Geometry Parallelogram Proofs

Geometry proofs are basic laws of geometry figures, using these geometry proofs the geometry problems are solved. The Geometry parallelogram proofs are the proofs for parallelogram that help the rules of solving parallelogram based problems. A parallelogram proofs form an argument that establishes the truth of the parallelogram figure. Let us see basic geometry parallelogram proofs.

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Geometry Parallelogram Proofs:

Properties of parallelogram:

We know that a quadrilateral is a closed plane figure formed by four line segments and a parallelogram is a quadrilateral in which the opposite sides are parallel to each other.

Proof in geometry 1: In a parallelogram, the opposite sides are of equal length.

Parallelogram proof

Proof: Let ABCD be a parallelogram. Link up BD. take the triangles ABD and BDC.

Since AB || CD and BD is a transversal of AB and CD, m∠ABD = m∠BDC

Since AD|| BC and BD is a transversal of AD and BC, m∠ADB = m∠DBC

The side BD is common to both ABD and BDC.

Hence, by AAS property, ΔABD ≡ ΔBDC.

∴ The particular sides are equal.

AD = BC and AB = CD

Proof in geometry 2: In a parallelogram, the opposite angles are of equal measure.
Parallelogram proof
Proof: Let ABCD be the parallelogram. Link up BD.

Since AB || DC and BD is a transversal to AB and DC,
m∠ABD = m∠BDC
Since AD || BC and BD is a transversal
to AD and BC,
m∠ADB =m∠CBD
∴ m∠ABC = m∠ABD + m∠DBC
= m∠BDC+ m∠ADB
= m∠ADC
Similarly, m∠BAD=m∠BCD
Geometry Parallelogram Proofs:

Proof in geometry 3: The diagonals of a parallelogram bisect each other.
Parallelogram proof
Proof: ABCD is a parallelogram. AC and BD form diagonals.
By ASA criterion, ΔAMB ≡ CMD
∴AM = CM, BM = DM.
∴This shows diagonals bisect each other.


Proof in geometry 4: If the opposite sides of a quadrilateral are of equal length, then the quadrilateral is a parallelogram.
Parallelogram proof
Proof: Let ABCD be a quadrilateral where AD = BC,AB = CD. Link up AC.

Consider the triangles from the figure, ACB and ADC. By SSS criterion, ΔABC ≡ ΔCDA

Then m∠ BAC = m∠ACD,m∠CAD=m∠ACB
∴ AB || CD and AD || BC. Hence ABCD form a parallelogram.

Quadratic Equation x Intercept

In coordinate geometry, the x intercept means the x-value of the point where the graph of a function or relation intercepts the x-axis of the coordinate system. It means the point at which the line cuts the x-axis.

The x intercept of a line is denoted as (x, 0).

In this article, we are going to learn finding x intercept for the quadratic equation.


Example Problems to Find X Intercept for Quadratic Equation:

Example 1:

Find x intercept for the quadratic equation y = x2 + x - 12

Solution:

Step 1: Given function

y = x2 + x - 12

Step 2: Plug y = 0 in the given function

0 = x2 + x - 12

We can write it as

x2 + x - 12 = 0

Step 3: Solve the above function for x to find x intercept

Find sum of two numbers and its product for x and - 12x2

x = 4x - 3x

-12x2 = (4x)(-3x)

Replace the term x by 4x -3x

x2 + 4x - 3x - 12 = 0

Take the common factor outside,

x(x + 4) - 3(x + 4) = 0

(x + 4)(x - 3) = 0

Factor each term,

x = - 4   and x = 3

Step 4: Solution

X intercepts for the given equation are (- 4, 0) and (3, 0)

Example 2:

Find x intercept for the quadratic equation y = 2x2 + 10x + 8

Solution:

Step 1: Given function

y = 2x2 + 10x + 8

Step 2: Plug y = 0 in the given function

0 = 2x2 + 10x + 8

We can write it as

2x2 + 10x + 8 = 0

Step 3: Solve the above function for x to find x intercept

Find sum of two numbers and its product for 10x and 16x2

10x = 8x + 2x

16x2 = (8x)(2x)

