Rectangle Geometry

Introduction to rectangle geometry:

Geometry is a theoretical subject, but easy to learn, and it has many real practical applications. Eventually, geometry has developed into a skillfully arranged and sensibly organized body of knowledge. Rectangle is defined as the four sided shape polygon which every angle is 90 degree or right angle. It opposite sides are parallel and also equal length. I like to share this Definition of Rectangle with you all through my article.

Formulas of Rectangle in Geometry:

Area of the rectangle formula as,

Area A= l*w

Where l is the length of the rectangle and w is the width of the rectangle.

Perimeter of a rectangle formula as,

Perimeter P= 2(l +w)

Circumradius of the rectangle formula as,

Circumradius R=`(sqrt(l^2+w^2))/(2)`

Inradius of the rectangle formula as,

Inradius r=`(lw)/(l+w)`

Examples of Rectangle in Geometry:

In geometry rectangle problem 1:

Find the area of the following figure:

Solution:

We can find the area of a rectangle by using the following formula

Area A= l*w

In the above figure, the value of length l=4 cm and width w= 6 cm

Substitute the values of l and w into the above formula. Then we get

Area A= 4* 6

=24

Answer: 24 cm2

In geometry rectangle problem 2:

Find the perimeter of the following figure:

Solution:

We can find the Perimeter of a rectangle by using the following formula

Perimeter P= 2(l+w)

In the above figure, the value of length l= 7cm and width w= 12cm

Substitute the values of l and w into the above formula. Then we get

Perimeter P= 2(7+12)

Here we can add the 7 and 12 and then multiplying with 2.

Then we get,

Perimeter P=2(19)

= 38

Answer: 38 cm

In geometry rectangle problem 3:

Find the circumradius of a rectangle with the values of length and width are 3 cm and 4 cm.

Solution:

We can find the circumradius of a rectangle by using the following formula

Circumradius R=`(sqrt(l^2+w^2))/(2)`

Substitute the values of l and w into the above formula. Then we get

Circumradius R =     `(sqrt(9+16))/(2)`


=    `(sqrt(25))/(2)`

=      ` 5/2`

= 2.5 cm

Answer: 2.5 cm


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Practice problems of rectangle in  geometry:

Find the area of rectangle with the length= 5 cm and width= 7 cm.
Find the perimeter of rectangle with the radius length= 6cm and width= 8 cm
Find the circumradius of a rectangle with the values of length and width are 6 cm and 8 cm.
Answer keys:

35 cm2
28 cm
10 cm

Integration Reduction Formula

Introduction to Integration Reduction Formula:

There are many functions whose integrals cannot be reduced to one or more the other of the well known standard forms of integration . However , in some cases these integrals can be connected algebraically with integrals of other expressions which can either be directly integrable or which may be easier to integrate than the original functions . Such connecting algebraic relations are called 'reduction formulae' . These formu;ae connect an integral with another which is of the same type , but is of lower degree or order or at any rate easier to integrate than the original one .

Example of Integration Reduction Formula

Ex:1  Find the reduction formula for  `int` xn eax dx , n being a positive integer and hence evaluate `int` x^3 eax dx .

Sol: Let  In  =  `int` xneax   dx   .

On using the formulae for integration by parts , we get

In   =   `(x^n e^(ax))/(a)`    -  `int` n xn-1  `(e^(ax))/(a)`   dx

=   `(x^n e^(ax))/(a)`    -   `(n)/(a)` `int` xn-1  eax  dx   .

=     `(x^n e^(ax))/(a)`    -  `(n)/(a)` In-1   .

This is called reduction formulae for  `int` xn eax dx   .   Now  In-1  in turn can be connected to  In-2   .  By successive reduction of  n  , the original integral In finally depends on I0  ,  where  I0 =  `int` eax dx    =   `(e^(ax))/(a)`     .

To evaluate `int` x^3 eax dx  , we  take a = 5  and use  the reduction formula  for  n = 3 , 2 , 1 in that order . Then we have I have recently faced lot of problem while learning how to solve linear equations word problems, But thank to online resources of math which helped me to learn myself easily on net.

I3  =   `int` x^3 e5x dx   =    `(x^3 5^(ax))/(5)`   -  `(3)/(5)` I2    .

I2   =   `(x^2 e^(5x))/(5)`   -  `(2)/(5)` I1

I1  =  `(xe^(5x))/(5)`   -  `(1)/(5)` I0

I0   =   `(e^(5x))/(5)`    +    c

Hence    I3   =  `(x^3 e^(5x))/(5)`   -  `(3)/(5^2)` x2e5x   +  `(6)/(5^3)` xe5x    -   `(6)/(5^4)` e5x  +  c

Integration Reduction Formula- Example

Q:1 Find the reduction formula for  `int` tannx  dx   for an integer  n`>=` 2  and hence  find `int` tan6x dx .

Sol : Let  In  =  `int` tannx  dx

=     `int` tann-2x  tan2x    dx

=    `int` tann-2x   sec2x   dx     -     `int` tann-2x   dx

=     `(tan^(n-1)x)/(n-1)`   -   In-2   ,

which is the required reduction formula .

When n is even  , In will finally depend on

I0   =   `int` dx      =    x + c1

When  n is odd  ,   In  will finally depend on

I1   =   `int` tanx  dx    =   log (secx)   +   c2

Now   ,    I6    =    `int` tan6x  dx

=     `(tan^5x)/(5)`   -  `int` tan4x  dx

=   `(tan^5x)/(5)`   -   `(tan^3x)/(3)`   +  `int` tan2x   dx

=     `(tan^5x)/(x)`   -   `(tan^3x)/(3)`   +  tanx  -  x  +  c .

Solving the Proportion Math

Introduction of solving the proportion math:

A part considered in relation to its whole. Statement of equality between 2 or more ratios like a/b=c/d. When we multiply we get ad=bc. The two ratios are equivalent. And we can say, two set of numbers are proportional, if one set is a constant times the other. Here we are going to see solving proportions with variable.I like to share this Perpendicular Lines Calculator with you all through my article.

Example Problems of Solving Proportion in Math:

Example 1:

Given:  `(45)/(h) = (35)/(21)`

Solution:

Step 1: We need to find h value.

Step 2: Here we are using cross multiplication method to solve h.

Step 3: So, when we cross multiply we get 945 = 35h

Step 4: Now we divide using 35 on both the sides.

Step 5: The value of h=27.

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Example 2:

Given:  `(15)/(t) = (20)/(44)`

Solution:

Step 1:  From this question we need to find t value

Step 2: Using cross multiplication we get, 660= 20t.

Step 3: Now we divide using 20 on both the sides

Step 4:  `(660)/(20)=(20t)/20`

Step 5: t = 33.

Example 3:

Given:  `(21)/(9) = (42)/(g)`

Solution:

Step 1:  From this question we need to find g value.

Step 2: Using cross multiplication we get, 21g = 378.

Step 3: Now we divide using 21 on both the sides

Step 4: `(21g)/(21) = (378)/(21)`

Step 5: 18 = g.

These are the simple examples of solving proportion in math.

Solving Word Problem Using Proportions in Math:

Example 1:

Sam read 40 pages of a book in 50 minutes. How many pages should he be able to read in 80 minutes?

Solution:

Step 1: We know that Sam read 40 pages of book in 50 minutes.

Step 2: And we need to find how many pages Sam read within 80 minutes. Let us assume the unknown number is x.

Step 5: So,  `(40)/(x) = (50)/80`

Step 6: Using cross multiplication we get, 3200 = 50x.

Step 7: Divide using 50 on both the sides. `(3200)/(50)=(50x)/(50)` .

Step 8: 64 =x. Therefore Sam read 64 pages within 80 minutes.



Example 2:

A car travels 48 miles in 20 minutes. And how many miles travel the car within 30 minutes?

Step 1: We know that a car travels 48 miles in 20 minutes.

Step 2: And we need to find how many miles travel the car in 30 minutes. Let us consider the unknown number is x.

Step 5: So,  `(48)/(20) = (x)/(30)`

Step 6: Using cross multiplication we get, 1440 = 20x.

Step 7: Divide using 50 on both the sides.  `(1440)/(20) = (20x)/(20)` .

Step 8: 72 =x. Therefore a car travels 72 miles in 30 minutes.

These are the example word problems of solving proportions in math.

Pattern Block Fractions

Introduction :

Fraction is one of a part in math numerical values. A fraction number is an integer value. Fraction number is a one branch of the whole number value in decimals.  A Fraction number is consisting of a two parts. That is numerator value and denominator value. In online few websites are providing fraction pattern tutoring. In this article we shall discuss the pattern block fractions. I like to share this Types of Fractions with you all through my article.

