Standard Position Learning

Standard learning is deals with the establishment of measurement of units, or reference instrument or component, suitable for use in calibrations of other instruments. Standard position learning is also deals with an angle in usual point pinched on a xy-plane with its vertex at the starting point and its primary side by the side of the positive x-axis.


About Standard position learning:

In standard Position an degree is determined, if the mortal side of an angle in standard position go rounds counterclockwise, then the angle is affirmative, which means (+ve).

If the mortal side of an angle in standard position go rounds clockwise, then the angle is depressing, it means (-ve).

By the way the above statements are explained the standard positions, and also in other way the standard position says an evaluating of an expression is also having some standard position in solving steps for the expression.

Standard position shown in positive axis,

Standard position shown in negative axis,


Example for standard position learning:

Example for standard position learning 1: Compare the fractions 9/10 and 5/8.

Solution:

Step 1: 9/10 and 5/8

Step 2: The LCM of 10 and 8 is 80,

So, the LCD of 9/10 and 5/8 are 80.

Step 3: 9/10 = 72/80 and 5/8 = 50/80,

Step 4: Compare the numerators as the denominators are equal.

Formula: 72 > 50,

Step 5: So, 40 / 72 > 27 / 72 and so 9/10 > 5/8.

Example for standard position learning 2: Compare the fractions 3/4 and 4/5.

Step 1: 3/4 and 4/5

Step 2: The L.C.M. of 4 and 5 is 20,

So, the LCD of 3/4 and 4/5 is 20,

Step 3: Formula: 3/4 = 15/20 and 4/5 = 16/20,


Step 4: Compare the numerators as the denominators are equal.

15/20 < 16/20,

Step 5:  So, 3/4 < 4/5.

Example for standard position learning 3: A destination point makes an angle of 130° from ship due north. Find the location of the destination point in terms of an angle in standard position.

Solution: The location of the destination point in terms of an angle in standard position is shown in diagram below,

The location of the destination point in terms of an angle in standard position is 360° - 40° = 320°.

Preparation for Calculus Function

Calculus (Latin, calculus, a small stone used for counting) is a branch in mathematics focused on limits, functions, derivatives, integrals, and infinite series. This subject constitutes a major part of modern mathematics education. It has two major branches, differential calculus and integral calculus. Calculus is the study of change, in the same way that geometry is the study of shape and algebra is the study of operations and their application to solving equations. In this article we hall discuss about preparation for calculus function.(Source: Wikipedia).


Preparation for operations of functions

A function exactly has four basic operations of algebra (addition, subtraction, multiplication, and division). Functions are combined by these fundamental operations; the domain of the new combined function is only the elements that were shared by the domains of the original functions.

Preparation of some formula for combining functions:

• The sum of two function f and function g: (f + g)(x) = f(x) + g(x) .

• The difference of two functions, f and function g: (f - g)(x) = f(x) - g(x) .

• The product of two functions, f and function g: (fg)(x) = f (x)×g(x) .

• The quotient of two functions f and g: (f/g) (x) = f(x)/g(x). If g(x) = 0

Preparation for calculus functions with example

Example 1:

The gradient at any point (x , y) of a curve is 3x2-12 and the curve passing point(, -7), find the equation of the curve.

Solve  `(dy)/(dx)` =3x2-12

Integration both sides, we get  y= `int` (3x2-12)dx or y=3 X `(x^3)/(3)` -12 + c.

Where c is a constant of integration. Or y=x3-12x+c

If you have problem on these topics Alternate Interior Angle Definition .

C can be found from the consideration that this equation is satisfied by the values x=2 and y=-7.

Therefore -7=8-24+c

Therefore c=9

Substituting in equation we get y=x3-12x+9m which is the required equation of the curve.

Lateral Area of Cylinder Formula

A cylinder is one of the most basic curvilinear geometric shapes, the surface formed by the points at a fixed distance from a given straight line, the axis of the cylinder. Two circles with radius "r" fixed at a perpendicular distance "h" is called as a cylinder. (Source: From Wikipedia).

Lateral area of a cylinder

Lateral area of a cylinder is defined as the surface area of the visible portion of a cylinder. Or the lateral surface area of a cylinder is defined as the area of the curved surface area excluding the area of the top and bottom circle faces.

