Non Linear Relationship

Non linear relationship deals with solving the linear equation problems which contain the polynomials. In non linear relationships the quadratic equation problems is also to be solved. Non Linear relationship has the relation with the families of vectors called vector spaces or non linear spaces, and with functions contains two vectors such as one input vector and output vector, according to certain rules. The example problems are discussed below with detailed solution which explains the non linear relationship.



Non linear relationship example problems:

Example 1:

Evaluate the non linear relationships for the given equation

y 2 - 24 y + 95 = 0

Solution:

In order to factor the above expression, we write the equation y 2 - 24 y + 95 in the form factored

y 2 - 24 y + 95 = (y + a)(y + b)

Hence that the sum of a and b is -24 and their product is 95. The numbers that satisfy these conditions are - 19 and - 5. Hence
y2 - 24 y+ 95 = (y- 19) (y- 5)

Substitute into the original equation and solve.
(y- 19)(y- 5) = 0

(y - 19)(y - 5) is equal to zero if
y - 19 = 0

or

y - 5 = 0

Solve the above equations to get the solution

y = 5

or

y = 19

Y = 5 or 19



Example 2:

Evaluate the non linear relationships for the given equation

Sqrt (y 2 – 16y+57) = 3

Solution:

Given equation is
Sqrt (y 2 – 16y+57) = 3

Squaring on both sides and solves it.
[Sqrt (y 2 – 12y+57)] 2 = (3) 2

And solve it.
y 2 – 12y+57= 9

Rewrite the equation with right side equation to 0.
y 2 - 16 y + 48 = 0

It is a quadratic equation with 2 solutions
y = 12 and y = 4

The above form is a quadratic equation with 2 solutions
y = 12 and y = 4

My forthcoming post is on taylor series lnx and cbse class 12 syllabus will give you more understanding about Algebra.

Non linear relationship practice problems:

1)  Evaluate the non linear relationships for the given equation

y 2 - 22 y + 72 = 0

Answer: y = 4 or 18

2)  Evaluate the non linear relationships for the given equation

Sqrt (y 2 – 12y+43) = 4

Answer: y = 3 and y = 9

Series Radius of Convergence

The radius of convergence of as series is defined as the quantity where it can be either a real number or an infinite number that indicates that within the particular radius the series will converge.

Thus the radius of convergence of the power series tends to converge evenly and exactly. If the power series is said to converge then it is called as the Taylor series of the function to which it converges inside the radius of convergence.

Having problem with taylor series convergence keep reading my upcoming posts, i will try to help you.

Mathematical definition of radius of convergence:

Let us consider the power series of a function which is given by


f (p) = `sum_(n=0)^oo` cn(p-a)n

Here a is called as the complex number constant, cn is the coefficient of the nth complex number and p is called as the complex number. The radius of convergence r is a positive real number or an infinite number and the series is said to converge if │p-a│< r and the series is said to diverge if │p-a│> r.

We can say in simple words that the series converge if the complex number p is nearer to the centre and it diverges if it is away. If the power series is given in the form of │p-a│ = r, then it is difficult to find the radius of convergence.

Tests to find radius of convergence:

We will see the two tests such as the root test and the ratio test that has been used to find the series radius of convergence.

Ratio test:

The ratio test to find the series radius of convergence is given by

r = `lim_(n->oo)` │cn / cn+1│

Thus it is clear that the series can converge if  `lim_(n->oo)`   │cn+1(p-a)n+1│/ │cn(p-a)n│is less than one

This can be equalized to

│p-a│< 1 / limn->∞ (│cn+1│/ │cn│)

= `lim_(n->oo)`   │cn / cn+1│


Root test:

The radius of convergence can also be determined by applying the root test to the series. Thus it is given by

C = `lim_(n->oo)` sup √(│cn(p-a)n│) = `lim_(n->oo)` sup   √(│cn││p-a│)

Thus it is made clear that the series converges if c < 1 and the series diverges if C >1.

Writing Parametric Equation

In mathematics, parametric equations are a method of defining a relation using parameters. A simple kinematical example is when one uses a time parameter to determine the position, velocity, and other information about a body in motion.
Abstractly, a Parametric Equation defines a relation as a set of equations. It is therefore somewhat more accurately defined as a parametric representation. It is part of regular parametric representation.  (Source: Wikipedia)


Definition for writing parametric equation of circle:

Parametric equation of circle:

It is the co-ordinates of a point on the circle in terms of a single variable θ. These single variables are called as parameter. The parametric equations of a circle with radius (r ≥ 0) and center (h, k) are written as

x = h + r cos θ          0   ≤ θ ≤ 2

y = k + r sin θ

Parametric equation of a circle with the radius r and centered at the origin.

x2 + y2 = r2

Equation of a circle with single variable formula

x = ±  `sqrt(r^2 - y^2)`

y = ±  `sqrt(r^2 - x^2)`

Each formula gives a portion of a circle

y=  `sqrt(r^2 - x^2)`          (Top)

y= - `sqrt(r^2 - x^2)`        (Bottom)

x=  `sqrt(r^2 - y^2)`          (Right side)

x = -`sqrt(r^2 - y^2)`           (Left side)

Example problem for writing parametric equations:

Example 1:

Determine the radius of parametric equation of the circle for the given x = 7 sin t and y = 7 cos t where 0 < t < `pi`

Solution:

Given  x = 7 sin t and  y = 7 cos t

Formula for the parametric equation of a circle can be writing as x2 + y2 = r2

x = r sin t

y = r cos t

(7 sin t)2 + (7 cos t)2 = r2

72 (sin2 t + cos2 t) = r2

49 (sin2 t + cos2 t) = r2                Hence (sin2 t + cos2 t = 1)

r2 = 49

Radius of the parametric circle is 7.

Example 2:

Determine the radius of parametric equation of the circle for the given x = 13 sin t and y = 13 cos t where 0 < t <`pi`

Solution:

Given  x = 13 sin t and y = 13 cos t

Formula for the parametric equation of a circle can be writing as x2 + y2 = r2

x = r sin t

y = r cos t

(13 sin t)2 + (13 cos t)2 = r2

132 (sin2 t + cos2 t) = r2

169 (sin2 t + cos2 t) = r2                Hence (sin2 t + cos2 t = 1)

r2 = 169

Radius of the parametric circle is 13.

Algebra is widely used in day to day activities watch out for my forthcoming posts on cbse syllabus for class 9th and cbse syllabus for class 1. I am sure they will be helpful.

Example 3:

Determine the parametric equation of a circle with radius = 5 and center (3, 7).

