Prime Model

Introduction to prime model:

In this article we shall discuss about prime model. Here, prime model is also denoted as prime number model. A Prime Number is a complete number, bigger than 1, so as to can be uniformly divided just by 1 otherwise itself. "Prime Factorization" is recognized which prime numbers need to multiply as one to get the original number.

Here, few prime numbers are:  1, 7, 13, and 19 etc...

The prime factorization model of a number is multiplying prime factors of a number.

Here, simple example is 38 = 2 x 19.

Examples of Prime Number Model:

The example of 1 - 100 prime number model is shown given below that,

Prime number of 1 - 10:

The prime number of 1 - 10 is 2, 3, 5 and 7.

Prime number of 10 - 20:

The prime number of 10 - 20 is 11, 13, 17 and 19.

Prime number of 20 - 30:

The prime number of 20 - 30 is 23 and 29.

Prime number of 30 - 40:

The prime number of 30 - 40 is 31 and 37.

Prime number of 40 - 50:

The prime number of 40 - 50 is 41, 43 and 47.

Prime number of 50 - 60:

The prime number of 50 - 60 is 53 and 59.

Prime number of 60 - 70:

The prime number of 60 - 70 is 61 and 67.

Prime number of 70 - 80:

The prime number of 70 - 80 is 71, 73 and 79.

Prime number of 80 - 90:

The prime number of 80 - 90 is 83 and 89.

Prime number of 90 - 100:

The prime number of 90 - 100 is 97. Please express your views of this topic Perfect Number by commenting on blog.

Example Problems Based on Prime Factorization Model:

Example problems based on prime factorization model is given below that,

Example:

How to find prime factorization model for 78?

Solution:

Step 1:

To create through the smallest amount prime number, which is 2, then the value is

`78/2 = 39`

Step 2:

Here 39 is not a prime number, therefore we require factoring it over again:

`39/3 = 13`

Step 3:

13 is a prime number,

78 = 13 × 3 × 2. This is the answer; all factors are a prime number. Since a result should be Correct.

Step 4:

The prime factorization for 78 is 13 × 3 × 2.

Standard Deviation Bars

Introduction about standard deviation bars:

In this article we are going to see about standard deviation bars. Bars are nothing but a bar chart or the pictorial representation of the data values; it should be plotted using the given data’s. Standard deviation is a statistical one and it be supposed to be measure the variability. The standard deviation is the root mean square deviation of the values from their mathematics mean. Let see how standard deviation bars represents.

Standard Deviation Bars

Standard deviation formula:

The formula for standard deviation is.

s.d= `sqrt((sum(X-M)^2)/(n-1)) `

Where,

S = sum of values

X = individual value

M = mean of all values

n = number of values

Problems on Standard Deviation Bars:

Example 1:

Find the Standard deviation for 1, 2, 3, 4, 5 and 6 then draw the standard deviation bars.

Solution:

Step 1:

Find the mean and deviation.

Mean =` (1+2+3+4+5+9)/6`

= `24/6`

M = 4

Step 2:

Find the sum of (X-M) 2

9+4+1+0+1+25= 40

Step 3:

Given n = 5, then find the total number of values using formula n-1.

n-1= 6-1

= 5

Step 4:

Find Standard Deviation using the formula:

S.D = `sqrt((sum(X-M)^2/(n-1)))`

= `sqrt(40/5)`

=` sqrt(8) `

S.D = 2.82

Example 2:

Find the Standard deviation for 3, 6 and 9 then draw the standard deviation bars?

Solution:

Step 1:

Find the mean and deviation.

Mean = `(3+6+9)/3`

= `18/3`

M = 6

Step 2:

Find the sum(X-M) 2

9+0+9= 18

Step 3:

Given n = 5, then find the total number of values using formula n-1.

n-1= 3-1

= 2

Step 4:

Find Standard Deviation using the formula:

S.D= `sqrt((sum(X-M)^2)/(n-1))`

= `sqrt(18/2)`

= `sqrt(9)`

S.D   = 3


Having problem with Differential Calculus keep reading my upcoming posts, i will try to help you.

Practice problems for standard deviation bars:

1. Find the Standard deviation for 4, 8 and 12 then draw the standard deviation bars?

Answer: 3.46

2. Find the Standard deviation for 4, 10 and 13 then draw the standard deviation bars?

Answer: 4.58

3.. Find the Standard deviation for 5, 10 and 15 then draw the standard deviation bars?

Answer: 5

Inverse Property of Addition and Multiplication

Introduction to inverse property of addition and multiplication:

For solving the expression in mathematics, the inverse properties of addition and multiplication are useful. The result of inverse property for addition is 0 and the result of inverse property for multiplication is 1. The inverse property requires a number and its inverse.

Example: Inverse of a term +a is –a for addition `1/a` for multiplication.

Explanation of Inverse Property of Addition and Multiplication:

Statement of inverse property of addition:

Inverse property of addition states that when a number or variable is added to its inverse number or variable, the result of this addition becomes zero. For addition the inverse of a given number or variable is obtained by changing the sign of given number or variable.

Example:

Consider a variable ‘t’. An inverse term of ‘t’ is –t. Addition of new opposite equal terms gives 0.

t + (-t) = t – t =0

Inverse property of multiplication:

Inverse property of multiplication states that when a number or variable is multiplied by its inverse, the result becomes one. For multiplication the inverse of a number or variable is the reciprocal of a given number or variable. Looking out for more help on Mathematical Expressions in algebra by visiting listed websites.

