Set Multiple Values

The definition of set is group of object. Collection of whole number, collection of natural numbers, and collection of fraction numbers it is called the some examples of set multiple values. The set symbol is represented the { }. The example of set multiple value is {4, 5, 7, 6}. Next we discuss about this articles set multiple values.

First Problem of Set Multiple Values

Problem 1: set multiple values

If A = {5, 3, 6, 2, 7, 13 ,15} and B  = {5, 3, 6, 2, 7}  find out n(A), n(B), n(A `uu` B)and n(A `nn` B) and to prove n (A `uu`B) = n (A) + n (B) – n (A `nn`B)

Solution:

A = {5, 3, 6, 2, 7, 13, 15}

B = {5, 3, 6, 2, 7}

A `uu` B = {5, 3, 6, 2, 7, 13, 15}

A `nn` B = {5, 3, 6, 2, 7}

n (A) = 7

n (B) = {5},

n (A`uu` B) = {7},

n (A`nn` B} = {5}

n (A) + n (B) – n (A `nn` B)

7 + 5 - 5

= 7

The A `uu` B = 7

So,

n (A`uu` B) = n (A) + n (B) – n (A `nn` B)

Second Problem of Set Multiple Values

Problem 2: set multiple values

If X = {7, 9, 4, 2, 5} and Y = {7, 9, 4, 2, 5, 13, 14} find n(X), n(Y) n(X `uu` Y), n(X `nn` Y) and to verify the identity n (X uu Y) = n (X) + n (Y) – n (X `nn` Y)

Solution:

X = {7, 9, 4, 2, 5}

Y = {7, 9, 4, 2, 5, 13, 14}

{X `uu` Y} = {7, 9, 4, 2, 5, 13, 14}

{X `nn` Y} = {7, 9, 4, 2, 5}

n (X) = {5}

n (Y) = {7}

n(X `uu` Y) = {7}

n(X `nn` Y) = {5}

n (X) + n (Y) – n (X `nn` Y)

= 5 + 7- 5

= 12 - 5

= 7

Here X `uu` Y = 7

So,

n (X`uu` Y) = n (X) + n (Y) – n (X `nn` Y)

Place Value Tenths Hundredths

Place value can describe the value of all numbers .The place value can be used to point out the position of a mathematical system. It is tremendously helps you to read the numerals by its places. Place value is the value which is given to the digit by popular quality .In this article we shall discuss the place value of tenths hundredths.

Place Value Tenths Hundredths:

The place values from one to ten thousand and to ten thousandth are as follows.

For example for the value 11111.11111

1 – Place value one.

10 = place value Ten.

100 – Place value Hundred.

1000 – Place value thousand.

10000 – Place value ten thousand.

1÷10 =0.1- place value tenth.

1÷100= 0.01- place value hundredth.

1÷1000=0.001- place value thousandth.

1÷10000=0.0001- place value ten thousandth.

1÷100000=0.00001- place value hundred thousandth.

Examples Problems on Place Value:

Ex:1 Round the following number to the hundredth place value 746.839256

Sol:

Given that the number 746.839256

From the given number on the left side from the decimal point,

7 indicates Place value Hundred.

4 indicate place value ten.

6 indicate Place value one.

From the given number on the right side from the decimal point,

8 indicate tenth place value.

3 indicate hundredth place value.

9 indicate thousandth place value.

2 indicate ten thousandth place value and so on.

As the question is to encircle the number 746.839256 to the hundredth place value.

The answer is 746.83

Ex:2 Add the numbers 1563.457 and 1999.336 and round the solution to the hundredth place value.

Sol:

Adding the numbers we get




The solution is  3562.793

From the given number on the left side from the decimal point,

3 indicate place value thousand.

5 indicate Place value Hundred.

6 indicate place value ten.

2 indicate Place value one.

From the given number on the right side from the decimal point,

7 indicate tenth place value.

9 indicate hundredth place value.

3 indicate thousandth place value.

As the question is to encircle the number 3562.793 to the tenth place value.

The answer is 3562.8 because the place value nearest to hundredth (7) is greater than 5(9).


Ex:3 Round the following number to the tenth place value 876.473957

Sol:

Given that the number 876.473957

From the given number on the left side from the decimal point,

8 indicates Place value Hundred.

7 indicate place value ten.

6 indicate Place value one.

From the given number on the right side from the decimal point,

4 indicate tenth place value.

7 indicate hundredth place value.

3 indicate thousandth place value.

9 indicate ten thousandth place value and so on.

As the question is to encircle the number 876.473957to the tenth place value.

The answer is 876.5

Ex:4 Add 388.8018 and 100.1111 and round the solution to tenth place value.

Sol:

Adding 388.8018 and 100.1111 we get,



The solution is  488.9129

From the given number on the left side from the decimal point,

4 indicate Place value Hundred.

8 indicate place value ten.

8 indicate Place value one.

From the given number on the right side from the decimal point,

9 indicate tenth place value.

1 indicate hundredth place value.

2 indicate thousandth place value.

9 indicate ten thousandth place value.

As the question is to encircle the number 488.9129 to the tenth place value.

