Histograms Graphs Math

Introduction:

       Let us learn about the histograms graphs in math. In mathematical concept of histogram is a graphical representation of given data records. It is easy way to understand the information by using different several of graph. A line graphs, bar graphs and pie graphs these are based on math Histogram graphs. Let us see graphs histogram functions.

Histograms graphs math:

Graphs are represented by title, bars, legend, axis (horizontal, vertical). These are important for graphs.

Title:Title is essential for graph because information (about histogram) is described by the title.
Bars: The bars have two essential characteristics, namely - height and width. The height explains the number of occurrences of values within an interval. The length of the interval covered by the bar is represented by the width.
Axis: axis is using for measured scale of values such as horizontal and vertical axis.
Legend: The legend provides extra information about the relation to the documents. These are important for histogram graphs.

Example for math histogram graphs:

        Let us learn about the frequency histogram graph in math. Frequency histogram graphs are easy methods to draw, at the same time we can clearly understand.

Column is represented for data range it divide into equal intervals.
Intervals are like (5-10, 11-15, 16-20……….) and another column for frequency.
Make a graph here data range intervals on horizontal axis but no space between intervals. And frequency percentages are we should mark on vertical axis this is also equal like horizontal axis.

Types of graphs:

1. Area graphs
2. Pie graphs
3.  Line graphs
4. Pictographs
5. Bar graphs
6. Frequency histogram graphs. These graphs are very clear to understand data records. Few of them as Follows.

Graphs:

X = 0-5  6-10  11-15  16 -20  21 -25  26 -30

Y =   10    20       30         40         50        60

Solution:

         Let us see bar frequency histogram graphs.



Line graphs:

X = 0    10  20  30  40  50

Y = 10  20  30  40  50  60

Solution: Let us see solution.

Tenth Grade Math Practice

Introduction to tenth grade math practice problem:

              Mathematics interacted well with all other branches of Science and Social Sciences and new fields such as Operations Research, Industrial Mathematics, Mathematics of Computation, Mathematical Statistics, Mathematical Biology, Mathematical Modeling, Cryptology, and Mathematical Economics etc. The tenth grade mathematics includes number theory, quadratic equation, greatest common divisor, least common divisor, factorization, and data handling etc. In this article we shall do some tenth grade practice problem.


Tenth grade math example practice problem


Example:

Find the L.C.M of 12(x–1)3 and 15(x–1) (x+2)2

Solution:

Lowest degree which is exactly divisible by the given polynomials and whose coefficient of the highest degree term has the same sign as the sign of the coefficient of the highest degree term in their product.

12(x–1)3 = 22 × 3 (x –1)3

15 (x–1) (x + 2)2 = 5 × 3 (x – 1) (x + 2)2

L.C.M. = 22 × 3 × 5 (x – 1)3 (x + 2)2 = 60 (x – 1)3 (x + 2)2

Example:

Find the L.C.M. of 6x2y, 9x2yz, 12x2y2z

Solution: 6x2y = 2 × 3 × x2 y

9xy2z = 32 x2 yz

12x2y2z= 22 × 3x2 y2z

L.C.M. = 22 × 32 × x2y2z = 4 × 9 x2y2z

L.C.M. = 36x2y2 z

Example:

Find the square root of

x2 – 4 = x2 – 22 = (x–2) (x+2)

x2 + x – 6 = x2 + 3x – 2x – 6 = x (x+3) – 2 (x+3) = (x+3) (x–2)

x2 + 5x + 6 = x2 + 3x + 2x – 6 = x (x+3) + 2 (x+3) = (x+3) (x+2)

Hence (x2 – 4) (x2 + x – 6) (x2 + 5x + 6)

= (x + 2) (x + 3) (x – 2) (x + 3) (x + 2) (x – 2)

= (x–2)2 (x + 2)2 (x + 3)2 = [(x – 2) (x + 2) (x + 3)]2

Therefore the required square root is (x – 2) (x + 2) (x + 3)


Tenth grade math example practice problem


Practice Problem 1:

Find the L.C.M. of x3 + 1, x2 – 1, (x + 1)2

Answer:

L.C.M. = (x + 1)2 (x – 1) (x2 – x + 1)

Practice Problem 2:

Find square root of x2 + 10x + 25

Answer:

x + 5

Get Help With Math Now

Introduction to get help with math now:

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.(source: From Wikipedia).

