Math Operations

Operator in math is a function which acts on two values ( operands). These operators are known as binar operators as they need two operands. There are four binary operators namely +  ,  -   ,  x   , /. Operator are mainly consists of number of terms. For example, the assignment operator are used for assigning the values to a particular variable.

The four fundamental operations on numbers are Addition, Subtraction, Multiplication and Division. The operator of Addition is ‘+’, Operator of Subtraction is ‘-’, Operator for multiplication is ‘x’ and Operator for division is ‘÷’. Let us see about arithmetic operations with appropriate operator.
Explanation for various types of operator statement:

1. '+' is an  operator statement:

For this operator, the addition operation and the string concatenation operation are included.
Addition operation is used for adding the two values that are given.
String concatenation are used for joining the two statements.

2. ++ is an increment operator:

For this, the values are incremented according to the given values.

3. -- is an decrement operator:

For this, the values are decrement according to the given values.

4. x  is a multiplication operator:

I like to share this Mode Math Definition with you all through my article.

For this, the multiplication function and the pointer function are included.
Multiplication function is used to multiply the two given values.
Pointer are used for denoting the reference variable.

5. / is a division operator:

For this, the values are divided for getting the remainder and the quotient factor.

6.  ^  exponential operator:

Exponential operator is used for raising the power to a function.

7. -  mathematical operator:

For this, the subtraction and the negation process are included in this operator statement.
Subtraction operation is used for finding the difference of the given two numbers.


Examples using math operations:

Addition Operator ‘+’: For adding two or more numbers we use the ‘+’ operator. This operator is known as plus.Normally additions have two operands and one operator. For example x + y here ‘x’ and ‘y’ is called as operator and the symbol’+’ is called as operand.

Ex: Add 5 and 6.

Sol:

5 + 6 = 11

Where, ‘+’ is the operator of addition.

5 and 6 are operands.

11 is the resultant of the operation of addition.

Subtraction Operator ‘-’: For subtracting two or more numbers, the operator used is ‘-’.  This operator is known as minus.

Ex : Subtract 11 and 5.

Sol:   11 – 5 = 6.

Where, ‘-’ is the operator of subtraction.

11 and 5 are operands.

6 is the resultant of the operation of subtraction.

Multiplication Operator ‘x’: For Multiplying two or more numbers, the operator used is ‘x’.  This operator is known as into.

Ex : Multiply 12 and 5.

Sol:             12 x 5 = 60.

Where, ‘x’ is the operator of multiplication.

12 and 5 are operands.

60 is the resultant of the operation of multiplication.

Division Operator ‘÷’: For dividing two or more numbers, the operator used is ‘÷’.  There is no special name for division operator.

Ex : Divide 10 by 5.

Sol:   10 ÷5 = 2.

Where, ‘÷’ is the operator of Division.

10 and 5 are operands.

2 is the resultant of the operation of division.

Additional Operators:

‘√’:

This operator is known as the square root.

Ex: Find Square root of 4?

Sol: √4 = 2.

Where, ‘√’ is the operator of square root.

4 is the operand.

2 is the resultant of the operation of square root.

Order of operations in math:

The order of operation is square root, division, multiplication, Subtraction and addition.

Ex: Calculate 5 + 6 – [(√4 ÷2) x 3].

Sol:  5 + 6 – [(2 ÷ 2) x 3]

5 + 6 – [1 x 3]

5 + 6 – 3

5 + 3

7

Steps and examples for addition operation in math:

Simple addition uses the following steps to adding the numbers.

Step 1: Add the first place number from the right (first number) first after that take over the carry to next step.

Step 2:   Add the ten’s place number (means second digit) as well as also add the carry from the first step. Now note down the answer and over the carry.

Step 3  :Carry on this process on until reach the end place of the given number.

The followings are the some of the types of addition

Add single digit number with single digit
Add single digit number with double digit
Add three one digit numbers
Add double digit number with three digit number
Add double digit number with double digit number
Add treble digit number with treble digit number etc….

Example for Simple Addition :

Add single digit number with single digit:

Ex: Add 5 with 9

Sol:

5
(+) 9
-----------     14

Ex 2: Adding the following three one digit numbers.

9, 6 and 3

Sol:

Given numbers 9, 6 and 3

Step 1:

Add the first two numbers

9

(+)      6

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

1 5

Step 2:

Add the next number with the previous step sum or total.

Previous step total =15

Third number =3

1 5

(+)        3

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

1 8

Therefore 9 + 6 +3 =18

Algebra is widely used in day to day activities watch out for my forthcoming posts on solve this math problem for me and cbse guide for class 10. I am sure they will be helpful.

