Quadratic Inequalities

In this page we are going to discuss about  quadratic inequalities concept.A quadratic inequality in one variable is in the form of an expression:

ax2 + bx + c ≤ 0  or   ax2 + bx + c < 0
ax2 + bx + c ≥ 0  or   ax2 + bx + c < 0

where a, b and c are real numbers, a ≠ 0. The values of x which satisfy the given inequality are called the solutions of the inequality.

A quadratic equation has only two roots. But a quadratic inequality has many roots.


Methods to solve quadratic Inequalities


There are two methods to solve quadratic inequalities -

Method 1: Finding the solution by dividing the given polynomial into factors. This method is called the 'Algebraic method'.

Method 2: Finding  the solutions by drawing the graph of the inequality. This method is called the 'Graphical method'.

Note:

• The trick in solving a quadratic inequality is to replace the inequality symbol with an equal sign and solve the resulting equation.  The solutions to the equation will allow us to establish intervals that will let you solve the inequality.
• Plot the solutions on number line creating the intervals for investigation. Pick any number from each interval and test it in original inequality. If the result is true, that interval is the solution to the inequality.

Solving quadratic inequalities


Below you can see the example on solving quadratic inequalities-

Example:1 Find the solution set of x2 – 4x – 21 ≥ 0 in

1. Algebraic Method
2. Graphical Method

Solution:

1.  Algebraic Method:

X2 - 4x - 21 ≥ 0
=> ( x + 3) ( x - 7 ) ≥ 0
Here the coefficient of x2 is > 0 and the sign of the quadratic expression is non-negative.
So, x ≥ -3 and x ≤ 7 or there are two cases for the product ( x + 3 ) ( x - 7 ) to be non-negative.

Case (1):
x + 3 ≥ 0 and x - 7 ≥ 0
=> x ≥ -3 and x ≥ 7
So it is necessary that  x≥ 7. Notice the line marked below with x ≥ -3 and x ≥ 7 and their intersection.



From the figure above, it is clear that the intersection is x ≥ 7.

Case (2):
x + 3 ≤ 0 and x - 7 ≤ 0
=>   x ≤ -3 and x ≤ 7
So it is clear x ≤ -3.
i.e., the values of x for which x2 - 4x - 21 ≥ 0 are given by these two cases.

All values of x satisfying the in equations x ≤ -3 or x ≥ 7 become the solution set for the given in equations. This solution set can  be shown in the graph given below.



From the figure above, it is clear that the intersection is x ≥ 7.


2. Graphical Method:

The given polynomial function is f(x) = x2 - 4x - 21

x	-3	-2	-1	0	1	2	3	4	…
f(x)	0	-9	-16	-21	-24	-25	-24	-21	…

Using the above chart, we can draw a graph given below





Now, we require the values of x satisfying x2 - 4x - 21 ≥ 0 , i.e., f(x) ≥ 0 ( y ≥ 0 ). That is we require the values of x for which f(x) lies on x-axis and above. But it is clearly understood from the figure that the values of  x are -3, left side to -3, +7, and right side to +7. Thus all values of x such that x ≤ 3 and x ≥ 7 become the solution set.

Review Geometry Learning

Geometry is learning a part of mathematics which deals with the dimensions of solids and surfaces, shapes, angles and lines etc., It reviews the deduction concepts and consequence logic's, which can be applied through out your life time . In this article, you can review the basic terms and shapes in geometry.

Learning Basics Terms in Geometry:

Learning the following definitions helps to review concepts in geometry.

Point:
A point have no dimensions, it just denotes a position.

Line:
A line can be defined as a straight curve which connects many points. A line has only one dimension, i.e., length.

Collinear points:
If two or more points lie on same line, then they are said to be as colinear points.

Plane:
A Plane can be defined as a set of points and line segments combined infinitely to form a flat surface, it can be extended infinitely in any directions. A plane has infinite length and width, but it doesn't have any height.

Midpoint:
A midpoint can be defined as a point, which divides a line segment equally from both the end points.

Line segment:
A line segment can be defined as a part of a line. It consists of two end points and it was represented by its end points.

Ray:
A ray can be defined as a line having one fixed end point and an infinite extension on the other end.

Angles:

An Angle can be formed by two rays having a common end point.These are the reviews of the basic terms in geometry learning.

Learning Basic Shapes in Geometry:


Square:
A Square can be defined as a regular quadrilateral with equal sides and 4 right angles.

Rectangle:
A Rectangle is a four sided geometrical figure with two pairs of same side lengths.

Triangle:
In geometry, only shape which has three sides and angles is a triangle.

Circle:
A circle is a geometric figure, formed by a locus of points which are all equidistant from a common point called as its center. The distance between the locus of points to the common point is known as radius of the circle.