Algebra, geometry and probability are the interesting topics and these are very helpful in many real life applications. Now we are going to get help with some math problems. From this we can get clear view about how to solve the math problems easily.


Get help with some math algebra problems now:


Example problem 1:

Solve for the variable x: x2 – 4 = 0

Solution:

x2 – 4 = 0

x2 – 22 = 0

Using the formula x2 - y2 = (x + y) (x – y)

From this formula, we get

(x + 2) (x – 2) = 0

x + 2 = 0 or x – 2 = 0

x = -2 or x= 2

So, the answer is x=-2 and x=2.

Example problem 2:

Solve for the value of x: 3x + 3 = 18

Solution:

3x + 3 = 18

Now, Subtract 3 on both sides of the equation, we get

3x + 3 - 3 = 18 – 3

3x = 15

Divide by 3 on both sides of the equation, we get

3x / 3 = 15 / 3

x = 5

So, the answer is x=3.


Get help with some geometry problems now:


Example problem 3:

Length and breadth of a rectangle are 15 cm and 10 cm respectively. Find area and perimeter of the given rectangle.

Solution:

(i) Area of the rectangle = Length × Breadth

= l × b

= 15 cm. × 10 cm.

= 150 Sq. cm.

So, the area of rectangle is 150 Sq. cm.

(ii) Perimeter of the rectangle = 2 (l + B)

= 2 (l + b)

= 2 (15 + 10)

= 2 × 25 = 50 cm.

So, the perimeter of rectangle = 50 cm.

Example problem 4:

Find the volume of a wooden plank 10 cm. long, 5 cm broad and 2 cm thick.

Solution:

The volume of the plank = Length × breadth × height

= 10 × 5 × 2 cm3

= 100 cm3

So, the volume of a wooden plank is 100 cm3.

Double Negatives In Math

Introduction of double negatives in math:

The double negative in math deals with the signed numbers in the math. The double negatives give some rule in which the math rules can be made while the summing of the numbers is made and results to find the solution of the numbers. The double negative can have the values in positive manner. The double negatives come under the arithmetic operations.

Double negatives in math:
The negative numbers are given with (-) signed in front of the numerals. The negative numbers are value lower than the zero. The decreases in the value are made in the negative numbers. The double negative numbers are also named as the addition of the signed numbers. The positive numbers are made to have in the way in which the values are represented that is the values are more than the zero. The signed numbers can be done through the some of the rules which are,

` (-)xx (-) = (+)`

`(-)xx (+) = (-)`

`(+)xx (-) = (-)`

The above are the some of the rules for the negative numbers. The double negative numbers are which the product of the two negative numbers results to the positive of the particular value. Then the summing of the double negatives gives us the summing process with the negative signed value. The subtracting of the double negatives results to the negative signed numbers.


Examples for the double negatives in math:


Example:

What is -6+ (-2)?

Solution:

From above: (- )+(-) becomes a negative sign.

-6+ (-2) = -6 - 2

Answer: -6+ (-2) = -8.

More example problems for adding signed numbers:

(-) 7+ (-2) = -7 - 2 = -9.

(-) 8-(+2) = -8 - 2 = -10.

Example:

What is -5+ (-2)?

Solution:

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

Answer: -5+ (-2) = -7.

Example: What is -6 - (+3)?

-6 - (+3) = -6 - 3

Answer: -6- (+3) = -9.

Boundary Line Math Definition

Introduction to boundary line math definition:

The boundary line is defined as the line or border around outside of a shape. The Boundary line defines the space or area. The limits of an area can be determined by the boundary line. The boundary line lies instantly inside the boundary. The boundary line indicating an edge of something. There is a boundary line for each and every shape. For each and every shape we can determine the area. The examples of boundary lines in math are given below. They are:

Triangle, Circle, Square, Rectangle, Parallelogram, Trapezoid, Rhombus, Cylinder, Cube and Cone


Examples problems- boundary line math definition:


Example 1: boundary line math definition

Find the area of the right- angled triangle whose height is 10cm and base is 5cm.



Solution:

Step 1: Given that base, b = 5cm and height, h = 10cm

Step 2: The formula for the area of the right-angled triangle is area= `1/2` (`bxx h)`

Step 3: Substitute the values of base and height

Step 4: area=`1/2(5 xx 10)`

Step 5: area=`1/2(50)`

Step 6: area=25cm

Answer: The area of the right- angled triangle is 25cm.

