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a calculus and preparation for calculus website, etc.

Online Volumes (Book Orders)
1,  Elements of Reason.
1A. Pattern Based Reason 
1B. Math Curriculum Notes
2. Three Skills for Algebra
3. Why Slopes & More Math

Mathematics Course Designers: LAMP offers food for thought.
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2. Solving Linear Equations
3. Fractions Ratios Rates Proportions & Units
4. Euclidean Geometry
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6. Number Theory
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||Définition d'une variable || Algèbre || Arithmetique || Logique ||La raison basée sur les règles et modelés||
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YOU are better than YOU think. Show yourself  how:  

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Read  logic chapters 1 to 5  in online volume Three Skills for Algebra  for greater skills & confidence in  work 
and study.

Learn to read notes and textbooks like a lawyer, so that no nuance, no subtlety and no clause escapes your attention.

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 Logic chapters 1 to 5  re- appear not in sequence, as is or longer,  in  Volume 1A,  Pattern Based Reason, Bon Appetite.

Logic Mastery
 Amazing, Amusing, Amorous,  Delicious, Delightful, Edifying, Strengthening Elixir. 
It eases work & learning difficulties Makes the hard easier. Opens eyes. Leads to greater precision.
in reading and
writing

Logic mastery makes the hard, easier. Logic mastery  leads to better, stronger and richer comprehension.  Logic mastery  improves reading and writing.  Logic mastery ease learning difficulties.  Logic mastery gives a headstart.  In sum, logic mastery  will develops critical thinking, improve reading and writing, and give a firmer base for work and studies at many levels. Good luck.


After logic  (a) continue reading Three Skills for Algebra, chapters 8 to 14  and do so alongside site area on solving liinear Equations ; or (b) see this calculus starter lesson and Volume 3, Why Slopes  & More Math, chapters 2 to 6;

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Caution: Site advice is approximately correct, for some circumstances, not all. That leaves room for thought

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What may be learnt and when depends on how skills and concepts are developed. Making the hard easier and clearer will allow earlier & richer development of skills and concepts.


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For online automated help in senior high school maths & calculus, visit  quickmath.com  For Automatic Calculus and Algebra Help with derivatives, integrals, graphs, linear equations, matrix algebra, visit calc101.com  With  overlap, each site quickmath & calc101offers a different range of services, some free, some not, all based on webmathematica. Good luck.

Chapter 14 
Compound Interest Calculations
(Compound Growth Calculations)

Previous:  14. Direct Use of Compound Interest Formula, Secoond Example

3  Using The Formula Backwards

The compound interest formula A = P(1+i)n involves four quantities, namely A, P, i and n. When any three are known, the fourth can be found. Properties of arithmetic and algebra say how this is done. Read on. The easiest quantity to find is A. In the following examples, we consider the cases where the fourth quantity is A or P or i. We can also consider the case where the fourth quantity is n.1

3.3  Indirect Use: Example 3

Problem (Finding the principal): Tom Oublier, lucky Tom, finds he has $1350 in an account today. For the past 21/2 years, the account has been paying Tom 9% compounded monthly. Tom Oublier has forgotten the initial amount he had in the account. What was the initial amount (principal) that he placed or deposited in the account?

There are two ways of getting the result. Both will be given. The advantages of each will be noted.


ARITHMETIC SOLUTION. The principal P is unknown. The principal is the initial amount in the account which we want to find. Now in the compound interest formula A = P(1+i)n,

  • the final amount A = $1350 is given or known,
  • the interest rate i = 9% ¸12 = .75% = .0075 is known, and
  • the number of compound interest periods (months) in 2.5 years n = 2.5 ×12 = 30.
Therefore replacement or substitution in the formula A = P(1+i)n gives
$1350 = P(1+.75%)30 = P(1+0.75× 1
100
)30
Calculation yields first
$1350 = P(1.0075)30
and then
$1350 = P×1.251272
Here (1.0075)30 = 1.251272 represents the number 1.0075 multiplied by itself 30 times. Suggestion: Use a calculator to check this multiplication result.

Now multiplying a quantity P by a nonzero number4 (or quantity) and then dividing by the same number yields the quantity P, no matter what P equals. In our situation,
P = (P×1.251272) ¸1.251272 = ($1350) ¸1.251272 = $1078.90
The second equality follows from the replacement of P×1.251272 by its equal $1350. The initial amount invested was $1078.90 to the nearest penny or cent. This yields the solution of one problem. In the next solution method, we will see how to algebraically describe the arithmetic solution of all similar problems.


ALGEBRAIC SHORTHAND SOLUTION. There is a pattern in the first solution. This pattern can be followed if we had the same problem again, but with different numbers. For instance, how would we solve the above problem if the 9% was replaced by 8%? We will try to capture the pattern using shorthand notation. This approach is given next. It requires a little more work. But it will give a formula for solving many similar problems.

On your first reading of the shorthand solution you may assume the letter have the values given above. On your second reading, pretend A, i and n have values not known to you - for instance they might be hidden in an sealed enveloped. This second viewpoint is a key to algebra.


