Trinomial Solution

Introduction to trinomial solution:

A polynomial with three terms is known as trinomial. One of the three terms is a constant and one of the remaining terms is a constant with a variable and another term is a constant with square of a variable. The constant is an optional. The roots of the trinomial are a solution for the given trinomial. The methods to solve a trinomial are by using quadratic formula or by using factoring method.


General form – Trinomial solution:


The general form of a trinomial is ax ^2+bx+c=0. In this x is a variable and a, b and c are constants.

The value of x is trinomial solution.

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Example problems – Trinomial solution:


Example 1 – Trinomial solution:

Solve the trinomial x ^2+8x+15=0.

Solution:

The given trinomial is x ^2+6x+9=0.

The given trinomial can be written as,

x ^2+3x+3x+9=0.

Take x as common from first two terms

x(x+3)+3x+9=0

Take 3 as common from last two terms

x(x+3)+3(x+3)=0

Take (x+3) as common

(x+3)(x+3)=0

The above product of monomials can be written as

x+3 = 0 and x+3=0

Now solve the above terms for x

So, `x=-3` and `x=-3`

The given trinomial’s solutions are -3 and -3.

So, the answer is -3.

Example 2 – Trinomial solution:

Solve the trinomial x ^2+12x+20=0.

Solution:

The given trinomial is x ^2+12x+20=0.

The given trinomial can be written as,

x ^2+10x+2x+20=0.

Take x as common from first two terms

x(x+10)+2x+20=0

Take 3 as common from last two terms

x(x+10)+2(x+10)=0

Take (x+10) as common

(x+10)(x+2)=0

The above product of monomials can be written as

x+10 = 0 and x+2=0

Now solve the above terms for x

So, `x=-10` and `x=-2`

The given trinomial’s solutions are -10 and -2.

Example 3 – Trinomial solution:

Solve the trinomial 2x ^2+20x+50=0.

Solution:

The given trinomial is 2x ^2+20x+50=0.

The given trinomial can be written as,

2x ^2+10x+10x+50=0.

Take 2x as common from first two terms

2x(x+5)+10x+50=0

Take 10 as common from last two terms

2x(x+5)+10(x+5)=0

Take (x+5) as common

(x+5)(2x+10)=0

The above product of monomials can be written as

x+5 = 0 and 2x+10=0

Now solve the above terms for x

So, `x=-5 ` and `x=-5`

The given trinomial’s solutions are -5 and -5.

So the solution is -5.

8th Grade Problem Solving

Introduction to 8th grade problem solving:

Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Here we are going to see about 8th grade problem solving and its example problems.                                                                                      Source: Wikipedia

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Example for 8th grade problem solving:


8th grade problem solving example: 1

Solve 3x + 3 = 12

Solution:

Given that 3x + 3 = 12

Subtract the -3 on both sides

3x + 3 – 3 = 12 – 3

3x = 9

Divide both side using 3

`(3x) / 3` = `9 / 3`

x = 3

The solution is x = 3.

8th grade problem solving example: 2

Solve 4x + 3y = 9

5x + 3y = 5

Solution:

Given that   4x + 3y = 9 ------------- (1)

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

Subtract the first equation and second equation

4x + 3y = 9

(`-` )  5x + 3y = 5

_______________

-x = 4

x = - 4

x = - 4 take the value and substitute in equation (1).

4x + 3y = 9 ------------- (1)

x = -4

4(-4) + 3y = 9

-16 + 3y = 9

Add the +16 on both sides

+16 – 16 + 3y = 9 + 16

3y = 25

Divide using 3 on both sides

` (3y) / 3` = `25 / 3`

y = 8.33

The solution is x = -4

y = 8.33

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Example for 8th grade problem solving:


8th grade problem solving example: 3

Solve 6x + 4y = 2

7x + 5y = 3

Solution:

Given that   6x + 4y = 2 ------------- (1)

7x + 5y = 3 ------------- (2)

From these two equations we cannot cancel particular variable

So we need to change this equation at least one variable equal on both equations

(1)   * 5 =   30x + 20y = 10

(2)   * 4 =   28x + 20y = 12

From this equation we can cancel one variable.

30x + 20y = 10

( - ) 28x + 20y = 12

Subtract you can get the value

2x = - 2

Divide using 2 on both sides

x = - 2 / 2

x = -1

x = - 1 take the value and substitute in equation (1).

6x + 4y = 2 ------------- (1)

6(-1) + 4y = 2

-6 + 4y = 2

Add +6 on both sides

+6 – 6 + 4y = 2 + 6

4y = 8

Divide using 4 on both sides

`y / 4` = `8 / 4`

y = 2

The solution is x = -1

y = 2

Logic Problem Solving

Introduction of solving logic problems:

Logic (from the Greek λογική logikē)[1] is the study of reasoning.[2] Logic is used in most intellectual activity, but is studied primarily in the disciplines of philosophy, mathematics, and computer science. Logic examines general forms which arguments may take, which forms are valid, and which are fallacies. It is one kind of critical thinking. In philosophy, the study of logic falls in the area of epistemology, which asks: "How do we know what we know?" In mathematics, it is the study of valid inferences within some formal language.[3]

-Source Wikipedia

The solving logic problems are given as follow:


Logic problem solving examples:


The examples of logic problem solving are given as follow:

Problem 1:

Roger can mow the lawn in 60 minutes and Sam can mow the lawn in 80 minutes. How long will it take for them to mow the lawn together?