Example:

Consider a variable is ‘t’. The inverse of this variable is t-1 which is `1/t`.

t × `1/t`  = 1

Example Problems to Inverse Property of Addition and Multiplication:

Example: 1

Which is the inverse element of 61 for addition?

a) +61

b) -(-61)

c) `1/61`

d) -61

Solution:

Given number is 61.

Inverse property result is 0.

61 + x = 0

61 is subtracted on both sides.

61 + x – 61 = 0 – 61

x = -61

Answer: d

Example: 2

Which of the following is the inverse element of 9 for multiplication?

a) 9-1

b) `(1/9)^-1`

c) +9

d) -9

Solution:

Inverse property of multiplication result is 1.

9  X x = 1

9 is divided on both sides.

`(9x)/9` = `1/9`  = 9-1

Answer: a

Practice Problems to Inverse Property of Addition and Multiplication:

Problem: 1

What is the inverse element of `7/8` for addition?

Answer: `-7/8`

Problem: 2

What is the inverse element of 100 for multiplication?

Answer: `1/100`

Post a Math Problem

Introduction to post a math problem:

Post a math problem is that several math problem posted on the internet for the students help. There are many websites that post a math problems to help the students. Among the entire website tutor vista is the famous and wonderful website which has excellent tutoring team to help the students any time with all the homework problems. The tutoring team will be online 24 x 7 to help the students.In the tutor vista website any student can post a math problem at any time and they can immediately get help. In this article let us see sample math problems posted in this website. Understanding Associative Property of Addition Definition is always challenging for me but thanks to all math help websites to help me out.

Post a Math Problem:

Example 1:

Find the solution of p in the equation 7p + 4p = 55

Solution

Given equation is 7p + 4 p = 55

Add leht side of the equation since both are in terms of p

7p+ 4p = 55 p

So 11p = 55

Divide both sides by 11

`(11p)/11` = `55/11`

p = 5.

Check the value by substituting p = 5 in the given equation

7(5)+4(5) = 55

35 + 20 = 55

55 = 55

So the answer p = 5 is correct.

Example 2:

Solve for x in the given equation 10x + 2 = 47

Solution

The given equation is 10x+2 = 47

Subtract 2 on both sides

So 10x + 2 -2 = 47 - 2

By subtracting 2 the equation is formed as 10x = 45

Now divide by 10 on both sides

`(10x)/10` = `45/10`

x = 4.5

So we get the answer as x = 4.5.

Now we can check this whether our answer is correct or not. For checking our answer substitute the value x = 4.5 in the given equation at the place of x. Is this topic What are Rational Numbers hard for you? Watch out for my coming posts.

So given equation is 10x + 2 = 47

Substitute x = 4.5

10(4.5) + 2 = 47

45 + 2 = 47

47 = 47

So both sides are equal and our answer x = 4.5 is correct.

Post a Math Problem:

Here are some of the question. Solve and find out the solution and check as shown above.

1. Find the value of k of the equation 3k = 27

2. Find the value of p of the equation 8p - 3p = 75

3. Find the value of p of the equation 12 p = 42 + 102

4. Find the value of m of the equation 15m - 7m = 2m +66

5. Find the value of s of the equation 5s + 50 - 3s = 60

Answers

1. k = 9

2. p = 15

3. p = 12

4. m =11

5. s = 5

Law of Sines Equation

Introduction to law of sines equation

The trigonometric law of sines equation is used to calculate the unknown side of the triangle. In triangle we recognize the either sides or angle we can calculate the sizes of the other sides and angles. If we can consider the right triangle, we can make use of simple trigonometric ratios to determine the unidentified lengths of the side of the triangle.  Within a general triangle, you need using other technique; they are the laws of sines and cosines formulas. Let us see about the law of sines equation.

Law of Sines Equation

Assume the following triangle

In the above triangle include the inside angles A, B and C.

If any three values are recognized subsequently we can obtain the other three values.

Total of the inside angles of the triangle equivalent to one eighty degree.

A+B+C = 1800

We can find the sides of triangle using sine rule. I have recently faced lot of problem while learning college math problems with answers, But thank to online resources of math which helped me to learn myself easily on net.

`a/sinA = b/sin B = c/sin C`

Examples for Law of Sines Equation

Example 1 for law of sines equation

Compute the unknown side of the triangle.

Solution

The total of the internal angles equal to one eighty degree

A + B + C = 180º

Therefore

C = 180º - (A+B)

= 180º - (30º+70º)

= 180º - 100º

C = 80º

To compute side c using the sine rule

`b/sinB = c/sinC`

`c = (38 sin 80^0)/sin 70^0`

= 38 x 1.0480

c = 39.824

To compute side a using the sine rule

`a/sinA = b/sin B`

`a = (38 sin 30^0)/sin 70^0 `

= 38 x 0.5320

a = 20.216

Thus the sides of the given triangle a = 20.216 and c = 39.824

Example 2 for law of sines equation

Compute the unknown side of the triangle.