The answer is 488.9

Practice Problems on Place Value:

Q:1 Round the following number to the hundredth place value 44776.76234

Ans:

The answer is 44776.76

Q:2 Round the following number to the tenth place value 56256353.3564847

Ans:

The answer is 56256353.4

Place Value Number System

The Place Value of a digit in a number system is the digit multiplied by 100, 101, 102, 103, 104, 105, 106, 107, 108, 109,... etc. according as the digit appears in the number system as once, tens, hundreds, thousands, ten thousand, hundred thousand, million, billion, trillion respectively.  Example for place value number system: 3333333333333. This can be represented as three trillion three hundred and thirty three billion three hundred and thirty three million three hundred and thirty three thousand and three hundred and thirty three. Here we are going to see in detail about place value number system.

Ones:

Ones is the word which means the number 1. The place value of ones is one. This is the right most number, For any number the right most number or the first number in the right side will be ones

Example problem:

1)  Find the ones place value in the number 85,494

Solution:

The ones place value in the number 85,494 is 4

2)  Find the ones place value in the number 856,490

Solution:

The ones place value in the number 856,490 is 0

Tens:

Tens is the word which means teh number 10. The place value of tens is two.

Example problem:
1)  Find the tens place value in the number 325,968

Solution:

The tens place value in the number 325,968 is 6

2)  Find the tens place value in the number 16,824

Solution:

The tens place value in the number 16,824 is 0

Hundreds:

Hundred is the word which mean the number 100, the word hundred is also known as centum. The place value of hundreds is three.

Example problem:

1)  Find the hundreds place value in the number 78,256

Solution:

The hundreds place value in the number 78,256 is 2

2)  Find the hundred place value in the number 954,263

Solution:

The hundred place value in the number 954,263 is 2

Thousands:

Thousand is the word which means the number 1,000. The place value of thousand is four. Ten thousand and hundred thousand comes under the category of thousand.

Example problem:

1) Find the thousands place value of the number 479,258

Solution:

The thousands place value of the number 479,258 is 9

2) Find the ten thousands place value of the number 479,258

Solution:

The ten thousands place value of the number 479,258 is 7

3) Find the hundred thousand place value of the number 479,258

Solution:

The hundred thousands place value of the number 479,258 is 4

Millions:

Million is the word which means the number 1,000,000. The place value of million is seven. Ten million and hundred million comes under the category of million.

Example problem:

1)    Find the million place value of 951,654,259

Solution:

The million place value of the number 951,654,259 is 1

2)    Find the ten million place value of 951,654,259

Solution:

The ten million place value of the number 951,654,259 is 5

3)    Find the hundred million place value of 951,654,259

Solution:

The hundred million place value of the number 951,654,259 is 9

Billions:

Billion is the word which means the number 1,000,000,000. The place value of billion is ten. Ten billion and hundred billion comes under the category of billion.

Example problem:

1)    Find the Billion place value of 907,789,246,846

Solution:

The Billion place value of the number 907,789,246,846 is 7

2)    Find the ten Billion place value of 907,789,246,846

Solution:

The ten Billion place value of the number 907,789,246,846 is 0

3)    Find the hundred Billion place value of 907,789,246,846

Solution:

The hundred Billion place value of the number 907,789,246,846 is 9

Trillions

Trillion is the word which means the number 1,000,000,000,000. The place value of trillion is thirteen. Ten Trillion and hundred Trillion comes under the category of Trillion.

Example problem:

1) Find the Trillion place value of 456,758,246,349,246

Solution:

The Trillion place value of the number 456,758,246,349,246 is 6

2)    Find the ten Trillion place value of 456,758,246,349,246

Solution:

The ten Trillion place value of the number 456,758,246,349,246 is 5

3)    Find the hundred Trillion place value of 456,758,246,349,246

Solution:
The hundred Trillion place value of the number 456,758,246,349,246 is 4

Statistics on College Students

Statistics should be help with formal science. These are generating well-organized help of algebraic data among the groups of individuals. In statistics, we are learning help on median, mode, mean and range. Measuring of these problems is extremely easy. For this we are using formulas. Now we will see statistics example problems on the college students.

Statistics Examples for College Students

Example 1:

Find the mean, median for the sequence of numbers, 560,247,281,396,185,160,288.

Solution:

The given numbers are 560,247,281,396,185,160,288.

Mean

Mean is the average of the given number. Calculate the total value of the given sequence.

Sum of the given numbers are = 560+247+281+396+185+160+288

= 2117

Total values are divided by 7 (7 is the total numbers) = `(2117)/(7)`

= 302.43

Median

Middle element of the sorting order of given series is a median.

The sorting order series is 160,185,247,281,288,396,560.

The middle element of the sorting sequence is 281.

So the median is 281.

Example 2:

Find the mode value for the following sequence of numbers, 291,358,640,291,305,291.

Solution:

Given series is, 291,358,640,291,305,291.

Mode is the most repeatedly occurring value of the series.

Here, the number ‘291’ should be occurring three times.

So, the mode value is 291.

Example 3:

Find the range value for the following sequence of numbers, 488,150,362,205,117,250.

Solution:

Given series is, 488,150,362,205,117,250.

Subtract the smallest value from the biggest value of the series is said to be range.

Range = 488 - 117

= 371

Standard Deviation Example for College Students

Let see the learning example for standard deviation in statistics on college students.

Example 4:

Find the standard deviation of the following numbers, 30,59,65,27,16,12,33,22.

Solution:

The given numbers are 30,59,65,27,16,12,33,22.

Establish the mean for the given data.