Solution:

Step 1: Let assume variables:

Assume y = Time to mow lawn together

Step 2: Using the formula:

1/s1+1/s2 =1/sa

1/60 + 1/80 = 1/y

Step 3: Solving the equation

Take LCM

60 x 80 =480

To multiply each sides with 480

1/60+1/80 = 1/y

480/60+480/80 = 480/y

8+6 =480/y

14y = 480

Y=480/14 = 34.28

Answer is: The time taken for each of them to mow the lawn together is 34.28 minutes.

Problem 2:

John, Richard and Alexander can complete painting the bench in 3 hours. If John does the job alone he can complete it in 6 hours. If Richard does the job alone he can complete it in 7 hours. How long will it take for Alexander to complete the job alone?

Solution:

Step 1: Let assume variables:

Assume y = time taken by Alexander

Step 2: Using the formula:

1/6 + 1/7 +1/y = 1/3

Step 3: Solving the equation

To multiply both sides with 42y

1/6+1/7+1/y = 1/3

42y/6+42y/7+42/y = 42y/3

7y+6y+42 = 14y

14y-13y = 42

Y = 42

Answer is: The time taken for Alexander to paint the bench alone is 42 hours.


Logic problem solving practice problem:


The logic problem solving practice problems are given as follow:

Problem 3:

C tank can be filled by pipe C in 4 hours and by pipe D in 6 hours. When the tank is full, it can be drained by pipe E in 5 hours. if the tank is initially empty and all 3 pipes are open, how many hours will it take to fill up the tank?

Solution:

Step 1: Let assume variables:

Assume y = time taken to fill up the tank

Step 2: Using the formula:

Pipe C drains the water it is subtracted.

¼+1/6-1/5 =1/y

Step 3: Solving the equation

Take LCM

Multiply both sides with 120

¼+1/6-1/5 = 1/y

120/4+120/6 -120/5 =120/y

30+20-24 = 120/y

26y = 120

Y= 120/26 = 4

Answer: The time taken to fill the tank is 4 x 8/13 hours.

Word Problems in Algebra with Solution

Introduction of word problems in algebra with solution:

Algebra is a very important terms in math. We have different concepts in algebra. Word problems are also one of the concepts in algebra. Word problems are commonly difficult to understand for kids. Basically word problems are very easy .If we understand the word problem clearly, we can easily find the answer.Here we are going to discuss with some word problems with solution.

Addition and subtraction word problems in algebra:


Example 1:

In a city there are 3,942 men, 4,2 14 women and 6,231 children. What is the population of the city ?

Solution:

Step 1: Men in the city             =3,942

Step 2: Women in the city        =4,214

Step 3: Children’s in the city    = 6,231.

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

Step 4: Total                             = 14,387.


Subtraction word problem:

Example 2:

Natalie has 53 marbles. Sam has 105 marbles and lost 8 when his bag of marbles dropped. Find the difference between the number of marbles that Sam has left and Natalie’s.

Solution:

Step 1: We know that Natalie has 53 marbles.

Step 2: Sam has 105 marbles and lost 8 marbles.

Step 3: Therefore same has 105 marbles – 8 marbles – Natalie had 53 marbles.

Step 4: Now we need to subtract 105-8-53 = 44 marbles.

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Multiplication and division word problems in algebra.


Example 3:

There are 9 desks in a class room and each desk has 6 legs. What is the total number of the legs of the desks?

Solution:

Step 1: We know that we have 9 desks in a class room. And each desk has 6 legs .

Step 2: Now we need to find total number of the legs of the desks.

Step 3: Here we need to multiply 9 times 6 = 54

Step 4: Therefore, the result is 54 legs.


Division word problems:

Example 4:

5 people can sit on a bench. How many benches are required for 45 people?

Solution:

Step 1: We know that, 5 people can sit on a bench.

Step 2: Now we need to find how many benches are required for 45 people.

Step 3: Here we need to divide 45/ 5 = 9 benches.

Step 4: Therefore 9 benches are required for 45 people.

Solving Math Eqations

Introduction to Math equations:

In math, Equation is a condition on a variable. Here the condition is meant that two expressions should have equal value. In an equation, we note that at least one of the two expressions must contain the variable.

There is always an equality sign in an equation in math. That is the equality sign looks that the value of the expression to the left of the sign must equal to the value of the expression to the right of the sign. Only equality sign is used in an equation.

Here the equation remains the same, if the expression on the left side and on the right side is interchanged.

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Solving Example Problems on Math equations:


Problem 1:

Solve x in this math equation:  x – 7 = 14

Solution:

The given equation is x – 7 = 14.

Solving x,

Add 7 to each side in the equation.

x – 7 + 7 = 14 + 7

x  = 21.

To check:

x – 7 = 14

21 – 7 = 14

14 = 14 .

Problem 2:

Solve x in this math equation:  3m + 7 = 1

Solution:

The given equation is 3m + 7 = 1.

Solving m,

Subtract from 7 to each side in the equation.

3m + 7 - 7 = 1 – 7

3m = - 6

Divide by 3 to each side.

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

m  = - 2.

To check:

3m + 7 = 1

3(-2) + 7 = 1

-6 + 7 = 1

1 = 1 .

Problem 3:

Solve the math equation : 5r = 20

Solution:

The given equation is 5r = 20

Solving r,

Divide by 5 to each side.

`(5r)/5` = `20/5`

r = 4.

To check:

5r = 20

5*4 = 20

20 = 20.

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Example word problem on math equations:


Problem 4:

Peter’s father’s age is 5 years more than three times Peter’s age. Peter’s father is 44 years old. Find the math equation to find Peter’s age?

Solution:

We do not know Peter’s age. So let us take peter’s age is A years.

