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 %

Place Value Chart Math

Introduction to place value math chart:

                           In math, the place value of numbers is established by its placed positions. The numbers are located in the particular number of a digit inside a numerical numbers. The place value specifies the position of a numerical system base value. The values are illustrates that numerical value in the standard forms in the identification. The values are established by place of the numerals accessibility.




Numbers on the place value chart in math:


Numbers- words

      The following chart graph is an example of place value chart in math.



In a given diagram, we have illustrate the place values of ones to hundred.

Decimal value of math chart:
 0.5 - 10 th

 0.05 -100 th

 0.005 -1000 th

 0.0005- 10000 th



Math example for decimal number place value:


Math chart Example 1:

       Place value is the sources of our whole number system. A place value in math system is one in which the position of a digit in a number establishes its value.

Look at number 584, the 5 in the hundreds place equal 500, The 8 in tens place equal 80, The 4 in the ones place equal 4.

Math chart Example 2:

 

The 6 is in the ten thousands place. It tells you there is 6 sets of ten thousand in the thousand place.

10,000+10,000+10,000+10,000+ 10,000+ 10,000 = 60,000   

The 5 is in the thousands place.  It tells you that there are 5 sets of thousand in the thousands place.1000+1000+1000+1000+1000 = 5000

The 2 is in the hundreds place. It tells you there are 2 sets of hundreds in the hundreds place.100+100=200

The 3 is in the tens place. It tells you there are 3 sets of tens in the tens place 10+10+10=30

The 4 is in the ones place. It tells you there is 4 ones in the ones place 1+1+1+1= 4

60,000 + 5000 + 220 + 30 + 4 = 65234

Fraction Subtraction Rules

Introduction For fraction subtraction rules:
A fraction is a part of a whole group or its region. A fraction written in a form of number with bottom part (denominator) denotes how many parts the whole divided into, and a top part (numerator). Fraction is in few types which can be written as many types which depend upon numerator and denominator by its value.

Fractions are subdivided into
Proper fractions
Improper fractions
Mixed fractions


Classification of Fractions:


Proper Fractions:
A proper fraction, fraction which a denominator shows in number of parts into whole divided and a numerator shows the number of parts which  we taken out. proper fraction also defined as numerator less than denominator.

Example:
`1/3, 2/5`

Improper Fractions:
Improper fractions is a fraction, whose numerator which is greater than the denominator are called improper fractions.

Example:
`3/2, 9/2`

Mixed Fractions:
A  mixed fraction is a fraction defined as which  has combination of a whole and its part.

Example:
2` 1/4,` 2` 2/9` ,


Rules for Fraction Subtraction:


In subtraction, rules for fraction numbers with same denominator, denominator remains same number and we subtract only numerator.
We can't do subtraction in fraction with different denominator rules for that we have to take LCM for all denominator and change different denominator into like denominator by taking LCM and adding fractions

Example Problems in Fraction subtraction rules:
Example 1:
subtract fractions  `1/3` from  `5/3`

Solution :
given fraction is a proper fraction, we have same denominator
`5/3 -1/3` =` (5-1)/3`

=`4/3`

Example 2:
subtract fractions  `2/5` from  `4/5`

Solution :
given fraction is a proper fraction, we have same denominator
`4/5 -2/5` =` (4-2)/5`

=`2/5`

Example 3:
subtract  fractions  `1/5` from `1/3`

Solution:
In this improper fraction we have different denominator  so,we take LCM
The LCM of 3 and 5 is 15.

Therefore, `1/3-1/5` =`(1xx5)/(3xx5)-(1xx3)/(5xx3)`

=`5/15-3/15`

=`2/15`
Example 4:
Subtract fractions  `1 2/5` from `3 3/6`

Solution:
`3 3/6 - 1 2/5` =  `(18+3)/6-(5+2)/5`

Now `21/6-7/5` =`(21xx5)/(6xx5)-(7xx6)/(5xx6) `    since LCM of 5,6 =30

=`105/30-42/30`

=`63/30`

Elementary Math Terms

Introduction to elementary math terms

     Elementary math terms are the basic form of algebra having little or no proper information of mathematics beyond arithmetic. In arithmetic numbers and their arithmetical operations (such as +, −, ×, ÷) occur, in algebra one also uses symbols (such as x and y, or a and b) to denote numbers. Elementary Algebra can be distinguished from abstract algebra, a more advanced field of study.And now we see about the elementary math terms below in simple problems.(Source in wikipedia )