These are the reviews of the basic shapes in geometry learning.

Long Integer Definition

Integer is a whole number which is not a fraction, that is it can be either positive, negative or zero. Thus the numbers -20, -10, 0, 10, 50, 98, 540 are integers. It can not have decimals. It is commonly used in computer programming as a data type. It can also be used to find out item's location in an array. If two integers are added, subtracted or multiplied, then the output is also an integer. Where as if one integer is divided with other integer, then the result may be an integer or a fraction.

When we think of the number, we first of all think of natural numbers: 1, 2, 3, 4, 5, ...... which are also called positive integers. By applying the operation of subtraction, we get the number zero and negative integers. The set {.....-4, -3, -2, -1, 0, 1, 2, 3, 4,........} is called the set of integers. By division operation, we get the positive and negative fractions. The integers and fractions together constitute the class of rational number p/q, where p and q are integers but q is not equal to 0.

Integers and long integers:


Numbers which includes positive integers{ 1, 2, 3, 4, .....} and negative integers{-1, -2, -3, -4,......} with zero{0} are called integers. There is no fractions or decimal points are included in the integers. They are represented by Z.

Therefore,

Z={ … -4, – 3, – 2, – 1, 0, + 1, + 2, + 3, +4 ….}

+ 1, + 2, + 3, … are positive integers.

– 1, – 2, – 3, … are negative integers.

Note:

Positive numbers can be written even without the ‘+’ sign.

For example, +5, +4, +3, +2, +1 are  written as 5, 4, 3, 2, 1.

Thus, Z = {…… – 3, – 2, – 1, 0, 1, 2, 3, ……}

The integers are represented on the number line as follows:



On the number line, the numbers which are to the right of zero are called positive integers. The numbers which are to the left of zero are called negative integers.

(1)  Every natural number or a whole number.

(2)  Every whole number is an integer.

Order in long integer:

• On the number line, we find that the number value increases as we move to the right and decreases as we move to the left.
• If we represent two integers on the number line, the integer on the right is greater than the integer on the left.
• In other words, the integer on the left is lesser than the integer on the right.

For example, consider the following points marked on the number line in the figure given below:


In the above figure,

4 is to the right of 2                `:.` 4 > 2

2 is to the right of -1               `:.` 2 > -1

-5 is to the left of  -1                 `:.` -5 < -1


Example problems:


Example 1: Which is smaller? -2 and -5

Solution:

First, mark the integers -2 and -5 on the number line.



On the number line, -5 is on the left side of -2.

Therefore, -5 is smaller than -2

That is -5 < -2.

Example 2: Write the following integers in ascending order.

3,-2,0,-4,-1,5

Solution:

First mark these integers on the number line.



Now arrange the integers from left to right to get them in ascending order

-4 < -2 < -1 < 0 < 3 < 5

Therefore, the ascending order is -4, -2, -1, 0, 3, 5.

Which Fraction is Larger

 Yes, which fraction is larger or smaller? The question has great importance. As in many of the competitive exams, we are asked to arrange a set of given fractions. Before learning, how to compare the fractions, we'll learn the basics of fractions. Then we'll learn two methods, that will help us to tell "which fraction is larger".

Fraction: A number of the form `a/b` , where a and b are integers and b`!=` 0,  is known as fraction. Here, a is called numerator and b is called denominator.

Example: `3/5` is a fraction, where Numerator = 3, and Denominator = 5.

7 is also a fraction, because it can be written as `7/1`

1st Method to find which fraction is larger


Method to compare two fractions: This method is known as Cross multiplication method. Steps are as follows:

Let `a/b` and  `c/d` be the two given fractions, that we have to compare and decide which fraction is larger?

Find the product of numerator of the first fraction and the denominator of the 2nd fraction. So, we get ad and bc.

i) If ad > bc, then `a/b`  >  `c/d` .

ii) If ad = bc, then   `a/b` = `c/d` .

iii) If ad < bc, then `a/b`  <  `c/d`

Ex: Compare, which fraction is larger `3/5` and `5/8`?

Products are 3X8= 24 and 5X5=25.

Here, 24 < 25,  So `3/5`` < ``5/8`

Now, we can say very easily, which fraction is larger.

Hence, `5/8` is larger fraction among the two.


2nd Method to know which fraction is larger:


This method is used to compare more than two fractions:

Step I: Find the LCM or LCD of the given fractions. Let it be m.

Step II: Convert all the given fractions into like fractions, each having m as denominator.

Step III: Now we compare the numerators. The fraction having greater numerator is greater and vice- versa.

Ex: Arrange the fractions  `2/5` , `3/10` and `9/14` in ascending order.

LCM of 5, 10 and 14 is 70.