Solution:

Given (h, k) =(3, 7)

Radius = 5

Let  x = h + r cos `theta`

y = k + r sin `theta`

Parametric equation of a circle with radius 5 and center (3, 7)

x = 3 + 5 cos `theta`

y = 7 + 5 sin `theta`

Rules for Signs in Algebra

Algebra is a part of mathematics deals with operations and rules of arithmetic of symbols and numbers, usually letters of the alphabet to express general relationships that hold for all members of the set. Rules for signs in algebra tell us what we are allowed to write. Operations are addition, subtraction, multiplication and division in algebra.

I like to share this Sig Fig Rules with you all through my article. 

Addition rules for signs in algebra:

If the expressions signs are the same, we can add and keep the sign. Otherwise expressions signs are different we can do subtract the expression and take the sign attached to the larger number value.  
Example
- 5 + - 8 = - 13

Subtraction:
Add the Opposite:

1. Change the subtraction sing into addition.
2. Change the sign of the second term to its opposite.

Example:
5 - (+ 2) = 3 {Since 2 is positive.}
5 + (- 2) =3  {Add the opposite.}
4 + (+ 2)= 6 {we can exchange - ( - ) with + +}.
           The subtraction rules for signs in algebra means that the operations of sequences - ⁺ and + ^- are interchanged.

Multiplication and division rules for signs in algebra:

If the expressions signs are the same, the result is opposite; otherwise negative.

Table for multiplication rules for signs in algebra:

+  x  +	=	+
-  x -	=	+
+  x  -	=	-
-x +	=	-

Table for division rules for signs in algebra:

+÷+	=	+
-÷-	=	+
+÷-	=	-
-÷+	=	-

Practice problems regarding rules for signs in algebra:

Problem 1:  -7 + -13
Solution:
   = -7 + (-13)
Add the two terms
   = -20 (answer)

Problem 2:  -`7/8` - -`5/8`
Solution:
   ⁼ - `7/8` + `5/8`
Change second term sign
    ⁼ - `7/8` + `5/8`
   = -`2/8`
   = - `1/4`

Problem 3:  -1 - 5
Solution:
=-1 - 5
Add the terms
=-6

Problem 4:  49 +( -15)
Solution:
=49 + -15
Subtract the terms
=34

Problem 5:
`1/2` - `1/8`
Solution:
=`1/2` - `1/8`
Subtract the terms:
  = `3/8`

Problem 6: 
`3/8` ÷ `1/8`
Solution:
= `3/8` ÷ `1/8`
Change the sing and second term
=`3/8` * `8/1`
= 3

Problem 7: -(-5)
Solution:
=+5

Problem 8: -2 + -2 + -2 + -2
Solution:
Add the all terms
=-8

Problem 9: 13 ÷-2
Solution:
=-7.5


My forthcoming post is on sample papers for class 12 cbse and sample papers for class 9 will give you more understanding about Algebra.

Problem 10:  -`3/8` (16)
Solution:
Multiply the terms
= -`48/8`
=-6

Problem 11: 3 + 3 + 3 + 3
Solution:
Add all term
=12

Problem 12:  -6(-3)
Solution:
=+18

Matrices Addition

In mathematics, a matrix (plural matrices, or less commonly matrixes) is a rectangular array of numbers, such as

[[1, 9, 13], [20, 55, 4]]

An item in a matrix is called an entry or an element. The example has entries 1, 9, 13, 20, 55, and 4. Entries are often denoted by a variable with two subscripts, as shown on the right. Matrices of the same size can be added and subtracted entry wise and matrices of compatible sizes can be multiplied. (Source: Wikipedia)

This article explains you about add two matrices, some solved problems on add two matrices and practice problems on adding two matrices.

Solved Problems on addition of matrices

Method to add matrices:

We can add the matrix with same order only.

By adding corresponding elements of two matrices we can get the sum of two matrix.

Problem 1:

Add the following two matrices

A = [[2,-5],[9,7]] B = [[9,6],[-4,0]]

Solution:

Given, Two matrix A = [[2,-5],[9,7]] ,  B = [[9,6],[-4,0]]

We need to find the addition .

That is A + B

A + B =  [[2,-5],[9,7]]  + [[9,6],[-4,0]]

We know that, By adding corresponding elements of two matrices we can get the sum of two matrix.

A + B =  [[2+9,-5+6],[9+(-4),7+0]]

= [[11,1],[5,7]]

Answer: A + B = [[11,1],[5,7]]

Problem 2:

Add the following two matrices,

P = [[23, 54, -31],[12, -72, 41]] and Q = [[ -9, 8, 0],[31,38,-45]]

Solution:

Given, Two matrix P = [[23, 54, -31],[12, -72, 41]] and Q = [[ -9, 8, 0],[31,38,-45]]

We need to add the above two matrix.

Both are 2 by 3 matrix. By adding corresponding elements of two matrices we can get the sum of two matrix.

P + Q =  [[23, 54, -31],[12, -72, 41]] + [[ -9, 8, 0],[31,38,-45]]

= [[23+ (-9), 54+8, -31+0],[12+31, -72+38, 41+(-45)]]

= [[14, 62, -31],[43, -34, -4]]

Answer: P + Q = [[14, 62, -31],[43, -34, -4]]

Problem 3:

Find the value of x, if  A = [[2,3x], [5,6]] , B = [[6,-3],[8,-2]] and A+B = [[8,12],[13,4]]

Solution:

Given, A = [[2,3x], [5,6]] , B = [[6,-3],[8,-2]] and A+B = [[8,12],[13,4]]

We need to find the x value.

A + B = [[8,12],[13,4]]

[[2,3x], [5,6]] + [[6,-3],[8,-2]] =  [[8,12],[13,4]]

[[2+6,3x-3],[5+8,6-2]] = [[8,12],[13,4]]

[[8, 3x-3],[13,4]] =  [[8,12],[13,4]]

In this corresponding elements are equal.

So we can take, 3x-3 = 12

Add 3 on both sides,

3x-3 + 3 = 12 + 3

3x = 15

Divide by 3 on both sides,

x = 5

Answer: The value of x = 5

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Practice problems on addition of matrices

Problems:

1. Add two matrices P = [[ -2,-4],[4,9]] and Q = [[8, 5],[2,-7]]

2. Find the value of  x , if  X = [[0,-2],[3,6]] , Y = [[ 2,-5],[x,4]] and X + Y = [[ 2,-7],[ 8, 10]]

Solution:

1. P + Q = [[6,1],[6,2]]

2. x = 5

How do you Multiply Trinomials

In basic algebra, a trinomial is consisting of three expressions. A trinomial is an equation concerning three expressions. An instance is the equation x = q + xm. In factoring trinomials technique is one important in elementary algebra. It is definite as the process of sum of three monomials. Given expression consists of three expressions that are within the normal form of ax2 + bx + c.