Replace the term 10x by 8x + 2x

2x2 + 8x + 2x + 8 = 0

Take the common factor outside,

2x(x + 4) + 2(x + 4) = 0

(2x + 2)(x + 4) = 0

Factor each term,

x = - 1   and x = - 4

Step 4: Solution

X intercepts for the given equation are (- 1, 0) and (- 4, 0)




Practice Problems to Find X Intercept for Quadratic Equation:

1) Find x intercepts for the quadratic equation y = x2 + 18x + 32

2) Find x intercepts for the quadratic equation y =2 x2 + 14x + 12

2) Find x intercepts for the quadratic equation y = 3x2 + 6x - 8

Solutions:

1) x intercepts are (-2, 0) and (-16, 0)

2) x intercepts are (-1, 0) and (-6, 0)

3) x intercepts are (0.914, 0) and (-2.914, 0)

Histogram Unequal Intervals

The Histogram Definition unequal intervals of applications are mostly used in statistics. The histogram unequal intervals are most effective way  to analysis the frequency distribution of grouped data. The concepts of histogram unequal intervals to show the quick way of understanding the grouped data in the graphical representations. The analyzing histograms diagrams are constructed from the two ways are depending on the uniform width, and varying widths.
Constructions for Histogram Unequal Intervals:

To analyzing  the histogram unequal intervals , we draw two perpendicular axes and choose a suitable scale for each axis. We blot class or group intervals of the continuous grouped or class of data on the horizontal axis and the respective number of  occurrence  of each class on the vertical axis. For each class, a rectangle is constructed with class interval as the base and height determined from the class frequency so that areas of the rectangles are proportional to the frequencies.

Let us analyzing histogram unequal intervals through an example, how a grouped frequency distribution of number of teachers in a city can be represented in histogram.


Histogram

From the above histogram unequal intervals diagrams, it is clear that the maximum number of students in the group 50 - 60 and the minimum number of students in the group 70 - 90.

Notes: In the diagram  (kink) before the class interval 30 - 40 on the horizontal axis. It shows that the full distance 0 - 30 is not shown.
Example of Histogram Unequal Intervals:

Another example of histogram unequal intervals for a math tutor wanted to analyse the performance of two class of students in a math test of 100 marks. Looking at their performances,  math tutor found that a few students got under 20 marks and a few got 70 marks or above. So the math tutor decided to analyzing histogram unequal intervals to group them into intervals of varying sizes as follows: 0 - 20, 20-30, ...., 60 - 70, 70 - 100. Then she formed the following table.
Marks                          Number of students
0 - 20                                         9
20 - 30                                       14
30 - 40                                       14
40 - 50                                       18
50 - 60                                       18
60 - 70                                       12
70 - above                                  4

Negative Fraction Calculator

Negative fraction is nothing but, the given input fractions are in negative values, and using negative fraction calculator will produce the output with in a second. Numerator and denominator to created the fraction, the numerator representing a number of equal parts and the denominator telling how many of those parts make up a whole.


Example is 4/5, the numerator value, 4, tells us that the fraction represents 4 equal parts, and the denominator, 5, tells us that 5 parts make up a total. An afterward growth be the common or "vulgar" fractions which are still used today like this (½, ⅝, ¾, etc.).
Examples Problems for Using Negative Fraction Calculator:

1. Subtract the fraction of -5/6 - 8/6, using negative fraction calculator.

Solution:

First enter the two fractions values in the negative fraction calculator box.
And we click the enter the equal to button in calculator.
Then we get the answer with in a second like this,

fraction calculator

Reduce button used to reduce the exact value in calculator.

Steps to calculate the negative fraction manually,

These two fractions have the same denominators, so we can directly subtract the numerators alone by keeping the denominator as it is.

Given value is,

= -5/5 - 2/6

= (-5 - 2) / 6

We get the answer is,

= -7 / 6.