Mean = `(30+59+65+27+16+12+33+22)/(8)`

= `(264)/(8)`

= 33

Construct the table for finding standard deviation.

x	x-33	(x-33)2
30	-3	9
59	26	676
65	32	1024
27	-6	36
16	-17	289
12	-21	441
33	0	0
22	-11	121
Total		2596

Find (x-m)2 / (n-1) = `(2596)/(8-1)`

= `(2596)/(7)`

= 370.86

Formula of the standard deviation = `sqrt((sum_(i=1)^n (x-m)^2) / (n-1))`

= `sqrt(370.86)`

= 19.26

Thus the standard deviation is 19.26.

These are the few statistics example problems that help on college students.

That’s all about the statistics on college students.

Derivative Calculator Step by Step

In calculus, the rate of change of curve is called derivative. Derivative Calculator is used to find the differentiation of given function. By Using the derivative calculator we can able to find the derivative value easily. Through manual calculator we can able to solve the derivative but it quite difficult to perform the calculation. But through calculator there is no other difficulty rather than entering the question, the answer will produced automatically. we have to enter the question in function place. then enter the value in which we derivative with respect to. then the order of the respected derivative, now click the derivative button. sample derivative calculator is given.

Derivative Calculus Formulas - Derivative Calculator Step by Step:

1. n th power derivative is `d/dx` (xn) = n x(n-1)

2. Exponential derivative `d/dx` (ex) = ex

3. Logarithmic derivative `d/dx` (log x) = `1/x` .

4. Trigonometry derivative `d/dx` (sin x) = cos x

5.Trigonometry derivative `d/dx` (cos x) = - sin x

6.Trigonometry derivative `d/dx` (tan x) = sec2x

7.Trigonometry derivative `d/dx` (sec x) = sec x tan x

8.Exponential derivative `d/dx` (eax) = aeax

9.Product derivative `d/dx` (uv) = u` (dv)/(dx)` + v `(du)/(dx)` .

10.Division derivative `d/dx(u/v)` = ` [v (du)/(dx) - u (dv)/(dx)]/v^2` .

Example Problems - Derivative Calculator Step by Step:

Derivative calculator step by step problem 1:

Find the derivative of given function      x6 + 3x4 + 5x2 + 3x.

Solution:

Given function is    x6 + 3x4 + 5x2 + 3x.

Step 1:      The derivative of given function can be expressed as  `d/dx` (x6 + 3x4 + 5x2 + 3x).

Step 2:      Separate the derivative function is   `d/dx` (x6) + `d/dx`(3x4) + `d/dx` (5x2 ) + `d/dx`(3x).

Step 3:      Now we know the n th power derivative rule. That is `d/dx` (xn) = n x(n-1)

Step 4:       In the first term n = 6, So derivative of   `d/dx` (x6) = 6 x(6-1)  = 6x5

Step 5:    Differentiate all the term, we get    =   `d/dx` (x6) + `d/dx`(3x4) + `d/dx` (5x2 ) + `d/dx`(3x).

=  6 x(6-1)  +  (3)(4) x(4-1)  + (5)(2) x(2-1)  + (3)(1) x(1-1)

=  6x5 + 12x3 + 10 x + 3

Answer:  6x5 + 12x3 + 10 x + 3.


Derivative calculator step by step problem 2:

Find the derivative of given function       5x2 -  3tanx .

Solution:

Given function is   5x2 -  3tanx .

Step 1:      The derivative of given function can be expressed as  `d/dx` (5x2 -  3tanx ).

Step 2:      Separate the derivative function is    `d/dx` (5x2 ) - `d/dx`(3 tan x).

Step 3:      Now we know the n th power derivative rule. That is `d/dx` (xn) = n x(n-1)

Step 4:       In the first term n = 2, So derivative of   `d/dx` (x2) = 2 x(2-1)  = 2x

Step 5:    Differentiate all the term, we get    =   `d/dx` (x2) - `d/dx`(3 tan x).

=  2x  - 3 sec2x .

=  2x  - 3 sec2x .

Answer:   2x  - 3 sec2x .

Study Online Limits of Trigonometric Functions

In online, trigonometric function are recognized as circular function of an angle. They are organizing to transmit the position of a triangle toward the lengths of the surface of a triangle. Let x, an angle in radians `x->0` `sin x->0`, so `lim_ (x->0)(sinx)` is of the in-between type of type 0. In online we acquire more information regarding all the problems contain step by step solution. Various websites are present in study online limits of trigonometric functions.

Study Online Limits of Trigonometric Functions:

The general rule of limit trigonometric function:

`"lim_(x->a) `  =>    `lim_(x->a) f(x) =0 `

Trigonometric functions acquire in a large variety of uses to include found indefinite lengths as well angles in triangles often right triangles.

Function of an angle expressed though ratios of the length of the surface of right-angled triangle contain the angle.

Trigonometric function is the proportion of the exterior of the right triangle. Function of an angle articulated while ratios of the length of the sides of right-angled triangle have the angle



Trigonometric function:

Sin = `(opposite)/("hypotenuse")`
Cos =` (adjacent)/("hypotenuse")`
tan =`(opposite) / (adjacent)`
cot  = `(adjacent) / (opposite)`
sec = `("hypotenuse") / (adjacent)`
csc =`("hypotenuse")/(opposite)`

There are three reciprocal function in trigonometric.

Csc x = `1/(sinx) `

Sec x =`1/(cosx) `

cot x =`1/(tanx) `

Examples for Study Online Limits of Trigonometric Functions:

Example 1 for  study online limits of trigonometric functions:

How to find `lim_(x->o) (sin 8x)/(8x)`

Solution:

Step 1: the given function is  `lim_(x->o) (sin 8x)/(8x)`

Step 2: let t = 4x we have `t->0` when `x->0`

Step 3:  `lim_(x->o) (sin 8x)/(8x)`

=    `lim_(t->o) (sin t)/(t)`

= 1.