Types of Pattern Block Fractions:

Proper Fractions:

In proper fractions having numerator value is less than the denominator value is called as proper fractions.

Example: `5/9` and `6/13`

Improper Fractions:

In improper fractions having numerator value is larger than or equal to their denominators value is called as improper fractions.

Example:  `6/5` and `9/7`

Mixed Fractions:

In mixed fraction Numbers having a two part, which is a whole number part and a fractional part is called as mixed numbers.

Example: `3(2/5)` and `4(1/3)`

Decimal Fractions:

In decimal fraction number having denominator value as 10, 100 or 1000 or any other higher power of 10 is called as decimal fractions.

Example: `7/10` and `65/100`

Simple Fractions:

In simple fraction number having both the numerator and denominator as whole numbers are called simple fractions.
Example: `4/5` , `12/95` .

Sample Problem for Pattern Block Fractions:

Pattern block fraction problem 1:

Evaluate the value of given fraction numbers `(7)/(9)` + `(5)/(9)`

Solution:

In the proper fraction number a denominator values are same. So we are directly add or subtract the numerator values.

Step 1: In the given problem denominator values are same.

Step 2: Add the numerator values of given numbers, we get.

`(7)/(9)` + `(5)/(9)` = `(7 + 5)/(9)`

=`(12)/(9)`

Step 3: Now we are simplify the fraction values.

= `(4)/(3)`

Pattern block fraction problem 2:

Evaluate the value of given fraction numbers`(9)/(11)` + `(8)/(11)`

Solution:

In the proper fraction number a denominator values are same. So we are directly add or subtract the numerator value.

Step 1: In the given problem denominator values are same.

Step 2: Add the numerator values of the given numbers, we get.

`(9)/(11)` + `(8)/(11)` = `(9+8)/(11)`

=  `(17)/(11)`

Definition of Division in Math

Introduction:

The definition of division steps are one of the easiest techniques in math. The Division is the one of the major arithmetic operation in math. Division is mathematically expressed by using of given symbol (/). In math division definition is inverse of the multiplications. In math division is the equal portion or group.  Consider X and Y are two value the product of two value is equal to Z means the mathematical expression is

`X*Y=Z`

The Y value is not equal to zero means

`X=Z/Y`

Types of Division Definition:

The given step is basic form of the divisions. These Division consists the basic three parts

Consider these equation `X=Z/Y`

In these

1] The upper part of the above equation is called as Dividend that is Z

2] The lower part of the above equation is called as Divisor that is Y

3] The X is called as quotient

Dividend/divisor= quotient

Division symbol definition specifies the various form some of them are given in below

1] `X=Z % Y`

2] `X=Z/X`

3] `X=Z) X` or `X=Z) bar(X)` [It is the US Notation]

Divisible:

The one number can be divided with another number and the result is whole number is called as divisible number.Understanding Writing Piecewise Functions is always challenging for me but thanks to all math help websites to help me out.

Division is used in various areas some of the names are given in below. That are

1] Division of Integers

2] Division of Rational Number

3] Division of Zero

4] Division of Complex Numbers

5] Division of Polynomials

6] Division of Matrices

7] Division of abstract algebra

8] Division and Calculus

Example for Division Definition:

Consider this example. There are 4 students in the class and 12 boxes are available then each student gets how many boxes?

Steps 1:

Write the all the given values

Total Students `(S) =4` and Total Boxes `(B) =12`

Finding Shared box (F)

Steps 2:

Form the equation using Division symbol

F=B/S

Steps 3:

Write the value based on above equation

`F=12/4`

Steps 4:

`F=3`

Each student gets the `3` Boxes.

Number Theory Research

Introduction :

In this we will study about number theory research. Number theory research is one of the most olden areas of research in math. Number theory is a study of all types of numbers in math  such as integers, whole number, and so on. Let us start our research in number theory.

Example Problems for Number Theory Research:

Example problem 1: How do you write 88,000 in scientific notation?

Solution:

Shift the decimal point to the left until the number is between 1 and 10. Calculate the places we move the decimal point.

88,000  ?  8.8

We moved the decimal point 4 places to the left. The power of 10 is 104.

88,000 = 8.8 × 104

Answer: 88,000 = 8.8 × 104

Example problem 2: Which of the following numbers are primes?

5, 18,17, 8

Solution:

The factors of 5 are 1 and 5. Since the only factors of 5 are 1 and itself, 5 is prime number.

The factors of 18 are given as1, 2, 3, 6, 9, and 18. Since 18 has more factors than just 1 and itself, 18 is not prime number.

The factors of 17 are given as 1 and 17. Since the only factors of 17 are 1 and itself, 17 is prime number.

The factors of 8 are given as1, 2, 4, and 8. Since 8 has more factors than just 1 and itself, 8 is not prime number.

Therefore, the prime numbers are 5 and 17.

Answer: The prime numbers are 5 and 17.

Practice Problems for Number Theory Research:

Practice problem 1: Which number is prime number? 20, 7, 16, 10

Practice problem 2: What is the prime factorization of 16?

Practice problem 3: Write is the greatest common factor of 10 and 6.

Solutions for number theory research:

Solution 1: The prime number is 7.

Solution 2: The prime factorization of 16 is 2 * 2* 2* 2.

Solution 3: The greatest common factor of 10 and 6 is 2.

Vectors Lines and Planes

Introduction to Vectors Line and Planes

Vector Lines

1) r  =  a + tb  ,  t  `in`  R

Cartesian Form : Let OXYZ be a right handed rectangular Cartesian coordinate framework . Let i , j , k be unit vectors along the positive X , Y , Z axes respectively .

We wish to find in Cartesian form the equation of the straight line L passing through a given point A (a1 , a2 , a3 ) and having direction cosines (b1 ,  b2 , b3) . We denote that L is the straight line passing through the point A with the position vector  a = a1i + a2j + a3k and parallel to the vector  b = b1i + b2j + b3k .

Hence P(x , y , z) is a point on L  `hArr`  `EE`   t  `in`  R such that  xi + yj - zk  =  OP   =  a + tb .

`hArr`  `EE`   t  `in`  R such that  xi + yj + zk   =  (a1 + tb1) i +  (a2 + tb2) j  +  (a3 + tb3) k

`hArr`  `EE`   t  `in`   R  such that   xi + yj - zk   =   a1 + tb1  ,  y  =  a2 + tb2  ,  z  =  a3 + tb3

`hArr`  x - a1  :  y - a2  :  z - a3  : : b1 : b2 : b3          ---------------->(1)

(1) is known as the Cartesian equation of a straight line in point - Direction cosines form .   It is usually expressed as  `(x-a_1)/b_1=(y-a_2)/b_2=(z-a_3)/b_3=t`    ,    `tin`

Since  kb1 , kb2 , kb3 , k  `!=` 0  are the direction ratios of the line the equation of a straight line in point - Direction ratios are  `(x-a_1)/(kb_1)=(y-a_2)/(kb_2)=(z-a_3)/(kb_3)`

2) r  =  ( 1 - t ) a + tb  ,  t  `in`  R

Cartesian Form:

Let OXYZ  be a right handed rectangular Cartesian coordinate framework . Let i , j , k be unit vectors along the positive X , Y, Z axes respectively . Let OP = r = (x , y , z) .

We wish to find the cartesian form of the equations of the straight line passing through two given points A(a1 , a2 , a3) and B(b1 , b2 , b3)

Let a = (a1 , a2 , a3) and  b  =  (b1 , b2 , b3)

P(x , y , z) is a point on L  `hArr` xi + yj + zk  =   OP  =  (1=i)a + tb for some  t  `in`  R .

`hArr`  (x-a1)i + (y-a2)j + (z-a3)k    =   t[ (b1-a1)i + (b2-a2)j + (b3-a3)k ]  ,    t  `in`  R

`hArr`  (x-a1) :  (y-a2) : (z-a3)  : :  (b1-a1) : (b2-a2) : (b3-a3)   --------------->(2)

(2) is known as the Cartesian form of the straight line passing through given points (a1 , a2 , a3) and (b1 , b2 , b3)  . It is usually written in form

`(x-a_1)/(b_1-a_1)=(y-a_2)/(b_2-a_2)=(z-a_3)/(b_3-a_3)`    =  `t`     ,    t  `in`   R

Properties of Vectors Lines and Planes

Properties

The vector equation of the plane passing through a poi A(a) and parallel to two non-collinear vectors b and c  is   r = a + sb + tc  ;   s , t  `in`  R
The vector equation of the plane passing through three non-collinear points A(a) , B(b)( , C(c)  is  r =  (1 - s - t)a + sb + tc  ;  s , t `in`  R
Let A and B  be the distinct points and c be a vector such that AB and C are non-collinear . Then the vector equation of the plane passing through A and B and parallel to c is  r  =  (1 - s)a + sb + tc   ;  s , t  `in`  R
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Solved Problems on Vectors Lines and Planes

1) Write the vector equation of the straight line passing through the points (2i + j + 3k) , (-4i + 3j - k) . Write also the Cartesian form of the equation .

