Constant Rate of Change Math

Introduction to constant rate of change math:

Rate on which reliant output modify by value to vary within independent input is describing the rate of change. Functional association be usually indicated through y = f(x). Alteration within y is indicated through ?y also alteration within x is indicated through ?x. Consequently, as of the function y = f(x), we contain y + ?y = f(x + ?x).  In addition to constant rates of alteration are known through method, ?y / ?x.

Constant Rate of Change in Math:

In math consider the function is the linear function that is graph of y also x is a line after that the function is known through y = mx + x. In linear function rates of change is a constant which is known through its slope, m. Slope of the line can be find out through the following

m = ?y / ?x =  (y2 – y1/(x2 – x1)       

In math consider the function is non-linear then its rates of alteration are dissimilar on dissimilar points. Therefore, for such functions derived are estimated through distinguish the function toward get the rates of modify. Derivative of y through value to x can be written as dy/dx.

Examples for Constant Rate of Change Math:

Example 1

Solve the rate of change for the line to exceeds through points (5, 4) and (7, 10)

Solution:

Here, x1= 5, x2=4, y1=7 and y2=10

so, slope m = (y2 -  y1)/(x2- x1)

m = (10 - 4)/(7 - 5) = 6/2 = 3

Consequently, rate of alteration otherwise slope of line is 3.

Example 2

Solve the rate of change of function, f(x) =4x2 + 10x at x = 6

Solution:

f(x) =4 x2 + 10x

Derivative f '(x) = 8x + 10

Substitute x = 6 in f '(x) which gives,

f '(6) = 8*6 + 10 = 48 + 10 = 58

Therefore, rate of change of function f(x) at x = 6 is 58.

Example 3

Solve the rate of change of function, f(x) = 3x3 + 3x2 - 7x at x = 3

Solution:

f(x) =3x3 + 3x2 - 7x

Derivative f '(x) = 9x2+6x -7

Substitute x = 3 in f '(x) which gives,

f '(6) = 9(32)+6(3) - 7 = 81 + 18-7 = 92

Therefore, rate of change of function f(x) at x = 3 is 92.

Precalculus Math Final Help

Introduction to Precalculus Math Final Help:

The Precalculus is one of the foundation class of mathematics. The topics involved in Precalculus test are real  complex numbers, binomial theorem, composite & polynomial functions, vectors, parametric equations, polar coordinates, matrices, rational functions, solving inequalities & equations and trigonometry. In this article we shall discuss about the examples involved in Precalculus math final help. The following are the examples involved in Precalculus math final help.

Precalculus Math Final Help Problems:

Example 1:  Express the real and imaginary parts for 28 - i `sqrt(14) `

Solution:        Let z= 28 - i `"sqrt(14)`      

Re(z) = 28     Im(z) = `sqrt(14)`

Example 2: For what value of m the vectors `veca` and `vecb` are perpendicular to each other

`veca` =m`veci` + `vecj`` + ``veck`   `vecb`= 7`veci` - 9 `vecj` + 2 `vecj`

Solution: Given `veca` `_|_``vecb `

So, `veca` `vecb` = 0

=> (m`veci` +  `vecj`` + ``veck`) . (7`veci` - 9 `vecj``veck` )

=>  7m - 9 + 2

=>  7m  = 7

m =`7/7`  =1

Example 3: Solve for x,  ln(x-3) + ln(x-2) = ln(2x -6 )

Solution:   Simplify left side, So

ln(x-3) + ln(x-2)= ln(x - 3) (x - 2)

Therefore the given equation can be rewritten as

ln(x - 3)(x - 2) =  ln (2x - 6)

Taking exponent on above equation

eln(x - 3)(x- 2) = eln(2x - 6)

Simplifying the above equation

(x - 3) (x - 2)  = (2x - 6)

x2  - 5x + 6 = 2x - 6

x2  - 7x +12 = 0

Factoring the above equation, we get

(x - 4) (x - 3) = 0

When products of two factors are equal to 0, then one of the factors must be equal to 0.

So, x-4 = 0 => x=4

x-3 = 0 => x=3

Here x = 4 & 3 is the solution.

Example 4:   If f(x) = 14x-1, find f-1(y)?

