A Simple Expression for Math

Introduction to a simple expression for math:

A simple expression for math involves the process of solving simple algebraic expression. The simple algebraic expression in math is mainly used to calculate the unknown variable value with the basis of known values. Generally the variables represented in the simple math expression are nothing but an alphabetic letters. The following are the example problems for simple math expression with detailed solution.


Simple math expression example problems:


Example 1:

Solve the simple math expression.

-5(n + 2) = n + 9

Solution:

Given expression is
-5(n + 2) = n + 9

Multiplying the factors in left term
-5n - 10 = n + 9

Add 10 on both sides
-5n - 10 + 10 = n + 9 + 10

Grouping the above terms
-5n = n + 19

Subtract n on both sides
-5n - n = n + 19 -n

Grouping the above terms
-6n = 19

Multiply -1/6 on both sides
n = -`19/6 `

n = -`19/6` is the solution for the given expression

Example 2:

Solve the simple math expression.

-2(n - 3) - 4n - 1 = 3(n + 4) - n

Solution:

Given expression is
-2(n - 3) - 4n - 1 = 3(n + 4) - n

Multiplying the integer terms
-2n + 6 - 4n - 1 = 3n + 12 - n

Grouping the above terms
-6n + 5 = 2n + 12

Subtract 5 on both sides
-6n + 5 - 5 = 2n + 12 -5

Grouping the above terms
-6n = 2n + 7

Subtract 2n on both sides
-7n - 2n = 2n + 7 -2n

Grouping the above terms
-9n = 7

Multiply -1/9 on both sides
n = - `7/9`

n = - `7/9` is the solution for the given expression


Simple math expression practice problems:


1) Solve the simple math expression.

-7(n - 2) - 2n - 2 = 5(n + 2) - 5n

Answer: N = 0 is the solution for the given expression.

2) Solve the simple math expression.

-5(n - 3) - 2n - 3 = 2(n + 1) - 3n

Answer: n = `13/4` is the solution for the above given expression.

Math Probability for Grade 4

Math probability for grade 4 Introduction:

The grade 4 math probability is number of possible events are divided into total number of possible events this is called the math probability for grade 4. This grade 4 math probability is contains two types of distributions. That is the discrete and continuous distribution.  The general formation of the math probability for grade 4.

The probability of Event P (A) =` ("No. of possible events n(a)")/("Total no. of events n(s)")`

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Math probability for grade 4 Examples:


Dice Problems:

Math probability for grade 4 Example 1:

Throw a single dice; what is the probability of getting number 1?

Solution:

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

Total Number of possibles = n (s) = 6

The number of outcomes = n (a) = {6}

The possible outcomes = n (a) = 1

The probability of getting value = `1/6.`

Math probability for grade 4 Example 2:

Throw a single dice; what is the probability of getting number 4?

Solution:

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

Total Number of possibles = n (s) = 6

The number of outcomes n (a) = {4}

The Number of possible outcomes = n (a) = 1

The probability of getting value =` 1/6.`

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Coins probability problems for grade 4:


Math probability for grade 4 Example 3:

Flip a coin and what is the probability of getting chance of one tail?

Solution:

Step 1:

Total number of outcomes = n (s) = {H, T} = 2

Step 2:

Tossing a coin with only one tail:

Number of possible event = n (a) = {T} = 1

Step 3:

Formula:

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

Answer:

P (A) = `1/2.`

Convert into a decimal 0.5

The probability of one tail is 0.5 or Rounded 50%.

Math probability for grade 4 Example 4:

If throw a coin what is the probability of getting one head? The possible outcomes are:

Solution:

Step 1:

Total Number of Possible Events = n (s) = {T, H} = 2

Step 2:

Tossing a coin with only one head:

Number of possible events = n (a) = {H} = 1

Step 3:

Formula:

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

Answer:

P (A) = `1/2` .

Convert into a decimal 0.5

The probability of one head is 0.5 or Rounded 50%.