Solution : If the position vectors of A and B with respect to O are OA = a , OB = b then the vector equation of a straight line passing through A and B is

r   =  (1 - t)a + tb  ,  t is real

Hence the vector equation of the straight line passing through the given points is

r   =  (1 - t) (2i+ j + 3k)  +  t(-4i + 3j - k)

Since r = (x , y , z) , a = (2 , 1 , 3) and b = (-4 , 3 , -1) the Cartesian form of the equation is

`(x-2)/(-4-2)=(y-1)/(3-1)=(z-3)/(-1-3)`

`(x-2)/(-6)=(y-1)/(2)=(z-3)/(-4)`

2) Find the vector equation of the plane passing through the points (1 , -2 , 5) , (0 , -5 , -1) and (-3 , 5 , 0)

Solution : If the position vectors of A , B and C with respect to O are OA = a , OB = b , OC = c then the vector equation of a straight line passing through A,B and C  is     (1 - s - t)a + sb + tc

Hence the vector equation of the straight line passing through the given three points is

r   =   (1 - s - t)(i - 2j + 5k)  +  s(-5j - k) + t(-3i + 5j)

Quartile Example

Introduction to quartile example:

The median calculate the central point of a distribution, bottom on the order, just about half of the information set falls under the median and half falls over the median.  The middle of the division is successful method to use. For the median, a usual measure of spreading can obtain from the lesser and higher quartile. The quartile contains the upper quartile and also the lower quartile. Let us see about the topic to quartile(upper quartile and lower quartile) example help are given below in the contents.

Example Problems to Solve the Method Quartile Example

let us see about the topic quartile example we can see bout different method to solve quartile (upper, lower, meadian and interquartile and range),

Example problem1:-using solve the method quartile example

Calculate the meadian, lesser quartile, higher quartile, interquartile and range of the given information of the following sequence.

15, 50, 31, 40, 18, 13, 24, 19, 25, 45, 12, 20.

Solution:

Initial, arrange the data in ascending order:

Given:  12, 13, 15, 18, 19, 20, 24, 25, 31, 40, 45, 50

Step :1

12, 13, 15 ,18,  19,  20,| 24, 25,  31, 40,45, 50

Lower quartile       Median         Upper quartile

Step :2 Find the meadian for lower quartile

Lower quartile = `(15+33)/2` = 16.5

Step :3

Median = `(20+24)/2` = 22

Step :4 Find the meadian for upper quartile

Upper quartile = `(31+40)/2` = 35.5

Step: 5

Interquartile range = Upper quartile – lower quartile

= 35.5 – 16.5 = 19

Step: 6

Range = largest value – smallest value

= 50 – 12 = 38

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Example Problem2:-using to Solve the Method Quartile Example

Estimate the median, lower quartile, upper quartile, interquartile and range of the given information of the following sequence.

15, 45, 35, 25, 65, 12, 30, 9, 20, 55, 95, 70.

Solution:

Initial, arrange the data in ascending order:

Given data:  9, 12, 15, 20, 25, 30, 35, 45, 55, 65, 70, 95

Step: 1

9, 12, 15, 20, 25, 30, | 35, 45, 55, 65, 70, 95

Lower quartile       Median              Upper quartile

Step :2 Find the meadian for lower quartile

Lower quartile = `(15 +20)/2` =17.5

Step :3

Median = `(30+35)/2` = 32.5

Step :4 Find the meadian for upper quartile

Upper quartile = `(55+65)/2` = 60

Step :5

Interquartile range = Upper quartile – lower quartile

= 60 – 17.5 = 77.5

Step :6

Range = largest value – smallest value

= 95 – 5 = 90

Triple Vector Product

In math, 3d vector is a vector with 3 dimensions. For example, 2`veci` + 2`vecj` + 4`veck` is a 3d vector. Cross product of a 3d vector gives resultant vector perpendicular to the two given vectors. A general formula is available for finding the cross product of 3d vectors. Let us study the explanation of 3d vector cross product with examples.

Let us take the following Triple Vector Product

a = a1 `veci ` + a2 `vecj` + a3 `veck ` and b = b1 `veci` + b2 `vecj` + b3 `veck`

Now, the cross product of the above two 3D vectors is as follows.

a x b = `|[veci,vecj, veck],[a_1,a_2, a_3],[b_1, b_2, b_3]|`

a x b = `veci` (a2b3 - a3b2) - `vecj` (a1b3 - a3b1) + `veck`(a1b3 - a3b1)

By using the above general description, we can find the cross product of two given 3D vectors.Looking out for more help on Algebra Tutor in algebra by visiting listed websites.

Derivation of cross product:

a x b = (a1 b1) `veci` + (a1 b2) `veci ` `vecj` + (a1 b3) `veci` `veck ` +

(a2 b1) `vecj ` `veci` + (a2 b2) `vecj` `vecj` + (a2 b3) `vecj` ` veck` +

(a3 b1) `veck`` veci` + (a3 b2)` veck` `vecj` + (a3 b2) `veck` `veck`

Some terms related to cross product:

`veci` x `vecj` = `veck`

`vecj` x `vec k` =` veci`

`veck` x `veci` =` vecj`

`vecj` x `veci` = -`veck`

`veck` x `vecj` = - `veci`

`veci ` x `veck` = -`vecj`

Based on the above terms, the formula of cross product becomes,

a x b = `veci` (a2b3 - a3b2) - `vecj` (a1b3 - a3b1) + `veck`(a1b3 - a3b1)

Inverse of Function Solver

Introduction to inverse function solver

If  ' f ' is a relation from A to B , then the relation  { (b,a) : (a,b) `in` f } is denoted as f-1 .

If ' f ' is a relation from A to B , then  (f   -1)-1  =  f   .

Theorem : If   f : A `|->` B is one-one , then f -1 is bijection from f(A) to A .

Proof : Let  f : A `|->` B be one-one .

Clearly f  -1  is a relation from   f(A) to A .

Let  b `in` f(A)  . then there exists  a `in` A such that   f(a) = b . Since f is one-one , a is the only element of A such that  f(a) = b . Thus given b `in` f(A) there is a  unique element a in A such that  (a,b) `in` f . Hence given b `in` f(A) there is unique element a `in` A such that  (b,a) `in` f-1 . Hence f-1 is function  from f(A) to A  and  f-1 (b) = a if and only  if f(a) = b . Clearly f-1 is an onto function . If  b1 , b2 `in` f(A) and f-1(b1) = f-1(b2) = a  say , then b1 = f(a) = b2 . Thus f-1 is one to one . Therfore  f-1 : f(A) `|->` A is bijection .

Definition : If  f : A `|->` B  is a bijection then the relation f -1 : { (b,a) : (a,b) `in` f } is a function from B to A and is called the inverse function of f .

Let us see some problems in inverse of function solver.

Examples on Inverse Function Solver
Problem: Find the inverse function of   f(x) = 4x +7 ?

Solution :   Given  f(x)  =  4x  + 7

Let   y  =  f(x)    `rArr`   x  =  f-1(y) .

`rArr`    y   =  4x   +   7  
`rArr`     x   =  `(y-7)/(4)`
`rArr`     f-1(y)   =  `(y-7)/(4)`

`:.`   f-1(x)  =  `(x-7)/(4)`

Answer :  f-1  =  `(x-7)/(4)` .

Problem: Find the inverse function of   f(x) = e4x+7  ?

Solution : Given  f(x) = e4x+7

Let    y  =  f(x)   `rArr`   x  =  f-1(y)

`rArr`    y   =  e4x+7  
`rArr`    log y =  4x + 7
`rArr`     x  =  `(logy-7)/(4)`    =  f-1(y)

f-1(x)  =  `(logx-7)/(4)` 

Answer : f-1(x)   =  `(logx-7)/(4)`

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Solved Problems on Inverse Function Solver

1) If A = { 1 , 2 , 3 } , B = { a , b , c } and  f = { (1,a) , (2,b) , (3,c) } , then find the inverse function of f ?

