Delayed Exponential Function Learning

Introduction of delayed exponential function learning:-

Delayed exponential function learning is the new way for the students. Student does learning the exponential delayed function definition and also solves the example problems. In math exponential decay function means decrease the rate of a value or delayed the rate of the value. It is modulated by a differential equation.

`(dN)/(dt) = -lambda N`

Where,

N – quantity

`lambda` – positive number

This is also called as decay constant.

Basic Formula for Delayed Exponential Function Learning:-

In the following basic formula for delayed exponential function learning

`N = N_o e^(kt) ` , where k<0 br="">
Where

N = population

`N_0` = initial population

k = delay rate

t = time

Example Problems for Delayed Exponential Function Learning:-

Problem 1:-

Solve the delayed exponential function relation `y=3^-x` and use approximate value of y

1.      -1.2
2.      -2.2
3.      -3.2

Solution:

Given: `y = 3^-x`

Put the value x = -1.2

`y = 3^(-x)`

= `3^(-(-1.2))`

= `3^(1.2)`

= 3.737

Put the value x = -2.2

`y = 3^(-x)`

= `3^(-(-2.2))`

= `3^(2.2)`

= 11.21

Put the value x = -3.2

`y = 3^(-x)`

= `3^(-(-3.2))`

= `3^(3.2)`

= 33.63

Here y values is decreased and x values is increased. So this type of function is called as delayed exponential function. I have recently faced lot of problem while learning basic math word problems, But thank to online resources of math which helped me to learn myself easily on net.

Problem 2:-

Solve the delayed exponential function relation `y = 2^-x. -4<=x<=4 `

Solution:

Find the ordered pairs to satisfy the equation.

Put the value x = -4

`y = 2^(-x)`

`= 2^(-(-4))`

`= 2^(4)`

= 16

Put the value x = -3

`y = 2^(-x)`

`= 2^(-(-3))`

` = 2^(3)`

= 8

Put the value x = -2

y = 2^(-x)

= 2^(-(-2))

= 2^(2)

= 4

Put the value x = -1

`y = 2^(-x)`

`= 2^(-(-1))`

` = 2^(1)`

= 2


Put the value x = 0

`y = 2^(-x)`

`= 2^(-0)`

= 1

Put the value x = 1

` y = 2^(-x)`

`= 2^(-(1))`

` = 2^(-1)`

= 0.5

Put the value x = 2

`y = 2^(-x)`

`= 2^(-(2))`

`= 2^(-2)`

= 0.3

Put the value x = 3

`y = 2^(-x)`

`= 2^(-(3))`

`= 2^(-3)`

= 0.1

Put the value x = 4

`y = 2^(-x)`

`= 2^(-(4))`

` = 2^(-4)`

= 0.0625

When x is 0, y is 1. So, the y–intercept is 1.

Here y values is decreased and x values is increased. So this type of function is called as delayed exponential function.

3 Systems of Linear Equations

Introduction to 3 systems of linear equations:

In mathematics, a system of linear equations (or linear system) is a collection of linear equations involving the same set of variables.  A solution to a linear system is an assignment of numbers to the variables such that all the equations are simultaneously satisfied.

In 3 systems of linear equations, there are 3 unknown variables. We have to find all 3 unknown variables. The example problems for 3 systems of linear equations are given below which helps you to learn solving system of 3 equations.

(Source: Wikipedia)

Example Problem of Solving 3 Systems of Linear Equations: 1

Solve the following system of 3 equations:

x + 2y +3 z = 1

x + 3y + 4z = 3

x + 4y + 6z = 5

Solution:

Step 1: Given equations

x + 2y + 3z = 1 .............. (1)

x + 3y + 4z = 3 ........... (2)

x + 4y + 6z = 5 ............(3)

Step 2: Subtract equation (2) from equation (1) to eliminate x

x  +  2y  +  3z  = 1

x  +  3y  +  4z  = 3     ( - )

