Fitted Regression Equation

In mathematics, regression is one of the most interesting topics in statistics. The process of determining the relationship along with the two variables is called as regression. It is also one of the statistical analysis methods that can be used to assessing the association between the two different variables. In this article, we shall discuss about the fitted regression equation. It is used to help the students for the study about the fitted regression equation and the step by step explanations for the student doubts. The following are the example problem in fitted regression equation.
Fitted Regression Equation - Formula:

Formula for regression:

Regression Equation(y) = a + bx
Slope (b) = `(NsumXY - (sumX) (sumY)) / (NsumX^2 - (sumX)^2)`
Intercept(a) = `(sumY - b(sumX)) / N`

Looking out for more help on percentage equation in algebra by visiting listed websites.

Where
x and y are the variables.
b = the slope of the regression line is called as regression coefficient
a = intercept point of the regression line which is in the y-axis.
N = Number of values or elements
X = First Score
Y = Second Score
`sumXY` = Sum of the product of the first and Second Scores
`sumX` = Sum of First Scores
`sumY` = Sum of Second Scores
`sumX^2` = Sum of square First Scores.
Fitted Regression Equation - Example Problem:

Example:

Plot the fitted regression scatter plot for the given set of data and also solving the regression equation by finding the regression coefficients and slope values.
x y
32 22
35 26
37 29
49 34
56 40
89 53


Solution:

Let us count the number of values.
N = 6

Determine the values for xy, x2
x y xy x2
32 22 704 1024
35 26 910 1225
37 29 1073 1369
49 34 1666 2401
56 40 2240 3136
89 53 4717 7921


Find the following values `sumX` , `sumY` , `sumXY` , `sumX^2`.
`sumX` = 298
`sumY ` = 204
`sumXY` = 11310
`sumX^2 ` = 17076


Substitute values in the slope formula
Slope (b) = `(NsumXY - (sumX) (sumY)) / (NsumX^2 - (sumX)^2)`
= `((6)*(11310)-(298)*(204))/((6)*(17076)-(298)^2)`
= `(67860 - 60790)/(102456 - 88804)`
=  `7070/13652`
b = 0.5178
Substitute the values in the intercept formula given.
Intercept (a) = `(sumY - b (sumX)) / N`
= `(204 - 0.5178 (298))/6`
= `(204 - 154.3261)/6`
= `49.6738/6`
a = 8.2789

Substitute the Regression coefficient value and intercept value in the regression equation
Regression Equation(y) = a + bx
= 8.2789 + 0.5178x

For the given set of data, we can plot the graph

1. Take the X and Y values in the given set of data

2. In the graph, take the scale as for x and y axis as follows

3. Mark the x and y values in the graph

4. Plot the scatter plot and fitted regression line as in the diagram

Fitted regression equation - Example

Variable Pairs

Variable pairs determines the concept of an ordered pair and it shows you to plot ordered pairs  on a graph. The variable determine the value of the particular terms of the equation.

The variable pair is a set of two variable like (x, y) written in parentheses. The variable in terms of graph represent the first value as x and the second value as y. (2,3) represents x axis is 2 units long and y axis is 3 units long.

Example Problem for Variable Pairs:

Solve for the ordered pair from the following equations:

4x + 3y = 12

2x + 3y =  8 solve for x and y

Solution:

Given that 4x + 3y = 12--------------------- (1)

2x + 3y =   8--------------------- (2)

Subtract the first equation and second equation,



4x + 3y = 12

2x + 3y =   8

________________

2x  = 4

________________

That is

4x – 2x = 2x

3y – 3y = 0

12 – 8   = 4

Now we have the equation 2x = 4 after subtraction.

2x = 4

Divide by 2 on both sides

x = `4 / 2`

x = 2

Substitute the x = 2 in the first equation, we will get the value of y variable.

First equation 4x + 3y = 12

4(2) + 3y = 12

8 + 3y = 12

3y = 12 – 8

3y = 4

y = `4 / 3`

The solution is x = 2 and y = `4 /3` .

My forthcoming post is on Exponential Data Sets and Exponential Differentiation will give you more understanding about Algebra.

Example Problem for Variable Pairs:

Solve for the ordered pair from the following equations:

8x + 2y = 4

x + 3y =   5

Solution:

Given that 8x + 2y = 4--------------------- (1)

x + 3y =   5--------------------- (2)

Subtract the first equation and second equation,

3 * (1)    24x + 6y = 12

2 * (2)      2x + 6y =-10

________________

22x   = 22

________________

That is

24x – 2x = 22x

6y – 6y = 0

12 – (-10)  = 22

Now we have the equation 22x = 22 after subtraction.

Divide by 22 on both sides. we get

x = 1

Substitute the x = 22 in the first equation,

First equation 8x + 3y = 4

8(22) + 3y = 4

176 + 3y = 4

3y = 4 – 176

3y = -172

y = `-172 / 3`

The solution is x = 1 and y = -`172 /3` .

The value of variable pairs (x,y) is (1,-`172/3` )

Linearly Independent Determinant

In mathematics, determinants are linearly independent if none of the determinants can be obtained from the others. When determinants are linearly independent, then each determinant contains new information about the variables. For example

A = [[0],[0],[2]] B = [[2],[-4],[2]] C = [[0],[4],[-4]] D = [[8],[4],[3]] .

Here, the determinants A, B and C are linearly independent but the determinant of D is not since determinant of D is equals to 9A + 4B + 5C. Following example and practice problems will help you to study about linearly independent determinant.
Example Problems of Linearly Independent Determinant:

Example problem 1:

Show that the determinant [[2,3],[4,5]] is linearly dependent or not using Wronskian determinant method.

Solution:

Step 1: Given determinant

[[2,3],[4,5]] .

Step 2: Condition of Wronskian determinant

If the value of the determinant is equal to 0, then the determinant is linearly dependent.

If the value of the determinant is not equal to 0, then the determinant is linearly independent.

Step 3: Calculate the value of the determinant.

[[2,3],[4,5]] = (2 * 5) - (3 * 4)

= -2

Since the value of the determinant is not equal to 0, the given determinant is linearly independent.

Step 4: Solution

Hence, the given determinant [[2,3],[4,5]] is linearly independent.

Example problem 2:

Show that the determinant [[1,4],[5,20]] is linearly dependent or not using Wronskian determinant method.

Solution:

Step 1: Given determinant

[[1,4],[5,20]] .

Step 2: Condition of Wronskian determinant

If the value of the determinant is equal to 0, then the determinant is linearly dependent.

If the value of the determinant is not equal to 0, then the determinant is linearly independent.

Step 3: Calculate the value of the determinant.

[[1,4],[5,20]] = (1 * 20) - (4 * 5)

= 0

Since the value of the determinant is equal to 0, the given determinant is linearly dependent.

Step 4: Solution

Hence, the given determinant [[1,4],[5,20]] is linearly dependent.
Practice Problems of Linearly Independent Determinant:

1) Show that the determinant [[4,3],[2,8]] is linearly dependent or not using Wronskian determinant method.

2) Show that the determinant [[2,5],[4,10]] is linearly dependent or not using Wronskian determinant method.

Solutions:

1) The given determinant [[4,3],[2,8]] is linearly independent.

2) The given determinant [[2,5],[4,10]] is linearly dependent.

Rules for Factoring Trinomials

Trinomial is nothing but if we are having the function with three terms we can say that is a trinomial. Here we are going to learn what the rules available in factoring the trianomial are. There are three rules available for factoring trinomials. We will see the rules for factoring the trinomials with appropriate examples. Factoring the trinomial is depends on the degree of polynomials.

Rules for Factoring Trinomials:

If we want to factor a trinomial we have to start from the highest exponents. For factoring the trinomial first we have to find the greatest common factor and we have to simplify.

Example:

Factor the following trinomial 6x8 + 30 x7 + 36 x6

Solution:

First we have to find the factors of the greatest value

6x8 = 2. 3.  x2. x6

30x7 = 2. 3. 5.  x. x6

36 x6 = 2. 3. 6. x6

And here GCF of the variables 6x8, 30 x7, 36 x6 is 6x6

So the factor value is 6x6 (x2 + 5x + 6) = 6x6 (x + 2) (x + 3)

Rule 2 for factoring trinomial:

In rule 2 we have to see the factoring rules in algebra based on grouping the value. Here we are going to group the given trinomial and we have to factor this.

Example:

Factor the trinomial 3y2 + y – 4

Solution:

3y2 + y – 4

This is 3y2 + 4y – 3y - 4

So we get 3y(y + 4) - (3y + 4)

From this y (3y - 1) (3y + 4)

My forthcoming post is on Triangular Prism Net and Formula for Volume of a Rectangular Prism will give you more understanding about Algebra.