Solution : Given    A = { 1 , 2 , 3 } , B = { a , b , c }

Also given the function  f = { (1,a) , (2,b) , (3,c) }

Now , inverse of f  i.e. f-1  =  { (a,1) , (b,2) , (c,3) }

and f-1 : B `|->` A is also a bisection .

Answer :   f-1  =  { (a,1) , (b,2) , (c,3) }

2) If A = { 1 , 2 , 3 } , B = { a , b , c, d }  and g = { (1,b) , (2,a) , (3,c) } . Find the inverse function ?

Solution : Given        A = { 1 , 2 , 3 } , B = { a , b , c, d }  and g = { (1,b) , (2,a) , (3,c) }

Here g-1 does not exist .

g : A `|->` B  is one to one but not onto .  g-1 is bisection from { a, b c }  to A .

Absolute Value Inequalities

Introduction :

Normally absolute value means it is nothing but the values without considering it sign. Here we are going to solve the absolute values for inequalities. If we solve the absolute value inequalities we will get two values. Using this we value we will find the limitations of the inequalities. We will see some example problems for absolute value inequalities. Normally absolute value of |a| = `+-` a

Problems Based on Absolute Values Inequalities.

Example 1 for absolute value inequalities:

Find the absolute values of the inequalities. |x - 2| >= 10

Solution:

Given inequality is |x - 2| `gt= ` 10

To find the absolute value for the inequality is

(x – 2) `gt=` 10 ………… (1) And –(x – 2) `gt=` 10 …………………. (2)

Equation 1:

(x – 2) `gt= ` 10 ………… (1)

Add +2 on both sides

x - 2 + 2 `gt=` 10 +2

x `gt=` 12

Equation 2:

-(x – 2) `gt=` 10

-x + 2 `gt=` 10

Add -2 on both sides.

-x + 2 – 2 `gt=` 10 – 2

-x `gt=` 8

Divide by -1.

x `lt=` -8

So x lies between -8 `gt=` x `gt=` 12

We will see some more example problems based on absolute value inequalities.


Example 2 for Absolute Value Inequalities:

Find the absolute values of the inequalities. |x - 5| >= 2

Solution:

Given inequality is |x - 5| `gt=` 2

To find the absolute value for the inequality is

(x – 5) `gt= ` 2 ………… (1) And –(x – 5) `gt= ` 2 …………………. (2)

Equation 1:

(x – 5) `gt= ` 2 ………… (1)

Add +5 on both sides

x - 5 + 5 `gt= ` 2 + 5

x `gt= ` 7

Equation 2:

-(x – 5) `gt= ` 2

-x + 5 `gt= ` 2

Add -5 on both sides.

-x + 5 - 5 `gt= ` 2 - 5

-x `gt= ` -3

Divide by -1.

x `lt= ` 1

So x lies between 7 `gt= ` x `gt= ` 1                                            

These are some of the examples for absolute value inequalities. Here we will treat the inequalities like absolute values. Normally the difference between the inequalities and linear equation is less than and greater than sign.

Add and Subtract Matrices

Introduction about matrices:

Matrix has a list of data. In matrix is a rectangular arrangement of the elements. The elements are shown in the rows and the columns. Array elements are put in the parenthesis or square brackets. In matrix is represented by capital letters for example A, B, C…… We can do addition, subtraction and multiplication in matrices. In this article we shall discuss about the addition and subtraction process of matrix.

Addition of Matrices:

If we add two matrices add the first element of first matrix with first element of second matrices repeat the process for all the elements. An addition matrix is similar to normal addition.

If we add two matrices they are in the same order. For example matrix A present in 3 `xx` 3 structure the matrix B also present in 3 `xx` 3 structure else ( if the matrix B present in 2 `xx` 3 form or etc) we cannot add the matrices.

Example sums for addition of matrices:

Example:

Add the following two matrices

A= `[[2,-3],[-5,4]]`

B= `[[8,2],[-1,7]]`

A + B=?

Solution:

Given

A= `[[2,-3],[-5,4]]`

B= `[[8,2],[-1,7]]`

A + B= `[[2,-3],[-5,4]]`+ `[[8,2],[-1,7]]`

A + B= `[[2+8,-3+2],[-5-1,4+7]]`

A + B= `[[10,-1],[-6,11]]`

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Subtraction of Matrices:

Subtraction of matrices are some more different from normal substitution. Subtraction of matrices follows some rules. The following steps are used to subtract the matrices.

Step 1:

First create the negative matrix for subtrahend matrix.

Step 2:

Add the negative matrix (previous step result) with next matrix

For example if we do the process A-B matrix first discover the negative of B matrix and then add –B matrix with A matrix  like A + (-B)

Example Problems for Subtraction of Matrices:

Example:

Subtract the following two matrices

A = `[[-5,-8],[4,6]]`

B= `[[2,5],[-8,-7]]`

A-B=?

Solution:

Given

A = `[[-5,-8],[4,6]]`

B= `[[2,5],[-8,-7]]`

Step 1:

First create the negative matrix for subtrahend matrix.

- B = - `[[2,5],[-8,-7]]`

- B = `[[-2,-5],[8,7]]`

Step 2:

Add the negative matrix (previous step result) with next matrix

A + (-B) =  `[[-5,-8],[4,6]]` + `[[-2,-5],[8,7]]`

A + (-B) = `[[-5 + (-2),-8+(-5)],[4+8,6+7]]`

A + (-B) = `[[-7,-13],[12,13]]`

Graphing Parabola Inequalities

Introduction for graphing parabola inequalities:
Parabola is a conic sections, the intersection of a right circular conical surface and a plane to a generating straight line of that surface.

It is a polynomial function of degree two. A function f : R ? R defined by  the f(x) = ax2 + bx + c, where a, b, c ? R, a ? 0 is called a quadratic function. The graph of a quadratic function is always a parabola inequalities  f(x) > ax2 +  2 , f(x) < ax2 + 2

The graphing for parabola inequalities is shown below.

Example Problem for Graphing Parabola Inequalities:

Ex 1:  Draw the graph of inequalities y > x2

Sol :  Draw x-axis and y-axis on the graph sheet. Mark the scale on x-axis 1 cm = 1 unit and  y-axis 1 cm = 2 units. Assigning value for x = –4 to 4 and we get the corresponding y values:

Plot the points (–4,16), (–3,9), (–2,4),(–1, 1), (0,0), (1,1), (2,4), (3,9), (4,16) on the graph sheet and join the points by smooth curve. This curve is called the parabola y >x2

x  -4  -3  -2 -1  0  1  2  3  4

y  16  9   4    1   0  1  4  9  16

graph of inequalities y > x2

Ex 2:    Draw the graph of  inequalities   y < x2 – 2x – 3.

Sol :    Draw x-axis and y-axis on the graph sheet and mark the scales on x-axis

1 cm = 1 unit and on y-axis 1 cm = 2 units. Assign values x = –4 to 5 and calculate the

Corresponding y values

We plot the points (–4,21), (–3,12), (–2,5), (–1,0), (0, –3), (1, –4), (2, –3), (3,0), (4,5) and

(5,12) on the graph sheet and join these points by a smooth curve. We get a required graph of the parabola y < x2 – 2x – 3

X   -4   -3   -2   -1   0    1   2   3   4  5

Y   21  12   5    0   -3   -4  -3  0   5  12

inequalities   y < x2 – 2x – 3

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Practice Problem for Graphing Parabola Inequalities:

Draw the graph of inequalities   y > x2-1
Draw the graph of  inequalities   y  <  3x2- 4x + 5
3.  Draw the graph of  inequalities   y  <  5x2- 4x + 9
4.  Draw the graph of  inequalities   y > x2+4

Radical Times a Radical Math

Introduction to radical times a radical math:
Radicals are nothing but a root which we also called as square root. Square root is indicated with the symbol (sqrt) and sqrt (). Normally, radicals are rewrite like; sqrt of (b) is rewrite as `b ^(1/2)` . Likewise radicals are expressed in various forms. The expression `sqrt (8)` is read as “radical eight”, or “the square root of eight”. Thus, we are going to discuss about the radical times radical in math which related to multiplication of two radicals in math.

Radical Times a Radical Math:

Radical times a radical is nothing but multiplying a radical with another radical in math. This is similar to the normal multiplication of radicals in math.

