(9) 2 x7 −2 x −2
(10) x-7
(11) 10x (x+10)
(12) (x-7) (x+11)
(13) Graph B; The coefficient of f(x) =0 when x=0
(14) 4 (c +2)(c+3)
3(c−3)( c+5)
(15) (a+b)12
16
(16) 8
15 ;implies that there are15 parts
(17) no real number; 2x+3=2, x=-1.5
(18) ∅
(19) Asymptotes of y= 1
8 x +24 −10

Find the point at which the expression is undefined; x=-3
Vertical asymptotes occur at the points of infinite discontinuity; x=-3; Horizontal asymptotes:
y=-10
(20) Removable discontinuity at x=2
(21) 1
5
(22) 2
(23) 2 √ 11
(24) −4 x12 y6
(25) 8 √ 2
(26) 3−2 √ 3 z
(27) 5 √ 33
3

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