Polynomials

The word ‘poly’ that is often used in math is actually derived from Greek. ‘Poly’ in that language means ‘many’. So, if an expression has many terms, then it is called as polynomial. But the number of terms has to be finite. Generally up to three terms expressions have their own identities like a single term is called a monomial, two terms as binomial and three terms as trinomial.  Polynomials (abbreviated as Pnml) can also form the core part of a function, in such cases they are called as polynomial functions.

From what we described any expression or any function which has many terms could be Pnmls. Is this statement correct? No, there is a limitation. Only expressions in which no term is divided by any variable and do not contain a non-negative exponent terms can be referred as Pnmls.

In this topic of polynomials help, we will see some examples of polynomials and how to simplify polynomials.The following are the polynomials examples.

2x^3 + 3x^2 – 7x + 2
6x4 – 2x^3 + 4x^2 + 2x + 9
4x^3y^2 – x^2y^2 + 2xy - 14
5x^3 + (x^2/2) + 4x –7

In the third example we find three terms contain two variables but it does not matter as far as the definition of Pnml  is concerned.

In the last example one can notice that one of the terms (the second term) is divided by 2. But it does not affect the Pnml status of the expression as 2 is only a constant. Only when any term is divided by any variable cannot qualify to be a Pnml.
The following expressions cannot be called as Pnmls.
x^3 + 4x^2 – 7x + (2/x)
6x4 – 2x-3 + 4x^2 + 2x + 9
7x^3– 2x^2 + 5x^2/3 + x + 1
1 + (x/1!) + (x^2/2!) + (x^3/3!) + …….

In the first expression, the last term is divided by a variable. In the second expression we find a negative exponent in the second term. The third term of the the third expression has a rational exponent which cannot be simplified as an integer. The fourth expression has infinite number of terms. Hence all of them cannot qualify as Pnmls.

It is possible that a Pnml may have exactly similar terms excepting that the coefficients may be different. Those terms are called like terms. Such expressions can be simplified by algebraically adding the like terms. Similarly a Pnml can be simplified and expressed in factored form by identifying common factors. Simplified Pnmls show a better presentation.

Middle School Math Practice

Introduction to middle school math day:

Middle school class books are designed to presents the practice with math skills and concept wise explanations such as whole numbers, fractions, decimals, percents, geometry, and square roots. Middle school math tells about the measurement, property, and relations about quantities and sets, with numbers and symbols. It is a function of expressions. In this article we shall discuss about middle school math day problems.


Sample problem for middle school math day:


Example 1:

Solve the given equation 4× (x + 8 + 11) = 88.

Solution:

Grouping the terms like, the left side of the equation becomes
4 × (x + 8 + 11) ==> 4 × (x + 19)

Using the distributive property,
4 × (x + 19) ==> 4 × x + 4 × 19

Carrying out multiplications,
4 × x + 4 × 19 ==> to 4x + 76

The equation now becomes
4 x + 76 = 88.

Subtracting a 76 (adding a -76) to each side gives us
4x + 76 + (-76) = 88 + (-76)

4x + (76 + (-76)) = 88 - 76

4x + 0 = 13

4x = 12

Since the x value is multiplied by 4, we divide both sides by 4 to solve for x:
4x = 12

4x ÷ 4 = 12 ÷ 4

(4x)/4 = 3

x = 3.

We get the answer value is X = 3

We can check the answer in the solution of a original equation:
4× (3 + 8 + 11) = 88

4 × 22 = 88

88 = 88 so our answer is correct.

Problem 2:

Find the diameter of the circle. A radius of the given circle is 9 cm.

Solution:

Given:

Radius of the circle is r = 9 cm

Diameter of the circle is d = 2r

d = 2 * 9

d = 18

So, the diameter of the circle is 18 cm.

Problem 3:

The perimeter of a rectangle is 1200 meters and its length L is 5 times its width W. Find the values of W and L, and the area of the rectangle.