General expression with exponent and radical:

`n (sqrt (a)) ^m =(n (sqrt (a))) ^m =(a1/n) ^m = a^m/n`                  

Multiplication property for radical expression: `n (sqrt (ab)) = n (sqrt (a)) n (sqrt (b))`

Division property for radical expression: `n (sqrt (a/b)) = (n (sqrt (a))) / (n (sqrt (b)))`

From, the above properties, we are going to learn about multiplication property in math is like,

`sqrt (a) xx sqrt (b) = sqrt (ab)`

Example for Radical Times a Radical Math:

Example for radical times radical math 1: Multiply the given radicals:  (2`sqrt 4` ) × (`root(2)(16)` )

(2`sqrt4` )  (`root(2)(16)` ) =2 × 2 × (`root(2)(16)` )

=2 × 2 × 4

=16

Answer: `16`

Example for radical times radical math 2: Multiplying the given radicals: (`sqrt (3)` ) (5`sqrt(5)` )

`sqrt (3)`.   (5`sqrt(9)` ) = `5``sqrt(3 xx 5)`

=`5sqrt(15)`

Answer: `5sqrt(5)`

Example for radical times radical math 3: Multiply the given radicals:   ( ` sqrt( 8)` ) `7sqrt(15)`

( ` sqrt( 8)` ) (`7` `sqrt(15)` )   = `sqrt(2 xx 4)` . `7` `sqrt(15)`

= `2``sqrt( 2)` `xx ` `7``sqrt(15)`

= `14` ` sqrt(30)`

Answer: `14sqrt (30)`

Example for radical times radical math 4: Multiply radicals:  `sqrt 16 and sqrt 81`

(`sqrt(16)` ) (`sqrt(81)`) =`sqrt(16)` `sqrt(81)`

= `sqrt(4 xx 4 xx 9 xx 9)`

= `4 xx 9`

= `36`

Answer: `36`

Post a Math Problem

Introduction to post a math problem:

Post a math problem is that several math problem posted on the internet for the students help. There are many websites that post a math problems to help the students. Among the entire website tutor vista is the famous and wonderful website which has excellent tutoring team to help the students any time with all the homework problems. The tutoring team will be online 24 x 7 to help the students.In the tutor vista website any student can post a math problem at any time and they can immediately get help. In this article let us see sample math problems posted in this website.

Post a Math Problem:

Example 1:

Find the solution of p in the equation 7p + 4p = 55

Solution

Given equation is 7p + 4 p = 55

Add leht side of the equation since both are in terms of p

7p+ 4p = 55 p

So 11p = 55

Divide both sides by 11

`(11p)/11` = `55/11`

p = 5.

Check the value by substituting p = 5 in the given equation

7(5)+4(5) = 55

35 + 20 = 55

55 = 55

So the answer p = 5 is correct.

Example 2:

Solve for x in the given equation 10x + 2 = 47

Solution

The given equation is 10x+2 = 47

Subtract 2 on both sides

So 10x + 2 -2 = 47 - 2

By subtracting 2 the equation is formed as 10x = 45

Now divide by 10 on both sides

`(10x)/10` = `45/10`

x = 4.5

So we get the answer as x = 4.5.

Now we can check this whether our answer is correct or not. For checking our answer substitute the value x = 4.5 in the given equation at the place of x

So given equation is 10x + 2 = 47

Substitute x = 4.5

10(4.5) + 2 = 47

45 + 2 = 47

47 = 47

So both sides are equal and our answer x = 4.5 is correct.

Post a Math Problem:

Here are some of the question. Solve and find out the solution and check as shown above.

1. Find the value of k of the equation 3k = 27

2. Find the value of p of the equation 8p - 3p = 75

3. Find the value of p of the equation 12 p = 42 + 102

4. Find the value of m of the equation 15m - 7m = 2m +66

5. Find the value of s of the equation 5s + 50 - 3s = 60

Answers

1. k = 9

2. p = 15

3. p = 12

4. m =11

5. s = 5

Easy Way to Simplify Fractions

Introduction:

   These articles we are discuss about simplify fractions in easy way. A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development were the common or "vulgar" fractions which are still used today (`1/2` , `5/8` , `3/4` etc.) and which consist of a numerator and a denominator. (Source – Wikipedia)

Easy way to simplify fractions solves the simple addition fraction, multiplication fraction and subtraction fraction.

Easy Way to Simplify Fractions-example Problems:

Example 1:

Add the fractions for given two fraction, `2/5` + `1/5`

Solution:

The given two fractions are `2/5` + `1/5`

The same denominators of the two fractions, so

                                         = `2/5` + `1/5`

Add the numerators the 2 and 1 = 2+1 = 3.

The same denominator is 5.

                                         = `3/5`

The addition fraction solution is `3/5` .

Example 2:

Subtract the fractions for given two fraction `4/6` - `3/3`

Solution:

The denominator is different so we take a (lcd) least common denominator

LCD = 6 x 3 = 18

So multiply and divide by 3 in first term we get

                                                           `(4 xx 3) / (6 xx 3)`

                                                         =`12/18`

Multiply and divide by 6 in second terms

                                                        = `(3 xx 6) / (3 xx 6)`

                                                       = `18/18`

The denominators are equals

So subtracting the numerator directly = `(12-18)/18`

Simplify the above equation we get = `-6/18`

Therefore the final answer is `-1/3`

Example 3:

Multiply the fractions for given two fraction, `4/6` x `5/6`

Solution:

The given two fractions are `4/6 ` x `5/6`

The same denominators of the two fractions, so

                                         = `4/6` x `5/6`

Multiply the numerators the 4 and 5 = 4 x 5 = 20.

Multiply the denominators the 6 and 6 = 6 x 6= 36

                                         =` 20/36`

The multiply fraction solution is `5/9`

Example 4:

Dividing fraction:

`5/7 ` divides `2/4`

Solution:

First we have to take the reciprocal of the second number

Reciprocal of `2/4` = `4/2`

Now we multiply with first term we get

`5/7` x `4/2`

Multiply the numerator and denominator

`(5 xx 4) / (7 xx 2)`

Simplify the above equation we get

= `20/14`

Therefore the final answer is `10/7`


Easy Way to Simplify Fractions-practice Problems:

Problem 1: Add the two fraction `3/9` + `2/9`

Solution: `5/9`

Problem 2: Subtract two fractions `10/9` – `6/9`

Solution: `4/9`

Problem 3: multiply two fractions `6/5` x `6/6`

Solution: `6/5`

Problem 4: Dividing two fractions `5/6` and `2/4`

Solution:` 5/3`

Solving Calculus Math Problem

Introduction to solving calculus math problem:
There are various branches exist in mathematics. The calculus is an important part of mathematics. It deals with the limits, functions, derivatives, integrals, and infinite series. In general calculus is divided into two parts,they are differential and integral calculus. Now we are going to discuss various problems related to calculus.

Example Problems for Solving Calculus Math Problem:

Solving calculus math problem – Example: 1

Solve `\int x^2e^xdx`

Solution:

Both side integration by the parts of twice to get this to an integral for which we know the formula.

`u=x^2`

`du=2xdx`

`dv=e^xdx`

`v=e^x`

`\int x^2e^xdx=x^2e^x-2\int xe^xdx`

`u=x`

`du=dx`

`dv=e^xdx`

`v=e^x`

`\int x^2e^x=x^2e^x-2\[xe^x-\int e^xdx\]=x^2e^x-2xe^x+2e^x+C`

Solving calculus math problem – Example: 2

Solve `\int \frac{\sqrt{x^2-4}}{x}dx`

Solution:

`x=2\sec \theta`

`dx=2\sec \theta\tan \thetad\theta`

`\int \frac{\sqrt{x^2-1}}{x}dx=\int \frac{\sqrt{4\sec^2\theta-4}}{2\sec \theta}(2)\sec \theta\tan \thetad\theta=2\int \sqrt{\tan^2\theta}\tan \thetad\theta=2\int \tan^2\thetad\theta`

We must use a trig substitution to get this in a form that is more easily integrable.