Other Rules for Factoring Trinomial:

Rule 3 for factoring trinomial:

If the given trinomial expression starts with the co – efficient of x2 we have to start from multiplication.

Example:

Factor the trinomial 10x2 - 14x + 4

Solution:

Given expression is 10x2 - 14x + 4

First we have to multiply 10 and 4

We get x2 - 13x + 40

Now if we divide this 10x2 - 4x – 10x + 4

x(10x –4) – (10x – 4)

(10x – 4) (x – 1)

These rules are helping us to factor the trinomial easily.

Measuring Angles Interactive

The interactive of measuring the exterior angles are the angles between any sides of shapes, and a line extended from the other side. Interactive figures exterior angles are any straight sides with flat shape. Polygon measuring exterior angle sum is 360°. In other words Polygon measuring of sum exterior angle is one full revolution.  The measuring angles are made with an extension of a side and its corresponding adjacent sides are the exterior angle.

Overview for Measuring Angles Interactive:

If you can measuring the interior and the exterior angle of straight line gives 180 degree. The Interactive regular polygons or objects exterior angles are the congruent, and polygon measuring exterior angle sum is 360°. So each polygon of measuring exterior angle is 360 divided by the n, the number of sides. Any figure interior angle is supplementary to an exterior angle at that vertex is 180.  Interior angles and exterior angles are always complementary.
Example Problems for Measuring Angles Interactive:

1) Measuring exterior angles of polygon, that the polygon angles are 80, x, 95, 125 degrees

Solution:

All polygons measuring of exterior angle sum is 360 degrees.

So add all the exterior angles and equate to the 360 degrees.

The given exterior angles are 80, x, 95, 125 degrees.

80+95+125+x = 360

300+x =x360

Subtracted both sides by 300

x = 360-300

x = 60`^@`

2) Measuring of exterior angles of polygon, that the polygon angles are 40, 2x, 3x, 2x, x degrees

Solution:

All polygon measuring exterior angle sums is 360 degrees.

So add up all the exterior angles and equate to the 360 degrees.

Given exterior angles are 40, 2x, 3x, 2x, x degrees.

40+x+2x+3x+2x = 360

40+8x = 360

Subtract both sides by 40 degrees

8x = 320

Divide both sides by 8

X = 40

Therefore each angle is

X = 40

2x = 2*40 = 80

3x = 3*40 =120

Each exterior angle is 40°, 40°, 80°, 120°, and 80°.


3) Measuring exterior angles of polygon, that the polygon angles are 80, x, 95, 125 degrees

Solution:

Given figure angle is four sided shape.

The sum of interior angle any four sided figure is 360 degree

All polygons measuring of interior angle sum is 360 degrees.

So add all the interior angles and equate to the 360 degrees.

The given interior angles are 80, x, 95, 115 degrees.

80+95+115+x  =360

290+x = 360

Subtracted both sides by 290

x = 360-290

x = 70`^@`

General Equation of an Ellipse

The general equation of an ellipse with major axis as the x- axis is

`((x-h)^2)/(a^2) + ((y-k)^2)/(b^2)= 1`

where (h,k) is the centre of the ellipse.

ellipse with major axis as x - axis

ellipse

The general equation of an ellipse with major axis as the y-axis is

`((x-h)^2)/(b^2) + ((y-k)^2)/(a^2)= 1`

ellipse with major axis as y - axis

ellipse

Lets solve some problems to find the general equation of ellipse.

Find the General Equation of an Ellipse - Examples

Ex : 1 Find the general equation of an ellipse whose vertices are (-1,4) and (-7,4) and eccentricity is `1/3` .

Sol : From the given data the major axis is parallel to x- axis.

`:.` The general equation form of an equation is `((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) =1`

The centre is the midpoint of AA'

example 1 to find general equation

`:.` Cis ( `(-1-7)/2,(4+4)/2` ) = (-4,4)

Thus the equation becomes

`((x+4)^2)/(a^2) +( (y-4)^2)/(b^2) = 1`

We know that AA' = 2a = 6 `rArr` a=3

b2 = a2(1-e2) = 9( 1- `1/9` ) = 8

The required equation is `((x+4)^2)/9 + ((y-4)^2)/8 = 1`

Ex 2: Find the equation of the ellipse whose foci are (1,3) and (1,9) and eccentricity is `1/2`

Sol : From the given data the major axis is parallel to y-axis

`:.` the general equation is of the form

`((x-h)^2)/(b^2) + ((y-k)^2)/(a^2) = 1`

example 2 to find general equation

The centre of the ellipse is the midpoint of F1F2

`:.` C is ( `(1+1)/2 , (3+9)/2` ) = (1,6)

F1F2 = 2ae = 6

`rArr` ae = 3

But e = `1/2`       `:.` a = 6

b2 = a2(1-e2) = 36 ( 1-`1/4` ) = 27

Thus required equation is

`((x-1)^2)/(27) + ((y-6)^2)/36 = 1

I am planning to write more post on Equation of a Hyperbola and factoring polynomials of degree 3. Keep checking my blog.

Some more Examples to Find the General Equation of an Ellipse:

Ex 3: Find the equation of the ellipse  given that the centre is (4,-1) and one of its focus is (1,-1) and passing through (8,0)

Sol : From the given data since the major axis is parallel to the x axis , the equation is of the form

`((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) =1`

The centre is C(h,k) is (4,-1).

example 3 to find general equation

`((x-4)^2)/(a^2) + ((y+1)^2)/(b^2) =1`

It passes through (8,0) `:. 16/(a^2) + 1/(b^2) = 1`

But CF1 = ae = 3

b2= a2(1-e2) = a2 - a2e2 = a2-9

(1) `rArr 16/(a^2) + 1/(a^2 -9) =1`

`rArr ` 16a2 - 144+a2 = a4 - 9a2

`rArr` a4 - 26a2 +144= 0

`rArr` a2 = 18 or a2 = 8

Case (i) a2 = 18 `rArr` b2 = a2 -9 = 18-9 = 9

Case (ii) a2 = 8  `rArr` b2 = 8-9 = -1 which is not possible

`:. ` a2 = 18 and b2 = 9

Thus the equation is `((x-4)^2)/18 + ((y+1)^2)/9 = 1`

Trig Problem Solver

Greek Mathematician Ptolemy, Father of trigonometry proved the equation sin2A+cos2A=1 using geometry involving a relationship between the chords of a circle. Trigonometry was mainly concerned with establishing the relations between sides and angles of a triangle. The trigonometry consists of angles, quadrants, ratios and Identities, Compound Angles, and trigonometrical Equations

Understanding trig function is always challenging for me but thanks to all math help websites to help me out.

Trig problem solver  can be used to  find the angle values without using the tables.

We have addition formlae Sin(A+B) = SinA Cos B + Cos A Sin B

Cos(A+B) = CosACos B - SinA Sin B

and    Tan (A+B) =  tanA + tan B

1 - TanA.Tan B
Solved Problems on Trigonometry:
Let us use trig problem solver  to  do this problem
Ex 1: Without using tables, find sin 75º
Sol:-
We use the first formula Sin(A + B) = sin A cos B + cos A sin B
Sin 75º = sin(45º + 30º)= sin 45º cos 30º + cos 45º sin 30º
=  1 . √3  +  1 . 1    =  √3 + 1  = 1.732 +1  =  2.732 =  0.97
√2   2      √2  2           2√2      2x1.414      2.818
Ans is 0.97
Checking with the tables we find sin 75º =  0.9659
Ex 2: Without using tables find cos 75º
Sol: Let us use the formula  Cos (A+B) = cos A cosB - sinA sinB
cos 75º= Cos(45º +  30º)= cos 45º cos30 -  sin 45º sin30º
=   1 . √3  -  1 .1 =   √3 -  1 =  1.732 - 1  =   0.732   =   0.26
√2   2    √2.2        2√2        2 x 1.414     2.818
Ans cos 75º= 0.26
Checking with the tables we  find sin 75º = 0.2588
Ex 3: Find   tan 90 º without using tables
Sol: Tan (A+B) =  tan A + tan B
1 - tan A tanB
Tan 90 = tan (60º + 30º) =  tan 60º + tan 30º  =  √3 +  1/√3 =   √3 + 1 /√3  = √3 + 1 / √3 =    infinity=  ∞
1 - tan 60ºtan 30º    1 - √3.(1)         1  -    1                 0
√3
Hence we get tan 90º = infinity.
More Problems on Trigonometry:
Ex 5:If a = 4 units  and angle A = 45° find the circum-radius.
Sol:Here we use the formula  'a' = 2 R sin A where 'a' is the length of the side opposite angle A and R is circum- radius.
Hence 4 =  2R sin 45º =  2R x 1 / √2
4  ÷ 1 / √2 = 2R
4√2 =  2 R  =>  2√2 = R
Ans : Circum- radius of the circle with  A = sin45º and 'a' = 4 units is 2√2 units.
Ex 5: In any triangle prove that a(sinB - sin C) + b(sinC - sin A ) +c (sinA - sinB ) = 0
Ans: a= 2R sin A and b = 2R sin B and c = 2R sin C
Hence 2R sinA ( sinB - sinC) + 2R sin B( sin C - sin A ) + 2R sin C ( sin A - sin B)
=  2R sinA sinB - 2R sinA sin C + 2R sin B sinC - 2R sinB sin A + 2 R sinC sinA - 2 R sinCsinB
= 0 since( +2RsinAsinB cancels - 2R sinA sinB )and ( + 2RsinB sin C cancels - 2R sinB sinc) and (+2RsinC sin A cancels - 2R SinCsinA)