Solution

The total of the internal angles equal to one eighty degree

A + B + C = 180º

Therefore

C = 180º - (A+B)

= 180º - (30º+70º)

= 180º - 100º

C = 80º

To compute side c using the sine rule

`b/sinB = c/sinC`

`c = (45 sin 80^0)/sin 70^0`

= 45 x 1.0480

c = 47.16

To compute side a using the sine rule

`a/sinA = b/sin B`

`a = (45 sin 30^0)/sin 70^0 `

= 45 x 0.5320

a = 23.94

Thus the sides of the given triangle a = 23.94 and c = 47.16

Practice Subtracting Fractions

Introduction to practice subtracting fractions:

To practice subtracting fractions, first we convert both denominators to the number. Fractions are  reduced to their lowest terms so that they would be easier to recognize. To find fraction’s lowest terms, you need to divide the numerator and the denominator by anycommon factors. Here we need to practice subtracting fractions that is learning subtracting fractions problems. In this article let us see problems to practice subtracting fractions.

Practice Subtracting Fractions:

Subtracting fractions is a process of collecting all  like fractions. If the denominators are the same then just subtract the numerators.

Example 1:

`9/ 8` - `4/ 8`

Solution:

If the denominators are the same then just subtract the numerators.

`9/ 8` - `4/ 8` =  `5/8`

Example 2:

`1/3` - `2/5`

Solution:

If the denominators are different then first find the LCM of denominator and make the denominator same then subtract the numerator.

`1/3` - `2/5`    LCM = 15  ie. Both 3 and 5 of denominator is multiple of 15.

`1/3`   - `2/5`  = `(1*5) / (3*5)` - `(2*3)/(5*3)`   = `5/15` - `6/15` = `-1/15` .

Example 3:

`11/ 8` - `4/ 8`

Solution:

If the denominators are the same then just subtract the numerators.

`11/ 8` - `4/ 8` =  `7/8`

Example 4:

`1/3` - `1/4`

Solution:

If the denominators are different then first find the LCM of denominator and make the denominator same then subtract the numerator.

`1/3` - `1/4`    LCM = 12  ie. Both 3 and 4 of denominator is multiple of 12.

`1/3`   - `1/4`   = `(1*4) / (3*4)` - `(1*3)/(4*3)`  =  `4/12` - `3/12` = `1/12` .

Practice Subtracting Fractions:

Example 5:

`12/ 8` - `5/ 8`

Solution:

If the denominators are the same then just subtract the numerators.

`12/ 8` - `5/ 8` =  `7/8`

Example 6:

`12/7` - `1/8`

Solution:

If the denominators are different then first find the LCM of denominator and make the denominator same then subtract the numerator.Please express your views of this topic Free math problem solver by commenting on blog.

`12/2` - `1/8`    LCM = 8  ie. Both 2 and 8 of denominator is multiple of 8.

`12/2` - `1/8`  = `(12*4) / (2*4)` - `1/ 8`   = `48/8` - `1/8` = `47/8` .

Example 7:

`18/ 8` - `9/ 8`

Solution:

If the denominators are the same then just subtract the numerators.

`18/ 8` - `9/ 8` =  `9/8`

Example 8:

`4/3` - `2/5`

Solution:

If the denominators are different then first find the LCM of denominator and make the denominator same then subtract the numerator.

`4/3` - `2/5`    LCM = 15  ie. Both 3 and 5 of denominator is multiple of 15.

`4/3` - `2/5`  = `(4*5) / (3*5)` - `2/15`  = `20/15` - `2/15` = `18/15` .

Two Step Equations with Integers

Introduction to two step equations with integers:

Solving two step equation with integers is nothing but solving the equations that has integers in two steps. just by simplifying the equations that involve integers and variables. here in resolving these two step equations with integers just we resolve the variables in the equations..The students will be able to solve one-step and two-step linear equations involving integers, fractions, and decimals. Let us see what is the two step equations with integers and the sample problems for two step equations.

Two Step Equations with Integers:

Solving Two-Step Equations - getting variable alone

1. Remove constant by adding opposite to both sides.

2. If variable is being divided by a number, multiply both sides by the number. If variable is being multiplied by a number, divide both sides by the number (remember dividing by a fraction is the same as multiplying by the reciprocal).Is this topic solving linear equations and inequalities hard for you? Watch out for my coming posts.

Example 1:

Simplify     `X/3` + 3 = 12

`X/3` + 3 – 3 = 12 – 3

`X/3` + 3 – 3 = 12 – 3

`X/3` = 9

`X/3` = 9

`X/3` · 3 = 9 ·3

X =  `9*3`

X = 27

Example 2:

Simplify     4X – 5 = 6

4X – 5 + 5 = 6 + 5

4X – 5 + 5 = 11

4X = 11

`(4X)/4` = `11/4`

`X` = `11/4`

X = 2.75.

Two Step Equations with Integers:

Example 3:

simplify -3- 9x = -21  keep the variables on one side and move the integers to other side.

-9x = -18  resolve by dividing

x  = 2

Example 4:

simplify 20 + 4x = 25

4x = 25 - 20

4x = 5

x = 20.

Example 5:

simplify

5x + 7 =14  keep the variables on one side and move the integers to other side.

5x = 7

x = `7/5 `

=1.4

Example 6:

simplify

`2/3` x - `1/(2x)` + 3 =15   keep the variables on one side and move the integers to other side.

`1/(6x)` + 3 =15

`1/(6x)` =12

x = 72

Example 7:

11=`2/(7m)` + 9

`11(7)` =`2/7` (7/1)m+9(7) Multiply each term by the Least Common Denominator (LCD)(7).