                              

Elementary math terms problems in solving the equations:

• Solve the given problem for  x: 1 - 3(x - 4) = 2(3x + 1) - 7

Solution:-

Now we allocate the  above equation:

1 - 3x +12 = 6x + 2 - 7

Collect like terms:

 -3x +13 = 6x - 5

Add 3x to both sides of the equal sign:

-3x +13 + 3x = 6x - 5 + 3x

13 = 9x - 5

Add 5 to both sides:

18 = 9x

Divide both sides by 9 in above equation :

2 = x

The final result of the given problem fro  x is 2.

• Solvethe given problem for y:   ax + by = c

solution:-

Step 1: Subtract ax from both sides:

Step 2:by = c - ax

Step 3:Divide both sides by b:

           C - ax
y =   -----------  
             b

The final result of the given problem are            C - ax
                                                                       y = -----------                                            
                                                                                b

• Solve the problem :X = 5*(20+5)

Solution:

Step 1:  5*(20+5) = 5*20+5*5

Step 2:  =100+25

               X  =125

The final result the given problem is X= 125

• Solve the problem :X= 9*(6 + 5)

Solution

Step 1: 9*(6 + 5) = 9*6 + 9 *5

Step2: 36 + 45 

Step 3: X =  81     

The final result of the given problem  is X = 81       

• Solve the problem : Y = 7+(4+6)

Solution:

Step 1:7+(4+6)=7+10

                  Y    =17

The final result of the given problem is Y =  17

The above discussed are the  element math terms like addition , subtraction ,Multiplication and division .

Practice problems in Elementary math terms :

• Solve algebraic equation       6(-2x - 3) - (x - 2) = -5(2x + 3) + 19  

Solution:    x = 10

• Subtract x³ – 3x² – 1 from 3x³ + 6x² – 4x – 8.

Solution:   2x³ + 9x² – 4x – 7

Intercept Theorem

Introduction:

                  Intercept theorem plays a vital role in elementary geometry. It deals with the ratios of various line segments, which are created if two intersecting lines are intercepted by a pair of parallels. This theorem is equivalent to ratios of similar triangles. Traditionally it is attributed. The intercept theorem is  also called as Thale's Theorem. If a transversal line makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts.

Concept of Intercept Theorem:

                  If a transversal line makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts. AP || BQ || CR.

Types of intercept Theorem

                  If a transversal line makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts. AP || BQ || CR.

What Does Transversal Mean in Math

Introduction to Transversal in Math:

Definition:

A line that cuts (passes through) across two or more (usually parallel) lines then it is said to be a transversal.
It can also be defined as a line that intersects two or more co-planar lines each at a different point is also termed as a Transversal.

Transversal Postulates:
If two lines are parallel to each other and intersected by a transversal then the corresponding angles are congruent.

Some important points:

• Supplementary angles are the angles whose sum is equal to 180⁰.
• Vertical angles are the angles that are not adjacent angles and are formed by two intersecting lines and are always equal to each other.

Properties of transversal of parallel lines:

• If two parallel lines are cut by a straight line (transversal), then the corresponding angles around each intersection are equal in measure or we can say mathematically as these angles are congruent.
• If two parallel lines are cut by a straight line (transversal) then the alternate interior angles are congruent.
• If two parallel lines are cut by a straight line (transversal) then the interior angles on the same side of the transversal are supplementary.

Proofs for transversal by solved examples

Solved Problems to prove the above given points:
Ex 1: In the given figure, If the angles 2 and 3 are congruent then prove that r and s are parallel.

parallel lines 

Sol: Step 1: Given in the problem that angle 2 = angle 3
       Step 2: From the data given above angle 1 and angle 2 are vertically opposite angles and are congruent.
                               therefore, angle 1 = angle 2
       Step 3:The transitive property says that if a=b and b=c then a=c
 Using this property we have,
          angle 2 = angle 3    (given in the problem)
          angle 1 = angle 2    (vertically opposite angles)
so ,    angle 1 = angle 3   
As the angles made by the transversal with the two lines are equal the two lines are parallel.
Ex 2: Show that the transversal shown in the figure cuts the parallel lines.

