Finding the Areas With Integrals

We use inegral method to evolve the area finding problems for the continous curves which bound on some points normally we said as the area. Sonetimes the integral functions are given on the which satisfying some lines the graph. These are all considered here. Now we are going to deal with finding the area on the continous curves and lines.

Explanation "finding areas with integrals"

On the first step here we going to separate the different areas according to the axis and the graphs present in the bounds which is drawn according to our problem.

By the way, we have to separate the integral limits.

And then we prosecute the execution.


Examples:

Find the area of the region bounded y = x2 − 5x + 4, x = 2, x = 3 and the x-axis.

solution:

For all x, 2 ≤ x ≤ 3 the curve lies below the x-axis.Integral area

Required area = int_2^3(-y)dx

=   int_2^3 -(x^2 - 5x + 4)dx

= - [x^3/3 - 5x^2/2 + 4x]_2^3

= - [(9 - 45/2 + 12) - (8/3 - 20/2 + 8)]

= - [-13/6 ] = 13/6   sq. units

This is the required area represented in the square units.

Find the regional area for the region for x = 1, x = 4 bounded by the 3x − 15y − 15 = 0 line in  x-axis.

Solution:

The line 3x − 15y − 15 = 0 appears in  the x-axis (below)  in the interval x = 1 and x = 4Integral area

∴Required area  = int_1^4(-y)dx

=   int_2^3 -1/15(3x - 15)dx

= 3/15int_2^3 (5 - x )dx

=  3/15[5x - x^2/2 ]_1^4

= 3/15 [5(4 - 1) -1/2 (16 - 1)]

= 3/15 [15 - 15/2]

= 3/2   sq. units.

This is the required area represented in the square units.

problems to explain "finding areas with integrals"

Find the area between the curves y = x2 − x − 2, x-axis and the lines x = − 2 and x = 4

Solution:

y = x2 − x − 2Integral area

= (x + 1) (x − 2)

This curve intersects x-axis at x = − 1 and x = 2

Required area = A + B + C

The part B lies below x-axis.

∴ B = = -int_-1^2(y)dx

Hence required area = int_-2^-1(y)dx  +  int_-1^2(-y)dx  + int_2^4(y)dx

= int_-2^-1(x^2 - x - 2)dx  +   int_-1^2(-(x^2 - x - 2))dx   +  int_2^4(x^2 - x - 2)dx

= [x^3/3 - x^2/2 - 2]_-2^-1 + [-[x^3/3 - x^2/2 - 2]]_-1^2 + [x^3/3 - x^2/2 - 2]_2^4

= 11/6 + 9/2 + 26/3  = 15 square units

This is the required area represented in the square units.

Mixed or Whole Numbers

A fraction is defined as the relation of a whole digit. For case 7/4 is a fraction. What does “4” position for? It is the integer part into which the total number division. What does “7” position for? It is the number of equivalent parts which contain taken out. A mixed fraction is said to be the whole integer and a fraction number joint into one number known as “mixed number”.

Example for mixed or whole numbers:

Change Mixed numbers to improper numbers.

To modify mixed numbers into an improper numbers.

Steps to be followed:

Multiply the complete number part by fraction’s denominator.
Calculation up to the numerator.
Then reminder to the on top of the denominator.

Example: Convert 4 2/6 to an improper fraction.

Multiply the complete number by the denominator: 4 × 6 = 24
Calculation up the numerator to that: 24 + 2 = 26.
Then note down that on top of the denominator, like this: 26/6 or 13/3.


Rounding Whole Numbers

Rounding whole number means to the closest to the means to evaluate the nearest integers ,having the zeros to the right position of the tens position. Note: when the number 5, 6, 7, 8, or 9 shows in the ones position, round up; when the numbers 0, 1, 2, 3, or 4 shows in the ones position, round down.

Examples:

Rounding 219 to the closest ten gives 220.
Rounding 355 to the closest ten gives 360.
Rounding 105 to the closest ten gives 100.

Similarly, to round a numeral to any position value, to find the digit with zeros in all of the positions to the right of the position value being rounded to is nearest in value to the new numbers.

Whole Number Portion:

The decimal number to the whole number portion is which of the numbers to the remaining of the decimal position.

Example:

In the numeral 48.65, the whole number portion is 48.