77-63 = 2m+63-63 Subtract 63 from both sides.

14/2 = `(2m)/2` Divide both coefficients by two

7=m

Properties of Limits

Introduction to properties of limits:

Let us see about the properties of limits. The majority fundamental concept of modern Calculus limit. Limit used to define the value that a function or order sequence approaches as the input or index approaches a few value. Properties of limits are important to calculus and also used to describe the continuity, the integrals and the derivates.

Definition:

The limits are the numerical values. The limit values are important for draw the specific diagram in the graph. Without the limits, the line cannot end at any session. The limits are may be positive or negative in the graphical representation. The limits are having some properties which are used in the mathematics.

Limit of numerical sequence:

Assume a mathematical sequence, a normal term of which advance to some number x at increasing an ordinal number n. In this case, that the mathematical series has a limit. This document has a more harsh definition: A number x is known as a limit of a numerical sequence.

This definition means, that x is a limit of a arithmetical order, if its general term approach unrestrictedly to x at increasing n. Geometrically it means, that for any > 0 it’s likely to find such a number N , that starting from n > N all terms of the sequence are located within an interval ( x – , x + ). A sequence, having a limit, is known as convergent; other hand - x divergent sequence.

First we will assume that `lim_(x->a)` f(x) and `lim_(x->a)g(x)` exist that c is any constant.

`lim_(x->a)[c f(x)] = clim_(x->a) f(x)`

In additional terms we can “factor” a multiplicative constant out of a limit.I have recently faced lot of problem while learning Calculus Solver, But thank to online resources of math which helped me to learn myself easily on net.

Properties of Limits:

The properties of limits are,

Assume,

`lim_(x->a)` f(x)= L and `lim_(x->a)` g(x) =M.

Where L and M indicates real numbers. Also assume c > 0 is a real number. Then

`lim_(x->a)` f(x) +g (x) =L+M.

`lim_(x->a)f(x) *g(x) = LM.`

`lim_(x->a)cf(x) = cL.`

Moreover M' 0 then,

`lim_(x->a)f(x) /g(x) = L/M.`

`lim_(x->a)1/g(x) = 1/L` .

These are the properties of limits used by the mathematics.

Decimal to Binary Formula

Introduction to decimal to binary formula:

In this section we have decimal to binary formula. Decimal to binary conversion is one of the most important conversions in math subject. Decimal to binary conversion is the very simple process compare than other conversion in math. Below we have some conversion based on decimal and binary numbers. Let us see about decimal to binary formula.

Example Problems for Decimal to Binary Formula:
Example problem 1: Can you convert the following decimal number into binary number with neat and clear explanation: 211

Solution:

Given 211

To convert the given number 211 into binary number we have to follow some conditions:

First we divide the given number by 2 until we get one or zero.

The neat step is given below:

Therefore, the binary value of 211 is 11010011.
Answer: The binary value of 211 is 11010011.

Example problem 2:  Can you convert the following binary number into decimal number with neat and clear explanation: 1101100

Solution:

Given 1101100

To convert the given binary number we have to multiply power of 2.

1101100 = 1 xx `2^6` + 1 xx `2^5` + 0 xx `2^4` + 1 xx `2^3 ` + 1 xx `2^2 ` + 0 xx `2^1` + 0 xx `2^0` = 64 + 32 +0 + 8 + 4+ 0 + 0 = 108

Therefore, the decimal number of 1101100 is 108.

Answer: The decimal number of 1101100 is 108.

Is this topic how to solve math problems hard for you? Watch out for my coming posts.

Practice Problems for Decimal to Binary Formula:

Practice problem 1: Can you convert the following decimal number into binary number: 56

Practice problem 2: Can you convert the following binary number into decimal number: 110011

Practice problem 3: Can you convert the following decimal number into binary number: 32

Practice problem 4: Can you convert the following binary number into decimal number: 100001

Practice problem 5: Can you convert the following decimal number into binary number: 78

Practice problem 6: Can you convert the following binary number into decimal number: 11111100

Solutions for decimal to binary formula:

Solution 1: 111000

Solution 2: 27

Solution 3: 100000

Solution 4: 33

Solution 5: 1001110

Solution 6: 252

Integral Calculus Calculator

Introduction to Integral Calculus Calculator:

The topics involved in calculus are functions, application of derivatives, limits, derivatives,  logarithm functions, trig functions and exponential functions. The two major sub divisions of calculus are differential calculus and integral calculus. In this article we shall discuss about Integral calculus calculator.  The following are the examples involved in Integral calculus calculator.