2. Subtract the fraction of -9/2 - 8/2, using negative fraction calculator.

Solution:

First enter the two fractions values in the negative fraction calculator box.
And we click enter the equal to button in calculator.
Then we get the answer with in a second like this,

fraction calculator

Reduce button used to reduce the exact value in calculator.

Steps to calculate the negative fraction manually,

These two fractions have the same denominators, so we can directly subtract the numerators alone by keeping the denominator as it is.

Given value is,

= -9/2 - 8/2

= (-9 - 8) / 2

We get the answer is,

= -17 / 2.

More Examples Problems for Using Negative Fraction Calculator:

3. Subtract the fraction of -2/7 - 4/7, using negative fraction calculator.

Solution:

First enter the two fractions values in the negative fraction calculator box.
And we click enter the equal to button in calculator.
Then we get the answer with in a second like this,

fraction calculator

Reduce button used to reduce the exact value in calculator.

Steps to calculate the negative fraction manually,

These two fractions have the same denominators, so we can directly subtract the numerators alone by keeping the denominator as it is.

Given value is,

= -2/7 - 4/7

= (-2- 4) / 7

We get the answer is,

= -7 / 6.

4. Subtract the fraction of -6/3 - 5/3, using negative fraction calculator.

Solution:

First enter the two fractions values in the negative fraction calculator box.
And we click enter the equal to button in calculator.
Then we get the answer with in a second like this,



fraction calculator

Reduce button used to reduce the exact value in calculator.

Steps to calculate the negative fraction manually,

These two fractions have the same denominators, so we can directly subtract the numerators alone by keeping the denominator as it is.

Given value is,

= -6/3 - 5/3

= (-6 - 5) / 3

We get the answer is,

= -11 / 3.

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.

Equation Of A Line

Equation of a line is the equation satisfied by all the points which lie on the line.

 So, we can call every point on the line as a solution of the equation.

 As there are infinitely many points on a line, the equation of a line has infinitely many solutions.

 For every value of x, you can find a value for y by substituting the x value in the equation of the line.

 That is, consider the line whose equation is 2x + y = 4.

 We can find the value of y, when x equals 2.

 Substituting x = 2 in 2x + y = 3, we have

 2(2) + y = 3

 ? 4 + y = 3

 ? y = -1

 So, (2, -1) is a point on the line 2x + y = 4.

 There are various approaches to derive equation of a line.
Point Slope Form of Equation of a Line

We can draw a line if we know its slope and a point through which it passes. And the line we get is unique. Is it? Try.

 Hence we can find the equation of a line when its slope m and a point P(x1, y1) through which passes are given.

 The equation of the line is given as (y – y1) = m(x – x1)
Slope Intercept Form of Equation of a Line :

Suppose if c is the y-intercept of the line, then (0, c) is a point on the line.

 So, the equation (y – y1) = m(x – x1) can be written as (y – c) = m(x – 0)

 That is, y = mx + c

 So, we can find the equation of the line if its slope and y-intercept are known.
Equation of a Line Passing through Two Given Points:

Equation of a  line passing through two given points is also unique.

 So, we can find the equation of a line, if we know any two points on it.

 Suppose line l passes through the points A(x1, y1) and B(x2, y2), then slope of the line l is

 m = y2 - y1 / x2 -x1

Now we know the slope and a point on the line (considering either A or B) and hence we can find the equation of the line using the formula (y – y1) = m(x – x1).

So, the equation of the line is

(y - y1) = y2 - y1 / x2 -x1  (x - x1)

OR

(y - y2) = y2 - y1 / x2 -x1  (x - x2)

That is y - y1/ x - x1 = y2 - y1 / x2 -x1

In general, we can rewrite the equation of a line in the form ax + by + c = 0, where a, b and c are real numbers.

LCM Using Prime Factorization

The expanded form of LCM is Lease common multiple which means find the common multiple with least value for a given set of numbers. This can be done by using either prime numbers or by using composite numbers or with a combination of prime numbers and composite numbers. The LCM can also be written with exponents.
Steps to Find the Lcm Using Prime Factorization:

Step 1: Write the given set of numbers.

