Independent Events

Independents Events is a key word often referred to in probability calculations. Knowingly or unknowingly , we use Independents Events often in the theory and application of probability. Two events are independent events, if the occurrence of one event  does not affect the probability of other. I like to share this Definition of Independent Events with you all through my article.

While calculating probability of COMPOUND EVENTS ( having two are more events in a sample or test ), it is necessary to consider whether the events are  INDEPENDENT EVENTS or not. A simple case study may make things better understandable. If a basket has 4 balls in which 2 are black and 2 are white. If a ball is taken and kept aside and then second ball is taken, the probability for the second ball being black depends on whether we have taken a black or white in the first chance. thus the probability in the second chance is depending on the first case or event, thus they become dependent events.

Same example can be slightly modified to make it INDEPENDENT EVENT. If in the example mentioned above, if the ball is not kept aside but placed in the basket itself. then the probability of getting a black is same always. the event of taking a black ball becomes INDEPENDENT EVENTS.


Independent events - Explained


The types of events that we have discussed so far are all independent events. By independent we mean that the first event does not affect the probability of the second event.

the probability of  independent events is product of each event.

If A and B are not independent, then the probability of A and B is

P(A and B) = P(A) × P(B|A)

where P(B|A) is the conditional probability of B given A.

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Examples of Independent Events


1) The event of getting a 6 the first time a die is rolled and the event of getting a 6 the second time are independent

2) If two cards are drawn with replacement from a deck of cards, the event of drawing a red card on the first trial and that of drawing a red card on the second trial are independent.

3) Landing on heads after tossing a coin and rolling a 5 on a single 6-sided die.

4)Choosing a marble from a jar and landing on heads after tossing a coin.

5)Spinning a number 6 and then spinning a number 5 on the same spinner.

6)Picking a marble from a jar, then picking another marble after replacing the first one.

7)Picking a red marble from one jar and picking a red ball from another jar.

Finding The Y Intercept

Introduction to finding the x and y intercept:

Intercepts are the points at which   the graph of a function crosses either x axis or y axis.

In coordinate geometry, the y-intercept is the y-value of the point where the graph of a function or relation intercepts the y-axis of the coordinate system . That is when x=0, the line crosses the y axis.

The x intercept is a point at which the line crosses the x axis. That is when y=0, the line crosses the x axis.The x intercept is denoted as (x, 0).

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Method for finding the x and y intercept:

To find the y intercepts of a given function or line, substitute the value of x as “0".The y intercept can be written as (0, y).
For finding the x intercept of a given function, Substitute the value of y=0.The x intercept will be written as (x, 0).
x intercept and y intercept are two separate points.
At the origin (0, 0) only both x intercept and y intercept are same.

Problems on finding the x and y intercept :


1. Finding the x and y intercept for y = 5x-15.

Sol:

y=5x-15

To find y-intercept substitute x=0 in the above equation.
y = 5(0) -15

y = 0 -15

y = -15

So the y-intercept of a given equation is (0, -15)

For finding the x-intercept substitute y=0 in the above equation.
0=5x-15

5x-15=0

Add 15 on both sides,

5x-15+15=15

5x=15

Divide by 5 on both sides

x=3

So the x-intercept of a given equation is (3, 0)

2. Finding the x and y intercept for 4x+7y-16=0

Sol:

4x+7y-16=0

To find y-intercept substitute x=0 in the above equation.
4(0) +7y-16=0

7y-16=0

Add 16 on both sides,

7y-16+16=16

7y=16

Divide by 7 on both sides

y=16/7

So the y-intercept of a given equation is (0, 16/7)

For  finding the x-intercept substitute y=0 in the above equation.
4x+7(0)-16=0

4x-16=0

Add 16 on both sides,

4x-16+16=16

4x=16

Divide by 4 on both sides

X=4

So the x-intercept of a given equation is (4, 0)

3. Finding the x and y intercept for -5x=-y

Sol:

-5x=-y

To find y-intercept substitute x=0 in the above equation.
-5(0) =-y

0=-y

Y=0

So the y-intercept of a given equation is (0, 0). Here the origin is the y- intercept.