Now, let us change each of the given fractions into an equivalent fraction having 70 as denominator.




Practice problems of which fraction is larger


1. Which fraction is larger: `5/6` or `6/7` ?

2. Which fraction is larger: `21/25` or `3/5` ?

Study graph functions

Graphs of functions f is the collection of the all ordered pairs (x, f(x)). In particular, if x is a real number, a graph means the graphical representation of this collection, in the form of a curve on a Cartesian plane, together with Cartesian axes, etc. Graphing on the Cartesian plane is sometimes are called curve sketching. If the function input x is an ordered pair (x₁, x₂) of real numbers, the graph is the collection of all ordered triples (x₁, x₂, f(x₁, x₂)), and its graphical representation is a surface .

Example problems for study graph functions :

Study linear functions graph:
If a function f   : R → R is defined in the form f(x) = dx + e then the function is called a linear function. Here d and e are constants.

Problem 1: 

Draw the graph of the linear function f : R → R defined by f(x) = 7x + 1.
Solution:
Draw the table of some pairs (x, f(x)) which satisfy f(x) = 7x + 1.

x	-2	-1	0	1	2
f(x)	15	8	1	-6	-13

Plot the points and draw the curve passing through these points. Note that, the curve is a straight line.



Study functions of graph:

Graph of a functions: The graph of a function f is a graph of the equation y = f(x)

Problem 2:
Draw the graph of the function f(x) =2x²

Solution:
Draw a table of some pairs (x, y) which satisfy y = 2x²

x	-3	-2	-1	0	-1	-2	-3
f(x)	18	8	2	0	2	8	18

Plot the points and draw the smooth curve passing through the plotted points.



Note:

Note that if we draw a vertical line to the above graph, it meets the curve at only one point
i.e. for every x there is a unique y

Example problem for study logarithmic graph functions :

Draw the graphs of the logarithmic functions (1) f(x) = log ₅2x (2) f(x) = log ₁₀ex (3) f(x) = log ₅3x
Sol:

The logarithmic function is the defined only for the positive real numbers. i.e. (0, ∞)
Domain: (0, ∞) Range: (− ∞, ∞)

Study Online Exponentiation

Exponentiation should be the operation, which is written as the form of an. Where a and n is said to be base and exponent as well as n is any positive integer. In general, exponentiation means that repetitive multiplication. Otherwise, exponentiation an is the product of n factors of a. Online study should be the topic-oriented or else technical help for clarifying doubts that are convey through the computer software. Let us study properties and example problems for exponentiation.


Properties - Study Online Exponentiation

We are having seven number of exponentiation properties that used for solving problems. In this properties, a, m and n are any integer values.

Product of like bases:

The product of powers with the same base means we can add the powers and keep the common base.

am an = am+n

Quotient of like bases:

To divide the powers with similar base, we can subtract the exponents and remain the common base.

`a^m/a^n` = am-n

Power to a power:

In case of raising the power to power, we need to keep same base and multiply the exponent values.

`(a^m)^(n)` = amn

Product to a power:

In case of raising the product to power, we need to raise each factor to the power value.

(ab)m = am bm

Quotient to a power:

In case of raising the quotient to power, we need to raise the numerator and denominator to the power value.

`(a/b)^n` = `a^n/b^n`

Zero exponent:

Any number that is raised with zero power should be equivalent to ‘1’.

a0 = 1

Negative exponent:

a-n = `1/a^n` or `1/a^(-n)` = an

These are the properties that are used for exponentiation problems in study math online.


Example Problems - Study Online Exponentiation


Example 1:

Solve 32 34.

Solution:

Given, 32 34.

This is in the structure of am an, so we need to use am an = am+n property.

Here, m = 2 and n = 4 and a = 3.

Thus, 32 34 = 32+4

= 36

= 3 × 3 × 3 × 3 × 3 × 3

= 9 × 9 × 9

= 729

Hence, the answer is 32 34 = 729.

Example 2:

Shorten the following `6^7/6^4` .

Solution:

Given, `6^7/6^4` .

This is in the structure of `a^m/a^n` , so we need to use `a^m/a^n` = am-n property.

Here, m = 7 and n = 4 and a = 6.

Thus, `6^7/6^4` = 67-3

= 64

= 6 × 6 × 6 × 6

= 36 × 36

= 1296

Hence, the answer is `6^7/6^4` = 1296.

That’s all about the study online exponentiation.

Simple Algebra Problems

Algebra is a branch of mathematics dealing with variables, equations , expressions.Here we are going to solve some of the simple algebra problems including some equations, expressions and algebraic identities.

Solve Simple algebra problems on numbers:

Ex 1: Subtract  – 4 from – 10.