Sol: In the above figure the angles 4 and 3 are known as alternate interior angles.Accoding to above discusssed points the alternate interior angles are congruent.
           `=>`    angle 4 = angle 3
So, the two lines are parallel lines.
Ex 3: Prove that the lines a and b shown in the figure below are parallel.

Sol: Clearly it is shown that the angles 1 and 2 are equal to` 90@` . That means that they are supplementary angles based upon the above discussion.
As the  angles made by the transversal with the two straigth are supplementary(equal) the two lines are parallel.Hence, proved.

Practice Problem on traversal line

Pro: Find the measures of the unknown angles in the following figure.Given r is parallel to s and angle 1 = `60@.`

Ans:  angle 2 =`120@.`
angle 3 = `60@` .
angle 4 = `120@`
angle 5 = angle 1 = 60⁰
angle 6 = angle 2 = 120⁰
angle 7 = angle 3 = 60⁰
angle 8 = angle 4 = 120⁰

Solving Genetics Probability Problems

In probability method, we have to decide the basic terms of probability.. The main aim of genetics probability subject is to give the fundamental properties the importance of their use with some basic examples of genetics. In a genetics probability, the basic term and properties is not easier to understand but work with sums we will start to get comfortable with in genetics probability.


Basic Definitions of genetics probability:

In a genetic probability method, an testing method is continuously repeated for infinite numbers of time and it happened by the expected probability in the genetics

Genetics probability solving problems:

Problem 1:

To solve consider a locus with two chromosome, C and c. If the frequency of CC is 0.64, what is the frequency of A under Hardy-Weinberg?

Solution:

Solving methods of genetic probability,

Under H-W, if q = freq(C) , then q2 = freq(CC), hence q2 =0.64 or q = 0.8.

Problem 2:

To solve if the genotypes CC, Cc, and cc have frequencies 0.7, 0.49, and 0.49 (respectively), what are q = freq (C)? r = freq (c)? After a single generation of random mating, what is the expected frequency of CC? Of Cc? Of cc?

Solution:

Solving methods of genetic probability,

p = freq (CC) + (1/2) freq (Cc) = 0.7 + (1/2)(0.49) = 0.945

q = 1-p = 0.945

freq (CC) = p2 = 0.9452 = 0.8930

freq (Cc) = 2qr = 2*0.945*0.8930 =1.6877

freq (cc) = r2= 0.89302= 0.797449

Problem 3:

To solve the suppose 60 out of 850 women’s are green eyes. What is frequency of green eyes? If a random women’s is chosen, what is the probability they are a green eyes?

Solution:

Solving methods of genetic probability,

Freq (Redheads) = 60/850 = 0.070 or 7.05 percent

Probability of a Redhead = 30/850, or 7.05 percent

Problem 4:

Solve the genotype cc is lethal and yet population has an equilibrium frequency for c of 30.  Here the Cc is the fitness and its value is one.  Find the CC genotype fitness?

Recall if the genotypes CC: Cc: cc have fitness 1-s : 1 : 1-t, then the equilibrium frequency of A is s/(r+s).

Here, s= 1 (as cc is lethal), so that freq(C) = 1- freq(c) =0.6 = 1/(1+s)

Solving gives 1+s = 1/0.8, or s = 1/0.8 -1 = 3.5

Hence fitness of CC = 1-s = 1-3.5=-2.5.

Learn Derivative at a Point

Learn derivative at a point involves the process solving derivative problems at an exam point view. The rate of change of the given function is determined with the help of calculus. To find the rate of change the calculus is divided into two types such as differential calculus and integral calculus. At exam point of view the derivative is easily carried out by differentiating the given function with respect to input variable. The following are the solved example problems in derivatives at exam point of view for learn.