For finding x-intercept substitute y=0 in the above equation.
-5x=-(0)

5x=0

Divide by 5 on both sides

x=0

So the x-intercept of a given equation is (0, 0).Here the origin is the x- intercept.

4. Finding the x and y intercept for 9y=3x+7

Sol:

9y=3x+7

To find y-intercept substitute x=0 in the above equation.

9y=3(0) +7

9y=0+7

9y=7

Divide by 9 on both sides

Y=7/9

So the y-intercept of a given equation is (0, 7/9)

For finding the x-intercept substitute y=0 in the above equation.
9(0)=3x+7

0=3x+7

Subtract 7 on both sides,

-7=3x+7-7

-7=3x

Divide by 3 on both sides

-7/3=x

x=-7/3

So the x-intercept of a given equation is (-7/3, 0)

5. Finding the y intercept for 8y= (7/3) +x

Sol:

8y= (7/3) +x

To find y-intercept substitute x=0 in the above equation

8y= (7/3) +0

8y=7/3

Divide by 8 on both sides,

Y=7/24

So the y-intercept of the given equation is (0, 7/24).

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Practice problems on finding the x and y intercept :


Finding the y-intercept  and x for the following equations.

1.  8x-5y=-6

2.  7y= (3/2) x+ (5/2)

3.  4x+8y-64=0

4.  8y=-6x

5.  Y=9x+3



Answer key:

1. (0, 6/5) and (-3/4,0)

2. (0, 5/14) and (-5/3,0)

3. (0, 8) and (16,0)

4. (0, 0)

5. (0, 3) and (-1/3,0)

Division Math Terms

Math Terms: Division

In math terms division is defined as equal sharing of things. Division method used to separate the equal things or parts or groups. Division can be denoted as '/' or '÷' symbol. Division is a inverse of multiplication. For example, if ‘4’ times ‘5’ equals ‘20’, written: 4 x 5 = 20. Then 20 divided by 5 equals 4, written: 20 / 5 = 4.

In math, terms involved in division has special name:

20 (dividend) ÷ 5 (divisor) = 4 (quotient).

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The basic rules for division are given as below:

Division of two numbers are same sign we get the answer  Positive
Division of two numbers are different sign mean we got the negative answer
Division of two numbers is same sign:

Positive number(+4)/ positive number(+2)=  (+2)positive number (+/+=+)
Negative number(-4) / negative number (-2)= (+2)positive number(-/-=+)
Division of two numbers is different sign:

Positive number(+4)/ negative number(-2)=(-2) negative number(+/-=-)
Negative number(-4)/ ÷ positive number(+2) =(-2)negative number(-/+=-)
Examples:

50 ÷ 5 = 610(same signs)

(-36) ÷ (-6) =+6  (same signs)

14 ÷ (-2) = -7 (different signs)

(-12) ÷ 3 = -4 (different signs)


Math Terms: Division - Examples


Example 1: There are 25 chocolates, and 5 children want to share them equally, how do they divide the chocolates?

Solution:

= Total number of chocolates / total members

= 25 ÷ 5 = 5

Therefore each member should get 5 chocolates.

Example 2: How many minutes are there in 720 seconds?

Solution:

x × 60 = 720 seconds where x represents number of minutes.

Just divide 720 by 60

720 ÷ 60 = 12, therefore there are 12 minutes in 720 seconds.

Example 3: In one day, a movie theater sells tickets for 7525 dollars. Each ticket costs 25 dollars. How many people purchased a ticket?

Solution:

x × 25 = 7525 where x represents the number of people who purchased a ticket.

7525 ÷ 25 = 301, therefore 301 people purchased a ticket.

Example 4: How many meters are there in 640 centimeters?