Ex 3: Add the following two numbers: 54 with 16

Sol:

Given numbers 54 and 16

Step 1:

Add the first digit (ones place) numbers (4+ 6 = 10) then over the carry (here 1 is the carry)

5 4
(+)   1 6
-----------
0

Step 2:

Add the second digit (tens place) numbers (5 +1 = 6) the add with previous step carry (6 +1 =7)

5 4
(+)   1 6
-----------
7 0

Finally we get 54 +16 = 70

Ex 4: Add the following two numbers: 15948 and 69741

Sol:

Given 15948 and 69741

1 5 9 4 8
(+)     6 9 7 4 1
------------------

Step 1:

Add the one place number then over the carry (8+1=9 carry =0)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
9

Step 2:

Add the tens place number then over the carry (4+4 = 8 carry 0)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
8 9

Step 3:

Add the hundreds place number then over the carry (9+7 =16 carry 1)

1 5 9 4 8
+)    6 9 7 4 1
------------------
6 8 9

Step 4:

Add the thousands place number (5+9 = 14) then add with carry from the above step (14 +1 = 15 carry 1)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
5 6 8 9

Step 5:

Add the ten thousands place number (1+ 6=7) then add with previous step carry (7 + 1 =8)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
8 5 6 8 9

Therefore 15948 + 69741 = 85689

Ex 5: Add 72 with 4.12

Sol:

Given numbers 72 and 4.12

Step 1:

Change the whole number into decimal form

Whole number =72

Decimal form = 72.00

Step 2:

7 2 .0 0

(+)       4. 1 2

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



Start the addition process from the right

Step 3:

Add the first place number from the right and over the carry to the next step (0 +2 =2 carry =0).

7 2 .0 0

(+)       4. 1 2

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

2



Step 4:

Add the next place number from the right and then add the carry with is step total. Finally over the carry to the next step (here 0 +1 =1 carry =0)

7 2 .0 0

(+)       4. 1 2

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

. 1 2

Step 5:

Add the next place number from the right and then add the carry with is step total. Finally over the carry to the next step (here 2 +4 =6 carry =0)

7 2 .0 0

(+)       4. 1 2

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

6. 1 2

Step 6:

Add the next place number from the right and then add the carry with is step total. Finally over the carry to the next step (here 7 +0 =7, carry =0)

7 2 .0 0

(+)       4. 1 2

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

7 6 .1 2

72 + 4.12 =76. 12

Examples on multiplaction operation in math:

Ex 1: Multiply 10 with 5.

Sol:     10 x 5 = 15.

Where, ‘x’ is the operator of multiplication.

10 and 5 are the operands of multiplication.

15 is the resultant of the multiplication operation.



Ex 2: Simplify 10(5) + 20(2) + 9(3) + 4(1).

Sol:     15 + 40+ 27+ 4

86

Ex 3: A train has 9 carriages. There are 42 seats in each carriage. How many seats are there on the train?

Sol:   Number of carriage = 9

Number of seats in each carriage = 42

Multiply number of carriages with number of seats in each carriage.

Number of seats in the train = 9 x 42

= 378.

Ex 4: There are five cupboards in a room. Each cupboard has 7 racks. Find how many racks totally the room has.

Sol:      Number of cupboards = 5.

Number of racks in each cupboard = 7.

To find total number of rack, multiply the number of cupboard with number of racks in each cupboard.

Total number of rack = 5 x 7

= 35.

Ex 5: There are six baskets full of apple in a lorry. 1st basket contains 35 apples, 2nd and 3rd basket contains 40 apples, 4th, 5thth basket contains 50 apples. Calculate the total number of apples in the lorry. and 6

Sol:      35 apples in the 1 basket

40 apples in the 2 basket

50 apples in the 3 basket

Total Number of Apples = 35(1) + 40(2) + 50(3)

= 35 + 80 +150

= 265

Number Percentage Calculator

Percent means ‘for every 100 ’. So, when we say P% .it means P  out of 100 . Thus, P%=P/100 . It is often denoted by symbol “% ”. Any percentage can be expressed as a fraction. For example, 40%=40/100=2/5 .

Percentages are used to find whether one quantity is large or small compared with another quantity. The first term usually represents a part of, or a change in the second term, which should be greater than zero.

Percentages are usually used to express numbers between zero and one, any dimensionless proportionality can be expressed as a percentage.

Please express your views of this topic Percentage of Difference by commenting on blog.

Let us now express some percentages as fractions:

a.       5%=5/100=1/20 .

b.      10%=10/100=1/10 .

c.      25%=25/100=1/4 .

d.     75%=75/100=3/4 .

e.      125%=125/100=5/4 .

f.        175%=175/100=7/4 .

g.       (3 1/8)%=25/800=1/32 .

h.     (6 1/4)%=25/400=1/16 .

i.        (8 1/3)%=25/300=1/12 .

j.        (16 2/3)%=50/300=1/6 .

k.       (66 2/3)%=200/300=2/3 .

l.        (87 1/2)%=175/200=7/8 .



Calculation of Percentage:

Calculation of percentage:

The percent symbol can be treated as being equivalent to the pure number constant 1/100=0.01,  while performing calculations with percentage.

If a number is first changed byP%  and then changed by Q% , then the net change in the number =[P+Q+((PQ)/100)] . Remember that any decreasing value in the formula should be taken as ‘negative’ and increasing value should be taken as ‘positive’.