So the solution is     `lim_(x->o) (sin 8x)/(8x)`      = 1.

Example 2 for study online limits of trigonometric functions:

How to find  `lim_(x->4) (sin (x-4))/(x^2-9x+20)`

Solution:

Step 1: The given function is `lim_(x->4) (sin (x-4))/(x^2-9x+20)`

Step 2:   `lim_(x->4) (sin (x-4))/(x^2-9x+20)`

`lim_(x->4) (sin (x-4))/((x-5)(x-4))`

Step 3:     `lim_(x->4)(sin(x-4)/(x-4)*1/(x-5))`

Step 4:  `1*-1 =-1`

So the solution is    `lim_(x->4) (sin (x-4))/(x^2-9x+20) = -1`

Example 3 for  study online limits of trigonometric functions:

How to find `lim_(x->pi/3) (sin x)`

Solution:

Step 1: the given function is `lim_(x->pi/3) (sin x)`

Step 2:   `lim_(x->pi/3) (sin x)` <br>

Step 3:   = 0.8660

So the solution is     `lim_(x->pi/3) (sin x) = 0.8660`

Example 4 for study online limits of trigonometric functions:

How to find  `lim_(x->-4) (sin (x+4))/(x^2+6x+8)`

Solution:

Step 1: The given function is   `lim_(x->-4) (sin (x+4))/(x^2+6x+8)`

Step 2:   `lim_(x->-4) (sin (x+4))/(x^2+6x+8)`

`lim_(x->-4) (sin (x+4))/((x+2)(x+4))`

Step 3:     `lim_(x->-4)(sin(x+4)/(x+4)*1/(x+2))`

Step 4:  `1*-1/2 `

Step 5:      =   `-1/2`

So the solution is   `-1/2`

Absolute Property Solutions

Basically absolute values mean a value without considering its sign. Here we are going to find the solutions of absolute properties. If we are having any number x mean absolute value of x is denoted like |x| and its value is +x. Likewise for |-x| = x. Here we are going to learn the properties of the absolute value and its solutions. If we know the properties of the absolute value it is easy to do the operations on the absolute values.

Absolute Property Solutions:

The first property is non negative property.

Absolute property solutions - Non negativity property:

Absolute value of the numbers is always greater than 0. if we are having a negative number its absolute value is a positive number.

Example:

What is the absolute value of the number -1.2?

Solution:

The given number is -1.2

We have to find the absolute value of the number -1.2

So |-1.2| = +1

So always |x| > 0

Absolute property solutions - Positive definiteness:

The second property is positive definiteness.

The absolute value of the number 0 is always 0. If |x| = 0 then x = 0 (always)

Example:

What is the absolute value of 0?

Solution:

Basically an absolute value is having two values. But for 0 the absolute value is |0| = 0

Absolute property solutions - Multiplicative property:

Multiplication of any two given absolute values equals to the individual absolute value multiplication.

|x y | = |x| |y|

Example:

|-5 x 3| = |-15| = + 15

|-5| x |3| = +5 x +3 = + 15

So both the values are same.

More Absolute Property Solutions:

Absolute property solutions - Subtraction addition property:

Addition of any two given absolute values is always less than the value of its individual addition.

|x + y| ≤ |x| + |y|

Example:

|4 + -3| = |1| = 1

|4| + |-3| = 4 + 3 = 7

1 < 7

Absolute property solutions - Symmetry property:

Symmetry property mean |-x| = |x|

Absolute value of the –x and absolute value of x is always same.

Example:

|-4| = |4| = +4

Absolute property solutions - Identity of indiscernible property:

The difference of any two absolute values of the number is 0 then these two absolute values are same.

|x - y| = 0 then x = y

Example:

Find the absolute value of |5 – (-5)|

Solution:

Absolute value of |5| = +5

Absolute value of |-5| = +5

So |5 – (-5)| = 0

Absolute property solutions - Preservation of division:

The division of any two absolute values and its individual absolute value division are equal.

|x / y | = |x| / |y| (where y ≠0)

Decimal to Percentage Conversion

The conversion of one model to another is the most important thing in the world, which one can never
deny. The above sentence is so assertive because of the underlying fact that learning is important at
every aspect, since it is important for the take away we get out of something. Conversion basically gives
a general perception that it is usually simplified to a simpler form to get a better understanding, so it is a
positive word which gives a positive meaning. This concept is all the same like above, decimal or fraction
conversion into the other.

The process of  Decimal to Percent conversion is done when one is good in basics and fundamentals of
mathematics. First of all one has to be good in decimal which will help in identification of the number
when posed for solving. Decimal can also be called as fraction, when a fraction is simplified the result
will be a decimal. The above will help in solving the question Convert Decimal to Percent. The process of
conversion is not rocket science, it is as simple as solving the basic addition and subtraction.

Converting Decimals to Percents basically involves basic multiplication. The multiplication of the decimal for the conversion is 100 and the symbol used for percent-age is %. This symbol signifies the percentage.

This conversion can be explained clearly using the following example; the question will be given as
convert 0.6 Decimal to Percentage, the answer for the question will be 60%, which is simplified by
multiplying the number 0.6 by 100. The answer will be 60% when multiplied. This can be further
explained by considering another example, convert 0.75 Decimals to Percents, the answer for this
example will 75%, which is resulted after multiplying 0.75 by 100.