Ans for the above problem is zero and we have proved it by using trig problem solver.
Ex 6:Prove that Σ a sin (B − C) = 0

Sol:

Σ a sin (B−C) = a sin(B − C) + b sin(C − A) + c sin (A − B)

= 2R sinA sin(B−C) + 2R sinB sin(C − A) + 2R sinC sin(A − B)

sinA = sin(B + C), sinB = sin(C + A) ; sinC = sin(A + B)

By using the above formula, we get

= 2R sin(B + C) sin(B − C) + 2R sin(C + A) sin(C − A) + 2R sin(A +B) sin(A − B)

= 2R [sin2B − sin2C + sin2C − sin2A + sin2A − sin2B]

After simplifying this, we get

= 0

Ex 7: Solve: 2sin2x + sin22x = 2

Sol:

2 sin2x + sin22x = 2

sin22x = 2 − 2sin2x

= 2(1 − sin2x)

sin22x = 2 cos2x

⇒ 4sin2x cos2x − 2 cos2x = 0

⇒ 2(1 − cos2x) cos2x − cos2x = 0

Simplify the above terms, we get

⇒ 2cos4x − cos2x = 0

⇒ cos2x (2 cos2x − 1) = 0

⇒ cos2x = 0 , cos2x =`(1)/(2)` = `(1/sqrt(2))^2`

After simplifying this, we get

⇒ cos2x = cos2`pi/2`

⇒ x = nπ ±`pi/2` , n ∈ Z

Ex 8: Solve : √3 sin x + cosx = 2

Sol:

This is of the form a cosx + b sinx = c,

So dividing the equation by √ ( √3)2 + 12  ) or 2

We get

√3/2 sin x +1/2 cosx = 1 ⇒ sinπ/3 . sinx + cosπ/3 . cos x = 1

cos( x - π/3) = 1

cos(x - π/3) = cos 0

x - π/3 = 2nπ ± 0

x = 2nπ + π/3; n ∈ Z

Ex 3: In a triangle ABC prove that a sinA − b sinB = c sin(A − B)

Sol:

By sine formulae we have

a/sinA =b/sinB = c/sinC = 2R

a = 2R sinA, b = 2R sinB, c = 2R sin C

a sinA − b sinB = 2RsinA sinA − 2R sinB sinB

= 2R (sin2A − sin2B)

= 2R sin(A + B) sin(A − B)

= 2R sin(180 − C) sin (A − B)

= 2R sinC sin(A − B)

= c sin (A − B)
Practice Problems Using Trig Problem Solver:

Pro 1: Solve the following:

(i) sin3x = sinx

(ii) sin 4x + sin2x = 0

Ans: (i) nπ ±`pi/4` , nπ (ii) `(npi)/3` , `[(2n + 1)pi]/2`

Pro 2: Solve the problem: sin2x + sin6x + sin4x = 0

Ans: x = nπ ± `pi/3` ; n ∈ Z

Pro 3:  Prove the following: cos−1 (2x2 − 1) = 2cos−1x

Pro 4:  Find the solution: sin[ cos-11/2]

Ans: √3/2

Logarithm

In mathematics, the logarithm for a given number for a given base is the power or an exponent in which the base should be raised so as to create that number. For example, the logarithm or log of 10000 to base 10 is 4, because 4 is the power in which ten must be raised to produce 1000: 104 = 1000..

The logarithm of x to the base b can be written as log b(x) or, if the base is implicit, as log(x). So, for a number x, a base b and an exponent y,

If x = b y, then y = log b(x)
Laws of Logarithms:

While logarithm is just an exponent which is simply being written down on the following line.

Laws of logarithms
Example and Practice Problems on Logarithms to the Base 10:

Example 1:

Find the logarithm of 568 to base 10. Write this in the exponential form.

Solution:

Log 568 = 2.754

This indicates that102.754 = 5623

Mind investigation: 102 = 100 and 103 = 1000.

Our number 568 is between these two values, so it is reasonable.

Example 2:

Find the logarithm of 999 to base 10. Write this in the exponential form.

Solution:

Log 999 = 2.9999

This indicates that102.9999 = 5623

Mind investigation: 102 = 100 and 103 = 1000.

Our number 999 is between these two values, so it is reasonable.

My forthcoming post is on Markov Chains and Curve Fitting will give you more understanding about Algebra.

Example 3:

Find the logarithm of 8778 to base 10. Write this in the exponential form.

Solution:

Log 8778 = 3.943

This indicates that103.943 = 5623

Mind investigation: 103 = 1000 and 104 = 10000.

Our number 568 is between these two values, so it is reasonable.

Example 4:

Find the antilogarithm of the value -6.978

Solution:

This indicates that "if log N= -6.978, what is N?"

Using the laws of logarithm N = 10-6.978 = 0.000 000 105.

Practice problems:

Problem 1:

Find the logarithm of 4545 to base 10. Write this in the exponential form.

Solution:

The answer is 3.657.

Problem 2:

Find the logarithm of  555555 to base 10. Write this in the exponential form.

Solution:

The answer is 5.744.

Is this topic physics problems hard for you? Watch out for my coming posts.

Problem 3:

Find the logarithm of 8969 to base 10. Write this in the exponential form.

Solution:

The answer is 3.952.

Partial Fraction and Rules

A rational fraction is one which is a ratio of two polynomials.  That is, a function of the form

$f(x) = \frac{P_m(x)}{Q_n(x)}$

where $P_m(x) = a_mx^m + a_{m-1}x^{m - 1} + \ldots + a_2x^2 + a_1x + a_0$ is a polynomial of degree $m$, and $Q_n(x) = b_nx^n  + b_{n-1} x^{n-1} + \ldots + b_2x^2 + b_1 x + b_0$ is a polynomial of degree $n$.  Here, $a_m, a_{m-1}, \ldots, a_2, a_1, b_m, b_{m-1}, \ldots, b_2, b_1, b_0$ are constants, and $a_m \not=0, b_n \not=0$.

Depending upon the degrees of the numerator and the denominator, we can categorize these rational fractions into two types :

The fraction $f(x)$ is said to be an improper fraction, if $m < n$.  For example,

$\frac{3x^3 + 4x^2 - 11}{x^2 + 3x -2}$

The fraction $f(x)$ is said to be a proper fraction, if $m < n$.  For example,

$\frac{3x +4}{x^2 - 7x + 1}$

If $f(x) = \frac{P_m(x)}{Q_n(x)}$ is an improper fraction (i.e., $m\gen), then it can be expressed as a sum of a polynomial and a proper fraction.  That is,

$\frac{P_m(x)}{Q_n(x)} = A_{m-n} (x) +  \frac{R}{Q_n(x)}$

where, $A_{m-n}(x)$ is an polynomial of degree $m -n$, and $R$ is another polynomial of degree less than $n$.  For illustration, consider the following improper fraction:

$\frac{2x^3 + 4x^2 -11}{x^2 + 3x + 2} $ $= (2x -2) +$ $\frac{2x - 7}{x^2 + 3x + 2}$

Note that we can perform addition and subtraction on a group of rational fractions to get a rational fraction whose denominator is the lowest common denominator of the given fractions.  But the converse process of separating a fraction into a group of simple, or partial fractions  is often required.  It is natural to inquire about how far the decomposition of a given rational fraction into partial fractions.

Since we can always, by ordinary division, represent an improper fraction as the sum of an integral function and a proper fraction, we need to consider the resolution of a proper fraction into partial fractions.

Remark:

If the denominator of a proper fraction can be factorized into two integral factors, which are algebraically prime to each other, then we can always decompose the fraction into sum of two proper fractions.