Similarly, if A’s salary is P%  less than B’s salary, then the percentage by which B’s salary is more than A’s salary is(100P)/(100-P) .

If expenditure also, then percentage change in expenditure or revenue=[P+Q+((PQ)/100)] . Where ‘P’ is the percentage change in price and ‘Q’ is the percentage change in consumption.

Problems on number percentages:

Ex1 : What percentage of 1600 is 40?

Sol:  Let 40  be "P% of 1600.

So, 40=P%   of 1600 =(P/100)(1600)=16P .

Thus, P=40/16=2.5% .

Ex2 : Calculate 40%  of 625 .

Sol: 40%  of a number =2/5  of the number =2/5  of 625=(2/5)(625)=250 .

Ex :3 A number is first increased by 30%  and then decreased by 20% . Find the net change in the number.

Sol:  Let the original number be 100 .

Increasing by 30%, " it becomes " 130 .

Now, if 130  is decreased by 20% , it becomes 104 .

Thus, the net change =(104-100)=4%  increase.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Divide Fractions by Whole Numbers and sample paper of class 9 cbse sa2. I am sure they will be helpful.

Practice problems on number percentages:

Q:1  A’s salary is 25%  more than B’s salary. By what percent is B’s salary less than A’s?

Sol:  Let B’s salary be .

Since A’s salary is 25%  more than that of B, his salary will be Rs. 125 .

Thus, B’s salary is Rs. 25 less than the A’s salary.

So, in percentage: (25125)(100)=20% . Hence, B’s salary is 20%  less than A’s salary.

Multiplication Properties

In this page we are going to discuss about multiplication properties concept. Integer used to measuring and counting in general things. Positive integer and negative integer is the two types of integers. Integer is nothing but using notational symbol represents numbers. Multiplying integer’s mean scaling a number by another.

3 x 4 = 3 + 3 + 3 + 3 = 12

3 x 4 = 4 + 4 + 4 = 12

Properties of multiplication

To study multiplying integers, we have to know five basic properties such as associative, commutative, distributive, and multiplicative identity. These multiplication properties are described as below.

Commutative property
Associative property
Multiplicative identity property
Distributive property
Zero property



Commutative property:

This is the product the two number is same even we change the order. That is, a * b = b * a.

Example: 5 * 4 = 4 * 5

= 20

Associative Property:

This is the product the three more number is same even we change the order. That is, (a * b) * c = a * (b * c).

Example: 5 *( 4 * 3 ) = (5 * 4) * 3

= 60

Identity Property:

This is the multiplication of any number by 1 is that number. That is, a * 1 = a.

Example: 9 * 1 = 9.

Distributive property:

Multiplication of a number with the addition of two numbers is same as addition of two products. That is, a* (b + c)=ab + ac.

Example: 5 * (4 + 3) = 5*4 + 5*3

= 35

Zero property:

Any thing multiply with zero is zero. That is, a * 0 = 0.

Example: 7 * 0 = 0

Multiplication rules

Positive * Positive = Positive

Negative * Negative =Positive

Positive * Negative = Negative

Negative * Positive = Negative

The following examples are to study multiplying integers with multiplication rule

5 * 4 = 20

-5 * -4 = 20

5 * -4 = -20

-5 * 4 = -20

Multiplication grid

Multiplication grid

Methods to multiplying integers

Below are the methods to multiplying integers-

Method 1:

Multiplying integers of 345 * 6

Solution:

Here, 5 is at unit place, so we do    5 * 1       =    5  * 6 =    30

4 is at 10s place, so we do   4 * 10     =  40  * 6 =   240

3 is at 100s place, so we do 3   * 100 = 300 * 6 = 1800
--------
The product result  =  2070
--------

Method 2:

Multiplying integers of 345 * 6

Solution:

345

x 6
--------------
30
240
1800
--------------
2070
--------------

Sample Space

In statistics, learning sample space basically represents the presentation and interpretation of the events of the possible outcomes that occur in a planned learning or scientific investigation. The learning sample space helps to refer all recording of information's are numerical or categorical, as an observation. The learning sample space assists in the following three cases, designed experiments, observational studies, and retrospective studies, the end result was a set of data that of course is subject to uncertainty.


Definition for Sample Space

The set of all favorable combinations of outcomes of a statistical experiment is labeled the sample space and is denoted by the mathematical symbol S.

Each possible outcome of a sample space is labeled a member of the sample space, in other words it is labeled the sample point. The trials of a sample space has a finite number of elements are separated by commas (,) and enclosed in braces ({}).

Thus the: sample space of possible outcomes when a coin is tossed, may be written S = {H, T),

Where, H means that “heads" and T means that “tails,”.


Examples for Sample Space:

Example 1:

Determine the sample space for the event of rolling a die, using the learning sample space.

Solution:

In rolling a die the number that shows on the top face.

The required sample space S1 = {1, 2, 3, 4, 5, 6}.

Example 2:

Determine the sample space for the number is even or odd.

