Roman Numerals Adding Answers

Introduction to Roman numerals:
Roman numerals are a numerals system of ancient Rome based on letters of the alphabet, which are combined to signify the sum (or in some cases, the difference) of their values. The first ten Roman numerals are as follows:

I, II, III, IV, V, VI, VII, VIII, IX, and X.

In this article we shall discuss the adding Roman numerals with answers.

(Source: Wikipedia)

Letters in Roman Numerals:

There are seven letters in Roman numerals. These letters are used for adding numerals.

They are the following:

I in decimal is one.

V in decimal is five.

X in decimal is Ten.

L in decimal is fifty.

C in decimal is hundred.

D in decimal is five hundred.

M in decimal is thousand.

Where I, V, X, L, C, D, M.

The rule for adding Roman numerals is:

In order to get answers for adding Roman numerals we must follow the following rules.

Add the amount of the Roman value if one or more letters are located after another letter of larger value.

Subtract the amount of the Roman value if one or more letters are located before another letter of larger value.

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Example Problems for Adding Roman Numerals with Answers:

Example 1:

Add the Roman numerals XV and IX.

Solution:

The given Roman numerals are XV and IX.

XV = 10+5 = 15.

IX = 10-1 = 9

Then adding the Roman numerals XV and IX we get

15+9 = 24

19 can be written as 10+5+ (5-1)

10 in Roman numerals as X

5 in Roman numerals as V

1 in Roman numerals as I

24 in Roman numerals as XXIV.

Finally the answers for adding XV and IX is  XXIV.

Example 2:

Add the Roman numerals XIV and CX.

Solution:

The given Roman numerals are XIV and CX.

XIV = 10+5-1 = 14.

CX = 100+10 = 110

Then adding the Roman numerals XIV and CX we get

14+110 = 124

124 can be written as 100+10+10+ (5-1)

100 in Roman numerals as C

10 in Roman numerals as X

5 in Roman numerals as V

1 in Roman numerals as I

124 in Roman numerals as CXXIV.

Finally the answers for adding XIV and CX is  CXVIV.

Line Segment Chart

Introduction to line segment chart

A line chart or line graph is a type of graph, which displays information as a series of data points connected by straight line segments. It is a basic type of chart common in many fields. It is an extension of a scatter graph, and is created by connecting a series of points that represent individual measurements with line segments (Source: Wikipedia)

Line Segments Chart

Line segments is a division of a line to be surrounded through two finish points. Line contains each point lying on the line among its finish points. Instance of line segments contain the faces of a triangle otherwise square. More usually, while the finish points are equally vertices of a polygon, the line part is any an edging if they are diagonal.

A line chart is over also over again utilized to imagine a development within data in excess of intervals of time; therefore the line is frequently haggard chronologically.Within the experimental sciences, data composed as of testing are frequently imagine through a graph to include an overlay arithmetical meaning representing the finest fit trend of the sprinkled data graph.

For instance, if one were to accumulate data lying on the speed of a body on assured points within time, one could visualize the data:

Example for Line graph

Time: 0 1 2 3 4 5

Speed: 0 4 8 12 14

Time can be represented within x-axis.

Time can be represented within y-axis.

Examples for Line Segment

Example 1

Within a line segment one endpoint is (-2, 6) also the midpoint of the line is (- 6, 10). Compute the co-ordinates of the other endpoint are (x1, y1).

Solution:

`(-6,10)=((x_(1)+(-2))/2),(y_(1)+6)/2`

To find for x1,

`-6=((x_(1)+(-2))/2)`

-12= x1-2

-10 = x1

To compute for x2,

`10=((x_(2)+(6))/2)`

20 = y1 + 6

Y1 = 14

So, the next endpoint of the line is (-10, 14)


Example 2

Within a line segment one endpoint is (-4, 8) also the midpoint of the line is (- 6, 8). Compute the co-ordinates of the other endpoint are (x1, y1).

Solution:

`(-6,8)=((x_(1)+(-4))/2),(y_(1)+8)/2`

To compute for x1,

`-6=((x_(1)+(-4))/2)`

-12= x1-4

-8 = x1

To solve for x2,

`8=((x_(2)+(8))/2)`

16 = y1 + 8

y1 = 8

So, the next endpoint of the line is (-8, 8)

Elementary Matrices

Introduction to Elementary Matrices:

Elementary matrix operation play an significant role in many matrix algebra application, such as finding the inverse of a matrix and solving instantaneous linear equations.

Elementary Operations

The three kind of basic matrix operations.