Now it is time to think about some of the real life scenarios, where we will need this concept for
applying. Of course every single thing that one studies or learnt should be applied sometime or
somewhere at some point of his / her life, if one does not uses or applies it, there is no point in learning
things. It is mostly used while forecasting the sales of any firm because it is always essential to use
percentile rather than using decimals while portraying the results, which will make the one to whom the
results are presented understand the numbers which resulted.

Intersection of Two Straight Lines

Two lines will intersect at a point. The point will have a pair of values as (x1, y1). The straight lines are represented by equation with two or one variable in x and y.  If those equations are solved to get the value of x and y, will represent the point intersection of those two lines. To solve the set of lines to get the value of x and y , we can use either the method of elimination or substitution. Now let us discuss few problems on this topic intersection of two straight lines.

Example Problems on Intersection of Two Straight Lines

Ex 1: Find the point of intersection of the following two lines

2x + 3y = 10; 2x + y = 6.

Sol: Given: 2x + 3y = 10 --------------(1)

2x + y = 6  --------------(2)

The point intersection of above two lines can be obtained by solving them as follows:

(1) – (2) implies: 2x + 3y = 10

2x + y = 6

We get, 2y = 4

y = 2.

From (2), 2x + (2) = 6

2x = 6 – 2 = 4

2x = 4

x = 2.

Therefore, the point of intersection is (2, 2).

Ex 2: Find the point of intersection of the following two lines

x + 2y = 1; 5x + 4y = -7.

Sol: Given: x + 2y = 1 ------------------(1)

5x + 4y = -7 -----------------(2)

The point of intersection of above two lines can be obtained by solving the as follows:

(1) x 5 – (2) implies: 5x + 10y = 5

5x + 4y = -7

We get, 6y = 12

y = 2.

Therefore, from (1), x + 2(2) = 1

Implies, x = 1 – 4 = -3.

Therefore the point of intersection is (-3, 2).

Ex 3: Find the point of intersection of the following two lines

3x + y = 10; y = 7.

Sol: Given: 3x + y = 10 -------------(1)

y = 7 -----------(2)

Since, y = 7 is one of the line, the value of the y coordinate will be 7.

Therefore, from (1), we get , 3x + 7 = 10

3x = 10 – 7 = 3

x = 1.

Therefore, the point of intersection is ( 1, 7).

Practice Problems on Intersection of Two Straight Lines

1. Find the point of intersection of lines 3x + 2y = 7 and x + y = 3.

[ Answer: (1, 2)]

2. Find the point of intersection of lines 3x - 2y = -2 and x - 2y = 6.

[ Answer: (-4, -5)]

3. Find the point of intersection of lines 4x - 3y = -10 and 3x + y = -1.

[ Answer: (-1, 2)]

Solving Binomials by Factoring

Introduction: 

In algebra, the polynomials which have two terms are called binomials. To factor binomials, we need to follow the following methods:

(i) 2a + ab = a(2 + b ) [ Here the given expression has two terms, where a is the common value]    
     = a( 2 + b)         
(ii) a² – b² = ( a + b) ( a – b)   [ This is the standard form]
(iii) (a + b)² = ( a + b)(a +b)
(iv) (a - b)² = ( a - b)(a - b)

Product of two polynomials will give three terms.
( a + b)² = ) a² +b² + 2ab
( a - b)² = ) a² +b² - 2ab

Let us see few problems on this topic solving binomials by factoring.

Example Problems on Solving Binomials by Factoring

Ex 1: Solve (x + 3) (x – 4) = 0

Soln: Given: (x + 3) (x – 4) = 0
This implies: x + 3 = 0 or x – 4 = 0
That is : x = -3, 4.
Therefore the solution is { -3, 4}.

Ex 2: Solve x + y = 7 and xy = 12, find x and y.

Soln: Given : x + y = 7 -----------(1)
                            xy = 12 ---------(2)
Therefore, x – y = sqrt [(x+y)² – 4xy]
                             = sqrt[ 7² – 4(12)]
                             = 1
Therefore, x – y = 1 ---------------(3)
From (1) and (3), we get:
x + y = 7 -----------(1)
x – y = 1 -----------(3)
2x = 8, this implies that x = 4.
Therefore, (1) implies 4 + y = 7
Hence y =3.
Therefore the solution is {4,3}.

Ex 3: Solve x - y = 5 and xy = 24, find the value of x + y.

Soln: x + y = sqrt [(x-y)² + 4xy]
                   = sqrt[ 5² – 4(24)]
                   = 11.
Therefore from, x + y = 11
                             x – y = 5,
We get 2x = 16.
Therefore, x = 8.
Hence from x + y = 11. y = 3.
Therefore the solution is { 8, 3}.

Ex 4: Solve x + y = 11 and xy = 24, find the value of x² – y².

Soln: x – y = sqrt [(x+y)² – 4xy]
                   = sqrt[ 11² – 4(24)]
                   = 5
Therefore, x² – y² = ( x + y )( x – y)
                             = 11 x 5 = 55.

Practice Problems on Solving Binomials by Factoring

1. If a + b = 9  and ab = 36, find a - b
[Ans: a – b = 5]

2. If a – b = 4 and ab = 12, find a² – b².
[Ans: a² – b² = 32]

Graph of Sinx

 The coordinate graph is called the Cartesian coordinate plane. The graph contains a couple of the vertical lines are called coordinate axes. The vertical axis of the y axis value and the horizontal axis value is the x axis value. The points of the intersection of those two axes values are called the origin of coordinate graphing pictures.  The trigonometry graph is a sin or cos waves. In this graph equation is in the form of y = mx + c. m is nothing but a sin or cos. In this article we shall discuss graph of sin x.

