Multiplying Integers Calculator

Introduction for multiplying integers calculator:

Calculator is a device which used to calculate the required process in that device and also we can substitute the many values in the calculator. Integer multiplying calculator we have the many integers value which has to be implies in that calculator and we find the value. An integer is a number that can be either greater than 0, called positive, or less than 0, called negative. Zero has a neither positive nor negative sign. Any integer on a number line has an exact value, which is distance from zero.

Multiplying Integers Calculator:

In first column enter the any integer and in the second column enter the any integer
when we press the enter we get the multiplied result in the third column.

In this above multiplying integer calculator works by the following rules are as follows.

The product of two positive integers will be a positive integers.
The product of a positive integer and a negative integer will be a negative integer.
The product of two negative integers will be a positive integer.
Positive number × positive number = positive number
Positive number × negative number = negative number
Negative number × positive number = negative number
Negative number × negative number = positive number

Example Problems with Multiplying Integers Using Calculator:

Example 1 :
Find the product of  (+ 15) × (– 2)    

Solution :
In first column enter the +15 and in the second column -2 enter the  -2
when we press the enter we get the multiplied result in the third column.

(+ 15) × (– 2) = – 30 (positive × negative = negative)

Example 2 :
Find the product of (– 12) × (+ 9)

Solution:
In first column enter the -12 and in the second column  enter the +9
when we press the enter we get the multiplied result in the third column – 108.
(– 12) × (+ 9)      = – 108 (negative × positive = negative)

Example 3 :
Find the product of (– 22) × (– 10) × (+7)

Solution:
In first column enter the -22 and in the second column  enter the -10 
First take two terms
(– 22) × (–10) = +220 (negative × negative = positive)

when we press the enter we get the multiplied result in the third column +220.
and now multiply +220  with the third term
+200 × (+7) = +1400 ( positive × positive= positive)

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Example 4 :
Find the product of (– 10) × (–10) × (+8)  × (-7)

Solution:
In first column enter the -10 and in the second column  enter the -10
First take two terms
(– 10) × (– 10) = +100 (negative × negative = positive)

when we press the enter we get the multiplied result in the third column +100.
now multiply +100  with the third term
+100 × (+8) = +800 ( positive × positive= positive)

In first column enter the +100 and in the second column  enter the -7
now multiply +800 with the third term
+800 × (-7) = -5600 ( positive × negative = negative)

Ratios Word Problems

Introduction to ratios word problems:

In mathematics education, the term word problem is often used to refer to any mathematical exercise where significant background information on the problem is presented as text rather than in mathematical notation. As word problems often involve a narrative of some sort, they are occasionally also referred to as story problems and may vary in the amount of language used.

Definition of Ratios Word Problems:

Ratios word problems:

Word problems can use ratios to relate the different items in the given question are called as Ratios problems.

Important things about ratio problems:

If necessary, change the quantities to the same unit.
Write the items in the ratio as a fraction.
Simplify if necessary.

Example Problems for Ratios Word Problems:

Example 1:

A bag contains full of red and green balls, the ratio of the red to green balls is 4:5. If the bag contains 140 green balls, how many red balls are there in the box?

Solution:

Step 1:

Assign the variables:

Let a = red balls

Write the given items in to the ratio as fraction

Red balls/ green balls = 4/5 = a/140

Step 2:

Solve the above equation

By using the cross multiplication

4 × 140 = 5 × a

560 = 5a      [ both side divided by 5 ]

560/5 = 5a/5

112 = a 

There are 112 red balls are in the bag.
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Example 2:

If the fruit stall sells a different fruits that are apple, orange and banana. The fruits are in the ratio of 4 to 5 to 6. If the fruit stall contains 30 Oranges, how many fruits does it have altogether?

Solution:

Step 1:

Assign the variables

Let x = Apples

y= Banana

Write the given items in the ratios as fractions

Apple/ orange = 4/5 = x/30

Banana/ orange = 6/5 = y/30

Step 2:

Solve the above two equations.

Using the cross multiply the two equations

4 × 30 = 5 × x

120 = 5x               [Both side can be divided by 5]

120/5 = 5x/5

24 = x

There are 24 apples are in the fruit stall

6 × 30 = 5 × y

180 = 5y               [Both side can be divided by 5]

180/5 = 5y/5

36 = y

There are 36 bananas are in the fruit stall.

Total number of fruits in the fruit stall = Number of fruits (apples + bananas + Oranges)

= 24 + 30 + 36

= 90

Therefore, the total numbers of fruit in the fruit stall are 90 fruits.

