Linear Functions and Matrices

Definition of Matrices:

A matrix is a group of numbers arranged into a fixed quantity of rows and columns. Usually a matrix is a two dimensional range of numbers or terms arranged in a set of rows and columns. m × n matrices A have m rows and n columns and it is written,

A = `[[a_(11),a_(12),a_(1n)],[a_(21),a_(22),a_(2n)],[a_(m1),a_(m2),a_(mn)]]`

Definition of Linear Functions:

A polynomial functions of single degree is defined as a linear functions. It relates a dependent variable with an independent variable in a simple way. Mathematical equation in which there is no independent-variable is raised to a power greater than one. A simple linear function with one independent variable traces a straight line when plotted on a graph. It is also called as linear equation.
Solving Linear Functions Using Matrices:

Steps to solve the system of linear equations using matrices,

Step 1: Let us consider the equation,

a1x + b1y = c1,

a2x + b2y = c2,

Step 2: If, A = `[[a_1,b_1],[a_2,b_2]]` , X = `[[x],[y]]` , and C = `[[c_1],[c_2]]` .

then, AX = C.

Step 3: Now multiply both sides by, A−1

=> A−1AX = A−1C

Step 4: We know that, A−1A = I, where I is identity matrices,

=> IX = A−1C.

Step 5: Now, Multiplication of any matrices with identity matrices gives the same matrices. So,

Hence, X = A−1C.
Example for Solving Linear Functions Using Matrices:

Example: 5x – y + 11 = 0 and x – y – 5 = 0, solve the linear equations for x and y using matrices.

Solution:           5x – y + 11 = 0     => 5x – y = – 11,

x – y – 5 = 0         => x – y = 5.

Let, A =` [[5,-1],[1,-1]]` , X = `[[x],[y]]` , and C = `[[-11],[5]]`

To find the value of A−1:

A = ` [[5,-1],[1,-1]]`

Swap the leading diagonal values,

=> ` [[-1,-1],[1,5]]`

Change the signs of the other two elements in the matrices,

=>` [[-1,1],[-1,5]]`

Now, find the value of |A|.

|A| = (–1 × 5) – ((–1) × 1)

=> |A| = –5 + 1

=> |A| = –4

Hence, A−1 = `1/(|A|) xx [[-1,1],[-1,5]]`

=> A−1 = `1/(-4) xx [[-1,1],[-1,5]]`

=> A−1 =` [[(-1)/(-4),1/(-4)],[(-1)/(-4),5/(-4)]]`

=> A−1 =` [[1/4,(-1)/4],[1/4,(-5)/4]]`

Therefore the solution is,

X = A–1C,

=> X =` [[1/4,(-1)/4],[1/4,(-5)/4]] xx [[-11],[5]]`

=> X =`[[(-5)/(15) -(10)/(15)],[(10)/5 - 5/5]]`

=> X = `[[(-15)/(15)],[5/5]]`

=> X =` [[-1],[1]] = [[x],[y]]`

Hence the solutions are, x = – 1, y = 1.

Solve Herons Formula

We are familiar with the general formula for finding the area of a triangle when the base and height of the triangle is given. For scalene traingle we do not have any area formula. Heron's formula proof is a general formuls used to find area of triangles of all types.

Heron a famous mathematician gave simple formula for finding the area of any triangle based on its three sides. Thus, this formula for area is known as Heron's formula and is stated as below,

If p, q, r, denote the lengths of sides of a `Delta` PQR, then,

Heron's formula of a triangle

Area of `Delta` PQR = `sqrt(s(s-p)(s-q)(s-r))` , where   s = `(p+q+r)/(2) = ("perimeter of triangle")/(2)`

Here, s = semiperimeter of `Delta` PQR.

Heron's formula can used to find area of all types of triangles, quadrilateral, trapezoid, etc.
Solved Examples Using Heron's Formula

Ex 1: Find the area of a triangle whose sides are respectively 150 cm, 120 cm, and 200 cm.

Sol: Given three sides, p = 150 cm, q = 120 cm and r = 200 cm.