---------------------------------------

0  -    y    -   z  = - 2

---------------------------------------

We get,

- y - z = - 2

Multiply the above equation by -1,

y + z = 2 .................. (4)

Step 3: Subtract equation (3) from equation (2) to eliminate x

x  +  3y  +  4z   =  3

x  +  4y  +  6z   =  5     ( - )

---------------------------------------

0  -   y    -  2z  =  - 2

---------------------------------------

We get,

- y - 2z = - 2

Multiply the above equation by -1,

y + 2z = 2 .................. (5)

Step 4: Subtract equation (4) from equation (5) to z value

y  +  2z  =  2

y  +    z  =  2     ( - )

---------------------------------------

0  +   z   =  0

---------------------------------------

Therefore,

z = 0

Step 5: Plug z = 0 in equation (4) to get y value

y + z = 2 .................. (4)

y + 0 = 2

y = 2

Step 6: Plug y = 2 and z = 0  in equation (1) to get x value

x + 2y + 3z = 1 .............. (1)

x + 2(2) + 3(0) = 1

x +  4 + 0 = 1

x = 1 - 4

x = - 3

Step 7: Solution

x = - 3, y = 2, z = 0

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Example Problem of Solving 3 Systems of Linear Equations: 2

Solve the following system of 3 equations:

2x + 3y + z = 2

4x + 5y + z = 3

3x + 2y + z = 5

Solution:

Step 1: Given equations

2x + 3y + z = 2 .............. (1)

4x + 5y + z = 3 ........... (2)

3x + 2y + z = 5 ............(3)

Step 2: Subtract equation (2) from equation (1) to eliminate z

2x  +  3y  +  z  = 2

4x  +  5y  +  z  = 3     ( - )

---------------------------------------

- 2x  -  2y    +  0  = - 1

---------------------------------------

We get,

- 2x - 2y = - 1

Multiply the above equation by -1,

2x + 2y = 1 .................. (4)

Step 3: Subtract equation (3) from equation (2) to eliminate z

4x  +  5y  +  z   =  3

3x  +  2y  +  z   =  5     ( - )

---------------------------------------

x   +  3y  +  0  =  - 2

---------------------------------------

We get,

x + 3y = - 2 .................. (5)

Step 4: Multiply the equation (5) by 2 and subtract from equation (4) to get y value

2x  +  2y  =  1

2x  +  6y  = - 4

---------------------------------------

0   -  4y   =  5

---------------------------------------

Therefore,

y = - 1.25

Step 5: Plug y = - 1.25 in equation (4) to get x value

2x + 2y = 1 .................. (4)

2x + 2(-1.25) = 1

2x - 2.5 = 1

2x = 3.5

x = 1.75

Step 6: Plug x = 1.75 and y = - 1.25  in equation (1) to get z value

2x + 3y + z = 2 .............. (1)

2(1.75)x + 3(- 1.25) + z = 2

3.5 - 3.75 + z = 2

- 0.25 + z = 2

z = 2.25

Step 7: Solution

x = 1.75, y = - 1.25, z = 2.25

Representing Functions as Power Series

Introduction to representing functions as power series:

A power series in one variable is an infinite sequence of the structure,

f(x) = `sum_(n=0)^oo a_(n)(x-c)^(m+n) = a_(0)+a_(1)(x-c)^(1)+a_(2)(x-c)^2+...`  where an correspond to the coefficient of the nth expression, c is a constant, and x varies about c. This series usually occur as the Taylor series of some recognized function the Taylor series.

In several situations c is equal to zero, for instance when allowing for a Maclaurin series. In such cases, the power series obtain the simpler structure   f(x) = `sum_(n=0)^oo a_(n)(x)^(n) = a_(0)+a_(1)(x)+a_(2)(x)^2+...`

Representing Functions as Power Series:

Power series arise in combinatory in the name of produce functions in the name of the Z-transform. The identifiable decimal details for actual numbers recognize how to also be analysis as an example of a power series, with integer coefficients, but with the case x set at 1⁄10.