A Parallelogram with Four Right Angles

Introduction for A Parallelogram with Four Right Angles:
Shapes is an important part in geometry. Basically, the shapes that are enclosed with its sides are said to be polygons. The polygons are classified based on their sides. Whenever any polygon which contain four sides, then it is said to be a quadrilateral. There are multiple types of quadrilateral. Parallelogram is one of the type of quadrilateral. In this article, we shall discuss about the parallelogram with four right angles. Also we shall solve problems based on Parallelogram with four right angles.

Parallelogram with Four Right Angles:

The parallelogram with four right angles are said to be a Rectangle.

Basically, Rectangle is same as the parallelogram. The opposite sides of the rectangle are equal and parallel. The four angles measures 90° each, which is the right angles.

Area of a Parallelogram with four right angles:

The area of the rectangle can be determined by using the formula,

Area of Rectangle =  l * w

where l is the length of the rectangle and w is the width of the rectangle.

Perimeter of the parallelogram with four right angles:

The perimeter of a rectangle can be determined by using the formula,

Perimeter of rectangle = 2(l + w)

where l is the length of the rectangle and w is the width of the rectangle

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Example Problems Regarding Parallelogram with Four Right Angles:
Example 1:

Find the area of the parallelogram with four right angles, whose length is 8cm and width is 5cm.

Solution:

The parallelogram with four right angles is nothing but the rectangle.

Area of rectangle = l * w , 

where l is the length of the rectangle and w is the width of the rectangle.

Area of rectangle = 8 * 5

= 40 cm2

Therefore the area of the parallelogram with four right angles is 40 cm2.

Example 2:

Find the perimeter of the parallelogram with four right angles, whose length is 10cm and width 5cm.

Solution:

The parallelogram with four right angles is nothing but the rectangle.

Perimeter of rectangle = 2 ( l + w)

where l is the length of the rectangle and w is the width of the rectangle.

Perimeter of rectangle = 2 ( 10 + 5 )

= 2 (15)

= 30 cm

Therefore, the perimeter of the parallelogram with four right angles is 30cm.

Probability Index Formula

Introduction to probability index formula:

Probability is defined as the chance of something going to happen in the future. It can be expressed as a number from zero (which means that will never happen) to 1 (which means that will happen certainly).

The maximum probability value will be one and the minimum probability formula value will be zero.

Example: There is a chance of only 67% that will rain tomorrow.

Probability index is defined as the sum of weighted values of five cranial and dental angles and appears to have important analytical value.

Probability Index Formula:

Formula for probability index can be shown as follows,

Probability index formula=  Odds given with an additional information x 100

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

Odds without that information.

Probability index is used to understand about the two given things  for the index of increased odds.

The Format sharing probability Index (I) is one which is equal to the Percent of Format A’s spectators also listening to Format B (S/A) in excess of the percent of people in the marketplace paying attention to the Format B (B/T)

Where,

A = persons who are using Format A.

B = persons who are using Format B.

T = persons who are in the market.

S = persons who are listening to both Format A and Format B.

In order to create an index we need to have awareness on the quotes shown above

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Probability Index

We can estimate the weibull modulus using the probability index with the help of least square method.

Other formulae in probability are as follows

Complement Formula:

This following formula shows the formula for complementary events.

P (X) + P (X’) = 1

Addition formula:

The rule of addition narrates the probability of event X or Y happening with the probability of X , the probability of Y  and the probability of X and Y .

Mathematically it is written as follows,

P(X U Y) = P(X) + P(Y) – P(XY)

Where,

XY represents X and Y.

Mutually Exclusive Formula:

If X and Y are mutually exclusive then,

P(X.Y) = 0

P(X U Y) = P(X) + P(Y) .

Multiplication Rule:

The multiplication formulae for probability can be given as follows
P(X & Y) = P(Y) P(X/Y)

P(X & Y) = P(X) P(Y/X)

Formulas for Independent Events:

In case if X and Y being independent to each other,

P(X/Y) = P(AX P(Y/X) = P(B)

P(A & B) = P(A)P(B)

Variable Expressions Practice

Introduction to variable expressions:
Expressions with variables are obtained by operations of addition, subtraction, multiplication and division on variables. For example, the expression 2n is formed by multiplying the variable n by 2; the expression (x + 10) is formed by adding 10 to the variable x and so on.

We know that variables can take different values; they have no fixed value. But they are nothing but numbers. That is why as in the case of numbers, arithmetic operations like addition, subtraction, multiplication and division can be done on them.

Simple Variable Expressions and their Formation
One important point must be noted regarding the expressions containing variables. A number expression like (4 × 3) + 5 can be immediately evaluated as (4 × 3) + 5 = 12 + 5 = 17.