Do My Math For Me

Introduction:

Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. Mathematics is used throughout the world as an essential tool in many fields, including natural science, engineering, medicine, and the social sciences. In this article we shall discuss about math problem with detailed solution.


Example problems:


Example 1: Isolate the variable x:

8x + 2y = 12.

Solution:

To isolate x in the given equation the following steps must be followed:

Step 1: Subtract the equation on both sides by 2y

8x + 2y - 2y = 12 - 2y

Step 2: Simplify the above equation it give the equation as:

8x = 12 - 2y

Step 3: Divide the equation both side by the value 8

8x / 8 = (12 - 2y) / 8

Now we get the isolated x value
x = (12 - 2y) / 8

Example 2: A car travels at the rate of 70 miles per hour for 3 hours and for next 3 hours at 80 miles per hour. Calculate the average speed of the whole journey?

Solution:

Step 1: To find the distance travel by the car use the formula

Distance of travel = Rate × Time
Total distance of travel = 70 × 3 + 80 × 3 = 210 + 240 = 450

Step 2:

Total hours of travel = 3+3= 6

Step 3: Using the formula for average speed find the average speed

Average speed of the car = Total distance travelled / Total time taken

=450/6

The average speed of the car journey was 75 miles per hour.


Practice problems:


Problem 1: Isolate the variable x in the equation

6x + 3y = 18.

Answer: x = (18 - 3y) / 6

Problem 2: A car travels at the rate of speed 60 miles per hour for 2 hours and for 3 hours at 70 miles per hour. Calculate the average speed of the whole journey?

Answer: The average speed of the journey was 66 miles per hour.

Geometric Math Templates

Introduction to geometric math templates:

In math, template defines model, shapes etc. In math, geometry is described as about the shapes and their properties. The geometry has a lengthy and legendary history and it is one of the fundamentals of the math. In mathematics, it plays a major role. In math, the geometry contains many templates. The geometric includes the shapes circle, triangle, sphere, cylinder, etc.. Now we are going to solve the problems for geometric math templates.


Examples – Geometric math templates:


Let see solve the problem using triangle shape in geometric math templates.

Example 1:

Work out the area of a triangle with height 8.8 cm and base of 17 cm.

Solution:

The formula for area of the triangle is  `1/2` (bxh)

= `1/2` (8.8 x 17)

= `1/2`(149.6)

=74.8

Hence the area of the triangle is 74.8 cm2.

Example 2:

Work out the volume of the sphere with the radius having 20.89 meter.

Solution:

We know the formula for finding volume of the sphere =` 4/3` П r3

Here П = 3.14, r = 20.89 meter and insert the value into formula we get

Volume =` 4/3` x 3.14 (20.89)3

Simplify the above we get

= ` 4/3`x 3.14 x 20.89 x 20.89 x 20.89

Simplify the above we get

=` 4/3` x28624.97

= 38166.20m3

These are examples in triangles and sphere for geometric math templates.


More examples – Geometric Math Templates:


Now we will solve the line shape problem in geometric math templates.

Work out is the slope of the geometric line which passes through (12.5, 10.2) and (15.6, 13.6)

Solution:

We know that the slope of a given line can be found as m =`(Y2-Y1)/(X2-X1)`

Here, x1 = 12.5 x2 = 15.6 y1= 10.2 y2 =13.6

m = `(13.6-10.2)/(15.6-12.5)`

= `3.4/3.1`

m =1.09

So the slope of the line is found to be 1.09.

Example 4:

The radius of a circle is 5.1cm2 inches.

Calculate is the area of the circle.

Formula= `Pi` r2

= 3.14*5.1*5.1

=81.67cm2

These are examples in line and circle for geometric math templates.

These are examples for geometric math templates.

That’s all about geometric math templates.

Free 4th Grade Math

Introduction about 4th grade math:

Study of basic math operations and math functions is called mathematics. In 4th grade math we can learn some basic math operation.

The basic arithmetic operations of mathematics are addition, subtraction, division, multiplication and placing values. The 4th grade math is deals with basic algebra.In this article we are discussing about 4th grade math with some example problems.