In the numeral 0.004, the whole number portion is 0.

Common Monomial Factor

Monomial means a single term of a polynomial. Ex : 2xy or 8mn or 3x2. Monomial factor is a factor which has a variable and a term with some exponents. Common monomial factor means finding the common factor from the given set of monomials.  The common monomial factor can be a combination of number and variable, a variable, a number, two or more numbers and two or more variables.

Steps to find the common monomial factor:

Step 1: write down all the given monomials.

Step 2: write the factors for each given monomial separately.

Step 3: Highlight the common factor from all the factors.

Step 4: The common factors are the common monomial factor.

Example based on Common monomial factor:

Ex 1: Find the common monomial factor for the monomials 9pq and 12pq.

Sol:

The given monomials are 9pq and 12pq

The factors of 9pq are 3, 3, p and q

The factors of 12pq are 2, 2, 3, p and q.

The common factors are 3, p and q.

The common monomial factor is 3pq.

Ex 2: Find the common monomial factor for the monomials 9x3y2 and 15xyz.

Sol:

The given monomials are 9x3y2 and 15xyz.

The factors of the monomial 9x3y2 is 3, 3, x, x, x, y and y and the factors 15xyz is 3, 5, x, y and z

The common factors are 3, x and y.

The common monomial factor is 3, x and y. It can be written as 3*x*y=> 3xy.

Ex 3:Find the common monomial factor for the monomials 25a2bc, 252a2b and 20 a2b2.

Sol:

The given monomials are 25a2bc, 252a2b and 20 ab2.

The factors of 25a2bc is 5, 5, a, a, b and c.

The factors of 252a2b is 5, 5, 5, 5, a, a and b

The factors of 20 a2b2  is 2, 2, 5, a, a, b and b.

The common factors are 5, a, a and b.

The common monomial factor is 5a^2b.



Ex 4:Find the common monomial factor for x, yz and xyz.

Sol:

The given monomials are x, yz and xyz.

The factors of x is x.

The factors of yz are y and z.

The factors of xyz are x, y and z.

The common monomial factor is 0 because there is no common factor from the given monomials.

Multiple Value Function

During mathematics, a multivalued function is an overall relation; specifically all input is related with one or else further outputs. Exactingly words, a well-defined function connections one, also only one, output toward at all particular input. The word "multiple value function" is, consequently, a misnomer while functions be single-valued.Multiple value functions frequently occur as of functions are not injective. Such functions perform not contain an inverse function, except they perform include an inverse relation. It is also called the set-valued function.

Examples for multiple value function

 

This figure do not symbolize a right function since the element 3 within x be related through two elements b with c within y.

Each real number larger than zero or else each complex numbers except for 0 have two square roots. The square roots of 9 be in the set {+3,−3}.

The square roots of 0 be describe through the multiset {0,0}, since 0 be a root of multiplicity 2 of the polynomial x².

Every complex number include three cube roots

Complex logarithm functions are multiple-valued. The values implicit with log (1) be 2πni for every integers n.

Inverse trigonometric function is multiple-valued since trigonometric function is periodic.

tan(π/4)=tan((5π/4)=tan((-3π/4)=tan(2(n+1)π/4)=...1

Therefore arctan(1) might be consideration of since contain multiple values such as π/4, 5π/4, −3π/4, and rapidly. We know how to treat arctan since a single-valued function through restrict the domain of to -π/2 < x < π/2. Therefore, the range of arctan(x) becomes -π/2 < x < π/2. These values as of a limited field are call principal values.

Example problems

Problem 1
Given function f(x)=7x-5, what is the value of f(1) and(f(2)?
solution:

Given function f(x)=7x-5)

We can  find the value of f(1), to substitute 1 for the given function,

f(1)=7x1-5

      =7-5

f(1)=2

We can  find the value of f(2), to substitute 2 for the given function,

f(2)=7x2-5

      =14-5

f(2)=9

Problem 2
Given function f(x)=5x²+6x-5, what is the value of f(5)?
solution:

Given function f(x)=5x²+6x-5

We can  find the value of f(5), to substitute 5 for the given function,

f(5)=5x5²+6x5-5

      =5x25+6x5-5

     =125+30-5

  f(5)=150

Rectangular Objects

Rectangular objects are one of the basis of mathematics. Rectangular objects are having six sides. One of the rectangular objects is cuboid. Cuboid are also having six sides. These rectangular cuboid are having same faces like the rectangular objects. Rectangular prism is also included in the rectangular objects. Length of the rectangular prism is equal to the length of the rectangular.