Three cases along with the method of solving a rational fraction into a sum of partial fractions is given hereunder:
Partial Fraction Rules 1:

Case 1:If

$f(x) = \frac{A_m(x)}{B_n(x)} = \frac{A_m(x)}{b (x - \alpha_1)(x - \alpha_2) \ldots \ldots (x - \alpha_n)}$

is a proper rational fraction (i.e., m
$f(x) = \frac{A_1}{x- \alpha_1} + \frac{A_2}{x - \alpha_2} + \ldots \ldots + \frac{A_n}{x - \alpha_n}$

where $A_1, A_2, \ldots \ldots, A_n$ are constants.

Illustrative Example 1: Resolving $\frac{8x-4}{3x^2 - 2x -1}$ into a sum of partial fractions.
Solution

We know that

$3x^2 - 2x -1 = 3x^2 - 3x + x -1$

$=(3x^2 - 3x) + (x -1)$

$= 3x(x -1) + (x -1)$

$= (3x + 1)(x -1)$

Thus,

$\frac{8x -4}{3x^2 - 2x -1} = \frac{8x -4}{(3x+1)(x - 1)}$

Let us assume that

$\frac{8x -4}{(3x+1)(x-1)} = \frac{A}{3x+1} + \frac{B}{x-1}$

$\Longrightarrow \quad \frac{8x -4}{(3x+1)(x-1)} = \frac{A(x-1) + B(3x + 1)}{(3x+1)(x -1)}$

$\Longrightarrow \quad \frac{8x -4}{(3x+1)(x - 1)} = \frac{(A + 3B)x + (B - A)}{(3x + 1)(x -1)}$

Since the denominators are equal, we can equate the numerators.  Thus, we have

$8x - 4 = (A + 3B)x + (B - A)$

I am planning to write more post on List of Fibonacci Numbers and What is a Fraction. Keep checking my blog.

Comparing x coefficients and constants on both sides, we have

$A + 3B = 8$ ---------------(1)

$A - B = 4$ ----------------(2)

Solving (1) and (2), we have

$A =  5$ and $B = 1$.

Thus,

$\frac{8x -4}{3x^2 - 2x -1} = \frac{5}{3x+1} + \frac{1}{x -1}$.
Partial Fraction Rules 2:

Case 2:

When some of the zeros are repeating, we can write

$f(x) = \frac{A_m(x)}{b (x - \alpha_1)^{k_1} (x - \alpha_2)^{k_2} \ldots \ldots (x - \alpha_r)^{k_r}}$

where $r \leq n$; and $k_1, k_2, \ldots \ldots, k_r$ are positive integers such that $k_1 + k_2 + \ldots + k_r = n$ and $\alpha_1, \alpha_2, \ldots \ldots, \alpha_r$ are distinct zeros of $B_n(x)$.  In this case, the decomposition takes the following form:

$f(x) = \frac{A_1}{x - \alpha_1} + \frac{A_2}{(x-\alpha_1)^2} + \ldots + \frac{A_{k_1}}{(x-\alpha_1)^{k_1}} + \frac{A_{k_1 +1}}{x - \alpha_2} + \ldots  + \frac{A_{n - k_r +1}}{x - \alpha_r}+ \ldots + \frac{A_n}{(x - \alpha_r)^{k_r}}$

Illustrative Example 2: Resolving $\frac{3x^2 + x -2}{(x-2)^2 (1-2x)}$ into a sum of partial fractions.
Solution

Assume that

$\frac{3x^2 + x -2}{(x-2)^2 (1-2x)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{1-2x}$

$\Longrightarrow \quad \frac{3x^2 + x -2}{(x-2)^2 (1-2x)} =\frac{A(x-2)(1-2x) + B(1-2x) + C(x-2)^2}{(x-2)^2 (1-2x)}$

Since the denominators are equal, we can equate the numerators.  Thus, we have

$3x^2 + x -2 = A (x-2x^2 -2 + 4x)+ B(1-2x) + C(x^2 - 4x + 4)$

$\Longrightarrow \quad 3x^2 + x -2 = (-2A + C) x^2 +(5A - 2B -4C) x -(2A - B-4C)$

Comparing on both sides, we  have

$-2A + C = 3   -------------------- (1)$

$5A - 2B - 4C = 1    --------------(2)$

$2A -B-4C = 2    ------------------(3)$

From (1), (2) and (3), we get

$ A = - \frac{5}{3},  B = -4,  and C = -\frac{1}{3}$

Thus, we have

$\frac{3x^3 + x -2}{(x-2)^2 (1 - 2x)} = - \frac{5}{3(x-2)} -\frac{4}{(x-2)^2} - \frac{1}{3(1-2x)}$.
Partial Fraction Rules 3:

When some of the roots of the denominator $B_n(x)$ of the proper rational fraction $f(x) = \frac{P_m(x)}{Q_n(x)}$ are complex, we can no longer factor $B_n(x)$ in the form of $b(x-\alpha_1)^{k_1} (x - \alpha_2)^{k_2} \ldots (x - \alpha_r)^{k_r}$ with real factors.  However, if the coefficients are real numbers, then the complex roots always come in conjuage pairs of the form $\alpha \pm \beta i$ and that, corresponding to each such a pair of complex conjugate zeros, there is a quadratic factor of $B_n(x)$.

Illustrative Example 3: Resolving $\frac{2x+1}{(x -1)(x^2+1)}$ into a sum of partial fractions.

Assume that

$\frac{2x+1}{(x -1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$

$\Longrightarrow \quad \frac{2x+1}{(x -1)(x^2+1)} =\frac{A(x^2 + 1) + (Bx + C)(x-1)}{(x -1)(x^2+1)}$

Since the denominators equal, we can equate the numerators.  Thus, we have

$2x+1= A(x^2 + 1) + (Bx + C)(x-1)$

$\Longrightarrow \quad 2x + 1 = (A+B)x^2 + (C- B)x + (A -C)$

Comparing on both sides, we have

$A + B = 0  -------------- (1)$

$C - B = 2 --------------- (2)$

$A - C = 1 --------------- (3)$

From (1), (2), (3), we have

$A = \frac{3}{2},  B = - \frac{3}{2}$, and $C = \frac{1}{2}$.

Thus, we have

$\frac{2x+1}{(x -1)(x^2+1)} = \frac{3}{2(x-1)} - \frac{3x-1}{2(x^2+1)}$

Numerical Variable

In mathematics the conception of a variable express the one quantity (the argument of the function also known as input) that determines the output. A function gives a unique value to each input of a specified type. A symbol is represented by a number is called variable. It is also called as numerical variable. We use letters for variables such as x, y, z, n, t.
Example for Numerical Variable:

Problem for numerical variable:1

Solve 5x + 3x = 4

Solution:

Given that 5x + 3x = 4

Add the two variables 5x and 3x is 5x + 3x = 8x

8x = 4

Take the 8 into after the equals

X = `4 / 8`

Divide the minimize the fraction `4 / 8` = `1 / 2`

So the answer will be x = `1 / 2` .

Problem for numerical variable: 2

Solve 19x – x + 2x = 19

Solution:

Given that 19x – x + 2x = 19

Subtract the 19x – x = 18x

18x + 2x = 19

20x = 19

Take the 20 to after the equals that will become `19 / 20`

x = `19 / 20` .

The answer will be   `19 / 20` .
Example for Numerical Variable:

I am planning to write more post on fourier transform, and real numbers. Keep checking my blog.

Problem for numerical variable:3

Solve 15x + 21x – 2x – 5 + 18 – x = 0

Solution:

15x + 21x – 2x – 5 + 18 – x = 0

Add the positive and negative symbols from the given equation

36x – 3x – 5 + 18 = 0

Subtract the variable 36x – 3x = 33x

Subtract the numerals -5 + 18 = 13

The equation will be 33x = 13

Take the 33 to after the equals x = 13 / 33

The solution will be x = 13 / 33

Problem for numerical variable :4

Solve x + 2x + x – 2x – 11 = 0

Solution:

Given that x + 2x + x – 2x – 11 = 0

Add the given variables x + 2x + x = 4x

4x – 2x – 11 = 0

Subtract the variable 4x -2x = 2x

2x – 11 = 0

Take the -11 into after the equal that will become +11

2x = 11

Take the 2 into after the equal

x = `11 / 2`

The answer will be 5.5.

Problem for numerical variable 5:

Solve -15x – 3x +2x – 4 = 0

Solution:

Given that -15x – 3x +2x – 4 = 0

Add the given variables – 15x – 3x = - 18x

- 18x + 2x – 4 = 0

Subtract the variable -18x + 2x = - 16x

- 16x – 4 = 0

Take the -4 to after the equals

-16x = 4

Take the – 16 to after the equals

-x = `4 / 16`

Minimize the fraction  - `4 / 16` to – `1 / 4`

x = – `1 / 4`

The answer will be – `1/ 4` .