Integral Calculus Calculator Problems:


Example 1:  `int` 71ey dy

Solution: `int` 71ey dy = 71`int` ey dy

Integrating the equation, we get

`int` 71 ey dy  = 71 ey + C

Example 2:  Find `int_(-1)^1` log `((29-x)/(29+x))` dx

Solution: Let f(x) = log `((29-x)/(29+x))`

f(-x) = log`((29+x)/(29-x)) `

= log (29 + x) - log (29 - x)

= - [log(29 + x) - log(29 - x)]

= - [log`((29+x)/(29-x))` ] = -f(x)
Hence  f(-x) = -f(x), we can say that f(x) is an odd function

So, `int_(-1)^1` log`((29-x)/(29+x))`dx = 0


Example 3: Integrate the given equation 3x2+5x + 10 dx

Solution: ∫3x2+5x +10dx = ∫3x2dx +∫5x+∫10 dx

Integrating the above equation,we get 

=> 3x3+  5x2 + 10x + C
3      2     

Example 4:  Evaluate `int_0^1x e^xdx` 

Solution: Using the method of integration by parts,

As we know `int` udv = uv - `int` v du

Here u=x, du=dx, dv= exdx, v=ex

`int_0^1` x ex  dx = (x ex)10 - `int_0^1` ex  dx

= e - (ex)10

= e - (e- 1)

= 1

Example 5:  Find the integral of the given equation  23x3+4x2+2x dx

Solution: ∫3x3+4x2+2x dx 

=>∫ 3x3dx + ∫4x2 dx +∫2x dx 

Integrating the above equation

=> 3x4 +4x3 + 2x2  + c
4     3       2

Integral Calculus Calculator Practice Problems:
Problem 1: Find the integral of the given equation 4x2+2x+20 dx

Solution: 4x3 + 2x2 + 20x + c
3       2

Problem 2: Find the integral of the given equation x3+4x+50 dx

Solution: x4 + 4x2 + 50x + c
4       2

Problem 3: Find the integral of the given equation x2+2x+8 dx

Solution: x3 + 2x2 + 8x + c
3      2

Solving Online Absolute Minimum

Learn solving absolute minimum online:

The function f given by f(x) = x, x (0,1) has neither a maximum nor  a minimum value. If we replace (0,1) by the closed interval [0,1], then the function has a maximum value 1=f(1) and the minimum value 0 = f(0). This reveals the emphasis to be put on the interval on which the given function is defined. We could also note that, in the interval [0,1], f has neither a point of local maxima nor a point of local minima and so f has neither a local maximum value nor a local minimum value even though maximum and minimum values of f exists.

The maximum value 1 of f at x = 1 is called the absolute maximum value (greatest value ) of f on [0,1] and the minimum value 0 of f at x = 0 is called the absolute minimum value (least value) of f on [0,1].

Learn Theorems Used in Solving Absolute Minimum Online:

Let f be a continuous function on an interval I = [a,b]. Then, f has the absolute maximum value and f attains it at least once in I. Also, f has the absolute value and attains it at least once in I.

Let f be a differentiable function on I and let x0 be any interior point of I. then

1. If f attains its absolute maximum value at x0, then f ‘ (x0) = 0

2. If f attains its, then  absolute minimum value at x0, then f ‘ (x0) = 0

Having problem with how to add fractions with different denominators keep reading my upcoming posts, i will try to help you.

Learn Rules in Solving Absolute Minimum Online:

We use the following rules to find the absolute maximum and minimum values of a function in a given interval.

Find all the points where f ’ takes the value zeros.
Take the endpoints of the interval
At all these points calculate the values of f.
Take the maximum and minimum values of f out of the values calculated in step 3. These will be the absolute maximum or minimum values.

Triangle with 3 Acute Angles

Introduction to triangle with 3 acute angles:

The triangle has three sides where the angles may differ and the name has been given to the triangle. The acute angle is nothing but the angle will be less than 90 degrees in its measure. The acute angle triangle contains all the three angles will be less than 90 degrees. Now we see about the triangle with 3 acute angles.

Triangle with 3 Acute Angles:

The triangle with three acute angles can be seen in an equilateral triangle. The triangles with three acute angles will be less than 90 degrees. The sum of these three acute angles in the triangle will be 180 degrees. In acute angled triangle, the angles will be less than 90 degree only. An example for acute angle will be shown below.

Problems on Triangle with 3 Acute Angles:

Example 1:

Find the third acute angle of the triangle where the other two acute angle measures are 57 and 77?

Solution:

The third acute angle of a triangle can be determined as follows,

The sum of the three acute angles will be 180 degrees in a triangle.

Let us assume the unknown triangle will be x

57 + 77 + x = 180

134 + x = 180

x = 180 - 134

x = 46.

Thus, the third acute angle measures about 46 degrees.

Is this topic 6th grade practice math test hard for you? Watch out for my coming posts.

Example 2:

Find the acute angle of a triangle where two angles of it are 55 degrees. Determine the third angle of an acute angled triangle?

Solution:

The two angles in the triangle are equal where its measurement is about 55 degrees.

Let us take the triangle measure and it can be calculated as explained below.

The angle which should be determine can be assumed as x or y.

55 + 55 + x = 180

110 + x = 180

x = 180 - 110

x = 70.

Since, this is an acute angle where its measure is less than 90 degrees.

Placing Fractions on a Number Line

Introduction of placing fraction on Number Line:

The number placed in the correct positions that have both positive and negative in a line is called as number line, where zero is placed at the center of the number line. Fractions can also be placed on the number line in correct place. Let see below the number line has been drawn show the number in correct place. Here we see how can point the correct placing of fractions on the number line.



Placing Fractions on the Number Line:

The fractions on the number line can mark the point in correct place easily.

Positive Fractions on the Number line:

Let take one positive fraction as `(13)/(4)` and see how to point that number in correct place on the number line.



Procedure for placing the fractions on the number line:

Step 1: Convert the fractions as mixed number.

The fraction  `(13)/(4)` = 3 ¼

Step 2: Keep the whole number 3 as it is.