Three times Peter’s age is 3A years.

Peter’s father’s age is 5 years more than 3A. that is Peter’s father is 3A + 5 years old.

Peter’s father age is 44 years old.

Therefore,

The equation is 3A + 5 = 44

Solving A,

Now subtract 5 from each side.

3A + 5 – 5 = 44 – 5

3A = 39

Divide by 3 to each side.

`(3A)/3` = `39/3`

A = 13.

So Peter’s age is 13 years old.

7th and 8th Grade Math

7th and 8th grade math:

In this article we discuss about solving problems in 7th and 8th grade math. In this article we are going to discuss for solving problems in 7th and 8th grade math topics. In our day to day we are using arithmetical concepts often. Solving problems in 7th and 8th grade math covers following chapters

Numbers
Measures
Algebra
Geometry

The solving problems in 7th and 8th grade math topics are given below.

Example problems in 7th and 8th grade math:

Example 1:

Simplify the given expression `((-80) * (3))/ (- 2)^2`

Solution:

Given expression `(-80) * (3)/ (- 2)^2`

`(-80) * (3) / (4)` =` - (80 * 3)/ (4)` =  -60

Solution is -60.

Example 2:

Simplify the given expression 8 + (5 – 2) =?

Solution:

Given expression 8+ (5-2)

8 + (5 – 2) = 8+3 = 11

Solution is 11.

Example 3:

Davit sells a house through a broker for Rs.5, 00,000 paying a brokerage of 2%. Find the amount of brokerage.

Solution:

The selling price of the house =Rs 500000

Rate of brokerage                     = 2%

Amount of brokerage                 = 2/100 * 500000

= 10000

The amount of brokerage         = Rs 10000

Example 4:

Radius of a sector is 10 cm. Its arc length is 18 cm. Find its perimeter.

Solution:

Given, r = 10cm and l = 18 cm

P = l+2r units.

= 18 + 2 * 10 cm.

= 18+20 = 38cm.

P = 38cm.

The perimeter is 38cm.

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Example 5:


Find the volume of the right prism whose are of the base is 350 cm2 and height is 24 cm.

Solution:

Given that area of the base, A = 350 cm2 and height (h) of the prism = 24 cm.

Volume of the right prism = Area of the base * height cu units.

= A h

= 350 * 24 = 8400

Volume = 8400 cm3

Example 6:

Find the circumference of a circle whose radius is 7cm.

Solution:

Radius r = 7cm

Circumference, C = 2 pi r

= 2 * 22/7 * 7

= 44

Circumference of the circle = 44cm.

Example 7:

One third of a certain number is 9. Find the number.

Solution:

Let the unknown number be m. One third of m is 1/3 * m

But 1/3 m = 9

m = 9 * 3 (by rule 3)

m = 27

The number is 27.

Example 8:

Sum of 3 consecutive odd numbers is 51. Find the numbers.

Solution:

Framing the equation:

Let the first odd numbers be x.

Then the second and third odd numbers are (x+2) and (x+4)

There sum is 51.

x+(x+2)+(x+4) = 51

x+x+2+x+4 = 51

3x + 6 = 51(adding the like terms)

3x+6 = 51 is the required equation.

Solving the equation:

3x+6 = 51

3x = 51 – 6 (by rule 1)

= 45

x = 45 * 1/3 (by rule 3)

x = 15

The consecutive odd numbers are x, (x=2), (x=4)

15, (15+2) and (15+4)

The required numbers are 15, 17 and 19.

Solving Solution Set

Introduction:

Information, lettering, any other article restricted in a set. Example, the solving solution elements of the set {p, q, r} are the letters p, q, and r.

These are used to solving the set problems.

Type 1: Empty set

Type 2: Equivalent sets

Type 3: Singleton set

Type 4: Universal Set

Type 5: Subset

Type 6: Proper Subset

Type 7: Power Set

We are leaving to learn in feature about the laws of set operations, Relations and Functions.


Examples:


1. Identify finite and infinite sets from the following in solving solution set:

(i) {All schools in Tamil Nadu}.

(ii) N.

(iii) The set of all prime numbers.

Solution:

(i) Every part of schools in Tamil Nadu can be counted one by one and we come to an end in the solution counting process. So the set {All schools in Tamil Nadu} is a finite set.

(ii) N = {1, 2, 3,…}. at what time we count up rudiments of N one by one as 1 for 1, 2 for 2, 3 for 3, 4 for 4, we are not able

to come to an end in the counting process. ? the set N is an infinite set.

(iii) When we write the prime numbers one by one as 2, 3, 5, 7, 11, 13, 17 and so on, we are unable to come to an

end in the counting process. ? the set of all solving prime numbers are an infinite set.

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More Problems:

Represent the following solving sets in Rule Form:


(i) The set of all natural numbers less than 6.


(ii) The set of vowels in English alphabet.


(iii) The set of the numbers 2, 4, 6, … .


Solution: (i) A natural number less than 6 can be described by the statement’s ? N, x < 6.

?the set is { x | x ? N, x < 6}.


(ii) A vowel in English alphabets can be described by the statement: x is a vowel in English alphabet.


? the set is {x | x is a vowel in English alphabet}.


(iii) A number x of the form 2, 4, 6, … can be described by the statement:x = 2n,n ? N.

? the set is { x | x = 2n, n ? N}.

Study Solution Set

Definition

A solution set is the set of values that satisfy a given set of equations or inequalities.

