Math 121 C2.pdf

DEFINITION OF INTEREST
Borrow P5,000 from the Bank
Marvin wants to borrow P5,000. The local bank says
"10% Interest". So to borrow the P5,000 for 1 year will
cost:
P5,000 × 10% = P500
Of course, Marvin will have to pay back the original
P5,000 after one year, so this is what happens:
Marvin Bank Marvin Bank
Today Next Year
P5,000 P5,000
P500

Lender or creditor- person(or institution) who invests
the money or makes the funds available.
Borrower or debtor- person(or institution) who owes the
money or avails of the fund from the lender.
Origin or loan date- date on which money is received by
the borrower.
Repayment date or maturity value- date on which the
money borrowed or loan is to be completely
repaid.
Time or term (t)- amount of time in years the money is
borrowed or invested; length of time between the
origin and maturity date.

Principal (P)- amount of money borrowed or invested on
the origin date.
Rate (r)- annual rate, usually in percent, charged by the
lender, or rate of increase of the investment.
Interest (I)- amount paid or earned for the use of money.
Simple Interest (𝐼𝑠 )- interest that is computed on the
principal and then added to it.
Compound Interest (𝐼𝑐)- interest that is computed from
the principal and also on the accumulated past
interests.
Maturity value or future value (F)- amount after t years;
that is the lender receives from the borrower on
the maturity date.

SIMPLE INTEREST VS
COMPOUND INTEREST
ANALYSIS:
Suppose you won P10, 000 and you plan to
invest it for 5 years. A cooperative group
offers 2% interest rate per year. A bank
offers 2% compounded annually. Which will
you choose and why?

SIMPLE INTEREST VS COMPOUND INTEREST
Investment 1: Simple Interest, with annual rate r
Time (t)
Principal
(P)
Simple Interest Amount after t years
(Maturity Value)
Solution Answer
1
10, 000
(10,000)(0.02)(1) 200 10 000 + 200 = 10 200
2 (10,000)(0.02)(2) 400 10 000 + 400 = 10 400
3 (10,000)(0.02)(3) 600 10 000 + 600 = 10 600
4 (10,000)(0.02)(4) 800 10 000 + 800 = 10 800
5 (10,000)(0.02)(5) 1 000 10 000 + 1 000 = 11 000
Investment 2: Compound Interest, with annual rate r
Time (t)
Amount at the
start of year t
Compound Interest Amount at the end of year t
(Maturity Value)
Solution Answer
1 10, 000 (10,000)(0.02)(1) 200 10 000 + 200 = 10 200
2 10 200 (10,200)(0.02)(1) 204 10 200 + 204 = 10 404
3 10 404 (10,404)(0.02)(1) 208.08 10 404 + 202.08 = 10 612.08
4 10 612.08 (10,612.08)(0.02)(1) 212.24 10 612.08 + 212.24 = 10 824.32
5 10 824.32 (10,824.32)(0.02)(1) 216.49 10 824.32 + 216.49 = 11 040.81

Simple interest (in pesos):
11 000 – 10 000 = 1 000
Compound interest (in pesos):
11, 041.81 – 10 000 = 1 040.81
Simple interest remains constant throughout the
investment term. In compound interest, the interest
from the previous year also earns interest. Thus, the
interest grows every year.
SIMPLE INTEREST VS COMPOUND INTEREST

SIMPLE INTEREST
Interest is the amount paid or earned for the use of
money. Simple interest means that the interest is
calculated only once for the entire rate of the loan. At
the end of the time period, the borrower repays the
principal plus the interest.

THE SIMPLE INTEREST FORMULA
Magic Triangle in a Circle of a Simple Interest

FINDING THE SIMPLE INTEREST
A bank offers 0.25% annual simple interest rate
for a particular deposit. How much interest will
be earned if 1 million pesos is deposited in this
savings for 1 year?

A bank offers 0.25% annual simple interest rate
for a particular deposit. How much interest will
be earned if 1 million pesos is deposited in this
savings for 1 year?
r = 0.25% P = 1 000 000 t = 1
= 0.0025
𝐼𝑠 = 𝑃𝑟𝑡 =(1 000 000)(0.0025)(1)
𝐼𝑠 = 2 500

How much interest is charged when P50 000 is borrowed for
9 months at an annual simple interest rate at 10%?