`2\int \tan^2\thetad\theta=2\int (\sec^2\theta-1)d\theta=2[\tan \theta-\theta]+C`

Now, we know `\sec \theta=\frac{x}{2} so \tan \theta=\frac{\sqrt{x^2-4}}{2} and \theta=\arcsec \frac{x}{2}.` So

`\int \frac{\sqrt{x^2-4}}{x}dx=2\ [\frac{\sqrt{x^2-4}}{2}-\arcsec \frac{x}{2}\ ]+C=\sqrt{x^2-4}-2\arcsec \frac{x}{2}+C`

Practice Problems for Solving Calculus Math Problem:

1. Solve `\int_0^\infty x^4 e^{-x^3} dx`

`Answer: \frac{2}{9}\Gamma\(\frac{2}{3}\)`

2. Simplify `\int_0^\infty 3^{-4z^2}dz`

`Answer: \sqrt{\frac{\pi}{16\ln3}}`

Consonant Math Definition

Introduction to consonant math definition
Let us discuss about the consonant math definition. The necessary of mathematics learn to the measurement, arrangement, space and modify. The consonant is called as they do not change the number. The consonant math is defined the fixed value of a problem. The example of the fixed value is maximum value of amount, minimum value of amount, dates, cost, title lines error messages. The consonant math numbers normally declare only real number does not used the imaginary numbers. Next we see the consonant math definition.

Consonant Math Definition

The addition math definitions are calculate the total value otherwise sum of the values. There are combine the many numbers. The addition example is 8+1+2=11. The addend are declare the any number, that is 8, 1, and 2 are addend.

The subtract math definitions are single value away from other value. The example is the value is 8 that are subtracting the 5, the answer is 3. The symbol is 8 - 5 = 3.

The multiplication math definitions are calculated the multiplying by the repeated value in addition. The example of multiplication is 4 x 2 = 4 + 4 = 8.

The math definitions of division are splitting into the same parts otherwise groups. The example is the 18 is divided the 3. The division answer is 6.

The math definitions of polygon is the many occurring similar line, all connected to the closed shape in the end to end format.

Other Consonant Math Definitions

The definition of circle is drawing the curve is always the similar distance of the curve, the line from the middle of circle to a position on the circle.

The definition of triangle is the two sides are same in the triangle length. The properties of triangle are vertex, area, and median. Scalene triangle is define the all sides are not equal.

The math definitions are square is each sides are same and to the every internal angels in 90o. The square is simply specified the regular polygon.

Variability in Math

Introduction to variability in math:

In statistics math, variability or variation is nothing but the spread in a variable. Some of the common measures of variability in statistics math are range, variation, and standard deviation. In this article variability in math , we are going to discuss about finding the measures of variability through the example problems, which will be very useful for the math students.

Measures of Variability - Range:

Definition:

Range is the difference between a greater value and a smaller value.

Steps involved to learn range:

Step 1: Arrange the given numbers from ascending to descending order.

Step 2: Identify the larger value in the set of data

Step 3: Identify the smaller value in the set of data

Step 4: We have to find the difference between larger and smaller value.

Example problems:

Example 1:

Find the range for the following set of data:

{ 13 , 14 , 15 , 16 , 17 }

Solution:

Range   =  larger value - smaller value

=  17 - 13

=  4

Example 2:

Find the range for the following set of data:

{ 8 , 14 , 24 , 34 , 45 }

Solution:

Range   =  larger value - smaller value

=  45 - 8

=  37

Measures of Variability - Variance and Standard Deviation:

Definition:

Variance is the mean of the squared deviation from its expected value. The standard deviation is the square root of its variance.

Step-by-step explanation:

Step 1: Mean:Find the average or mean value of the given data

Step 2: Variance:To find the variance, get each difference from the mean value and square the each value and finally

average the result.

Step 3: Standard deviation: Find the square root of variance to calculate the standard deviation.

Example problem:

Find the variance and standard deviation for the given set of data:

{ 2, 9, 3, 1, 5 }

Solution:

Step 1: Mean

Mean   =  ` ( 2 + 9 + 3 + 1 + 5 ) / 5`

=  ` 20 / 5`

= 4

Step 2: Variance

Variance  =   `( (5-4)^2 + (9-4)^2 + (3-4)^2 + (1-4)^2 + (5-4)^2 )/5`

=   `( (1)^2 + (5)^2 + (-1)^2 + (-3)^2 + (1)^2 )/5`

=   `(1+25+1+9 + 1)/5`

=   `37/5`

=  7.4

Step 3:Standard deviation

Standard deviation  =  `sqrt ( 7.4 )`

=  2.72

Rules Algebraic Expressions

Introduction :

In mathematics, an expression is a finite combination of symbols that are well formed according to the rules applicable in the context at hand. Symbols can designate values (constants), variables, operations, relations, or can constitute punctuation or other syntactic entities. The use of expressions can range from simple arithmetic operations like

`3+5 xx ((-2)^(7)- (3)/(2))`

Source: Wikipedia

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Rules for Solving Algebraic Expressions:

Step 1: Group the terms containing the identical variable collectively in algebraic expressions.

Step 2: carry out the operation inside the parentheses for the variable and other.

Step 3: revise the algebra 2 simplify expressions and solving algebraic expressions.

Step 4: To make sure the equation, if there is able to simplify the algebraic simplify expressions and then repeat the step 1 to 4.

Example problems for solving algebraic expressions using rules:

Solving algebraic expressions using rules, 12x+5y+10+10z-3y+6x+10z-2

Solution:

Step 1: Is to group like terms. Group the terms containing the same variable together. Group constants together.

(12x+6x)+(5y-3y)+(10z+10z)+(10-2)

Step 2: Is to carry out the operation inside the parentheses for the variable x.

(12x+6x)=18`xx` x= 18x

Step 3: Is to carry out the operation inside the parentheses for the variable y.

(5y-3y)=2y

Step 4: Is to carry out the operation inside the parentheses for the variable z.

(10z+10z)=20z

Step 5: Is to carry out the operation inside the parentheses for the constants.

10-2=8

Step 6: Is to revise the problem.

18x+2y+20z+8

Since the left over terms are not like terms, the problem cannot be any further.

The correct answer is 18x+2y+20z+8

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Rules for Solving Algebraic Expressions in Order of Operation:

In long math problems with +,-,x,%,(), and exponents in them, you have to identify what to do first. Without follow the same rules, you may get unlike answers. You can easily keep in mind the silly sentence, Big Elephants Destroy Mice And Snails, you can commit to memory the order of operations, and you must follow.

Big                   “B” means Brackets. We need to carry out operation in side parentheses first. 

Elephants          “E” means an exponent, you must calculate exponents next.

Destroy            “D” means division. Begin on the left of the equation and perform all multiplication and divisions in the order in which they appear.

Mice                “M” means multiply

And                 “A” means addition.

Snails              “S” means subtract. For all time on the left hand side and carry out additions and subtractions operation.

Example for solving algebraic expressions using order of operations rules:

To solving 16`-:`4(2-5)+5-2 using order of operations rules in algebra.

Solution:

=16`-:`4(2-5)+5-2

=16`-:`4(-3)+5-2   where 16`-:`4 are `16/4`

=`16/4` `xx`(-3)+5-2

=4`xx`-3+5-2

=-12+5-2

=-7-2

=-9

Answer is -9

Probability Number Line

Definition of probability number line:
The probability number line is one of the major topics in mathematics. Probability number line determines how likely some event will be happening. Probability distribution is consists the list of all the number random variables and their corresponding probabilities.To valid probability must be 0 and 1. And sums of probability number must be equal to 1.

P (A) =n/N

n specifies occurrence event of A

Example of Intersection Probability Number Line :

The reverse or set off an occurrence event A is the occurrence events of not A. It specifies the event A not occurring.

So mathematical expression of probability is given by P (not A) = `1 - P (A).`

An example probability of not rolled a 6 on a 6side die is [1 – (possibility of rising and falling a six)].

So probability is=1-1/6

Answer=5/6

Join probability:

Consider A and B are Events. In this A and B are occur in a single experiments are called as the Join probability. Join probability is some time called as intersection probability. That is denoted as `P (AnnB).` Suppose A and B are not depending means join probability is given in below

`P (A and B) =P (AnB) =P (A) P (B)`

The given probable sums are example for intersection probability.

The 2 coins are tossed. The probability number line for both coin get Heads=1/2*1/2

The Answer is=1/4

If flipping a single coin 3 times means find the probability number line

If 3 times flipping then probabilities are

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

Events are

M =Head on first flip, N =Tails on First Flip, O=Heads on Second flip

S=Three heads, Q=Three Tails, R=zero Heads
Probability number lines are

Between, if you have problem on these topics all prime numbers from 1 to 100, please browse expert math related websites for more help on the prime number theorem.

Example for Union Probability:

Suppose the A event or B event or both events occurred in first time of an experiment that is called as union of events. The union events A and B denoted as P(AUB). If two events are commonly selected then the probability is

`P (A or B) =P (AuuB) =P (A) +P (B)`

The given probable sums is the chance for rolling a die is 1 or 2 on a six-sided die is

`P (A or B) =P (A) +P (B)`

`P (1 or 2) =P (1) +P (2)`

`P(1)=1/6 and p(2)=1/6`

`=1/6+1/6`

`Answer is=1/3`

Find probability for even number obtain when a die was rolled.