Step 2: Take the first number from the given set.

Step 3: Write the number with the multiples of prime numbers.

Step 4: Repeat step 2 and Step 3 for all the numbers in a given set.

Step 5: Choose the highest exponent term form all and multiply it to get LCM of given numbers.
Example Problems – Lcm Using Prime Factorization:

Example 1- LCM using prime factorization:

Find the LCM of 30 and 10.

Solution:

The given numbers are 30 and 10.

First it needs to find the prime factors of 30.

30 = 2*15

     = 2 * 3  * 5

The prime factors of 30 are 21, 31 and 51.

Next it needs to find the prime factors of 10.

10 = 2 * 5.

The prime factors of 10 are 21 and 51.

The highest exponent of all terms is 21, 31 and 51.

The LCM of 30 and 10 is 2 * 3* 5 `=>` 30.

My forthcoming post is on prime factorization algorithm, adding and subtracting rational expressions with unlike denominators will give you more understanding about Algebra

Example 2- LCM using prime factorization:

Find the LCM of 42 and 60.

Solution:

The given numbers are 42 and 60.

First it needs to find the prime factors of 42.

42 = 2*21

     = 2 * 3  * 7

The prime factors of 30 are 21, 31 and 71.

Next it needs to find the prime factors of 60.

60 = 2 * 30

     = 2 * 2 * 15

     = 2* 2 * 3* 5

The prime factors of 10 is 22 , 31 and 51.

The highest exponent of all terms is 22, 31, 51 and 71.

The LCM of 30 and 10 is 22 * 3* 5* 7 `=>` 420.

Example 3- LCM using prime factorization:

Find the LCM of 50 and 100.

Solution:

The given numbers are 50 and 100.

First it needs to find the prime factors of 50.

50 = 2*25

     = 2 * 5  * 5

The prime factors of 50 are 21 and 52

Next it needs to find the prime factors of 60.

100 = 2 * 50

       = 2 * 2 * 25

       = 2* 2 * 5* 5

The prime factors of 10 is 22 and 52

The highest exponent of all terms is 22 and 52

The LCM of 30 and 10 is 22 * 52 `=>` 100

Equivalent Percentage

Percentage is a number represents fraction of 100. In percentage, numerator is any number but denominator is only 100. So that it is called as Percentage. We can denote percentage of a number by using symbol %.

Example of Percentage: `50/100` (or) 50%

Equivalent percentage is a percentage for equivalent fraction or decimal.

In this article, we shall see about how to find equivalent percentage for given values.
Example Problems – Equivalent Percentage:

Problem 1:

What is the equivalent percentage of 5 out of 10.

Solution:

Step 1:

First, we divide numerator by denominator.

So `5/10` = 0.5

Step 2:

Multiply and divide the above decimal value by 100.

 `(0.5*100)/100`

Step 3:

So we get, `50/100` is equal to 50%.

Answer:

Equivalent percentage of `5/6` is 50%.

Problem 2:

What is the equivalent percentage of 6 out of 8.

Solution:

Step 1:

First, we divide numerator by denominator.

So `6/8` = 0.75

Step 2:

Multiply and divide the above decimal value by 100.

 `(0.75*100)/100`

Step 3:

So we get, `75/100` is equal to 75%.

Answer:

Equivalent percentage of `6/8` is 75%.

Problem 3:

What is the equivalent percentage of fraction `4/5` .

Solution:

Step 1:

First, we divide numerator by denominator.

So `4/5` = 0.8

Step 2:


Multiply and divide the above decimal value by 100.

`(0.8*100)/100 `

Step 3:

So we get, `80/100` is equal to 80%.

Answer:

Equivalent percentage of `4/5` is 80%.

Problem 4:

What is the equivalent percentage of 0.06?

Solution:

Multiply and divide the given decimal value by 100

`(0.06*100)/100 `

So we get, `6/100` is equal to 6%.

Answer:

Equivalent percentage of 0.06 is 6%.

