Learn Linear Functions Slope

A linear function is defined as the polynomial function contains the degree of one (y = mx + b). One can learn the linear equation relates a dependent variable with an independent variable in a simple way. The power of the linear function which is not always greater than one where there is no independent variable. A simple linear function with one independent variable (Ax + By + C = 0) traces a straight line when plotted on a graph. It is also called as linear equation. Learning the concept of slope using linear equations is known as learning linear functions slope.

Please express your views of this topic Linear Approximation Equation by commenting on blog.

Learning Linear Functions Forms:

The function is defined by,

f = { ( X, Y)/ Y = mX + b }

where m and b are constants, x and y is called a linear functions. The function derives a straight line while graphing.

Functions such as these gives graph that are straight lines, and, thus, the name linear. Linear functions come in three main forms.

Point Slope Form is given by the equation, m = (y - y1) / ( x – x1)
Slope-Intercept Form is given by the equation, y = mx +b
General Form is given by the equation, Ax + By + C = 0.


Slope of the Linear Functions Learning:

Calculations of rate at which the change takes place can be done under the concept of slope. Slope calculates the rate of change in the dependent variable as the independent variable changes. The slope is denoted by m.

Consider the linear function:

y = mx + b

where, m is the slope of the line and b is the y-intercept. Slope is defined as the ratio of unit change in y to the change in x.

slope m = Change in y / Change in x

=> m = (y_(2) - y_(1))/(x_(2) - x_(1))

Learning Linear Function Slope - Examples:

Find the slope of the line segment relating the following points:
(-1,-2) and (1, 6)

Sol:

Here, x1 = -1        

y1 = -2

x2 = 1        

y2 = 6

slope m = (y2 – y1) / (x2 – x1)

=> m = (6 – (-2) / (1 – (-1))

=> m = (6 + 2) / (1 + 1)

=> m = 8 / 2

=> Slope m = 4


Find the slope of the equation, 9x - 3y = 6

Sol:

9x - 3y = 6

=> -3y = 6 – 9x

=> y = (-1/3)(6 – 9x)

=> y = 3x – 2

It is in the general form, y = mx + b

Therefore, slope m = 3 and y-intercept b = -2.

Continuity Learning

Let 'a' be an aggregate and a in A, A function f : A |-> R is said to be continuous at "A". If each E > 0 EE   delta >0 such that x in A, || < delta  x - a

rArr   |  f ( x ) - f ( a ) | < E
lim_(x->a) f ( x ) = f ( a )
f is continuous at a    hArr lim_(x->a) f ( x ) = f ( a )

A function f : A |-> Ris said to be continuous on the aggregate 'a'. If ' f ' is continuous at every point a in A
A function f : A -> R is said to be discontinuous at a A. If ' f ' is not continuous at ' a '.


Continuity on the left and right at 'a':-

Let 'a' be an aggregate and a in A. A function f : A|-> R is said to be continuous on the left at 'a'. If each E>0 EE delta >0 such that x in A , a - delta < x <= a  rArr | f ( x ) - f ( a ) | < E

lim_(x->a-)  f ( x ) = f ( a )

Let 'a' be an aggregate and a in A. A function f : A|-> R is said to be continuous on the left at 'a'. If each E>0 EE delta >0 such that x in A , a <= x < a + delta   rArr | f ( x ) - f ( a ) | < E

lim_(x->a+) f ( x ) = f ( a )

Continuity of a function at open and closed intervals:-

A function ' f ' is said to be continuous on an open interval ( a, b ). If ' f ' is continuous at ' x ' AA x in ( a, b ).

A function ' f ' is said to be continuous on an closed interval [ a, b ].

' f ' is continuous at 'x' AA x in ( a, b )
' f ' is right continuous at 'a'.
' f ' is left continuous at ' b'.


Types of discontinuity:-

Let f : A -> Rand a in A be a point of discontinuity of ' f '.