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Free examples problem for 4th grade math:

Free addition problems for 4th grade math:

1. Find the add value of the given numbers, using addition operation, 5 + 2 + 6

Solution:

Given numbers, 5 + 2 + 6

First step, we are going to add the first two numbers,

5 + 2 = 7

Then add third number with first two numbers of sum values,

7 + 6 = 13

Finally we get the answer for given numbers are 13.

2. Find the add value of the given numbers, using addition operation, 8 + 9 + 7

Solution:

Given numbers, 8 + 9 + 7

First step, we are going to add the first two numbers,

8 + 9 = 17

Then add third number with first two numbers of sum values,

17 + 7 = 24

Finally we get the answer for given numbers are 24.


Free subtraction problems for 4th grade math:


3. Find the subtract value of the given numbers, using subtraction operation, 5 - 2 - 6

Solution:

Given numbers, 5 - 2 - 6

First step, we are going to subtract the first two numbers,

5 - 2 = 3

Then subtract third number with first two numbers of subtracted values,

3 - 6 = -3

Finally we get the answer for given numbers are -3.

4. Find the subtract value of the given numbers, using subtraction operation, 2 - 8 - 8

Solution:

Given numbers, 2 - 8 - 8

First step, we are going to subtract the first two numbers,

2 - 8 = -5

Then subtract third number with first two numbers of subtracted values,

-5 - 8 = -13

Finally we get the answer for given numbers are -13.

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Free multiplication problems for 4th grade math:


5. Find the multiply value of the given numbers, using multiplication operation, 6 * 4 * 5

Solution:

Given numbers, 6 * 4 * 5

First step, we are going to multiply the first two numbers,

6 * 4 = 24

Then multiply the third number with first two numbers of multiplied values,

24 * 5 = 120

Finally we get the answer for given numbers are 120.

6. Find the multiply value of the given numbers, using multiplication operation, 2 * 8 * 3

Solution:

Given numbers, 2 * 8 * 3

First step, we are going to multiply the first two numbers,

2 * 8 = 16

Then multiply the third number with first two numbers of multiplied values,

16 * 3 = 48

Finally we get the answer for given numbers are 48.

Reciprocal Math Terms

Introduction for reciprocal:

The reciprocal number is usually specified the following technique. The number is n it is normally indicated the reciprocal is `1/n` . Another technique for the indicated the reciprocal number is x/y the multiplicative opposite of a fraction is `y/x` . The example reciprocal of 19 is `1/19` . We converse the description of reciprocal significance. Let us notice about reciprocal significance in this article.

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Reciprocal math terms:


1) Reciprocal equation on math terms:

Reciprocal equation one which remains unchanged in appearance when the reciprocal of the unidentified quantity is replacement for that quantity.

2) Reciprocal proportion on math terms:

Proportions such that, of four terms obtain in order, the first have to the second the similar ratio which the fourth has to the third, or the first has to the second the same ratio which the reciprocal of the third has to the reciprocal of the fourth. Thus, 2:5:: 20:8 appearance a reciprocal proportion, because 2:5:: 1/20:1/8.

3) Reciprocal quantities on math terms:

Reciprocal quantities are some two numbers which create unity as soon as multiplied collectively.

4) Reciprocal ratio on math terms:

The reciprocal ration is the ratio among the reciprocals of two numbers; as, the reciprocal ratio of 5 to 7 is that of `3/5` to `1/7` .

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Example problems for reciprocal math terms:


1) Solve the reciprocal term: `4/x` =`4/2`

Solution:

Here, taking reciprocals on both sides,

`x / 4` = `2/4`

x= 2*`4/4`

x=`8/4`

Answer: x=2

2) Calculate the reciprocal value for `12/9` .

Solution for reciprocal of a number:

Reciprocal is `12/9` = `9/12`

3) Calculate the Reciprocal value for `17/5` .