Solution:Let y = 14x - 1

The given equation can be rewritten as

y + 1 =14x

=> x = y + 1
14

Therefore the f-1(y) = y + 1
14

Example 5: Write the given number in complex form `sqrt(-69)`

Solution: `sqrt(-69)` = `sqrt((-1)(69))`

= `sqrt(-1)` * `sqrt(69)`

= i `sqrt(69)`

Precalculus Math Final Help Practice Problems:

Problem 1:Write the given number in complex form `sqrt(-1415)`

Answer: = i `sqrt(1415)`

Problem 2:Write the real and imaginary parts for 89 - i `sqrt(75) `

Answer:   Re(z) = 89     Im(z) =`sqrt(75)`

Problem 3:   If f(x) = x-9, find f-1(y)?

Answer: y + 9

Radical Perfect Prime Numbers

Introduction to radical perfect prime numbers:
A radical perfect prime number is a symbol and it’s denoted as root of a number, the thought of a radical the root of a number can best be understood by first tackling the design of exponentiation, or raising a number to a given power. We point to a number raised to the nth power by writing xn. This expression denotes that we are multiplying x by itself n number of times. Example for radical perfect prime numbers, v5 =2.236, and v3 = 1.732. a radical perfect prime number is a number that has exactly 2 factors...1 and itself , and one thing is 1 is not a prime number ,so  that 2,3,5,7,11,13,17,19,23 etc  are all primes.

Examples for Radical Perfect Prime Numbers:

Addition for radical perfect prime number:

Example 1: Find the value of 2v3 +3 v3

Solution:

=v3 (2+ 3)
=5v3

Here the root is square root, only take the base are same so that we can add the roots easily.

So the result is =5v3

Example 2: Find the value of v11 +6 v23+5v11 +7 v23

Solution:

= v11 (1+5)+ v23 (6+7)

= v11 (6)+ v23 (13)

= 6 v11 + 13 v23

Here the root is square root, the base are same in different terms so that we can add the roots easily.

So the result is 6 v11 + 13 v23

Examples for different radical perfect prime numbers:

Example for square root:

v3 = 1.732

v5 = 2.23

v7 =  2.64

v11 = 3.31                              

Some other Examples for Radical Perfect Prime Numbers:

Multiplication for radical perfect prime numbers:

Example 1:

Solution:

To find multiplication of v3x . v1y    

= v(3 x y)            

=  v3x y   

The result is =  v3x y    

Both the radicals have an same index of two, so we are multiply mutually under one radical keeping the 2 as its index number.

Example 2:

Solution:

To find multiplication of v(23xy2). vy    

= v(23 x y3)            

= v(23 x y3)

The result is v(23 x y3)

Both the radicals have an same index of two, so we are multiply mutually under one radical keeping the 2 as its index number.


Factorization for Radical Perfect Prime Numbers Examples:

Example 1:

To find the value of v75

v75 = v5*3*3

= 3v5

Radical perfect prime factors for v75 = 3v5

So the result is in the form of prime numbers and radical.

Integral Trapezoidal Rule

Introduction to integral trapezoidal rule :

The integral trapezoidal rule is a method which has been used to calculate the definite integral that has been given by

a
∫ f(x) dx
b

A definite integral is defined as the integral which has both the upper and the lower limits. In this topic we are going to see about the integral trapezoidal rule. The integral trapezoidal rule is also called as the trapezium rule or the trapezoid rule.

Integral Trapezoidal Rule:

The trapezoidal rule has been used by considering the region that has been under the graph of the given function f(x) as a trapezoid and then finding its area.

a
∫ f(x) dx ≈ (b-a) f(a)+f(b)
b                        2

The next step is to split the interval of the integration (a, b) into a number of smaller intervals and then apply the trapezoidal rule on every subintervals. This is given by

a                                            n-1
∫ f(x) dx ≈ (b-a)   (f(a)+f(b))/2   + ∑ f(a+k((b-a)/n)
b                 n                         k=1

This integration can also be written as

a
∫ f(x) dx ≈ ((b-a)/2n)(f(x0)+2f(x1)+2f(x2)+…+2f(xn-1)+f(xn))
b

Where the value of xk = a + k(b-a)/n, for the values of k = 0, 1, …. N

Between, if you have problem on these topics Derivative of Parametric Equations, please browse expert math related websites for more help on Integration by Parts Table.

Error Analysis of the Integral Trapezoidal Rule:
The error of the integral trapezoidal rule is given by the difference of the integral and the result.

b                                                n-1     
Error = ∫ f(x) dx -  b-a   (f(a)+f(b)/2) + 2   + ∑ f(a+k((b-a)/n)
a                 n                             k=1

This can be simply written as

Error = - (b-a)3    f′′(ξ)
12n2

Integral trapezoidal rule for non uniform intervals:

Let us consider the given values are x1 ….xn and y1,…..yn, then according to the integral trapezoidal rule

n
∫ydx ≈ ½ ∑ (xi – xi-1)(yi+yi-1)
i=2

The integral trapezoidal rules are said to be exact when the periodic functions are said to be integrated with their periods and the trapezoidal rule converges very rapidly for the periodic functions. When the function is said to be periodic and it integrates over one complete period then there exists a number of graphs which makes the error of the trapezoidal rule to be cancelled.