Explanation for rectangular objects

There are many rectangular objects are present. They are defined as,

1. Rectangular cuboid

2.Rectangular prism

Rectangular cuboid:

The volume of the rectangular cuboid are find using the formula that are shown below,

Volume = Height `xx` Width `xx` Length

This can also be written as,

V = h `xx` w `xx` l.

Surface area of the rectangular cuboid are find using the formula,

Surface area = 2wl `xx` 2lh `xx` 2hw

The diagrammatic representation of the cuboid are shown below,



Example problem for rectangular objects

Example problem 1: Find the surface area and volume of the cuboid by using the formula, where the height = 4, length = 3, width = 2.

Solution:

Step 1: The volume of the rectangular cuboid are,

Volume = Height `xx` Width `xx` Length

from given h= 4, l =3, w =2

Step 2: Therefore by substituting the given information, we get,

V = 4 `xx` 3 `xx` 2

= 24.

Therefore, the volume of the cuboid is 24units.

Step 3: The surface area of the cuboid is given by,

Surface area = 2wl `xx` 2lh `xx` 2hw

Step 4: by substituting the above information given, we get,

Surface area = 2 ( 3 `xx` 2 ) `xx` 2( 4 `xx` 3 ) `xx` 2 ( 3 `xx` 2 )

= 2(6) `xx` 2(12) `xx` 2(6)

= 12 `xx` 24 `xx` 12

= 3456

This is the required area for cuboid.

Example problem 2: Find the surface area and volume of the cuboid by using the formula, where the height = 3, length = 4, width = 5.

Solution:

Step 1: The volume of the rectangular cuboid are,

Volume = Height `xx` Width `xx` Length

from given h= 3, l =4, w =5

Step 2: Therefore by substituting the given information, we get,

V = 3 `xx` 4 `xx` 5

= 60

Therefore, the volume of the cuboid is 60 units.

Step 3: The surface area of the cuboid is given by,

Surface area = 2wl `xx` 2lh `xx` 2hw

Step 4: by substituting the above information given, we get,

Surface area = 2 ( 6 `xx` 4 ) `xx` 2( 4 `xx` 3 ) `xx` 2 ( 3 `xx` 6 )

= 2( 24 ) `xx` 2( 12 ) `xx` 2( 18 )

= 288 `xx` 24 `xx` 36

= 248832

This is the required area for cuboid.

Adding Fractions Grade Six

Fractions:

A fraction 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 was the common or "vulgar" fractions which are still used today (½, ⅝, ¾, etc.) and which consist of a numerator and a denominator.(Source: Wikipedia)

This article will help for grade six students. Through this article grade six students learn about adding fractions.We are going to see some solved problems on adding fractions and some practice problems on adding fractions.

Solved problems on adding fractions grade six:

Problem 1:

Add the fractions 1/8 and 1/8

Solution:

Given, 1/8 + 1/8

Here both fractions have the same common denominator.

So we Can add the numerator normally and keep the denominator as it is.

1/8 + 1/8 = (1+1)/8

= 2/8

We can also simplify it further,

(2÷2)/(8÷2) = 1/4

Answer: 1/8 + 1/8 = 1/4

Problem 2:

Adding the fractions 1/8 + 1/16

Solution:

Given, 1/8  + 1/16

Both fractions have different denominator,

So we need to find least common denominator

Multiple of 8 = 8, 16, 24 ,32...

Multiple of 16 = 16 , 32 ....

Least common multiple = 16.

So to make a common denominator, multiply 1/8 by 2 on both  numerator and denominator,

(1 * 2) / (8 * 2) = 2/16

Now we can subtract,

1/8 + 1/16 = 2/16 + 1/16

= (2 + 1) / 16

= 3/ 16

Answer: 1/8 + 1/16 =3/16

Problem 3:

Adding the fraction 12 / 32 + 6 / 18

Solution:

Given , 12 / 32 + 6 / 18

Both fractions have different denominator,

So we need to find least common denominator

Multiple of  32 = 32, 64, 96, 128, 160, 192, 224 32, 64, 96, 128, 160, 192, 224, 256, 288, 320.........