Power series are calculation a generality of polynomials as formal substance, wherever the number of expressions is allowed to exist unlimited. This involve give up the option to reserve subjective values for indefinite.

This analysis contrast with of power series, whose variables assign arithmetical values, and to series so only include a specific value if junction knows how to be recognized. Is this topic how many faces does a cylinder have hard for you? Watch out for my coming posts.

Examples for Representing Functions as Power Series:

Example 1:

How to solve representing function as power series `1/(1-x^2)`

Solution:

Step 1: the given function is `1/(1-x^2)`

Step 2: to evaluate the function is

`sum_(n=0)^oo(x^2)^n`

Step 3:   `|x^2| <1 br="br">
Step 4:   `|x|^2 <1 br="br">
Step 5: so the solution is `-1
Example 2:

How to solve representing function as power series `1/(1-9x^2)`

Solution:

Step 1: the given function is `1/(1-9x^2)`

Step 2: to evaluate the function is

`sum_(n=0)^oo(9x^2)^n`

Step 3:   `|9x^2| <1 br="br">
Step 4:   `|x|^2 <1 br="br">
Step 5: so the solution is `-1/3
Example 3:

How to solve representing function as power series `x/(4x-1)`

Solution:

Step 1: the given function is   `x/(4x-1)`

`x(1/(1-4x))`

Step 2: to evaluate the function is

`xsum_(n=0)^oo(4x)^n`

Step 3:     `xsum_(n=0)^oo(4)^n(x)^n`

Step 4:      `sum_(n=0)^oo(4)^n(x)^n-x`

Step 5:           `sum_(n=0)^oo(4)^n(x)^(n+1)`

So the solution is      `sum_(n=0)^oo(4)^n(x)^(n+1)`

Example 4:

How to solve representing function as power series `1/(1-16x^2)`

Solution:

Step 1: the given function is `1/(1-16x^2)`

Step 2: to evaluate the function is

`sum_(n=0)^oo(16x^2)^n`

Step 3:   `|16x^2| <1 br="br">
Step 4:   `|x|^2 <1 br="br">
Step 5: so the solution is `-1/4

Six Grade Probability

Introduction to Six Grade Probability:

Probability is a way of expressing knowledge or belief that an event will occur or has occurred. In mathematics the concept has been given an exact meaning in probability theory, that is used extensively in such areas of study as mathematics, statistics, finance, gambling, science, and philosophy to draw conclusions about the likelihood of potential events and the underlying mechanics of complex systems.

(Source: Wikipedia)

Example Problems for Six Grade Probability:

Six grade probability – Example: 1

Two coins are tossed. Find the probability of getting two heads.

Solution:

Step 1:

n (s) = {HH, HT, TH, TT}= 4

Step 2:

Tossing a coin with two heads:

n (a) = {HH}= 1

Step 3:

Formula:

P (A) = n(a)/n(s)

Answer:

P (A) = 1/4

Six grade probability – Example: 2

Three coins are tossed. Find the probability of getting two tails.

Solution:

Step 1:

n(s) = {TTT, TTH, THT, THH, HTT, HTH,HHT,HHH }= 8

Step 2:

There are 3 tosses with only two tails:

n (a) = { TTT, TTH, THT, HTT}=4

Step 3:

Formula:

P (A) = n(a)/n(s)

Answer:

P (A) = 4/8

P(A) = 1/2.

Six grade probability - Example: 3

When the die is rolled. What is the probability of occurring five?

Solution:

Total Number of possible = n (s) = {1, 2, 3, 4, 5, 6}

n (s) = 6

The number of outcomes n (a) = {5}

n (a) = 1

Formula:

P (A) = n(a)/n(s)

Therefore the probability of getting value = 1/6.

Six grade probability – Example: 4

When the two dice are rolled. What is the probability of occurring four or four?