Solution:

We know, 1 meter = 100 centimeters

x × 100 = 640 centimeters where x represents number of meters

Just divide 640 by 100

640 ÷ 100 = 6.4, therefore there are 6.4 meters in 640 centimeters.


Example 5: Fleming has 50 ice creams. If he gave his friends 5 ice creams each, how many friends can he share his ice creams with?

Solution:

50 ice creams ÷ 5 for each friend

50 ÷ 5 = 10 friends

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Math Terms: Division – Practice problems:


Problem 1: Harry has 121 ice creams. If he gave his friends 11 ice creams each, how many friends can he share his ice creams with?

Answer: 11

Problem 2: How many meters are there in 480 centimeters?

Answer: 4.8 meters

Problem 3: In two days, a movie theater sells tickets for 4248 dollars. Each ticket costs 12 dollars. How many people purchased a ticket in two days and find the average of one day.

Answer: 354, 177

Basic Fraction Math

Introduction to basic fractions math:

In algebra, fractions are commonly used in mathematics, science and the world something like us. They have been used by human ruler while the time of the earliest Egyptians and are still used by public to solve problems in their daily lives.

In basic fractions math came into frequent use in the 16th century. Look at this basic fraction:

`5/10`

Observe that each of the digits from 1 to 9 has been used in this fraction in math. When this basic fraction math is simplified, we see that:

`5/10`=`1/2`


Types of basic fractions math:


It is very important and dissimilar types of fractions in algebra.

Proper fraction in math.

Improper fraction in math.

Mixed fraction in math.

Proper fraction in math:

Fraction, where the denominator is bigger than the numerator.

For example: `3/5`

Here numerator (3) is smaller than the denominator (5)

Improper fractions in math:

Fraction, where the denominator is less than the numerator

For example: `3/2`

Here numerator (3) is greater than the denominator (2)

Mixed fractions in math:

Fractions, where there is a number merge with a fraction.

For example:3`1/5`(means 3+`1/5`)

We change in mixed fraction to an improper fraction as follows:

3`1/5` =(3x5)+`1/5`=`(15+1)/(2)`=`13/2`

We change in an improper fraction to a mixed fraction as follows:

`5/3` means there are three halves (1 x`2/3` ).

This means the same as 1+ `2/3` = 1`2/3`

Basic algebraic fractions in math:

Fractions, which include algebraic terms in their numerators and denominators.

For example: `(2ab)/5` ;`3/y`


Example for basic fractions math:


It takes `1/3` hour to get ready a salad for breakfast, another `1/4` hour to make a main dish and `1/4` hour to make dessert. Can the entire snack be made in no more than 1 hour? Explain how you can decide.

Solution:

Step 1:

Think about what the problem asks you to find:

“No more than 1 hour” means 1 hour or less.

Is the addition of `1/3`+`1/4`+`1/4` less than or greater than 1?

Step 2:

Use estimation to solve the problem:

`1/3`+`1/4`>`1/4`

It follows that `1/3`+`1/4`+`1/4`>1.

Answer: so, the complete breakfast cannot be ready in 1 hour.

Math Prime and Composite

Introduction:

In math, a number is described as a arithmetical thing and used to measure and calculate. Also it can name as numeral and contains zero, off-putting (negative) numbers, rational numbers, irrational numbers, and so on. The formula of numerical function consist of one or more numerical as input and produce its related numerical output. This method involves arithmetic function for example addition, subtraction, multiplication, division, and exponentiation. I like to share this Laplace Transform Calculator with you all through my article.



Definition for math Prime and Composite numbers:


Math Prime number:

A natural number other than 1 divisible only by itself and 1 or any natural number which has precisely two divisors. The following numbers are prime 2, 3, 7, 11, 13, .... Every natural number greater than 1 may be resolved uniquely into a product of prime numbers example: 180 = 2 x 2 x 3 x 3 x 5. In the case of a prime number p, the product has to be interrupted as p itself

Math Composite number:

It is an integer that has more than one prime factors. They can be expressed as unique set of prime numbers. The first composite number is 4. Besides 1 every other number is either prime or composite number. Besides 1 each other natural number is either prime or composite number. Having problem with solving proportions with fractions keep reading my upcoming posts, i will try to help you.