Identify the sample space values that is s and number line are given below

`S= {1, 2, 3, 4, 5, 6}`

Write only the even number line in S

`E= {2, 4, 6}`

Use Classical probability formula

`P (E) = n (E) / n(S)`

`P (E) = 3 / 6`

`P (E) = 1 / 2`

Ordered Pair Math

Introduction to ordered pair math:

An ordered pair in math is the pair of two objects which occur in a particular order.

That is, here that order is very important.

Ex: Let 7 and 9 be two numbers. Then (7, 9) is one ordered pair and (9, 7) is another ordered pair. Hence in (7, 9) we call 7 as the first component and 9 as the second component.

Two ordered pairs are said to be equal if they have the same first components and the same second components.

Ex: (- 7, 11) = (- 7, 11) and (5, 8) = (5, 8).

Suppose (x, y) = (4, 3). Then x = 4, y =3.

Now let us see few problems on these concepts.

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Example Problems on Ordered Pair Math:

Ex 1: By using the set A = {a, b}, form all possible ordered pairs.

Solution: Given:  A = {a, b}.

Therefore the ordered pairs are {(a, a), (a, b), (b, a), (b, b)}.

Ex 2: Find all possible ordered pair from X = {1, 2, 3}.

Solution: Given: X = {1, 2, 3}.

Therefore the possible ordered pairs are given by { (1,1), (1,2 ), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3) }

Ex 3: If (3a – 5, 8b +7) = (7, 23), Find the value of ‘a’ and ‘b’.

Solution: Given: (3a – 5, 8b +7) = (7, 23)

`implies` 3a -5 = 7 and 8b + 7 = 23

`implies` 3a = 5+7 and 8b = 23 – 7

`implies ` 3a = 12 and 8b = 16

`implies` a = `12/3` = 4 and b = `16/8` = 2.

Therefore a = 4 and b = 2.

Ex 4: If (3a, - 5) = (a -2, b+3), find the values of ‘a’ and ‘b’.

Solution: Given: (3a, -5) = (a- 2, b+3)

`implies` 3a = a – 2 and -5 = b+3

`implies ` 3a – a = - 2   and -5 – 3 = b

`implies` 2a = - 2   and b = - 8

`implies` a = `-2/2` = - 1 and b = -8.

Therefore a = -1, b = - 8.

Practice Problems on Ordered Pair Math:

1. If (x – 3, x + 2y) = (3x -1, 3), Find the value of x and y.

[Ans: x = -1, y = 5/2]

2. If (3a – 2) 5 – b) = (4, - 1), find a and b.

[Ans: a = 2, b = 6].

Median Even Number

Introduction for Median of numbers:

Median is the center value of the given numbers or allocation in their ascending order. Median is the average value of the two center elements when the size of the allocation is even. Half the numbers in the list are fewer, and half the numbers are greater. To locate the Median, place the numbers you are given in value order and find the center number. But there are two center numbers (as happens when there is an even amount of numbers) then average those two numbers.

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Median of Numbers:

Odd number:

Example:

Find the Median of {8, 2 and 7}.

Put them in order: {2, 7, and 8},

The center number is 7,

So the median is 7.

Even numbers:

Example:

Find the Median of {8, 2, 7 and 3}.

Put them in order: {2, 3, 7, and 8},

The center numbers are 3 and 7;

The average of 3 and 4 is 3.5,

So the median is 3.5.

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Examples for Median of Numbers:

Example 1:

To find the median of numbers 5, 8, 7, 3, 4?

Solution:

Step 1:
Count the total numbers given.
There are 5 elements or numbers in the allocation.

Step 2:
Arrange the numbers in ascending order.
3, 4, 5, 7, 8

Step 3:
The total element in the allocation (5) is odd.
The middle position can be calculated using the formula. (n + 1) / 2.
So the middle position is (5 + 1) / 2 = 6 / 2 = 3.
The number at 3rd position is = Median = 5.

Example 2:

To find the median of numbers 6, 4, 7, 2, 1, 8?

Solution:

Step 1:
Count the total numbers given.

There are 6 elements or numbers in the allocation.

Step 2:
Arrange the numbers in ascending order.

1, 2, 4, 6, 7, 8

Step 3:
The total element in the allocation (6) is even.

As the whole is even, we have to take average of number at n / 2 and (n / 2) + 1

So the position are n / 2 = 6 / 2 = 3 and 4

The number at 3rd and 4th position are 4, 6

Step 4:
Find the median.

The average is (4 + 6) / 2 = Median = 5

Interest Growth Formula

Introduction to Interest growth formula:

Interest growth can calculated in two ways, they are

Simple interest growth

Compound interest growth

Simple interest growth:

The interest that is paid for only the borrowed amount not to interest calculated on the original amount.

Compound interest growth:

The interest that is paid not only for the borrowed amount but also to interest calculated on the original amount.

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Formula to Find Simple and Compound Interest Growth

Formula for calculating simple interest growth:

Interest= P*I*N

Here p is the original amount

I is the rate of interest

N is the number of years.

Formula for calculating compound interest growth:

A = P (1 + `r/q` ) nq  

Here (here the interest is compounded q times a year)

P is amount borrowed

r is the rate of interest

n is the number of year

q is the number time the interest is compounded.

A= P (1 + r) n 

Here the interest is compounded only one time  in a year

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Model Problems on Interest Growth Formula

Example: 1

Find the simple interest growth for the amount of 500$ and the rate of interest is 5% for 3 years

Solution:

Formula for calculating simple interest growth:

Interest= p*I*N

Here p is the original amount = $500

I is the rate of interest = 5%

N is the number of years = 3

Interest = 500*5%*3

Interest = 500*(5/100)*3

Interest = 5*5*3

Interest = $75

So the simple interest growth is $75

Example: 2

Find the amount using compound interest growth for the amount of 100$ and the rate of interest is 3.5% for 5 years if it is compounded 4 times?

Solution:

Formula for calculating compound interest growth:

A = P (1 + `r/q` ) nq

Here

P is amount borrowed = $1000

r is the rate of interest = 3.5%

n is the number of year =5

q is the number time the interest is compounded, q=4

Interest = 1000(1+ `(.035)/4` )20

=$1190.34

Amount after the 5years is $1190.34

Definition of Triangular Pyramid

Introduction to the definition of triangular pyramid:
In this article we see about definition of triangular pyramid. Pyramid is a polyhedron 3 – dimensional geometric shape. The pyramid has 4 – vertices. Out of them 3 are base of the pyramid and one is top of the pyramid.

Types of pyramid:

Square pyramid
Rectangular pyramid
Triangular pyramid
Pentagonal pyramid
These are some of the types of pyramid. In this section we see about definition of triangular pyramid.

Definition of Triangular Pyramid:
A pyramid which has a triangular base is said to be triangular pyramid.

Types of triangular pyramid:

Equilateral triangular pyramid – base of this pyramid is equilateral triangle
Isosceles triangular pyramid – base of this pyramid is isosceles triangle
Scalene triangular pyramid – base of this pyramid is scalene triangle
Volume of a triangular pyramid = `1/3` area of the triangle x length

Volume = `1/3` x `1/2 ` x base x height x length

That is volume = `1/6 xx b xx h xx l`

Let example we see some of the example problems for triangular pyramid.

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Example Problems – Definition of Triangular Pyramid:

Example problems 1 – definition of triangular pyramid:

Calculate the volume of triangular pyramid where the base is 6.5 m, height is 14.6 m and length is 15.4m.

Solution:

Given:

Base b = 6.5 m

Height h = 14.6 m

Length l = 15.4 m

Volume of triangular pyramid = `1/6` x b x h x l

= `1/6` x 6.5 x 14.6 x 15.4

= `1/6 ` 1461.46

= 243.57

Answer: volume of a triangular pyramid = 243.57 cubic meter.

Example problem 2 – definition of triangular pyramid:

Calculate the volume of triangular pyramid where the base is 12 m, height is 20 m and length is 15m.

Solution:

Given:

Base b = 12 m

Height h = 20 m

Length l = 15 m

Volume of triangular pyramid =` 1/6` x b x h x l

= `1/6` x 12 x 20 x 15

= `1/6` 3600

= 600

Answer: volume of a triangular pyramid = 600 cubic meter.