Sol :
=– 10 – (– 4)

= – 10 + (additive inverse of – 4)

= – 10 + 4

= – 6 (see addition of two integers)

Ex 2: Find the product of  (a) (+ 6) × (– 5) (b) (– 12) × (+ 12) (c) (– 15) × (– 4)

Sol : 

(+ 6) × (– 5) = – 30 (positive × negative= negative)

(– 12) × (+ 12) = – 144 (negative × positive = negative) 

(– 15) × (– 4)= + 60 (negative × negative = positive) 

Ex 3: Find the sum of given algebraic expression 2x4 – 3x2 + 5x + 3 and 4x + 6x3 – 6x2 – 1.

Sol:

Using the associative and distributive property of the real numbers, we obtain (2x4 – 3x2 + 5x + 3) + (6x3 – 6x2 + 4x – 1)

= 2x4 + 6x3 – 3x2 – 6x2 + 5x + 4x + 3 – 1

= 2x4 + 6x3 – (3+6)x2 + (5+4)x + 2

= 2x4 + 6x3 – 9x2 + 9x + 2.

  The following scheme is helpful in adding two polynomials 2x4 + 0x3 – 3x2 + 5x + 3 0x4 + 6x3 – 6x2 + 4x – 1 2x4 + 6x3– 9x2 + 9x + 2 


Ex 4: Find the subtraction of given simple algebraic expression 2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6.

Sol: Using associative and distributive properties, we have (x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1)

= x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1

= x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1

= (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1)

= –x3 + 8x2 – 4x – 5.

The subtraction can also be performed in the following way:

Line (1): x3 + 5x2 – 4x – 6.

Line (2): 2x3 – 3x2 – 1.

Changing the sign of the polynomial in Line (2), we get Line (3): –2x3 + 3x2 + 1.

Adding the polynomial in Line (1) and Line (3), we get –x3 + 8x2 – 4x – 5 

Ex 5: Find the product of   given simple algebraic expression x3 – 2x2 – 4 and 2x2 + 3x – 1.

Sol:

(x3 – 2x2 – 4) (2x2 + 3x – 1)

= x3 (2x2 + 3x – 1) + (–2x2) (2x2 + 3x – 1) + (–4) (2x2 + 3x – 1)

= (2x5 + 3x4 – x3) + (–4x4 – 6x3 + 2x2) + (–8x2 – 12x + 4)

= 2x5 + 3x4 – x3 – 4x4 – 6x3 + 2x2 – 8x2 – 12x + 4

= 2x5 + (3x4 – 4x4) + (–x3 – 6x3) + (2x2 – 8x2) + (–12x) +4

= 2x5 – x4 – 7x3 – 6x2 – 12x + 4.


Algebra Identities to solve some simple problems :


 Algebraic identity for (x + a)(x + b)

 By using the distributive properties of numbers,

(x + a )(x + b ) = x(x + b) + a(x + b) = x2 + xb + ax + ab

= x2 + ax + bx + ab= x2 + (a + b)x + ab.

 (x + a)(x + b) ≡ x2 + (a + b)x + ab

(x – a)(x + b) ≡ x2 + (b – a)x – ab

(x + a)(x – b) ≡ x2 + (a – b)x – ab

(x – a)(x – b) ≡ x2 – (a + b)x + ab

(a + b)2 ≡ a2 + 2ab + b2

(a – b)2 ≡ a2 – 2ab + b2

(a + b)(a – b) ≡ a2 – b2

(a + b)3 ≡ a3 + 3a2b + 3ab2 + b3

(a − b)3 ≡a3−3a2b+ 3ab2−b3

a3 + b3 ≡ (a + b)3−3ab(a + b)

a3 − b3 ≡ (a −b)3+ 3ab(a−b)

Solve the following simple algebra problems using above identities:


Pro 1: Find the product:  (x + 3) (x + 5)

Sol: (x + 3) (x + 5) = x2 + (3 + 5) x + 3 × 5 = x2 + 8x + 15.


Pro 2:  Find the product:  (p + 9) (p – 2)

Sol: (p + 9) (p – 2) = p2 + (9 – 2) p – 9 × 2

= p2 + 7p – 18.


Solve simple algebra problems on the cubic identity


Algebra problems on cubic identity:

(i) (3x+2y)3 = (3x)3 + 3(3x)2(2y) + 3(3x)(2y)2 + (2y)3

= 27x3 + 3(9x2)(2y) + 3(3x)(4y2) + 8y3

= 27x3 + 54x2y + 36xy2 + 8y3.