Problem for Math Prime and Composite numbers:


Problem-1:

Write the prime numbers up to 500

Solution:

Prime numbers up to 500

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499.

Problem-2:

what are the first 100 composite numbers

Solution:

The first 100 composite numbers are

4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48,  49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82,  84,  85,  86,  87, 88,  90,  91,  92,  93,  94,  95, 96, 98, 99, 100, 102, 104, 105, 106, 108, 110, 111, 112, 114, 115,  116, 117, 118, 119, 120, 121, 122, 123,  124,  125, 126, 128, 129,  130, 132, 133.

Angle Bisector Theorem Proof

Exterior angle bisector theorem proof :

Statement :   THE EXTERNAL  BISECTOR OF AN ANGLE OF A TRIANGLE  DIVIDES THE OPPOSITE SIDE  EXTERNALLY  IN THE RATIO OF SIDES CONTAINING THE ANGLE.

GIVEN :  a triangle ABC   in which AD  is the bi sector of  the exterior angle A  and intersects  BC produced in D

REQUIRED TO PROVE : BD /CD = AB /AC

CONSTRUCTION ; draw   CE parallel to  DA  meeting AB  in   E

PROOF ;  since  CE parallel to DA  and  AC   intersects  them

angle  CAD =  angle ACE   (alternate angles)

also CE is parallel to  DA  and    BK  intersects them

so   angle KAD = angle AEC   (corresponding angles)

therfore    angle  ACE  = angle AEC

thus in  triangle ACE     angle  ACE = angle AEC

SO   AE = AC (  sides opposite to equal angles are equal )

in triangle BAD   EC is parallel to AD

THERE FORE        IF A LINE DRAWN PARALLEL TO ONE SIDE OF A TRIANGLE IT DIVIDES THE OTHER TWO SIDES IN THE SAME RATIO.                     BD/CD = BA/EA

BD/CD=AB/AE        reason  BA=AB  AND     EA= AE

BD/CD = AB/AC      reason  AE= AC

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Exterior angle bisector theorem proof : Example

in triangle ABC  AE is  the bisector of exterior angleCAD   meeting  BC  produced  in E  if  AB = 10 CM  AC =6 CE BC = 12 CE  find CE ?

solution;     according to exterior angle bisector theorem   BE / CE= AB/AC

let  CE = X

(  12+X ) / X =  10 / 6  CROSS MULTIPLECATION

5X = 36 + 3X

5X - 3X = 36

2X = 36

X=  36/2

X = 18

CE =  18 cm

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ASSIGNMENT

AE is the bisector of the exterior angle CAD  meeting BC  produced  in E  if AB = 12 CM AC = 6 CM  AND BC = 14 cm  find CE ?

LET   CE =  X

BE / CE  =AB / AC  (according to external angle bisector theorem )

14 + x /  x  = 12 / 6

14 + x  / x =    2

by  cross multiplection

2x - x =  14

x =   14

CE = 14

Frayer Model Math

Frayer model:

Frayer model consisting of four parts which gives the definition for the object, some facts about the object, examples, and non-examples for the particular object.  Frayer model gives the clear understanding of the object. Flayer model is also known as map or chart. There are five parts available in the Flayer model map. They are:
Concept word,
Definition,
Characteristics of the concept word,
Example’s for the concept word, and
Non-Example’s for that concept word. Please express your views of this topic calculus tutor work at home by commenting on blog.

It can be used for all the subjects. So, the student uses the Frayer model in math to analyses the concepts.