Looking out for more help on Finding the Volume of a Cylinder in algebra by visiting listed websites.

Formula for the lateral surface area of a cylinder

Here we are going to learn how to derive the formula for the lateral area of a cylinder.

As we already know, the lateral surface area of a cylinder is the area of the curved surface.

the above diagram shows the net of a cylinder. So the area of the curved surface is equal to the area of a rectangle. The area of the rectangle is the product of the length and breadth. Here the breadth is equal to the height of the cylinder and the length of the cylinder is equal to the circumference of the circles.

so the formula of the lateral area of a cylinder = height of the cylinder * circumference of one of the base circle

= h * 2 `pi` r

The formula of the lateral area of a cylinder  = 2 `pi` r h square units

Example problem to find the lateral area of a cylinder

Example 1

What is the lateral area of cylindrical can, if the radius and height of the can is 4 cm and 8 cm respectively?

Solution

Formula to find the lateral area a cylinder = 2 `pi` r h square units

=  2 * 3.14 * 4 * 8

= 200.96 square centimeters

So the lateral area of the cylindrical can is 200.96 square centimeters.

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Perpendicular Bisectors

Two lines are considered perpendicular to each other if they form congruent adjacent angles.

A perpendicular line or a segment which passes through the midpoint of the line is referred as the perpendicular bisector.

Dividing a line into two equal segments is done by perpendicular bisector.
Circumcentre refers the point of intersection for the three bisectors which is perpendicular.

Bisector is a term which refers cutting, which is it means cutting into two equal parts.

The Related Terms for Perpendicular Bisectors are given below:

Bisector
Bisector of a line
Line
Line Segment
Midpoint
Perpendicular Lines


Functions of perpendicular bisectors:

When two lines or plane form the conjugant angle it is referred as perpendicular. Through the point B, the line AB is perpendicular to CD. In the definition a line may be infinitely long for representation. In the example AB and CD are considered as the infinitely long lines. Hence the line AB is not intersected with CD which is to be considered as the perpendicular lines since these segments meet the stage of infinity which is used to form similar adjacent angles.

Properties of Perpendicular bisectors:

A line L is detained to be a perpendicular bisector of a line segment AB if L divides AB into two equal parts at right angle. Here, C is the middle point of the line division AB and at this point the line L intersect the line segment.

The line that is perpendicular to the known line segment and that which bisects it is the perpendicular bisector of a line segment.

Lines are held to be perpendicular to each other if they intersect a right angle.

A bisector divides the octave like in half and may be used in place of a generator to derive collections for which structure implies multiplicity is not true such as the ascending melodic trivial, harmonic minor, and octatonic scales.

A line or curve that bisects or divide a line segment, angle, or other build into two equal parts.


Example:

Define the equation of a line which is perpendicular to 4y - 4x = 8 and passes across the point (3, -6) and intersect to form right angles .

Solution:

Given 2y - 2x = 4 and the point (3,-6)

To detect the perpendicular line, we have to find slope first.

Step 1: For finding the slope, we need to change the given equation into slope intercept form.

2y - 2x = 4

Add 2x on both side,

2y - 2x = 4

+ 2x = +2x

2y = 2x + 4

Step 2: Divide by 2 on both sides,

y = (x + 2)

Compared it with y = mx + b we get slope of this equation is 1.

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Step 3 : We know that multiplication of slope of perpendicular lines is equals to -1

Let slope of required line is m then

1 * m = -1

m = -1

Step 4 : The equation of required line is,

(y - y1) = m(x - x1)

(y - (-6)) = -1 (x - 3)

Y + 6 = -1x + 3

Subtract 6 on both sides,

y = -x -3

Probability

The probability is number of possible outcomes is divided in to the total number of possibles. This is probability. The probability contains two types of distributions these are the continuous probability and the discrete probability. The continuous probability is contains Binomial distribution, normal distribution, continuous uniform distribution, and Gamma distribution. The general formation of the probability is

Probability P (A) = "Number of possible outcomes n (a)" / ("Total Number of outcomes n(s)")

I like to share this Probability Simulations with you all through my article.


Examples of Probabilities:

Probability - Example 1:

Roll a dice; find the probability of exactly 2.