Interchange two columns.

Multiply each element in a column by a non-zero number.
Multiply a column by a non-zero number and add the result to another column.
While these operation are performed on rows, they are called basic  row operations; and when they are perform on columns, they are called basic column operations.

Types of Elementary Matrices Operations

There are three types of elementary matrices, which match to three types of row operations .

Row switching                RI—>RJ
A row within the matrix can be switched with a new row.

Row multiplication           K RI—>RJ, where k?0            
Each element in a row can be multiplied by a non-zero even.

Row addition              RI+ KRj—> RI
A line can be replaced by the sum of the  row and a multiple of a new row.

The elementary matrix for any row operation is obtained by execute the operation on an identity matrix.

Elementary Operation Notation

In many reference, you resolve encounter a compacted notation to describe elementary operations. That notation is given by

Row operations

1. Interchange rows i and j Ri <--> Rj

2. Multiply row i by s, where s ? 0 sRi --> Ri

3. Add s times row i to row j sRi + Rj --> Rj

Column operations

1. Interchange columns i and j Ci <--> Cj

2. Multiply column i by s, where s ? 0 sCi --> Ci

3. Add s times column i to column j sCi + Cj --> Cj

Rules on Elementary Matrices

Elementary Matrices

Now, imagine R has 1 ,n>n elementary square matrix obtained from the identity matrix by performing arts an basic operation. A result, we have three types of elementary matrices, every subsequent to a type of elementary operations:

Transposition matrix Tij : an matrix obtain from In with rows i and j switch,

Basic diagonal matrix Di(r) : a diagonal matrix whose entries are 1 except in cell (i' > i)  , whose entry is a non-zero element r of R

Row substitution matrix Eij(s) : In+sUij , where s > R and Uij is a matrix unit with i=j  .



Examples based on elementary matrices:

Ex 1: Construct a 3 × 2 matrix whose entries are given by aij = i - 2j

Sol : The general 3 x 2 matrix is of the form,

A= [aij]  =`[[a_11,a_12],[a_21,a_22],[a_31,a_32]]`

where ,i= 1,2,3(rows),j= 1,2 (columns)

it is given that aij = i-2j

a11 =1-2 = -1

a12 =1-4 = -3

a21  =2-2 =0

a22 =2-4 = -2

a31 = 3-2 =1

a32 = 3-4 = -1

Therefore , The required matrix  is  A = `[[-1,-3],[0,-2],[1,-1]]`

Ex 2: Elementary column operation on the term “A”, where the identity is used in matrix   `A= [[1,6], [4,1], [0,1]]` and identity matrix `x = [[1,0], [0 ,1]]`

Sol:

Given: Identity matrix `x = [[1,0], [0 ,1]]`

`A= [[1,6], [4,1], [0,1]]`

Step 1:  The given identity matrix x is interchanged of first and second column and the finding result is named it   (x1).

Step 2: To do the elementary column operation on given `A= [[1,6], [4,1], [0,1]]`

Step 3: The operation in matrix is done by [row *column]

`X= [[1,0], [0,1]]`

The first and second column is interchanged and the result is,

`X_1= [[0,1], [1,0]].`

Now, the found X1 is calculated with given “A”

`A= [[1,6], [4,1], [0,1]]`

`X_1= [[0,1], [1,0]]`

`A*X_1` for calculation elementary column operation on

`A*X_1`

` [[1,6], [4,1], [0,1]] * [[0,1], [1,0]]`

` [[1*0+6*1 , 1*1+6*0],[4*0+1*1 , 4*1+1*0],[0*0+1*1, 0*1+1*0]]` 

`[[0+6, 1+0], [0+1,4+1], [0+1, 0+0]]`

`[[6,1],[1,4],[1,0]]`

Hence ,the elementary column operation for the given value `A=[[6,1],[1,4],[1,0]]`

Practice arithmetic word problems

Learning arithmetic/arithmetics (from the Greek word ἀριθμός = number) is the very oldest and most basic branch of mathematics, used by almost everyone, for tasks collection from simple day-to-day counting to highly developed science and business computations. It refers to the simpler properties while using the conventional operations of addition, subtraction, multiplication and division with lesser values of numbers. Let us see practice arithmetic word problems in this article.

Practice Arithmetic Word Problems - Addition and Subtraction:

Addition Problem :

Abraham has 20 ice creams; Antony has 15 ice creams. Find out the total number of ice creams from both of them?