Solution for reciprocal of a number:

Reciprocal is `17/5` = `5/17`

4) Calculate the Reciprocal value for `14/9`

Solution for reciprocal of a number:

Reciprocal is `14/9` = `9/14`

5) Calculate the Reciprocal of the fraction `5/2`

Solution:

Reciprocal is

`5/2` =`2/5`

6) Calculate the Reciprocal of the fraction `4/7`

Solution:

Reciprocal is

`4/7` = `7/4`

7) Reciprocal of 75

Solution:

1 divided by a number specifies the reciprocal of that number.

Reciprocal of 75 = `1/75`

8) Calculate the Reciprocal of `1/39` .

Solution:

1 divided by a number specifies the reciprocal of that number.

Reciprocal of `1/39` = 39

9) Calculate the reciprocal of `63/59` .

Solution:

1 divided by a number specifies the reciprocal of that number.

Reciprocal of `63/59` = `59/63` .

Boolean Algebra

The two-valued logic in algebra is called the Boolean Algebra. It was first given by an English mathematician by name George Boole. The arithmetic operations which are performed on the Boolean quantities have only two outcomes either true or false, 0 or 1. And hence the Boolean logic forms the basis for the computation in binary computer systems.

Any algorithm or computer circuit can be represented using a system of Boolean equations. In the Boolean system the two possible values are zero and one. The Boolean algebra symbols used in Boolean operations are,  “.” which represents the logical operation “AND”, it is also denoted by ‘^’.  The symbol ‘+’ is used to represent the logical operation ‘OR’ which is also denoted by ‘v’. Logical negation or complement or not is denoted by  (~)or  (‘), for instance NOT A is denoted as A’ or  ~A (read as ‘negation A’)
The Boolean algebra Properties are:
Commutative law: A+B=B+A; A.B=B.A
Associative Law: (A+B)+C=A+(B+C); A.(B.C)= (A.B).C
Distributive Law: A(B+C)=(A.B)+(A.C); A+(B.C) = (A+B)(A+C)
Idempotence property: A+A=A; A.A=A
Involution Law: (A’)’= A
Law of Complements (negation): A+A’=1; A.A’=0
Boolean algebra theorems are as given below:
Simplification Theorems:
A.B+A.B’=0
(A+B)(A+B’)=A
A+AB=A
A(A+B)=A
(A+B’)B=AB
AB’+B=A+B
De-Morgan’s Theorems:
(A+B+C…..)’= A’B’C’…
(ABC…..)’=A’+B’+C’….
[f(A1,A2, A3…An,0,1,+,.)]’=f(A1’,A2’,A3’….An’,0,1,+,.)
Duality Property:
(A+B+C+….)D= ABC….
(ABC…)D= A+B+C…..
[f(A1,A2, A3…An,0,1,+,.)]D = f(A1’,A2’,A3’….An’,0,1,+,.)
Multiplying out and factoring theorem:
(A+B)(A’+C)=AC+A’B
AB+A’C=(A+C)(A’+B)
Consensus Theorem:
AB+BC+A’C=AB+A’C
(A+B)(B+C)(A’+C)=(A+B)(A’+C)
Absorption: A(A+B)=A; A+AB=A
Prove that A+BC=(A+B)(A+C)
Proof: Here to prove the given Boolean equation we expand the left hand side expression, apply distributive law, absorption law and idempotence (AA=A=A+A)
(A+B)(A+C)=AA+BA+AC+BC (using distributive law)
=A+BA+AC+BC   (using idempotence property)
=A+BC           (using absorption)


A Boolean function can be represented using a Truth table. The truth table for the logical operation AND would be,
AND    0      1
0       0      0
1       0       1
The truth table for the logical operation OR would be,
OR   0      1
0     0      1
1    1       1

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Boolean algebra Problems
Simplify: [[A(A+B)]’+B’A’]’
Solution: [[A(A+B)]’+B’A’]’
=[A.(A+B)]’’. [B.A’]’
=[A.(A+B)].(B’+A)
=[(A+AB)(B’+A)]
=A(1+B)(B’+A)
=A(1)(B’+A)
=A(B’+A)
=AB’+ AA
=AB’+A
=A[B’+1]
=A(1)
=A