Integers Rational Irrational Numbers

Introduction to integers rational irrational numbers:

Let us study about integers rational irrational numbers. Integer numbers are commonly seemed as the set of numbers that starts from infinite negative side to the infinite positive side.
Rational numbers are defined as numbers that can express themselves in the form of ratios as `m/n` where ‘n’ should not be equal to equal to ‘0’.
The numbers which cannot express themselves in the simple ratio are named as irrational numbers.

Integers Rational Irrational Numbers:

Example 1: Add the following numbers: 0.27, `3/2` , 1.52, -3, 2.5, 7.


Solution:

The given numbers are 0.27, `3/2` , 1.52, 3, 2.5, 7.
To add the given numbers the following steps are to be considered:
= 0.27 + `3/2` + 1.52 - 3 + 2.5 + 7

= 0.27 + 1.3 + 1.52 - 3 + 2.5 + 7 (do conversions of the fractions to decimal numbers)

= 9.59

Example 2: Subtract the following numbers by using same given order: 3.9, `4/5` , 0.13, 4, 1


Solution:

The given numbers are 3.9, `4/5` , 0.13, 4, 1.
To subtract the given numbers the following steps are to be considered:
= 3.9 – `4/5` – 0.13 – 4 – 1.

= 3.9 – 0.8 – 0.13 – 4 – 1. (do conversions of the fractions to decimal numbers)

= -2.03

Example 3: Do the multiplication process for the following numbers:  1.1, 0.4, 3, 9, `6/5` .


Solution:

The given numbers are 1.1, 0.4, 3, 9, `6/5` .
To add the given numbers the following steps are to be considered:
= 1.1 * 0.4 * 3 * 9 * `6/5` .

= 1.1 * 0.4 * 3 * 9 * 1.2 (do conversions of the fractions to decimal numbers)

= 14.256

Example 4:Do the division process for the numbers: `9/4.5` .


Solution:

= `9/4.5`
= 2


Exercises:

Add the following numbers: 0.2, 6, `11/5` , 4 (Answer: 12.4)
Subtract the following numbers by using same given order: 7, `5/8` , 1.3, 2. (Answer: 3.075)
Do the multiplication process for the following numbers:  0.09 0.3, `15/2` , 1, 7 (Answer: 1.4175)
Do the division process for the numbers: `2/0.5` (Answer: 4)

Adding Radicals Calculator

Introduction to adding radicals calculator:

Radicals:

The symbol used to indicate root of a number is known as radicals.  The square root, cube root can be expressed by using the radical symbol.  The radicals also mean that the root.  To understand the radicals, first we need to know about exponentiation. That is exponential xn = x * x * x * x …….n times x. The root of xn can be expressed as root (n)(x^n) = x .

The radical symbol is,

Fig(i) Radicals

Calculator:

A calculator is a small (often pocket-sized), usually inexpensive electronic device used to perform the basic operations of arithmetic.(source : Wikipedia).

Fig(ii) Adding radicals calculator

In this article we are going to learn about how to adding radicals using calculator.

Rules to Adding Radicals :

Before we get into problems , we need to know the rules for adding radicals.

Rules :

We can add the radicals with the same index and radicand . For example the radicals of index 2 can only add. We can add the radical of index 2 with the radical of index 3.

For example , `sqrt(4)` + `sqrt (4)` = ( 1 + 1) `sqrt(4)`

= 2 `sqrt(4)`

Let us see some problems on adding radicals by using calculator.

I am planning to write more post on Factorization Calculator, Equivalent Fractions Calculator. Keep checking my blog.

Problems on Adding Radicals Calculator:

Problem 1:

Add the radicals `5sqrt(36)` + `2sqrt(81) `

Solution:

Given, `5sqrt(36)` + `2sqrt(81) `

We need to add the given radicals.

Before that we can simplify the given radicals.

We know that, `root(n)(x^n) ` = x

`sqrt(36)` = `sqrt(6^2)` = 6

`sqrt(81)` = `sqrt(9^2) ` = 9

`5sqrt(36)` + `2sqrt(81)` = 5*6 + 2*9

= 30 + 18

= 48

Answer: 5`sqrt(36)` + 2`sqrt(81)` = 48

Problem 2:

Add the radicals `8sqrt(8)` + `sqrt(8) `

Solution:

Given, `8sqrt(8)` + `sqrt(8) `

We need to add the given radicals.