How much interest is charged when P50 000 is borrowed for
9 months at an annual simple interest rate at 10%?
r = 10% P = 50 000 t = 9
12 = 3
4year
= 0.10
𝐼𝑠= 𝑃𝑟𝑡 = (50 000)(0.10)(3
4)
𝐼𝑠 = 3 750
When the term is
expressed in months
or days it should be
converted to years
The simple interest charged is P3, 750.

FINDING THE PRINCIPAL AMOUNT
When invested at an annual rate of 7%, an amount earned
P11, 200 of simple interest. How much money was originally
invested?

FINDING THE PRINCIPAL AMOUNT
When invested at an annual rate of 7%, an amount earned
P11, 200 of simple interest. How much money was originally
invested?
r = 7% 𝑰𝒔 = 11 200 t = 2
= 0.07
𝑃 =
𝐼𝑠
𝑟𝑡
=
11 200
(0.07)(2)
P = 80 000
The amount invested is P80 000.

FINDING THE SIMPLE INTEREST RATE
If an entrepreneur applies for a loan amounting to P500,000 in
a bank, the simple interest of which is P157,500 for 3 years,
what interest rate is being charged?

FINDING THE SIMPLE INTEREST RATE
If an entrepreneur applies for a loan amounting to P500,000 in
a bank, the simple interest of which is P157,500 for 3 years,
what interest rate is being charged?
P = 500 000 𝑰𝒔 = 157 500 t = 3
𝑟 =
𝐼𝑠
𝑃𝑡
=
157 500
(500 000)(3)
r = 0.105 = 10.5%
The bank charged an annual simple interest rate of 10.5%.

FINDING THE TIME OF A SIMPLE INTEREST
How long will a principal earn an interest equal to half of it at
5% simple interest?

FINDING THE TIME OF A SIMPLE INTEREST
How long will a principal earn an interest equal to half of it at
5% simple interest?
P = P 𝑰𝒔 = 1
2 (𝑃) r = 5%
= 0.05
𝑡 =
𝐼𝑠
𝑃𝑟
=
1
2 𝑃
(𝑃)(0.05)
t = 10 years
It will take 10 years for a principal to earn half of its
value at 5% simple interest rate

MATURITY VALUE
Maturity (future) Value is the total payback of
principal and interest of an investment or a loan.
𝑭 = 𝑷 + 𝑰𝒔
F – maturity (future) value
P – principal
𝐼𝑠 - simple interest
𝑭 = 𝑷 𝟏 + 𝒓𝒕
F – maturity (future) value
P – principal
r – rate
t - time

FINDING THE MATURITY VALUE
Find the maturity value if 1 million pesos is deposited at an
annual simple interest rate of 0.25% after
(a) 1 year (b) 5 years
P = 1 000 000 r = 0.0025 t = 1
F = P(1 + rt)
= 1 000 000(1 + (0.0025)(1))
F = 1 002 500
The future or maturity value after 1 year is P1,002,500.

FINDING THE MATURITY VALUE
Find the maturity value if 1 million pesos is deposited at an
annual simple interest rate of 0.25% after
(a) 1 year (b) 5 years
P = 1 000 000 r = 0.0025 t = 5
F = P(1 + rt)
= 1 000 000(1 + (0.0025)(5))
F = 1 012 500
The future or maturity value after 5 years is P1,012,500.

SIMPLE INTEREST AND MATURITY VALUE
EXERCISE
A. Complete the table below by finding the unknown.
Principal (P) Rate (R) Time (T) Interest (I)
(1) 2% 5 P10, 000
P360, 000 (2) 2 P3, 00
P500,000 10.5% (3) P175 000
P880, 000 9.25% 2.5 (4)

EXERCISES
B. Solve the following problems.
1. What are the amount of interest and maturity value of a loan
for P150,000 at 6.5% simple interest rate for 3 years?
2. At what simple interest rate per annum will P25,000
accumulate to P33,000 in 5 years?
3. In order to have P200,000 in 3 years, how much should you
invest if the simple interest is 5.5%?
4. Steven deposited P20,000 in a bank that pays 0.5% simple
interest. How much will be her money after 6 years?