For example, for a set {fi} of polynomials over a ring R, the solution set is the subset of R on which the polynomials all vanish (evaluate to 0).i

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Expressing the Requirements


a solution set is a set of possible values that a variable can take on in order to satisfy a given set of conditions.

When you say that the solution set of two parallel lines.  I assume you mean the intersection between two parallel lines y=m*x+b and y=m*x+c. That is, we seek m*x+b = m*x+c.

Expressing this requirement in the terms of a polynomial f(x) =0, where f(x) =b-c, first, we observe that this has no solution unless b=c. i.e., unless the two parallel lines are in fact one and the same. In this case, the variable x is could not constrained and we have an infinite set of solutions since f(x)=b-c=0 is independent of x. In general, however, there is no solution to f(x)=b-c=0 when b is different from c. No x-value in R will satisfy this condition. This means two parallel lines do not intersect to the irrespective of the x value.



The solution set of the single equation f(x) = x is the set {0}.

1. The solution set of the single equation f(x) = x is the set {0}.

2. For any non-zero polynomial f can be over the complex numbers in one variable, the solution set is made up of finitely many points.

3. However, for that the complex polynomial in more than one variable the solution set has no isolated points.

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If the solution set is empty, then there are no such xi such that

f(x0,...,xn) = c

Becomes true for a given c.

He solution set of an equation in the form ax + by = c with a, b, and c real-valued constants, this forms a line in the vector space R^2. However, it cannot always be easy to graphically depict solutions sets – for example, the solution set to an equation in the form ax + by + cz + dw = k (with a, b, c, d, and k real-valued constants) is a hyper plane

Math Pattern Solutions

Introduction of math pattern solutions:

A pattern, from the French patron, is a type of theme of recurring events or objects, sometimes referred to as elements of a set. These elements repeat in a predictable manner. It can be a template or model which can be used to generate things or parts of a thing, especially if the things that are created have enough in common for the underlying pattern to be inferred, in which case the things are said to exhibit the unique pattern.

(Source:Wikipedia)


Concept of math pattern solutions:


In math, numeric patterns in algebra are patterning prepared from numbers.

The numbers can be in a listing. Any math operation likes addition, subtraction, multiplication, or division, you makes the pattern.

Many of the patterns in math you watch will use addition. The same number will be additional to each number in the list to make the next number in the math list.

Example

2, 4, 6, 8, 10, 12, 14…

The pattern in math is to add 2 every time. The subsequently number is 16, then 18.

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Example for math pattern solutions:


6, 14, 30, 62, …

In the math series on top of, each term after the primary is determined by multiplying the previous term by m and then adding n. What is the value of n?

Solution:

Method 1:

The best way to solve this would be if you see that the pattern:

6 × 2 + 2 = 14
14 × 2 + 2 =28

The value of n is 2.

Method 2:

If you were not able to see the pattern then you can get nearer with two equations and then resolve for n.

6m + n =14             (equation 1)
14m + n = 30          (equation 2)

Use substitution method

n in equation 1

n = 14 – 6m

Substitute into equation 2

14m + 14 – 6m = 30
8m = 16
m = 2

Substitute m = 2 into equation 1

6(2) + n = 14
n = 2

Final solution is n = 2

Balanced Equation Math

Introduction of balanced equation math:

We know that algebra is very important term in math. Algebra has a different type of equations. Normally we can do addition , subtraction, multiplication and division. An algebra equation contains variables, symbols etc. Here we are going to learn about balanced equation in math. It’s also called as verification in math. Using this method we can also verify if the solution is correct or not.

Example problems of balanced equation in math:

Example 1:

Y+9=15

Solution:

Step 1: y-9+9= 15-9 (Subtract 9 on both the sides)

Step 2: After simplification we get y = 6.

Balanced equation:

Step 1: y+9= 15

Step 2: 6+9= 15

Step 3: Therefore 15=15.

Here we get the value of y = 6.Now we need to substitute y value in the question. When we substitute we get, the same value on both the sides. Its called balanced equation in math.



Example 2:

Given:

Solve: 5z-10=130

Solution:

We need to find the z value

Step 1: 5z-10=130 (first we need to add 10 on both the sides)

Step 2: 5z=140 (Now we divide using 5 on both the sides)

Step 3: z=28.

Balanced equation:

Step 1: 5z-10=130

Step 2: 5(28)-10= 130.

Step 3: Therefore 130 = 130.

Here we get the value of z = 28.Now we need to substitute z value in the question. When we substitute we get, the same value on both the sides. Its called balanced equation in math.

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More about balanced equation math:


Example 3:

Solving equations using elimination method

Solve: 4p+3q =21 and -4p-6q=12

Solution:

Step 1: It question is 4p+3q =21 and -4p-6q=12

Step 2: first we need to change the sign and simplify the equations. Here also we have the opposite sign. So we need not change the sign

Step 3: When we simplify we get -3q= 33.So q= -11

Now we need to calculate the value of p.

Step 6: so, we plug in q value in any equation to find p.

Step 7: here the equation one is 4p+3q =21

Step 8: So, 4p+3(-11)=21

Step 9: 4p-33=21 (Add 33 on both the sides).

Step 10:4p-33+33=21+33. 4p=54

Step 10: So the value of p =13.5.

Balanced equation:

For, the past question we have 2 equations. We can take any one of the equation and substitute p and q value.

Step 1: Let us take the first equation, 4p+3q =21

Step 2: 4(13.5)+3(-11)= 21

Step 3:54-33=21

Step 4:21=21

Here we get the value of p = 13.5 and q= -11.Now we need to substitute y value in the question. When we substitute we get, the same value on both the sides. Its called balanced equation in math.