Framework of Frayer model in Math:

The Frayer model is a way of concept map. The framework of the Frayer model includes: the concept word at center of the map, the definition for the concept word at top of the model, characteristics of the concept word next to the definition part of the Frayer model, example for the concept word on down side of the definition for the concept word, non-example for the concept word on the other side of the examples for the concept word. Both examples and non-examples help the students to have a clear idea on the concept word. In math the Frayer model used to give the key ideas for given word and makes easy to understand by the students. Frayer model will lead students to a deeper understanding of a word and its relationship for the given keyword in math. I have recently faced lot of problem while learning Median Calculator, But thank to online resources of math which helped me to learn myself easily on net.


Sample of Frayer model in math:

The Frayer model is a way of concept map. The framework of the Frayer model includes: the concept word at center of the map, the definition for the concept word at top of the model, characteristics of the concept word next to the definition part of the Frayer model, example for the concept word on down side of the definition for the concept word, non-example for the concept word on the other side of the examples for the concept word. Both examples and non-examples help the students to have a clear idea on the concept word. In math the Frayer model used to give the key ideas for given word and makes easy to understand by the students. Frayer model will lead students to a deeper understanding of a word and its relationship for the given keyword in math.

Technical Math Problems

Introduction to technical math problems:

Mathematics is basically the study about quantity, structure, space, change, etc... It is the systematic study of the shapes and motions of physical objects. Mathematics is applied widely in all areas. Mathematics has been used throught the world which is a basic and essential tool for all fields .In this article we shall discuss about technical math problems. Understanding Do Math Problems is always challenging for me but thanks to all math help websites to help me out.


Technical math Problems:


The problems which are derived  from mathematical techniques are known as technical math problems. There are various techniques related with mathematics .Let us see some math problems which are solved technically by various techniques.


Technical Problem 1:


A clock is started at noon. By 20 minutes past 5, the hour hand has turned

Through________ degrees

Explanation:

Angle traced by hour hand in 12 hrs = 360 degrees

Angle traced by hour hand in 5 hrs 20 min ie)  16/3hrs=((360/12)*(16/3))

=160 degrees

Technical  Problem2:

If a quarter kg of potato costs $1200 , how many paise will 200 gm cost?

Explanation:

Let the required weight be X kg.

hence   250 : 200 :: 80 :  X      250 *X= 200*160

X=200*160/250

X= $96

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Technical Problem 3:


Mike started a business investing the amount about seven thousand rupees. Lincoln joined him after six   months with  an amount of  $10500and Sami joined them with fourteen thousand after another  six months. The profit should be shared in what ratio for Mike, Lincoln and  Sami respectively, 3 years after Mike started the business?

Explanation:

Mike : Lincoln: Sami = (36 x 7000) : (10500 x 30) : (14000 x 24) = 12 : 15 : 16.

Technical  Problem 4:

Adam can do a certain work in the same time in which Bollinger and Clarke together can do it. If Adam and Bolllinger together could do it in 10 days and Clarke alone in 30 days, then Adam.Bollinger and clarke combined together and they could do it in ____days:

Explanation:

Let us take Adam as A,Bollinger as B and Clarke as C

(A + B)'s 1 day's work=1/10

C's 1 day's work=1/30

(A + B + C)'s 1 day's work =((1/10)+(1/30))=4/30……(1)

A's 1 day's work = (B + C)'s 1 day's work ………….... (2)

From (1) and (2), we get: 2 x (A's 1 day's work)

So, Adam.Bollinger and clarke combined together and they could do it  in 7and half a days.

Advanced Algebra Problems

Introduction Advanced algebra problems:

Advanced Algebra and Basic Algebra systematically develops the concepts and tools in algebra that are vital to every mathematician, whether pure or applied, aspiring or established. However, the books give the reader as a global view of algebra and its role in mathematics as a whole.

Advanced Algebra is used in a forward-looking way that takes into account the historical development of the subject. It is suitable for the more advanced parts of a two-semester first-year graduate sequence in algebra. It is requires as the reader only a familiarity with the topics developed in Basic Algebra regarding those problems.


Advanced algebra problems Binomial Theorem:


The Binomial Theorem is a quick way of expanding (or multiplying out) a binomial expression that has been raised to some (generally inconveniently large) power. For instance, the expression (3x – 2)10 would be very painful to multiply out by hand. somebody see  a formula for this expansion, and where we can plug the binomial 3x – 2 and the power 10 into that formula to get that expanded (multiplied-out) form.