Learn derivative at a point example problems:


Example 1:

Find the derivative for the given differential function.

f(e) = 3e 3 -4 e 4  - 5e

Solution:

The given function is

f(e) = 3e 3 -4 e 4  - 5e

The above function is differentiated with respect to e to find the derivative

f '(e) = 3(3e 2 )-4(4e 3  ) - 5

By solving above terms

f '(e) = 9e 2 -  8e 3 - 5

Example 2:

Compute the derivative for the given differential function.

f(e) = 6e6 - 5 e5 - 4 e4 - e

Solution:

The given equation is

f(e) = 6e6 - 5 e5 - 4 e4 - e

The above function is differentiated with respect to e to find the derivative

f '(e) =  6(6e 5)  -5 (5 e4 ) -4(4 e3) - 1

By solving above terms

f '(e) =  36e 5  - 25 e4  - 16 e3 - 1

Example 3:

Compute the derivative for the given differential function.

f(e) = 2e 2 -4 e 4  - 15

Solution:

The given function is

f(e) = 2e 2 -4e 4  - 15

The above function is differentiated with respect to e to find the derivative

f '(e) = 2(2e  )-4(4 e 3 ) - 0

By solving above terms

f '(e) = 4e -16e3

Example 4:

Compute the derivative for the given differential function.

f(e) = 5e5 -4e 4 -3e 3  - 2

Solution:

The given function is

f(e) = 5e5 -4e 4 -3e 3  - 2

The above function is differentiated with respect to e to find the derivative

f '(e) = 5(5e 4 )-4(4e 3 ) -3( 3e 2) -0

By solving above terms

f '(e) = 25e 4 -16e 3  -9 e 2


Learn derivative at a point practice problems:


1) Compute the derivative for the given differential function.

f(e) = 2e 3 -3 e 4  - 4 e 5

Answer: f '(e) = 6e 2 -12 e3 - 20 e 4

2) Compute the derivative for the given differential function.

f(e) = e 3-e5 - 4 e 6

Answer: f '(e) = 3e2 - 5e4 - 24 e 5

Linear Algebra Test

Linear function is a function with polynomial degree 1. This linear function matches a dependent variable with independent variable and maintains a relation in a simpler way. The linear algebra is the important method which is used in the linear function, variable function, linear equation, vector, matrix etc. This linear function is used in the high school and college students are using the linear function by using the linear algebra.

Sample Problem for linear algebra:


f is a linear function. Value of x and f(x) are given in the table below; complete the table using linear algebra

X                           f(x)

-1                          11

1                           –

–                           1

6                          -12

7                         --

–                          -35

Solution:

f is a linear function in linear algebra whose formula has the form to test is

f(x) = a x + b

where a and b are constants to be test and found. Note that 2 ordered pairs (-2,18) and (5,-19) are given in the table. These are two ordered pairs to test are used to write a system of linear equations as follows

11 = - 1 a + b and -12 = 6 a + b

Solve the above system to obtain a = - 3 and b = 2 and write the formula for function f as follows

f(x) = - 3 x + 2

We now use the formula for the f to find f(x) given x or find x given f(x).

for x = 0 , f(0) = -3(0) + 2 = 2

for f(x) = 1 , 1 = -3 x + 2 which gives x = 1/3

for x = 7 , f(7) = -3(7) + 2 = - 19

for f(x) = - 30 , -30 = -3 x + 2 which gives x = 32/3

We now put the values of the calculated above in the answer.

X                           f(x)

-1                          11

1                           2

1/3                         1

6                         -12

7                         -19

32/3                      -35


Practice problem:


Problem 1:
f is a linear function in linear algebra. Value of x and f(x) to test are given in the table below; complete the table.

X                              f(x)

-5                             15

1                               –

–                              1

10                         -14

8                             --

–                           -35


Problem 2:
f is a linear function. Value of x and f(x) to test are given in the table below; complete the table.

X                               f(x)

-6                             18

1                              –

–                             1

9                           -20

10                             --

–                             -40

Exponentiation Tutoring

This article is a online exponentiation tutoring.

