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.

Algebra Properties of Real Numbers

Introduction to algebra properties of real numbers:

In algebra, numbers that are used to measure the real-world quantities like area, speed, length, probability, temperature, volume, rate and so on. Those numbers should be called as real numbers. Few examples for that number should be 9, -12, 0, 2.523, `sqrt(3)` , `(sqrt(5))/(3)`, 3×106, `pi` so on. Algebra properties of real numbers should express in terms of addition as well as multiplication. Here we will see about the algebra properties of real numbers.

List of Properties:

The lists of properties in algebra for the real numbers are shown below:

Commutative properties
Associative properties
Distributive properties
Identity properties
Inverse properties

Explanation of Properties:

Let us consider the numbers a, b and c are the real numbers.

Commutative properties:

i) a + b = b + a

ii) a × b = b × a

This property says that the order in which we add or multiplication with real number that should not affect the solution.

Example:

2 + 5 = 7 = 5 + 2
3 × 4 = 12 = 4 × 3


Associative properties:

i) (a+b)+c = a+(b+c)

ii) (a×b)×c = a×(b×c)

This property tells that when the real numbers are grouping in terms of adding or multiplied could not matter. Here, the parentheses should not make a sense.

Example:

(5+6)+2 = 13 = 5+(6+2)
(2×3)×4 = 24 = 2×(3×4)


Distributive properties:

i) a×(b+c) = (a×b) + (a×c)

ii) (b+c)×a = (b×a) + (c×a)

Example:

3×(2+5) = (3×2) + (3×5) = 6 + 15 = 21
(2+5)×3 = (2×3) + (5×3) = 6 + 15 = 21


Identity properties:

i) a+0 = 0+a = a

ii) a×1 = 1×a = a

Here, 0 -> Additive identity and 1 -> Multiplicative identity.

Example:

3+0 = 0+3 = 3
4×1 = 1×4 = 4

Stuck on any of these topics prime number 1-100, and  solve my math problem for me try out some best online tutoring math website.

Inverse properties:

i) For every real number a. -a is called additive inverse of a, when, a+(-a) = (-a)+a = 0.

ii) For every real number a`!=` 0. `(1)/(a)` is called multiplicative inverse of a, when a × `(1)/(a)` = `(1)/(a)` × a.

Example:

5+(-5) = (-5)+5 = 0
2 × `(1)/(2)` = `(1)/(2)` × 2 = 1
These are the properties of real numbers in algebra.

Sat Math Practice Questions

Introduction to sat math practice questions:

Scholastic Assessment Test or Sat reasoning test is a test which is used to get admissions in US universities and colleges. Sat test deals with quantitative questions and English skills. Sat test is conducted for 3 hrs and 45 minutes. It consists of three parts; they are i) Critical reasoning ii) Math aptitude and iii) Writing.

Math section of Sat questions deal with quantitative questions and logical reasoning questions. Math questions are given with multiple choices. Students should practice sat questions to get more marks in tests.

Sat Math Practice Questions:examples

Example 1:

What is the remainder when 16^3 + 17^3 + 18^3 + 19^3 is divided by 70?

Solution:

The given expression can be written as,

= 16^3 + 19^3 + 17^3 + 18^3

The given numbers is written in the form (a^3 + b^3) form,

= (16 + 19) (16^2 - 16 * 19 + 19^2) + (17 + 18) (17^2 - 17*18 + 18^2)

On solving this, we get,

= 35 (16^2 - 16 * 19 + 19^2 + 17^2 - 17*18 + 18^2)

= 35 (even number).

When 35 is multiplied by any even number, the product will be divisible of 70.

Therefore the remainder becomes 0.

Answer is 0.



Example 2:

Find the smallest positive integer that is divided by 4, 6 and 8 leaves a remainder of '2' in each case, but the number is perfectly divisible by 37?

Solution:

The least positive integer = x.

The given condition is, the smallest positive integer "x" leaves a remainder of 2 when divided by 4, 6 and 8.
Hence, x - 2 will be divisible by all the three numbers 4, 6 and 8.

The LCM for 4, 6 and 8 is 24.
The other values that x - 2, will be multiples of 24 since all the multiples of 24 will be multiples common to 4, 6 and 8.

Now, if x - 2 = 24, then x = 26. It is not divisible by 37.

If x - 2 = 48, then x = 50. It is also not divisible by 37.
If x - 2 = 72, then x = 74. It is divisible of 37.

The answer is 74.

Sat Math Practice Questions:

Practice question 1:

Of the 80 students in class, 25 are studying English, 15 Tamil and 13 Hindi. 3 are studying English and Tami;; 4 are studying Tamil and Hindi; 2 are studying English and Hindi; and none is studying all 3 languages at the same time. Find the number of students who are not studying any of the three languages?

A) 27
B) 18
C) 53
D) 62
E) 36

Answer is option E.



Practice questions 2:

A car travels at fifteen feet per minute. How many feet does it travel in 24 min and 40 seconds?


A) 360
B) 370
C) 367.5
D) 600
E) 960

Answer is option B.



Practice question 3:

Rounding (101)2 to the nearest hundred.

A) 10,000
B) 18,000
C) 10,200
D) 17,000
E) 19,000

Answer is option C.