Step 1:  s = `(p + q + r)/(2) = (150 + 120 + 200)/(2)`

= `(470)/(2) = 235`

Step 2:  (s-p) = (235 -150) = 85

(s-q) = (235 - 120) = 115

(s -r ) = (235 - 200) = 35

Step 3: Area of triangle = `sqrt(s(s-p)(s-q)(s-r))`

= `sqrt(235xx 85xx 115xx 35)`

= `sqrt(5xx 47 xx 5xx 17 xx 5 xx 23 xx 5 xx 7)`

=  25`sqrt(47 xx 17 xx 23 xx 7)`

= 8966.56 cm2

Ex 2: Find the area of a triangle whose two sides are 18 cm and 10 cm and the perimeter is 42 cm.

Sol:  Given p = 18 cm, q = 10 cm and perimeter = 42 cm

Step 1:  perimeter = p + q + r

42 = 18 + 10 + r

42 = 28 + r

subtract 28 on both sides

42 - 28 = 28 - 28 + r

14 = r

Step 2 :  s = `("perimeter)/(2) = (42)/(2)`

s = 21

Step 3:  (s - p) = 21 - 18 = 3

(s - q) = 21 - 10 = 11

(s - r)  = 21 - 14 = 7

Step 4:  Area of triangle = `sqrt(s(s-p)(s-q)(s-r))`

= `sqrt ( 21 xx 3 xx 11 xx 7)`

= `sqrt( 3 xx 7 xx 3 xx 7 xx 11)`

= 3 x 7 `sqrt(11)`

= 21`sqrt(11)` cm2
Practice Problems on Heron's Formula

Pro 1: Find the area of a triangle whose sides are respectively 100 cm, 80 cm, and 60 cm

Ans: 2400 cm2

Pro 2: Find the area of a triangle whose two sides are 12 cm and 8 cm and the perimeter is 30 cm.

Ans: 39.68 cm2

Adding Cosine Functions

Trigonometry is one of the part of mathematics managing with angles, triangles and trigonometry functions such as sine, cosine, and tangent which are abbreviated as sin, cos, and tan respectively. There are several formulas for adding trigonometric functions such as sine, cosine, and tangents. The formulas for adding cosine functions are used to find the ratios when two angles are given. Those formulas for adding cosine functions are shown below.


cosine (A + B) = cosine A cosine B − sine A sine B.
Proof for Adding Cosine Functions:

To prove: cosine (A + B) = cosine A cosine B − sine A sine B.

Proof: We know that, cosine (A − B) = cosine A cosine B + sine A sine B.

Replacing B with (−B), these identities becomes,

sine (A − (− B)) = cosine A cosine (− B) + sine A sine (− B)

Since we know that, cosine (−A) = cosine A, and

sine (−A) = − sine A.

cosine (A + B) = cosine A cosine B − sine A sine B.

Hence proved that, sine (A – B) = sine A cosine B – cosine A sine B.
Solved Examples for Adding Cosine Functions:

Example 1: If sineA = 14, sineB = 6, cosineA = 12, and cosineB = 8, then find cosine(A + B).

Solution: Given that, sineA = 14, sineB = 6, cosineA = 12, and cosineB = 8.

We know that, cosine (A + B) = cosineA cosineB − sineA sineB,

=> cosine (A + B) = (12 × 8) − (14 × 6),

=> cosine (A + B) = 96 − 84,

=> cosine (A + B) = 12.

Answer: cosine (A + B) = 12.

Example 2: If sineA = 5, sineB = 9, cosineA = 17, and cosineB = 3,  then find cosine(A + B).

Solution: Given that, sineA = 5, sineB = 9, cosineA = 17, and cosineB = 3.

We know that, cosine (A + B) = cosineA cosineB − sineA sineB,

=> cosine (A + B) = (17 × 3) − (5 × 9),

=> cosine (A + B) = 51 − 45,

=> cosine (A + B) = 6.

Answer: cosine (A + B) = 6.

Rules for Division

In mathematics, especially in elementary arithmetic, division (÷) is the arithmetic operation that is the inverse of multiplication.

Understanding Rules of Divisibility is always challenging for me but thanks to all math help websites to help me out

Specifically, if c times b equals a, written:

C = b x a

where b is not zero, then a divided by b equals c, written:

a/b = c

For instance,

6/3 = 2

since

2 x 3 = 6

In the above expression, a is called the dividend, b the divisor and c the quotient.
Basic Rules for Division:

The basic rules for division are given as below:

Rule 1:

Division of 2 numbers with the same sign should be ‘+’ ve (positive sign) sign.

