Same Perimeter Different Area

Perimeter is nothing but the path around the shape. And area is nothing but the space occupied by the 2 dimensional object. Here we are going to deal with the same perimeter and different area of the shape. Every shape is having different formulas for area. And for all shape if we want to find the perimeter we have to add the length of all sides. We will see some example problems for same perimeter and different area.

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Example Problems for same Perimeter and Different Area:

Example 1 for same perimeter and different area:



Find the area and perimeter of the following shapes and compare the area and perimeter.

Solution:

Shape 1:

The first shape is a rectangle. We know the area of the rectangle is l X w

Here length l = 3 cm and the width w = 4 cm

So the area of the rectangle = 3 X 4 = 12 cm2

Perimeter of the rectangle is = 2 (l + w) = 2 (3 + 4) = 2 X 7 =14 cm.

Shape 2:

The shape 2 is a triangle. To find the area of the triangle we have to use the following formula.

Area of the triangle = (`1 / 2` ) bh.

From the above b = 7 cm and h = 2 cm.

So the area =` (1 / 2) xx 7 xx 2 ` = 7 cm2

Perimeter of the triangle   = sum of all sides = 7 + 4 +3 = 14 cm.

From the above bath shape is having the same perimeter and different area.

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Example 2 for same Perimeter and Different Area:

Find the area and perimeter of the following shapes and compare the area and perimeter.



Solution:

Shape 1:

The first shape is a trapezoid. We know the area of the trapezoid is `(1/2) h (a + b)`

Here a and b are the base lengths. A = 6 cm. b = 4 cm and the height is 5 cm.

So the area of the rectangle = `(1/2) 5 (6 + 4) ` = 25 cm2

Perimeter of the rectangle is = sum of all sides = 6 + 5 + 4 + 3 = 18 cm.

Shape 2:

The second shape is a rectangle. We know the area of the rectangle is l X w

Here length l = 6 cm and the width w = 3 cm

So the area of the rectangle = 6 `xx` 3 = 18 cm2

Perimeter of the rectangle is = 2 (l + w) = 2 (6 + 3) = 2 X 9 =18 cm.

From the above bath shape is having the same perimeter and different area.

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Pictures of Division Sets

In this article we shall discuss  pictures of division sets. Here, division is also meant by fraction of a whole. A division can be creation over to a decimal through dividing the upper digit, or numerator, during the lower digit, or denominator. Division is instead of as ratios, and significance for fraction which is one of the important math processes. Thus the division `3/5` is also used to point out the ratio 3:5 and the division 3 ÷ 5 as well.

How to do Pictures of Division Sets:

The pictures of division sets are shown given below that,



 
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Example Problems Based on Learn Pictures of Division Sets:

The example problems based on learn about pictures of division sets are given below that,

Example 1:

How to learn about pictures of division for 277 divide by 8 sets?

Solution:

Step 1:

The given value is 277 divide by 8 sets.

Step 2:

Here, 277 divide by 8 sets is also denoted as 277/8.

Step 3:

Now, 277 divide by 8 sets is explain about using long division procedure pictures.

Here, using long division procedure pictures are shown given below that,



Step 4:

The final answer for pictures of division sets is 34.625.

Example 2:

How to learn about pictures of division for 197 divide by 6 sets?

Solution:

Step 1:

The given value is 197 divide by 6 sets.

Step 2:

Here, 197 divide by 6 sets is also denoted as 197/6.

Step 3:

Now, 197 divide by 6 sets is explain about using long division procedure pictures.

Here, using long division procedure pictures are shown given below that,



Step 4:

The final answer for pictures of division sets is 32.833







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Practice problems based on learn about pictures of division sets:

The practice problems based on learn about pictures of division sets are given below that,

Problem 1:

How to learn about pictures of division for 163 divide by 4 sets?

Answer: The final answer for pictures of division sets is 40.75

Problem 2:

How to learn about pictures of division for 52 divide by 4 sets?

Answer: The final answer for pictures of division sets is 13.

Least to Greatest Calculator

The order of the least number to the greatest number is the called the ascending order. The ascending order is doing on arrange the give number for the given order and also the least to greater calculator is used to give a input to mixed order. Then click the calculate button to arrange the given order. In this article id discuss about the least to greatest calculator.