Problem 5:

What is the equivalent percentage of 0.25?

Solution:

Multiply and divide the given decimal value by 100

`(0.25*100)/100 `

So we get, `25/100` is equal to 25%.

Answer:

Equivalent percentage of 0.25 is 25%.
Practice Problems: Equivalent Percentage

Problem 1:

What is the equivalent percentage of `8/5` ?

Answer:

160%

 Problem 2:

What is the equivalent percentage of `10/1000` ?

Answer:

1%

Problem 3:

What is the equivalent percentage of 0.81?

Answer:

81%

Problem 4:

What is the equivalent percentage of 0.025?

Answer:

2.5%

How to Solve polynomials

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An algebraic expression of the form a₀+a₁x+a₂x²+….+a_nxⁿ where a₀, a₁, a₂,….a_n are real numbers, n is a positive integer is called a polynomial in x.

Polynomial consists of:-

Constants: Like -8, 3/5, 20

Variables: Like a and b or x and y

Exponents: Like the 2 in a²

Purpose of Polynomials:

Listed below is the purpose of polynomials.

• Polynomials show in a wide mixture of area of mathematics and science.
• They are used to form polynomial equations, which instruct a wide range of troubles, from simple word tribulations to complex harms in the sciences.
• The purpose of polynomials is to describe polynomial functions, which emerge in setting range from fundamental chemistry and physics to economics and social science.
• Polynomials are used in calculus and arithmetic examination to estimate extra functions.
• In advanced mathematics, polynomials are used to build polynomial rings, a mid concept in intangible algebra and reckoning geometry.

Introduction of distributive property problems

An operation is distributive if the result of applying it to a sum of terms equals the sum of the results of applying it to the terms individually.
a ( b + c ) = ( a x b ) + ( a x c )
Here, ‘a’ is multiplied with the sum of two terms ‘b and c’ in the left hand side which, gives the same answer when ‘a’ is multiplied individually with ‘b’ and ‘c’ and then added.

Distributive Property Problems Example Part - 1:

1) Solve the problem using distributive property 9(9 + x).

Solution:

=9(9 + x)

=(9 * 9 + x * 9)

=(81 + 9x)


2) Solve the problem using distributive property 2(4 + 9x)

Solution:

2(4 + 9x)

(4 * 2 + 9x * 2)

(8 + 18x)


3) Solve the problem using distributive property 7(-1 + x)

Solution:

=7(-1 + x)

=(-1 * 7 + x * 7)

=(-7 + 7x)


4) Solve the problem using distributive property 12(a + b + c)

Solution:

=12(a + b + c)

=(a * 12 + b * 12 + c * 12)

=(12a + 12b + 12c)


5) Solve the problem using distributive property 7(a + c + b)

Solution:

=7(a + b + c)

=(a * 7 + b * 7 + c * 7)

=(7a + 7b + 7c)

Distributive Property Example Problems Part - 2:


6) Solve the problem using distributive property -10(3 + 2 + 7x)

Solution:

-10(3 + 2 + 7x)
Combine like terms: 3 + 2 = 5
-10(5 + 7x)
(5 * -10 + 7x * -10)
(-50 + -70x)


7) Solve the problem using distributive property -1(3w + 3x + -2z)

Solution:

=-1(3w + 3x + -2z)
=(3w * -1 + 3x * -1 + -2z * -1)
=(-3w + -3x + 2z)


8) Solve the problem using distributive property 1(-2 + 2x2y3 + 3y2)

Solution:

=1(-2 + 2x2y3 + 3y2)

=(-2 * 1 + 2x2y3 * 1 + 3y2 * 1)

=(-2 + 2x2 y3 + 3y2)


9) Solve the problem using distributive property 5(5 + 5x)

Solution:

=5(5 + 5x)

=(5 * 5 + 5x * 5)

=(25 + 25x)


 10) Solve the problem using distributive property y(1 + x)

Solution:

=(y + yx)

=(y + xy)

11) Solve the problem using distributive property 5(x + 10).