Solution:

Abraham has 20 Ice creams

Antony has 15 ice creams

Totally 20 + 15 = 35 ice creams

Subtraction Problem :

There are 40 students in a class room. 32 students got pass in computer science. How many students failed?

Solution:

Total students in the class room = 40

Pass in computer science students = 32

Therefore failed students = 40 – 32 = 8.

Practice arithmetic word problems - Multiplication and Division:

Multiplication Problem :

Victoria delivers 80 newspapers in a day. How many newspapers does she deliver in 22 days?

Solution:

Per day Victoria delivers newspapers = 80

If she delivers for 22 days, then 80 * 22 = 1760

Hence she can deliver 1760 newspapers in 22 days.

Division Problem:

Davina has 72 sweets. If she shares with her friends 4 sweets each, how many friends can share her sweets with?

Solution:

Davina has 72 sweets.

Each friend shares 4 sweets.

Total friends shares with her sweets = `72 / 4` = 18.

Hence 18 friends can share her sweets with.

Practice Problems – Practice Arithmetic Word Problems:

Problem 1:

There are 14 apples and 20 oranges in a fruit basket. Totally how many fruits are there in the basket?

Answer:

Total fruits in the basket = 34 fruits

Problem 2:

There are 67 students in a class. 12 students were not wear uniform. How many students are wearing uniform?

Answer:

Number of students wearing uniform = 55

Problem 3:

A balloon cost is $2.What is the cost of 12 balloons?

Answer:

Cost of 12 balloons = $24

Problem 4:

5 toys cost is $25. What is the cost of 1 toy?

Answer:

Cost of 1 toy = $5

Basics about the ogive statistics

We consider the following types of graphical representation of frequency distribution in statistics.
(1) Histogram or Area diagram
(2) Frequency polygon
(3) Ogive or cumulative frequency distribution graph
We shall discuss a bit in detail about the ogive graph in this article. The curve that is obtained by plotting cumulative frequencies against their respective class boundaries is called an cumulative line graph. As such there are two types of cumulative line graph – the less than the cumulative line graph and the more than cumulative line graph. The less than cumulative line graph is obtained by using the less than type cumulative frequency on the vertical axis to plot the cumulative line graph. The more than cumulative line graph is obtained by using the more than type of cumulative frequency on the vertical axis to plot the cumulative line graph. The points thus plotted are then joined successively by short line segments. The graph thus obtained is called an ogive graph.

Ogives may be considered useful for obtaining quartiles graphically. The more than type and the less than type cumulative line graph intersect in a point. If a perpendicular line is drawn from the point of intersection of the two cumulative line graph (more than type and less than type) on the horizontal axis, then the x co ordinate of this point gives us the value of the median, also called the second or the middle quartile.
Q gives are very useful in making short term projections.

Example:
The ages of students in the anatomy class are shown in the following frequency table.
Age Interval(yrs) Frequency
15 - 17                    13
18 – 20                    15
21 – 23                    20
24 – 26                    10
27 – 29                     8
30 – 32                     4
Draw an ogive for the above data.
Solution: First lets add a cumulative frequency column to that table:
Age Interval(yrs) Frequency Cumulative frequency
15 - 17                     13                    13
18 – 20                     15                    28
21 – 23                     20                    48
24 – 26                     10                    58
27 – 29                      8                            66
30 – 32                      4                            70

The cumulative line graph can now be plotted using the upper limit of class interval on the x axis and the cumulative frequency on the y axis as follows:

Ogive Graph

The above graph represents the cumulative line graph of the given example problem question.

Introduction to 4th grade geometry problems

In general, 4th grade geometry problems includes types of polygons, area and perimeter of the polygons. Here we are going to see about  4th grade geometry problems on area and perimeter.  Area (A) is a two-dimensional measure. The area is measured in terms of square units such as square inches, square feet and square centimeters. Hectares and acres are also considered in some special cases. The entire boundary of a figure or the distance around the figure is called as the perimeter. For circle it is called as the circumference, such lengths are measured with inches, feet, and centimeters.

4th Grade Geometry Problems in Square:

Formula to find area and perimeter:

Area of square = side x side square unit.

Area of square (A) = a^2 square units,   (a is the side length of the square)

Perimeter of the square = 4 x side length.

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

1. Find the area and perimeter of the square, whose side length is 5 meters.

Sol:

Area of square =a^2

= 5 x 5

Area of square =25m^2

Perimeter of the square = 4 x a

=4 x 5

Perimeter of the square  = 20 meters

2. Find the area and perimeter of the square, whose side length is 12 feet.

Sol:

Area of square   =a^2

= 12 x 12

Area of square  =144ft^2

Perimeter of the square    = 4 x a

=4 x 12

Perimeter of the square = 48 feet ft.