Solution:

Total Number of possible = n (a) = {1, 2, 3, 4, 5, 6}

n (s) = 6

The number of outcomes n (a) ={2}

n (a) = 1

The probability of getting value = 1/6 .

Probability - Example 2:

Tossing three coins and finds the probability of two tails and so one head. The possible outcomes are:

Solution:

Step 1:

n (s) = {TTT, TTH, THT, THH, HTT, HTH, HHT, HHH}=8

Step 2:

There are 3 tosses with only one head:

n (a) = {TTH, THT, HTT}=3

Step 3:

Formula:

P (A) = (n(a))/(n(s))

Answer:

P (A) = 3/8 .

Probability - Example 3:

Three cabin van colors red and green. Cabin 1 has 10 vans of red and 10 vans of green, cabin 2 has 7 vans red and 3 green vans and cabin 3 contains 6 red vans and 4 green vans. The respective probabilities of choosing a cabin are 1/4, 1/6, 1/8. What is the probability that the van chosen is green?

Solution

We begin by defining the following sets. Let,

G = the ball chosen is green.
C1 = Cabin 1 is selected
C2 = Cabin 2 is selected
C3 = Cabin 3 is selected

Then P (G|C1) = 10/20 , P(G|C2) = 3/10 and P(G|C3) = 4/10 .

P(G) = ((10/20)xx(1/4)) + ((3/10)xx(1/6))+((4/10)xx(1/8)) = 9/(40)

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Probability - Example 4:

The odds in possible of occurrence of an event 8: 17. Find the probability of the occurrences of this event

Solution:

Number of possible outcomes = 8.

Number of non-possible outcomes = 17.

Total number of outcomes = (8+17) = 25.

P (E) = (Number of the favorable outcomes)/(Number of the unfavorable outcomes)

= 8 / 25.

Learning Parallelograms

Learning definition of parallelograms :

A parallelogram is a four-sided figure so as to have two pair of parallel sides. Opposite sides are similar and identical in length, and opposite angles are identical (angles "a" are the same, and angles "b" are the same). Every plane surfaces as of the floor of a space or church or of the ground surrounded by an enclosure an open space in a structure. The quantity of breathing space within the limit of a flat (2-dimensional) objects such as a triangle or circle

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Learning Properties and Conditions of Parallelograms:

Properties of Parallelograms:

The opposite sides are equal in length.

The opposite angles are similar.

The diagonals intersect all other.

NOTE: Squares, Rectangles and Rhombuses are each and every one Parallelogram.

Conditions to be a Parallelogram:

If quadrilaterals have two pair of similar sides, then it is a parallelogram.

If quadrilaterals have two pair of opposite sides of the equivalent lengths, then it is a parallelogram.

If two pair of opposite angle of a quadrilateral is the equivalent, then the figure is a parallelogram.

If the diagonal of a quadrilateral intersect each other, the figure is a parallelogram.

If a quadrilateral has one pair of opposed sides which are similar and equal in length, then the figure is a parallelogram.

Learning example for Area and Perimeter of Parallelograms:

Learning formulas of parallelograms :

Area of Parallelogram: = b × h , where b = breadth and h = height
Perimeter of Parallelogram:= 2 (b + h) and b=breadth and h=height.


Ex 1:Find the area of a parallelogram with a bottom of 12 centimeters and a height of 5 centimeters.

Solution:Area of Parallelogram = b x h = (12) * (5)

Area of parallelogram = 60 cm2

Ex 2:Find the area of a parallelogram with a bottom of 9 m and a height of 5 m.

Solution:Area of parallelogram = b x h = 9 x 5

Area of parallelogram = 45 m2

My forthcoming post is on Probability Distribution Examples and cbse 9th class books will give you more understanding about Algebra.

Problems based on perimeter of parallelogram

Ex 1:Find the Perimeter of a parallelogram with a bottom of 12 centimeters and a height of 5 centimeters.

Solution: Perimeter of parallelogram = 2 (b + h)  = 2 (12 + 5)= 2 (17)

Perimeter of parallelogram = 34 centimeter

Ex 2:Find the Perimeter of a parallelogram with a bottom of 9 m and a height of 5 m.