 (ii) (2x2– 3y)3 = (2x2)3 – 3(2x2)2 (3y) + 3(2x2) (3y)2 – (3y)3

= 8x6 – 3(4x4)(3y) + 3(2x2)(9y2) – 27y3

= 8x6– 36x4y + 54x2y2 – 27y3


Pro 1: If the values of a+b and ab are 4 and 1 respectively, find the value of a3 + b3.

Solution:a3+ b3 = (a+b)3 – 3ab(a+b) = (4)3 – 3(1)(4)

= 64 – 12

= 52.


Pro 2: If a – b = 4 and ab = 2, find a3 – b3.

Solution:a3 – b3 = (a–b)3+3ab(a–b)

= (4)3+3(2)(4) =64+24

=88.

Median Test Learning

Median Definition :If the generally number of values within the case is even, next the median is the mean of the two center numbers. The median is a supportive number within cases wherever the distribution contain extremely large huge values which would or else twist the data. Represent the center value of a given set of values is called median.

To locate of values set within lower value toward higher value (rising order). If the collections of data contain an odd number of ways; the median is the middle value of the place behind arrangement the listing into rising order.


Median test learning


Method for median computation

`M=(n+1)/2`

Wherever,

M- Median,

n – Entirety number of set

But here is an odd number of values, then the median is the center value of the set.

If an even number to find median first

For example the given set{2,6,8,10}

Total number of value (n) =4

Median`M=(n+1)/2`

`M=(4+1)/2`

`M=(5)/2`

M=2.5

Consequently median must appear among second and third value.

Second value=6

Third value =8

Mean of second and third value is the Median of this set

Median`M=(14)/2`

`M=(7)/2`

M=3.5


Examples for median test learning


Example 1 for median test learning

Compute the median of this set {20, 10, 12, 40, 60,50,30}.

Solution:

First values can be arranged in order

Therefore the given set={10, 12, 20, 30, 40, 50, 60}

Here, total number of value (n) =7.

Median`M=(n+1)/2`

`M=(7+1)/2`

`M=(8)/2`

M=4

Therefore the fourth value is median that is 30.

Example 2 for median test learning

Compute the median of this set {10, 12, 13, 17, 18, and 20}

Solution:

Total number of value (n) =6

Median`M=(n+1)/2`

`M=(6+1)/2`

`M=(7)/2`

M=3.5

Consequently median must appear among Third and Fourth value.

Third value=13

Fourth value =17

Mean of third and fourth value is the Median of this set

Median`M=(13+10)/2`

`M=(30)/2`

M=15

Example 3 for median test learning
Compute the median of this set {26, 32, 36, and 55}

Solution:

Total number of value (n) =4

Median`M=(n+1)/2`

`M=(4+1)/2`

`M=(5)/2`

M=2.5

Consequently median must appear among second and third value.

Second value=32

Third value =36

Mean of second and third value is the Median of this set

Median`M=(32+36)/2`

`M=(68)/2`

M=34

Learning Weighted Median

In calculation of weighted median, the importance of all the items was considered to be equal. However, there may be situations in which all the items under considerations are not equal importance. For example, we want to find average number of marks per subject who appeared in different subjects like Mathematics, Statistics, Physics and Biology. These subjects do not have equal importance. Learning the formula to find weighted median by giving Median.


Learning Median Definition:


The arithmetic median computed by specific importance of every object is known weighted arithmetic median. To consider the every importance object, we can assume number known as weight to every object is directly proportional to its specific importance.

Weighted Arithmetic Median is computed by using following formula:

`sum` Yw = `(sum ty) / (sum t)`

Where:
Yw Stands for weighted arithmetic median.
y   Stands for values of the items and
t   Stands for weight of the item

Learning the important three methods of weighted median:

1.Arithmetic Median

2. Weighted Average

3. Average speed.

The formula for weighted average is:

Weighted Average = Sum of weighted objects / total number of objects


Learning to solve examples of Weighted median:


Ex 1:A student obtained 30, 40, 50, 60, and 35 marks in the subjects of Math, Statistics, Physics, Chemistry and Biology respectively. Assuming weights 1, 3, 5, 4, and 2 respectively for the above mentioned subjects. Find Weighted Arithmetic Mean per subject.

Solution:

Subjects	Marks Obtained       y	Weight    t	`sum` ty
Math	30	1	30
Statistics	40	3	120
Physics	50	5	250
Chemistry	60	4	240
Biology	35	2	70
Total		`sum` t = 15	`sum` y= 710

Now we will find weighted arithmetic mean as:
`sum` Yw = `( sum ty)/(sum t) = 710/ 15`   =  47.33marks/subject

Ex 2: A class of 30 students took a math test. 15 students had an average (arithmetic median) score of 90. The other students had an average score of 70. What is the average score of the whole class?

Solution:Step 1: To get the sum of weighted terms, multiply each average by the number of students that had that average and then sum them up.