Add Improper Fractions

Introduction about fraction:
If two real numbers are in a / b form, then it said to be a fraction notation,

Here,   a (numerator)/ b (denominator) --- > both ‘a’ and ‘b’ not equal to zero.

Types of fraction:

1. Proper Fraction.

2. Improper Fraction.

3. Mixed Fraction.

Proper Fraction:

In fraction the numerator is lesser than the denominator means, such a fraction is called proper fraction.    

Example:

1/3, 2/5, 5/9.

Improper Fraction:

In fraction the numerator is greater than the denominator means, such a fraction is called improper fraction

Example:

5/3, 7/2, 9/5.

Mixed Fraction:

Mixed Fraction is the mixture of whole number and proper fraction.

Example 2 3/5, 1 ¾

Rules for Adding Improper Fraction:
Step 1: Initial step is the check whether the denominator is same or different.

Step 2: If the denominators are same add the numerators

Step 3: If the denominators are different then we have to make the common denominator by taking the Least common denominator

Step 4: Convert the denominator as a least common denominator

Step 5: Finally add the numerators

Note: When we add two improper fractions the result is also an improper fraction

Problems on add improper fraction

7/5+11/7                

Solution:

Step 1: Here the denominators are different

Step 2: Take the LCD for 5,7

LCD of 5,7 = 35

Step 3: To make the common denominator as 35

Multiply and divide by 7 for the first term and 5 for the second term

7/5*7/7 + 11/7 * 5/5

49/35+55/35

Step 4: Add the numerator,

49+55/35 = 104/35

Hence the answer for 7/5+11/7 =104/35

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Problems on Add Improper Fraction

9/7+14/11

Solution:

Step 1: Here the denominators are different

Step 2: Take the LCD for 7,11

LCD of 7,11=77

Step 3: To make the common denominator as 77

Multiply and divide by 11 for the first term and 7 for the second term

9/7*11/11 + 14/11 * 7/7

99/77+98/77

Step 4: Add the numerator,

98+99/77 = 197/77

Hence the answer for 9/7+14/11 =197/77

Problems on add improper fraction


6/5+12/10

Solution:

Step 1: Here the denominators are different

Step 2: Take the LCD for 5,10

LCD of 5,10 = 10

Step 3: To make the common denominator as 10

Multiply and divide by 1 for the first term

6/5*2/2 + 12/10

12/10+12/10

Step 4: Add the numerator,

12+12/10 = 24/10

Hence the answer for 6/5+12/10 =24/10

Tangent Angle Circle

Introduction to tangent angle circle:

The tangents to a circle are the lines that are from a point outside the circle and intersect the circle at only one point on the circle. When two tangents of a circle are from a single point and angle is formed between the two tangents of the circle. In the following article we will see in detail about the topic tangent angle circle.

More about Tangent Angle Circle:
The tangent of a circle are the lines that intersect the circle at only one point on the circle. If two tangents of a circle arises from a single point outside the circle then there is an angle formed between the tangents of the circle. This angle between the tangents of the circle can be related to the angle covered by the arcs that are formed by the intersection of the tangents with the circle.

Angle between the two tangents from a single point to the circle:

The angle x is the measure between the two tangents is half of the difference between the major arc and the minor arc.

Angle x = (Angle covered by the major arc – Angle covered by the minor arc)/2

Angle between a tangent and a secant from a single point to the circle:

In some cases there can be a tangent and a secant arising from a common point outside the circle in those cases the angle between a tangent and a secant from a point to the circle is half of the difference between the major arc angle and the minor arc angle.

Angle x = (Angle covered by major arc – Angle covered by the minor arc)/2

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Example Problem on Tangent Angle Circle:

1. Find the angle between the two tangents in the given diagram.

Solution:

The angle x = (Angle covered by major arc – angle covered by minor arc)/2

`= (245-115)/2`

`= 130/2`

`= 65` `degrees`

2. Find the angle between the two tangents to the circle in the given diagram.

Solution:

The angle x = (Angle covered by major arc – angle covered by minor arc)/2

`= (275-85)/2`

`= 190/2`

`= 95 degrees`

Practice problem on tangent angle circle:

1. Find the angle x between the tangent and the secant to the circle in the diagram.

Answer: 50 degrees

Real Numbers Integers Rational

Introduction to real numbers integers rational:

A real numbers integers rational number is a number that can be represented as a/b, where an a and b are integers and b ≠ 0. From the definition of a rational number, we see that fraction such as -1/2, 3/4 and 2/5 are rational numbers. If we replace b in a/b by an integer 1, we have a/b = a/1 =a. Hence the integers b = b/1, -8 = -8/1, 0 = 0/1 and so on are real number integers rational.

Properties of the Real Numbers Integers Rational:

In real numbers integers rational the division of two integers does not necessarily result in an integer. Hence, there is a need to include fractions to form a bigger set of numbers known as set of rational numbers. The rational numbers and the irrational numbers completely fill the number line and form the set of real numbers. We can summarise the properties of the rational number as given below. Let a, b and c be any rational numbers.

1) Commutative:

a + b = b + a.

a * b = b * a.

2) Associative:

(a + b)+c = a + (b + c)

(a * b)*c = a * (b * c)

3) Identity:

a + 0 = 0 + a = a

a * 1= 1 * a = a

4) Distributive:

Multiplication can be distributed over addition.

a * (b + c) = (a * b) + (a * c)          

5) Closure:

The system of rational numbers is closed under addition, subtraction, multiplication and division (except by 0).

a + b, a – b, a * b, a / b, (b ≠ 0) are all rational numbers.

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Examples for Real Numbers Integers Rational:

1) Find three real numbers integers rational between 1/5 and 1/3.

Solution:

Let q1, q2, q3, be the three required the rational numbers. Then

q1 = 1/2 (1/5 + 1/3) = 1/2 (3 + 5/ 15) = 1/2 * 8/15 = 4/15

q2 = 1/2 (1/2 + 1/3) = 1/2 (4 + 5/15) = 1/2 * 9/15 = 3/10

q3 = 1/2 (3/10 + 1/3) = 1/2 (9 + 10/30) = 1/2 * 19/30 = 19/60

1/5 < 4/15 < 3/10 < 19/60 < 1/3. 

Hence, three rational numbers between 1/5 and 1/3 are 4/15, 3/10 and 19/60.

2) Find nine real numbers integers rational numbers between 2/5 and 1/2.

Solution:

2/5 = 0.4, 1/2 = 0.5, clearly, we have

0.4 < 0.41 < 0.42 < 0.43 < 0.44 < 0.45 < 0.46 < 0.47 < 0.48 < 0.49 < 0.5

Therefore, nine rational numbers between 2/5 and 1/2 are

41/100, 42/100, 43/100, 44 /100, 45/100, 46/100, 47/100, 48/100, 49/100.

Types of Compound Inequalities

Introduction to types of Compound inequalities:

Compound Inequalities are is defined as the terms of condition availability in the particular inequalities conditions are satisfied in between two or more simple inequalities joined by the terms 'and' or 'or'.

Types of compound inequalities are

There are three types of inequalities

1.  And - this is the inequality that has separated the two inequalities.

2.  Or   - this is the inequality that has separated the two inequalities.

3.  a < bx + c < d – this is the single inequality form.

And, or Types of Compound Inequalities with Examples:
And compound inequalities examples:

This is like an intersection of the particular two solutions. It is easy to solve. To find the intersection of particular two sets.

Ex:1    2x + 2 < 8 and 3x – 4 > -4

Sol:          2x < 8 – 2          3x > -4 + 4

2x < 6                3x > 0

x < 3                 x > 0          

The intersection and the inequalities of the solution is (x | 0 < x < 3)

Ex: 2 Solve    4x +5 > 1 and 3x – 4 < -1

Solution:4x > 1 – 5 and 3x < -1 + 4

4x > -4     and 3x < 3

x > -1      and   x < 1        

The intersection and the inequalities of the solution is ( x| -1 < x < 1)

Or compound inequalities :

This type is like a union of the particular two solutions inequality. It is easy to solve. To find the union of particular two points.

Ex: 2x - 3 > 7  or  3x – 2 < -5

2x > 4     or   3x < -3

X > 2    or    x < -1   

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Types of Compound Inequalities Form:

Single compound inequalities form:


This type inequalities are the as long as the variable has to be appeared only in the middle part of the variable in their inequality.


Ex: -2 < 5x -7 < 3


Sol:Given       -2 < 5x – 7 < 3                        add the 7 both side of 3 parts


5 < 5x < 10                            divide each of the 3 parts by 5


1 < x < 2


The answer is(x | 1 < x < 2)