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Least to Greatest Calculator - Examples:

Least to greatest calculator - Example 1:

          75, 710, 15, 425, 235, 145, 505 Arrange the order of least to greatest order

Solution:

Alter input as fractions if necessary:

          75/1, 710/1, 15/1, 425/1, 235/1, 145/1, 505/1

The least common denominator (LCD) is: 1.

Alter as equivalent fractions with the LCD:

75/1, 710/1, 15/1, 425/1, 235/1, 145/1, 505/1

Ordering these fractions by the numerator:

          15/1  <   75/1  <   145/1  <   235/1  <   425/1  <   505/1  <   710/1

Therefore, the order of your input is:

          15  <   75  <   145  <   235  <   425  <   505  <   710

Least to greatest calculator - Example 2:

705, 7110, 115, 4025, 2135, 1435, 5705

Solution:

Alter input as fractions if necessary:

          705/1, 7110/1, 115/1, 4025/1, 2135/1, 1435/1, 5705/1

The least common denominator (LCD) is: 1.

Alter as equivalent fractions with the LCD:

          705/1, 7110/1, 115/1, 4025/1, 2135/1, 1435/1, 5705/1

Ordering these fractions by the numerator:

          115/1  <  705/1  <  1435/1  <  2135/1  <  4025/1  <  5705/1  <  7110/1

Therefore, the order of your input is:

          115  <  705  <  1435  <  2135  <  4025  <  5705  <  7110

Least to Greatest Calculator - more Examples:

Least to greatest calculator - Example 1:

13, 19, 16, 112, 118, 211, 214, 218, 310, 411

Solution:

Alter input as fractions if necessary:

          13/1, 19/1, 16/1, 112/1, 118/1, 211/1, 214/1, 218/1, 310/1, 411/1\

The least common denominator (LCD) is: 1.

Alter as equivalents fractions with the LCD:

          13/1, 19/1, 16/1, 112/1, 118/1, 211/1, 214/1, 218/1, 310/1, 411/1

Ordering these fractions by the numerator:

          13/1  <  16/1  <  19/1  <  112/1  <  118/1  <  211/1  <  214/1  <  218/1  <  310/1  <  411/1

Therefore, the order of your input is:

          13  <  16  <  19  <  112  <  118  <  211  <  214  <  218  <  310  <  411

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Least to greatest calculator - Example 2:

`2/3, 5/2, 6/7, 7/4, 2/8, 6/5, 8/6, 2/3`

Solution:

 Alter input as fractions if necessary:

          `2/3, 5/2, 6/7, 7/4, 2/8, 6/5, 8/6, 2/3`

The least common denominator (LCD) is: 840.

Alter as equivalents fractions with the LCD:

          `560/840, 2100/840, 720/840, 1470/840, 210/840, 1008/840, 1120/840, 560/840`

Ordering these fractions by the numerator:

         ` 210/840<560/840=560/840<720/840<1008/840<1120/840<1470/840<2100/840`       

Therefore, the order of your input is:

         ` 2/8<2/3=2/3<6/7<6/5<8/6<7/4<5/2`

Value of an Integral Type Expected

Expected value is one of an important concept in probability. In probability, expected value of a given real-values are chance the variables as present a compute of the center of the distribution of the variable. In online, few websites are providing math tutoring. Tutor, will give step by step explanation for the expected value problems. Expected problems are deals with, probability, geometry distribution, etc. in this article we shall discuss for value of an integral type expected.


Sample Problem for Value of an Integral Type Expected:

Value of an integral type expected problem 1:

Evaluate the expected value from the given continuous random variable using uniform distribution. Value is expected from the interval value is 3 < x < 7.

Solution:

Given:

A given interval value for the uniform distribution is a = 3 and b = 7

Formula for finding the expected value for uniform distribution is

E(X) = `int_a^bx f(x)dx` --------------- (1)

Here, the value of f(x) is `1/(b - a)` , for 3 < x < 7

= `1/(7 - 3)`

= `1/4`

f(x) = `1/4`

In the next step we put the f(x) value in the above equation, we get

= `int_3^7x (1/4) dx`

= `1/4int_3^7x dx`

= `1/4[ x^2/2]_3^7`

= `1/4[(7)^2/2-(3)^2/2 ]`

= `1/4 ` [1/2 (7)2 - (3)2]

= `1/8` [49 - 9]

= `1/8` (40)

= `40/8`

= 5

We get expected value E(x) is 5.