List of Real Numbers

Let us study about list of real numbers. The various kinds of numbers together form the math problems for different methods.
Real numbers are also seemed to be as one type of numbers of math world. These real numbers are defined as the numbers that are with the combination of both definition of rational and irrational numbers.
Some of the examples for the list of real numbers are discussed in detail as below.

List of Real Numbers:

List of real numbers – example 1:

Add the following real rational numbers `1/2, 3/1, 36/6, 7/2, sqrt(4) and 5.2`

Solution:

The real rational numbers are `1/2, 3/1, 36/6, 7/2, sqrt(4) and 5.2`
Adding the given real rational numbers as follows:
`1/2 + 3/1 + 36/6 + 7/2 + sqrt(4) + 5.2`
0.5 + 3 + 6 + 3.5 + 2 + 5.2
Therefore the total value of the given real rational numbers is found to be as ’20.2’

List of real numbers – example 2:

Add the following real irrational numbers 23.232…, `sqrt(3)` , 3.3432…, `sqrt(7) and sqrt(5)`
Solution:

The real irrational numbers are 23.232…, `sqrt(3)` , 3.3432…, `sqrt(7) and sqrt(5)`
Adding the given real irrational numbers as follows:
23.232.. + `sqrt(3) ` + 3.3432.. + `sqrt(7) + sqrt(5)`
23.232.. + 3.3432..+ `sqrt(3) + sqrt(7) + sqrt(5)`
23.232.. + 3.3432.. + `sqrt(15)` (first adding all the square root values together)
23.232.. + 3.3432.. + 3.8729..
Therefore the total value of the given real irrational numbers is found to be as ’30.4481..’

List of real numbers – example 3:

Multiply the following real rational numbers `1/2, 3/1, 36/6, 7/2, sqrt(4) and 5.2`

Solution:

The real rational numbers are `1/2, 3/1, 36/6, 7/2, sqrt(4) and 5.2`
Multiplying the given real rational numbers as follows:
`1/2 * 3/1 * 36/6 * 7/2 * sqrt(4) ` `* 5.2`
0.5 * 3 * 6 * 3.5 * 2 * 5.2
Therefore the multiplied value of the given real rational numbers is found to be as ’327.6’

List of real numbers – exercises:

Add the following real rational numbers `12/3, 4/1, 18/6, 9/2, sqrt(36)` and 0.1 (Answer: 21.6)
Multiply the following real rational numbers `12/3, 4/1, 18/6, 9/2, sqrt(36)` and 0.1 (Answer: 129.6)
Add the following real irrational numbers 3.8782…, `sqrt(11)` , 0.372…, `sqrt(17)` and `sqrt(2)` (Answer: 9.3554…)

Simple Algebraic Expressions

Expressions are central concept in the algebra. A variable can take several values. Its values are not fixed. Otherwise, a constant has a fixed value. We are combining variables and constants to make algebraic expressions.

Operations of Simple algebraic expressions:

  For this, we can use the operations of addition, subtraction, multiplication and division. We have already come across simple algebraic expressions like 7x 10, 4y-3, 5+2zand so on. The above expressions were obtained by combining variables and constants. The expressions are also obtained by combining variables with themselves.
Simple Algebraic Expressions Example Problems:

1. Find the coefficients of y in the following expressions?

    4x – 3y –Coefficient of y is -3
     8 + yz, –Coefficient of y is z
    my + m- –Coefficient of y is m

2. State with reasons, which of the following pairs of terms are of like terms and which are of unlike terms:

    – 4ab, 7ba- Like terms Here ab=ba
    3xy, 3 x- Unlike Terms. The variable y is only in one term

Algebraic Expressions Practice Problems:

1. Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.

    Subtraction of z from y.
    One-half of the sum of numbers x and y.
    The number z multiplied by itself.
    One-fourth of the product of numbers p and q.
    Numbers x and y both are squared and added.
    Number 5 added to three times the product of numbers m and n.

Algebra is widely used in day to day activities watch out for my forthcoming posts on physics problems solutions and practice pre algebra. I am sure they will be helpful.

2.  Identify terms which contain x and give the coefficient of x.

(i) y2x + y        (ii) 13y2 – 8yx (iii) x + y + 2

(iv) 5 + z + zx  (v) 1 + x + xy  (vi) 12xy2 + 25

3. State whether a given pair of terms is of like or unlike terms.

(i) 1, 100          (ii) –7x, 5/2 x      (iii) – 29x, – 29y

(iv) 14xy, 42yx (v) 4m2p, 4mp2 (vi) 12xz, 12x2z2

Types Of Inequalities Geometry

An inequality is a statement about the relative size or order of two objects, or about whether they are the same or not (source: WIKIPEDIA). These inequalities will allow us to formulate several important geometric relationships. The types of inequalities geometry are transitive property, Substitution postulate, and trichotomy postulate. All the types of inequalities are used for proving inequality geometry.

Types of Inequalities Geometry
In geomeric inequality:
          A whole is equal to the sum of all its parts. A whole is greater than any of its parts.
         In geometry: The lengths of line segments and the measures of angle are positive number.
Consider these two applications:
                             
   If ACB is a line segment, then AB = AC + CB, AB > AC, and AB > CB.
   If ?DEF and ?FEG are adjacent angles, m?DEG = m?DEF + m?FEG,

m?DEG > m?DEF, and m?DEG > m?FEG                    
Types of inequalities:
Type 1: Transitive property of inequality:
      if a,b, and c are real numbers such that a > b and b > c, then a > c.
Then in geometry: If BA > BD and BD > BC, then BA > BC. Also, if m?BCA > m?BCD > m?BAC, then

m?BCA > m?BAC.
Type 2: Substitution postulate of inequality:
    A quantity may be substituted for its equal in any statement of inequality.
In geometry: If AB . BC and BC = AC, then AB > AC. Also, if m?C > m?A and m?A = m?B, then m?C > m?B.
Type 3: The trichotomy postulate Inequality:
Given  any two quantities, a and b, one and only one of the following is true;
                         a < b   or  a = b     or a > b.
Proving Inequalities and its Types in Geometry

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

Given: m?DAC = m?DAB + m?BAC and m?DAB > m?ABC

Prove: m?DAC > m?ABC
        Statements                                                               Reasons
1. m?DAC = m?DAB + m?BAC                                         1. Given.
2. m?DAC > m?DAB                                                       2. A whole is greater than any of its parts.
3. m?DAB > m?ABC                                                       3. Given.
4. m?DAC > m?ABC                                                       4. Transitive property of inequality.
This proof shows the inequalities geometry.

Opposite Of Tangent

In Trigonometry, we have six important trigonometric functions. The main three trigonometric functions equations are sine, cosine, tangent. The opposite three trigonometric functions are cosecant, secant, cotangent.

So, Opposite of tangent is Cotangent.

Cotangent is shortly called as ctg.

Cotangent is also referred as reciprocal of tangent trigonometric function.

The cotangent of angle ? in a right-angle triangle is the ratio of adjacent side and opposite side.

Cotangent is used to find unknown angle and unknown side.

The notation of Cotangent is cot ? (or) ctg ?.

Let us see some important formulas of opposite of tangent.
Formulas of Opposite of Tangent:

Formula for cotangent in a right triangle:

Cot ? = `("adjacent side")/("opposite side") `

Pythagorean identities for ctg:

1 + cot2 ? = csc2 ?

We can also rewritten the above one.

Cot2 ? = csc2 ? – 1

Derivative of ctg:

`d/dx` cot x   = - csc2 x

Integral of ctg:

`int`cot x dx = ln(sin x) + C

The cotangent value table for standard angles is shown below.

Angle/Function


   0°


30° (or) `pi/6`


45° (or) `pi/4`


60° (or) `pi/3`


90° (or) `pi/2`

    Cot ?


   8


  `sqrt(3)`


     1


  `1/sqrt(3)`


  0



Negative argument of ctg:

Cot(-?) = - cot(?)

Double angle of ctg:

Cot(2?) =` (cot^2 theta-1)/(2cot theta)` = `1/2` (cot ? – tan ?)

Triple angle of ctg:

Cot(3?) = `(cot^3 theta-3 cot theta)/(3 cot^2 theta-1)`

Half angle of ctg:

Cot(`theta/2`) = cot ? + csc ?

Cot(`theta/2`) = `(sin theta)/(1-cos theta)`

Cot(`theta/2`) = `sqrt((1+cos theta)/(1-cos theta))`

Relation between cotangent and inverse cotangent:

Cot(cot-1(?)) = ?