' f ' is said to have Removable discontinuity at 'a'. If lim_(x->a) f ( x ) exists and lim_(x->a) f ( x ) != f ( a ) or f ( a )  is not defined.
' f ' is said to have Jump discontinuity or discontinuity of first kind at 'a'. If lim_(x->a-) f ( x ), lim_(x->a+) f ( x ) both exists and

lim_(x->a-)    f ( x )        !=  lim_(x->a+) f ( x )

' f ' is said to have sample discontinuity at 'a'. If ' f ' has removable discontinuity ( or ) jump discontinuity at 'a'.
' f ' is said to have finite discontinuity at 'a'. If ' f ' is not continuous at 'a' and ' f ' is bounded at 'a'.
' f ' is said to have Infinite discontinuity at 'a'.

If lim_(x->a) f ( x ) = oo   ( or ) lim_(x->a) f ( x ) = -oo
f ( x ) is unbounded in every neighbourhood of 'a'.

Example problems on Continuity

1)Examining the continuity of ' f ' defined by f ( x ) = | x | + | x - 1|

Solution:-    | x| = x if x > 0
= - x if x <= 0
| x - 1| = x if x > 1
= - x if x <= 1

If x <= 0
f ( x ) = | x | + | x - 1|
= - x - ( x -1 )
= 1 - 2x

If 0 < x < 1
f ( x ) =  | x | + | x - 1|
= x  - ( x - 1 )
= x - x + 1
= 1

If x >= 1
f ( x ) = | x | + | x - 1|
= x + x -1
= 2x -1

f ( x ) = 1 - 2x ; If x <= 0
= 1        ;  if 0 < x < 1
= 2x -1  ; if x >= 1

Continuity of ' f ' at x =0

Left Hand Limit            lim_(x->0-) f ( x ) = lim_(x->0-) 1 - 2x = 1

Right Hand Limit          lim_(x->0+) f ( x ) = lim_(x->0+) 1  = 1

f ( x ) = 1 - 2x if x <= 0
f ( 0 ) = 1

lim_(x->0) f ( x ) = f ( 0 ) = 1

Continuity of ' f ' at x =1

Left Hand Limit             lim_(x->1-) f ( x ) = lim_(x->1-) 1  = 1

Right Hand Limit            lim_(x->1+) f ( x ) = lim_(x->1+) 2x - 1  = 1

f ( x ) = 2x -1 if x >= 1
f ( 1 ) = 2 ( 1 )  - 1 = 1

lim_(x->1) f ( x ) = f ( 1 ) = 1

The function is continuous at x = 1

Intersecting Acute Angles

The intersecting acute angles measure the 0° and 90°. A sketch of angles is the main topic in geometry. Two lines join at a single point to form an angle.

If P and R are two straight lines they intersect at a point Q and make an angle Q. The angle Q is also represented as PQR. Here center point represents the vertex of the angle.

Example problems for intersecting acute angles:

Problem 1:-

Solving the intersecting acute angle from the where slopes  m1= 4 and m2=14

Solution:

In above diagram show the acute angles intersecting the two lines

tan phi = (m2- m1)/ (1+m1m2)

= (14-4)/(1+(14x4))

=10/ 57

= 0.175

Hence we find the phi value.

phi = arc tan(0.175)

from the tangent table,we get
phi =1o  (rounded value)



from the given example we can find the obtuse angle.

To subtract the acute angle between the the two values L,L2 from the 1800 straight angle.

That is,

phi = 180- phi

phi = 180 -1o

phi = 179o

Hence we find the obtuse angle.

Problem 2:-

Solving the intersecting acute angle from the where slopes  m1= 20 and m2=28

Solution:

In above diagram show the acute angles intersecting the two lines

tan phi = (m2- m1)/(1+m1m2)

= (28-20)/(1+(28x20)

= 8/ 561

= 0.0142

Hence we find the phi value.

phi   = arc tan(0.0142)

from the tangent table,we get
phi   = 1o  (rounded value)

from the given example we can find the obtuse angle.

the obtuse angle is calculated by subtracting the acute angle between the value of L and L2 from the straight angle 180 0

ie phi = 180- phi

phi = 180 -1o

phi = 179o

Hence we find the obtuse angle.

My forthcoming post is on algebra 2 formulas sheet and sample papers for class 10 cbse will give you more understanding about Algebra.

