Multiple Coefficients of Determination

In this topic we will discuss about multiple coefficients determination. The multiple coefficients are the coefficients that have the many term in an expression. In a variable, a coefficient is a number in front of a variable. For example, in the expression `2x^3-7x+19` , the coefficient of the `2x^2` is 1 and the coefficient of the x is -7. The third term, 19, is known as a constant. Here we see about multiple coefficients and problems.

Additional Information about Multiple Coefficients of Determination:

In an expression the coefficients may be positive or negative or zero. When addition or subtract the polynomials, just we want to add or subtract the coefficients in the same terms. The Coefficient in an expression is the number which is multiplied by one or more variables or powers of variables in the term. The following some of the related terms in multiple coefficients determination.

1) Variable

2) Expression

3) Polynomial

4) Term
Example Problem for Multiple Coefficients of Determination:

Multiple coefficient of determination – Example 1

Identify the number of terms in the expression, 2x2 + 4x + 7y + 6xy + 5xy2 + 3x2y

Solution:

Step 1:

When an expression is writing as an addition, the parts that are added are the terms of the expression.

Step 2:

2x2 the variable has the coefficient of 2.

4x the variable have the coefficient of 4.

7y this variable has the coefficient of 7.

6xy this variable’s having the coefficient of 6.

5xy2 this variable’s having the coefficient of 5.

3x2y this variable’s having the coefficient of 3.

Therefore, there are multiple terms in the expression.

Multiple coefficient of determination – Example 2

2) Identify the number of terms in the expression, x3 + 4x2 + 3x + 6y3 + 2y2 + 7y

Solution:

Step 1:

When an expression is writing as an addition, the parts that are added are the terms of the expression.

Step 2:

x3 the variable having the coefficient of 1.

4x2 the variable having the coefficient of 4.

3x this variable have the coefficient of 3.

6y3 this variable’s having the coefficient of 6.

2y2 this variable’s having the coefficient of 2.

7y this variable’s having the coefficient of 7.

Therefore, there are multiple terms in the expression.

Symmetric Equation of a Line

There are several ways to denote a distinct line in R3. Symmetric equation is one such way of representing distinct lines. Initially to deal with symmetric equation familiarity with parametric equation is necessary. Obviously this notation is considered compact too. Getting into the concept of symmetric equation, it sets each component of the line equal to a common parameter. Following this all the components are set equal to one another. Knowledge in vector equation of line and parametric equation would help us in solving tutorials.

More about Symmetric Matrix Equation of a Line:

The parametric equation of the line is stated as follows:

`(x-a_x)/(m_1)=(y-a_y)/(m_2)=(z-a_z)/(m_3)`

To arrive at this equation, each expression in the parametric equation of the line should be treated equal. For example

`x: a_x + (t)(m_1)`

It should be taken as

`x=a_x + (t)(m_1)`

The expression should be solved for each value of t and ultimately the resultant expression would be equal to the constant t and therefore would be equal to each other. Interestingly this expression of line violates one of the common doctrines of mathematics which limits us to use only one equality sign per expression. It is this factor that helps us to use this mode of representation in fewer or more dimensions. In case we are to use it in R2 form then the z expression would be ignored thus containing only one equality sign per expression. Similarly if we are expected to express a line that is parallel to a co ordinate plane then the term of the axis to which it is parallel to is also ignored.
Example Problems – Symmetric Equation of a Line:

Example 1 – Symmetric equation of a line:

Find the symmetric equation of the line through point (6,-7,20) & perpendicular to the plane 2x+3y-6z-8=0.

Solution:

Let us consider a normal vector n

2x + 3y - 6z - 8 = 0 is n = <2 -6="-6" 3="3">

The vector n is also the directional vector of the line thru point P(6, -7, 20) and perpendicular to the given plane.

The equation of the line L is:

L = P + tn = <6 -7="-7" 20="20"> + t<2 -6="-6" 3="3">

L = <6 -7="-7" -="-" 20="20" 2="2" 3t="3t" 6t="6t" t="t">

By solving for t convert the equation of the line to symmetric form

L:

x = 6 + 2t

y = -7 + 3t

z = 20 - 6t

Symmetric form of equation.