EXERCISE
C. Complete the table below by finding the unknown.
Principal
(P)
Rate
(r)
Time
(t)
Interest
(I)
Maturity
Value (F)
60 000 4% 15 (1) (2)
(3) 12% 5 15 000 (4)
50 000 (5) 2 (6) 59 500
(7) 10.5% (8) 157 500 457 500
1 000 000 0.25% 6.5 (9) (10)

EXERCISES
D. Solve the following problems.
1. Bien invested a certain amount at 8% simple interest rate per year.
After 6 years, the interest he received amounted to P48,000. How
much did he invest?
2. Apollo Air Taxi Service borrowed P3,450,000 at 8% simple interest
for 9 months to purchase a new airplane. Find the maturity value
of the loan.
3. Kyle invests P25, 000 in an investment fund for 3 years. At the end
of the investment period, his investment will be worth P87,450.
Determine the simple interest rate that is offered.

EXACT AND ORDINARY INTEREST RATE
Ordinary interest is calculated on the basis of a 360-day year or
a 30-day month; exact interest is calculated on a 365-day year.
The interest formulas for both ordinary and exact interest are
actually the same, with time slightly differing when given as
number of days.
Ordinary Interest, for D days
𝑇𝑖𝑚𝑒 =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑎𝑦𝑠 𝑜𝑓 𝑎 𝑙𝑜𝑎𝑛
360
; 𝐼𝑂 = Pr
𝐷
360
Exact Interest , for D days
𝑇𝑖𝑚𝑒 =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑎𝑦𝑠 𝑜𝑓 𝑎 𝑙𝑜𝑎𝑛
365
; 𝐼𝑒 = 𝑃𝑟
𝐷
365

THE HAND CALCULATOR
Spaces are ALL 30 days except for February which can be either
28 or 29 days in a Leap year!
Knuckles are ALL 31 days

COMPUTING THE NUMBER OF DAYS BETWEEN TWO DATES
1.Actual time- the number of days is obtained by
counting all the days, inclusive between
the 2 given dates including the last day
but not the first day.
2. Approximate time- assume that every month counts
30 days.

EXACT AND ORDINARY INTEREST OF ACTUAL AND APPROXIMATE DAYS
There are four varieties of interest between two dates:
1. Ordinary interest at actual no. of days “Banker’s rule”
𝐼𝑜𝑖 = 𝑃𝑟
𝑎𝑐𝑡𝑢𝑎𝑙 𝑡𝑖𝑚𝑒
360
2. Ordinary interest at approximate no.of days
𝐼𝑜𝑎 = 𝑃𝑟
𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 𝑡𝑖𝑚𝑒
360
3. Exact interest at actual no. of days
𝐼𝑒𝑖 = 𝑃𝑟
𝑎𝑐𝑡𝑢𝑎𝑙 𝑡𝑖𝑚𝑒
365
4. Exact interest at approximate no. of days
𝐼𝑒𝑎 = 𝑃𝑟
𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 𝑡𝑖𝑚𝑒
365

EXACT AND ORDINARY INTEREST
Solve for the following:
1.Find the actual and approximate number of days from Oct.3,2017 and
March 30, 2019
2. Find the ordinary and exact interest of the amount P15,000 using the
approximate and actual number of days from Oct 3, 2017 to March
30,2019 at 12.5% simple interest.
3. Jun invested P20,000 on November 30,2017 to March 1, 2020 at 14.5%
interest rate. How much will he earn and how much is the final amount?
4.Ryan McPherson borrowed $3,500 on June 15 at 11% interest. If the loan
was due on October 9, what was the amount of interest on Ryan’s loan
using the exact interest method?

COMPOUND INTEREST
Compound interest means that the interest is calculated
more than once during the time period of the loan. When
compound interest is applied to a loan, each succeeding
time period accumulates interest on the previous interest
in addition to interest on the principal. Compound interest
loans are generally for time periods of a year or longer.