Solution:Perimeter of parallelogram = 2 (b + h)    = 2 (9 + 5)    = 2 (14)

Perimeter of parallelogram = 28 m

Polynomials Solver Online

An Algebraic expression of the form axn is called a monomial in x, where a is a known number, x is a variable and n is a non-negative integer. The number a is called the coefficient of xn and n, the degree of the monomial.

For example, 7x3 is a monomial in x of degree 3 and 7 is the coefficient of x3. The sum of a finite number of monomials in x is called a polynomial in x. The coefficients of the monomials in a polynomial are called the coefficients of the polynomial. If all the coefficients of a polynomial  is zero, then the polynomial is called the zero polynomial. Here we are going to see some polynomials solver online.

Polynomials solver online for different operations

Polynomials solver online for Addition

Add the polynomials based on the coefficients of the like powers.

Example 1:

Find the sum of 2x4 – 3x2 + 5x + 3 and 4x + 6x3 – 6x2 – 1.

Solution:

Use the properties of real numbers (distributive and associative)

(2x4 – 3x2 + 5x + 3) + (6x3 – 6x2 + 4x – 1) = 2x4 + 6x3 – 3x2 – 6x2 + 5x + 4x + 3 – 1

= 2x4 + 6x3 – (3+6) x2 + (5+4)x + 2

= 2x4 + 6x3 – 9x2 + 9x + 2.

Polynomials solver online for Subtraction:

We subtract polynomials like addition solver

Example 2:

Subtract 2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6.

Solution:

Using properties of real numbers which is used in addition solver, we have

(x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1) = x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1

= x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1

= (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1)

= –x3 + 8x2 – 4x – 5.

Polynomials solver online for Multiplication

To find the multiplication or product of two polynomials, we use the distributive properties and the law of exponents.

Example 3:

Find the product of x3 – 2x2 – 4 and 2x2 + 3x – 1 .

Solution:

(x3 – 2x2 – 4) (2x2 + 3x – 1)

= x3 (2x2 + 3x – 1) + (–2x2) (2x2 + 3x – 1) + (–4) (2x2 + 3x – 1)

= (2x5 + 3x4 – x3) + (–4x4 – 6x3 + 2x2) + (–8x2 – 12x + 4)

= 2x5 + 3x4 – x3 – 4x4 – 6x3 + 2x2 – 8x2 – 12x + 4

= 2x5 + (3x4 – 4x4) + (–x3 – 6x3) + (2x2 – 8x2) + (–12x) +4

= 2x5 – x4 – 7x3 – 6x2 – 12x + 4.

When finding the product of two polynomials, we multiply each term of one polynomial by each term of the other polynomial and then the products are added (summed). The following scheme may be helpful for the beginners.

Polynomials solver online for Quadratic equation

Definition
It is a polynomial equation the general form is Ax2+Bx+c=0 and the general formula is,

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

Example 1:

To find x2+4x+4

Solution:

= x2+4x+4

= x2+2x+2x+4

= (x+2)(x+2)

Equating zero so that the answer is x=-2,-2

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Example 2

To find x2+3x+2

Solution:

The formula is, x = `(-b+-sqrt(b^2 - 4ac))/(2a)`

Here a=1,b=3,c=2

x = `(-3+-sqrt(3^2 - 4.1.2))/(2(1))`

x = `(-3+-sqrt(9 - 8))/(2)`

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

= `(-2)/(2)`, `(-4)/(2)`

x = -1,-2

Irrational Square Root Answers

In or article we are going to discuss about the term irrational square root. The term square root of a number is that particular value that, once multiplied by itself, gives the number. Then the given number is to be checked whether it is irrational. The term irrational defines when the produced answer or number which cannot be written in simple fraction it means the decimal go on forever without any reputation of numbers. Then we call that number to be irrational. The result that we get in final is called as answer. The term square root in math is defined by the symbol (sqrt). It is always defined that pi is always irrational. Now in this article we are going to discuss about the irrational square root answers in detail with suitable examples and detail explanations.



Example problems on irrational square root answers:

Example 1:

Find the sqrt 17 and produce the answers by checking whether the number 17 is irrational?