Different type of methods using for multiply trinomials:

Factoring trinomial is the method to generate arithmetic operation of multiplication. Let us determine some instance problems for factoring trinomials are x2 +3x – 7.  Different kind of technique for multiply trinomials that are,

Distributive method:

It is the operation of multiplying every expression in the initial trinomial is multiplied among the each expression in the second trinomials. Multiply trinomial using distributive method we know how to divide into two types,

Horizontal method
Vertical method


Horizontal method:

Horizontal technique is definite as the process of multiplying the specified trinomials in horizontal also combines the terms.

Vertical method:

A vertical technique is the definite as the process of multiplying the specified trinomials in vertical and adds the terms


Example for multiply trinomials:

Example 1:

how do you multiply trinomials (x+2) (x2+6x+4)

Solution:

Step 1: the given trinomial factors are (x+2) (x2+6x+4)

Step 2: using vertical method to multiply the given trinomials

x2+6x+4

x+2

---------------------

x3+6x2+4x

2x2+12x+8

----------------------

x3+8x2+16x+8     (do you add the terms)

----------------------

so the solution to the given trinomial is x3+8x2+16x+8

Example 2:

how do you multiply trinomials (2x+4) (x2+3x+6)

Solution:

Step 1:  the given factors are (2x+4) (x2+3x+6)

Step 2: using horizondal method to multiply the given trinomials

Step 3:  to multiply the first binomial term 2x

= 2x(x2+3x+6)

= 2x3+6x2+12x

Step 4: to multiply the second binomial term 4

=  4(x2+3x+6)

=  4x2+12x+24

Step 5:  do you add the group

=   2x3+6x2+12x+4x2+12x+24

=  2x3+8x2+24x+24

so the solution is 2x3+8x2+24x+24

Example 3:

how do you multiply trinomials (x-3) (x2+3x+1)

Solution:

Step 1:  the given factors are (x-3) (x2+3x+1)

Step 2: using horizondal method to multiply the given trinomials

Step 3:  to multiply the first binomial term x

= x(x2+3x+1)

= x3+3x2+x

Step 4: to multiply the second binomial term -3

=  -3(x2+3x+1)

= -3x2-9x-3

Step 5:  do you add the group

=  x3+3x2+x-3x2-9x-3

=  x3-8x-3

so the solution is x3-8x-3

Equivalence Properties of Equality Solving Online

In mathematics, equality is a term used for representing the equal quantities by a symbol ‘=’(equal). The main property used for solving the equality is equivalence property. Online gives three equivalence properties with neat definition. Online also gives some example expressions with equality having equivalence properties. Now by this online we are going to learn some problems solving using equivalence properties.

Explanation to equivalence properties of equality solving online:

The three properties of equivalence are,

Reflexive property of equality
Symmetric property of equality
Transitive property of equality

Reflexive property:

The reflexive property of equality says that a every element is equal to the element itself.

y = y

Symmetric property:

The symmetric property of equality says that if a we have equal elements x = y then it is equal to y = x.

Transitive property:

The transitive property of equality says that if a we have equal elements x = y and y = z then  we can take it as y = z.

Example problems to equivalence properties of equality solving online:

Example: 1

Solve: m + 8 = n and n = 12.

Solution:

Given,

m + 8 = n and n = 12

By definition of transitive property of equality,

m + 8 = 12

m = 12 – 8

m = 4

Answer: m = 4 and n = 12

Example: 2

Solve: 12 + 6 = a + 12

Solution:

Given 12 + 6 = a + 12

By definition of symmetric property of equality,

x + y = y + x

12 + 6 = a + 12

a = 6

Answer: a = 6

My forthcoming post is on T Distribution Tables and example of algebra will give you more understanding about Algebra.

Practice problems to equivalence properties of equality solving online:

Problem: 1

Solve: n + 56 = 56 + 12

Answer: n = 12

Problem: 2

Solve: Solve: 5 + m = n + 6 and n + 6 = 12.

Answer: m = 7 and n = 6

Area Word Problem

The area of rectangle parallelogram, triangle, trapezium, quadrilateral and circle formulas are given below.

Area of rectangle = Length * Breath (Sq. units)

Area of parallelogram = Breath * Height (Sq.units)

Area of triangle = 1/2 * b * h (sq.units)

Area of trapezium = `(1)/(2)` * (a + b) * h

Area of Quadrilateral = `(1)/(2)` *  d * (h1 + h2)

Area of the circle = πr2 sq. units

I like to share this Surface Area of Cones with you all through my article.

Example problems for Area word problem:

Word problem 1:

A thread of length 264 cm is cut into two equal portions. One part is bent in the form of a circle and the other part in the form of an equilateral triangle. Find the ratio of the areas enclosed by them.

Solution:


Perimeter of the circle = 264/2=132 cm.

But perimeter of the circle = 2πr.

2 × 22/7 × r = 132 or r = 21 cm.

Area of the circle = πr2 = 22/7 × 21 × 21 = 1386 cm2.

Perimeter of the equilateral triangle = 3a

But perimeter = 132 cm.  3a = 132 or a = 44 cm.

Area of the equilateral triangle = √3/4*a2 = √3/4 * 442  = 484 √3 cm2


The ratio of the area of circle to that of the equilateral triangle

= 1386 : 484√3= 21√3: 22

Word problem 2:

Cost of levelling a land is Rs. 12 per square metre. A ground is in the form of a trapezium whose parallel sides are of lengths 18m and 12 m. If its other two sides are each of  5m length, find the total cost incurred in levelling the ground.

Solution:

ABCD is the given trapezium figure, where AB = 18m, CD = 12 m, AD = BC = 5 m. Draw CE parallel to DA (see Figure). ΔEBC is isosceles whose height


h = `sqrt(5^2 - 3^2)` = 16 = 4cm. Now, Figure the area of the trapezium ABCD

= 1/2 (a + b) × h = 1/2 (18 + 12) × 4

= 2 × 30= 60 sq. metre.

The cost of leveling 1 sq.metre is Rs. 12.

So the cost of levelling the entire land = 60 × 12 = Rs. 720.

Example problem 3:

The dimensions of a rectangular metal sheet are 4m × 3m. The sheet is to be cut into square sheets each of side 4 cm. Find that area.


Solution:

Area of the metal sheet = 400 × 300 = 12,0000 cm2.

Area of a square sheet = 4 × 4 = 16 cm2.

Practice area word problems:

Practice problem 1:

Find the area of a trapezium. The parallel sides have lengths 7 cm and 8 cm is 30 cm2. Find the distance between the parallel sides.

Answer: 4cm


Between, if you have problem on these topics Subtraction of Fractions, please browse expert math related websites for more help on Examples of integration.