COMPOUND INTEREST VS SIMPLE INTEREST
Simple interest Compound interest
Based
on
original
principal
principal that grows
from 1 interest
interval to another
Function Linear Exponential
Simple versus
Compound
Interest

COMPOUND INTEREST FORMULA
𝐹 = 𝑃 1 + 𝑟 𝑡
where
P – principal or present value
F – maturity value at the end of the term
r – rate
t – term/time in years
The compound interest 𝐼𝑐 is given by
𝐼𝑐 = 𝐹 − 𝑃

FINDING THE MATURITY VALUE OF A
COMPOUND INTEREST
Find the maturity value and the compound interest if P10, 000
is compounded annually at an interest rate of 2% in 5 years.

FINDING THE MATURITY VALUE OF A
COMPOUND INTEREST
Find the maturity value and the compound interest if P10, 000
is compounded annually at an interest rate of 2% in 5 years.
P = 10 000 r = 0.02 t = 5
A. 𝐹 = 𝑃 1 + 𝑟 𝑡
= 10 000 1 + 0.02 5
𝐹 = 11 040.81
The future value F is P11,040.81 and the compound interest is P1,040.81.
B. 𝐼𝑐 = 𝐹 − 𝑃
= 11 040.81 − 10 000
𝐼𝑐 = 1 040.81

PRESENT VALUE P at COMPOUND INTEREST
𝑃 =
𝐹
1 + 𝑟 𝑡
where
r – rate

FINDING THE PRESENT VALUE P at COMPOUND INTEREST
What is the present value of P50,000 due in 7 years if money is
worth 10% compounded annually?
F = 50 000 r = 0.10 t = 7
𝑃 =
𝐹
1+𝑟 𝑡
=
50 000
1+0.10 7
𝑃 = 25 657.91
The present value is P25, 657.91.

FINDING THE PRESENT VALUE P at COMPOUND INTEREST
How much money should a student place in a time deposit
in a bank that pays 1.1% compounded annually so that he
will have P200,000 after 6 years?
F = 200 000 r = 0.011 t = 6
𝑃 =
𝐹
1+𝑟 𝑡
=
200 000
1+0.011 6
𝑃 = 187 293.65
The student should deposit P187,293.65 in the bank.

COMPOUND INTEREST
EXERCISE
A. Complete the table below by finding the unknown.
Principal
(P)
Rate
(r)
Time
(t)
Compound
Interest
(𝑰𝒄)
Maturity
Value (F)
10 000 8% 15 (1) (2)
3 000 5% 6 (3) (4)
50 000 10.5% 10 (5) (6)
(7) 2% 5 157 500 50 000
(9) 9.25% 2.5 (10) 100 000

COMPOUND INTEREST
EXERCISE
B. Solve the following problems.
1. On the 7th birthday of her daughter, Anna deposited an amount in a
bank peso bond that pays 1% interest compounded annually. How
much should she deposit if she wants to have P100,000 on her
daughter’s 18th birthday?
2. In a certain bank, Francis invested P88,000 in a time deposit that
pays 0.5% compound interest in a year. How much will be his money
after 6 years? How much interest will he gain?

COMPOUNDING MORE THAT ONCE A YEAR
𝐹 = 𝑃 1 +
𝑖(𝑚)
𝑚
𝑚𝑡
where
𝑖(𝑚) - nominal rate of interest (annual rate)
m – frequency of conversion

COMPOUNDING MORE THAT ONCE A YEAR
DEFINITION OF TERMS
• Conversion of interest period – time between
successive conversion of interests.
• Frequency conversion (m) – number of conversion
periods in one year
• Nominal rate (𝒊(𝒎)) – annual rate of interest
• Rate (j) of interest for each conversion period
𝒋 =
𝒊(𝒎)
𝒎
=
𝒂𝒏𝒏𝒖𝒂𝒍 𝒊𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝒓𝒂𝒕𝒆
𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 𝒐𝒇 𝒄𝒐𝒏𝒗𝒆𝒓𝒔𝒊𝒐𝒏
• Total number of conversion periods (n)
n = tm = (frequency of conversion) x (time in years)