Example 2: boundary line math definition

Find the area of the equilateral triangle whose side is 5cm.



Solution:

Step 1: Given that the side is 5cm.

Step 2: The formula for the area of the equilateral triangle is area= `sqrt(3)/4` ` xx` a2

Step 3: Substitute the values of the side.

Step 4: area= `sqrt(3)/4` ` xx` 5 2

Step 5: area= `sqrt(3)/4 xx` 25

Step 6: area=6.25`xxsqrt(3)`

Step 7: area=6.25 `xx` 1.732 (The value of `sqrt(3) ` is 1.732)

Step 8: area=10.825cm

Answer: The area of the equilateral triangle is 10.825cm.


One more example problem- boundary line math definition:


Example 1: boundary line math definition

Find the area of the circle whose radius is given as 2cm.

      

Solution:

Step 1: Given that the radius is 2cm.

Step 2: The formula for the area of the circle is area= `pi `r2

Step 3: Substitute the values of the radius

Step 4: We know that the value of `pi ` is `22/7` (or) 3.14

Step 5: area= 3.14 `xx` 22

Step 6: area= 3.14 `xx` 4

Step 7: area=12.56cm

Answer: The area of the circle is 12.56cm

Arithmetic math Diagrams

Introduction to arithmetic math diagram:

Arithmetic or arithmetic (from the Greek word ????տ? = number) is the oldest and most elementary branch of mathematics, used by almost everyone, for tasks ranging from simple day-to-day counting to advanced science and business calculations.

In common usage, it refers to the simpler properties when using the traditional operations of addition, subtraction, multiplication and division with smaller values of numbers.




Arithmetic math diagram and problems:


Example problems for arithmetic math diagram:

Addition problems:

Question 1:

Solve: From the given problem, perform addition operation for 4 red color balls and 6 blue color balls.

Solution:

The given problem is

4 red balls and 6 blue color balls



Total number of balls = red color balls + blue color balls.

= 4 + 6

= 10

Total number of balls = 10 Balls.

Question 2: Solve 41 + 6 + 7 + 3

Solution: The given problem is
= 41 + 6 + 7 + 3
= 54 + 3
= 57

Question 3: Solve 5x + 2x + 4x

Solution: The given problem is

= 7x + 4x
= 11x

Subtraction problems:

Question 1:

Solve: From the given problem, perform subtraction operation for 4 black color balls and 2 green color balls

Solution:

The given problem is

4 black balls and 2 green color balls



Total number of balls = black  color balls - green color balls = red color balls.

= 4 -2

= 2

Total number of balls =2 red balls.

Question 2: Solve 40 - 4 - 2 -3

Solution: The given problem is
= 40 - 4 - 2 -3
= 36 -2 - 3
= 34 -1

= 33

Question 3: Solve 8x - 6y -3x -4y

Solution: The given problem is

= 8x - 6y -3x - 4y

= 8x - 3x - 6y - 4y

= 5x - 2y

Question 4: Solve 14x - 7x - 2x

Solution: The given problem is
=14x -7x -2x

= 7x - 2x

= 5x

Multiplication problems:

Question 1:

Solve: From the given problem, perform Multiplication operation for 2 red color balls and 2 blue color balls.

Solution:

The given problem is

2 red balls and 2 blue color balls



Total number of balls = red color balls * blue color balls = Yellow color balls.

= 2 *2

= 4

Total number of balls = 4 Yellow color balls.

Question 2:

Multiply 381 by 102

Solution:

381 נ102 means 102 times 381

Or 100 times 381 + two times 380

Or 38100 + 760

Or 38860

Therefore 381 נ102 = 38860

Division problems:

Question 1:

Solve: From the given problem, perform Division operation for 4 brown color balls and 2 pink color balls.

Solution:

The given problem is

4 brown balls and 2 pink color balls




Total number of balls = brown  color balls / pink color balls = Green color balls.

= `4 / 2 `

= 2

Total number of balls = 2 Green color balls.

Question 2:  solve (6/10)-(5/10)

Solution:
= (`6/10` )-(`5/10` )
= (6-5) / 10
=  `1 / 10`


Practice problems for arithmetic math diagram:


Practice problems are given below for test preparation,

1) Answer the math expression

32 + (3 * 3) -205

Solution: - 187

2) Answer the math expression

(ֳ)2 + 202 -10 -1

Solution: 200

3) Answer the math expression

3x + 2y -20 + 4y -2x

Solution: x + 6y -2

Problems Involving Coins : 2

Introduction to problems involving coins

The purpose of the article is to help readers understand:

• Represent the value of a given combination of coins.
• Represent a word problem in to an equation.
• How to solve an equation.
• Various denominations (as per US Standards)
• Solve problems consisting probability of coins.
• Representation of worth of amount in various denominations.