`t` =` (x - 6)/2` = `(y + 7)/3` = `(z - 20)/-6`


Example 2 – Symmetric equation of a line:

Find the symmetric equation of the line through point (2, 3, 4) & perpendicular to the plane x+2y-3z-4=0.

Solution:

Let us consider a normal vector n

x+2y-3z-4=0 is n = <1 -3="-3" 2="2">

The vector n is also the directional vector of the line thru point P(2, 3, 4) and perpendicular to the given plane.

The equation of the line L is:

L = P + tn = <2 3="3" 4="4"> + t<1 -3="-3" 2="2">

L = <2 -="-" 2t="2t" 3="3" 3t="3t" 4="4" t="t">

By solving for t convert the equation of the line to symmetric form

L:

x = 2 + t

y = 3 + 2t

z = 4 - 3t

Symmetric form of equation.

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

Solving for x With Polynomials

Polynomial defined as the function p(x) of the form p(x) =a0 +a1x+a2x2+……….anxn. Where a0, a1…an real numbers and n is the non negative integer is called polynomial in x over reals.For example 4x2-7x+3 is a polynomial over integers. If one of the powers of x in p(x) is either a negative integer of a fraction (either positive or negative), then p(x) is not a polynomial. For example x+2/x  is not a polynomial. The highest exponent of the variable in polynomial is called the degree of the polynomial.. Here we are going to study about how to solving for x with polynomial and its example problems.

Solving for X with Polynomials - Example Problems

Example: 1

Solve for x in the following polynomial expression 3x+5+6x +7 = 3x+4

Solving steps:    In the left hand side combine the like term first

3x+ 6x+ 7 + 5 = 3x + 4

9x + 12 = 3x + 4

Add both sides -4 we get

9x + 12 - 3 = 3x + 4 - 4

In right hand side 4 - 4 will be cancelling

9x+ 9 = 3x

Add both sides -3x

9x -3x + 9 = 3x -3x

6x + 9 =0

Add both sides -9 we get

6x + 9 -9 = -9

6x = -9

Divide both sides 6

x = - 9/6

The simplest form is - 3/2

Therefore the value of x = - 3/2

Solving for X with Polynomials - Example: 3

Solve for x in the following polynomial x2 + 9x +18 =0

Solving steps:

First we have to find the factor for a given polynomial

We can write

x2 + 9x +18  = (x+3)(x+6)

These are the two factors the equation

Now we solve the both equation.

Both terms equating to zero we get

First x+3=0

Add both sides -3 we get

x = -3

Next term is x+6 = 0

Add both sides -6

x = -6

Therefore the value of x is -1,-6

Polynomials Algebra Tiles

In mathematics, a polynomial is an expression of finite length constructed from variables (also known as indeterminates) and constants, using only the operations of addition, subtraction, multiplication, and non-negative, whole-number exponents. For example, x2 − 4x + 7 is a polynomial, but x2 − 4/x + 7x3/2 is not, because its second term involves division by the variable x and because its third term contains an exponent that is not a whole number. In this article we shall discuss about polynomials algebra tiles. (Source: Wikipedia)


Sample Problem for Polynomials Algebra Tiles:

Solved polynomials algebra tiles problem:

Example 1:

Factorize the polynomial 2x3 – 10x2 – 24x+ 72

Solution

Sum of the coefficients of terms: 2–10–24 + 72 = 40 ≠ 0.

Therefore the value (x–1) is not a factor.

Sum of the coefficients of even degree terms = –10 + 72 = 62

Sum of the coefficients of odd degree terms = 2 – 24 = –22

Since they are not equivalent we estimate that (x + 1) is also not a factor. Let us check whether x – 2 is a factor. By synthetic division method

2 | 2       -10     -24        +72

|

|          +4      -12          -72
________________________

2        -6     -36     |    0

_________________________

Since the remainder is 0, (x – 2) is a factor. To find other factors

2x2 – 6x – 36 = 2x2 – 12 x + 6x –36

= 2x(x –6) + 6(x–6) = (2x + 6) (x – 6)

Therefore, x3 – 5x2 – 12x + 36 = (x–2) (x–6) (x+3)


Example 2:

Factorize 4x3 + 2x2 – 10x + 4

Solution:

Since the addition of the coefficients of all the terms: 4 + 2 – 10 + 4 = 10 – 10 = 0

We guess that (x – 1) is a factor.