Solution:

Perimeter of rectangle=2L+2W,

2 L + 2 W = 1200

We now rewrite the statement. Its length L is 5 times its width into a mathematical equation as follows:

L = 5 W

We have to substitute L =3W in the equation 2 L + 2 W = 1200

2(5 W) + 2 W = 1200

12 W = 1200

W =100 meters

Use the equation L = 5 W to find L.

L = 5 W = 500 meters

Use the formula of the area.

Area = L x W = 500 * 100 = 50000 meters 2.

So, the area of the rectangle=50000 meters 2.


Practice problem for middle school math day:

Find the diameter of the circle. The radius of the given circle is 12 cm.

Answer: d = 24

Solve the given equation 9× (x + 3 + 11) = 99.

Answer: x = -3

Rectangle Word Problem

Introduction to Rectangle:

A Rectangleis any quadrilateral with four right angles. The term "oblong" is occasionally used to refer to a non-square rectangle. A rectangle with vertexes ABCD would be denoted as  ABCD.

We shall work out some problems on rectangles


Rectangle word problem 1:


A rectangle has a perimeter of 120 meters and its length L is 2 times its width W. Calculate the dimensions W and L, and the area of the rectangle.

Solution to rectangle word problem 1:

The formula to find the perimeter of rectangle is

2 L + 2 W = 120

Now, we rewrite the statement, length L is 2 times its width “w” into a mathematical equation as follows,

L = 2 W

We substitute L in the equation 2 L + 2 W = 120 by 2 W.

2(2 W) + 2 W = 120

Expand and group like terms.

6 W = 120

Solve for W.

W = 20 meters

Use the equation L = 2 W to find L.

L = 2 W = 40 meters

Use the formula of the area.

Area = L W = 40 * 20 = 800 meters 2


Rectangle word problem 2:


The perimeter of a rectangle is 70 feet and its area is 304 feet 2. Find the length L and the width W of the rectangle, such that L > W.

Solution for rectangle word problem 2:

The formula for perimeter of a rectangle is given by

2 L + 2 W = 70

and the formula area of a rectangle is given by

L W = 304

Divide all terms in the equation 2 L + 2 W = 35 by 2 to obtain

L + W = 35

Solve the above for W

W = 35 - L

Substitute W by 35 - L in the equation L W = 304

L(35 - L) = 304

Expand the above equation and rewrite with right term equal to zero.

-L 2 + 35 L - 304 = 0

The above is a quadratic equation with two solutions.

L = 16 and L = 19

Use W = 35 - L to find the corresponding values of W.

W = 19 and W = 16

Since L > W, the rectangle has the dimensions

L = 19 feet and W = 16 feet.


Practice rectangle word problem 1:


The perimeter of a rectangle is 90 feet and its area is 504 feet 2. Find the length L and the width W of the rectangle, such that L > W.

Practice rectangle word problem 2:

A rectangle has a perimeter of 240 meters and its length L is 5 times its width W. Calculate the dimensions W and L, and the area of the rectangle

Type in Word Problem And Solve

Introduction for 'type in word problem and solve':

In general, word problem refers to the mathematical exercise where the information on the given problem can be written in mathematical expressions. If we type a word problem in online, the online tutors will solve the problems. In this article type in word problem and solve, we are going to discuss few basic word problems. Please express your views of this topic Practice Probability Problems by commenting on blog.


Example problems for 'type in word problem and solve':


The example word problems are given below:

Example 1:

Totally, there are 1000 seats in a theatre. Out of which, 700 seats are occupied. Calculate the percentage of seats that are occupied.

Solution:

Total seats    =  1000

Occupied seats  =  700

Percentage   = `<< 700 / 1000>>`   x 100

= `<< 7/10>>`   x 100

= 70

Example 2:

Randy bought a doll for `$` 45 but he sold it for `$` 70. Calculate his gain amount.