Multiple of  18 = 18 36 54 72 90 108 126 144 162 180 198 216 234 252 270 288 306 324........

The least common multiple of 32 and 18 is 288.

So we need to make the denominator as 288.

(12 * 9) / ( 32 * 9 ) = 108 / 288

( 6 * 16) / ( 18 * 16) = 96 / 288

12 / 32  + 6 / 18 = 108 / 288 + 96 / 288

= ( 108 + 96) / 288

= 204 / 288

= 51 / 72

Practice problems on adding fractions grade six :

Problems:

1. Adding the fractions  6/7 + 15 /7

2. Adding the fractions 3/4 + 5/6

Answer:

1.3

2.19 / 12

Probability and Chance

Probability is a way of expressing knowledge or belief that an event will occur or has occurred. In mathematics the concept has been given an exact meaning in probability theory, which is used extensively in such areas of study as mathematics, statistics, finance, gambling, science, and philosophy to draw conclusions about the likelihood of potential events and the underlying mechanics of complex systems. In this article we shall discuss about probability and chance problems.(Source: wikipedia)

Probability and chance example problem

Example 1:

If the odds in favor of an event be 3/5 find the chance of the occurrence of the event.

Solution:

Let the given event be E and let P(E)=x then

Odds in favor of E =`(P(E))/(1-P(E))`

= `(P(E))/(1-P(E))` =`3/5`

= `(x)/(1-x)` =`3/5`

5x=3-3x

8x=3

x= `3/8`

Therefore required probability = `3/8`

Example 2:

There dice are thrown together. Find the chance of getting a total of at least 6.

Solution:

In throwing 3 dice together the number of all possible outcomes is (6x6x6)=216

Let E = event of getting a total of at least 6.

Then, E = event of getting a total of less than 6.

= event of getting a total of 3 or or 5.

={(1, 1, 1),(1,1,2),(1,2,1),(2,1,1)(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2)(2,2,1)}

Now , n(`barE` )=10= P(not E)=P(`barE` )=`(n(barE))/(n(S))` =`10/216` =`5/108`

=P(E)=1-P(not E)=1-`5/108` =`103/108`

Hence the required probability is `103/108`

Example 3:

A bag contains 12 red and 15 white balloon. One ball is drawn at random. Find the chance that the ball drawn is red

Solution

Total number of balls=(12+15)=27

Let S be the sample space. Then

n(S)= number of ways of selecting 1 balloon out of 27=27

Let E be the event of drawing a red balloon. Then

n(E)= number of ways of selecting 1 red balloon out of 12=12

Therefore P(getting a /red balloon)=P(E)=`(n(E))/(n(S))` =`12/27` = `4/9`

Probability and chance practice problem

Problem 1:

The odds in favor of the outcomes of an event are 8:13 find the chance that the event will occur

Answer:

`8/21`

Problem 2:

If the odds against the occurrences of an event be 4:7 find the chance of the occurrences of the event?

Answer:

`7/11`

Graph Systems of Inequalities

Algebra is a subdivision in mathematics in which comprises of infinite number of operations on equations, polynomials, inequalities, radicals, rational numbers, logarithms, etc. Graphing system algebra inequalities is also a part of algebra. It is similar to the graphing of ordinary equations, but here we got to graph the inequality given either greater or less than. In graphing system of inequalities , the area overlapping is the solution set.. The following sections helps us to learn much better about graphing system of inequalities.

Steps to solve system of inequalities:

The steps involved in graphing system of inequalities are as follows,

Consider the inequality y > ax+c
Step 1: Convert the given equation as y =ax+c.
Step 2: Since the given inequality is a function of x, let y =f(x).
Step 3: Therefore f(x) = ax+c.
Step 4: Substitute various values for ‘x’ and find corresponding f(x).
Step 5: Table the values as follows x  &  f(x)  the values of x as -3, -2, -1, 0,1,2,3. and for f(x) their corresponding values.
Step 6: The values in the table are the co-ordinates, graph them.
Step 7: Shade the inequality range above the line, since greater than symbol (>) is given.
Step 8: Shade the inequality range below the line, if less than symbol (>) is given.
Step9: Repeat the same steps for next equation also.
Step 10: Shade the region which is overlapped, which is the solution region for the system of inequalities.