Practice problem 2:

Find the area of the quadrilateral one of whose diagonals is of length 15 cm and the lengths of the altitudes to this diagonal are 3 cm and 5 cm.

Answer: 60cm2

Parallelogram Bisect

In geometry, a parallelogram is a quadrilateral with two pairs of parallel sides. In Euclidean Geometry, the opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

Opposite sides of a parallelogram will never intersect.

The diagonals of a parallelogram bisect each other.(Source: From Wikipedia).

The area of a parallelogram is twice the area of a triangle created by one of its diagonals.

Area of the parallelogram is given by,

A=b*h

b is the base and h is the height.

Now, we are going to see the proof of parallelogram bisect and some of the problems. From these problems, we can get clear view about the diagonals of the parallelogram bisect each other.

Prove: The diagonals of a parallelogram bisect each other.

Proof: ABCD is a parallelogram. AC and BD are diagonals.

By ASA (two corresponding angles and the included side) criterion, ΔAMB ≡ CMD

Therefore, AM = CM, BM = DM.

Since the diagonals AC and BD divide the line segment of equal length.

∴The diagonals bisect each other at point M, point M is a midpoint of each diagonal.

Parallelogram bisect problems:

Example problem 1:

From the given diagram, find the value of R and Y of the diagonal of the parallelogram?

Solution:

We know that,

In parallelogram, the diagonals bisects each other.

So, AM=MC

DM=MB

Given: AM= 5cm

MB= 6cm

So, the value of MC=Y is 5cm

So, the value of DM=R is 6cm

Example problem 2:

From the given diagram, find the length of the diagonal when AM= 4x+5 and MC = 3x+7 of the parallelogram?

Solution:

We know that,

In parallelogram, the diagonals bisects each other

So, AM=MC

DM=MB

Here, 4x+5=3x+7

Subtract 3x on both sides of the equation

4x+5-3x=3x+7-3x

1x+5=7

Subtract 5 on both sides of the equation

1x+5-5=7-5

1x=2



Plug in x=2 in the AM and MC

AM= 4(2) +5=8+5=13

MC=3(2) +7= 6+7= 13

So, the length of the diagonals of the parallelogram = 13+13=26 cm.

Adding Intervals

Adding intervals is nothing but the interval of the frequency of data. The adding is the summing of the data. The intervals are made to have in the statistics in the form of the frequency data. The sum of the intervals are made to find through the summing of the terms and divided through the number of the data in the table.



Adding intervals:

The adding intervals gives the mean in which the values of the intervals are made to given with the two values, the two values are made to have along the digits that has the various intervals of the data. These adding intervals are made to have in the changes that the summing of the data presents and the number of data are made to divided with the summing of the total value and the summing of the data.

The adding intervals can also be said as the mean of the given values. The given values are made to have the changes that obtained through the group of the data. The data that provided with a intervals are found with the mean. Some of the examples are made to given below which provides some ideas for adding the intervals.


Examples for adding intervals:

Example 1:

Heights in cms of ten people are:

126, 145, 150, 157, 149, 136, 166, 129, 143, 134

Solution:

Mean: sum of all values/total numbers

Mean= 1435/10

=143.5

Answer: 143.5

Example 2:

The highest scores of the five players in twenty twenty are:

198,192,182,190,179.

Solution:

Mean = sum of all values/total numbers.

= 941/5

Answer: 188.2

Example 3:

Heights in cms of thirteen players are:

110,115,120,125,130,135,150,145,150,155.

Solution:

Mean: sum of all values/total numbers

Mean = 1335/10

= 133.5

Answer: 133.307


Example 4:

The highest scores of the ten players one day are:

198,195,192,190,186,194,182,180,179,178.

Solution:

Mean = sum of all values/total numbers.

= 1874/10

Answer: 187.4

Quadratic Irrational

An equation which consists of more than one terms are squared but no higher power in terms, having the syntax, ax2+bx+c where a represents the numerical coefficient of x2, b represents the numerical coefficient of x, and c represents the constant term

Example: 2x2+5x+20

The roots of the quadratic equation having the irrational number, such an equation is called quadratic irrational. In this content we shall discuss about quadratic irrational with suitable example problems.

Identify types of roots in quadratic equation:

A quadratic equation is in the form of ax2+bx+c,

First we need to find the discernment d = b2- 4ac

If d > 0, the roots are real roots and unequal

If d = 0, the roots are real and equal

If d < 0, the roots are irrational.

Example problem of quadratic irrational:

Solve the quadratic equation

x2+3x+5

Solution:

The equation is in the form of ax2+ bx +c

Where a = 1; b = 3; c = 5

Find the discernment d = b2- 4ac

d = (3)2 - 4* 1* 5

d = 9 -20

d = -11 < 0

If d < 0, the roots are irrational.

To Factor this expression we use quadratic formula method,

`x = ((-b)+-sqrt(b^2-4 * a*c)/(2*a))`

`x = (-3+-sqrt((3)^2-4*1*5))/(2*1)`

`x= (-3+-sqrt(9-20))/2`

`x = (-3+-sqrt(-11))/2`

`x= (-3+-i(3.31))/2` we know that `sqrt(-1) = i`

` x = (-3+-i3.31)/2 ` and `x = (-3-i3.31)/2`

Hence the answer is `x = (-3+-i3.31)/2` and `x = (-3-i3.31)/2`

If you have problem on these topics Decimal Place values.

Example problem of quadratic irrational:

Solve the quadratic equation

x2+5x+9

Solution:

The equation is in the form of ax2+ bx +c

Where a = 1; b = 5; c = 9

Find the discernment d = b2- 4ac

d = (5)2 - 4* 1* 9

d = 25 -36

d = -11 < 0

If d < 0, the roots are irrational.

To Factor this expression we use quadratic formula method,

`x = ((-b)+-sqrt(b^2-4 * a*c)/(2*a))`

`x = (-5+-sqrt((5)^2-4*1*9))/(2*1)`

`x= (-5+-sqrt(25-36))/2`

`x = (-5+-sqrt(-11))/2`

`x= (-5+-i(3.31))/2` we know that `sqrt(-1) = i`

`x = (-5+-i3.31)/2` and `x = (-5-i3.31)/2`

Hence the answer is `x = (-5+-i3.31)/2` and `x = (-5-i3.31)/2`

Example problem of quadratic irrational:

Solve the quadratic equation

x2+7x+13

Solution:

The equation is in the form of ax2+ bx +c

Where a = 1; b = 7; c = 13

Find the discernment d = b2- 4ac

d = (7)2 - 4* 1* 13

d = 49 -52

d = -3 < 0

If d < 0, the roots are irrational.