‘+’ve (Positive number) ÷’+’ve ( positive number) = ‘+’ve ( positive number)
‘-‘ve (Negative number) ÷ ‘-‘ ve (negative number) = =’+’ ve (positive number)

Rule 2:

Division of 2 numbers with different signs should be ‘-’ve negative

‘+’ve (Positive number) ÷(‘-‘ve negative number) = ‘-‘ve (negative number)
‘-‘ve (Negative number) ÷ ‘+’ve (positive number) = ‘-‘ve (negative number)

Examples for division:

Examples:

54 ÷ 9 = 6 (same signs)

(-32) ÷ (-2) = 16 (same signs)

10 ÷ (-2) = -5 (different signs)

(-12) ÷ 12 = -1 (different signs)


Rules for Division on Dividing Variable:

In this operation division represents dividing the variables based on the presence of values.

Example:

Solve the following division: 49 ÷ p, given that p = -7

Solution:

49 ÷ p

Substitute p = -7,

= 49 ÷ (-7)

= -7 (dissimilar signs).
Division on Dividing Decimal on Rules for Division:

Rules for decimal division solving problems,

Rule 1:

To create the decimal divisor as whole number by changing the decimal point to the right side.

Rule 2:

To change the same decimal point in the dividend to the right side to create as whole number

Rule 3:

After divide the new dividend or whole number by new divisor or whole number


Example:

Divide the following decimal function:  24.24 ÷ 0.20

Solution:

24.24 ÷ 0.20 = 24.24 / 0.20

= 242.4 / 2 (Take the decimal divisor as whole number)

= 2424 / 20 (Take the decimal dividend as whole number)

= 121.2 (Divide the new dividend by new divisor).

The Quotient

Quotients are one of the basic part of division operator in mathematics. A quotient is the result of a division. For example, when dividing 6 by 3, the quotient is 2, while 6 is called the dividend, and 3 the divisor. The quotient can also be expressed as the number of times the divisor divides into the dividend.

A relationship between dividend, quotient, divisor and remainder is :

Dividend = Quotient x Divisor + Remainder
Example Problems:

Example: Solve Quotient solver in 325 ÷ 15 by long division method.

Solution: Step 1:

-------                   15 ) 325

In the above equation 15 is divisor and 325 is dividend. In the divisor has two decimal numbers put the value dividend of 32.

Step 2:

2
-------
15 ) 325
30
---------
25

In 15 x 2 = 30 the divisor number 15 is multiplied with 2 to get an answer 30. In 30 is less than from 32.So use the value then subtract the value and get 25.

Step 3:

21
-------
15) 325
30
-------
25
15
--------
10

In 15 x 1 = 15 the divisor number 15 is multiplied with 1 to get an answer 15. In 15 is less than from 32.So use the value then subtract the value and get 10.

Step 4:

21.6
-------
15| 325
30
-------
25
15
---------
100
90
---------
100

The value 10 has no more value in the right side. So put 0 to get 100 and put decimal point on the quotient. Then normal divison 15 x 6 = 90 the divisor number 15 is multiplied with 6 to get an answer 90. In 90 is less than from 100.So use the value then subtract the value again we get 10.

Step 4:

21.66
--------
15| 325
30
--------
25
15
---------
100
90
----------
100
90
-----------
10 (continued)

Repeat the step again the value 10 have no more value in the right side. So put 0 to get 100 and put decimal point on the quotient. Then normal divison 15 x 6 = 90 the divisor number 15 is multiplied with 6 to get an answer 90. In 90 is less than from 100.So use the value then subtract the value again we get 10. Finally we get an answer 21.66.
Practice Problems:

Problem 1: Solve quotient solver in 425÷15

Answer: - 17

Problem 2: Solve quotient solver in 22÷11

Answer: - 2

Problem 3: Solve quotient solver in 58÷4

Answer: - 14.5

Problem 4: Solve quotient solver in 63÷2

Answer:  31.5

Problem 5: Solve quotient solver in 110÷5

Answer: - 22

Gaussian Function Equation

Gaussian function is defined as a function of the form. The graph of a Gaussian is a attribute symmetric "bell curve" shape that promptly falls off towards plus/minus infinity. Height of the curve's peak is a. Position of the centre of the peak is b. width of the bell is c. This function is commonly used in statistics, heat equations and diffusion equations and to define the Weierstrass transform. This function arises by the exponential and quadratic functions.In this article, we shall discuss about Gaussian function equation.

Gaussian Function Formulas:

 ( x) = a e^(-(x-a)^2/(2c^2))

here, a, b, c > 0 and  e = 2.718(approximate).