Solution:

=5( x + 10)
=(5* x + 5 * 10)
=(5x + 50)

Calculus problems solver

All the algebra, functions, relations etc that we study in math can be termed as pre-calculus. The next huge topic in math is calculus. Calculus is distinct from pre-calculus. In calculus there is less of static properties and more of dynamic properties. We talk of one quantity tending to another in calculus.
Calculus is a wide topic that encompasses various subtopics. The most common of which are: Limits, continuity, differential calculus and integral calculus.

The origin of calculus goes back at least 2500 years to the ancient Greeks. They use to find help with calculus problems of how to find area of any polygon.  The method they used was like this: the polygon was divided into triangles and the area of the triangles was added to find the area of the entire polygon, which is very similar to what we do in integration.

In solving calculus problems like finding the area of closed region bounded by curves, it is not possible to split the region to triangles as the boundaries are curved. The Greek method was to inscribe polygons in the figure the circumscribe polygons about the figure and then let the number of sides of the polygon increase. The limiting position of these areas was the area under consideration. Eudoxus used this method to solve calculus problems pertaining to area of circle and to prove the famous formula of A = pi r^2.

The problem of trying to find the slope of the tangent line to a curve gave rise to the branch of calculus called differential calculus. In problems of finding rate of increase of quantities, velocity, acceleration etc, the idea of limits is useful.

Sample calculus problems:

1. The price of a box of pencils is as follows: 5x/(x-1). Where x is the number of boxes one buys. So that means the more then number boxes that a person buys, the lower is the per box cost. What would the price tend to if a person buys lesser and lesser number of boxes?
Solution: The above problem is same as solving the limit: lim(x->1)(5x/(x-1)). We can see that x tends to 1, the term tends to infinity. So if someone wants to buy just one box, the cost would be pretty high.

2. Radius of a circle increases at a constant rate of 4 cm/s. Find the rate of change of area of circle when radius is 6cm.
Solution: This is application of derivatives.
A = pir^2
dA/dt = 2pir.dr/dt
dr/dt=4 and r=6
so, dA/dt = 2*pi*6*4 = 48pi (cm)^2/s

Improper Fractions

Fractions form an integral part of our lives. Fraction is part of a whole. Hence, in a fraction we have two numbers. One on the top called numerator and the other on the bottom called denominator.
We classify fractions as proper fractions and improper fractions based on the values of numerator.

Improper fractions: 
Fractions with numerator greater than the denominator are called improper fractions.
Example: 13/4, 12/5
How to simplify improper fractions:
This is nothing but reducing improper fractions.
As the numerator is greater than the denominator in improper fractions, when we do the long division of this fraction, we will be getting a quotient greater than 1 and a remainder.
We can generalize the simplification by an algorithm:
Step 1: Perform the long division
Step 2:
Example: Simplify the improper fraction 18/7

Step 2:  Simplified form of the mixed fraction is


Improper fractions:

Improper fractions have two parts on simplification. The part which does not have denominator is called whole part and the part with denominator is called fractional part. The other name for the simplified improper fraction is mixed fraction.
Mixed fraction has both whole number part and a fractional part.

Adding improper fractions:
We adopt four steps to add improper fractions.
Step 1: Find the Least common multiple of the denominators
Step 2: Find the equivalent fractions of individual fractions such that the denominators of both the fractions are same.
Step 3: Add both the numerators and write the common denominator
Step 4: Simplify the improper fraction if need be
Example: 11/3 + 5/4
(11 x 4)/(3 x 4) + (5 x 3)/ (4 x 3)
= (44/12) + (15/12)
= 59/12
=4 11/12

Multiplying Improper Fractions
In this, the method adopted is as same as in multiplication of proper fractions. We follow three steps to evaluate the product of two improper fractions.
Step 1: Multiply both the numerators and write the result in numerator
Step 2: Multiply both the denominators and write the result in denominator
Step 3: Simplify if need be
Multiply 15/7 and 4/3
Step 1: Numerator = 15 x 4 = 60
Step 2: Denominator = 7 x 3 = 21
Step 3: Required product = 60/21 = 3 x 20/ 3 x7 = 20/7 = 2 6/7