To Factor this expression we use quadratic formula method,

`x = ((-b)+-sqrt(b^2-4 * a*c)/(2*a))`

`x = (-7+-sqrt((7)^2-4*1*13))/(2*1)`

`x= (-7+-sqrt(49-52))/2`

`x = (-7+-sqrt(-3))/2`

`x= (-7+-i(1.732))/2` we know that `sqrt(-1) = i`

`x = (-7+-i1.732)/2` and `x = (-7-i1.732)/2`

Hence the answer is `x = (-7+-i1.732)/2` and `x = (-7-i1.732)/2`

Example problem of quadratic irrational:

Solve the quadratic equation

x2+8x+17

Solution:

The equation is in the form of ax2+ bx +c

Where a = 1; b = 8; c = 17

Find the discernment d = b2- 4ac

d = (8)2 - 4* 1* 17

d = 64 -68

d = -4 < 0

If d < 0, the roots are irrational.

To Factor this expression we use quadratic formula method,

`x = ((-b)+-sqrt(b^2-4 * a*c)/(2*a))`

`x = (-8+-sqrt((8)^2-4*1*17))/(2*1)`

`x= (-8+-sqrt(64-68))/2`

` x = (-8+-sqrt(-4))/2`

`x= (-8+-i(2))/2` we know that `sqrt(-1) = i`

`x = (-8+-i2)/2` and `x = (-8-i2)/2`

Hence the answer is `x = (-8+-i2)/2` and `x = (-8-i2)/2`

Geometric Translations

Translation:

    The changes of position of a geometrical figure without any change in its direction.

Translation of axes:

    Shifting the coordinates of points to coordinates refered to new set of axes parallel to the original axes. If (x, y) be the coordinates of a point  P with respect to given axes, (x', y') coordinates of the same point P with respect to the new set of axes O'X, O'Y, drawn parallel to the original axes at the point O', whose coordinates are (h, k), then the translation formulae giving the translation of axes are given by

      x = x' + h, y = y" + k

Examples for geometric translations:

Example 1:
Find the new coordinates of the given data below by using geometric translations

(a). Initial coordinates (3, 7) and distance of coordinates (3, 4).

Solution:
Let (x, y) be (3, 7) and (h, k) be (3, 4).
To find the translations coordinates or new co-ordinates:
Formula:
            x' = h + x,   and  y' = k + y
           So,
            x' = 3 + 3,   and  y' = 7 + 4
            x' = 6   and  y' = 11
Therefore, the required translation coordinates are (6, 11) in fig below.

Example 2:
Find the line of translation of the given data below by using geometric translations
(a). Initial coordinates of line  (0, 0) and (2, 6) and distance of coordinates (6, 6).
Solution:
Given:
Initial states of line co-ordinates are,
(0, 0) and (2, 6)
The distance of moving the object line co-ordinates, (6, 6).
To find the translations coordinates or new co-ordinates:
Formula:
            x' = h + x,   and  y' = k + y
           Here, (h, k) = (6, 6)

The translations of  co-ordinates of (0, 0):  (0 + 6, 0 + 6) = (6, 6).

The translations of  co-ordinates of (2, 6):  (2 + 6, 6 + 6) = (8, 12).

Therefore, the required translations of line of coordinates are (6, 6) and (8, 12) in figure below.

Example 2:

    Find the translations co-ordinates of given initial coordinates of shape and distance co-ordinates below by using geometric translations
    (a). Initial coordinates of polygon, (2, 2), (2, 4), (4, 2) and (4, 4)  and distance of coordinates (3, 5).
Solution:
Given:
          Initial state of rectangle co-ordinates are,
                         (2,2), (2, 4), (4, 2) and (4, 4)
          The distance of moving object coordinates, (3, 5)

My forthcoming post is on substitution method with fractions and karnataka pu board will give you more understanding about Algebra.

To find the new rectangle coordinates:
          Formula:
                      x' = h + x,   and  y' = k + y

The translations of  co-ordinates of (2, 2):  (2 + 3, 2 + 5) = (5, 7).

The translations of  co-ordinates of (2, 4):  (2 + 3, 4 + 5) = (5, 9).

The translations of  co-ordinates of (4, 2):  (4 + 3, 2 + 5) = (7, 7).

The translations of  co-ordinates of (4, 4):  (4 + 3, 4 + 5) = (7, 9).

Therefore the required translations of coordinates of new rectangle are, (5, 7), (5, 9), (7,7) and (7, 9), shown in figure below.

Preparation for Recursion

Recursion is a function which is defined with its own definition; it is defined for infinite statements for finite components. The process of repeating objects is a self-similar way is said to be recursion. Recursive formula is one of the concepts in recursion. Finding the next term of the sequence with one or more preceding terms is said to be recursive formula. Students learn recursive formula for the preparation of exam. Let us see about preparation for recursion in this article.

Worked Example Problems - Preparation for Recursion:

Example 1 for Preparation for Recursion:
The first term in a sequence is 27. Each term proceeded after first term is 8 more than the term before it. Find the recursive formula for the above sequence.
Solution:
Step 1:
Given first term `a_1 = 27`

Step 2:
Finding the next term `a_2` of the sequence
We have to add 8 to each term, we get,
`a_2 = a_1 + 8`
`a_2 = 27 + 8 = 35`

Step 3:
Finding the next term `a_3` of the sequence
We have to add 8 to each term of the sequence, we get,
`a_3 = a_2 + 8`
`a_3 = 35 + 8 = 43`

Step 4:
Then form the formula for the above sequence, we get,
`a_n = a_(n-1) + 8`
Step 5:
Therefore, the recursive formula for the sequence is `a_n = a_(n-1) + 8` .

Example 2 for Preparation for Recursion:
Write the first five terms of the sequence `a_1 = 3` and the recursive formula is `a_n = a_(n-1) + 4` .
Solution:
Step 1:
Given
The first term of the sequence is `a_1 = 3`
Recursive formula is `a_n = a_(n-1) + 4`

Step 2:
Second term:
Finding next term of the sequence using above recursive formula, we get,
`a_1 = 3`
Given formula is `a_2 = a_1 + 4`
`a_2 = 3 + 4`
`a_2 = 7`

Step 3:
Third term:
Finding next term of the sequence using above recursive formula, we get,
`a_2 = 7`
Given formula is `a_3 = a_2 + 4`
`a_3 = 7 + 4`
`a_3 = 11`

Step 4:
Fourth term:
Finding next term of the sequence using above recursive formula, we get,
`a_3 = 11`
Given formula is `a_4 = a_3 + 4`
`a_4 = 11 + 4`
`a_4 = 15`

Step 5:
Fifth term:
Finding next term of the sequence using above recursive formula, we get,
`a_4 = 15`
Given formula is `a_5 = a_4 + 4`
`a_5 = 15 + 4`
`a_5 = 19`
Therefore, the first five terms in the sequence are 3, 7, 11, 15, and 19.