FINDING THE MATURITY VALUE,
COMPOUNDING m TIMES A YEAR
Find the maturity value and interest if P10,000 is deposited in
a bank at 2% compounded quarterly for 5 years?
P = 10 000 𝒊(𝟒)
= 0.02m = 4 t = 5
A. 𝐹 = 𝑃 1 +
𝑖(𝑚)
𝑚
𝑚𝑡
= 10 000 1 +
0.02
4
(4)(5)
𝐹 = 11 048.96
B. 𝐼𝑐 = 𝐹 − 𝑃
= 11 048.96 − 10 000
𝐼𝑐 = 1 048.96

FINDING THE MATURITY VALUE,
COMPOUNDING m TIMES A YEAR
Find the maturity value and interest if P10,000 is deposited in a
bank at 2% compounded monthly for 5 years?
P = 10 000 𝒊(𝟏𝟐)= 0.02 m = 12 t = 5
A. 𝐹 = 𝑃 1 +
𝑖(𝑚)
𝑚
𝑚𝑡
= 10 000 1 +
0.02
12
(12)(5)
𝐹 = 11 050.79
B. 𝐼𝑐 = 𝐹 − 𝑃
= 11 050.79 − 10 000
𝐼𝑐 = 1 050.79

(COMPOUNDING m TIMES A YEAR)
𝑃 =
𝐹
1 +
𝑖(𝑚)
𝑚
𝑚𝑡
where
𝑖(𝑚)
- nominal rate of interest (annual rate)
m – frequency of conversion

Find the present value of P50,000 due in 4 years if money is invested
at 12% compounded semi-annually.
𝑃 =
𝐹
1+
𝑖(𝑚)
𝑚
𝑚𝑡
=
50 000
1+
0.12
2
(2)(4)
𝑃 = 31 370.62
F = 50 000 𝒊(𝟐)= 0.12 m = 2 t = 4

What is the present value of P25,000 due in 2 years and 6
months if money is worth 10% compounded quarterly?
𝑃 =
𝐹
1+
𝑖(𝑚)
𝑚
𝑚𝑡
=
25 000
1+
0.10
4
(4)(2.5)
𝑃 = 19 529.96
F = 25 000 𝒊(𝟒)
= 0.10 m = 4 t = 2.5

FINDING THE PRESENT VALUE
(Compounding m times a year)
EXERCISE
Solve the following problem.
1. How much should Apollo set aside and invest in
fund earning 2.5% compounded quarterly if she
needs P75,000 in 15 months?
2. Find the compound amount due in 8 years if
P200,000 is invested at 12% compounded
quarterly.
3. Adriel borrowed P15,000 payable with interest
that is compounded semi-annually at 9%. How
much must he pay after 3 years?

FINDING THE PRESENT VALUE
(Compounding m times a year)
EXERCISE
Solve the following problem.
4. Julie is planning to invest P100,000. Bank A is offering 5%
compounded semi-annually while Bank B is offering 4.5%
compounded monthly. If he plans to invest this amount for 5
years, in which bank should she invest?
5. What present value, compounded daily at 0.6% will amount to
P59,780.91 in 3 years?

COMPOUND INTEREST
EXERCISE
Solve the problems completely.
1. Calculate the present values of P40,000 due in 4 years at the given rate
of interest
a. 6% compounded semi – annually
b. 8% compounded quarterly
c. 7% compounded monthly
d. 9% compounded daily.
2. In how many years will it take P18,000 accumulate to P20,000 when
deposited in a savings account that earns 0.33% compounded monthly?

COMPOUND INTEREST
EXERCISE
3. Harvey borrowed an amount of P40,000 which she paid with an interest
of P2,000 at the end of 3 years. At what nominal rate compounded
semi-annually was it invested?
4. A debt of P8,000 will mature in four years’ time. By assuming that the
money is worth 9% compounded quarterly, calculate :
a)the present value of this debt
b)the value of this debt at the end of the two years
c)the value of this debt at the end of the five years

COMPOUND INTEREST
EXERCISE
5. Martha invested P400,000 in a boutique 5 years ago. Her investment is
worth P700,000 today. What is the effective rate of her investment?
6. In the last 5 year, ALB Mutual Fund grew at the rate of 10.4% per year
compounded quarterly. Over the same period, SCG Mutual Fund grew
at the rate of 10.6% per year compounded semi – annually. Which
mutual fund has a better rate of return?

Annuity A sequence of equal payments
made regularly or periodically.