These are the example problems of balanced equation in math.

Math Integer Solver

Introduction to math integer solver:

In mathematics, integer solver is one interesting topics in number representation. Integer has a set of numbers in which positive numbers, negative numbers and zero. It contains complete entity or unit. In integer, there is no fractional part. Integer performs different arithmetic operations such as addition, subtraction, multiplication and division. Let us solve some example problems in math integer solver.

Example for integers:

112, -252, 0 etc,

Positive integer: {1, 2, 3, 4, 5 …..}

Negative integer: {…….., -5, -4, -3, -2, -1}


Example problems for math integer solver:


Let us see some of the example problems for math integer solver.

Example 1:

Perform arithmetic operation addition for the given two integer numbers

256 + 386

Solution:

Given two integer numbers are

256 + 386

Both are two positive integers so, the result is also a positive numbers

Here we add 256 into 386, and then we get the result

256 + 386

642

Solution to the given two integers is 642.

Example 2:

Perform arithmetic operation subtraction for the given two integer numbers

857 and 420

Solution:

Given two integer numbers are

857 - 420

Here we subtract 857 into 420, and then we get the result

857 - 420

437

Solution to the given two integers is 437.



Example 3:

Perform arithmetic operation multiplication for the given two integer numbers

45 × 28

Solution:

Given two integer numbers are

45 × 28

Both are positive integers so, the result is also a positive numbers

Here we multiply 45 into 28, and then we get the result

45 × 28

1260

Solution to the given two integers is 1260.

Example 4:

Perform arithmetic operation division for the given two integer numbers

1250 / 25

Solution:

Given two integer numbers are

1250 / 25

Both are positive integers so, the result is also a positive numbers

Here we divide 1250 by 25, and then we get the result

1250/ 25

10

Solution to the given two integers is 10.

Example 5:

Perform arithmetic operation addition for the given two integer numbers

- 36 + 42

Solution:

Given two integer numbers are

- 36 + 42

- 36 is negative number and 42 is positive number

Here we add - 36 into 42, and then we get the result

- 36 + 42

6

Solution to the given two integers is 6.

Example 6:

Perform arithmetic operation subtraction for the given two integer numbers

22 - 0

Solution:

Given two integer numbers are

22 - 0

22 is positive numbers and 0 has no sign

Here we subtract 22 into 0, and then we get the result

22 – 0

22

Solution to the given two integers is 22.


Practice problems for math integer solver:


Some practice problems for math integer solver are,

Perform arithmetic operation for the given two integer numbers

i). 56 + 23

ii). 85 – 34

iii). 32 × 12

iv). 64 / 8

Solution:

i). 79

ii). 51

iii). 384

iv). 8

Patterns in Math

Introduction for Patterns in Math:

In math, under a certain conditions, the numbers are listed, which is said to be a pattern. Basically patterns in math are classified into three types. Every pattern has some properties. In this article, we shall discuss about various kinds of patterns in math. Also we shall solve some problems regarding pattern in math.

Types of Patterns in math:

Arithmetic Pattern
Alphabetic Pattern
Geometric Pattern
These are the 3 different types of patterns in math.

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Arithmetic patterns:

Typically the patterns are indicated as sequences. There are two types of possibilities for the occurrence of sequence. They are finite and infinite sequences. We can articulate a number pattern using some unique symbols. We can clarify the number patterns in several ways. Let us consider a number pattern {2, 4, 6, 8, 10,….}. For the given pattern, the first term of this pattern is 2; and the second term can be attaining by adding 2 with the first term.

Alphabetic Patterns:

Patterns based on alphabetical lettering are called as alphabetic pattern. In addition, the alphabets in a sequence are the alphabetical pattern for the exacting sequence.

Geometric Patterns:

Whenever the geometric shapes involved in some patterns, then they are said to be geometric patterns. For instance, Ellipses are the geometric shapes. Those ellipses are developed from the basic geometric shape called circles. Is this topic T Test Example hard for you? Watch out for my coming posts.


Examples for patterns in math:


Example 1:

Find the missing terms from the pattern given below.

2, 4, 8, 16, 32, 64, ___, ____

Solution:

The first term of the pattern is 2

The second term of the pattern = 2 * 2 = 4

The third term is 4 * 2 = 8

The fourth term is 8 * 2 = 16

Similarly, the missing terms can be determined as follows.

The seventh term will be 64 * 2 = 128

The eighth term is 128 * 2 = 256

So the correct pattern is 2, 4, 8, 16, 32, 64, 128 and 256.

Example 2:

Find the next two terms in the pattern given below.

1, 3, 5, 7, 9, 11, ___, ____

Solution:

The first term of the pattern is 1

The second term of the pattern = 1 + 2 = 3

The third term is 3 + 2 = 5

The fourth term is 5 + 2 = 7

Similarly, the missing terms can be determined as follows.

The seventh term will be 11 + 2 = 13

The eighth term is 13 + 2 = 15

This is an odd sequence of number.

So the exact pattern is 1, 3, 5, 7, 9, 11, 13 and 15.

Math Problems Quizzes

Introduction of math problem quizzes:

In this lesson we can show the math problem quizzes. It very helpful and you can develop your math skills. These quizzes are really exercising your brain at the same time carefully handle the number terms in word problems. The quizzes are showing word problem, fraction, decimal, ect. Let we see math quizzes.

Math problem quizzes:

1) At restaurant, the men drunk 4 1/2 cup of coffee and women drunk 5 ½ cup of coffee. How many coffees were drunk at restaurant?