My forthcoming post is on Solving System of Inequalities and cbse syllabus for class 9th will give you more understanding about Algebra.

Practice Problems - Preparation for Recursion

Problem 1:
   The first term in a sequence is 15. Each term proceeded after first term is 7 more than the term before it. Find recursive formula for the sequence.
Solution: Sequence is `a_n = a_(n-1) + 7`

Problem 2:
   Write the first five terms of the sequence `a_1 = 4` and the recursive formula is `a_n = a_(n-1) + 2` .
Solution: First five terms are 4, 6, 8, 10, and 12

Rational Symmetry

Rational symmetry is general solution of the system corresponding to its Janet base is rational. In the former case, there is only a two parameter group of rational generators, the remaining ones require a larger function field. As it has been explained, the largest rational symmetry group of an ode, in particular if the symmetry group is the projective one.

Brief Explanation of Rational Theory

Formally, rational symmetry is symmetry with respect to some or all rotations in m-dimensional Euclidean space. Rotations are direct isometrics that is isometrics preserving orientation. Therefore a symmetry group of rational symmetry is a subgroup of E+ (m) (see Euclidean group).
 Symmetry with respect to all this rotations of all points which mean translational symmetry with respect to all translations, so space is identical, and the symmetry group is the whole E(m). With the modified this notion of symmetry for vector fields of the symmetry group can also be E+(m).
 With respect to rotations of the point we can take that point as origin. These rotations form the special orthogonal group SO(m) and the group of m×m orthogonal matrices with determinant 1. For m=3 this is the regular changing group.

Rational Theory Example

Examples without extra reflection symmetry:
  * n = 2, 180°: the dyad, quadrilaterals by this symmetry are the parallelograms; extra examples: letters Z, N, S; apart from the colors: yin and yang

•  n = 3, 120°: triad, triskelion, Borromean rings; sometimes the term trilateral regularly is used;
•  Here n = 4, 90°: tetrad, swastika
•  n = 6, 60°: hexad, raelian symbol, new and original version
•  n = 8, 45°: octad, Octagonal muqarnas, computer-generated (CG), and ceiling

My forthcoming post is on Sinusoidal Function Examples and 6th grade math problems and answers will give you more understanding about Algebra.

Summary of Rational symmetry

Finally Rational symmetry at one point and 2-fold at another one (or ditto in 3D with respect to parallel axes) implies rotation group p6, that is double translational symmetry and 6-fold rational symmetry at some point (or, in 3D, parallel axis). The translation distance of this rational symmetry generated by one such pair of rotocenters is 2√3 times their distance.

Linear Algebra Identities

In general, identiites is defined as the condition, that satisfies for all the values of the involved  variables. The linear algebra identities deals with one variable and two variable solutions. The identities are involved in certain condition, by using this condition we can find the values in the given equation. The identities are involved in two variable or variable with numbers. Linear algebra identities is useful to work the problem quick and easy. But the identities must be read or memorize.

Understanding Simplifying Trigonometric Identities is always challenging for me but thanks to all math help websites to help me out. 

The linear algebra identities as shown in below:

Problems using the linear algebra identities

Example 1:

Simplify, (3x + 7y)² 

Solution:

Using the linear algebra identities simplify the (3x + 7y)²

(3x + 7y)² using the identities

(a+b)² = a²+2ab+b².

Here 3x is substitute to a and  7y is substitute in b

(3x + 7y)²= (3x)² + 2(3x)(7y) + (7y)²

= 9x²+ 42xy + 49y²

(3x + 7y)²  = 9x²+ 42xy + 49y²

Example 2:

simplify, (11a – 7b)²

Solution:

using the linear algebra identities simplify the (11a – 7b)²

(11a – 7b)² using the identities

(a-b)² = a²-2ab+b².

Here 11a is substitute to a and 7b is substitute in b

= (11a)² – 2(11a) (7b) + (7b)²

= 121a² – 154ab + 49b².

(11a – 7b)² = 121a² – 154ab + 49b².

Example 3:

Simplify, (2p + 5q)(2p – 5q)

Solution:

using the linear algebra identities simplify the (2p + 5q)(2p – 5q)

(2p + 5q)(2p – 5q) using the identities

(a+b)(a-b) = (a²-b²).

Here substitute a for 2p and b for 5q.

= (2p)² – (5q)²

= 4p² – 25q²

(2p + 5q)(2p – 5q) =  4p² – 25q²

Problem solving linear algebra identities with three variable

Simplify the function using the three variable  linear algebra identities

a² + b² + c² ≡ (a + b + c)² – 2(ab + bc + ca)

(i) (2x + y + 2z)²

(ii) (x – 2y + z)²

(iii) (2p – 3q – r)²

(iv) (2a + 3b − 2c)²

Solution:
simplify problem in the given three variable linear algebra identities

(i) (2x + y + 2z)² = [(2x) + y + (2z)]² = (2x)² + y² + (2z)² + 2(2x)y + 2y(2z) + 2(2z)(2x)

= 4x² + y² + 4z² + 4xy + 4yz + 8zx.

(ii) (x – 2y + z)² = [x + (–2y) + z]²= x² + (–2y)² + z² + 2x(–2y) +2 (–2y)z + 2zx

= x² + 4y² + z² – 4xy – 4yz + 2zx.

(iii) (2p – 3q – r)² = [(2p) + (–3q) + (–r)]²

= (2p)² + (–3q)² + (–r)² + 2(2p) (–3q) + 2(–3q) (–r) + 2(–r)(2p).

= 4p² + 9q² + r² – 12pq + 6qr – 4rp.

(iv) (2a + 3b – 2c)² = [(2a) + (3b) + (–2c)]²

= (2a)² + (3b)² + (–2c)² + 2(2a)(3b) + 2(3b)(–2c) + 2(–2c)(2a)

= 4a² + 9b² + 4c² + 12ab – 12bc – 8ca. 