Simple Relation between cotangent and other trigonometric functions:

Cot(?) = tan(`pi/2` – ?)

Cot(?) = `(cos theta)/(sin theta)`

Let us see sample problem of opposite of tangent.
Example Problem - Opposite of Tangent:

My forthcoming post is on Definition of Continuity, and  Precalculus Problems will give you more understanding about Algebra

Example Problem 1:

Find the angle using cotangent if adjacent side is 5 cm and opposite side is 12 cm in a right triangle. (note: ? be angle)

Solution:

As we know that,

Cot ? = `("adjacent side")/("opposite side") `

cot ? = `5/12`

cot ? = 0.4167

Take inverse cotangent both sides. we get,

? = cot -1(0.4167)        

? = 67.4°

Therefore, angle of right triangle is 67.4°.

Example Problem 2:

Find the value of cot 45° using sin 45° = `1/sqrt(2)` and cos 45° = `1/sqrt(2)` .

Solution:

As we know that,

Cot ? = `(cos theta)/(sin theta)`

So, cot 45° = `cos 45/sin 45`

cot 45° = `(1/sqrt(2))/(1/sqrt(2))`

Cot 45° = 1.

Solving Complex Rational Expressions

Rational Expressions: Rational expressions are defined as one of the basis of mathematics. All the rational are having polynomials terms in both of the numerator function and the denominator function. The polynomials are present along with the variables.

 Complex rational expressions: Complex rational expressions are having the polynomials with the fraction format. The expressions should have atleast one fraction with the polynomials.
Explanations for Solving Complex Rational Expressions

There are many steps are followed for solving the complex rational expressions. They are defined as follows,

Step 1: Write the given rational expressions.

Step 2: Bring the denominator terms to the multiplication format by taking inverse.

Step 3: Then in the next step, we have to solve the obtained result.
Example Problem for Solving Complex Rational Expressions

Problem 1: Solve the given rational expressions, `((4a)/3)/((3a)/2)` .

Solution:

Step 1: Write the given complex rational expressions,

`((4a)/3)/((3a)/2)`

Step 2: Bring the denominator terms to the numerator by taking the inverse, we get,

`((4a)/3)` `xx` `(2/(3a))`

Step 3: In the next step, we have to simplify the obtained terms,we get,

`(8a)/(9a)`

Step 4: By eliminating the like terms, we get,

`8/9`

This is the obtained result for solving the complex rational expressions.

Algebra is widely used in day to day activities watch out for my forthcoming posts on algebra rational expressions and multiplying rational expressions solver. I am sure they will be helpful.

Problem 2: Solve the given rational expressions, `((6a)/3)/((9a)/4)` .

Solution:

Step 1: Write the given complex rational expressions,

`((6a)/3)/((9a)/4)`

Step 2: Bring the denominator terms to the numerator by taking the inverse, we get,

`((6a)/3)` `xx` `(4/(9a))`

Step 3: In the next step, we have to simplify the obtained terms,we get,

`(8a)/(9a)`

Step 4: By eliminating the like terms, we get,

`8/9`

This is the obtained result for solving the complex rational expressions.

Problem 3: Solve the given rational expressions, `((8a)/2)/((4a)/2)` .

Solution:

Step 1: Write the given complex rational expressions,

`((8a)/2)/((4a)/2)`

Step 2: Bring the denominator terms to the numerator by taking the inverse, we get,

`((8a)/2)` `xx` `(2/(4a))`

Step 3: In the next step, we have to simplify the obtained terms,we get,

= 2

This is the obtained result for solving the complex rational expressions.
Practice Problem for Solving Complex Rational Expressions

Problem 1: Solve the given rational expressions, `((5a)/10)/((25a)/5)` .

Answer: `1/10`

Problem 1: Solve the given rational expressions, `((5a)/4)/((15a)/8)` .

Answer: `2/3`

Introduction to preparation for differentiation strategies

The preparation process for differentiation strategies represents the process of differentiation under the polar coordinates, variables in equations. The differential equations may be present in the ordinary differential equations with different functions like algebraic functions, exponential functions, etc.. In this article we deal with the differential equations with the variables to differentiate for the strategy in the differentiation.

Preparation for Differentiation Strategies with Examples

Preparation for the differentiation strategies on the differential equations  `[x+y]^2 dy/dx = 25`  
Solution:

The equation given as   `[x+y]^2 dy/dx = 25`      

Put x + y = z

Differentiating with respect to 'x' we get,

  `1 + dy/dx` =  `dz/dx`    

  `=>`   ` dy/dx` =  `dz/dx -1`            

The given equation becomes

  `z^2[dz/dx - 1] = 25`  

  `=>`  `dz/dx -1` = `25/z^2`                          

  `=>`  `dz/dx` = `1 + 25/z^2`                          

  `=>`  `dz/dx` = `[z^2+25]/z^2`                          

  `=>`  `dz` = `[z^2+25]/z^2 dx`                          

  `=>`  `dz``z^2/[z^2+25]` =   `dx`                            

Integrating we have

  `=>` `int z^2/[z^2+25]dz` =   `int dx`                            

  `=>` `int [z^2+25 - 25]/[z^2+25]dz` =   `int dx`                            

  `=>` `int [[z^2+25]/[z^2+25] - 25/[z^2+25]]dz` =   `int dx`                  

  `=>` `int [1 - 25/[z^2+25]]dz` =   `int dx`                  

  `=>` `int dz - int 25/[z^2+25]dz` =   `int dx`                            

  `=>` `z - 25 1/5 tan^-1z/5` =   `x + c`                            

Put x + y = z

  `=>` `x+y - 5. tan^-1[[x+y]/5]` =   `x + c`                

Reduce 'x' on both sides,

  `=>` `y - 5. tan^-1[[x+y]/5]` =   ` c`   is the required solution.                

Problems for the Preparation for Differentiation Strategies

Homework or practice problems on the preparation for differentiation strategies:

Solve `x dy = ( y + 4x^5 e^[x^4]) dx`  
Solution:

The answer is `y/x = e^[x^4] + c`    

Solve    `[x^2 -y] dy + [y^2-x]dy =0`  
Solution:

The answer is    `x^3 + y^3 = 3xy`

Introduction to algebra formulas for volume

Introduction to algebra formulas for volume

Algebra is one of the ancient syllabuses in Mathematics. Modern algebra is very much different from ancient algebra. Algebra is the subdivision of mathematics in which variables and numbers are used to express the equation.In algebra, variables are sometimes used in place of numbers. Algebra uses the property of commutative and associative property.  In algebra exams a variable is a symbol that represents a number. Algebra exams are mainly covers the linear equations, factorization, and polynomial equations.

Algebaric Geometry Formulas for Volume

There is more number of shapes in algebraic geometry. The geometry shapes are square, rectangle, triangle, prism, cylinder and cone, etc.. some of the volume formulas for geometry shapes are given below.

Application of algebraic geometry:

 1) Statistics

 2) Control theory

 3) Robotics

 4) Geometric modeling

 5) Game theory

Formulas for volume:

Formula fro volume of the cylinder:

      Volume = pi * r^2 * h. Here Pi = 3.14, r = radius, h = height.

Formula for volume of cube:

     Volume = a3. Here, a = side length of the cube.

Formula for volume of Sphere:

   Volume = (4/3) * pi * r^3. Here, pi =3.14, r = radius.

Algebra Formulas for Volume

Formula for volume of the Rectangular prism:

        Volume = a1 * a2 * a3. Here, a1 = Side 1, a2 = Side 2, a3 = Side 3.

Formula for volume of the cone:

       Volume = (1/3) * Pi * r^2 * h. Here, Pi = 3.14, r = Radius, h = Height.

Volume of the pyramid:

       Volume = (1/3) * base area * h.

Here, h = height, Base area = Depends on the base of the pyramid.

Formula for volume of the Ellipsoid:

      Volume = (4/3) * pi * r^1 * r^2 * r^3.

Here, Pi= 3.14, r1 = Radius , r^2 = Radius 2, r^3 = Radius 3.

Example Problems on Algebra Formulas for Volume

Ex 1:

Find the volume of cube with the side length is 5 cm.

Sol:

Given, side length a = 5 cm

              Volume of cube = a3

                                       = 53

                                       = 125 cm^3

Answer :

The final answer is 125 cm^3

Ex 2:

Find the volume of cylinder with the radius is 4 cm and height is 7 cm.