Step 3: Dive the part in between 3 to 4 as 4 and mark the point as per the numerator of the fraction

Another Method:

Step 1: Keep the whole number as it

Step 2: Convert the fraction  `(1)/(4)` to decimal form.

The fraction of ¼ is equal to 0.25 decimal forms.

Step 3: Mark the point 0.2 after the whole number 3 as 3.25    

Placing Fractions on the Number Line:

Negative Fractions on the Number Line

Let take another example of negative fractions `(-7)/(2)`

Now we are not bothering about the negative sign, just placing the fraction in correct point on the number line of negative integer side. By placing fractions on the number line, as like as follow positive fraction on the number line as mentioned above.



Procedure to point the mixed numbers on the number line:

Step 1: Convert the fraction to mixed number.

`(-7)/(2)` = -3 ½

Step 2: Keep the whole number 3 as it is.

Step 3: Dive the part in between 3 to 4 as 2 and mark the point as per the number consists in numerator of the fraction

Another Method:

Step 1: Keep the whole number as it

Step 2: Convert the fraction `(1)/(2)` to decimal form.

The fraction of ½ is equal to 0.5 decimal forms.

Step 3: Placing the point 0.5 after the whole number 3 as 3.5           

Adding Fractions with Negative Numbers

Introduction for Adding Fractions with Negative numbers:

A fraction (from the Latin fractus, broken) is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development were the common or "vulgar" fractions which are still used today (½, ?, ¾, etc.) and which consist of a numerator and a denominator.

Source – Wikipedia.

Adding Fractions with Negative Numbers – Method 1:
The examples show the adding the fractions with negative numbers.

This is the formula to calculate the algebraic fractions for addition is ` a/b + c/d` = `(ad + bc) / (bd)` .

Adding the fractions: `-3/5 + 2/8 `

Solution:

Multiply the numerator and denominator

= `(-3xx8 + 2xx5) / (5xx8)`

Then adding the numerator

= `(-24 + 10) / 40`

= `-14/40` .
By reducing this fraction we get `-7/20` .

Adding the fractions: `4/6 + (-3)/7`

Solution:

Multiply the numerator and denominator

= `(4xx7 + 6xx-3)/(6xx7)`

Then adding the numerator

= `(28 - 18)/42`

= `10/42`

By reducing this fraction we get `5/21` .

Adding the fractions: `15/-6 + 10/-2`

Solution:

Multiply the numerator and denominator

= `(15xx-2 + (-6)xx10)/(-6xx-2)`

Then adding the numerator

= ` (-30 - 60)/-12`

= `-90/-12`

By reducing this fraction we get `15/2` .

I am planning to write more post on rational numbers definition, rounding to the nearest tenth. Keep checking my blog.

Adding Fractions with Negative Numbers – Method 2:
The examples show the adding the fractions with negative numbers.

Adding the algebra fractions: `-3/4 + 9/-7` .

Solution:

LCM for the denominators in the fractions are

The LCM of 4, -7 = -28

So `(-3xx-7)/(4xx-7)` = `-21/-28`

`(9xx4)/(-7xx4)` = `36/-28`

Denominators are equal and adding the numerator in the fractions.

= `-21/-28+36/-28`

= `(-21+36)/-28`

= `15/-28` is the solution.

Adding the algebra fractions: `(-8x)/11 + (-10x)/4` .

Solution:

The given fractions are

= `(-8x)/11 + (-10x)/4`

LCD for the denominators 11 and 4 is 44.

= `(-8xxx4)/44+((-10x)xx11)/44`

To express the denominators in the LCD form as

= `(-32x)/44+(-110x)/44`

We can simplify the numerators.

=` (-32x-110x)/44`

We can add the numerators.

= `(-142x)/44`

Summary of Complement of Set

Introduction to summary of complement set

A universal set is the set of all elements, denoted by the alphabet capital U or occasionally capital E. Set is generally represented by a rectangle and labeled U in Venn diagrams.

Let A be a given set and U be the universal set. The set of all elements of U which are not present in A is called the complement of the set A and is denoted by A' or Ac or A. In this article for Summary of complement set we shall discuss about the  examples for summary of complement set.

Note: Ac = U - A.

Examples on Summary of Complement of Set:

Examples on summary of complement of set are:

The complement of set is one of the operations of the set. Let Set A, the complement of set A represented by A’, is the set of all elements in the set that are not in A.

The shaded area outside A represent complement of A.

The number of elements in the A and the number of elements in the A’ construct the total number of elements in U .

n(A) + n(A’ ) = n( U ).

Example:

Let U = {x : x is an integer, –4 = x = 6}, Q = {–4, –2, 0, 2, 4, 5} and

R ’ = {–3, –2, –1, 2, 3}.

a) List the elements of set Q’ or complement of  Q

b) Find n(R’)

c) Draw a Venn diagram to the sets U, Q and complement of Q or (Q ’)

Solution:

a) First, list out the members of U.

U = {–4, –3, –2, –1, 0, 1, 2, 3, 4, 5, 6}

Q ’ = {–3, –1, 1, 3, 6} ? in U but not in Q

b) Find n(R)

n( U ) = 11, n(R ’ ) = 5

Use the formula:

n(R) + n(R ’ ) = n( U )

n(R) = n( U ) – n(R ’ ) = 11 – 5 = 6

c) Draw a Venn diagram to display the sets U , Q and Q ’

Practice Problems on Summary of Complement of Set:

Practice problem on summary of complement of set are:

1. Let U = {x : x is an integer, –4 = x = 7}, R = {–4, –2, 0, 2, 4, 5, 6} and S’ = {–3, –2, –1, 2, 3}.

a) List the elements of set R’ or complement of  R.