Generally the exponentiation is used in all sciences including the mathematical, physical, chemical or even in biological sciences but it is assumed or preferably called a mathematical operation which generally written as an it is clear from this notation that the exponentiation involves two numbers one is called the base i.e. a and other is called the exponent i.e. n. When n is any positive integer then the exponentiation simply means the repeated multiplication of the number ‘a’ to itself n times which are written as:

 

Generally, exponent is the superscript to the base. Exponentiation above will be read as a raised to the power n or exponent of n, or a to the n. A few exponents have their pronunciation: for instance, a2 is read as a square and a3 as a cube.

Exponentiation comes to use in economics or biology or chemistry or physics or computer science to calculate compound interest or population growth.


Example problems on exponentiation tutoring:


Positive integer exponents:

a2 is pronounced as a square because any square with side a has area equal to a2. The expression a3 is referred to as the cube of a since the volume of a cube with side length a is a3.

So 32 is called as "three squared", and 23 is "two cubed".

The numbers with positive integer exponents may be defined by the initial condition

a1 = a

thus in general the recurrence relation

an+1 = a·an.


Exponents of one and zero:


If a is not zero and n an m are two positive integer exponents such that the exponent (n – m) is also positive then we can say that



If we consider a special case in which the exponent’s n and m are equal then we can wrote the equality as

 

Thus we lead to the following rule: Any number with raised to the power 1 is the number itself.

Solving Quadratic Formula

Quadratic Equation:

An equation which has one or more terms are squared but no higher power in terms, having the syntax, ax2+bx+c =0 where a represents the numerical coefficient of x2, b represents the numerical coefficient of x, and c represents the constant numerical term.

Types of quadratic equation

Pure quadratic equation:

The numerical coefficient cannot be zero. If b=0 then the quadratic equation is called as a ‘pure’ quadratic equation

Complete quadratic equation:

If the equation having x and x2 terms such an equation is called a ‘complete’ quadratic equation. The constant numerical term ‘c’ may or may not be zero in a complete quadratic equation. Example, x2 + 5x + 6 = 0 and 2x2 - 5x = 0 are complete quadratic equations.


Quadratic Equation Formula


The quadratic equation has the solutions ax2+bx+c =0

x =√(b2-4ac)/2a

Consider the general quadratic equation

ax2+bx+c =0

With a`!=` 0. First divide both sides of the equation by a to get

x2+b/a x + c/a =0

which leads to

x2+ b/a x = - c/a

Next complete the square by adding ((b)/(2a) )2to both sides

X2+ ((b)/(a) )x+((b)/(2a) )2 = -((c)/(a) )+ ((b)/(2a) )2

(x+(b)/(2a) )2=-((c)/(a) ) + ((b^2)/(4a^2) )

(x+(b)/(2a) )2 = (b^2-4ac)/(4a^2)

Finally we take the square root of both sides:

x+(b)/(2a) = +-(sqrt(b^2-4ac))/(2a)

or

x =-(b)/(2a)  +-(sqrt(b^2-4ac))/(2a)

The final form of Quadratic Formula is

x =-b+-sqrt(b^2-4ac)/(2a)

The two roots of the equation is

``          -b-(sqrt(b^2-4ac))/(2a)

-b+(sqrt(b^2-4ac))/(2a)


Example Problem on solving Quadratic formula


Example:

Find the roots of the equation by quadratic formula method, x2-10x+25=0

Solution:

Step 1:  From the equation, a = 1, b = - 10 and c = 25.

Step 2:  To Find X:
plug-in the values in the formula below
x = ``

Step 3:  We get the roots, x = ``
x = 5 and x = 5
which means x1 = 5 and x2 = 5.

Here x = 5 is root of the equation.

learning problems in probability

Introduction:

Learning Probability is also one part of mathematics subject. Probability disturbed with investigation of approximate phenomena. Learning the fundamental things of the probability is random variables, stochastic processes, and events. Mathematical abstraction of the non-deterministic events or calculated quantities that may moreover be single occurrences or evolve over time in an apparently random fashion. While a coin toss or roll of a die is a approximate event.

Let us some example problems in probability with solved solutions.

Learning example problems in probability:

Learning Example problem 1:

Three dice are rolled once. What is the chance that the sum of the expression numbers on the three dice is greater than 15?