Solution:

Cost price of doll   =  $ 45

Selling price of Clock  =  $ 70

Profit  or  Gain   =  Selling price - Cost Price

=  70 - 45

=  $ 25

Example 3:

Julie buys an ornament, which costs `$` 810. Suppose the sales tax rate is 8%, find the total amount she have to pay for the ornament?

Solution:

Sales tax  =  8% of the price tax

= 8%  x  810

= 0.08 x 810

= 64.8

Final price = price before the tax + sales tax

= 810 + 64.8

= $ 874.8

I have recently faced lot of problem while learning how to solve calculus problems, But thank to online resources of math which helped me to learn myself easily on net.

Practice problems for 'type in word problem and solve':


1) Totally, there are 830 seats in a theatre. Out of which, 700 seats are occupied. Calculate the percentage of seats that are occupied.

Answer: 84.3 percent

2) Robert bought a doll for `$` 40 but he sold it for `$` 60. Calculate his gain amount.

Answer: 20 dollars

3) Serena buys an ornament, which costs `$` 800. Suppose the sales tax rate is 4%, find the total amount she have to pay for the ornament?

Answer: 832 dollars

Solution Set Online

Introduction of solution set online:

In the solution set online, a collection of well defined objects is called a set. For example, the collection of all natural numbers, the collection of all equilateral triangles in a plane, the collection of all real numbers, the collection of all vowels in English alphabet are some examples of sets since we can definitely say what objects are there in each of the collections. Consider the following statements for online solution set:

(i) The set of all tall students in your class.

(ii) The set of good books you have studied.


Example of solution set online:


The examples of solution set online is given as follows:

1. If A = {1, 3, 4, 5, 6, 7, 8, 9} and B = {1, 2, 3, 5, 7}, find n(A), n(B), n(AUB) and n(A∩B) and verify the identity

n(AU B) ≡ n(A) +n(B) − n(A∩B).

Solution: We observe that AUB = {1, 2, 3, 4, 5, 6, 7, 8, 9} A∩B ={1, 3, 5, 7}.

n(A) = 8, n(B) = 5, n(A UB) = 9 and n(A∩B) = 4.

We find n(A) + n(B) − n(A∩B) = 8 + 5 − 4 = 9.

Here, n(AUB) = 9. So n(AUB) = n(A) + n(B) − n(A∩B).

In fact, this result is true for any two finite sets.

2. If X = {a, c, d, e, f, g, h, i} and Y = {a, b, c, d, g}, find n(X), n(Y), n(XUY) and n(X∩Y) and verify the identity

n(XU Y) ≡ n(X) +n(Y) − n(X∩Y).

Solution: We observe that XUY = {a, b, c, d, e, f, g, h, i} A∩B ={a, c, e, g}.

n(X) = 8, n(Y) = 5, n(X UY) = 9 and n(X∩Y) = 4.

We find n(X) +n(Y) − n(X∩Y) = 8 + 5 − 4 = 9.

Here, n(XUY) = 9. So n(XUY) = n(X) + n(Y) − n(X∩Y).

In fact, this result is true for any two finite sets. Understanding Subtracting Complex Numbers is always challenging for me but thanks to all math help websites to help me out.


Exercise problems of solution set online:


1. If A = {1, 2, 3} and B = {2, 3, 4}, find A ∩B.

Answer:  A∩B = {2, 3}.

The exercise problem of solution set online is given as follow:

2. If A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 7}, find A − B and B − A.

Answer: A−B = {2, 4, 5, 6}. B −A = {7}.

3. If A = {1, 2, 3, 4} and B = {2, 4, 6}, find A UB.

Answer: AUB = {1, 2, 3, 4, 6}.

Algebra Geometric Problem

Introduction(algebra geometric problem):

The description in a course guide: "Introduces the basic notions and techniques of modern algebraic geometry. Algebraic sets, Hilbert's Nullstellensatz and varieties over the algebraically closed fields. We relate varieties is the over of complex numbers to complex analytic manifolds. For varieties of dimension one (i.e. curves) we discuss is the genus, divisors, linear series, line bundles and the Riemann-Roch theorem." Johan de Jong will be teaching of the follow-up course in the spring.