By synthetic division,

1 | 4         +2        -10        +4

|

|              4        +6        -4

________________________

4         6         -4     |   0

_________________________

Remainder is 0. Quotient is 4x2 + 6x – 4

To find other factors, factorize the quotient,

4x2 + 6x – 4 = 4x2 + 8x – 2x – 4

= 4x (x + 2) – 2 (x + 2)

= (x + 2) (4x – 2)

∴ 2x3 + x2 – 5x + 2 = (x – 1) (x + 2) (2x – 1)
Practice Problem for Polynomials Algebra Tiles:

Factorize 2x2 + 3x -2

Answer: (x + 2) (2x – 1)

Factorize X2 – 3x – 18

Answer: (x + 3) (x – 6)

Add and Subtract Fractions Calculator

Add and subtract calculator which is used for finding the addition and subtraction for various value of a fraction. Calculator is a process of calculating fractions with different values.In general fraction contains two parts numerator and denominator. the upper part is the numerator and lower part is a denominator which depends upon the this two parts. there are types of fractions such as proper,improper fraction and mixed fraction.

Rules for Add and Subtract on Fractions Calculator:

fraction calculator

First choose the operations in the column shown(add,subtract,multiply and divide)
then,fraction #1 enter the fraction value and in the fraction#2 then press the calculate
We get final result in column below the calculate.
For the next operation press reset and follow the steps.

Fraction shown above will be the calculator for the fraction addition and subtraction steps in the calculator will be explained below.
Addition on fractions:

For adding same denominator fractions just add all the numerator and keep same denominator.


For addition with different denominators fraction:

Find the l.c.m. for all the denominators given .
Change into equivalent fraction with the same LCM denominator
By taking  the LCM common in  the denominator and add all the numerators.


Subtraction on fractions :

For  subtract same denominator fractions subtracting  all numerator and keep same denominator.


For subtraction with different  denominator:

Finding  l.c.m. for all the denominators given.
Change equivalent fraction with same l.c.m denominator and then subtract


Problems for Add and Subtract Calculator:
Example 1:
Add   1/4 and 2/4

Solution:
here we have the same denominator so, just add the numerator keep the denominator same.
1/4 +2/4 = (1+2) /4

=3/4

Example 2:
Add  4/5  and 1/3

Solution :
The LCM of 5 and 3 is 15.
Therefore,4/5+1/3 = (4xx3)/(5xx3)+(1xx5)/(3xx5)

=12/15+5/15

=17/15

Example 3:
Add 2 4/5 and 3 5/6

Solution:
The given fraction is a mixed fraction first change into proper fraction and add
 2 4/5 +3 5/6=(2xx5)+4/5+(3xx6)+5/6=14/5+23/6

so, the denominator is different LCM of 5,6 =30
=(14xx6)/(5xx6)+(23xx5)/(6xx5)

=84/30 +115/30

=119/30

Example 4:
Subtract  3/5 -1/5

Solution:
here, same denominator just subtract numerator and keep denominator is same.

 3/5-1/5 =(3-1)/5
=2/5
Example 5:
subtract 4 2/5-2 1/5

Solution :
The given fraction is a mixed fraction first change into proper fraction and subtract
(4xx5)+1/5 -(2xx5)+1/5

so, 21/5 -11/5

=(21-11)/5

=10/5

Example 6:
Subtract  3/4   from  5/6

Solution :
We need to find equivalent fractions of 3/4 and 5/6 , which have the same denominator.
This denominator is given by the LCM of 4 and 6. The required LCM is 12.