Solution:

The required sample space S2 = {even, odd}.


More than one sample space:

If we know which element in S1 occurs, we can tell which outcome in S2 occurs; however, knowledge of what happens in S2 is of little help in determining which element in S1 occurs. Provides more information than S

Ex:  A coin is tossed twice. What is the Sample space?

Sol: The sample space; for this experiment is

S= {HH, HT, TH, TT}.

My previous blog post was on Multiplication Rule please express your views on the post by commenting.

Binary Number Representation

The binary numeral system, or base-2 number system, represents numeric values using two symbols, 0 and 1. More specifically, the usual base-2 system is a positional notation with a radix of 2. Owing to its straightforward implementation in digital electronic circuitry using logic gates, the binary system is used internally by all modern computers.

Understanding Binary Logistic Regression is always challenging for me but thanks to all math help websites to help me out.

The representaion of binary numbers are uses in mathematics are defined as follow,

Binary numbers representation in math:

In the binary number representation consists of octal, decimal, and hexa decimal numbers in the column. We can be represent the binary numbers by use of its operation. In the binary number representation, the decimal is easiest method of understanding binary numbers.

For example we can represent: 4567,

4 is represent the 1000’s

5 is represent the 100’s

6 is represent the 10’s

7 is represent the 1’s

Which means the representaion of 4567 is given as follow,
4567 = 1x1000 + 2x100 + 3x10 + 4x1

Given binary number representation,

1000


= 103 = 10x10x10

100


= 102 = 10x10

10


= 101 = 10

1


= 100 (any number to the exponent zero is 1)

The table above can be represented as the binary numbers,

Such that,

4567 = 4x1000 + 5x100    + 6x10     + 7x1

= 4x103 + 5x102 + 6x101 + 7x100

Examples for binary number representation in math:

The examples of binary number representation in math is given as follow:

Example:1

To determine the decimal number in 10102?

Solution:

Step 1: 1 => 1×2×2×2 = 8

Step 2:  0 => 0×2×2 = 0

Step 3: 1 => 1×2 (=2)

Step 4: 0 => 0

Answer is: 1010 = 8+0+2+0 = 10.

Example 2:

To determine the decimal number in 10112?

Solution:

Step 1:  1=> 1×2×2×2 (=8)

Step 2: 0 => 0×2×2 (=0)

Step 3: 1 => 1×2 (=2)

Step 4: 1 => 1

Answer is: 1001 = 8+0+2+1 = 11.

My forthcoming post is on how to find the prime factorization of a number and neet entrance exam syllabus will give you more understanding about Algebra.

Example 3:

To determine the decimal number in 1.112?

Solution:

Step 1:  1 => 1

Step 2:  1 => 1×(1/2)

Step 3: 1 => 1×(1/4)

Answer is :1.75.

Example 4:

To determine the decimal number in 11.112?

Solution:

Step 1: 1 => 1×2 (=2)

Step 2: 1 => 1

Step 3:  1 => 1×(1/2)

Step 4: 1 =>  1×(1/4)

So, 11.11 is 2+1+1/2+1/4 = 3.75 in Decimal

Answer is: 3.75.

Examples of Poisson Distribution

Definition: A random variable X is  a Poisson distribution if the probability mass function of X is P(X = x) =e−λ  λx / x!,                                       x = 0,1,2, …for some λ > 0

The mean of  Poisson Distribution denoted by λ, and the variance is denoted by λ.
The parameter of Poisson distribution is λ.

The Poisson distribution is a restrictive case of Binomial distribution under the following conditions.
(i)   Number of trials(n) is indefinitely huge(large), that is n → ∞.
(ii)  The constant probability(p) of success in each trial is very less.
ie., p → 0.
(iii) np = λ is finite where λ is a positive real number. When an event occurs rarely, the distribution of such event may be assumed to follow a Poisson distribution. The example problems of poisson distribution is   given below

Examples of Poisson distribution:

Some Examples of poisson distributions are given below

(1) The number of gamma particles emitted by a radio active source in a given time interval.
(2) The number of phone calls received at a telephone exchange in a given time interval.
(3) The number of defective articles in a packet of 250, produced by a good industries limited.
(4) The number of printing errors at each page of a book by a good publication centre.
(5) The number of road accidents reported in a city at a particular time.

Example Problems for Poission distribution:

Example problem 1: If a publisher of  technical books takes a great pain to ensure that his books are free of typological errors, so that the probability of any given page containing atleast one such error is 0.005 and errors are independent from page to page

(i) what is the probability of its 400 page novels will contain exactly one page with error.

(ii) atmost three pages with errors.
[e−2 = 0.1353 ; e−0.2. = 0.819].