Sol:

Given, radius r = 4 cm, height h = 7 cm

          Volume of cylinder = pi * r^2 * h

                                            = 3.14 * 42 * 7 cm^3

                                            = 352 cm^3

Answer:

The final answer is 352 cm^3

Ex 3:

Find the volume of sphere with the radius is 12 cm.

Sol:

Given radius r = 12 cm

  Volume of sphere = (4 / 3) * pi * r^3

                            = (4 / 3) * pi * 123 cm^3

                            = 7234.56 cm^3

Answer:

The final answer is 7234.56 cm^3

Introduction to Orleans-Hanna Algebra Prognosis Test

Introduction to Orleans-Hanna Algebra Prognosis Test

The Orleans-Hanna Algebra Prognosis Test verifies teachers’ attitude regarding students’ keenness intended for algebra. Problem-solving items replicate NCLB Standards.

The following are things you need to cover for preparing Orleans-Hanna Algebra Prognosis Test.

The things cover algebraic topics such as

Exponents
Integers
Algebraic expressions
Graphs and charts
General middle-school mathematics.
Orleans-hanna Algebra Prognosis Test Solved Problems

Pro 1:-Find the sum of two algebraic expression that are given below

4x^2+3x+4 and 5x^2 +2x+12.

Solution:-The given algebraic expressions are the following.

4x^2+3x+4.

5x^2 +2x+12.

The sum of two expressions is

4x^2+3x+4 + 5x^2 +2x+12.

Now group the common terms together.

4x^2 + 5x^2 +3x + 2x + 4 + 12.

9x^2 + 5x + 16.

The simplified expression is 9x^2 + 5x + 16.

Pro 2:-Find the value of x in the given algebraic expression.

4x + 12 + x + 3x = 12x + 3x + 2

Solution:-The given algebraic expression is

4x + 12 + x + 3x = 12x + 3x + 2

Group the like terms together

4x +x + 3x + 12 = 12x + 3x + 2

Now add like terms together

4x +x + 3x gives 8x.
12x + 3x gives 15x.

Now,8x + 12 = 15x + 2

In next step subtract 2 on both side.

8x + 12 -2 = 15x + 2 - 2.

We get 8x +10 = 15

Now subtract 10 on both side

8x +10 – 10 = 15x – 10

By simplifying it we get

8x = 15x -10

Now subtract 15x on both sides of the above equation.

8x – 15x = 15x – 15x -10

-7x  = -10

Now divide by -7 on both sides
`(-7x)/(-7) = (-10)/(-7)`

X  =`(10)/(7)`

The answer for x is `(10)/(7)`

Orleans-hanna Algebra Prognosis Test Practice Problems

Pro 1:-Find the sum of two algebraic expression that are given below

x^2+13x+24 and x^2 +12x+12.

Ans:- 2x^2+25x+36.

Pro 2:-Find the value of x in the given algebraic expression.

x + 24 +2x + 3x = 12x + 6x + 12.

Ans:- x = 1

Having problem with Pr-algebra Practice keep reading my upcoming posts, i will try to help you.

Integration by Trignometric substitutions

Introduction to trigonometric substitutions :  Trignometric substitutions enable us to replace the binomials a^2 +x^2 , a^2-x^2 , and x^2-a^2  by single squared terms and thereby transform a number of integrals containing square root s into integrals we can evaluate directly .
The most common substitutions are x = a  tan (theta ) ,
x = a sin (theta) and
x = a sec (theta).
With  x = a tan (theta) ,
a^2 + x^2 = a^2 + a^2 tan ^2(theta)
= a^2 (1 + tan ^2 (theta))
= a^2 vsec^2 (theta).

With a = a sin (theta) ,
a^2-x^2 = a^2 – a^2 sin^2(theta)
= a^2 (1-sin^2(theta))
= a^2cos^2(theta).

With x =a sec(theta) ,
x^2 – a^2 = a^2 sec^2(theta) – a^2
= a^2(sec^2(theta) -1)
= a^2tan^2(theta).

Integration by Trigonometric Substitution :  we want any substitution we use  in an integration to be reversible so that we can change back to the original variable afterward .
for example if x = a tan (theta) ,
we want be able to set (theta) = tan ^-1(x/a) after the integration takes place .
If x = a sin (theta) , we want be able to  set (theta) = sin^-1 (x/a) ,
where we are done similarly for x = a sec (theta)To simplify calculations with the substitution x = a sec(theta) , we will restrict its use to integrals in which (x/a)=1.

This will place (theta) in and make tan (theta)= 0 .
We will  then have
sqrt (x^2 –a^2) = sqrt (a^2 tan^2(theta))
= mod(a tan (theta))= a tan (theta),
free of absolute values , provide a >0.

Trigonometric Substitution Examples :Let us take a  Trigonometric Substitution Problems to understand Integration by Trigonometric Substitution  Here we have to evaluate  integration (dx / sqrt (4 +x^2)).

we can see from the  problem we have to use Trigonometric Substitution Integration .
let us  set x = 2 tan (theta) ,
dx = 2 sec^2 (theta) d(theta), -pi/2< theta4 +x^2 = 4+4 tan^2 (theta)
= 4 (1+ tan^2 (theta) )= 4 (sec^2 (theta)).

Then  integration dx / sqrt (4+x^2) = integration 2sec^2
(theta)d(theta) / sqrt 4sec^2 (theta) = integration sec^2(theta)d(theta)
=  mod (sec (theta)

and we know sqrt sec^2 (theta) = mod (sec(theta) , sec (theta)> 0  for –pi/2 < theta < pi/2 .
In mod (sec (theta)+ tan (theta) +c
= In mod (sqrt (4 +x^2 )/2 + x/2) + c
= In mod (sqrt (4 +x^2) +x )+C’

Second Order Differential Equation

Second Order Homogeneous Differential Equation
The general form of Second order differential equation is given by, a d^y/dx^2 + b dy/dx + cy= f(x) where a, b, c are some constants and f(x) is a function of x.  The equation is homogenous when f(x) equals zero, ay’’ + by’ + cy = 0 is a second order homogenous differential equation. For instance, d^2y/dx^2 -2 dy/dx + 3y = 0 is an example for a second order homogenous differential equation.

Second Order Non Linear differential Equation
The general form of second order nonlinear differential equation is given by,  y'' + p(t)y'' + q(t) y = f(x). The general solution is given by y = y(p) x + y(h)x where the particular solution of non-homogenous is y(p) x and the general solution of the associated homogenous equation is y(h)x.  Second order nonlinear differential equation can also be written as (a) d^2y/dx^2 + (b) dy/dx + (c)y = f(x) where a, b, c are constants and f(x) is the function of x.

In general, a y’’ + b y’ + c y = 0 is the homogeneous second order differential equation where the constants are a, b and c and also a does not equal zero. Using this equation we can arrive to an auxiliary equation or the characteristic equation that has roots which helps in solving second order differential equations. Let us consider the function y=e^(rx) an exponential function with r as constant. We know that the derivative of an exponential function is a product of itself and the constant, using this property we get, y’= r.e^(rx) and y’’= r^2.e^(rx). Let us substitute these expressions in the general form, that gives us, a[r^2.e^(rx)]+ b.[r. e^(rx)] + c [e^(rx)]= 0. Here e^(rx) never equals zero and hence y=e^(rx) is the solution of the equation. The auxillary equation is ar^2 +b r + c = 0 which is an algebraic equation for which the roots r1 and r2 can be found either by factoring or by using the quadratic formula. If the roots are unequal and real then the general solution is given by, y = c1e^(r1x) + c2e^(r2x)

Second order differential equation examples
Solve the given equation 3y’’ +y’- y = 0
Here the auxiliary equation would be,  3r^2 + r -1 =0.
Using the quadratic formula, [-b (+/-) sqrt(b^2-4ac)]/2a ; a = 3, b = 1, c = -1
We get, {-1(+/-)sqrt[1^2 – (4.3.-1)]}/2.1 = [-1(+/-)sqrt(13)]/6

The roots being real and distinct, the general solution is,
 y = c1e^[x/6(-1+sqrt(13)] + c2 e^[x/6(1+sqrt(13)]

Multiplicative Progression

Progression:
When there is a definite relationship between the two successive terms of a series, we call the series as progressing series. The relation between the two successive terms decides the nature of the progression.
When the successive term is obtained on multiplying a given term with a constant number, the sequence so obtained is called geometric sequence. Example: 2, 4, 8, 16…

Convention:
Common ratio: The constant term that is multiplied with a given term of the sequence to obtain the successive term is called common ratio.