Therefore, 5/6-3/4 =(5xx2)/(6xx2)-(3xx3)/(4xx3)
=10/12-9/12

=1/12

Definition of Subset Learning

We come across set of different kinds in everyday life. A football team is set of players, a class is a set of students, a set of books in mathematics and so on. Basically, the set is an undefined term.  Anyhow, it can be a well defined  collection of objects. Here the word 'objects' has been taken in wider and broader sense. Mathematically, the numbers, words, letters, signs, symbols, thoughts, etc. all are the objects. Let us learn about a set and  definition of a subset in this article.

Learning Definition of a Subset

Definition

If each element of A is an element of B, then the set A is said to be a sub set of the set B.

Symbolically, we represent this by A `sub` B and read it as ' A is a subset of B'.

Thus A `sub` B `hArr` (  x `in` A   `rArr` x `in` B ).

If A `sub` B and A `!=` B then the set A is said to a proper subset of B.

learning examples of subset

1. {1, 2, 3, 4, 5 } `sub` { x : x is a natural number }

2. { a, b, c, d } `sub` { a, b, c, d, e}

3. Set of all vowels is a subset of set of all alphabets

4. A = { set of all integers}

B = { set of all positive integers}

B is a subset of A.

5. T = { x : x is a student of class 9 in your school }

W = { x : x is a student in your school}

T is a subset of W.



Definition of Subset Learning - Points to Remember

1. `phi` `sub` A ( since `phi` has no elements, it is a subset of every set)

2. A `sube` A ( every set is a subset of itself)

3. A `sube` B and B `sube` A `hArr`    A = B

4. A `sub` B and B `sub` D `=>`   A `sub` D

5. If A `sub` B, then x `!in` B `rArr` x `!in` A.

Adding Exponents Worksheet

Exponents are nothing but multiplying the number by itself.  For example, y2 is the same as y * y. Exponent denotes the number of times the number needed to be multiplied by itself. An exponent is nothing but a superscript, or small number that can be written at the top right edge of a number, variable, or set of parentheses. In this article we shall discuss the adding exponents worksheet.

Problems for Adding Exponents Worksheet:

Adding exponents worksheet:

Problem 1:

To find x3y4. x5y3

Solution:

Rearrange the factors and multiply the exponent terms, using the rule

Here,

am.an = am+n

= (x3. x5)( y3. y4)

= x8. Y7

X and y are different terms,

So the final result is x8. Y7

Problem 2 for Adding exponents worksheet:

To find  (32)3

Solution:

(32)3 = 93 =  729 using power rule,

The answer is 729

Problem 3:

To find  (22)3

Solution:

(22)3 =(2x2)3 =  43=64 using power rule,

The answer is 64

Mixed variables for multiply exponents worksheet:

Have a mix of variables:

Example 1:

=xy2z y3z

=x y2+3z1+1

=xy5z2

so the result is =x3y5z2

Example 2:

=x3y3z3 .  x2yz

=x3+2 y3+1 z3+1

=x5y4z4

so the result is =x5y4z4

With constant examples:

Example 1:

=5xyz .4xyz

=(5 4)x1+1 y1+1 z1+1

=20x2y2z2

The result is =20x2y2z2

Example 2:

=7xyz.3 x2yz

=(73 ) x1+2y1+1z1+1

=21x3 y2z2

The result is  =21x3 y2z2
More Examples for Multiply Exponents Worksheet:

Example 1:

=5xyz `xx` 4xyz

= (5`*` 5)x1+1 y1+1 z1+1

=25x2y2z2

The result is =25x2y2z2

Example 2:

=7xyz `xx` 4 x2yz

= (7`*` 4) x1+2y1+1z1+1

=28x3 y2z2

The result is  =28x3 y2z2

Example 3:

=5xyz `xx` 7xyz

=(7`*` 5)x1+1 y1+1 z1+1

=35x2y2z2

The result is =35x2y2z2

Example 4:

=7xyz  ` xx` 2 x2yz

= (7`*` 2 ) x1+2y1+1z1+1

=14x3 y2z2

The result is  =14x3 y2z2