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Value of an Integral Type Expected Problem 2:

Evaluate the expected value from the given probability density function using exponential distribution `e^-2x` with the interval of [0, `oo` ].

Solution:

Given:

A given probability density function f(x) is `e^-2x` .

Formula for finding the expected value for uniform distribution is

E(X) =` int_0^ooxf(x)dx` --------------- (1)

In the first we find out the expected value of the exponential distribution function


=` int_0^oox(e^(-2x))dx`

=` int_0^ooxe^(-2x)dx`

Here we use `int udv = uv - int vdu`

u = x       dv = `e^(-2x)`

u' = 1      v = `(e^(-2x))/-2 `

u'' = 0      v' = `(e^-2x)/4`

in the next step we substitute the above values, we get

= ` [(xe^(-2x))/-2- (e^(-2x))/4]_0^oo`

= ` [((ooe^(-2oo))/-2-(e^(-2oo))/4 )- ((0 - e^(0))/4)]`

Here we use `e^-oo` = 0, e0 = 1/4

= 0.25

We get the expected value E(X) is 0.25.

Solve Graphing Trigonometric Functions

Graphs of Trigonometric Functions


The relation (variation) between the angles and the values of the trigonometric ratios at them are plotted by graphs .

Pro 1:    Graph of  y  =  sinx

x    -`pi`    `-pi/2`    0    `pi/2`    `pi`    `(3pi)/2`    `2pi`    `(5pi)/2`    `3pi`
y    0    -1    0    1    0    -1    0    1    0
Plot the graph in the coordinate plane by taking angles x in radian measure on X-axis and the values of sinx  =  y on Y-axis .

Sol :



By choosing a suitable scale , plot and join the points of  y = sin x with a smooth curve to get the graph .

This curve passes through the origin . The values of sin x vary between -1 and +1 which are respectively the minimum and the maximum . It is in the shape of a wave whose wave length is 2`pi` . This wavelength is nothing but the period .

SInce  `-1<=sinx<=1`   `AA`  x  `in`  R , the sine function is bounded . It can be proved that it is a continuous function on  R .

Pro 2:  Graph of  y  =  cosx

x    `-pi`    -`pi/2`    0    `pi/2`    `pi`    `(3pi)/2`    `2pi`    `(5pi)/2`
y    -1    0    1    0    -1    0    1    0

Sol :



By choosing a suitable scale , plot and join the points of y = cos x with a smooth curve to get the graph .

This curve does not pass through the origin . It is evident that the maximum and the minimum values are  +1 and -1 respectively .

Since  `-1<=cosx<=1`   `AA`  x  `in`  R  ,  the cosine function is bounded . It can be shown that is a contnous function and periodic with `2pi` as the period .

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Solve Graphing Trigonometric Functions : Tanx and Cotx

Pro 3:  Graph of  y  =  tanx

x    `-pi/2`    0    `pi/2`    `pi`    `(3pi)/2`
y    not defined    0    not defined    0    not defined

Sol :



The curve nearly  touches the vertical lines  at  x  =  ...........`-pi/2`  , `pi/2` , `(3pi)/2` , . .. . . . .

The curve has nreakes at  x  =  (2n + 1)`pi/2`   ,  n  `in`  Z  and passes through the origin . it is not bounded .

The tan function is periodic and `pi`  is the period of it .

Pro 4:  Graph of  y  =  cotx



The curve nearly touches the vertical lines at  x  =  . . . . . . . `-pi` , 0 , `pi`  . . . . . . .  .

The curve has breakes at  x  =  n`pi`   ,    n  `in`   Z   and does not pass through the origin . It is not bounded .

The cot function is periodic and  `pi`  is the period of it .

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Solve Graphing Trigonometric Functions : Secx and Cosecx

Pro 5:   Graph of  y = secx



The curve nearly touches the vertical lines at x  =  . . . . . . . . `-pi/2` , 0 , `pi/2` , `pi` , `(3pi)/2` . . . . . .

The curve has breakes at  x  `in`   (2n+1) `pi/2` ,  n `in`  Z . It is not bounded . The values  of  secx  lie  in  `(oo,-1]uu[1,oo)`

The secant function is periodic and 2`pi`  is the period of it .

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Pro 6:   Graph of  y  =  cosecx



The curve nearly  touches the vertical lines at  x  =  . . . . . . 0 , `pi/2` , `pi` , `(3pi)/2` , 2`pi` , . . . . . .

