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Main Category => Homework Board => Topic started by: Revival Jam on June 10, 2008, 02:32:29 am



Title: Mathamatic Formulas And Additional Help.
Post by: Revival Jam on June 10, 2008, 02:32:29 am
Mathamatic Formulas

Exponential Identities

Powers
x a x b = x (a + b)
x a y a = (xy) a

(x a) b = x (ab)

x (a/b) = bth root of (x a) = ( bth (x) ) a

x (-a) = 1 / x a

x (a - b) = x a / x b
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Logarithms

y = logb(x) if and only if x=b y
logb(1) = 0

logb(b) = 1

logb(x*y) = logb(x) + logb(y)

logb(x/y) = logb(x) - logb(y)

logb(x n) = n logb(x)

logb(x) = logb(c) * logc(x) = logc(x) / logc(b)

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Interest and Exponential Growth

The Compound Interest Equation

P = C (1 + r/n) nt

where
    P = future value
    C = initial deposit
    r = interest rate (expressed as a fraction: eg. 0.06)
    n = # of times per year interest in compounded
    t = number of years invested

Simplified Compound Interest Equation

When interest is only compounded once per yer (n=1), the equation simplifies to:

P = C (1 + r) t

Continuous Compound Interest
When interest is compounded continually (i.e. n --> ), the compound interest equation takes the form:
P = C e rt

Demonstration of Various Compounding
The following table shows the final principal (P), after t = 1 year, of an account initally with C = $10000, at 6% interest rate, with the given compounding (n). As is shown, the method of compounding has little effect. n P
1 (yearly) $ 10600.00
2 (semi-anually) $ 10609.00
4 (quarterly) $ 10613.64
12 (monthly) $ 10616.78
52 (weekly) $ 10618.00
365 (daily) $ 10618.31
continuous $ 10618.37

Loan Balance
Situation: A person initially borrows an amount A and in return agrees to make n repayments per year, each of an amount P. While the person is repaying the loan, interest is accumulating at an annual percentage rate of r, and this interest is compounded n times a year (along with each payment). Therefore, the person must continue paying these installments of amount P until the original amount and any accumulated interest is repayed. This equation gives the amount B that the person still needs to repay after t years.
B = A (1 + r/n)nt - P  (1 + r/n)nt - 1
--------------------------------------------------------------------------------
(1 + r/n) - 1 

where
B = balance after t years
A = amount borrowed
n = number of payments per year
P = amount paid per payment
r = annual percentage rate (APR)


Multiplication Tablex 0 1 2 3 4 5 6 7 8 9 10 11 12
0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6 7 8 9 10 11 12
2 0 2 4 6 8 10 12 14 16 18 20 22 24
3 0 3 6 9 12 15 18 21 24 27 30 33 36
4 0 4 8 12 16 20 24 28 32 36 40 44 48
5 0 5 10 15 20 25 30 35 40 45 50 55 60
6 0 6 12 18 24 30 36 42 48 54 60 66 72
7 0 7 14 21 28 35 42 49 56 63 70 77 84
8 0 8 16 24 32 40 48 56 64 72 80 88 96
9 0 9 18 27 36 45 54 63 72 81 90 99 108
10 0 10 20 30 40 50 60 70 80 90 100 110 120
11 0 11 22 33 44 55 66 77 88 99 110 121 132
12 0 12 24 36 48 60 72 84 96 108 120 132 144


Remembering 9's
What's 9 x 7 ? Use the 9-method! Hold out all 10 fingers, and lower the 7th finger. There are 6 fingers to the left and 3 fingers on the right.
The answer is 63!

If there is anything else,or forumals,please PM me.

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Title: Re: Mathamatic Formulas And Additional Help.
Post by: Supernovadelta on June 10, 2008, 02:45:41 am
In the
Code:
Logarithms

y = logb(x) if and only if x=b y
logb(1) = 0

logb(b) = 1

logb(x*y) = logb(x) + logb(y)

logb(x/y) = logb(x) - logb(y)

logb(x n) = n logb(x)

logb(x) = logb(c) * logc(x) = logc(x) / logc(b)


The logb,can also be changed in refrence to logb(p x) = n lopx (np)
The formula here tells you that the addition complexcity is defined as multiplication not devision.


Title: Re: Mathamatic Formulas And Additional Help.
Post by: Revival Jam on June 10, 2008, 02:50:33 am
In the
Code:
Logarithms

y = logb(x) if and only if x=b y
logb(1) = 0

logb(b) = 1

logb(x*y) = logb(x) + logb(y)

logb(x/y) = logb(x) - logb(y)

logb(x n) = n logb(x)

logb(x) = logb(c) * logc(x) = logc(x) / logc(b)


The logb,can also be changed in refrence to logb(p x) = n lopx (np)
The formula here tells you that the addition complexcity is defined as multiplication not devision.

In some ways yes,and you can always change the (PX) to any variable in the equation.


Title: Re: Mathamatic Formulas And Additional Help.
Post by: Ghost Rider on June 10, 2008, 03:12:07 am
This looks realy hard.


Title: Re: Mathamatic Formulas And Additional Help.
Post by: Y8razer on June 11, 2008, 12:40:00 am
Looks nice and clean,good job! ;D ;D ;D


Title: Re: Mathamatic Formulas And Additional Help.
Post by: Supernovadelta on June 11, 2008, 01:59:48 am
This looks realy hard.

Not realy,if you know your algebra.


Title: Re: Mathamatic Formulas And Additional Help.
Post by: History on June 20, 2008, 08:52:10 pm
This looks realy hard.

Not realy,if you know your algebra.
I agree,you got to know your algebra.


Title: Re: Mathamatic Formulas And Additional Help.
Post by: Revival Jam on August 09, 2008, 12:39:23 am
This looks realy hard.

Not realy,if you know your algebra.
I agree,you got to know your algebra.
If you want to succeed in life,I'll have to agree with you people.


Title: Re: Mathamatic Formulas And Additional Help.
Post by: thepure_ofswizy on January 27, 2010, 06:01:40 pm
lol whats with the multiplication table  :o