Solution:

Given: The sqrt 17 is given

To check whether they answer is irrational

Step 1: As the first step we have to find the sqrt 17 . We use scientific calculator for calculating square root.

= sqrt 17

= 4.12310

Step 2: As found answer the decimal goes on without end without any reputation of numbers. Then we call as irrational number.

sqrt17 = 4.12310

=" irrational".

Answers: 17 are irrational.

Example 2:

Find the sqrt pi and produce the answers by checking whether the number pi is irrational?

Solution:

Given: The sqrt pi is given

To check whether they answer is irrational

Step 1: As the first step we have to find the sqrt pi.  We use scientific calculator for calculating square root.

= sqrt pi

= 1.77245


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Step 2: As found answer the decimal goes on without end without any reputation of numbers. Then we call as irrational number.  Hence according to pi we cannot get an approximate answer.

sqrt pi = 1.77245

= "irrational."

Answers: pi is irrational.

Feasible Solution Definition

The definition of feasible solution In linear programming, any group of values for the variables xj,j= 1, 2, …,n, that (1) accomplish the set of restrictions where them bi  are numerical constants known collectively as the right-hand side and the aij are coefficients of the variables xj, and (2) satisfy the restrictions xj≥ 0. Simple problem in linear programming definition in which it is necessary to find the maximum or minimum value of a simple function subject to certain constraints.

Conditionas

Conditions

Consider Gauss Jordan elimination on a matrix and then group of all variables that are not part of the identity sub matrix equal to zero. We are read out values for the remaining variables. Basic feasible solution all the variables are positive. Definition of basic feasible solutions equate to corners. Pre-multiplying by the matrix AB−1 is equivalent to doing the row operations that transform AB into the identity matrix.

At once fix a matrix A ∈ Rmxn with rank m ≤ n and a vector b ∈ Rm. A function B from 1…..m to 1…..n is indicates which columns to find the identity matrix. In very particular B(i) = j means , the jth column is intended to be all zeros except for a one in row i. The function N from 1….m-n to 1..n indicates which columns are not used in the identity matrix. Each column should be used exactly once, so assume that the union of the ranges of N and M is 1..n. Let AB denote the sub matrix of A corresponding to the columns indicated by B, that is AB = (aB(1) …. aB(m)  ) where at denotes the tth column of A. In order for row operations to produce the identity matrix it is necessary and sufficient that AB be invertible.

Example

Given a combination of functions significant these conditions one can build a definition and basic solution as follows. Set xN = 0 and xB = AB−1  b, where xB and xN are defined analogously to AB. Let x be the vector with component xi equal to the appropriate component of xB or xN depending on which the corresponding column belongs to. Functioning a vector is a basic solution. If it is nonnegative it is feasible and is therefore called a definition of basic feasible solution.

Equations de Algebra

Algebra equation is one of the essential topic used in algebra. Algebra equations are used to determine the missing variable in the equation. In algebra equation, we have one equal sign. The equation should be same to both sides. We will solve the algebra equation using arithmetic operations so that we will get a value for missing variable. This concept of solving algebra equation problems are used in real life applications. There are different forms of equations in algebra. They are linear equation, quadratic equation, equation containing radicals, Equation containing absolute value.

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Let us solve each form of algebra equations.

Example equations de Algebra Problems:

Linear equations:

Problem 1:

Solve for x in the given equation:  8x – 16 = 4x – 24

Solution:

The given equation is 8x – 16 = 4x – 24

Add 16 to each side.

8x – 16 + 16 = 4x – 24 + 16

8x              = 4x – 8

Subtract 4x from each side.

8x – 4x      = 4x – 4x – 8

4x              = - 8

Divide by 4 each side.

`(4x)/4`         = `-8/4`

X               = `-2` .

Problem 2:

Solve for x in the following equation:  4(2x – 6) + 5(4x + 6) = 8(x + 2) + 6

Solution:

The given equation is 4(2x – 6) + 5(4x + 6) = 8(x + 2) + 6

Applying distributive properties to expand the equation

8x – 24 + 20x + 30 = 8x + 16 + 6

Grouping alike terms.

28x + 6  =  8x + 22

Subtract 8x from each side.

28x – 8x + 6 = 8x - 8x + 22

20x + 6  =  22

Subtract 6 from each side.