Simplify (A+C)[AD+AD’]+AC+C
Solution: Given,
(A+C)[AD+AD’]+AC+C
= (A+C)A(D+D’) + AC + C       [using distributive law]
=(A+C)A+AC+C            [using complement identity]
=A[(A+C)+C] + C            [using associative law]
=AA+AC+C            [using distributive law]
=A+(A+1)C            [using idempotent, identity and distributive laws]
= A +C                [using identity twice]

Simplify (AB)’(A’+B)(B’+B)
Solution: Given,
(AB)’(A’+B)(B’+B)
=(AB)’(A’+B)            [using complement law and identity law]
=(A’+B’)(A’+B)            [using De-Morgan’s law]
=A’A’+A’B+B’A’+B’B        [using distributive law]
=A’+B’B                [on simplification, OR distributes over AND]
=A’                [Complement and identity laws]

7th Grade Math Need Help

Introduction to  7th grade math need help

7th grade math need help is about the problems and chapters of math that are studied and revised in the grade 7. 7th grade math need help  involves chapters like addition, subtraction, multiplication and division, fractions, decimals, measurements, area, volume, surface area, statistics, word problems, scientific notation etc.These topics will be simpler and easier one to learn and to perform calculation with the problems . 7th grade math need help will be more interesting one to work out the problems.


Examples on 7th grade math need help


Add the six digit number 8 4 57 1 2 + 1 2 4 2 5 1
Solution

8 4 5 7 1 2

1 2 4 2 5 1 +

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

9 6 9 9 6 3

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

2. Subtract `3/7`  - `1/3`

Solution

Here the denominators are different

So we have to find the LCM

The LCM of 7, 3 = 21

` (3xx3)/ (3xx7)` = `9/21`

`(1xx7) / (3xx7)` = `7/21`

Now we subtract

`9/21- 7/21`

`(9-7)/21`

`2/ 21` is the solution.

3. Find the area and perimeter of the rectangle if length is 9 cm and width is 6 cm.

Solution

Length of the rectangle = 9 cm

Width of the rectangle = 6 cm

Formula to calculate the area of the rectangle = length * width

So

Area of the rectangle given = 9 * 6

= 54 square centimeter

Perimeter of the rectangle = 2(length + width)

= 2 (9 + 6)

= 2(15)

= 30 cm

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More problems on 7th grade math need help


4. Find the area and perimeter of a square of the side a = 10cm

Solution

The side of the square = 10 cm

Formula to calculate the area of the square = `a^2` or (side * side)

Area of the square = 10 * 10

= 100 square centimeter.

Formula to calculate the perimeter of the square is 4* side

Perimeter of the given square = 4 * 10

= 40 cm

5. Find the number suffixes of the series given 35, 40, 45, 50, 55

Solution

The given series is 35, 40, 45, 50, 55

In this series the number is increased with 5

So the number suffixes of this series is 60, 65 and goes on

Math 8 Inequalities

Introduction to math 8 inequalities:

In mathematics, an inequalities is a statement about the relative size or order of two objects or about whether they are the same or not.

· The notation a < b means that a is less than b.
· The notation a > b means that a is greater than b.
· The notation a ≠ b means that a is not equal to b.
In this article we shall discuss about math 8 inequalities. (Source: wikipedia)

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Math inequalities example problem


Example:

Solving the inequalities 4x – 8 > 2x +10

Solution:

The given inequalities is

4x – 8 > 2x +10

Adding the 8 on both side of inequality equation

4x -8+8> 2x +10+8

4x>2x+18

Subtract 2x on both side of the inequality equation

2x>18

Simplifying the x

x>18/2

x>9

Example:

Solving the following inequality equation -3< 4(x+3)-3<16

Solution:

Given inequality equation is

-3< 4(x+3)-3<16

Multiply the factor values for given equation

-3<4x+12-3<16

-3<4x+9<16

Subtracting nine on both sides of inequality equation

-3-9<4x+9-9<16-9

-12<4x<7

Divide by 4 for all terms

-3<x<1.75

Conclusion:

The solution includes all real number value the interval is (-3, 1.75)

Example:

Solving the following inequality equation -2< 4(x+6)-3<10

Solution:

Given inequality equation is

-2< 4(x+6)-3<10

Multiply the factor values for given equation

-2<4x+24-3<10

-2<4x+21<10

Subtracting 21 on both sides of inequality equation

-3-21<4x+21-21<10-21

-24<4x<-11

Divide by 4 for all terms in equation

-6<x<-2.75

Conclusion:

The solution includes all real number value the interval is (-6, -2.75)

Example:

Solving the inequality 2x – 6 > 2x +12

Solution:

The given inequality is

2x – 6 > 2x +12

Adding the 6 on both side of inequality equation

2x -6+6> 2x +12+6

2x>2x+18

Subtract 2x on both side of the inequality equation

2x>18

Simplifying the x

x>18/2

x>9

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Math inequalities practice problem

Problem:

Solving the inequalities 2x – 8 > 2x +8

Answer:

x>10

Problem:

Solving the inequality 2x – 8 > 2x +8

Answer:

The solution includes all real number value the interval is (-2, 1.75)

Transposing Math Functions

Introduction to transposing math functions:

            Transposing math functions involves the process of interchanging or converting one form of the given function into another form. For example the non linear function is transposed into linear form and the polynomial function is converted into normal function form. The math functions mainly deals with linear and non linear functions. Transposing of linear and nonlinear functions is carried out below with the help of certain operations. The following are the example problems for transposing math functions.

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Transposing math functions example problems:

Example 1:

Transpose the given math functions and find the coefficient value.

Sqrt (3 u + 1) = u - 3

 

Solution:

Given equation is
sqrt (3 u + 1) = u - 3

Let us squaring on both sides, then the above equation becomes 
[sqrt (3 u + 1) ] ² = (u - 3) ²

Solve the above equation 
3 u + 1 = u ² - 6 u + 9

Write the above equation in factor form. 
u ² - 9 u + 8 = 0

The above form is a quadratic equation with 2 solutions 
u = 8 and u = 1

Example 2:

Transpose the given math functions and find the coefficient value.

                    f(u) = u ³ – 48u + 5

Solution:

Given function is

f(u) = u ³ – 48u + 5

The math function of f is the set of all real numbers. The derivative f ' is given as

f '(u) = 3 u ² - 48

f '(u) is defined for all real numbers. Let us now solve f '(u) = 0 

3 u ² - 48 = 0

Add 48 on both sides,

 3 u ² – 48 + 48 = 48

    3 u ² = 48

       u ² = 16
u = 4 or u = -4

Since u = 4 and u = -4 are the coefficient value.

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Transposing math functions practice problems:

1) Transpose the given math functions and finds the coefficient value.

                      u ² – 4u + 13 =  0

Answer: u = 2 + 3i or u = 2- 3i.

2)  Transpose the given math functions and finds the coefficient value.

               2 u ² + u - 6 = 0

Answer: u = -2 or u = 3/2

Perimeter Problems in Math

Introduction about perimeter:

In math, perimeter can be calculated for 2D and 3D shapes. The length of the boundary of any closed figure is called perimeter of that shape. The perimeter is expressed in units. In this article we shall see how to calculate the perimeter of basic 2D shapes with example problems.

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Math perimeter Formulas to solve problems:


Square:

Math Formula:

Perimeter of the square (P) =4 x a units

Rectangle:

Math Formula:

Perimeter (p) =2(l x w) units

Circle:

Math Formula:

Circumference (perimeter) of the circle = 2pr units

( r is the radius of the circle)

Triangle:

Math formula:

Perimeter of equilateral triangle (P) =3 a units

a – side length

Kite:

Math formula:

Perimeter of kite (P) = 2a + 2b units

a , b – side length.

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Perimeter problems in math -Example problems:


Square:

The square has the side length 7cm.find the perimeter of the square.