The general expression of the Binomial Theorem is as follows: served

n

(a+b)^n = ?    (n k) a^n-k b^k

K=0

(n k)=nCk = n! / ((n-k)! k!)

Then the factorial notation represented as "n!" means “the product of all the whole numbers between 1 and n", so, for instance, 6! = 1×2×3×4×5×6. Then the notation "10C7" (often pronounced as "ten, choose seven") means:

10C7 = 10! / ((10-7)! 7!)

= 10! / (3!7!)

= (1.2.3.4.5.6.7.8.9.10)/(1.2.3.4.5.6.7)

= 4.3.10

= 120

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Example for advanced algebra problems


Here we will explain the algebra problem for three variable system of linear equations.

Problem:

Solve x+y+z=10;  2x-y+z=2;  -x+2y-z=5

Solution:
x + y + z = 10                       (1)
2x - y + z = 2                        (2)
-x + 2y - z = 5                       (3)

equating(1) and (3) equation, we get

x + y + z = 10
-x + 2y - z = 5

3y = 15
y = 5

Substitute y=5 in equation (1)
x + 5 + z = 10
x + z = 5
z = -x + 5                 (4)
Substitute y=5 in equation (2)
2x - 5 + z = 2
2x + z = 7
z = -2x + 7                (5)
by equating (4) & (5) we get,
-x + 5 = -2x + 7
x  = 2
z  = -2(2) + 7

= -4 + 7

= 3

Answer:
x = 2
y = 5
z = 3

Advanced Calculus Homework

Introduction to Advanced calculus homework:

Calculus was first discovered  by Isaac Newton and another mathematician named Gottfried Leibniz.Calculus is concerned with comparing quantities which differ in a non-linear way. It is used widely in science and engineering since many of the things we are studying (like velocity, acceleration and current in a circuit) do not perform in a simple, linear fashion. If quantities are recurrently changing, we need calculus to study what is going on. I like to share this Integral Power Rule with you all through my article.


Two branches of advanced calculus homework:-

For the homework on advance calculus, it is necessary to know about the two main branches of calculus. They are:

1) Differential Calculus.

2) Integral Calculus.

Differential calculus:-

In advance calculus, the term Differential calculus is used to determine the rate of change where the function is known.

Integral calculus:-

In advance calculus, the term Integral calculus is used to determine function where the rate of change is known to us. Understanding Quartile Calculator is always challenging for me but thanks to all math help websites to help me out.


Advanced Differential Calculus Homework Problem:-


The advanced calculus homework problems are given below:

Problem :- 1

Differentiate the function y = (1+arctanx)/ (2-3arctanx)

Solution:-

The given equation is

y = (1+arctanx)/ (2-3arctanx)

y’ = { (2-3arctanx) D{1+arctanx} – (1+arctanx){2-3arctanx} } / (2-3arctanx)2

= (2-3arctanx) {1/(1+x^2)} - (1+arctanx)(-3) {1/(1+x^2)} / (2-3arctanx)2.

= { (2 – 3arctanx )/(1+x^2) – (1 + arctanx ) (-3)/(1+x^2) } * 1/ (2-3arctanx)2

= (2-3arctanx+3+3arctanx ) / (1+x^2) * 1/ (2-3arctanx)2

= 5 /  (1+x^2)(2-3arctanx)2



Problem :- 2

Differentiate  f(x) = 2x +10 arc cot x solve f(x) = 0 for x.

solution :-

Given function is f(x) =2x +10 arc cot .

f’(x)   = 2+ 10` ((-1)/(1+x^2)).`

= 2 - 10 / (1+x^2)

=2(1+x^2)/(1+x^2) - 10/(1+x^2)

= (2 + 2x^2-10) / (1+x^2)

= (2x^2  - 8)/(1-x^2)

=2(x^2- 4)/(1+x^2)

= 2(x-2)(x+2)/(1+x^2).