Solution :

n = 400 , p = 0.005
np = 2 = λ
(i)  P(one page with error) = P(X = 1) = e−λ λ1/1! = e-2 21/1!
= 0.1363 × 2 = 0.2726
(ii)  P(atmost 3 pages with error) = P(X ≤ 3)
= Σ e−λ λx / x!    [limits 0 to 3]
= `sum` e−2 (2)x  /  x!
= e2 [1 +2/1! + 22/2! + 23/3!]
= e−2 (19/3 )= 0.8569

Learning Simple and Compound Events

Probability of any event  is the ratio of several favorable outcomes of an event to the total number of outcomes of an event. Otherwise if we toss a single dice then it is called as a simple event. If we toss a two dice then it is called as a compound event. Inclusive means when two events are happen at a same time. Mutually exclusive means when two events cannot happen at a same time.

Learning Simple and Compound Events Formulas :

learning  simple and compound events formulas in the probability of an event can be expressed as

P(a) = number of a favorable outcomes / total number of a possible outcomes

learning  simple and compound events formulas in the probability of two independent events(A and B) are multiplied by the probability of the first event by the probability of the second event. Two events are said to be independent. if P(A and B) = P(A) P(B). P(A)and P(B) are are non zero

P(A and B) = P(A) . P(B)

learning  simple and compound events formulas in the probability of two dependent events (A and B ) are multiplied by the probability of A and the probability of B after A occurs.

P(A and B) = P(A) . P(B following A)

learning  simple and compound events formulas in the probability of one or other of two mutually exclusive events (A or B) are added to the probability of the first event to the probability of the second event.Two events are said to be disjoint. if and only if P(A and B) = 0

P(A or B) = P(A) + P(B)

learning  simple and compound events formulas in the probability of one or the other of two inclusive events(A or B) are added to the probability of the first event to the probability of the second event and subtract the probability of both events happening.

P(A or B) = P(A) + P(B) - P(A and B).

practice problems in learning simple and compound events

Example 1:Abraham is going to the Super market to pick a new sports bats. Today, the shopkeeper has 25 cricket bats and 50 Tennis bats are available for sales. If Abraham randomly picks the bats, then what is the probability that the bat would be a Cricket bat?

Solution:

Probability(randomly choosing a Cricket bat) = Number of Cricket Bats / Number of cricket bats + number of tennis bats

= 25 / (25 + 50 )

= 25 / 75

= 1 / 3

= 0.33

if Abraham randomly picks a bat (which is a Cricket bat) having a probability  0.33.

Example 2:Lenin is going to the Super market to pick a new sports bats. Today, the shopkeeper has 20 cricket bats and 40 Tennis bats are available for sales. If Lenin randomly picks the bats, then what is the probability that the bat would be a Cricket bat?

Solution:

Probability(randomly choosing a Cricket bat) = Number of Cricket Bats / Number of cricket bats + number of tennis bats

= 20 / (20 + 40 )

= 20 / 60

= 1/ 3

if Lenin randomly picks a bat (which is a Cricket bat) having a probability  = 0.33

Exponential Growth Formula

A function is said to be Exponential growth that including exponential decay when the growth rate of that mathematical function is proportional to the function's current value. In a discrete domain of definition with equal intervals of the function is called as geometric growth or geometric decay. The exponential growth model is also called as the Malthusian growth model.


Exponential growth formula:

Exponential formula defines the  X as exponentially on time t.

X(t) = a . b(t/r)

"a" denotes the initial value

a = x,

X(0) = a,

b= a

It denotes the positive growth of the factor, t = time required

Example for exponential growth formula:

If a power doubles every 5 minutes, initially there’s only one doubles, how many powers would be there after 2 hours?

Here, a= 1, b= 2, t = 5 minutes.

X(t) = a . b(t/r) = 1 . 2{(120 minutes)/(5 minutes)}

X(2 hour) = 1 . 2 24 = 16777216

After two hours, there would be 16777216 powers.


Exponential growth Problem:

A microbiologist is researching a newly-discovered species of fungi. At time t = 0 hours, he puts one hundred fungi into what he has determined to be a favorable growth medium. Six hours later, he measures 200 fungi. Assuming exponential growth, what is the growth constant "i" for the fungi? (Round i to two decimal places.)

For the given problem, the units on time t will be hours, because the growth is being measured in terms of hours. The starting amount P is the amount at time t = 0, so, for this problem, P = 100. The ending amount is A = 200 at t = 6. The only variable we don't have a value for is the growth constant i, which also happens to be what I'm looking for. So I'll plug in all the values we know, and then solve for the growth constant:
A = Peit
200 = 100e6i
2 = e6i
ln(2) = 6i
`ln(2)/6` = i = 0.11552453

Derivative Step by Step

erivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity.-  Source : Wikipedia

Calculus is used to solve differentiation of any subjective equation and the equivalent output result. we need to find f(x) where,     d/dx f(x) = g(x).The derivative of constant is zero in calculus. The linear equations are also compared using calculus. Integration is considered as one of most important study of calculus in mathematics. British mathematician, Isaac Newton and the German man Gottfried Leibnitz, invented the calculus. The methods that are applied in continous graphs, curves or fucntions  is calculus.

Understanding Derivative of Cot x is always challenging for me but thanks to all math help websites to help me out.