But an expression like (4x + 5), which contains the variable x, cannot be evaluated. Only if x is given some value, an expression like (4x + 5) can be evaluated. For example, when x = 3, 4x + 5 = (4 × 3) + 5 = 17 as found above.

Formation of  Variable Expressions

y + 5                              5 added to y

t – 7                               7 subtracted from t

3 x + 2                           first x multiplied by 3, then 2 added to the product

10 a                               a multiplied by 10

x / 3                               x divided by 3

– 5 q                              q multiplied by –5

2 y – 5                           first y multiplied by 2, then 5 subtracted from the product

10y+7                            y multiplied by 10 and then 7 added to the product

2n-1                              n multiplied by 2 and 1 subtracted from the product

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Evaluating Variable Expressions

Example 1: Evaluate the variable expressions.

x+2y;

x=2, y=3

Solution:

Substitute the x, y and z values in the given expression.

x+2y=2+2(3)

=2+6

=8

Example 2: Evaluate the variable expressions.

(x + y)/x y;

x=7, y=6

Solution:

(x + y)/x y= (7+6) / (7*6)

= 13/42

Example 3: Evaluate the expressions.

7xy+5yz+3zx-40;

x=-1, y=1, z=-3

Solution:

7xy+5yz+3zx-40=7(-1) (1) +5(1) (-3) +3(-3) (-1)-40

= (-7) + (-15) + (9) – 40

=-7-15+9-40

=-53

Example 4: Evaluate the expressions.

(2x+y+z)-(x+y-2z); x=8, y=7, z= 4

Solution:

(2x+y+z)-(x+y-2z) = (2(8) + (7) + 4)-((8) + 7-2(4))

= (16+7+4)-(8+7-8)

= 27-7

= 20

Example 5: Evaluate the variable expressions.

(x+1) + (y+1) + (z+1)-(z-1);

x=6, y=7, b z=3

Solution:

(x+1) + (y+1) + (z+1)-(z-1) = (6+1) + (7+1) + (3+1)-(3-1)

= 7+8+4-2

= 17

Odd Number of Factors

Introduction to odd number factors:

The integers are also called as number and each number has divisors. The divisors are called as factors. The factors do not give any remainder in divisible of number. Each number has more than one factors. We can list out how many number of factors are present in each number. Now we are going to see about odd  number factors.

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Explanation for Odd Number Factors in Math

Some notes about factors:

The factors are create the number and the two types of factors are present in math. One is prime factor and composite factor. The prime number is created by prime factor that is two factors are present. The composite number is created by composite factor that is two or more factors are present.

We can list out the numbers based on odd number  that is counting of factors are in odd. Listing method is used for find the odd number factors.

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More about Odd Number Factor in Math
Example problems for odd number factors:

Problem 1: List out the factors for given numbers and find out the odd number factors.

17, 63, 16, 81, 21.

Answer:

List the factors:

17  - 1 and 17 ( The number of factors is even 2).

63 -  1, 3, 7, 9, 21 and 63 ( The number of factors is even 6).

16 – 1, 2, 4, 8 and 16 ( The number of factors is odd 5 ).

81 – 1, 3, 9, 27 and 81 ( The number of factors is odd 5 ).

21 – 1, 3, 7 and 21 ( The number of factors is even 4 ).

From the list, the odd number factors are 16 and 81. Because both has count in odd number that is 5.

Problem 2: Check the given number has odd number factors or not.

49.

Answer:

Given number is 49.

The factors are 1, 7 and 49.

Therefore, the given number has odd number factor as 3.

Exercise problems for odd number factors:

1. List out the odd number factors of given number.

100

Answer: The odd number factors are 1, 2, 4, 5, 10, 20, 25, 50 and 100.

2. Check whether given number has odd  number factor or not.

64

Answer: The given number has odd number factor because it has 7 factors.

Simplify Logarithm

Introduction for simplify logarithms:

In arithmetic, the logarithms of a number to be a identified base is the power or exponent to which the base have to be raised in order to generate that number. For example, the logarithms of 10000 to the base 100 is 4, because 4 is the power to the which ten must be raised to produce 10000: 104 = 10000, so log1010000  = 4  
                                                          
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Simplify Logarithms Rules:
Product rule: If n, A and B are positive numbers and x ?1, then
Logn(AB) = lognA +lognB

Quotient rule: If A, B and n is positive numbers and n ? 1, then,
log n(`A/B` ) = log nA –logn B

Power rule: If n and A are positive numbers, n ? 1 and m is a real number, then
LognAB =BlognA

Change of base rule: If A, B and n are positive numbers and A ? 1, x ? 1, then
log AB = lognA   *  logBn

Reciprocal rule: If A and B be the positive numbers other than 1, then
logAB =`1/(log_BA) `

Examples for Simplify Logarithms:

Example 1:

Simplify : 56log25+56log26 =56 log2 (6x)

Solution:

logarithmic function56log25+56log26 =56 log2 (6x)

56 log2(5*6) = 56log2(6x)

56log2(30) =56 log2(6x)      by logarithm rule

Equate both sides, s the base are same 2, 6x = 30

x =30/6

x = 5

The answer is x = 5.