(Answer:10)

2) Yesterday I bought 6.25ml paint. And I used 1.50ml paint. Now, how much paint I have? Or Find rest of paint?

(Answer: 4.75)

3) There were 40 questions on a science test. William got 70% of them correct. How many questions did he answer correctly?

(Answer: 28)

Find Missing number:

4)  --- , 5 , 4 ,10,--- , 15, 8 ,20, 10 ,25, ---- ,---- , ----- ,----, 16 ,40, 18 ,45, 20 ,----, ----,----,

24,60 26, ---- ,----,

(Answer: 2,6,12,30,14,35,50,22,55,65,28)

5)  Suppose 5 dogs take 5 minutes to eat 5 bones.

How many minutes would it take 4 dogs to eat 4 bones?

How many bones would it take to eat 24 bones in 24 minutes?

(Answer: 4 and 24)

To find with formula:

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6) A pillar 8 feet tall casts a shadow 4 feet long on the ground. If the pillar was 26 feet tall, how many feet in length would the shadow be?

(Answer: 13)

7) Which of the following are the prime factors of 96?

(Answer: 2 and 3)

8) A wall has an area of 96 square inches and a width of 8 inches. What is its length?

(Answer: 12inches)

9) A florist buys roses at $1.50 a piece and sells them for $3.00 a piece. If there are no other expenses, how many roses should be sold in order to make a profit of $300?

(Answer: 200)

10) If the average person throws away 2.5 pounds of garbage every day, how much garbage would the average person throw away in one week?

(Answer: 17.5 pounds)                                                                                                                                                                                                                                                                         

Learning Solution Set

Introduction for learning solution set:

Theory:

Set is an achievement of our recent mathematics. It appears in all branches.  It originated when mathematicians attempted to axiomatize mathematics within the frame work of logic. Mathematicians developed the theory of sets and solutions of set. It becomes a milestone in the growth of mathematics. Now we proceed to introduce the concept of a set. We are learning the solution of set by operations. Let us see about learning solution set in this article. Please express your views of this topic Union Set Theory by commenting on blog.


Learning Solution Set : Set Operations


We shall now study about the set operations

(i) Union of two sets

(ii) Intersection of two sets

(iii) complements

(i) Union of two sets

Let A and B be two given sets. The set of all elements that belong either to A to either to B or to A or to both is called the union of A and B. We denote the union of A and B by A U B. We are learning the union by following expression:

A U B = { x | x ∈ A or x ∈ B or x ∈ A and B}.

We write A U B = {x| x ∈ A or x ∈ B} where it is understood that the word or is used in the inclusive sense; that is, x ∈ A or x ∈ B stands for x ∈ A or x ∈ B or x ∈ A and B.


(ii) Intersection of two sets

Let A and B be two given sets. The set formed by the elements that are common to both A and B is called the intersection of A and B. We denote the intersection of A and B by A ∩B. We are learning intersection by following expression:

A ∩ B = {x | x ∈ A and x ∈ B}.

(iii) Complements:

Subtract the given sets A and B. Learning solution by A - B.

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Learning Example Solutions for Set Operations:


Find the solution for following examples:

1). If A = {1, 2, 3, 4} and B = {3,4, 6}, find A UB.

Solution:

List the elements of both A and B and avoid duplication.

Thus, A UB = {1, 2, 3, 4, 6}.

2). If A = {4,5,6,7} and B = {5,6,8}, find A∩B.

Solution:

All elements in A and B: 4,5,6,7,5,6,8.

Common elements in A and B:5,6

∴ A ∩B = {5,6}.

3). If A  = {2,3,4,5}, B={3,4,5}, find A - B.

A - B = {2}.

Rotational Symmetry

If an object retains the original shape when transformed, it is said to have symmetry. For example a vertical parabola when flipped across the vertical line passing through the vertex, it retains its shape. So we call that a vertical parabola is symmetrical over the axis through the vertex which is also called as axis of symmetry of the parabola.

Same way if an object retains the original shape when rotated over the center of the object by certain angle, it is said to have a rotational symmetry. It is also called as angular symmetry. This concept is dealt in the topic of rotational symmetry definition geometry. Let us take a closer look.

In real life there are many examples of rotational-symmetry. For example, a fan with certain number of blades has an angular symmetry because the angle between any two consecutive blades is constant. Rotating the blade by that much of angle we can observe that the shape of the fan appears to be same as it was originally.
In the above example, it was fairly easier to observe the angular symmetry. But, in general, how to find rotational symmetry? One has to keenly note that if the given shape looks same as before even when undergoes the transformation of rotation by some angle.
In certain cases, the original shape is retained repeatedly. That is, before completing one full revolution, the shape exhibits the originality a number of times. This ‘number of times’, is known as order of rotational symmetry of the shape. Consider an equilateral triangle. The shape remains the same after every rotation of 120o and hence in one complete revolution the shape of the equilateral triangle remains the same three times. Therefore, the order here of the angular symmetry in this case is 1. A square has that order as 4 whereas a rectangle has only two. A circle has the same order infinitely.

Now let us discuss an important point. An object comes back to the original shape only when it undergoes a complete rotation. Can we say that 1 is the order of angular symmetry for this object? No! A rotational symmetry for any object is said to exist only when it exhibits the original shape more than once in one complete revolution. For that matter, all the objects come back to the original shape after a rotation of 360o. Therefore, 1 as the order of angular symmetry does not exist and also it does not make sense. Thus, the minimum order for an angular symmetry is 2.