Sample Problem for Graph of Sin X:

Graph of sin x problem 1:

Solve the given trigonometry functions 3sin 5x - y = 0 and draw the graph for the given function.

Solution:
In the first step we find the plotting point of the given trigonometry functions. The given function is
                                      3sin 5x - y = 0
We are going to find out the plotting points for a given equation. In the first step we are going to change equation in the form of y = mx + c, we get the following term
                                  3sin 5x – y = 0
                                                y = 3sin 5x
In the next step we are find out the plotting points of the above equation.
In the above equation we put x = -5 we get
           y = 3sin 5(-5)
           y = -0.4
In the above equation we put x = -4 we get
            y = 3sin 5(-4)    
           y = -2.7
In the above equation we put x = 0 we get
            y = 3sin 5(0)
           y = 0
In the above equation we put x = 2 we get
            y = 3sin 5(2)
           y = -1.6
From equation (1) we get the following value

X	-5	-4	0	2
y	0.4	-2.7	0	-1.6

Graph:

Graph of Sin X Problem 2:

Solve the given trigonometry functions y = 2sin 3x and the draw graph for the given function.
Solution:
In the first step we find the plotting point of the given trigonometry functions. The given function is
                      y = 2sin 3x
We are going to find out the plotting points for a given equation. In the first step we are going to change equation in the form of y = mx + c, we get the following term
                        2sin 3x – y = 0
                              2sin 3x = y                                   
In the above equation we put x = -4 we get
           y = 2sin 3(-4)
           y = 1
In the above equation we put x = -3 we get
            y = 2sin 3(-3)   
           y = -0.82
In the above equation we put x = 0 we get
            y = 2sin 0
           y = 0
In the above equation we put x = 3 we get
            y = 2sin 3(3)
           y = 0.82
From equation (1) we get the following value

x	-4	-3	0	3
y	1	-0.82	0	0.82

Graph:

Multiply Complex Numbers

The complex number consists of imaginary part and real part is defined as the complex number.

The complex number is written the form x + yi where x and y are real numbers and i is defined as the imaginary unit.

The normal numbers in the complex numbers is extended by using extra numbers.

The complex numbers are used in quantum physics, engineering and applied mathematics.

Multiply Complex Number:

Multiplication: (a + bi)(c + di) = (ac - bd) + (bc + ad)i

Examples on Multiply Complex Numbers:

Example 1:

Solve (2 + 2i)(5 + 3i)

Solution:

The multiplication of the complex numbers in the arithmetic form is given as,

(2 + 2i)(5 + 3i) = 10 + 6i + 10i + 6i^2

= 10 + 16i - 6   { Since i^2  = -1 }

= 4 + 16i.

The complex number multiplication answer is 4 + 16i.

Example 2:

Solve

(3 + 7i)(4 + 3i)

Multiplication of the complex number in algebraic form is given as,

(3 + 7i)(4 + 3i) = 12 + 9i + 28i + 21i^2

= 12 + 37i - 21

= -9 + 37i.

These are examples of multiplication of complex numbers.

Hence the complex number multiplications answer is -9+37i.

Example 3:

Solve

(4 + 7i)(3 + 3i)

Multiplication of the complex number in algebraic form is given as,

(4 + 7i)(3 + 3i) = 12 + 21i + 12i + 21i^2

= 12 + 33i - 21

= -9 + 33i.

These are examples of multiplication of complex numbers.

Hence the complex number multiplications answer is -9+33i.

Example 4:

Solve

(4 + 3i)(3 + 7i)

Multiplication of the complex number in algebraic form is given as,

(4 + 3i)(3 + 7i) = 12 + 9i + 28i + 21i^2

= 12 + 37i - 21

= -9 + 37i.

These are examples of multiplication of complex numbers.

Hence the complex number multiplications answer is -9+37i.

Example 5:

Solve

(2 + 5i)(4 + 6i)

Multiplication of the complex number in algebraic form is given as,

(2 + 5i)(4 + 6i) = 8 + 20i + 12i + 30i^2

= 8 + 32i - 30

= -28 + 32i.

These are examples of multiplication of complex numbers.

Hence the complex number multiplications answer is -28+32i.

Practice Algebra Calculus

Just as algebra introduces students to new ways of thinking about arithmetic problems (by way of variables, equations, functions and graphs), calculus may be seen as introducing new ways of thinking about algebra problems.

A branch of mathematics, algebra calculus  mainly deals with derivatives, integrals, limits, infinite series, differential equations, and functions, is called calculus. It is used for computing rate of change of velocity and rate of change of acceleration, the slope of a curve, and optimization

Types of Calculus and Problems in Algebra Problems:


Calculus can be classified into two major types as follows

Differential calculus
Integral calculus

Ex 1: Compute the values of Δy and dy if y = f(x) = x^3 + x^2 − 2x + 1  where x changes (i) from 2 to 2.05 and (ii) from 2 to 2.01

Sol :    (i) We have f(2) = 2^3 + 2^2 − 2(2) + 1 = 9

f(2.05) = (2.05)^3 + (2.05)^2 − 2(2.05) + 1 = 9.717625.

And Δy = f(2.05) − f(2) = 0.717625.