Example 2:

Simplify : 44log8512-44log88

Solution:

44log8512 - 44log88 = 44[log8(512/8)]          { by logarithms rule}

= 44log883

= 3*44                         { log88 = 1 }

= 132

The answer is 72.

Example 3:

Simplify logarithms :  12 log289

Solution:

convert the logarithmic equation

=  12log28  { by logarithms rule}

= 12log22 3

= 12*3log22

= 36

The answer is 36.

Example 4:

Simplify : (i) 67 log3 27 - 67log3 9

(ii)56 log5 25 +56  log5 5

Solution:

(i) Since  expression is a sum of two logarithms as well as the bases are equal, we can apply the product rule

(i) 67 log3 27 -48log3 729 = 67[log 3 (27/9)]

= 67 [ log3 (32)]

= 67 log3 32

= 67*2 log33

= 67* 2

= 134

(ii) 56log525+56log5(5)  = 56log5 (25*5)

= 56 log 5(625)

= 56log 5(5^4)

=56log5(54)

= 4*56log55 = 224

Example 5:

Simplify : log6 9x – log6(7x+1)

Solution:

by quotient law, we can write the equations as log6 `((9x) /( 7x+1))` Changing into exponential form, we get

`((9x) / (7x+1))` = 60  = 1

9x = 7x + 1

x = `1/2`

Factoring Trinomials with Coefficients

Introduction to factoring trinomials with coefficients:

In mathematics, a rational function is any function which can be written as the ratio of two polynomial functions. Factorization (alsofactorisation in British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. (Source: Wikipedia)

Trinomial:

It consists of three variables or three terms in its equation. For example x^2 - 3x + 9 is trinomial equation.

Example Problems for Factoring Trinomials with Coefficients

Factoring trinomials with coefficients example problem 1:

Factorize the given polynomial expression x^2 + 2x + 1

Solution:

Given rational expression is x^2 + 2x + 1

Factorize numerator and denominator values, we get

First factorize the numerator value, we get

(x^2 + 2x + 1) = (x^2 + x + x + 1)

Grouping the first two terms and second two terms, we get

= (x^2 + x) + (x + 1)

= x (x + 1) + 1 (x + 1)

= (x + 1) (x + 1)


The factors of the given polynomial expression is (x + 1) and (x + 1)

Answer:

The final answer is (x + 1) and (x + 1)

Factoring trinomials with coefficients example problem 2:

Factorize the given polynomial expression x^2 + 19x + 60

Solution:

Given rational expression is x^2 + 19x + 60

Factorize numerator and denominator values, we get

First factorize the numerator value, we get

(x^2 + 19x + 60) = (x^2 + 15x + 4x + 60)

Grouping the first two terms and second two terms, we get

= (x^2 + 15x) + (4x + 60)

= x (x + 15) + 4 (x + 15)

= (x + 15) (x + 4)


The factors of the given polynomial expression is (x + 15) and (x + 4)

Answer:

The final answer is (x + 15) and (x + 4)

Factoring trinomials with coefficients example problem 3:

Factorize the given polynomial expression x^2 - 21x + 20

Solution:

Given rational expression is x^2 - 21x + 20

Factorize numerator and denominator values, we get

First factorize the numerator value, we get

(x^2 - 21x + 20) = (x^2 - 20x - x + 20)

Grouping the first two terms and second two terms, we get

= (x^2 - 20x) - (x - 20)

= x (x - 20) - 1 (x - 20)

= (x - 20) (x - 1)


The factors of the given polynomial expression is (x - 20) and (x - 1)

Answer:

The final answer is (x - 20) and (x - 1)

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Practice Problems for Factoring Trinomials with Coefficients

Factoring trinomials with coefficients practice problem 1:

Factorize the given polynomial expression 2x^2 + 17x + 15

Answer:

The final answer is (2x + 15) and (x + 1)

Factoring trinomials with coefficients practice problem 2:

Factorize the given polynomial expression x^2 + 18x - 19

Answer:

The final answer is (x - 1) and (x + 19)

Factoring trinomials with coefficients practice problem 3:

Factorize the given polynomial expression x^2 - 15x + 44

Answer:

The final answer is (x - 11) and (x - 4)