Practice problems for intersecting acute angles:

1. Solving the intersecting acute angle from the where slopes  m1= 2 and m2=4

Answer:- 168o

2. Solving the intersecting acute angle from the where slopes  m1= 4 and m2= 8

Answer:- 173o

3. Solving the intersecting acute angle from the where slopes  m1= 8 and m2=12

Answer:- 178 o

4. Solving the intersecting acute angle from the where slopes  m1= 12 and m2=14

Answer:- 174 o

Time Value of Money

Time value of money is one of the most important concepts in mathematics, also widely used in real life finance and economical scenarios. Time value of money is taught sometime in high school mathematics that plays a major role in the practical life. Let’s discuss about the same in this post.

What is Time Value of Money or TVM?
Time value of money is popularly abbreviated as TVM. The concept of Time value of money defines that the value of money keeps changing with time. Most of the time, the value of money today is lesser than the value of money tomorrow. For example: The popular parenting magazine cost was Rs.50 till last year. Today, the cost of the same parenting magazine is Rs.75. This demonstrates that in the period of one year, the value of money has changed. Understanding the concept of Time value of money requires understanding in related concepts like Present value and Future value.Present Value and Example:

The current worth of money is termed as the present value. For example: Only for today they have offered the scheme of Rs.699 for post pregnancy weight loss program. From tomorrow the rate of post pregnancy weight loss program will be again Rs.1799. Therefore, the present value here is Rs.699. Based on the present value and the difference in future, one can find out the rate of increase or decrease in TVM.

Future Value and Example:
Future value is the amount or value of money on a specified date in future with respect to the same in today’s date. For example: The book on develop reading habit cost is Rs.100 today but the cost of same book on develop read habit will increase to Rs.150 by next month, owing to its popularity. Thus, Rs.150 is the future value in this situation.
These are the basics on Time Value of Money.

Coordinate Plane

What is a Coordinate Plane?

The coordinate plane is a two-dimensional surface on which we can plot points, lines and curves. It has two scales, called the x-axis and y-axis, at right angles to each other. The plural of axis is 'axes' (pronounced "AXE-ease").


Definitions

X axis

The horizontal scale is called the x-axis and is usually drawn with the zero point in the middle. As you go to the right on the scale the values get larger and are positive. As you go to the left, they get larger but are negative.
Y axis

The vertical scale is called the y-axis and is also usually drawn with the zero point in the middle. As you go up from zero the numbers are increasing in a positive direction. As you go down from zero they increase but are negative.


Origin

The point where the two axes cross (at zero on both scales) is called the origin. In the figure above you can drag the origin point to reposition it to a more suitable location at any time.


Quadrants



Quadrants of the coordinate plane When the origin is in the center of the plane, they divide it into four areas called quadrants. The first quadrant, by convention, is the top right, and then they go around counter-clockwise. It is conventional to label them with numerals but we talk about them as "first, second, third, and fourth quadrant".

Coordinates of a point

The coordinates of a point are a pair of numbers that define its exact location on a two-dimensional plane. Recall that the coordinate plane has two axes at right angles to each other, called the x and y axis. The coordinates of a given point represent how far along each axis the point is located.
Ordered Pair

The two numbers in parentheses are the x and y coordinate of the point. The first number (x) specifies how far along the x (horizontal) axis the point is. The second is the y coordinate and specifies how far up or down the y axis to go. It is called an ordered pair because the order of the two numbers matters - the first is always the x (horizontal) coordinate.

The sign of the coordinate is important. A positive number means to go to the right (x) or up(y). Negative numbers mean to go left (x) or down (y). (The figure at the top of the page has the values of the axes labelled with the appropriate sign).
Abscissa

The abscissa is another name for the x (horizontal) coordinate of a point. Pronounced "ab-SISS-ah" (the 'c;' is silent). Not used very much. Most commonly, the term "x-coordinate" is used.
Ordinate



The ordinate is another name for the y (vertical) coordinate of a point. Pronounced "ORD-inet". Not used very much. Most commonly, the term "y-coordinate" is used.

Pan Balance Math Problems

A balance is a beam balance or laboratory balance or pan balance used to measure the accurate value of mass of an object. A weighting pan and scale pan span is hold above the machine. Pan balance is used to distribute weight between two containers in a machine balance.  By using pan balance we have to calculate the weight of the objects and evenly distribute the objects. Let us see about pan balance math problems in this article.