90 × 15 + 70 × 15 = 1350 + 1050 = 2400

Step 2: Total number of objects = Total number of students = 30

Step 3: Using the formula

Weighted Average = Sum of Weighted objects / Total Numbers of objects.

=  `2400 / 30`

= 80.

Answer: The average score of the whole class is 80.

Learn Linear Functions in Real Life

In real life learning, linear function has single degree polynomial in its equation. The degree of the linear function always one. It cannot be increased. Linear function is  very useful in real life. In learning, linear function does not dependent any variable degrees. The linear function with the degree of one, always gives the straight line. In real life, linear function has the different variable equations. 'In learning of linear function in real life', we learn about linear equations and its degrees.

Forms of Linear function in real life


The function, which has straight line graph and that line function is known as linear function. In learning, linear functions are in three main forms.

1. Slope-Intercept Form is given by y = mx +b.

Example: y = 3x + 2

Here, slope m = 3 and y intercept b = 2

2. Point Slope Form is given by m = (y - y1) / ( x – x1)

Example: (y – 3) = 2(x – 1)

3. General Form is given by ax + by + c = 0

Example: 2x + 3y -1 = 0

Examples of set of real life linear function:

Example 1:

Solve for x and y. where, y = 3x + 4 and y = 7x.

Solution:

Plug, y = 3x + 4 in y = 7x

3x + 4 = 7x

x = 1

Now plug, x = 1 in y = 3x + 4

y = 3 + 4

y = 2

Answer:

The solutions are x = 1 and y = 2.

Example 2:

5x - y + 11 = 0 and x - y - 5 = 0, solve for x and y.

Solution:

5x - y + 11 = 0     (1)

x - y - 5 = 0         (2)

Subtract Equation (2) form Equation (1), we get,

4x + 16 = 0

4x = - 16

x = - 4

plug x = - 4 in equation (1)

- 4 – y + 13 = 0

- y + 9 = 0

y = - 9

The solutions are x = - 4 and y = - 9.


Learn linear function in real life - Practice problems:


Problem 1:

3x - y = 9 and 4x + y = 5. Solve for x, y.

A) (2, 3) B) (- 2, 3) C) (2, -3) D) (3, -2)

Answer: C

Problem 2:

2x + y = 12 and y = 3x - 3. Solve for x, y.

A) (3, 6) B) (3, - 6) C) (4, 5) D) (6, - 3)

Answer: A

Problem 3:

Solve the linear function 2x - y = 12 and y = 4x - 2

A) (- 5, 22) B) (5, - 22) C) (- 5, - 22) D) (5, 22)

Answer: C

solving online area of a circle

CIRCLE:

A line forming a closed loop, every point on which is a fixed distance from a center point.

A circle is a plane continuous figure connecting points which are equidistant from a given point, known as the center.

The word circle originates from the Latin word 'circus'. Chariot races, very popular in the Roman era were either circular or oblong and eventually the word was used to describe the shape as well.


Area of a circle - Important terms

In order to find the area of a circle, it is essential that we first understand a few important terms related to circles.

The outermost portion of the circle which separates the interior of the circle from the exterior is known as the circumference.

The center, as already mentioned, is the point inside the circle which is equidistant from all points lying on the circumference.

The distance between any point on the circumference from the center is known as the radius.

A line segment connecting any two points of lying on the circumference on the circle is called a chord.

The longest possible chord of a circle is known as a diameter. The diameter equals twice the radius.


Solving the area of a circle

Consider a circle with center P. Let us denote the radius of the circle by 'r' and the diameter by 'd'. Pi (denoted as π) is a numerical value, which is a crucial number used to calculate various attributes of circular figures. Its value up to 2 decimal points is 3.14. In terms of the radius.

we define the area (A) of the circle as:                  A = π r²

In terms of the diameter, which is twice the radius, this equation can be re-written as:

A = 0.25 π d²

The area in terms of the circumference (C) of the circle is given as:

A = C²/ 4π

Examples and exercises of area of circle.


Example 1: The radius of a circle is 3 inches. What is the area?

A = π r²

A  =  (  3.14  ) ( 3 ) 2

A = (3.14) (9)

A =  28.26 in 2

Answer  the following :

1) The diameter of a circle is 8 centimeters. What is the area?

2) The radius of a circle is 5 feet. What is the area?

Half Angle Properties

Introduction:

Let we will discuss about the half angle properties. The half angle is the angle that is half of original angle. Tha is, the product of two half angles at an edge is equal to the original(full) angle at that same edge. We will develop the properties formulas for half angle of both sine and cosine.


