ALGEBRA 2B SEMESTER B EXAM 2022

Semester B Exam Part 1 ALG 2B 2022

Polynomial of 4 terms
3,-2,2
Graph 1. (for mine, may be switched) On x) -1 and y) 3 .
-6,-2
1,-1,2i,-2i
x^4 - 3x^3 - 7x^2 - 27x -18
Three real roots, two of which are equal value
1.41
y=-3x+22
neither
$24.00
7|x^3|y^4
5
Graph 3. (for mine, may be switched) On x) 3
f(x)=298(1.08)^x; 375
5
2.1
0.62
y=4/x-7 + 6
Graph 1. (for mine, may be switched)
x=9, x=3
-(x+1)/(x-1)(x+2), x =-3, x=-2,x=1,x=6
x^2/ 2x+2
-63/(x-3)(x+6)
x=-4,x=16
720
5/36
5/6
0.389
Mean; 87.2: median: 85.5; mode: 83
5.1
The range of hours worked is the same for juniors and seniors.
⅕
16/225
1/66
160
period=⅓ pi, amplitude = 5
pi3/2
47.1 in.
Graph 4. (for mine, may be switched) On pi and 2pi only.
y=0.5 cos 3 (theta)
-1
- pi 2
15.7 ft
50

Part 2 Semester Exam Part 2
1.
8 and 10
Step-by-step explanation:
Given
+ = ← combine the fractions on the left side
=
= ( cross- multiply )
9(x² + 2x) = 40(2x + 2) ← distribute both sides
9x² + 18x = 80x + 80 ( subtract 80x + 80 from both sides )
9x² - 62x - 80 = 0
Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.
product = 9 × - 80 = - 720 and sum = - 62
The factors are - 72 and + 10
Use these factors to split the x- term
9x² - 72x + 10x - 80 = 0 ( factor the first/second and third/fourth terms )
9x(x - 8) + 10(x - 8) = 0 ← factor out (x - 8) from each term
(x - 8)(9x + 10) = 0
Equate each factor to zero and solve for x
x - 8 = 0 ⇒ x = 8
9x + 10 = 0 ⇒ 9x = - 10 ⇒ x = -
x is an integer , then x = 8 and x + 2 = 8 + 2 = 10
The integers are 8 and 10
2.
x =2
Step-by-step explanation:
To solve this we will follow the step below;
8^2x = 4096
Simply find 8 to the power of what number will give 4096 at the left-hand side of the equation
8⁴ = 4096
Substitute 8⁴ by 4096 at the right hand side of the equation
8²ˣ = 8⁴
The 8 at the right-hand side will cancel-out the 8 at the left-hand side
2x = 4
Divide both-side of the equation by 2
2x/2 = 4/2
x=2