4th grade geometry problems in Rectangle:

Formula to find the Area and perimeter of the rectangle:

Area of the rectangle (A) = length x width

Area (A) = l x  w (l is the length and w is the width of the rectangle )

perimeter of the rectangle = 2(length + width)

perimeter (p) =2(l x w)

1. Find the area and perimeter of rectangle, whose length and width are 12meter and 6 meter respectively.

Sol:

Area of rectangle = l x w  square unit.

Given:    Length= 12 meters, Width =6 meters

=12x6

Area of rectangle  = 72 m^2                

Perimeter of the rectangle   = 2(l + w)

=2(12 + 6)

= 2 (18)

Perimeter of the rectangle  = 36 meter

2. Find the area and perimeter of rectangle, whose length and width are 8.5meter and 3 meter respectively.

Sol:

Given: Length= 8.5 meters, Width =3 meters

Area of rectangle   = l x w square unit.

=8.5x3

Area of rectangle = 25.5m^2                

Perimeter of the rectangle    = 2(l + w)

=2(8.5 + 3)

= 2 (11.5)

Perimeter of the rectangle = 23 meter

4th Grage Geometry Problems in Circle:

Formula to find the area and circumference of the circle:

Area of the circle = πr^2

( r is the radius of the circle)

Circumference of the circle = 2πr


1. The radius(r) of a circle is 5 inches. Find the area and circumference of that circle?

Sol:

Given: r = 5inches

Area of the circle= π x r^2

π = 3.14

A = 3.14 x (5)^2

=3.14 x 25 inch^2.

Area  = 78.5 in^2

Circumference of the circle = 2πr.

π=3.14, r = 5 inches

Circumference of the circle = 2 x 3.14 x 5

= 31.4 inches


2. The radius(r) of a circle is 7 inches. Find the area and circumference of that circle?

Sol:

Area of the circle= π x r^2

Given:  r= 7inches,            

value of π=3.14

Area = 3.14 x (7)^2

= 3.14 x 49 inch^2.

Area  =  153.86 in^2

Circumference of the circle = 2πr.

π=3.14 , Given r = 7 inches

=2 x 3.14 x 7

Circumference of the circle = 43.96inches

Introduction to Word Problems based on Ratio and Proportion

A ratio is the comparison of two quantities by division. It is a relation that one quantity bears to another with respect to magnitude. If A and B are two numbers, than the ratio of A to B is A/B and is denoted by A:B. The ratio does not have any unit.

The equality of two ratios is called Proportion. If (A/B) = (C/D) , then A,B,C,D are said to be in proportion and can be written as

A : B :: C:D

The Ratio and Proportion world problems are a kind of problems in which , the relation between different quantity has to be determined and then using the definition of ratio and proportion, the unknown has to be calculated.

Here are some of the ratio and proportion word problems:

Ratio and Proportion: Word Problems.

Problem: Find the value of k that must be added to 7, 16, 43, 79 so that they are in proportion. (Answer: 5)

Problem: Find the fourth proportional to the numbers 60, 48, 30. (Answer: 24)

Problem: The Income of Alex and Bob are in the ratio of 3:2 and their expenditure in the ratio of 5:3. Find the income of Alex if each saves dollars 1000. (Answer:  $6000)

Problem: A mixture contains alcohol and water in the ratio of 12:5. On adding 14 litres of water, the ratio of alcohol to water becomes 1:1. Find the quantity of alcohol in the mixture. (Answer: 24 litres)

Ratio and Proportion: Multiple Choice Word Problems:

Problem 1: If the ratio of ages of Alex and Bob is 6:5 at present and fifteen years from now, the ratio will get changed to 9:8, then find Alex's age.

(A). 24 years

(B) 30 years

(C) 18 years

(D) 33 years

(Answer: (B) 30 years)

Problem 2: If dollars 58 is divided among 150 children such that each girl and each boy gets 25 dollars and 50 dollars respectively. Then how many girls are?

(A) 52

(B) 54

(C) 68

(D) 62

(Answer: (C) 68)

Problem 3: The number that must be added to each of the numbers 8, 21, 13 and 31 to make the ratio of first two numbers equal to the ratio of last two numbers is

(A) 5

(B) 7

(C) 9

(D) None of these.

(Answer: (A) 5)