It is the fact that if A/B =0 then A = 0 thus

2(x-2)(x+2) = 0

It is the fact that if AB = 0 ,then A=0 or B = 0,it follows

x - 2 = 0 or x + 2 = 0

That is the only solution to f(x) = 0 are

x = 2 or x = - 2


Advanced Integral Calculus Homework Problem:-


Problem 1

Integrate the following equation:-

? 1/(1+?x) dx

Solution:-

The given equation is ? 1/(1+?x) dx

Now use the power of substitution

x = u2

so that

?x = ?u2 = u.

and

dx = (2u) du.

Plugin it into the problem, replacing all forms of x, getting

? 1/( 1+?x) dx   = ?1/(1+u) (2u) du.

= ?2u /(u+1) du.

= ?(2 - 2/(u+1)) du.

= 2u - 2 ln ?u+1? + c.

= 2?x -2ln ??x+1? + c.

adding decimals learning

Introduction to learning adding decimals :

The decimal numeral system (also called base ten ) has ten as its base. It is the numerical base most widely used by modern civilizations. Decimal notation often refers to the base-10 positional notation such as the Hindu-Arabic numeral system; however it can also be used more generally to refer to non-positional systems such as Roman or Chinese numerals which are also based on powers of ten.

Adding decimals learning is very similar to adding whole numbers. The most important point to remember while adding decimals is to line up all the decimal points

Write down the numbers, one under the other.
Add zeros at the end so the numbers have the same length
Then add normally, remember to put the decimal point in the answer.

learning Adding decimal :

Learning of adding decimal numbers:

As usual in addition, start from the right, and add each column in turn.

The numbers we are adding do not have the same number of digits to the right of the decimal point, we still have to line up the decimal points before adding decimal values. If there is a carry while adding (that is, if a column adds up to more than, we remember to add the tens digit of that column to the next column.

Adding decimals is same like the simple addition. Adding decimal is simple one for all the decimal place such as two decimal place, three decimal place etc.

Adding decimals:

Line up the place values and decimals. Arrange the decimals so that the decimal points align. Add zeros as needed to make equivalent decimals. Then place the decimal point in the sum (straight down).

Adding as with whole numbers.

Learning step by step for adding decimals:

Step 1: Write the numbers in a column. Line up the decimal points.

Step 2: For whole numbers, place a decimal point after the number. Then add one or more zeros to the right side of the decimal point.

Step 3: Add (or subtract) the numbers the same way as for numbers without decimal points.

Step 4: Transport the decimal point directly down into the answer.


learning Examples for decimal addition


Example 1:

Add the two decimal values i) 37.3 and 53.6 ii) 41.6 and 34.5


Example 2:

Add the two decimal numbers 39.344 and 63


learning Practice problems for adding decimal:

1) Add the decimal values 132.09 and 32.70            Answer: 164.79

2) Add the decimal values 65 and 43.05                    Answer: 108.05

Undefined Slope

Introduction:

Slope is defined as the alteration in the height of the horizontal distance between two points on a line. The condition for undefined slope is used when the denominator value of the slope becomes 0. Any number divisible by zero gives infinity, therefore the slope is defined as undefined slope. The chart for the undefined slope looks like vertical line. There is no existence of y axis or y coordinate value. Through online learning, students can get help with all math problems. I like to share this Find Volume of a Cylinder with you all through my article.

Formula for Undefined Slope:

The required state for the slope to be undefined slope in equation is, the denominator value of the slope becomes zero therefore the slope becomes undefined.

The formula for finding the slope is given as,

`"m = (y_2-y_1)/(x_2-x_1)`

Once the value of `x_2-x_1 = 0` , then the slope is said to be undefined. The undefined slope can also be defined as, when the value of x-axis or coordinate is the same for both points, the slope is undefined. Understanding Trig Identities Solver is always challenging for me but thanks to all math help websites to help me out.