Gaussian integral is      int_(-oo)^oo e^(-x^2) dx = sqrtpi

nt_(-oo)^oo a e^(-(x-a)^2/(2c^2)) dx = a c sqrt (2pi)
Definition of Gaussian Function Equation :

Probability function of the normal distribution is,

f(x) = 1/(sigma sqrt(2pi)) e^ [(- (x-mu)^2)/(2sigma^2)] .

Full width at half maximum is finding the points x0 . So, we must solve

^ [(- (x_0-mu)^2)/(2sigma^2)] . = (1/2) (xmax).

But  f(xmax)  = mu

^ [(- (x_0-mu)^2)/(2sigma^2)] . = (1/2) (mu. = (1/2)

Now   solving the above equation ,        ^ [(- (x_0-mu)^2)/(2sigma^2)] . = 2^(-1)

(- (x_0-mu)^2)/(2sigma^2)] . = - ln2

(- (x_0-mu)^2)] . = - 2sigma^2 ln2

 (x_0-mu)^2] . = 2sigma^2 ln2 .

 (x_0-mu)] . = +-sqrt(2sigma^2 ln2)             .

_0 . = +-sqrt(2sigma^2 ln2) + mu             .

Therefore full width at half maximum is FWHM = 2 sqrt(2 ln2) sigma ~~ 2.35        .

Gaussian function equation for two dimensions:


The bi variate normal distribution is sigma = sigma_x = sigma_y

So,  f(x, y) = 1/(sigma^2 sqrt(2pi)) e^ [(- ((x-mu_x)^2 + (y - mu_y)^2))/(2sigma^2)] .

in elliptical function  sigma_x != sigma_y

f(x, y) = 1/(sigma_x sigma_y sqrt(2pi)) e^ - [(x-mu_x)^2/(2(sigma_x)^2) + (y - mu_y)^2/(2(sigma_y)^2)] .

Gaussian function

Gaussian function also used to find   A(x)   =  ^[(-x^2)/(2sigma^2)]

Instrument function is  I(k) = ^(-2pi^2 k^2 sigma^2) sigma sqrt(pi/2)[erf((a - 2pi i k sigma^2)/(sigmasqrt2)) + erf((a+2pi i k sigma^2)/(sigmasqrt2))]   .

I max    =   sigma sqrt (2pi) erf(a/(sigmasqrt2)) .

As  a -> oo  instrument  equation reduced as

im_(a->oo) I(k) = sigma sqrt(2pi)e^(-2pi^2 k^2 sigma^2).

Inequalities Lesson Plan

Inequalities are the main concepts in mathematical algebra.Usually inequalities are the combination of variables and constants  greater than and less than sign.  For example 2y -10 < 20 is a less than inequality. In this article we are going to study about inequalities lesson plan through suitable example problems.
Brief Summary about Inequalities Lesson Plan


General form of an inequality:
Inequalities take the following form,

     ax +by <= c

     Here x and y are variables and then ‘a’ and ‘b’ are coefficients of x and y and then c is the constant term.  <= is denote the inequality sign.

     The following signs are used to represent the inequalities.  They are

    < = less than
    > = Greater than
    <= less than or equal to
    >= Greater than or equal to

Examples on Solving Inequalities Lesson Plan

Example 1:

     Solve the following inequality 5p + 15 > 45

Solution:

     The given inequality is 5p + 15 > 45

     Now we have to subtract the value 15 on both sides, then we have to get the following inequality,

    5p + 15 -15 > 45 - 15

     That is 5p > 30

     Dividing the value 5 on both sides, then we have to the following inequality,

     That is p>6

     That is p value is greater than 6.

Example 2:

     Solving the following inequality 2r + 18 `<=` 36 and represent it on the number line

Solution:

     The given inequality is 2r + 18 `<=` 36

     Now we have to subtract the value 18 on both sides, then we have to get the following inequality,

     2r + 18 -18 `<=` 36 -18

     That is 2r `<= ` 18

     Dividing the value 2 on both sides, then we have to the following inequality,

     That is r`<=` 9

     That is r value is less than or equal to 9.

     That is r=9 and r is less than 9.


Representation of number line:

 Inequalities on a number line

     First we have to draw the number line and mark the values.  Here r is less than or equal to 9 so we can first shade the value 9 and then draw the arrow in the direction of negative direction.  Because it is the less then (<) value.

    These are the important problems in the inequality lesson plan.