Solution:

Total Number of possible = n (a) = { 1,1}{1,2}{1,3}{1,4}{1,5}{1,6}

{2,1}{2,2}{2,3}{2,4}{2,5}{2,6}

{3,1}{3,2}{3,3}{3,4}{3,5}{3,6}

{4,1}{4,2}{4,3}{4,4}{4,5}{4,6}

{5,1}{5,2}{5,3}{5,4}{5,5}{5,6}

{6,1}{6,2}{6,3}{6,4}{6,5}{6,6}

n (s) = 36

The number of outcomes n (a) = {1,4}{2,4}{3,4}{4,1}{4,2}{4,3}{4,4}

{4, 5}{4, 6}{5, 4}{6, 4}

n (a) = 11

Formula:

P (A) = n(a)/n(s)

Therefore the probability of getting value = 11/36. I have recently faced lot of problem while learning what is an acute angle, But thank to online resources of math which helped me to learn myself easily on net.

Practice Problems for Six Grade Probability:

1. Three coins are tossed and find the probability of all heads.

[Answer: P(A) = (1)/(8)]

2. Two dice are rolled then What is the probability of occurring six or six?

[Answer: (11)/(36)]

Metric System Distance

Introduction to metric system distance:
The distance is the unit used to represent how far two objects are apart. There are various units for the measurement of the distance like the international standard unit, metric unit, and other conventional units. The metric system of the distance measurement is one among them and the lengths are represented as the 10th multiples of the base unit. In the following article we will discuss in detail about the metric system of distance and the units in the metric system of distance. I like to share this Distance Time Formula with you all through my article.

More about Metric System Distance:
As described before, in the metric system of the distance the units of the distance are represented as the 10th multiples of the base unit in the metric system of distance. The base unit in the metric system of distance is the meter. The various units of the distance measurement in the metric system are,

1 Kilometer = `1000` meters

1 Hectometer = `100` meters

1 Decameter = `10` meters

1 Decimeter = `1/10` meters

1 Centimeter = `1/100` meters

1 Millimeter = `1/1000` meters

The above metric system distance relations can also be used for the conversion from the meters to the other units by reversing the above relations. Please express your views of this topic Frequency Polygon by commenting on blog.

Example Problems on Metric System Distance:

1. Convert the distance of 1.26 meters into centimeters and decimeters.

Solution:

1 Centimeter = `1/100` meters

1 meter = 100 centimeters

1.26 meters = 1.26*100 centimeters

1.26 meters = 126 centimeters

1 Decimeter = `1/10` meters

1 meter = 10 decimeters

1.26 meters = 1.26*10 decimeters

1.26 meters = 12.6 decimeters

2. Convert the distance of 0.25 kilometers and 0.75 decameters into meters.

Solution:

1 Kilometer = 1000 meters

0.25 Kilometer = 0.25*1000 meters

0.25 Kilometer = 250 meters

1 Decameter = 10 meters

0.75 Decameter = 0.75*10 meters

0.75 Decameter = 7.5 meters

Practice problems on metric system distance:

1. Convert the given distance of 14.56 decimeters into meters.

Answer: 145.6 meters

2. Convert the given distance of 0.98 kilometers into meters.

Answer: 980 meters.

3. Convert the given distance of 15600 millimeters into meters.

Answer: 1.56 meters

Standard Exponential Form

Introduction to Standard Exponential Form

Notation : `a^b` , where a is known as base and b is known as exponent which must be a positive and integral value. I like to share this Exponential Function Definition with you all through my article.

Explanation of `a^b` : The standard exponential form, `a^b` can mathematically be expressed as `a`   multiplied by itself `b` times. That is to say

`axxaxxaxxaxx.... b "times"`. This yeilds a single number if we know the value of a and b.

eg. Assuming a = 2 and b = 3 , we can express `a^b` = `2^3` = `2xx2xx2`  = 8(as b = 3 so we multiply a = 2 with itself 3 times). Note that the answer 8 results in single number.