Annuity Example
A smartphone is purchased with a down payment of
P1,000 and the balance will be paid at P1,075.83 per
month for a year. What is its cash price if the interest
rate is 6% compounded monthly?

Annuity Example
A smartphone is purchased with a down payment of
P1,000 and the balance will be paid at P1,075.83 per
month for a year. What is its cash price if the interest
rate is 6% compounded monthly?
Given:
P = ? r = 6% A = 1,075.83 t = 1
𝑃 = 𝐴
1 + 𝑖 𝑛
− 1
𝑖 1 + 𝑖 𝑛 𝑃 = 𝐴
1 +
𝑖
𝑚
𝑚𝑛
− 1
𝑖
𝑚
1 +
𝑖
𝑚
(𝑚𝑛)

Annuity Example
A smartphone is purchased with a down payment of P1,000 and the balance will be paid at
P1,075.83 per month for a year. What is its cash price if the interest rate is 6%
compounded monthly?
Given:
P = ? r = 6% A = 1,075.83 t = 1
𝑃 = 𝐴
1 +
𝑖
𝑚
𝑚𝑛
− 1
𝑖
𝑚
1 +
𝑖
𝑚
(𝑚𝑛)
𝑃 = 1075.83
1 +
0.06
12
1𝑥12
− 1
0.06
12
1 +
0.06
12
(1𝑥12)
𝑷 = 𝟏𝟐𝟓𝟎𝟎 + 𝟏, 𝟎𝟎𝟎 = 𝟏𝟑, 𝟓𝟎𝟎

Amortization
A debt repayment scheme in
which the original amount
borrowed is repaid by making
equal payments periodically.

Example
Find the monthly amortization for a P150,000 debt which is to be repaid in
2 years at 7% interest compounded monthly.

Example
Given:
A = ? r = 7% P = 150,000 t = 2
𝑃 = 𝐴
1 + 𝑖 𝑛 − 1
𝑖 1 + 𝑖 𝑛
𝐴 =
𝑃
1 + 𝑖 𝑛 − 1
𝑖 1 + 𝑖 𝑛
𝐴 =
𝑃
1 +
𝑖
𝑚
𝑚𝑛
− 1
𝑖
𝑚
1 +
𝑖
𝑚
𝑚𝑛

Example
Given:
A = ? r = 7% P = 150,000 t = 2
𝐴 =
𝑃
1 +
𝑖
𝑚
𝑚𝑛
− 1
𝑖
𝑚 1 +
𝑖
𝑚
𝑚𝑛
𝐴 =
150,000
1 +
0.07
12
2𝑥12
− 1
0.07
12
1 +
0.07
12
2𝑥12
6,715.89

Annuity and Future
Value Relationship

𝐹 = 𝐴
1 + 𝑖 𝑛 − 1
𝑖
𝐴 =
𝐹
1 + 𝑖 𝑛 − 1
𝑖

Example
Simon starts saving for his retirement. He opens an investment and
deposits 800 USD into the account which earns 10% per annum
compounded monthly. He deposits monthly for 10 years. What is the
value of his retirement savings at the end of 10 year period?

Solution
Given:
F = ? i = 10% t = 10 A = 800
𝐹 = 𝐴
1 + 𝑖 𝑛 − 1
𝑖
𝐹 = 𝐴
1 +
𝑖
𝑚
𝑛𝑥𝑚
− 1
𝑖
𝑚

Solution
Given:
F = ? i = 10% t = 10 A = 800
𝐹 = 𝐴
1 +
𝑖
𝑚
𝑛𝑥𝑚
− 1
𝑖
𝑚
𝐹 = 800
1 +
0.10
12
10𝑥12
− 1
0.10
12
𝐹 =163,875.98

Stocks and Bonds
Stocks represents shares and ownership in a
company. A share is a unit of ownership of a
corporation’s profits and assets. Ownership can be
quantified by dividing the number of shares owned by
the number of shares issued.

Credit Cards
A plastic card which contains information and authorizes the person whose name
appears on it to charge purchases or services to his account for which he will be later be
billed. The information on the cards can be read by store readers and ATMs.

Math 121 C2.pdf

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Math 121 C2.pdf