The curve has taken

Multiplication and Division of Whole Numbers

Division:

For numbers, the operation of assert how many times one number, the divisor, is contained in a second, the dividend. The result is known as quotient and if the divisor is not contained an integral number of times in the dividend; any number left over is called the remainder. The symbol of division is / or ÷.

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Multiplication:

The product of two numbers is known as multiplication. Here we are going to learn about some example problems of multiplication and division whole numbers.

Example Problems of Multiplication Whole Numbers:

Example 1:

12*14

Step 1: First we need to take 4 and multiply with 12 = 48.

Step 2: Now we need to take 1 and multiply with 12 = 12.

Step 3:

Step 4: Therefore the answer is 168.

Example 2:

23*42

Step 1: First we need to take 2 and multiply with 23 =46.

Step 2: Now we need to take 4 and multiply with 23 =92.

Step 3:



Step 4: Therefore the answer is 966.

Example 3:

231* 14

Step 1: First we need to take 4 and multiply with 231 = 924.

Step 2: Now we need to take 1 and multiply with 231 = 231

Step 3:



Step 4: Therefore, the answer is 3234.

These are the examples problems of multiplication of whole numbers.

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Examples of Dividing Whole Numbers:

Example 1:

Solve: `(143)/(3)`

Solution:

Step 1: Here we need to perform division operation `(143)/(3)` .

Step 2: First we need to take first two digits that is 14.  And we know that 4 times 3 = 12.

Step 3: Now we need to minus 12 from 14 = 2.

Step 4: Here we need to get the next term.

Step 5: The next term is 3. So presently we have 23.

Step 6: We know that 7 times 3 = 21.

Step 7: So the quotient is 47 and remainder is 2.



Example 2:

Solve:  `(276)/(12)`

Solution:

Step 1: Here we need to perform division operation `(276)/(12)` .

Step 2: First we need to take first two digits that is 27.  And we know that 2 times 12 = 24.

Step 3: Now we need to minus 24 from 27 = 6.

Step 4: Here we need to get the next term.

Step 5: The next term is 6. So presently we have 36.

Step 6: We know that 3 times 12 = 36.

Step 7: So the quotient is 23 and remainder is 0.



These are the examples of division of whole numbers.

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Triangles Venn Diagram

Venn diagrams are used to describe the relationship of the sets. The description of some sets is given and you are asked to draw a Venn diagram to illustrate the sets. You are well known about the triangles that they are having three sides. Based on their sides and the angles, it is classified into different types. The relationship of the triangles can also be described using Venn diagrams.

Triangles and Venn Diagrams

Venn diagrams:

Venn diagrams to determine the relationships between the sets such as subsets and intersections.

Venn diagram for A within B :



Here, all members of A belongs to B or A ⊂ B or A ∪ B = B or A ∩ B = A or n(A ∩ B) = n(A).

Venn diagram for  A overlap B:



Here, some members of A belongs to B or A ∩ B ≠ Ø or n(A ∩ B ) ≠ 0

Venn diagram for disjoint sets of A and B :



Here, no members of A belongs to B or A ∩ B = Ø or n(A ∩ B ) = 0

Triangles:

scalene triangle:

A triangle is said to be scalene triangle, if all the sides are of different lengths.Some scalene triangles are also right triangles.



Right triangle:

A triangle is said to be right triangle, if one interior angle is exactly 90°. Some right triangles are also scalene triangles.



Equilateral triangle:

A triangle is said to be equilateral triangle, if all the sides are of same length.



Isosceles triangle:

A triangle is said to be isosceles triangle, if two sides are of same length and two angles are equal.



Examples

Let U is the set of triangles, A is the set of isosceles triangles, B is the set of equilateral triangles and C is the set of right-angled triangles.

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Draw a Venn diagram.

Solution:

We have to determine the relationships between the sets.

All the equilateral triangles are said to be isosceles triangles, so B ⊂ A. (within)

Some of the right-angled triangles are said to be isosceles triangle. C ∩ A ≠ Ø (overlap)

No right-angled triangles are equilateral triangles. C ∩ B = Ø (disjoint)

Venn diagram for the triangles:



Let U is the set of triangles, S is the set of scalene triangles and R is the set of right triangles.

Draw the Venn diagram

Solution:

The relationship between scalene and right triangle to be defined.

Some of the right triangles are scalene triangles.