Steps to solve the Derivative function:

Derivative  problem 1:

Find the first derivative and second derivative of

f(x) = x4 + 7x3 - 3x2 - 9x +3

Solution:

Step 1:  First differentiate f(x) = x4 + 7x3 - 3x2 - 9x +3

Step 2:  when we differentiate the first term we get  (1 × 4) x (4-1)

Step 3:  The above equation can above simplified as  4 x 3

Step 4:  Like wise we can differentiate  and simplify all the terms in the equation

Step 5:  Finally after the first derivative we will get  4 x 3 + 21 x2 - 6x - 9

Step 6:  Now we will go for Second derivative
End of the second derivative we will get   12x2 + 42 x - 6


First derivative:

f' = (df )/ (dx)

= (1 × 4) x (4-1) + (7 × 3)x(3-1) - (3 × 2)x(2-1) - (9 × 1)x (1-1)

= 4 x 3 + 21 x2 – 6x - 9

Second derivative:

f '' = (df ' )/ (dx)

= 4 x 3 + 21 x2 – 6x - 9

= 12x2 + 42 x - 6

Answer:

(d^2)/(dx^2)  (x4 + 7x3 - 3x2 - 9x +3) = 12x2 + 42 x - 6

Derivative problem 2:

Find the differential for y = sqrt(2x^3 - 9x)

Solution:

Step 1:  First turn sqrt(2x^3 - 9x)to (2x3 - 9x)1/2
Step 2:  Differentiate in the usual method
Step 3:  Then finally we get,

d/(dx) ((2x3 - 9x)1/2 ) = ½ ( 2x3 - 9x ) ½ -1( 6x2 - 9 )

= ½ ( 2x3 - 9x ) -1/2( 6x2 - 9 )

= ( 6x^2 - 9 ) / (2 sqrt (2x^3 - 9x ))

Answer of given calculus test exam problem is

d/(dx) ((2x3 - 9x)1/2 ) = ( 6x^2 - 9 ) / (2 sqrt (2x^3 - 9x ))

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Derivative practice problem:

Derivative practice problem 1:

Find the derivative of f(x) = ( x - 5 ) ( 2x + 1 )

Answer:   d/(dx) f(x) = 4x - 9

Derivative practice problem 2:

Differentiate the given equation with respect to t. y =  8t3 + 12t

Answer:  24t2   + 12

Division with Remainders Word Problems

Division is an arithmetic operation which is the opposite of multiplication. The symbol used for division is (÷).

Notation:

Division is often represented by inserting fraction bar which is nothing but the horizontal line between divisor and dividend. Dividend is placed in numerator and divisor are placed in denominator.  For example, a divided by b is written as,

`a/b`

Key terms - division with remainders word problems

Dividend:

The number which is being divided in a division is known as dividend.

Divisor:

The number which divides the dividend is known as divisor.

Quotient:

Quotient is the solution / answer for the problem

Remainder:

The remainder is the amount "left over" after the division

Example problems on division with remainders problems

Division problem 1:

What is division value and remainder of the given fraction 13 / 4

Solution:

When we divide 13 by 4 it gives 3 ( 3 * 4 = 12) as a quotient and 1 is the (13 - 12) remainder value

Quotient = 3

Remainder = 1

Division problem 2:

What is division value and remainder of the given fraction 6 / 4

Solution:

6 / 4 =   There is one 4 in 6

Hence Quotient = 1

Remainder = 2

Solving Division with remainders Word problems:

Division with remainders word problem 1:

A 40 in stick is cut into pieces of 7 in length. How many pieces we can get out of this stick? What is the length of the scrap? Solve the word problem.

Solution:

The problem describes how to divide a 40 in stick into pieces of 7 in each. The remaining piece of the stick is called the scrap.

Therefore, the dividend is 40in., the divisor is 7 in. The quotient of this is the number of pieces and the remainder is the length of the scrap. When we divide 40 by 7, there are fivet 7’s in 40 (5*7= 35) and the remainder is 5.

Hence, the number of pieces of 7 in = 5(quotient)

The length of the scrap = 5 (remainder)

Division with remainders word problem 2:

There are 11 apples and 4 kids . We have to give equal number of apples to each kids. How many apples will each kids get and find the amount of apple left over?

Solution:

Here we have to divide number of apples  10 by number of kids  4.  Hence ,

11 / 4 = 2 with remainder 3

Number of apples each kid will get = 2

The apples left over = 3

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Division with remainders word problem 3:

Consider there are 10 boys in a class. Divide them into 3 groups. How many boys each group will consists and how many boys are left over?

Solution:

Here the dividend is total number of boys 10 and the divisor is 3. We have to divide total number of boys by number of groups to find the number boys each group having .

10 / 3 = 3

Remainder = 1

Hence each group gets 3 boys and 1 boy is left over.

Basic Geometrical Concepts Solve Online

Geometrical is a study of structural mathematics .Basic geometrical has the concept of shapes and structure of an object. It contains two dimensional and three dimensional objects. In this article basic geometrical concepts solve online we are going to see about some basic geometric problems through online.
Generally online help means that here tutor and student should communicate directly with help of online chatting. Student can get more geometrical help and idea from tutor through online with help of chatting. It may be in audio or non audio.