Solution:

Given:

a= 7 meters

Perimeter of the square = 4 x a

=4 x 7

Perimeter of the square = 28 meters

Rectangle:

The rectangle has length and width 5cm and 4cm respectively. Find the perimeter of the rectangle.

Solution:

Given:    Length= 5 meters, Width =4 meters

Formula:

Perimeter of the rectangle   = 2(l + w)

=2(5+ 4)

= 2 (9)

Perimeter of the rectangle = 18 meters

Circle:

The radius(r) of a circle is 10 inches. Find the circumference (perimeter) of that circle?

Solution:

Given: r = 10 inches

Formula:

Circumference of the circle = 2pr.

= 2 x 3.14 x 10

Circumference of the circle = 62.8 inches

some more example problems:

Triangle:

Find the perimeter of equilateral triangle whose side 14 cm.

Solution:

All sides of equilateral triangle area equal.

Side of triangle, a = 14 cm

Formula:

Perimeter of equilateral triangle (P) = 3a

= 3 x 14

Perimeter of equilateral triangle = 42 centimeters

Kite:

The kite has length 8cm and width 4 cm. find the perimeter of the kite.

Solution:

Given:

Length (a) = 8 cm

Width (b) = 4 cm

Perimeter of kite = 2a + 2b

= 2 x 8 + 2 x 4

= 16 + 8

= 24 centimeters

I Need Help on 4th Math

Introduction to I need help on 4th math:

The topics involved in four grade math help are  patterns, addition, measurement, subtraction, number sense, multiplication, functions, fractions & mixed numbers, division, algebra, decimals, adding and subtraction of decimals and probability & statistics. In this article "I need help on 4th math" we shall deal with expanded form, median & mode, and measurements. The following are the examples related to I need help on 4th math grade.

I like to share this Subtracting Mixed Fractions with you all through my article.

I Need Help On 4th Math Problems:


Example 1: Find the value form the expanded form 9 × 10000 + 3× 1000 + 9 × 10 + 20

Solution:  9 × 10000 + 4 × 1000 + 9 × 10 + 20

= 90000 + 4000 + 90 + 20

=> 94110.

Example 2: Find the mode for given set of data. 63, 31, 28, 29, 31, 62, 88

Solution: Mode:The mode is the number that occurs most often in a set of data.

Form the given data set we can say that 31 is the number that has occurred twice.

For the given set of data Median = 31.

Example 3: Find perimeter of a square with length 12.

Solution: The perimeter of a square = 4 * a ( a-> length)

Here length =12, So perimeter of square = 4 * 12 = 48

Example 4: Find the circumference of a circle with diameter 24.

Solution:  The circumference of a circle = 3.14(pi) * diameter of the circle.

Here diameter = 10, So Circumference of a circle = 3.14 * 24

Therefore the Circumference of a circle =   75.36

Example 5: Find the median of 8, 2, 7, 5, 3, 9

Solution:Evaluate the total numbers from the given set of values.

From the given set of values, there are totally 6 numbers, Which is even,

Now arrange the total numbers in ascending order,

2, 3, 5, 7, 8, 9

As the given set of numbers is even , Sort out two middle numbers from the above given list of numbers,

Which is  5 & 7 are the two middle numbers,

In order to find the median, add both middle terms and divide by 2,

Adding both numbers `(5+7)/(2)` = 6,  So,median=6.


I Need Help On 4th Math Practice Problems:


Problem 1:Find the mode for given set of data. 1, 2, 5, 2, 8, 4, 3

Solution:: 2


Problem 2:Find the median of 3, 1, 2, 8 5, 6, 7 and 4.

Solution:  Median =4.5

Problem 3: Find the perimeter of a square with length 28.

Solution: The perimeter of area = 112.

Problem 4: Find the circumference of a circle with diameter 15.

Solution: The Circumference of a circle =47.1

Problem 5: Find the value form the expanded form 9 × 10000 + 1000 + 10 × 10 + 22

Answer: 91122