-6 + 6 + 20x = 22 - 6

20x = 16

Divide by 20 each side.

`(20x)/20 ` =`16/20`

X  = ` 16/20`

Simplifying it.

X = ` 4/5`

Thus we have solved algebra equations.


Additional Problems on equations de Algebra:

Equation containing absolute value:

Problem 3:

Solve for m in the given equation: |4m – 7| + 2 = 20

Solution:

The given equation is |4m – 7| +  2 = 20

Subtract 2 from each sides

|4m – 7| + 2 – 2 = 20 - 2

|4m – 7| = 18

There are two solutions for this equation.

4m – 7 = 18             ,            4m – 7 = -18

4m = 25                 ,             4m = -11

m = `25/4`               ,             m = `-11/4 ` .

Quadratic equation:

Problem 4:

Solve y in the given equation: y2 + 3y + 2 = 0

Solution:

y2 + 3y + 2 = 0

y2 + 2y + y + 2 = 0

Y(y+2) + 1(y+2) = 0

(y+2) (y+1) = 0

Y = -2  and -1 .

Isosceles Trapezoid

An isosceles trapezoid is a type of quadrilateral with a line of symmetry bisecting one pair of opposite sides, making it automatically a trapezoid. Two opposite sides (bases) are parallel, the two other sides (legs) are of equal length. The diagonals are of equal length. An isosceles trapezoid's base angles are equal in measure. Any quadrilateral with exactly one axis of symmetry must be either an isosceles trapezoid or a kite [Source: wikipedia]

Example 1:

Find the area of an isosceles trapezoid with sides is 6 cm, bases 4 cm and 10 cm, and altitude 8 cm?

Solution:

Area of the trapezoidal = h* (b1+b2)/2

Where h = altitude of the trapezoidal

b1, b2 = base of the trapezoidal

From the problem h=8 cm b1= 4 cm and b2 = 10 cm

Area = 8 * (4+10)/2

Area = 8 * 14 /2

Area = 56 cm2

Example 2:

Find the area of an isosceles trapezoid with bases 6 cm and 8 cm Side AB= 12 cm?

Solution:

Area of the trapezoidal =  `sqrt((s-a)*(s-b)*(s-c))`

Where s = `(a+b+2c)/(2)` (or) s =`(a+b)/(2)` + c

Area of the trapezoidal =   `sqrt(((a+b)^2 (a-b+2c)(b-a+2c))/(16))`

From the problem a= 6 cm b= 8 cm c= 12 cm

S = 6+8+2(12)/2

S= 14+24/2

S=38/2

S= 19 cm

S-a = 19-6 = 13 cm

S-b=19-8=11 cm

S-c=19-12 = 7cm

Area = `sqrt(13*11*7^2)`

=`sqrt(13*11*49)`

=`sqrt(7007)`

=83.70 cm2

Example 3:

Find the area of an isosceles trapezoid with sides is 18 cm, bases 12 cm and 14 cm, and altitude 16cm?

Solution:

Area of the trapezoidal = h* (b1+b2)/2

Where h = altitude of the trapezoidal

B1, b2 = base of the trapezoidal

From the problem h=16 cm b1= 12 cm and b2 = 14 cm

Area = 16 * (12+14)/2

Area = 16 * 26 /2

Area = 208 cm2

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Example 4:

Find the altitude of an isosceles trapezoid with sides is 4 cm, bases 5cm and 8cm, and area = 160cm2?

Solution:

Area of the trapezoidal = h* (b1+b2)/2

Where h = altitude of the trapezoidal

B1, b2 = base of the trapezoidal

From the problem b1= 5 cm and b2 = 8 cm, area = 160cm2

160 = h * (5+8)/2

160= h * 13/2

H=160*2/13

H=24.62 cm

Solving Non-Collinear Points

In this section we have solving non-collinear points. In three dimensional geometry, collinear is one of the important topics to measure any of three points are collinear, planes, and so on. In below we will some problems in collinear and non-collinear. We also have some practice problems with answers without explanation. Let us see about solving non-collinear points.

Example problems for solving non-collinear points:

Example problem 1: Check that following points are collinear or non- collinear: P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1).