Denominations of coins used in solving problems

S.No	COIN	Worth (in Cents)
1	Quarter	25 cents
2	Dime	10 cents
3	Nickel	5 cents
4	Cent	1 cent

We will use the values in the above chart to solve problems involving coins


Examples of word problems involving coins


Ex 1: If a pencil costs 4 quarters and cost of an eraser is 5 dimes then how much will it cost (in dollars) to purchase 2 pencils and 1 eraser?

Sol:  Cost of 1 pencil = 4 quarters = 4 * 25 cents= 100 cents

Cost of an eraser = 5 dimes = 5*10 cents = 50 cents

Total cost of 2 pencils and 1 eraser (in cents) = 2*100 + 50 = 250 cents

Total cost of 2 pencils and 1 eraser (in dollars) = 250/100 = $2.5 ANSWER.

Ex 2: In a fundraiser program a total of $534 were collected in dimes and quarters. If the total number coins were 3000, then how many quarters were in the collection?

Sol: Suppose the number of dimes be equal to‘d’ and number of quarters be equal to ‘q’ in final collection.

So now we can write the following 2 equations:

1)      Equation for matching the total number of coins for both the denominations

d+q=3000 (Equation A)

2)      Equation for matching the total worth of coins and total sum collected (in cents)----10d+25q=534*100 (Equation B)

Representing in table form the problem can be summarized as:

Kind of coin	Number of coins	Value of each coin (in cents)	Total value  (in cents)
Dimes	d	10 cents	10*d
Quarters	q	25 cents	25*q
Total	3000		53400

Solving the two equations:

Multiplying equation A by 2 and dividing equation B by 5 gives

2d+2q = 6000 (Equation C) and 2d+5q= 10680 (Equation D)

Now subtracting Equation C from Equation D gives 3q=4680 and hence q=1560

Putting value of q in Equation A fives value of d= 1440

Hence,

                Number of dimes = 1440 Answer

Number of quarters = 1560 Answer

Ex:3 In a charity box, there is a collection of nickels, dimes, and quarters which amount to $5.90. There are 3 times as many quarters as nickels, and 5 more dimes than nickels. How many coins of each kind are there?

Sol: Let n = the number of nickels.

Then 3n = the number of quarters.

(n + 5) = the number of dimes.

Kind of coin	Number of coins	Value of each coin (in cents)	Total value  (in cents)
Nickels	n	5	5*n
Quarter	3n	25	25*3n
Dimes	n+5	10	10*(n+5)

The total value of all the coins is 590 cents.

5n+25*3n+10*(n+5)=590

5n+75n+10n+50=590

90n=540

n = 6 hence ,

Number of Nickels= 6 Answer

Number of Quarter = 3*6 = 18 Answer   and

Number of Dimes = n+5 = 6+5 = 11 Answer


Problems Based on Probability involving Coins


Concept :  Probabilities are represented as numbers between 0 and 1 (both including) to reflect the chances of occurrence of an event. A probability of 1 represents that the event is certain and a probability of 0 represents that the event is impossible.

When a coin is tossed probability of occurrence of Head or Tail is equal and since sum of probabilities of an event is 1 (at max), we say that probability of occurrence of Head will be equal to probability of occurrence of Tail and both will b equal to half(0.5).

Possible outcomes useful when solving problems involving coins

1) Toss of one coin: T or H

2) Toss of 2 coins: TT, TH, HT, TT

3) Toss of 3 coins: TTT, TTH, THT, THH, HTT, HTH, HHT, HHH

And similarly total outcomes of toss of n coins will be equal to 2n.

Ex 4: A coin is tossed. What is the probability of getting a Head?

Solution : As also discussed in concept above, probability of occurrence of Head in a single toss in a fair coin will be equal to 0.5 OR 50% .

Ex 5: What is the probability of getting three tails, in each flip, if a coin is tossed three times?

Solution : Since the probability of getting tails is 0.5, the probability of three tails would be 0.5*0.5*0.5=0.125 or 12.5 %