Half angle properties formula for sine:


We should start with formula for cosine of double angle. That is,
cos 2θ = 1 - 2sin^2 θ

Half angle properties formula - sine:

Let us consider,


Then 2θ = α and the formula becomes:


Solve for,


That is, we acquire sin(α/2) lying on the left of equation and everything else on right,



Solving of equation gives the following sine of half-angle identity:


The sign of sin α/2 represents on quadrant in which α/2 lies.
If α/2 is in the first or second quadrants then formula uses the positive case:


If α/2 is in the third or fourth quadrants, the formula uses the negative case:


Example:

Find the value of sin 45o using the sine half-angle properties.

Solution:

The sine half angle formula is ,



Therefore,

sin 45o = `sqrt((1 - cos 90 )/(2))`

= `sqrt((1 - 0 )/(2))`

= `sqrt((1)/(2))`

= 0.707


Half angle properties formula for cosine:


Half angle properties formula - cosine:

Using same process, with the similar replacement of θ = α / 2. We want to substitute into the identity,


We get,


Reverse process of the equation should be,


Addition of 1 with both sides of an equation. We get,


Making division process on both sides of equation by 2


Solving for cos(α/2), we get,


As before, the sign we need depends on quadrant.
If α/2 is in first or fourth quadrants that formula uses in positive case:


If α/2 is in the second or third quadrants, the formula uses the negative case:


Example:

Find the value of cos 115o using the cosine half-angle formula.

Solution:

We need to find cos 115o

That is, α = 230°, and so α/2 = 115°.

Therefore, cos 115o = `sqrt((1 + cos 230 )/(2))`

= `sqrt((1-0.643)/(2))`

= `sqrt((0.357)/(2))`

= 0.422

We have seen about the half angle properties.

Geometric Area

The geometric area refers to the size of the interior of a planar (flat) figure. Area (A) is a two-dimensional measure. The square units that  an area is measured with square inches, square feet and square centimeters. Hectares and acres are also considered in some special cases. In the International System of Units (SI), the standard unit of area is the meter squared (m ²).

Geometric area of Square

Formula to find the area of the square:

Area of the square is calculated by multiplying the base times itself.

Area of square = side x side square unit.

Area = a²

1. The side length of the square is 10 feet; find the area of the square.

Solution:

Area = a²

Given: Side length= 10feet.

                               = 10 x 10

                   Area    =100 feet²

2. The side length of the square is 4.5 m; find the area of the square.

Solution:

Area = a²

Given: Side length= 4.5m.

                              = 4.5 x 4.5

                      Area =20.25 m²

Geometric area of Rectangle

Area of the rectangle is calculated by multiplying the base times the height.

 Area of rectangle     =  (length x width) square unit.

Area = l x w square unit

3. The length and breadth of the rectangle are 10 meters and 5 meters respectively .find the area of that rectangle.

Solution:

Area = l x w square unit

           
Given: Length= 10 meters

                        Breadth=5 meters

                                    =10x5

                        Area    = 50 m²      

     

4. The length and breadth of the rectangle are 11 feet and 6 feet respectively .find the area of   that rectangle

Solution:

Area = l x w square unit.

            Given: Length= 11feet

                        Breadth=6feet

                                    =11x 6

                        Area    = 66 ft²

Geometric area of Triangle

Area of the triangle is calculated by multiplying the base times the one-half the height.

Area of triangle= `(1)/(2)` (base x height)

Area = `(1)/(2)bh`

5. What is the area of triangle with base 5 and height 10 feet?

Solution:

Area = `(1)/(2)bh`

Given: Base= 5 feet.

         Heigh t= 10feet

                    = 1/2 (5x 10)

                    =1/2(50)

            Area =25 ft²

6. What is the area of triangle with base 7 and height 14 feet?

Solution:

Area = `(1)/(2)bh`

Given: Base= 7 feet.

         Height=14feet

                    =1/2 (7 x 14)

           Area =1/2 (98)

            Area =49 ft²

Geometric area of Circle

Area of the circle is calculated by multiplying pi ( π = 3.14) by the square of the radius

Area of the circle= π x r²

7. The radius(r) of a circle is 3 inches. Find the area of that circle?

Solution:

Area of the circle= π x r²

π=3.14

Given: r=3inches

          A=3.14 x (3)²

                   =3.14 x 9 inche^2.

    Area= 28.26 in^2.

8. The radius(r) of a circle is 7.5 meters. Find that area of the circle.

Solution:

Area of the circle is calculated by multiplying pi ( π = 3.14) by the square of the radius

Area of the circle= π x r²

π=3.14

Given: r= 7.5 meters

          A= 3.14 x (7.5)²

            =3.14 x 56.25

  Area =176.625 m²

circumference circle

In our real life, we could see many circular objects. For example, wheel of a vehicle, shape of a pizza, tennikoit ring,.. Take a rope or an inelastic thread and measure the circular path of the given object and that forms the circumference of the circular object. As the size of the circle increases, the circumference of circle also increases.