Basic Geometrical Concepts Shapes Solve Online :

Basic geometrical concepts are plane, point, line, angle, ray, two dimensional objects, and three dimensional objects. Main purpose of using geometry is finding area, volume and perimeter of the given shape.

Basic geometrical concepts:

Two dimensional objects:

Square
Rectangle
Triangle
Circle
Parallelogram
Trapezoid

Three dimensional objects:

Cylinder
Cone
Cube
Sphere
Hemisphere
Prism
Pyramid
3d geometry shapes

Solve Example Problems in Basic Geometrical Concepts:

Solve Example problems in basic geometrical concepts:

Example problem 1:

Find the angle of triangle from the given figure?

Geometry-triangle

Solution:

Given data:

Two angles of triangle is 550 and 380

Solve the third angle of triangle

Total angle of triangle=1800

55+38+x=180

93+x=180

X=180-93

X=870

Third angle of triangle is 870

Example problem 2:

Which one is 3D shape of geometrical object?

Geometry- shapes

Solution:

Option (a)-Two dimensional object(square)

Option (b)-Two dimensional objects(Circle)

Option (c)-Three dimensional object(Parallelogram)

Option(d)-Three dimensional object(Cone)

Example problem 3:

How many edges and vertices in the cube?

Cube

Solution:

Cube having 12 edges

Cube having 8 vertices

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So the cube having top portion of four edges

Side portion of four edges

Bottom portion of four edges

So totally have 12 edges

Vertices mean nothing but corner point of the given shapes. When the two edges or lines are meeting It will produce the vertices

Solve Proving Proofs

Mathematical statement considered the sequence of statements by proofs, every statement being adjusted with some definition or a proposition or an axiom that is before launched by the method of deduction using only the permitted logical rules. Repeatedly we prove a proposition straightly from what is given in the proposition. But so many times it is easier to prove a same proposition not proving the proposition by itself,

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Solve Proving Proofs - Types of Proofs:

Solve Proving Proofs Indirect Method:

Solve proving proofs - Indirect method:

Prove the given function f : R  =>  R defined by f(x)  =  2x  +  5 is one – one.

Sol :  A function is one – one if  f(x1)  =  f(x2)  => x1  =  x2.

Using this we have to show that “ 2x1 + 5  = 2x2  +  5” => “  x1  =  x2”.This is of the form p  =>  q, where , p is 2x1  +  5 =  2x2  +  5 and q : x1  =  x2. so this is the direct method proofing.

We can also prove the same by using contrapositive of the statement. Now contrapositive of this statement is ~ q=>  ~ p, is, contrapositive of “if f(x1)  =  f(x2), then  x1  =  x2” is “if x1  =  x2, then f(x1) =  f(x2)”.

Now                 x1  !=   x2

=>                   2x1  !=   2x2

=>                   2x1  +  5  !=   2x2  + 5

=>                   f(x1)  !=   f(x2).
Solve Proving Proofs of Direct Approach:

Solve Proving Proofs -direct Proof:


It  is the proof of  a proposition in which we directly start the proof  with what is given in the proposition.

1)Straight forward approach.

2) Mathematical Induction approach.

3)Proofs By cases or by   exhaustion.

Solve Proving Proofs -Indirect Proof:

Instead of proving the proposition directly, we establish the proof of the proposition through  proving a proposition which is equivalent to the given proposition.

1)Proofs  by contradiction.

2)Proofs by using contrapositive statement.

3)Proofs by a by a counter example.


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Solving Proving Proofs - some Examples for Proving Proofs:

Solving proving proofs - Direct Method:

Prove the given function f:  R  =>  R

Defined by f(x) = 2x + 5 is one – one.

Sol :  Note that a function f is one – one if

f(x1)  =  f(x2)    x1  =  x2 (definition of one – one function)

Now given that   f(x1)  =  f(x2),

=> 2 x1 +  5 = 2 x2 +  5

=>                 2 x1  +  5  –  5  = 2 x2  +  5  –  5(adding the same quantity on both sides)

=>                 2 x1  +  0  =  2 x2  +  0

=>                 2 x1  =  2 x2 (using additive identity of real number)

=>                 (2)/(2) x1  =  (2)/(2) x2 (dividing by the same non zero quantity)

=>                 x1  =  x2.

Sets and Operations

Sets and relations are two of the most important concepts in mathematics that is taught in middle school and is widely used in practical mathematics. A set is a collection of things such as letters, numbers, words, things etc. For example: {woodcraft construction toys, building blocks, construction sets}. This is a set of educational toys such as woodcraft construction toys and so on. Sets also functions different relationship in mathematics. Let’s have a look at the same in this post.

Union of Sets: When two sets are compared and a final set is created with the elements of set A and set B, the set created is called Union of sets. It is referred as A U B. While creating the union of sets, a single element if occurs in both the sets is used only once in the final set.