Answer: R’ = {–3, –1, 1, 3, 7}

b) Find n(S)

Answer: 7

Inverse Chain Rule

Introduction to inverse chain rule:

The chain rule is one of the most common rules used in the calculus for the calculation of the derivatives of the complex functions and the inverse chain rule is different that it is used for the calculation of the antiderivatives and the integrals of the functions. In the following article we will see in detail about the topic inverse chain rule.I like to share this Chain Rule Derivative with you all through my article.

More about the Topic Inverse Chain Rule:

As described before the chain rule is one of the most common rules for the calculation of the derivatives of the complex functions and the inverse chain rule is the inverse of the chain rule and is used for the calculation of the antiderivatives and the integrals of the functions. Here the Leibniz notation of the substitution is used. For the function described below,

`int_a^b f(g(t))g'(t) dt = int_{g(a)}^{g(b)} f(x)dx`

We substitute g(t) as x. And the differentiation of x with respect to t yields,

`x = g(t)`

`dx/dt = g(t)`

The above relation can be written as,

`dx = g'(t)dt`

This way of integration by the substitution is called as the inverse chain rule.

I am planning to write more post on Acitivity Application Of Derivatives, how to solve a math word problem. Keep checking my blog.

Example Problems on Inverse Chain Rule:

1. Find the integrals of the function `int_0^2 x*cos(x^2+1) dx` using the inverse chain rule method.

Solution:

The given function is `int_0^2 x*cos(x^2+1) dx`

Here we substitute `x^2+1 = v`

And now differentiating v with respect to x we get,

dv/dx = 2x

dv = 2x dx

x dx = dv/2

Also the limits changes to 1 to 5.

So substituting the values in the function,

`int_0^2 x*cos(x^2+1) dx = int_0^2 cos(x^2+1) (x dx)`

= `int_0^2 x*cos(v) dv/2`

= `1/2 int_1^5 cos(v) dv`

= `1/2(sin(5)-sin(1))`

2. Find the integrals of the function `int_0^1 sqrt{1-x^2} dx` using the inverse chain rule method.

Solution:

The given function is `int_0^1 sqrt{1-x^2} dx`

Here we substitute `x = sin (v)`

And now differentiating v with respect to x we get,

dx/dv = cos (v)

dx = cos(v) dv

And also the limits changes to 0 to pi/2.

So substituting the values in the function,

`int_0^(pi/2) (sqrt{1-sin^2 v}) dx` = `int_0^(pi/2) sqrt{1-sin^2(v)} cos(v) dv`

= `int_0^(pi/2) cos(v) cos(v) dv`

= `int_0^(pi/2) cos^2(v) dv`

= `pi/4`

Binary Number System

Introduction to binary number system:

The binary number system, or base-2 number system, signifies numeric values using two numbers, 0 and 1.The usual binary number system or base-2 system is a positional symbol with a radix of 2.

For example 102 is a binary number.

This article is entirely about binary number system.

Binary Number System:

Binary number system is based on powers of 2, comparing to the decimal number system, which is based on powers of 102.

In the binary number system, only the digits 0 and 1 are used.

Therefore, the first 10 numbers in binary notation 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 in decimal notation, are 02, 12, 102, 112, 1002, 1012, 1102, 1112, 10002, and 10012.

Since every position indicates a specific power of 2, the number 123 means (1 × 102) + (2 × 101) + (3 × 100), the decimal equivalent of a binary number can be calculated by adding together each digit multiplied by its power of 2.

For example, the binary number 10111102 can be written in decimal number system as follows,

(1 × 26) + (0 × 25) + (1 × 24) + (1 × 23) + (1 × 22) + (1 × 21) + (0 × 20) = 64 + 0 + 16 + 8 + 4 + 2 + 0 = 94.

Between, if you have problem on these topics is pie a rational number, please browse expert math related websites for more help on why are unequal class intervals sometimes used in a frequency distribution.

Example Problems- Binary Number System:

Example 1:

Convert the number 5 to the binary number system.

Solution:

The given decimal number is 5.
We have to convert 5 into binary number system by dividing the number 2.
Divide 5 by 2 that is 5 ÷2 =1(remainder) and the quotient is 2.
Divide 2 by 2 that is 2 ÷2 =0(remainder) and the quotient is 1.

We have to start from the remainder that is 5 = 1012.

Check:

1012 = 1 x 22+0 x 21+1 x 20 =4+0+1 =5

Example 2:

Convert the number 5 to the binary number system.

Solution:

The given decimal number is 9.
We have to convert 9 into binary number system by dividing the number 2.
Divide 9 by 2 that is 9 ÷2 =1(remainder) and 4 (quotient).
Then divide the quotient 4 by 2 that is 4÷2=0(remainder) and 2(quotient).
Then divide the quotient 2 by 2 that is 2÷2=0(remainder) and 1(quotient)
We have to start from the remainder that is 9 = 10012.  