Polynomials

The word ‘poly’ that is often used in math is actually derived from Greek. ‘Poly’ in that language means ‘many’. So, if an expression has many terms, then it is called as polynomial. But the number of terms has to be finite. Generally up to three terms expressions have their own identities like a single term is called a monomial, two terms as binomial and three terms as trinomial.  Polynomials (abbreviated as Pnml) can also form the core part of a function, in such cases they are called as polynomial functions.

From what we described any expression or any function which has many terms could be Pnmls. Is this statement correct? No, there is a limitation. Only expressions in which no term is divided by any variable and do not contain a non-negative exponent terms can be referred as Pnmls.

In this topic of polynomials help, we will see some examples of polynomials and how to simplify polynomials.The following are the polynomials examples.

2x^3 + 3x^2 – 7x + 2
6x4 – 2x^3 + 4x^2 + 2x + 9
4x^3y^2 – x^2y^2 + 2xy - 14
5x^3 + (x^2/2) + 4x –7

In the third example we find three terms contain two variables but it does not matter as far as the definition of Pnml  is concerned.

In the last example one can notice that one of the terms (the second term) is divided by 2. But it does not affect the Pnml status of the expression as 2 is only a constant. Only when any term is divided by any variable cannot qualify to be a Pnml.
The following expressions cannot be called as Pnmls.
x^3 + 4x^2 – 7x + (2/x)
6x4 – 2x-3 + 4x^2 + 2x + 9
7x^3– 2x^2 + 5x^2/3 + x + 1
1 + (x/1!) + (x^2/2!) + (x^3/3!) + …….

In the first expression, the last term is divided by a variable. In the second expression we find a negative exponent in the second term. The third term of the the third expression has a rational exponent which cannot be simplified as an integer. The fourth expression has infinite number of terms. Hence all of them cannot qualify as Pnmls.

It is possible that a Pnml may have exactly similar terms excepting that the coefficients may be different. Those terms are called like terms. Such expressions can be simplified by algebraically adding the like terms. Similarly a Pnml can be simplified and expressed in factored form by identifying common factors. Simplified Pnmls show a better presentation.

Middle School Math Practice

Introduction to middle school math day:

Middle school class books are designed to presents the practice with math skills and concept wise explanations such as whole numbers, fractions, decimals, percents, geometry, and square roots. Middle school math tells about the measurement, property, and relations about quantities and sets, with numbers and symbols. It is a function of expressions. In this article we shall discuss about middle school math day problems.


Sample problem for middle school math day:


Example 1:

Solve the given equation 4× (x + 8 + 11) = 88.

Solution:

Grouping the terms like, the left side of the equation becomes
4 × (x + 8 + 11) ==> 4 × (x + 19)

Using the distributive property,
4 × (x + 19) ==> 4 × x + 4 × 19

Carrying out multiplications,
4 × x + 4 × 19 ==> to 4x + 76

The equation now becomes
4 x + 76 = 88.

Subtracting a 76 (adding a -76) to each side gives us
4x + 76 + (-76) = 88 + (-76)

4x + (76 + (-76)) = 88 - 76

4x + 0 = 13

4x = 12

Since the x value is multiplied by 4, we divide both sides by 4 to solve for x:
4x = 12

4x ÷ 4 = 12 ÷ 4

(4x)/4 = 3

x = 3.

We get the answer value is X = 3

We can check the answer in the solution of a original equation:
4× (3 + 8 + 11) = 88

4 × 22 = 88

88 = 88 so our answer is correct.

Problem 2:

Find the diameter of the circle. A radius of the given circle is 9 cm.

Solution:

Given:

Radius of the circle is r = 9 cm

Diameter of the circle is d = 2r

d = 2 * 9

d = 18

So, the diameter of the circle is 18 cm.

Problem 3:

The perimeter of a rectangle is 1200 meters and its length L is 5 times its width W. Find the values of W and L, and the area of the rectangle.

Solution:

Perimeter of rectangle=2L+2W,

2 L + 2 W = 1200

We now rewrite the statement. Its length L is 5 times its width into a mathematical equation as follows:

L = 5 W

We have to substitute L =3W in the equation 2 L + 2 W = 1200

2(5 W) + 2 W = 1200

12 W = 1200

W =100 meters

Use the equation L = 5 W to find L.

L = 5 W = 500 meters

Use the formula of the area.

Area = L x W = 500 * 100 = 50000 meters 2.

So, the area of the rectangle=50000 meters 2.


Practice problem for middle school math day:

Find the diameter of the circle. The radius of the given circle is 12 cm.

Answer: d = 24

Solve the given equation 9× (x + 3 + 11) = 99.

Answer: x = -3

Rectangle Word Problem

Introduction to Rectangle:

A Rectangleis any quadrilateral with four right angles. The term "oblong" is occasionally used to refer to a non-square rectangle. A rectangle with vertexes ABCD would be denoted as  ABCD.

We shall work out some problems on rectangles


Rectangle word problem 1:


A rectangle has a perimeter of 120 meters and its length L is 2 times its width W. Calculate the dimensions W and L, and the area of the rectangle.

Solution to rectangle word problem 1:

The formula to find the perimeter of rectangle is

2 L + 2 W = 120

Now, we rewrite the statement, length L is 2 times its width “w” into a mathematical equation as follows,

L = 2 W

We substitute L in the equation 2 L + 2 W = 120 by 2 W.

2(2 W) + 2 W = 120

Expand and group like terms.

6 W = 120

Solve for W.

W = 20 meters

Use the equation L = 2 W to find L.

L = 2 W = 40 meters

Use the formula of the area.