Sum of a geometric progression:
To find the sum of a geometric progression we adopt the following steps:
Step 1: Assume the sum of the given geometric progression as ‘S’.
Step 2: Find the common ratio of the progression.
Step 3: Multiply every term of the progression with ‘r’ to obtain S x r.
Step 4: Find the value of (Sr – S)
Step 5: Solve for S.
Example: Find the sum of 2 + 4 + 8 + 16 …256
Step 1: S = 2 + 4 + 8 + 16 …256
Step 2: Common ratio is 2
Step 3: 2S = 4 + 8 + 16 …512
Step 4: 2S – S = 512 -2
Step 5: S= 510.

Sigma Notation Series
If the first term and the common ratio of the geometric progression are represented by ‘a’ and ‘r’ respectively, then the geometric progression can be shown as
a + ar + ar^2 + ar^3… ar^(n-1)
It is a common practice in mathematics to represent the summation using sigma notation. The corresponding sigma notation for the above geometric summation is

Observe that, on substituting the values of k varying from zero to (n-1), we get the requisite geometric summation.

Special Series
Series can be obtained by finding the successive terms following a certain mathematical expression. Such series are called special series. The mathematical expression that is followed decides the algorithm to find the summation of the series.

Case 1: When the modulus value of the common ratio of the geometric sequence is less than ‘1’, the sequence will converge.
Example: 1 + 0.5 +0.25 …

Case 2: When the sequence is a rational function of ‘n’ with the modulus value less than ‘1’ for any value of n, the sum of the sequence will converge.
Example:

Case 3: When the denominator of the sequence is linear function of ‘n’, the summation of the sequence can be a convergent or divergent one.

Sum of Geometric progression:
By following the algorithm that is given below, we obtain the general formula for the summation of a geometrical sequence whose first terms is ’a’ and common ratio is ‘r’.
Step 1: Assume the sum of the given geometric progression as ‘S’.
Step 2: Find the common ratio of the progression.
Step 3: Multiply every term of the progression with ‘r’ to obtain S x r.
Step 4: Find the value of (Sr – S)
Step 5: Solve for S.

Nth term formula:
By applying the iterative method we obtain the nth term of a geometric sequence whose first terms is ‘a’ and common ratio is ‘r’ as an = a(r) (n-1).

Geometric progression examples:
Find the 15 term of the sequence 3, 6, 12…
First term = a = 3
Common ratio = r = 2.
15th term of the sequence is ar14 = 3(2)14 = 49152.

Median in Statistics

In statistics we use Mean Median and Mode to obtain information about the given data from the sample set of observed values. Let us consider a data, 26, 15, 33, 25, 28, 17, 37 is the data of the number of marbles a set of boys have with them. How do we find the middle value or the number which comes exactly in the middle of the list? To find the middle value, we need to find the Median Value which is the value exactly in the middle of the data when the data is arranged in the ascending order that is the smallest to greatest value. Let us arrange the data in the numerical order, 15,17,25,26,28,33,37. The Median Value or the middle value is 26.

Median in Math is one type of average in the branch of statistics in which we consider the middle value of the given data which is arranged in a numerical order in other words from smallest to greatest value.Median Math that is Median in statistics a branch of mathematics is the middle value of the data, which separates the data into two equal halves; which means fifty percent of the numbers are above the median and fifty percent of the numbers are below the median. And if there is even number of values in the data then the median is the mean of the two middle values. The Median in Math is a very useful number in cases when the distribution has very large extreme values which would otherwise skew the data.

Let us consider a data and learn the methods used in Finding Median.
39, 26, 7, 21, 3, 48, 67, 36
First we need to arrange the given data in the increasing order.
3,7,21,26,36,39,48,67
As you can see, the number of values in the data is even. We know that if the number of values in the given data is odd we can easily find the Median; the value exactly in the middle of the data. You might wonder, if the number of values is even then how to go about? We use a simple method in Finding Median in such cases; here, we get two middle values or two medians, so we just take the average of these two to get the actual median of the data.

There are 8 values in the given data; the two middle values are 26 and 36. To find the Median we find the mean of these values
(26 +36)/2 = 31
The required Median is 31

Frequency table

Frequency of data occurs when we classify statistical data in respect of either a variable or an attribute. What is a frequency table may be defined as a tabular representation of statistical data, usually in an ascending order, relating to a measurable characteristic according to individual value or a group of values of the characteristic under study.

In case, the characteristic under study is an attribute, say nationality, and then the tabulation is made by allotting numerical figures to the different classes the attribute may belong like, in this Example of a Frequency Table or Frequency Table Example, counting the number of Americans, Indians, British, French, German and so on.  The qualitative characteristic is divided into a number of categories or classes in the frequency table which are mutually exclusive and exhaustive and the figures against all these classes are recorded. The figure corresponding to a particular class, signifying the number of times or how frequently a particular class occurs is known as the frequency of that class. Thus, the number of Americans, as found from the given data, signifies the frequency of Americans. So we define frequency table is a statistical table that distributes the total frequency to a number of classes.

A frequency table is useful in other statistical applications such as finding the frequency polygon, plotting a histogram, bar charts and pie charts. The frequency tables are used both for qualitative and quantitative data.

To manage and operate on a date that is put into a frequency table is much easier than operating on raw or unclassified data. The table can be used to calculate mean, median and mode. Which in turn are useful in coming to many business or scientific decisions.  Frequency tables also find application in the field of code cracking, science, marketing surveys, census surveys etc.

Testing of hypothesis is dependent on the similarities and differences between the frequency tables involved.  Frequency table is also useful in finding the standard deviations and other measures of central tendencies. A frequency table is said to be skewed when the mean and median are different. The study of this skewedness forms the basis of many important decisions in business and stock market.

If we have the test scores of 35 students from total score of 10, then we can make a frequency table of the data. The first column would be the score from 0,1, to 10 and the second column would be the number of students in each score. That would make it a simple frequency distribution.

Significant figures rules and properties

What are significant figures?
Significant figures are important and interesting figures. Significant figure is the first non – zero digit of the number.
Rules of Significant figures
Following are the significant digits rules: -
1. Leading zeros in a number are not the significant figures. For example 0.008 has only one significant figure and that is 8.
2. Zeroes which are imbedded in between are always significant figures. For example, 24.07 have four significant figures.
3. If the decimal point is specified then only the trailing zeros are significant figures. For example, 400.00 have five significant figures and 400 has only one significant figure and that is 4.

Learn more about Definite Integral Examples also for better understanding.
Properties of Significant figures
1. Adding and Subtracting Significant figures:
Adding Significant figures - When the two numbers are added, the number of decimal places of the sum is always equal to the least number of decimal places in any of the numbers added.

Subtracting Significant figures - When the two numbers are subtracted, the number of decimal places of the difference is always equal to the least number of decimal places in any of the numbers subtracted.

2. Multiplication and Division of Significant figures:
Multiplying significant figures - When the two numbers are multiplied, the number of decimal places of the product is always equal to the least number of decimal places in any of the numbers multiplied.
Dividing significant figures - When the two numbers are divided, the number of decimal places of the product is always equal to the least number of decimal places in any of the numbers divided.

3. Rounding Significant figures: We should round off the figures if we have too many significant digits. We can round off any number using either of two ways: -
• If digit is smaller than 5 then the digit remains unchanged, such as 2.43 can be rounded off to 2.4
• If the digit is greater than 5 or then we should drop that digit and add one to the preceding digit, such as 2.47 can be rounded off to 2.5

Number Lines

In Math, number line is a line of any length which is used to place the real numbers in order. Real numbers include the positive numbers, negative numbers, zero, decimal numbers and even fractions.

What is a number line? As the name suggests, a number line is a line with numbers, numbers here mean, real numbers which consist of positive, negative, decimals, fractions including zero. Zero is called the origin. The line can be of any length according to the numbers we need to place on it.
There are many points on the number line, each of the point relates to a real number and again each number has a unique point that corresponds to it.
For example, 2.5 (2 ½) corresponds to a point on a number line that is halfway between 2 and 3.

Rules to remember:
On the number line,
1. the point zero is the origin
2. all the points to the right of the origin are positive numbers and
3. all the points to the left of the origin are negative numbers
4. The number is smaller if it is left of another
5. The number is greater if it is right of another
6. The distance between two numbers is the difference between the greater and the smaller number.
A number line is used for learning as to how to compare the negative and positive numbers
As we start moving from left to right on a number line, the number gets bigger and as we move from right to left, the number gets smaller

Number line template:

Integer Number line: On the number line, the point zero is the origin, all the points right to the origin are positive integers and all the points left to the origin are negative integers.

Decimal number line: 
We can place decimals on a number line. For which, each segment of the number line is divided into ten equal parts.

Let us place the point 3.5 on the number line.

  For a better understanding, let us place the following points on a number line,
A=1.8, B=2.6, C=4.2, D= -1.4