Dividing Improper Fractions
In this as well, we follow the general rules of division of fractions.
Step 1: Write the first fraction
Step 2: Find the multiplicative inverse of the second fraction
Step 3: Multiply the results of steps 1 and 2 following the algorithm of multiplication of fractions.
Example:
Divide 15/ 7 by 4/3
Step 1: 15/7
Step 2: Multiplicative inverse of 4/3 is ¾
Step 3: 15/7 x 3/4 = (15 x 3)/ (4 x 7) = 45/28.
What is an improper fraction?
Improper fractions are those whose numerators are greater than denominators. The value of improper fractions is always greater than one.

Unit Conversion

Unit conversion is converting from one unit to another of the same quantity. Unit conversion is used when we have to compare or convert any quantity in any particular form. Unit conversion can be done for length, volume, time, temperature, area, energy etc. It is done using conversion factors.
For example: -
I kilogram = 1000 grams
1 foot = 12 inches
1 meter = 100 centimeters
1 minute = 60 seconds
1 day = 24 hours
What is the Metric Unit for Length?
The metric unit of length is meter.
Unit conversion length
1 meter = 1000 millimeter
1 meter = 100 centimeter
1 meter = 10 decimeter
1 meter = 0.001 kilometers
How to do unit conversions?
Unit conversions can be done by using the following steps: -
Write the given value.
For example: - Convert 582 cm to m.
Find conversion factor for the given and the desired units
In the above example the conversion factor is 100 cm = 1m.
Write it as a fraction with the given units as a denominator or in the opposite direction.
582 cm (1 m)/(100 cm)
Cancel the ‘like’ units that is cm in the numerator can be cancelled by cm in the denominator.
582 cm (1 m)/(100 cm )=582  (1 m)/100

Multiply odd units, we will be left with 582 times 1 m in numerator which gives 582 m and 100 in denominator which gives 582 m/100 = 5.82 m
5.82 m
For example: -
If we have to convert 12 millimeter to kilometer, then the steps would be: -
Convert 12 mm to km
Conversion factors are: -
1000 mm = 1 m
1000 m = 1 km
1.2 mm(1m/1000mm)((1 km)/(1000 m))
=0.0000012 km
This is how unit conversion is done when we have any quantity in a unit and we need to convert it in different unit.

Introduction to Trinomials

What is a trinomial?
To understand the concept of trinomial, we should know what a monomial is. A monomial is an algebraic expression that has only one term.

For example: - 2a, 4x, 6z are all monomials. A trinomial is a polynomial which consists of three monomial terms. For example: -
• 2a+3b-4c
• 4x-84-6z
They both are the examples of trinomials.
Factoring trinomials
Rules to factor trinomials: (how to factor trinomials)
A trinomial expression is an algebraic expression which has exactly three terms. Trinomials can either be quadratic equation or a higher order equation. In case it is a quadratic equation, it can be simplified either by factoring or using quadratic formula.

For example: - 2x^2-7x+6 can be written as 2x^2-4x-3x+6 which can be factored as (2x-3) (x-2) so this is how we factor trinomials if they are in quadratic form.
A higher order trinomial can be turned into a quadratic equation by factoring common terms and then can be factored again.

For example: - 6x^2y + 14xy + 4y
In this expression, we have 2y common in all terms, so we can take that common, so that becomes
2y (3x^2+7x+2)
And that can further be factored as 2x (3x+1) (x+2).
Following instructions should be followed for solving trinomials: -
• First, factor the common factors from all the terms. For example if we have 3x^2+24x+45 then we can take 3 common from all the terms and write the expression like this, 3(x^2+8x+15).
• Check the trinomial equation you are left with.
• If the highest power of the trinomial equation is 2, then it can be factored like a quadratic equation
• If the highest power is the higher degree, then we should look for a pattern that allows you to solve it like a quadratic equation.
• Solve the quadratic part of the equation and make the factors.