Solve Reimann Sums and Integrals

We are already introduced to finding the value of definite integral by the anti-derivative approach
b
∫ f(x) dx = F(b) - F(a)
a
In this  approach we find  the anti-derivative F(x) of f(x)  and then subtract F(a) from F(b) to get the value of the definite integral
Reimann Sums is yet another method to compute the definite integral value from first principles
Let f(x)  be  a function defined on the closed interval [a,b] then the definite interval of f(x) from a to b is given by
      `int_a^bf(x)dx`          = `lim_(n->oo)(b-a)/n`     `sum_(r=1)^n f(a+r` Δx)
where Δx = `(b - a)/n`  ( provided the limit exits ) 

Ilustrative example to understand Reimann Sums as a method to determine the definite integral value

Examples for solve riemann sums and integrals
`int_1^2(2x+5)dx`
    Let  f(x) = 2x+5  and [a,b] = [1,2]
       Δx = `(b-a)/n`  = `(2-1)/n`  =`1/n`
       f(x) = 2x+5
f(a+rΔ) = f(1 +  r`(1)/n` )  =2 (1 +  r`(1)/n` )  + 5
By the Formula
      `int_a^bf(x)dx`          = `lim_(n->oo)(b-a)/n`     `sum_(r=1)^n f(a+r` Δx)
where Δx = `(b - a)/n`  ( provided the limit exits )
 `int_1^2(2x+5)dx`  = `lim_(n->oo)(1)/n`     `sum_(r=1)^n ``(2(1+ r/n)+5)`
= `lim_(n->oo)(1)/n`     `sum_(r=1)^n (7+ 2r/n)`
`=lim_(n->oo)(1)/n sum_(r=1)^n``(7) +lim_(n->oo)(1)/n (2/n)sum_(r=1)^n (r)`
=`lim_(x->oo)1/n(7n)+1/n(2/n)lim_(n->oo) (n)(n+1)/2'`

`='lim_(n->oo)(7) +lim_(n->oo) 1/n(2/n)(n)(n+1)/2'`

`= 7 + lim_(n->oo)(n+1)/n`
`= 7 + lim_(n->oo)(1+ 1/n)`
`= 7 + 1`
`= 8`
Ans is 8
Let us Validate the Answer by using the antiderivative Approach
`int_1^2(2x+5)dx= (2(x/2)^2 +5x)1^2`
`='(2^2-1^2) +5(2-1)'`
`= (4-1) + 5`
`= 3 +5`
`= 8`

My forthcoming post is on algebra math problem solver will give you more understanding about Algebra.

Conclusion for solve riemann sums and integrals

Thus Reimann Sums method divides the area under the curve into numerous rectangles and does the summation of the area of each rectangle to give the total  area under the curve which is essentially the value of the definite intergal

How to Use Ratios

When two quantity of same unit are compared or divided, it is called as a ratio.  The ratio will give a relationship between the values in terms of the quantity.  Suppose Jason has $50 and Mike has $150.  Then the ratio of their money can be written as `50/150` = `1/3` `implies` 1: 3.  This shows that Mike is having an amount that is 3 times of Jason.



This concept will help in dividing certain values among some body else.

Now let us see few problems of this kind.

Example problems ratios:

Ex 1: The sum of the salaries of two persons A and B for a month is $8625.

Their salaries are in the ratio 25: 44.  Find their salaries.

Sol: The total sum is $8625.

Total ratio is 25 + 44 = 69

Therefore The salary for A is` 25/69 xx` 8625 = $3125.

The salary for B is `44/69 xx` 8625 = $5500.

Ex 2: Two numbers are in the ratio 7:17.  The first number is 441.

Find the second number.

Sol: Let the total value of the two numbers be x.  Total ratio is 7 + 17 = 24.

Given    :  `7/24 xx` x = 441

`implies` x = 441` xx 24/7` = 1512.

Therefore the second number is `17/24 xx ` 1512 = 1071.

More example problems on ratios:

Ex 3: Two groups A and B are of 850 is strength.  They are in the ratio 7:10. If  50 of them added to group A and some more to group B, the new ratio will be 8:13.  Find number added to B.

Sol: Given: The total is 850.

Ratio is 7:10

`implies` Total ratio is 7 + 10 = 17.

Therefore A group = `7/17 xx` 850 = 350

B group = `10/17 xx` 850 = 500

Now, also given, the new ratio is `[350+50] /[500 +x]` = `8/13` .

`implies ` 13(400) = 8 (500+x)

Therefore 5200 = 4000+8x

Therefore 8x = 5200 – 4000 = 1200

Therefore x = `1200/8` = 150.

Therefore 150 added to B.

Algebra is widely used in day to day activities watch out for my forthcoming posts on pre algebra tutorials and algebra 2 solver free. I am sure they will be helpful.

Large Confidence Interval

In statistics large confidence intervals for a single unidentified population mean μ where the population standard deviation is recognized. Here, the margin of error is known as the error bound for a population mean (shortened EBM). The margin of error depends on the confidence level (shortened CL). Confidence interval is answered by using the values for example the sample mean and the standard deviation. In this article we will discuss large confidence interval.

Definition – large confidence interval:

Statistics confidence interval is distinct as the function of calculating of sample mean subtracted by error with this for the population mean of the function

Formula for calculating the confidence interval by using the function is given by,

Confidence interval =` barx` -` EBM`

Error bounded for the sample mean is distinct as the calculation of t-score value for the confidence interval that is multiply to the standard deviation value divided by the total values given in the data set.

Formula for evaluating the error bounded for the sample mean is given by,

`EBM = t_(alpha/2) (S/(sqrtn))`

Here

` t_(alpha/2)` , It is the t-score that is  right equal to`(a/2)` with the confidence interval .

`S` , is the Variable for standard deviation of sample mean for the certain data.

`n` , is the variable for entire number of values in the data set.

Example problem – large confidence interval:

Large confidence interval problem:

Find out the large confidence interval value for the given data is that sample mean is given that as 25, observed mean value for the given trial is 10.5981, observed standard deviation value is given that  is 2.1960 which includes the confidence interval of 95%.

Solution: Given

Sample mean = 25

Observed mean = 10.5981

Standard deviation = 2.1960

Confidence level for the confidence interval is 95%

`EBM = t_(alpha/2) (S/(sqrtn))`

`(a/2)` = `0.025`

`t_(alpha/2) = 2.14`

`EBM = t_(alpha/2) (S/(sqrtn))`

` = 2.14xx((2.1960) / sqrt(25))`

` EBM = 0.939888`

` barx - EBM` ` = 10.5981 - 0.939888`

` = 9.658212`

` barx + EBM = 10.5981 + 0.939888`

` = 11.537988`

The confidence interval with the confidence level 95% is given by (9.658212, 11.537988)

Between, if you have problem on these topics Simplify Fraction, please browse expert math related websites for more help on iseet 2013 sample papers.

Fractions Application

A fraction is a way of expressing  a number in equal parts.A fraction consists two numbers, a numerator which gives the number of equal parts and a denominator which gives the number of those parts that make up a whole. For example, the fraction ¾ could be used to represent three equal parts of a cake, where the whole cake is divided into four equal parts. Because it is impossible to divide something into zero equal parts ,zero can never be the denominator of a fraction. Application of fraction is widely used in many field. Fraction applications are very useful. Fraction application is used in engineering, science and in many other fields.