Example problem for graphing inequalities:

Problem 1:

Graph the given system of inequalities and find the solution region,

3x - y < 4

2x + y < 3
Convert the given equation as

y >3x+4
Since the given inequality is a function of x,
Let y =f(x).
Therefore
f(x) = 3x+4.

Substitute various values for ‘x’ and find corresponding f(x).
When x= -3
f(-3) = (-3)3 +4,
-9+4,
-5, therefore the co-ordinates are (-3, -5)

When x= -2
f(-2) = (-2)3 +4,
-6+4,
-2, therefore the co-ordinates are (-2, -2)

When x= -1,
f(-1) = (-1)3 +4,
-3+4,
1, therefore the co-ordinates are (-1, 1)

When x= 0
f(0) = (0)3 +4,
0+4,
4, therefore the co-ordinates are (0, 4)

When x= 1
f(1) = (1)3 +4,
3+4,
7, therefore the co-ordinates are (1, 7)

When x= 2
f(2) = (2)3 +4,
6+4,
10, therefore the co-ordinates are (2, 10)
Tabulate and graph them, and mark the inequality
Following the same procedure for the second equation and graph them. the graph looks like,

On Graphing them combined, we get the solution region as ,

Problem 2:

Graph the system of inequalities

4x +3y <2

2x+y >8

Solution:

Following the above procedure and graphing them, the solution range is as follows,

Radical Rules Math

Radical symbol used to indicate the square root or nth root. Radical of an algebraic group, a concept in algebraic group theory. Radical of a ring, in ring theory, a branch of mathematics, a radical of a ring is an ideal of "bad" elements of the ring. Radical of a module, in the theory of modules, the radical of a module is a component in the theory of structure and classification. Radical of an ideal, an important concept in abstract algebra. The radical symbol is ' √ ' . The cubic root of a can be expressed as `root(3)(a)`   and nth root of x can be expressed as  ` rootn x `                                                                                                                                                                                                   

                                                                                                                                                                                                                   Source Wikipedia.

I like to share this sine rules with you all through my article. 

Radical rules in math:

                    Math radical has product rule, Division rule and exponential rule.
Math Product rules for radical:
              Square root product rule:         `sqrt(ab)`   =  ` sqrta * sqrt b`
              Cube root product rule:          `root3 (ab) `   = ` root3 (a) * root3 (b)`
              n^th root product rule                ` rootn (ab) = rootn (a) * rootn (b)`
Examples for radical product rule math problem 1:
              Multiply the two math radical `sqrt5 * sqrt14`
        Solution:
                            Given radicals`sqrt5 * sqrt14`        
               We know the math radical product rule  `sqrt(ab)`   =  ` sqrta * sqrt b` 
                                     So, `sqrt5 * sqrt14`  =`sqrt(5 * 14)`
                                                                = `sqrt70`
        Answer: `sqrt70` 
Examples for radical product rule math problem 2:
              Multiply the two math radical `sqrt 18 * sqrt 3`
        Solution:
                            Given radicals `sqrt18 * sqrt3`           
               We know the math radical product rule  `sqrt(ab)`   =  ` sqrta * sqrt b` 
                                     So,  `sqrt18 * sqrt3` =`sqrt(18 * 3)`
                                                                = `sqrt54`
        Answer:  `sqrt54`
Math Division rules for radical:
                Square root division rule:        ` sqrt(a/b)` = `sqrta / sqrt b`
               Cube root division rule:          ` root3 (a/b)` =  `root3 (a) / root3 (b)`
               n^th root division rule                 `rootn (a/b)` = `rootn (a) / rootn (b)`
Examples for radical division rule math problem 1:
              Simplify  the math radical ` root3 (66) / root3 (11)`
       Solution:
                            Given math radicals  ` root3 (66) / root3 (11)`      
               We know the math radical division rule    ` root3 (a/b)` =  `root3 (a) / root3 (b)`
                                                So, ` root3 (66) / root3 (11)`   = `root3 (66/11)`
                                                                      = `root3 6`
        Answer: `root3 6`
Examples for radical division rule math problem 2:
              Simplify  the math radical ` root4 (16) / root4 (24)`
       Solution:
                            Given radicals ` root4 (16) / root4 (24)`      
               We know the math radical division rule    `rootn (a/b)` = `rootn (a) / rootn (b)`
                                              So, ` root4 (16) / root4 (24)`   = `root4 (16/24)`
                                                                      = `root4 (2/3)`
        Answer:  `root4 (2/3)`