Worked Examples to Pan Balance Math Problems

Example 1 to Pan Balance Math Problems:

Solve the pan balance math problem.

Pan Balance Math Problems

Find the weight of the two rectangles in the above problem?

Solution:

Step 1:

Given weight in left side object = 6 kg

Given weight in the right side object = 11 kg

Step 2:

In left side, the pan is not balanced if we remove the two rectangles.

So, subtract the right side object weight – left side present object weight, we will get the result.

Step 3:

Subtract the amount on left side object from the right side object, we get,

11 – 6 = 5 kg

Step 4:

Two rectangles have the total of 5 kg.

Step 5:

Two rectangles in the left side of the pan balance are of the same size.

Therefore, each rectangle is of 2.5 kg.

Example 2 to Pan Balance Math Problems:

Solve the pan balance math problem.

Pan Balance Math Problems

Find the weight of the diamond object in the above problem?

Solution:

Step 1:

Total weight in the right side of the pan balance = 17 kg

Triangle object = 4 kg

Square object = 8 kg

Diamond object =?

Step 2:

In left side, the pan is not balanced if we remove the diamond object.

So, subtract the right side object weight – left side present object weight, we will get the result.

Step 3:

Adding the triangle and square shaped object weight = 4 + 8 = 12

Left side = 12 kg

Step 4:

Subtracting the right side objects weight – left side objects weight = 17 – 12 = 5 kg

Step 5:

Therefore, the weight of the diamond object is 5 kg.



Practice Problems to Pan Balance Math Problems

Solve the pan balance math problem.

Pan Balance Math Problems

Find the weight of the diamond object in the above problem?

Solution:

1. Pan balance weight in the right side is 10 kg.

Proving Identities Calculator

An identity is a relative which is tautologically factual. These are typically taken to denote something that is factual by meaning, either straight by the definition, or as a result of it. For model, algebraically, this occurs if an equation is content for all values of the concerned variables. Definitions are frequently indicated by the 'triple bar' sign ≡, such as A2 ≡ x·x. A calculator is a little (often small), typically cheap electronic machine used to do the basic operations of mathematics. Present calculators are additional moveable than most computers, although most PDAs are similar in size to handheld calculators.


Proving Algebra and Trigonometric Identities Calculator:

proving identities calculator

c + 0 = c (additive identity)
c + (-c) =0 (additive inverse)
c x 1 = c (the multiplicative identity)
c  (1/c )= 1 (the multiplicative inverse)
sin2a≡ 1 – cos2a
cos2a≡ 1 – sin2a
tan2a ≡ sec2a − 1

Examples of Proving Algebra Identities:

Study example 1:

21+ 0 = 21

Solution:

(We know that the additive identity in algebra (c+0 = c). Therefore the answer is 21)

Study example 2:

23+(-23) = 0

Solution:

(We know that the additive inverse in algebra (c+(-c) = 0) this problem answer 0.)

Study example 3:

1x14= 14

Solution:

(We know that the multiplicative identity  (1x c = c). So this problem answer 14.)

study example 4:

23 (1/23) = 1

Solution:

(We know that the multiplicative inverse  (c x 1/c = 1). So this problem answer is 23.)

Between, if you have problem on these topics math 2nd grade word problems, please browse expert math related websites for more help on how to prepare for iit jee 2013.

Examples for Proving Trigonometric Identities:

Study example 1:

Prove the following identity: (sin M)4 +2(sin M)2(cos M)2 + (cos M)4 = tan M cot M

Solution:

(sin M)4 +2(sin M)2(cos M)2 + (cos M)4 = tan M cot M

(sin M)4 +2(sin M)2(cos M)2 + (cos M)4 =(sin M/cos M)xx(cos M/sinM)

(sin M)4 +2(sin M)2(cos M)2 + (cos M)4 = 1
((sin2M)2 + (cos2M)2)((sin2M)2 + (cos2M)2) = 1       { (a+b)2=a2+b2+2ab}

( (Sin2M)+cos2M )2 =1

(1)(1) = 1

Study example 2:

Prove the following identity: tan(p)cos(p) = sin(p)

Solution:

tan(p)cos(p) = sin(p)

(sinP/cosP) cos p= sin(p)

sin(p) = sin(p)