Check:

10012 = 1 x 23+0 x 22+0 x 22+1 x 20 = 8+0+0+1 = 9       

Related Rates Calculus Problems

Introduction to related rates calculus problems:
The study of rates of change is called as calculus. Calculus related rates is dealing with finding the rates of the changes of the two variables, the two variables are related to each other in a certain way. Using the rate of change of one variable we can find the rate of change of another variable usually with respect to time.

Examples for Related Rates Calculus Problems

Q :     The given problem is to solve the calculus related rates. A man and a woman start from the same point. The man walks S60o E at 1.3 m/sec. The women walks south at 1.6 m/sec. At what rate is the distance between the man and women changing after 35 minutes?

Sol :          Let x be the distance the man has traveled, y the distance the women has traveled and z the distance between the man and the women.

Speed of the man, (dx/dt)=1.3 m/sec and the speed of the women, (dy/dt)=1.6 m/sec. The rate of change of the distance between them, (dz/dt) after 35 minutes have passed. After t seconds, the distance the man has traveled is 1.3t while the women’s distance is 1.6t. So after 35 minutes, or 2100 seconds, the man’s distance is x=1.3(2100)=2730m and the women’s distance y=1.6(2100)=3360m.

Use the cosine law to equate the variables.

z^2=x^2+y^2-2xycos?

The angle between the paths of the man and the women will always be 60o. Since it is a constant, substitute it into the equation before differentiating.

z^2=x^2+y^2- 2xycos(60o)

z^2=x^2+y^2 -xy

Using the chain rule, differentiate the equation with respect to time.

2z(`dz/dt` )=2x(`dx/dt` )+2y(`dy/dt` )-(y(`dx/dt` )+x(`dy/dt` ))

2z(`dz/dt` )=2x(`dx/dt` )+2y(`dy/dt` )-y(`dy/dt` )-x(`dy/dt` )

2z(`dz/dt` )=(2x-y)(`dx/dt` )+(2y-x)(`dy/dt` ) 

` dz/dt` = ((2x-y)(`dx/dt` )+(2y-x)(`dy/dt` ))/2z

Before substitute the information into the equation, must find the value of z after 35 minutes. x=2730 and y=3360 after 35 minutes, substitutes these values into the Cosine Law formula to find the value of z at t=35 minutes.

z^2=x^2+y^2-xy

z=sqrt(x^2+y^2-xy)

=sqrt((2730)2+(3360)2-(2730)(3360))

˜ 3093.5 m

Substitute the information given in the equation into the equation for dz/dt.

`dz/dt` = (2(2730)-(3360)(1.3))+(2(2360)-(2730)(1.6))/2(3093.5)

= ((5460-3360)(1.3)+(6720-2730)(1.6))/6187

= ((2100)(1.3)+(3990)(1.6))/6187

= (2730+6384)/6187

= `9114/6187`

˜1.47 m/sec

The calculus related rates problem are therefore 35 minutes, the distance between the man and the women is increasing at 1.47 m/sec.

Practice Problem for Related Rates Calculus Problems:

Q :        The given problem is to find the calculus related rates a car leaves a gas station at an intersection at 4.30 p.m and travels east at 80 km/h. A truck traveling north at 100 km/h arrives at the same gas station at6.30 p.m. At what time are the two vehicles closest together?

Answer: Approximately 5.43.p.m

Point Slope Formula

Introduction

Point slope form of a line is mainly used to graph the linear equation. The point slope form mainly needs slope m and a point P ( X1, Y1) Any straight line can be obtained if we know the slope and the any point which lies on it. Generally point slope form is used when we are aware of only single point and a slope m. Any straight line can be graphed on the X and Y axis using the Point slope form.  Many equations can be graphed with the help of this form of line.  And several other points may also be traced which lie on the same line.

Students can learn about it from the tutors online. Students can get also help with their homework problems from the geometry homework help page.Students can get online math homework help for free.

Examples Based on Point Slope Formula

Standard  point slope equation

The standard point slope of equation is given by :

Y  - Y 1    =    m (  X -X1 )

(x, y) = coordinate of any point  and  m is the slope,  (X1, Y1)  = x and y coordinates of the given point P (X1, Y1)

When we graph it we get a straight line


Example : You are given the point (4,3) and a slope of 2. Find the equation for this line in point slope form.

solution :   Just plug the given values into your point-slope formula above. The point (4,3) is in the form of (x1,y1). That means in pleace of  y1, use 3. In place of x1, use 4. Slope is given as  2.So m = 2. So


point slope formula :  Y - Y1  =  m (X-X1)


Y  - 3  = 2 (X-4)


More Practice:  Your point is (-1,5). The slope is 1/2. write the equation that describes this line in point-slope form


point slope formula : Y - Y1 = m (X  - X1)


y - 5  =  1/2 (X-(-1))


y - 5   = 1/2 (X + 1)

Students can also learn about x and y intercepts also.

Stuck on any of these topics Slope Intercept Form, geometry problems try out some best online tutoring math website.

Point Slope Form of Line - Summary

Point slope form is all about a given point and slope and framing out an equation that describes the straight line which is to be graphed,  its not too complicated to learn. Just the one line formula is required.

This equation is often also written as :   y- y1 = m(X - X1)

The above equation is called the point slope form of line.
The resulted graph is called the straight line..
A given point P (x1, y1)
The slope of line = m

Students can get more help on the topic from Geometry help available online.