Before that we can simplify the given radicals.

We know that,` root(n)(x^n)` = x

`sqrt(8)` = `sqrt(4 * 2 )`

= `sqrt(4)` * `sqrt(2)`

= `sqrt(2^2)` * `sqrt(2)`

= `2sqrt(2)`

`sqrt(8)` = `2sqrt(2)`

8`sqrt(8)` + `sqrt(8)` = 8 * `(2sqrt(2)) ` + 2`sqrt(2)`

= 16 `sqrt(2)` + 2`sqrt(2)`

= (16 + 2)`sqrt(2)`

=  18`sqrt(2)`

Answer: 8`sqrt(8)` + ` sqrt(8) ` = 18`sqrt(2)`

Problem 3:

Add the radicals 3`root(3)(9)` + 2`root(2)(8) `

Solution:

Given,3`root(3)(9)` + 2`root(2)(8)`

We need to add the given radicals.

Here , given two radicals are has two different index as well as different radicand.

So we can add it further.

My Previous Blog :- http://online-math-problems-solved.blogspot.in/2012/10/solving-literal-equation.html

Polynomials in One Variable

Introduction to polynomials in one variable
Polynomials in one variable, here we need to understand each terms very clearly. Polynomial here means, a mathematical expression consisting of  a sum/subtraction/multiplication but not division of variables , constant sand exponents.One variable means  expression consisting of one letter. Polynomial with one term is called as a "Monomial", two terms as "Binomial" and three terms as "Trinomials". Terms can be separated by addition sign or subtraction sign only but not by multiplication sign.So we can say in otherwords that polynomial is a sum of monomials . And each monomial in polynomial is called as a Term.


Examples of Polynomials with One Variable
Let us see some examples to understand better the polynomials with one variable:

1. 4x

here we have one term only with single variable x

2. 2x^3 + 4x^2 - x

Here we have three term with one variable x.

3. 2x^2

Here we have one term with variable x and the exponent as 2.

Polynomial the word poly means "many" and the word nomial means "terms". so polynomial means many terms. But we are learning here polynomials with one variable.

In polynomial we have constants , variables and exponents. we need to understand these with examples.

Constants can be any number as 3 , 2 , -20 , 3/2 etc.

Variables can be any letter as x , y , z , a , b etc

Exponents will be always positive integers such as 2x^2 , 2a3, 4y5 etc

Algebra is widely used in day to day activities watch out for my forthcoming posts on what s an algebraic expression and what is an example of an algebraic expression. I am sure they will be helpful.

Degree of a Polynomial with One Variable

The largest exponent of the variable is called the degree of a polynomial with one variable.

Let us take an example to understand better:

1. 2x4 + 4x^3 - x^2 + 8

This is a polynomial with one variable with the highest degree as 4 which is also called as the highest power or exponent of the given variable.

2. 3a^2 - 5

This  is a polynomial with one variable with degree 2.

3. x

Here the degree is 1. A variable without a power is considered to be as 1 always.

Standard Form of a Polynomial in One Variable
Standard form means writing the polynomial in one variable with the highest degree first.

Let us take an example:

3x^3 - 2x^2 +5

here the polynomial in one variable is written on that order of degree( highest degree first).

It is not necessary to write it in standard form , but it helps us in doing simplification.

Let us take few more examples:

1.Add: 3x + 4x^2+5 and 8x^2 +2x

writing it in standard form

=(4x^2+3x+5) + (8x^2+2x)

add the like terms

=4x^2+8x^2+3x+2x+5

=12x^2+5x+5

2. subtract: (3a3+ 8a^2+3) -(2a^2- 3a3+8)

write it in standard form

=(3a3+8a^2+3)-(-3a3+2a^2+8)

Reverse the sign

=3a3+8a^2+3 +3a3-2a^2-8

now add/ subtract the like terms

=3a3+3a3+8a^2-2a^2+3-8

=6a3+6a^2-5

3. Multiply: 3x^2 by 2x^3

Multiply the constant and add the exponent of the variable

=6x5

Conclusion

Polynomials in one variable is easy to simplify.

If we add / multiply polynomials in one variable we get the results as a polynomial itself.

Always while adding polynomials simply add the like terms( variable with same degree) together.

In case of multiplication , multiply the constants first then the variable and then add the degree of the variable.

While doing the subtraction change the sign to reverse sign of each term you are subtracting.