Definition of Algebra geometric:


A geometric algebra Gn(Vn) is an algebra constructed over a vector space Vn in which a geometric product is defined. The elements of geometric algebra are multi vectors. The original vector space V is constructed over the real numbers as scalars. From now on, a vector is something in V itself. Vectors will be represented by boldface, small case letters (e.g. a), and multi vectors by boldface, upper case letters. Understanding Definition of Right Angle is always challenging for me but thanks to all math help websites to help me out.


Problem:


You have a square where is the distance from 1 corner to its opposite corner is 2 cm. What are the dimensions of this square correct to be the 3 decimal places?

Solution: Call the distance from one corner to the other D. Call one side X. Using the Pythagorean theorem we have D^2 = X^2 + X^2 or D^2 = 2X^2

Since D = 2, we have

2^2 = 2X^2
4 = 2X^2
2 = X^2

by dividing both sides by 2. Taking the square root of both sides, we have X = 1.414 rounding to 3 decimal places.

In classical algebraic geometry, is the main objects of interest are the vanishing sets of collections of polynomials, meaning the set of all points that simultaneously satisfy one or more polynomial equations. For instance, the two-dimensional sphere in three-dimensional Euclidean space R3 could be defined as the set of all points (x,y,z) with

x^2+y^2+z^2-1=0

A "slanted" circle in R3 can be defined as the set of all points (x,y,z) which satisfy the two polynomial equations:

x^2+y^2+z^2-1=0

x+y+z=0

Geometry Problem Solver

Introduction to Geometry problem solver:

Geometry problem solver has been very carefully selected to bridge the gap between the exposition and the regular exercise set. By doing these exercise and checking the complete solutions provided, we are able to test their comprehension. We have learned about the areas and perimeters of some plane geometrical figures such as triangles, quadrilaterals and circles, parallelogram.Let us see some example and practice problems for geometry. I like to share this What are Quadrilaterals with you all through my article.


Sample Geometry problem solver:


Some of the sample geometry problem solver are as follows:

Example 1:

Find the base of a parallelogram if its area is 40 cm^2 and altitude is 15 cm.

Solution:

Area = b × h.

40  =  b × 15.

b  =  40 / 15  =  8 / 3

Base  = 8 / 3 cm

Example 2:

A house in the form of a rectangle has base 15m and height 10m.find the Area of the house?

Solution:

Let  b  = 15 and h = 10.

Then the area of the rectangle  =  b × h  = 15 × 10

= 150 sq. meters

Example 3:

Find the area of the trapezium for the given bases and height a=10, b = 8, h = 6.

Solution:

Area = 1/2 (a  +  b) h

=1 / 2 (10 + 8) 6

= 54 sq. units.

Example 4:

Find the area of the quadrilateral given in d = 50m, h1 = 10m, h2 = 20m.

Solution:

Area = 1/2d (h1 + h2) = 1/ 2 *50(10+20)

= 25 × 30

= 750 m2

Example 5:

Find the area circle and given perimeter is 264 cm    (use`pi`=22/7 )

Solution:

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

But perimeter of the circle = 2`pi` r.

2 × 22/7 × r = 132 or r = 21 cm.

Area of the circle =`pi` r2 = 22/7 × 21 × 21

= 1386 cm^2.

Having problem with Calculate Area keep reading my upcoming posts, i will try to help you.

Practice Geometry Problem solver:


Some of the practice geometry problem solver are as follows:

1. Find the area of a triangle when base length = 24 cm, height = 3 cm.

Answer: Area = 48cm^2

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

Answer: Area = 60cm^2

3. Find the area of the quadrilateral ABCD where the diagonal AC is of length 44 cm and the lengths of the perpendicular from B and D to AC are 20 cm and 12 cm respectively.

Answer: Area = 704cm^2

4. Find the area of the trapezium for the given bases and height a=40, b=20,  h=50

Answer: Area = 1500cm^2