Introduction on circumference of a circle:

Circle is one of the most important shapes in geometry. Circumference is defined as the measure needed to make the closed curve. Circle has a fixed center. Distance between the center and a point on the circumference of the circle is called as radius. A line passing through center and any two points on the circumference of the circle is called as diameter.  If the radius or diameter is given then we can easily calculate the circumference of the circle. This article helps us to find the circumference of the circle with the given details.


Circumference of a circle:


The diagram below shows the circumference of circle.



The Circumference of a circle can be calculated using the following formula:

Circumference = 2  `pi`r  or `pi`

where r ----> radius of circle

d ----> diameter of circle

π---- > constant and value of π is  3.14 or 22/7

Diameter is twice the radius. So d = 2r and   r = `d/2`


Problems on finding circumference of the circle:

Ex 1: Find the Circumference of the circle with radius 20 cm.

Sol:

Step 1: Write the formula

Circumference of a circle  = 2  `pi`r units

Step 2: Plug the known values and calculate circumference

2π r               = 2 (22/7) 20

= 2 (3.14) 20

= 125.6 cm

Ex 2: Find the Circumference of the circle with radius 32 cm.

Sol:

Step 1: Formula for Circumference of a Circle

Circumference of a circle  = 2  `pi`r units

Step 2: Plug the known values and calculate circumference

2πr                  = 2 (22/7) 32

= 2 (3.14) 32

= 200.96 cm

Ex 3: Find the Circumference of the circle with diameter 52 in.

Sol:

Step 1: Write the formula

Circumference of a circle  = `pi` d

Step 2: Plug the known values and calculate circumference

πd             = (22/7) 52

= (3.14) 52

= 163.28 in

Ex 4: Find the Circumference of the circle with diameter 74 in.

Sol:

Step 1: Write the formula

Circumference of a circle  = `pi` d

Step 2: Plug the known values and calculate circumference

πd            = (22/7) 74

= (3.14) 74

= 232.36 in

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Practice problems on circumference of the circle:


Find the Circumference of the circle with radius 18 cm. .
Answer: 113.04 cm

Find the Circumference of the circle with radius 28 cm.
Answer: 175.84 cm

Find the Circumference of the circle with diameter 63 in.
Answer: 197.82 in

Find the Circumference of the circle with diameter 81 in.
Answer: 254.34 in

Different Number Combinations

In combinatorial mathematics, a k-combination of a finite set S is a subset of k different numbers of S. Specifying a subset does not arrange them in a particular order; by contrast, producing the k different numbers in a specific order defines a sequence without repetition, also called k-permutation (but which is not a permutation of S in the usual sense of that term. SOURCE: WIKIPEDIA



Example problems of different number combinations:

Different number combinations example 1:

How many lines can you draw using THREE non collinear (not in a single line) points X, Y and Z on a plane?

Solution:

You need two points to draw a line. The order is not important. Line XY is the same as line YX. The problem is to select TWO points out of THREE to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

XY, XZ

YX, YZ

ZX, ZY

There is a problem: line XY is the same as line YX, same for lines XZ and ZX and YZ and ZY.

The lines are: XY, YZ and XZ; three lines only.

So in fact we can draw THREE lines and not SIX and that's because in this problem the order of the points X, Y and Z is not important.

This is a combination problem: combining TWO items out of THREE and is written the general form as follows:

n C r = n! / [ (n - r)! r! ]

Special case:

n C 0 = n C n = 1

The number of combinations (nCr) is equal to the number of permutations divided by r! to removes those counted more than once because the order is not mainly use.

Different number combinations example 2:

Calculate 4C3

Solution:

We can find the combination value by using the following formula:

n C r = n! / [(n - r)! r!]

Substitute the value of n and r into the above formula, then we get

4C3= 4! / [(4-3)! 3!]

=4! / [1!*3!]

=24/[1*6]

=24/6

=4

Answer: 4

Different number combinations example 3:

Calculate 6C6

Solution:

We can find the combination value by using the following formula:

n C r = n! / [(n - r)! r!]

Substitute the value of n and r into the above formula, then we get

6C6= 6! / [(6-6)! 6!]

=6! / [0!*6!]

=6! / [1*(6!)]

=6! / 6!

= 1

Answer: 1

Having problem with Probability Sets Read my upcoming post, i will try to help you.

Practice problems of different number combinations:


1) In how many ways can you select a committee of 2 students out of 5 students?

Answer: 20 (5C2)

2)  How many triangles can you make using 4 non collinear points on a plane?

Answer: 24 (4C3)