Solution:

Given P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1)

To check that given is collinear or non-collinear, we have to calculate the distance between the points.

If there is any relation between the points means collinear, otherwise non-collinear.

Now,

PQ = sqrt((1+ 2)^2+ (2 - 3)^2+ (3- 5)^2) =sqrt( 9 +1+ 4) = sqrt(14)

QR = sqrt((7 -1)^2+ (0 - 2)^2+ (-1-3)^2) = sqrt(36 + 4 +16) = sqrt(56) = 2 sqrt(14)

PR = sqrt((7 + 2)^2+ (0 - 3)^2+ (-1- 5)^2)= sqrt(81+ 9 + 36) = sqrt(126) = 3sqrt(14)

From the above we can see that, PQ + QR = PR.

Therefore, the P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1) are collinear.

Answer: P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1) are collinear.

Example problem 2: Are the following points A (3, 6, 9), B (10, 20, 30) and C (25, – 41, 5), the non-collinear?

Solution:

Given A (3, 6, 9), B (10, 20, 30) and C (25, – 41, 5)

To check that given is collinear or non-collinear, we have to calculate the distance between the points.

If there is any relation between the points means collinear, otherwise non-collinear.

 AB = sqrt((10 -3)^2 + (20- 6 )^2 + (30-9)^2) = sqrt(49 + 324 +441) = sqrt(814)

BC = sqrt((25 - 3)^2 + (-41- 20)^2 + (5 - 30)^2 )= sqrt(225 + 3721 + 625) = sqrt( 4571)

AC = sqrt((25 -3)^2 +(-41 - 6)^2 + (5 -9)^2) = sqrt(484 + 2209 + 16) = sqrt(2709) = 3sqrt(301)

Form the above analysis we can see that there is no relation between the three points. So the given are non-collinear.

Answer: The A (3, 6, 9), B (10, 20, 30) and C (25, – 41, 5) non-collinear.

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Practice problems for solving non-collinear points:

Practice problem 1: Show that the following points are non-collinear (2, –1, 3) (0, 1, -1) and (-2, 1, 3).

Practice problem 2: Check that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear or non-collinear.

Solutions for solving non-collinear points:

Solution 1: Non-collinear

Solution 2: Collinear

Scientific Notation

Scientific notation is clear as shorter method to state very big numbers into simplest form. This method is called as scientific notation. Scientific notation is based on powers of the base integer 10 have expressed in the scientific notation. Example for scientific notation of 5670000000 this can be note down as 5.67 x 109. And the one more method in decimal is 0.0000081 = 8.1 x 10 -6

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Steps to find scientific notation:

Steps to find scientific notation is given below that:

The scientific notation (number) has one number to the left of the decimal point.

We contain to put the decimal point following the first digit.

How many places the decimal point was moved point out through the power of ten (10).

Numbers in Scientific notation is writing as a common form:

P × 10q (where P is the coefficient that is in between 1 and 10. And q is the exponents)

Coefficient × 10exponents

A positive exponent in scientific notation demonstrates that the decimal point is moved that number of spaces to the right. A negative exponent demonstrate that the decimal point is shifted that number of spaces to the left.

Example problems based on 4 in scientific notation:

Example 1:

4 change into scientific notation.

Solution:

Specified that 4

Specified number is not a big number, but it is simple to convert scientific notation.

First note down 4 x 100 = 4 [100 = 1]

After that it can be note down in scientific notation = 4 x 100

Example 2:

4444 change into scientific notation.

Solution:

Specified that 4444

Specified number is a big number, but it is simple to convert scientific notation.

First note down 4.444 this is not equivalent to specified number

Calculate how many numbers available following the decimal point in earlier step

Present are 3 digits following the decimal point so you can place 8 zero

4.444 × (1000)

Therefore the notation will be 103

The scientific notation can be note down as 4.444 x 103

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Example 3:

0.0044 change into scientific notation.

Solution:

Specified that 0.0044

Specified number is not a big number, but it is simple to convert scientific notation.