In general dy = f ′(x) dx = (3x^2 + 2x − 2)dx

When x = 2, dx = Δx = 0.05 and dy = [(3(2)^2+2(2)−2] 0.05 = 0.7

(ii) f(2.01) = (2.01)^3 − (2.01)^2 − 2(2.01) + 1 = 9.140701

Δy = f(2.01) − f(2) = 0.140701

When dx = Δx = 0.01, dy = [3(2)^2 + 2(2) − 2]0.01 = 0.14

Remark :   The approximation Δy ≈ dy becomes better as Δx becomes smaller . Also dy was easier than to compute Δy. For more complicatedfunctions it may be impossible to compute Δy exactly. In such cases theapproximation by calculus is especially useful.
Example 2 Algebra Calculus Problems:

Ex 2 :  Use differentials to find an approximate value for

`root(3)65` .

Let y = f(x) =`3sqrtx`=x.^1/3 Then dy =1/3x ^-2/3 dx

Since f(64) = 4. We take x = 64 and dx = Δx = 1

This gives dy =1/3 (64) ^−2/3 (1)

=1/3(16)       = 1/48

365      = f(64 + 1) ≈ f(64) + dy = 4 +1/48

≈ 4.021

Practice Problems

Pro 1:    If f(x) = x^4, then find f'(x)

Sol :              f(x) = x^4

f'(x) = 4x^3

Pro 2:    If f(x) = 5x3^ + 2x^2, then find f'(x).

Sol :              f(x) = 5x^3 + 2x^2

f'(x) = 15x^2 + 4x

Pro 3:   x7  dx  =  (x^7+1) / (7 + 1) + c

Sol :                   = x^8/8 + c

Algebra with Negative Numbers

The algebra is a division of mathematics, which can be explained as calculation of arithmetic operations. It is also referred to that character type of abstract algebra pattern. The negative number is indicating the symbol "-". For example the number -6, is the negative number. The elementary algebra is the division of the algebra.  In the subsequent we see in detail about algebra with negative numbers.

Example Problem for Algebra with Negative Number

Example 1:

Given two number -23x and -43x, adding these negative number

Solution:

In the problem given two numbers are the negative numbers.

Adding these two number

= -23x + (-43x)

= - (23x+43x)

= - (66x)

Answer for this problem is -66x.

Example 2:

Given two number -120x and -56x, subtracting these negative number

Solution:

In the problem given two numbers are the negative numbers.

Subtracting these two number

= -102x - (-56x)

= -102x+56x

= - 46x

Answer for this problem is -46x.


More over Problem for Algebra with Negative Number

Example 3:

Given two number -430x and -123x, adding these negative number

Solution:

In the problem given two numbers are the negative numbers.

Adding these two number

= -430x + (-123x)

= - (430x+123x)

= - (553x)

Answer for this problem is -553x.

Example 4:

Given two number -1230x and -1405x, subtracting these negative number

Solution:

In the problem given two numbers are the negative numbers.

Subtracting these two number

= -1230x - (-1405x)

= - 1230x + 1405x

= 175x

Answer for this problem is 175x.

Example 5:

Given two number -2345x and -450x, adding these negative number

Solution:

In the problem given two numbers are the negative numbers.

Adding these two number

= -2345x + (-450x)

= - (2345x+450x)

= - (2795x)

Answer for this problem is -2795x.

Example 6:

Given two number -5623x and -2345x, subtracting these negative number

Solution:

In the problem given two numbers are the negative numbers.

Subtracting these two number

= -5623x - (-2345x)

= - 5623x + 2345x

= - 3278x

Answer for this problem is -3278x.

Multiplication Problems

Introduction to multiplication problems:
Multiplication – It is a repeated addition.
For example 2 * 3 = 6 this is nothing but 2+2+2 = 6 or 3+ 3 =6
The order of multiplication problems does not affect the answer.The answer will not be changed even though the performing order changed.
Distribution in multiplication problems 
            To multiply 4(10+5), we multiply 4*10 and 4*5, adding the result. Thus,
                             4(10+5) = (4*10) + (4*5)
                                           =40+20
                                           =60
This is nothing but the distributive property of multiplication.
It will be easy to do multiplication if we are well known with multiplication tables.
When we multiply two numbers, there is no matter which number is first or second but the answer will be same.

More on Multiplication Problems

Example 1:

Multiply 46 * 3
Solution:
First multiply the ones digit with 3 and next tens digit with 3
If we have any remainder carry to next digit and add with it
46 * 3 =138

Example 2:

Multiply 45 * 5
Solution:
First multiply the ones digit with 5 and next tens digit with 5
If we have any remainder carry to next digit and add with it
45 * 5 = 225

Example 3:

Multiply  100 * 39
Solution:
This multiplication problem is very simple
Just multiply 1*39 and add two zero at the end
So 3900 is the answer

Example 4:

Multiply  100 * 65
Solution:
This multiplication problem is very simple
Just multiply 1*65 and add two zero at the end
So 6500 is the answer.

Example 5:

Multiply:   66 1* 23
Solution:
      6 6 1
(*)    2 3
-------------
     1983
     1322
-------------
  1 5 203
------------

Example 6:

Multiply:  2546 * 121
Solution:
                      2 5 4 6
              (*)       1 2 1
          -------------------------
                  2 5 4 6
               5 0 9 2
           2 5 4 6
--------------------------------
          1 0 1 8 4
----------------------------------

Example 7:

Multiply 45 * 9
Solution:
First multiply the ones digit with 9 and next tens digit with 9
If we have any remainder carry to next digit and add with it
45 * 9 = 405

Example 8:

Multiply  100 * 65
Solution:
This multiplication problem is very simple
Just multiply 1*65 and add two zero at the end
So 6500 is the answer.