Area = L W = 40 * 20 = 800 meters 2


Rectangle word problem 2:


The perimeter of a rectangle is 70 feet and its area is 304 feet 2. Find the length L and the width W of the rectangle, such that L > W.

Solution for rectangle word problem 2:

The formula for perimeter of a rectangle is given by

2 L + 2 W = 70

and the formula area of a rectangle is given by

L W = 304

Divide all terms in the equation 2 L + 2 W = 35 by 2 to obtain

L + W = 35

Solve the above for W

W = 35 - L

Substitute W by 35 - L in the equation L W = 304

L(35 - L) = 304

Expand the above equation and rewrite with right term equal to zero.

-L 2 + 35 L - 304 = 0

The above is a quadratic equation with two solutions.

L = 16 and L = 19

Use W = 35 - L to find the corresponding values of W.

W = 19 and W = 16

Since L > W, the rectangle has the dimensions

L = 19 feet and W = 16 feet.


Practice rectangle word problem 1:


The perimeter of a rectangle is 90 feet and its area is 504 feet 2. Find the length L and the width W of the rectangle, such that L > W.

Practice rectangle word problem 2:

A rectangle has a perimeter of 240 meters and its length L is 5 times its width W. Calculate the dimensions W and L, and the area of the rectangle

Type in Word Problem And Solve

Introduction for 'type in word problem and solve':

In general, word problem refers to the mathematical exercise where the information on the given problem can be written in mathematical expressions. If we type a word problem in online, the online tutors will solve the problems. In this article type in word problem and solve, we are going to discuss few basic word problems. Please express your views of this topic Practice Probability Problems by commenting on blog.


Example problems for 'type in word problem and solve':


The example word problems are given below:

Example 1:

Totally, there are 1000 seats in a theatre. Out of which, 700 seats are occupied. Calculate the percentage of seats that are occupied.

Solution:

Total seats    =  1000

Occupied seats  =  700

Percentage   = `<< 700 / 1000>>`   x 100

= `<< 7/10>>`   x 100

= 70

Example 2:

Randy bought a doll for `$` 45 but he sold it for `$` 70. Calculate his gain amount.

Solution:

Cost price of doll   =  $ 45

Selling price of Clock  =  $ 70

Profit  or  Gain   =  Selling price - Cost Price

=  70 - 45

=  $ 25

Example 3:

Julie buys an ornament, which costs `$` 810. Suppose the sales tax rate is 8%, find the total amount she have to pay for the ornament?

Solution:

Sales tax  =  8% of the price tax

= 8%  x  810

= 0.08 x 810

= 64.8

Final price = price before the tax + sales tax

= 810 + 64.8

= $ 874.8

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Practice problems for 'type in word problem and solve':


1) Totally, there are 830 seats in a theatre. Out of which, 700 seats are occupied. Calculate the percentage of seats that are occupied.

Answer: 84.3 percent

2) Robert bought a doll for `$` 40 but he sold it for `$` 60. Calculate his gain amount.

Answer: 20 dollars

3) Serena buys an ornament, which costs `$` 800. Suppose the sales tax rate is 4%, find the total amount she have to pay for the ornament?

Answer: 832 dollars

Solution Set Online

Introduction of solution set online:

In the solution set online, a collection of well defined objects is called a set. For example, the collection of all natural numbers, the collection of all equilateral triangles in a plane, the collection of all real numbers, the collection of all vowels in English alphabet are some examples of sets since we can definitely say what objects are there in each of the collections. Consider the following statements for online solution set:

(i) The set of all tall students in your class.

(ii) The set of good books you have studied.


Example of solution set online:


The examples of solution set online is given as follows:

1. If A = {1, 3, 4, 5, 6, 7, 8, 9} and B = {1, 2, 3, 5, 7}, find n(A), n(B), n(AUB) and n(A∩B) and verify the identity

n(AU B) ≡ n(A) +n(B) − n(A∩B).

Solution: We observe that AUB = {1, 2, 3, 4, 5, 6, 7, 8, 9} A∩B ={1, 3, 5, 7}.

n(A) = 8, n(B) = 5, n(A UB) = 9 and n(A∩B) = 4.

We find n(A) + n(B) − n(A∩B) = 8 + 5 − 4 = 9.

Here, n(AUB) = 9. So n(AUB) = n(A) + n(B) − n(A∩B).

In fact, this result is true for any two finite sets.

2. If X = {a, c, d, e, f, g, h, i} and Y = {a, b, c, d, g}, find n(X), n(Y), n(XUY) and n(X∩Y) and verify the identity

n(XU Y) ≡ n(X) +n(Y) − n(X∩Y).

Solution: We observe that XUY = {a, b, c, d, e, f, g, h, i} A∩B ={a, c, e, g}.

n(X) = 8, n(Y) = 5, n(X UY) = 9 and n(X∩Y) = 4.

We find n(X) +n(Y) − n(X∩Y) = 8 + 5 − 4 = 9.

Here, n(XUY) = 9. So n(XUY) = n(X) + n(Y) − n(X∩Y).

In fact, this result is true for any two finite sets. Understanding Subtracting Complex Numbers is always challenging for me but thanks to all math help websites to help me out.


Exercise problems of solution set online:


1. If A = {1, 2, 3} and B = {2, 3, 4}, find A ∩B.

Answer:  A∩B = {2, 3}.

The exercise problem of solution set online is given as follow:

2. If A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 7}, find A − B and B − A.

Answer: A−B = {2, 4, 5, 6}. B −A = {7}.

3. If A = {1, 2, 3, 4} and B = {2, 4, 6}, find A UB.

Answer: AUB = {1, 2, 3, 4, 6}.