First note down the number as 4.4

There are 3 digit following the number is there so we can write 10-3

The scientific notation can be note down as 4.4 x 10-3

Inverse Sine Function

In this page we are going to ndiscuss about inverse sine function concept. In trigonometric we are having sine, cosine and tan functions and we have the inverse function for all these. Sine inverse function is the inverse function of sine. We will learn about the sine inverse function. Let us take an example function f its inverse function is F-1. F (x) = y then x = F-1(y). The inverse trigonometric functions are important for the integrals.


Explanation for inverse sine function

Normally if we have a function we can find the value of inverse by taking the inverse of the given function. Sin inverse function is nothing but the inverse function of the sine. We can represent this like the following sin-1. If we have a function like y = sin x then the sin inverse function will be like the following.  = sin-1 y. We have to take the sin inverse for both side then we will get the variables one side and the inverse function of the sine another side. Let us see the graph for the sin function

Inverse of sine function
Examples for inverse of sine function

Here we are going to see some problems to find the sin inverse value.

Example 1:

Find the sin inverse value of y where y = Sin (0.9898)

Solution:

y = Sin (0.9898)

To find the sin inverse we have to take the sin inverse for both sides.

Sin-1 (y) = Sin-1 (sin (0.9898))

Sin-1 (y) = 0.9898

Example 2:

Find the sin inverse of the given value. X = 0.2588

Solution:

x = 0.2588

To find the sin inverse we have to take the sin inverse for both sides.

Sin-1 (x) = Sin-1 (0.2588)

Sin-1 (x) = 14.99



Example 3:

Find the sin inverse of the given value. X = 0. 1234

Solution:

x = 0.1234

To find the sin inverse we have to take the sin inverse for both sides.

Sin-1 (x) = Sin-1 (0.1234)

Sin-1 (x) = 7.08

Learn Nonlinear Functions

The Nonlinear Functions Examples function  f(x)  satisfies both of the following properties:

• Additivity, f (x+y) = f(x) + f(y)

• Homogeneity, f (ax) = a f(x)

The following functions are the non-linear functions,

Exponential Functions

The function defined by  f (x) = am (a > 0, a!=1) is called an exponential function with base a and exponent m. The domain of f is the set of all the real numbers.

learn nonlinear functions-Laws of Exponents:

If m and n are real numbers and a and b are positive numbers.

• (am) (an) = am+n Ex: 22.23=22+3=25

• (ab)m = ambm Ex: (2x3)2=2 2x2 3

• (am)n = amn Ex: (23)2 = 23(2) = 2 6

• a0 = 1

• (am) / (an) = am-n Ex:55/52=55-2=53

Quadratic Functions are examples for nonlinear functions

• General formula: f(x) = ax2+bx+c, where a, b and c are constants .

• The graph of a quadratic function f(x)=ax2+bx+c is called a parabola.

• If a>0 we say the parabola opens upward.

• If a<0 downward.="" opens="" p="" parabola="" say="" the="" we="">
Note: vertex is the “lowest” or the “highest” point on the parabola.

• Let f(x)=ax2+bx+c , the vertex point is when
X= - b ,  f( - b ) = 4ac-b2
2a        2a         4a
Thus, the vertex is. X = -b , f (- b)
2a      2a

Logarithm Functions

• If ax=y, where a is any positive number other than 1, then we define x as the logarithm of y to the base a, denoted by loga, and is defined by loga y = x <=> ax=y, a > 0, a ≠ 1

• The logarithm function is closely related with the exponential function.

• Logarithm of any negative number with base a (a > 0, a != 1) is undefined.

Common Logarithms

• The logs used are to base 10, i.e. a = 10.

• Denoted by log10 or log or lg.

Natural Logarithms

• The logs used are to base e, i.e. a = e

• Denoted by ln or loge.


Example problems on learn non-linear functions

Ex:1 Sketch the graph of

f(x)=-x2-10x.

Sol:

1) a=-1, so a < 0 then the parabola opens downwards.

2) a=-1, b=-10 the vertex is when
b            (-10)
x= - ------- = - --------- = - 5
2a          2 x (-1)

f(-5)=-(-5)2 -10(-5) = -25+50=25

Thus, the vertex point is (-5, 25).

3) C=0, so the y-intercept is at (0,0)

4) f(x)=0; -x2-10x=0

x(-x-10)=0

x=0 or –x-10 = 0

x=-10

The x-intercept points are (0,0) and (-10, 0).