Sample Examples of the Exponential Form:

1. `3^2` = `3xx3` = `9` (3 multiplied by itself 2 times)

2. `4^3` = `4xx4xx4` = `64` (4 multiplied by itself 3 times)

[Note: Here in standard exponential form, if we assume a and/or b are variables, they do not yeild any number. Instead they remain in the variable format. like

3. `x^3` = `x xx x xx x` (`x` multiplied by itself 3 times)

4. `x^a` = `x xx x xx x xx ... ` (`x` multiplied by itself a times) ]

[ Notes 2. If we take `b` as a decimal value such as 0.3, it comes under the section of nth root of a number, while here we are discussing how to compute standard exponential form ]

Abstracts of Standard Exponential Form:

Following are the rules and abstracts to calculate expressions involvint standard exponential forms.

1. `(a^b)^c` = `a^(bc)`

eg. a. `(2^3)^2` = `2^(3xx2)` = `2^6` = `2xx2xx2xx2xx2xx2` = 64

b. `(3^2)^4` = `3^(2xx4)` = `3^8` = `3times3times3times3times3times3times3times3`

Abstracts of Standard Exponential Form (continued)

2. `a^b xx a^c` = `a^(b+c)` (constraints : Both exponential form must have base that are equal)

`a^b//a^d` = `a^(b-d)` (For multiplication, both exponents are added, while for division, exponent for numerator is subtracted by exponent of denominator.Please express your views of this topic 7th grade math problems online by commenting on blog.

Examples :

1. `5^2xx5^3` = `5^(2+3)`= `5^5` = 3125

2. `3^5//3^2` = `3^(5-2)` = `3^3` = 27

3. `a^(-c) = 1/a^c` (when exponent is a negative number, one can make it positive by reciprocating the expression.)

such as `2^-3` = `1/2^3` = `1/8`

Example Problems on Standard Exponential Form

Can you compute these expressions

`3^5 = `
`x^3=`
`x^b=`
`(2^2)^4 = `
`2^5 xx 2^3 = `
`2^8//2^5 =`
`2^-3 = "(convert to positive exponential form)"`
Compute `(2^5xx2^7)/2^4`

Distributive property

The Distributive Property is one of the number properties. It says that when number is multiplied to an addition of two or more numbers, the result is the same as the sum of the products of the same number and each addend. To define distributive property algebraically, it is expressed in formula form as, a*(b + c) = a*b + a*c. That is, the number is ‘distributed’ to each of the addend and then the addition can be done. Thus, it can also be referred as distributive property of addition.But one must clearly understand that while a*(b + c) = a*b + a*c is true, a/(b + c) is ?  (a/b) + (a/c). Hence to stress this point some emphatically refer this property as ‘Distributive Property of Multiplication over Addition’.

Thus in general, the definition of distributive property is when a term is multiplied to the sum of group of terms, then the result is same as the sum of the products of the first term with each of the terms of the sum. It may be noted this property is applicable to subtraction of terms also, because in the general formula any of the three ‘a’. ‘b’ and ‘c’ can also be negative.

The distributive property greatly helps calculations and provides easier methods of solution. For example I need to multiply 51*101. If one tries the actual multiplication he/she has to take a paper and pencil and do the work. But the easiest way is by applying the property we discussed and it may be amazing to note that you can find the answer by mental calculation!. That is,51*101 = 51*(100 + 1) = 51*100 + 51*1 = 5100 + 51 = 5151 which is as fast as a calculator. Please express your views of this topic cbse 10th question papers by commenting on blog.

This property is also a great tool in factorization. In such cases, we use the property the other way round. That is we do, a*b + a*c = a*(b + c). For example let us take a quadratic expressionx2 + 6x + 8. Let us study how the distributive property helps in factoring.
Splitting the middle term, x2 + 6x + 8 = x2 + 4x + 2x + 8
Identifying the common factor x in the first two terms and the common factor 2 in the last two terms and using this property, x2 + 4x + 2x + 8 = x(x + 4) + 2(x + 4)
Again finding a common factor (x + 4) and applying the property once more,
x(x + 4) + 2(x + 4) = (x + 4)(x + 2)