For example:
A = { Cloth toys for babies , Wooden play toys for kids , Soft toys for babies, Barbie Doll}
B = {Ludo, Chess, Snakes and Ladder, Barbie Doll}
A U B = {

Cloth toys for babies , Wooden play toys for kids, Soft toys for babies, Barbie Doll, Ludo, Chess, Snakes and Ladder}Intersection of Sets: When two sets are compared and a final set is created using the only common elements of set A and set B, then it is called as intersection of sets. It is referred as A ? B.

For example:
A = {Hair, Eye, Nose, Ear, Lips, Hands, Legs}
B = [Hair, Eye, Nose, Ear, Lips, Teeth, Tongue}
A ? B = {Hair, Eye, Nose, Lips}
Difference of Sets: The difference of sets is nothing but the set that includes the unique elements of set A and B. It is referred as A –B. For example:
A = {1, 2, 3, 4, 5}
B = {2, 4, 6, 8, 10}
A – B = {1, 3, 5, 8, 10}
These are some of the basic relations in sets.

Geometric Ray

We know that, a ray is an important topic of line, and geometrical chapter in mathematics subject. In mathematics, a ray is defined as several ways, when we have seen a vector, a ray vector is going to one point to other point, and it is called as rays. In this geometrical part, ray is used in mostly as the lines and it’s relevant. In this article we are going to explain about the geometric ray, and its definitions.

Definition of the Geometric Ray:

Generally a ray is defined as many ways; it is one part of line.
It has a straight line with one end point and other direction (side) extends to infinity.
And a ray is start with end point, and then other point on the ray next.

For example:

Picture representation for geometric ray:

Geometric rays

If suppose the above ray (figure) is vector ray means, a ray is vector (xy) form of point x to point y.
And a ray is start with end point x, and other side goes to infinity on next. It is named ray.

Note:

If suppose the given ray (see figure) is vector ray means, a ray is vector (xy) form of point x to point y.

Ray is` vec (XY)` .

Exact definition for ray:

Ray is straight line that starts at a point and continues outward in another direction. And a ray is start with end point, and then other point on the ray next.

Geometric rays
Examples Pictures and Explain in Geometric Rays:

Example problem 1: Draw a straight line, with two points ray figure.

Figure:

example figure for ray

Example problem 2: Draw the both opposite side rays.

The ray figures:

example figure for ray, and    example figure for ray

These rays are mostly used in the sphere intersection problems.

Example problem 3: Draw the angle figure used on 2 ray figures with 1 common end point:

Angle figure in ray:

example figure for ray

Example problem 4: Draw the parallel line, and use the ray with intersect (cross) the parallel line.

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ray example

These all above the general explanations and example figures are making clear about the geometric ray definition and concept.

Sequence Listing

In this article, we are going to see about the sequence listing. There are two types of sequences

1. Arithmetic sequence and

2. Geometric sequence.

Arithmetic sequence means, the sequence of numbers such that the difference between two consecutive members of the sequence is a constant. Geometric sequence means, the sequence of numbers such that the ratio between two consecutive members of the sequence is a constant. The sequence listing formulas and example problems are given below.



Formula for arithmetic sequence:

nth term of the sequence : an = a1 + (n - 1)d

Series of the sequence: sn = (n(a_1 + a_n))/2

Formula for geometric sequence:

nth term of the sequence: an = a1 * rn-1

Series of the sequence: sn = (a_1(1-r^n))/(1 - r)
Example Problems for Sequence Listing:

Example problem 1:

Find the 15th term of the series 2, 4, 6, 8, 10,. ....

Solution:

First term of the series, a1 = 2

Difference of two consecutive terms, d = 4 - 2 = 2

n = 15

The formula to find the nth term of an arithmetic series, a_n = a_1 + (n-1)d

So, the 15th term of the series 2, 4, 6, 8, 10....... = 2 + (15 - 1) 2

= 2 + 14 * 2

After simplify this, we get

= 2 + 28

= 30

So, the 15th term of the sequence 2, 4, 6, 8, 10,.... is 30

Series of the arithmetic sequence: sn = (n(a_1 + a_n))/2

 s_15 = [15(2 + 30)]/(2)

 s_15 = [15(32)]/(2)

 s_15 = (480)/(2)

After simplify this, we get

 s_15 = 240



Example problem 2:

Find out the 11th term of a geometric sequence if a1 = 4 and the common ratio (C.R) r = 2

Solution:

Use the formula a_n = a_1 * r^(n-1) that gives the nth term to find a_11 as follows

a_11 = a_1 * r^(11-1)

= 4 * (2)10

= 4 * 1024

After simplify this, we get

= 4096

The 11th term of a geometric sequence is

Series of the sequence: sn = (a_1(1-r^n))/(1 - r)

 s_11 = [4(1 - 2^11)]/(1 - 2)

= [4(-2048)]/(-1)

After simplify this, we get

s_11 = 8192

The above examples are helpful study of sequences listing.