Example Problems:

Example 1:

Find the slope from the given points (6, 3), (6, 10).

Solution:

The formula for calculating the slope is given as,

`m = (y_2-y_1)/(x_2-x_1)`

`m = (10-3)/(6-6)`

`m = 7/0`

Any number divided by 0 becomes infinity or undefined therefore the slope is undefined.

Example 2:

Find the slope from the given points (7, 4), (7, 5).

Solution:

The formula for calculating the slope is given as,

`m = (y_2-y_1)/(x_2-x_1)`

`m = (5-4)/(7-7)`

`m = 1/0`

Any number divided by 0 becomes infinity or undefined therefore the slope is undefined.

Example 3:

Find the slope from the given points (-4, -6), (-4, -2).

Solution:

The formula for calculating the slope is given as,

`m = (y_2-y_1)/(x_2-x_1)`

`m = (-2-(-6))/(-4-(-4))`

`m = (-2+6)/(-4+4)`

`m = 4/0`

Any number divided by 0 becomes infinity or undefined therefore the slope is undefined.

Plug in Math Problem

Introduction for plug in math problem:

We are going to see about the topic of math in plug concept with some examples and their problems.  Mathematics provides the easiest way to deal with the concept of plug.  The plug concept is used in all branches of math like algebra, trigonometry, and integral calculus and so on. Understanding Trigonometric Functions Graphs is always challenging for me but thanks to all math help websites to help me out.

Example Problems for Plug in Math

Example problem for plug in math 1:

Solve: 2(8x - 12) + 3 (5x + 8) if the value of x is 4.

Solution:

The given equation is 2(8x - 12) + 3 (5x + 8)

Applying the value of x in the given above equation

= 2(8(4) - 12) + 3 (5(4) + 8)

= 2(32 - 12) + 3 (20 + 8)

= 2(20) + 3 (28)

= 40 + 84

= 124

Answer: 124.

Example problem for plug in math 2:

Find out the value of f (a) = 15a + 5, if the value of ‘a’ is 1, 2, and 3.

Solution:

The given equation is f (a) = 15a + 5

Substitute the ‘a’ value in the given equation

Now, substitute the value of a is 1

f (1) = 15(1) + 5

= 15 + 5

= 20.

f (1) = 20

Now, substitute the value of a is 2

f (2) = 15(2) + 5

= 30 + 5

= 35.

f (2) = 35

Now, substitute the value of a is 3

f (3) = 15(3) + 5

= 45 + 5

= 50.

f (3) = 50

The answer is

a = 1; f (a) = 20

a = 2; f (a) = 35

a = 3; f (a) = 50

Example problem for plug in math 3:

Prove that the condition is true (a + b)2 = a^2 + b^2 + 2ab.  The value of a is 15 and b is 12.

Solution:

Given (a + b)^2 = a^2 + b^2 + 2ab

Substitute the value of a = 15 and b = 12 in the given formula.

First doing in the left side

(15 + 12)2                                                                                                        (15 + 12 = 27)

(27)2 = 729                                                                              (the value of 272 is 729)

(a + b)2 = (15 + 12)2 = 729

Now, doing in the right side

= 152 + 122 + 2 (15) (12)                                                          (The value of 152 is 225 and 122 is 144)

= 225 + 144 + 360

= 729

a^2 + b^2 + 2ab = 152 + 122 + 2 (15) (12) = 729.

729 = 729

Hence the proof (a + b)2 = a^2 + b^2 + 2ab. Having problem with The Simplex Method keep reading my upcoming posts, i will try to help you.

Practice Problems for Plug in Math

Practice Problem 1:

Find out the value of f (x) = 100x + 50, if the value of ‘x’ is 5, 10, and 15.

Answer:

x = 5; f (x) = 550

x = 10; f (x) = 1050

x = 15; f (x) = 1550.

Practice Problem 2:

Find out the value of f (p) = p (p - 1) + 10, if the value of p is 4.

Answer: 22