Solution:

In probability problems while three dice are rolled,

the trial space S = {(1,1,1), (1,1,2), (1,1,3) ...(6,6,6)}.

S contains 6 × 6 × 6 = 216 outcomes.

Let A be the event of receiving the sum of appearance numbers greater than 15.

A = { (4,6,6), (6,4,6), (6,6,4), (5,5,6), (5,6,5), (6,5,5), (5,6,6), (6,5,6),

(6,6,5), (6,6,6)}.

N (S) = 216 and n (A) = 10.

Therefore P (A) equal to n (A) / n(S)

= 10/ 216

=5 / 108.


Probability Problem 2:


Learning Example problem 2:

In a vehicle Stand there are 100 vehicles, 60 of which are cars, 30 are mini truck and the rest are Lorries. If each vehicle is uniformly likely to leave, find out the probability in this problem.

a) Mini truck departure first.

b) Lorry departure first.

c) Car leaving second if either a lorry or mini truck had gone first.

Solution:

a) Let S be the model space and A be the event of a mini truck leaving first.

n (S) = 100

n (A) = 30

Probability of a mini truck departure first:

P (A) = 30 / 100 = 3 / 10.

b) Let B be the event of a lorry departure first.

n (B) = 100 – 60 – 30 = 10

Probability of a lorry departure first:

P (B) = 10 / 100 = 1 / 10.

c) If moreover a lorry or mini truck had gone first, then there would be 99 vehicles totally left there, 60 of which are cars. Let T be the trial space and C be the occurrence of a car departure.

n(T) = 99

n(C) = 60

Probability of a car departure after a lorry or mini truck has gone:

P(C) = 60 / 99 = 20 / 33.

Learning Probability Density Functions

The Learning Probability density functions are known as Gaussian functions. A probability density function of a continues random variable is defined as a function that describes the virtual probability for that random variable to occur at a given point within the using space. The Learning of probability density functions is same for the probability distribution functions. In this article we see the learning  the definition of probability density functions and some example problem for the probability density functions.

Definition of Learning the Probability Density Functions:


Probability density function is defined as continues random variable functions f(x). this functions satisfies the following properties.

Probability functions limits between a and b

P(a≤x≤b) = ∫ a to b  f(x) . dx

A probability density function has only real for the real value.

F(x) ≥ 0

Integral of the probability density functions is 1.

∫ -oo to oo f(x) dx  = 1

Learning the fundamental properties of probability density functions:

F(x) is continues random functions

P(a<= x <b) =  P(a < x<b) = P( a< x <= b)

= P(a<=X<b) =

Where f is probability density distribution functions

In the differentional functional of the probability density functions, we have

P(x<X<=x+dx) = F(x+dx)-F(x) = dF(x)=F'(x)dx = f(x)dx

Where

X is Known as probability differential functions.


Learning the some example problems for the probability density functions:


Example 1:

If the probability density function of a random variable is given by f(x) = H (x – x^3);  0 < x <1 .find the H.

Solution:

Since ,

F(x) is the probability density functions means,

F(x)= ∫ a to b f(x) dx

Substitute the values,

H [ x^2/2 – x^4/4] limits 0 to1 = 1

H [ 1/2- 1/4] = 1

1 After simplify, We get

H= 4

Example-2

If the probability density function of a random variable is given by f(x) = H (1– x^5);  0 < x <1 .find the H.

Solution:

Since ,

F(x) is the probability density functions means,

F(x)= ∫ a to b f(x) dx

Substitute the values,

H [ x – x^6/6] limits 0 to1 = 1

H [ 1- 1/6] = 1

1 After simplify, We get

H= 6/5 = 1.2

Practice problem of  learning the probability density functions:

Problem -1:

A continuous random variable X follows the probability law, f(x) =H x (x – x^2 ), 0 < x < 1and 0 for elsewhere. Find k

Answer: H = 12

Problem -2:

A continuous random variable X follows the probability law, f(x) =H x^2 (x– x^4 ), 0 < x < 1and 0 for elsewhere. Find k

Answer: H = 9.33