Looking out for more help on Division of Fractions in algebra by visiting listed websites.

Application of fractions:

Fractions:

Proper fraction

Improper fraction

Proper fraction:

In a proper fraction all the numerators should be  smaller than denominator,then it said to be a proper fraction.

For example,

¼,2/6,8/9

Improper fraction:

In a improper fraction all the numerators should be greater  than denominator, then it said to be a improper fraction.

13/2,7/2,9/2

Equivalent fractions:

Multiplying both the denominator and numerator as a same number that is called equivalent fractions.

Example problems:

1)3/2

Multiplying both the denominator and numerator by 2

3*2/2*2 = 6/4

The result is 6/4

2)2/5

Multiplying both the denominator and numerator by 3

2*3/5*3 =  6/15

The result is 6/15

3)40/90

Dividing both the denominator and numerator by 10

(40/10)/(90/10) =  4/9

The result is 4/9


Application of fraction in Addition:

Problems:

1)      (2/3)+(1/2)

Sol:

Take l.c.m for both denominators

So the common factor is 6

=(4+3)/6

=7/6

So the result is 7/6

2)(2/5)+(2/4)

Sol:

Take l.c.m for both denominators

So the common factor is 20

=(8+10)/20

=18/20

So the result is 18/20

subtraction of fractions:

Problems:

1)      (3/4)-(2/3)

Sol:

Take l.c.m for both denominators

So the common factor is 12

= (9-8)/12

=1/12

So the result is = 1/12

2)      (2/4)-(1/3)

Sol:

Take l.c.m for both denominators

So the common factor is 12

= (6-4)/12

=2/12

So the result is = 2/12


Multiplication of fractions:

Problems:

1)    (2/3)*(2/5)

Sol:

Multiplying both denominators and numerators,

=(4/15)

The answer is  =(4/15)

2)    (4/3)*(6/7)

Sol:

Multiplying both denominators and numerators,

=(24/21)

The answer is  =(24/21)

Manipulative in Math

A math manipulative is an object which is designed so that a student can learn some mathematical concept by manipulating it. The use of manipulative provides a way for children to learn concepts in developmentally appropriate, hands-on ways. Mathematical manipulative are used in the first step of teaching mathematical concepts, that of concrete representation. The second and third step is representational and abstract, respectively. (Source: Wikipedia)


Having problem with Difficult Math Problems keep reading my upcoming posts, i will try to help you.

Manipulatives in math - Example problems:

Here we will discuss the some of the examples of manipulative problems are:

Manipulatives in math - Example 1:

Find the x value: X+20=10

Solution:

X+20=10

Subtract both sides on -20

x+20-20=10-20

x= -10

Answer:-x = -10

Manipulatives in math - Example 2:

Find the x value: X-25=11

Solution:

X-25=11

Both sides adding by 25

X-25+25=11+25

X = 36

Answer:-x= 36

Manipulatives in math - Example 3:

Find the x value: 3x-9=18

Solution:

3x-9=18

Both side adding by 9

3x-9+9=18+9

3x=27

3x/2=27/3

X = 9

Answer:-X=9

Manipulatives in math - Example 4:

Find the x value: (3x-15)/2=18

Solution:

(3x-15)/2=18

Both sides multiply by 3

((3x-15)/2)*3=18*3

3x-15=54

Both sides adding by 15

3x-15+15=54+15

3x=69

3x/3=69/3

Both sides dividing by 3

X = 23

Answer:-X=23

Manipulatives in math - Example 5:

Solve: 6x2 +9 x2 +8 y3-3y3

Solution:

6x2 +9x2 +8y3-3y3

Adding the x and y terms

15x2+5 y3

Answer:-  15x2+5 y3

Manipulatives in math - Example 6:

8x2 * 5x2 +7 y3*5y3

Solution:

8x2 * 5x2 +7 y3*5y3

Multiply the x and y terms

40x2+2+35y3+3

40x4+35y6

Answer:- 40x4+35y6

My forthcoming post is on solve math problems step by step and cbse sample papers for class 12 will give you more understanding about Algebra.

Manipulatives in math - Example 6:

9x3 * 5x2 +8 y3/2y4

Solution:

9x3 * 5x2 +8 y3/2y4

Multiply the x terms

45x3+2+8 y3/2y4

Dividing the y terms

45x5+4y -1

Answer:- 45x5+4y -1

Manipulatives in math - Practice problems:

Manipulative in math – practice problem 1:

Find the value of x: x+8=15

Answer:-x= 7

Manipulative in math – practice problem 2:

Find the value of x: 5x-55=30

Answer:-X=17

Manipulative in math – practice problem 3:

Find the value of x: (5x-65)/2=8

Answer:- X = 21

Manipulative in math – practice problem 4:

9x2 +7x2 +5y3-y3

Answer: - 1 6x2+4y3

Manipulative in math – practice problem 5:

8x2 * 5x3 +6y3*7y3

Answer: - 40x5+42y6

Product of Consecutive Numbers

The numbers which has same difference between each other is called as consecutive numbers. The another name of consecutive numbers is continuous numbers in algebra.The consecutive numbers has 1 as difference. The product operation is simple one. Now we are going to see about consecutive numbers and product of consecutive numbers.

Explanation for product of consecutive numbers

Some notes about consecutive numbers:
                           In math, the consecutive numbers are classified into two types. They are odd consecutive numbers and even consecutive numbers. Both types are different in expression.       
Product of consecutive numbers in math:
                                         The product of consecutive numbers are simple operation. The consecutive numbers are give the same result as normal multiplication.
Product of odd consecutive numbers:
                             An odd number which has difference between other number as two. The series of odd number is x, x + 2, x + 4 and so on. Here x is odd number. For example, 3 and 5 are odd consecutive numbers.
Product of even consecutive numbers:
                               A number which has difference between other numbers as two. The series of even consecutive numbers are x, x + 2, x + 4 and so on. Here x is even number.For example, 2 and 4 is even consecutive numbers.

More about product of consecutive numbers

Example problems for product of consecutive numbers:
Problem 1: Find the product value of given two odd consecutive numbers 21 and 23.
Answer:
The given odd consecutive numbers are 21 and 23.
Multiply the given consecutive numbers as 21 x 23 = 483.
Therefore, the product of given odd consecutive numbers is 483.
Problem 2: Find out the product value of even consecutive numbers between 5 to 9.
Answer:
The even consecutive numbers are 6 and 8.
The product of even numbers as 6 x 8 = 48.
Exercise problem for product of consecutive numbers:
The product value of two consecutive numbers is 15. Find that two consecutive values.
Answer: The two consecutive values are 3 and 5.
Find out the product of given two consecutive numbers 12 and 14.
Answer: The product value of given consecutive numbers is 168.