Algebra Exam Credit Recovery Assignment: Solutions and Explanations

This document presents a comprehensive set of solutions to an Algebra assignment designed for exam credit recovery. The solutions cover a variety of topics, including calculating the time it takes for two accountants to complete a task, finding the inverse of a function and verifying it through composition, determining the domain and range of functions, finding real and imaginary roots of a polynomial using the rational root theorem, sketching the graph of a rational function, solving a radioactive decay problem using the half-life model, working with arithmetic sequences, and calculating the total distance traveled by a pendulum using infinite series. Each problem is solved with detailed explanations, making it a valuable resource for students seeking to improve their understanding of Algebra and boost their exam scores.

Q 1.
Answer
An experienced accountant (E) prepares tax return in 16 hours.There-
fore, in 1 hour E prepares:1
16 tax return.
An novice accountant (N ) prepares tax return in 21 hours.Therefore,in 1
hour N prepares:1
21 tax return.
If E and N work together, in one hour they will prepare:
1
16+ 1
21= 37
336tax return
Suppose, together they require n hours to prepare one complete tax return:
=⇒ n × 37
336= 1
or,
n = 336
37 ≈ 9.081 hours
Answer:Together they will prepare a tax return in 9.081 hours
Q 2.
Answer
f (x) = 3x + 4
5x + 6
Inverse of f (x):(Dawkins, 2018)
y = 3x + 4
5x + 6
Interchange x and y:
x = 3y + 4
5y + 6
Solve for y:
x(5y + 6) = 3y + 4 =⇒ 5xy − 3y = 4 − 6x =⇒ y =
4 − 6x
5x − 3
1

Replace y with f−1(x):
f −1(x) = 4 − 6x
5x − 3
Composition test:
(f −1 o f )(x) = x
Therefore,
(f −1 o f )(x) =4 − 6f (x)
5f (x) − 3
= 4 − 6 3x+4
5x+6
5 3x+4
5x+6 − 3
= 20x + 24 − 18x − 24
15x + 20 − 15x − 18
= 2x
2 = 2
hence verified.
Domain of f (x) :It is defined for all realnumbers except for the condition
5x + 6 = 0 which occurs at x = −6
5. Therefore, the domain of f (x) is set of
all real numbers except x = −6
5: {x, x ∈ R | x 6= −6
5}
Domain of f−1(x) : It is defined for all real numbers except for the condition
5x − 3 = 0 which occurs at x =3
5. Therefore, the domain of f−1(x) is set of
all real numbers except x =3
5: {x, x ∈ R | x 6=3
5}
Range of f (x) :It is the domain of f−1(x), that is the range of f (x) is the
set of all real numbers except y =3
5
Range of f−1(x) : It is the domain of f (x),that is the range of f (x) is the
set of all real numbers except y = −6
5
Q 3.
Answer
x4 + 18x3 + 71x2 − 18x − 72 = 0
The rationalroot theorem states that if P (x) is a polynomialwith integer
coefficients and ifp
q is a root of P (x), then p is a factor of constant term of
P (x) and q is a factor of the leading coefficient of P (x).
2

For the given polynomial P (x) = x4 +18x3 +71x2 − 18x − 72, using the ratio-
nal theorem we have, p:a factor of -72 = ±(1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72)
and q:a factor of 1 = ±1
Possible valuesp
q = ±(1,2,3,4,6,8,9,12,18,24,36,72)
±1 = ±(1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72)
From the possible roots of P (x), the exact roots are selected by synthetic di-
vision of P (x) with each factor.The roots which perfectly divide P (x) are
x = ±1, −6, −12 As P (x) is a polynomialwith degree 4,all its roots are
found.
P (x) has a positive leading term with even degree (x4), therefore the ends of
the graph go to +∞ as x goes to +∞ or −∞
Q 4.
Answer
f (x) = 2x2 − 8
x2 + 3x − 10
3

Simplify f (x) by factorizing numerator and denominator:
f (x) = 2(x + 2)(x − 2)
(x + 5)(x − 2)
Horizontalasymptote:Since the numerator and denominator are ofequal
degrees,the equation for horizontalasymptote is y = a/b where a and b
are coefficients ofhighest degree terms in the numerator and denominator
respectively.Here,a = 2 and b = 1. Therefore,y = 2 is the horizontal
asymptote.
Verticalasymptote:The denominator is 0 when x = −5 or x = 2.x = −5
is therefore a vertical asymptote.x − 2 is common for both numerator and
denominator,so it won’t be considered.Infact x = 2 is a singular point.
Data points to plot f (x) on the graph:
Graph of the function f (x) is shown in figure below:
4

Q 5.
Answer
Half life radio active decay model is given by:
At = A0
1
2
t
t0.5
where,At is the amount decayed to after t years,A0 is the initialamount,
t0.5 is the half life period which is 2020 years.
5

Time required for the isotope to decay to 60% to its amount therefore is:
0.6A0 = A0
1
2
t
2020
0.6 = 1
2
t
2020
log 0.6 = t
2020
log 0.5
∴ t = 2020 ×
log 0.6
log 0.5≈ 1489 years
Q 6.
Answer
The given system of equations can be written in matrix form in the following
way:


6 −1 −3
−3 0 9
0 2 1


| {z }
A


x
y
z


|{z}
¯x
=


−23
60
23


| {z }
B
Cramer’s rule can be used to solve¯x only if the determinant of A is not zero.
|A| = det


6 −1 −3
−3 0 9
0 2 1

 = 6(1×0−2×9)−(−1)(−3×1−0×9)+(−3)(−3×2−0×0) = −10
Using Cramer’s rule,the components of¯x are determined in the following
way:
x =
det


b1 −1 −3
b2 0 9
b3 2 1


|A| , y =
det


6 b1 −3
−3 b2 9
0 b3 1


|A| , z =
det


6 −1 b1
−3 0 b2
0 2 b3


|A|
6

where, b1, b2 and b3 are elements of column matrix B.
Therefore,
x =
det


−23 −1 −3
60 0 9
23 2 1


|A| = (−23(0 − 18) − (−1)(60 − 9 × 23) + (−3)(120 − 0))
−93 = −93
−93= 1
Similarly,
y =
det


6 −23 −3
−3 60 9
0 23 1


|A| = −744
−93 = 8
z =
det

 6 −1 −23
−3 0 60
0 2 23


|A| = −651
−93 = 7
Q 7.
Answer
Roof of the building is described by an hyperbolic equation:
4y2 − x2 = 70
As shown in the figure,the side wallof the building are constructed at a
distance of 7 m from the center (0, 0).
The height of the side wall (on either side) will be the distance (y- coordinate)
from the ground to a point where it meets the roof (hyperbola),vertically.
As this point lies on the hyperbola, it satisfies its equation.
Therefore, height (h) of the side wall on the right side (x = 7) is determined
by plugging point (7, h) in the equation:
4h2 − 72 = 70 =⇒ 4h2 = 119 =⇒ h =
r 119
4 ≈ 5.4544
7

Similarly, height (h) of the side wall on the left side (x = −7) is determined
by plugging point (−7, h) in the equation:
4h2 − (−7)2 = 70 =⇒ 4h2 = 119 =⇒ h =
r 119
4 ≈ 5.4544
Thus, height of side wall on either side is 5.4544 m.
Q 8.
Answer
The given arithmetic sequence is:
1
6, 1
2, 5
6, 7
6, . . . ,
7
2
The difference between successive terms (d) of the sequence is1
3
The formula for nth term of the sequence therefore is:
an = a0 + d(n − 1)
where a0 is the starting term of the sequence.For the given sequence a0 = 1
6
and d =1
3. Therefore,
an = 1
6 + 1
3(n − 1)
Now, 7
2 = 1
6 + 1
3(n − 1)=⇒ n = 11
that is,7
2 is the 11th term of the sequence.
Sum of n terms of an arithmetic sequence is given by:
n−1X
k=0
a0 + k · d =
n
2(2a0 + (n − 1)d)
Therefore,
1
6+ 1
2+ 5
6+ 7
6+ · · · +
7
2 =
11−1X
k=0
1
6+ k ·
1
3 = 11
2 2 ×1
6 + (11 − 1) ×
1
3 = 121
6
8

Q 9.
Answer
In the first pass the pendulum passes through a distance of 15 inches.
In the second pass it passes through a distance which is2
5 of the first pass,
that is2
5 × 15 inches.
In the third pass it passes through a distance which is2
5 of the second pass,
that is2
5 × 2
5 × 15 inches.
In the next pass it passes through a distance which is2
5 of the previous pass
and so on.The distance the pendulum travels in each pass can be expressed
a sequence of the form:
15, 2
5 × 15,2
5 × 2
5 × 15,. . .
or more precisely,
15 2
5
0
, 15 2
5
1
, 15 2
5
2
, 15 2
5
3
, . . .
The total distance covered is the sum of distance travelled in each pass:
S = 15 2
5
0
+ 15 2
5
1
+ 15 2
5
2
+ 15 2
5
3
+ · · · =
∞X
i=0
15 · 2
5
i
Let, x =2
5. Therefore,
S = 15
∞X
i=0
xi
Now, this infinite series converges for −1 < x < 1:
∞X
i=0
xi = 1
1 − x
Since, x =2
5 satisfies the condition −1 < x < 1, we have:
S = 15
∞X
i=0
xi = 15 × 1
1 − x= 15 × 1
1 −2
5
= 25
9

Thus, the pendulum travels a total distance of 25 inches before it stops.
Q 10.
Answer
S =
∞X
i=1
10 1
2
i
Let, x = 12.Therefore,
S =
∞X
i=1
10 · xi = 10(1 + x2 + x3 + x4 + . . . )
The infinite series (1 + x2 + x3 + x4 + . . . ) converges for −1 < x < 1 and,
∞X
i=1
xi = 1 + x2 + x3 + x4 + · · · = 1
1 − x
Since x =1
2 satisfies the condition −1 < x < 1:
S =
∞X
i=1
10· 1
2
i
= 10(1+x2+x3+x4+. . . ) = 10×1
1 − x= 10× 1
1 −1
2
= 20
References :
1. Dawkins Paul, Lamar University, 2018.
http://tutorial.math.lamar.edu/Classes/CalcI/InverseFunctions.aspx
10