Example 9:

Multiply 83 * 6
Solution:
First multiply the ones digit with 6 and next tens digit with 6
If we have any remainder carry to next digit and add with it
83 * 6 = 498

Example 10:

Multiply  1000 * 869
Solution:
This multiplication problem is very simple
Just multiply 1*869 and add three zero at the end
So 869000 is the answer.

Equations with Exponent

Introduction to solving equations with exponents

Exponent equations are the equations in which variable appear as an exponent.
To solve these equations rules and laws of exponents are used. Exponent equations are of two types
(1) Exponent equations in which bases are same
(2) Exponent equations in which bases are different.

Steps to Solve Equations with Exponent

Solving Exponential Equations of the same base

1) Ignore the bases, and simply set the exponents equal to each other

2) Solve for the variable

 When the bases of the terms are different

1) Ignore the exponents; rewrite both of the bases as powers of same number. For    example if there are 2 and 4 in the bases, then convert base 4, in to base 2 by writing it again as (2)^2

2) once the bases are same , ignore them

3) Equalize the exponents

4) Solve for variable

Simple Problems of Equations with Exponents

Solve for variable Answer 1. 3m  =  35 Since the bases are the same, set the exponents equal to one another: m = 5 2. 5t   = 125 125can be expressed as a power of 5: 5t  = 53 t = 3  3.  493y=343 49 and 343 can be expressed as a power of 7: [(7)2]3y = 73 76y = 73 6y = 3 y = 1/2

More Problems of Equations with Exponents

Solve for x. Answer 1.  52x+1  =  53x-2 Since the bases are the same, set the exponents equal to one another: 2x + 1 = 3x - 2 3 = x 2.  32x-1  = 27x  27 can be expressed as a power of 3: 32x-1  = 33x 2x - 1 = 3x -1 = x 3.   43x-8  = 162x 16 can be expressed as a power of 4: 43x-8= [(4)2]2x 3x - 8 = 4x   -8 = x

Practicing Integers

Introduction to Practicing Integers:
                 An integer is a set of whole numbers. Whole numbers above zero is said to be positive integers denoted as ‘+’ sign and whole numbers below zero is said to be negative integers denoted as ‘-‘ in a number line. Zero is said to be neither negative nor positive integer and it does not constant. Here, integers can be performed with four basic operations such as addition, subtraction, multiplication, and division. The positive integers can be written with or without the sign. Let us see practicing integers in this article.

Practicing Integer Problems - Practicing Adding Integers in Math

Adding same signed Integers:

Example 1:

4 + 2

Solution:

The absolute value of 4 and 2 is 4 and 2. Put the positive sign before the sum of two integers.

4 + 2 = 6

Therefore, the solution for adding 4 and 2 is 6.

Example 2:

(-4) + (-8)

Solution:

The absolute value of -4 and -8 is 4 and 8. Put the negative sign before the sum of two integers.

(-4) + (-8) = - (4 + 8) = - 12

Therefore, the solution for adding -4 and  -8 is -12.

Adding different signed Integers:

Example 3:

5 + (-5)

Solution:

The absolute value of 5 and -5 is 5 and 5. Put the sign of larger number sign before the sum of two integers.

5 – 5 = 0

Therefore, the solution for adding 5 + (-5) is 0.

Practicing Subtracting Integers in Math

Example 4:

7 - (-2)

Solution:

The absolute value of 7 and -2 is 7 and 2. Subtract the integers and put sign of larger integer before the sum.

7 – (-2) = 7 + 2 = 9

Therefore, the solution for subtracting the above integers are 9.

Practicing Multiplying Integers in Math

Multiplying same signed Integers

Example 5:

4 × 3

Solution:

The absolute value of 4 and 3 is 4 and 3. Place the same sign before the answer as it is in the given problem.

4 × 3 = 12

Therefore, the oslution for multiplying 4 and 3 is 12.

Example 6:

(-5) × (-6)

Solution:

The absolute value of -5 and -6 is 5 and 6.

- 5 × - 6 = - 30

Place the same sign before the answer as it is in the given problem -30.

Multiplying different signed Integers:

Example 7:

(-5) × (4)

Solution:

The absolute value of -5 and 4 is 5 and 4.

-5 × 4 = - 20

There is a negative sign before one of the integers in a given problem so put the negative sign before the product.

Therefore, the solution for multiplying -5 and 4 is -20.

Practicing Dividing Integers in Math

Dividing same signed Integers:

Example 8:

16 ÷ 4

Solution:

The absolute value of 16 and 4 is same.

The integer 16 is the multiple of 4 and 4.

Therefore, the solution for dividing 16 ÷ 4 is 4.

Dividing different signed Integers:

Example 9:

28 ÷ -7

Solution:

The absolute value of 28 and -7 is 28 and 7. There is a negative sign before the integers in a given problem so put the negative sign before the quotient.

The integer 28 is the multiple of 4 and 7.

Therefore, the solution for dividing 28 ÷ -7 is -4.

Problems to Practicing Integers

1. Add the two integers 3 + 5

Key: 8

2. Subtract the two integers (– 5) – (- 4)

Key: -1

3. Multiply the two integers 7 × -3

Key: -21

4. Divide the integer 15 by -5

Key: - 3