First, note the following simple idea. The compound interest formula A = P(1+i)n says the symbol A and the expression P(1+i)n both stand for, represent or give the final amount in the account.5

5More precisely we can say:

  • The symbol A is shorthand for a quantity.
  • The expression P(1+i)n when computed gives the same quantity.
Here the symbol A and the more complicated expression or symbol P(1+i)n both represent the same quantity. So we take the liberty of using one in place of the other, as convenient.

Second, we can assume and use the rule: when B is a nonzero number or quantity, then
P ×B
B
= (P ×B)¸B = P
This rule holds no matter what the number B might be, provided this number B is nonzero. We are now ready (after all this preparation) for a one-line shorthand solution.

We now use the above ideas. We will apply this rule with B = (1+i)n. That is, the value of B is given by (1+i)n whenever the latter is computed. So
P = P×B
B
= P×(1+i)n
(1+i)n
= A
(1+i)n
.
The second equality follows by replacing P×(1+i)n by its equal A. This gives us the formula
P = A
(1+i)n
for P in terms of the other three quantities A, i and n. So we can compute P whenever the values of A, i and n are given. This formula solves many forgotten principal problems at once. This formula for the principal P, the initial amount in a compound interest account, is known as the present value formula. It gives the value Pnow of getting the amount A in n periods from now if you investment Pnow compounds at the rate i for each of these n periods.

using the formula - number substitution. The substitutions A = $1305 , i = [.09/12] = 0.0075, and n = 30 leads to
P = A
(1+i)n
= $1350
(1+.75%)30
This gives
P = $1350
(1.0075)30
= $1350
1.251272
or
P = ($1350) ¸1.251272 = $1078.90
The result is the same as before. The arithmetic is also the same as before.

Remark (Algebraic Viewpoint). The formula we have found, namely
P = A
(1+i)n
for P in terms of the other three quantities A, i and n can be used whenever the last three quantities are given. We can further use the formula without us repeating the above reasoning. The above reasoning with shorthand notation solves many problems at once. Shorthand notation helps us describe calculations. It also leads us to new ones.

 


More Chapter Sections: Up ] 14 The Formula ] 14. Direct Use - First Example ] 14. Direct Use, Second Example ] [ 14 Indirect Use - First Example I ] 14 Indirect Use - Second Example ] 14 Going Further ]

 

www.whyslopes.com
2. Three Skills for Algebra 

Foreword, Chapters 
& Appendices 

Foreword
1. Introduction
2. Implication Rules
3. Chains of Reason
4. Romeo and Juliet
4. Induction Mathematical
5 Knowledge Islands
6  Old Language
7  Arith Skill Check
7. The Next Chapters
8 The Three Skills
8 VNR-Concise-Encyclopedia
PS. What is a Variable
9. Algebra Talk
10 Two More Skills
11 Why Shorthand
12 Shorthand Usage
13 What's Next
14 Compound Interest
15 Linear Equations
PS I.  Distributive Law
PS II. Polynomials
16 Painless Proofs
17 Pythagoras
18 Rules of Algebra
19  Functions & Sets
20 Degrees & Radians
21 What's Next
22. Arith & Geometric Sums
23 Summation Notation
24 Your Money
25 Induction & Recursion
26 What's Next
27 Pronouns in Logic
28 Occurrence Tables
29 Contrapositive
30 Truth Tables
31 Indirect Reason
A. Advice For Learning

Real Player Videos

Perfect arithmetic skills with whole numbers & fractions
after or besides chapters 1 to 14.

Arithmetic Videos Summary
Addition with Decimals
Subtraction with Decimals
Multiplication with Decimals
Fraction Arithmetic
Recognizing Primes
Long Division for Decimals
Square Root Simplification
Greatest Common Divisors
Least Common Multiples

Words Before Symbols: 
What is a Variable?
Introduction
Variation between Examples

Variation of Letters

A letter denotes a variable

Cases of Double Variation

Three Notions of a Variable

Constants, Parameters
& Variables

Talking about numbers
Dependent or Independent
Variable, a Matter of Choice

Complex number: starter lesson  

Solving Linear Equations:

A. Letters and Lengths

B. & C. Solving Linear Eq'ns
with stick diagrams.

(i) x + 20 = 29
(ii) 2x + 5 = 20
(iii) 3x + 10 = 32
(iv) 5a + 16 = 3a+ 24

(v)  (½)x + 8 = 24½
(vI)  (¾)a + 16 = (¼)a+ 24
(vii) (¾)q + 17 = 32
(viii) 13 =[2/3]x +7 twice
(x) Animated Examples
(i) Integral Coefficients (A)
(ii) Integral Coefficients (B)
(iii) Fractional Coefficients

(iv) With Parameters

Problem Solving with Linear
Equations in one or many
unknowns, and in essentially 
one unknown - Symbols before
words. 


C. Solving Linear Eq'ns 
without
Stick Diagrams

D. Problems in 
essentially one unknown

E: 2D Systems - Sub Methods.
F. Larger Systems



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