Other Radical rules in math:

Relation rules with exponential term:
                      `root(n)(x)`^m =( `root(n)(x)` )^m = (x^1/n )^m = x^m/n
Examples for radical with exponent math problem 1:
              Simplify the math radical `"(root5 4)^5 * `
        Solution:
                     Given math radicals `(root5 4)^5 * (sqrt27)^2`
                                                = `(root5 4)^5 * (sqrt27)^2`
              we know the math radical rule with an exponents  ( `root(n)(x)` )^m = x^m/n
                          So,     = 4^5/5 * 27^2/2
                                  = 4¹ * 27¹
                                              = 4 * 27
                                              = 108
        Answer:   108
Examples for radical with exponent math problem 2:
              Simplify the math radical `(root4 5)^3 * (sqrt5)^3`
        Solution:
                     Given radicals `(root4 5)^3 * (sqrt5)^3`
                                              = `(root4 5)^3 * (sqrt5)^3`
              we know the math radical rule with an exponents  ( `root(n)(x)` )^m = x^m/n
                          So,     = 5^3/4 * 5^3/2
                                  = `5` ^{(`3/4` + `3/2` )}
                                              = 5^9/4
             Answer:  5^9/4

Calculus Help with Assignment

Calculus (Latin, calculus, a small stone used for counting) is a branch in mathematics focused on limits, functions, derivatives, integrals, and infinite series. This subject constitutes a major part of modern mathematics education. It has two major branches, differential calculus and integral calculus, which are related by the fundamental theorem of calculus. Calculus is the study of change, in the same way that geometry is the study of shape and algebra is the study of operations and their application to solving equations.

(Source: Wikipedia)

Example problems for calculus help with assignment

Calculus help with assignment problem 1:

Differentiate the given function with respect y and find the value of g'(y). The given function is g(y) = 0.45y3 + 3.3t2 - 4y4 + y5

Solution:

The given function is g(y) = 0.45y3 + 3.3t2 - 4y4 + y5

Differentiate the given function with respect to x, we get

g'(y) = (0.45 * 3)y2 + (3.3 * 2)y - (4 * 4)y3 + 4y4

= 4y4 - 16y3 + 1.35y2 - 6.6y

Answer:

The final answer is 4y4 - 16y3 + 1.35y2 - 6.6y

Calculus help with assignment problem 2:

Differentiate the given equation with respect to t, y = 23t5 + 42t3 - 9t2 - 16

Sol:

Given y = 23t5 + 42t3 - 9t2 - 16

Differentiating both sides, we have

`(dy / dt)` = (23 * 5)t4 + (42 * 3)t2 - (9 * 2)t - 0

= 115t4 + 126t2 - 18t

`(dy / dt)` = 115t4 + 126t2 - 18t

Answer:

The final answer is 115t4 + 126t2 - 18t

Calculus help with assignment problem 3:

Integrate the given function ∫ (9x2) dx

Solution:

The given function is 9x2 dx

Integrate the given function with respect to x, we get

∫ (9x2) dx = 9 `(x^3 / 3)` + c

= 3x3 + c.

Answer:

The final answer is 3x3 + c

Calculus help with assignment problem 4:

Integrate the given function g(x) and find the value of G(3). The given function is g(x) = 14x6 - 5x4 + 28x + 40

Solution:

The given function is g(x) = 14x6 - 5x4 + 28x + 40

Integrate the given function with respect to x, we get

G(x) = 14 `(x^7 / 7)` - 5 `(x^5 / 5)` + 28 `(x^2 / 2) ` + 40x + c

= 2x7 - x5 + 14x2 + 40x + c

Find the value of G(3):

Substitute x = 3 in the above equation, we get

G(3) = 2 (3)7 - (3)5 + 14 (3)2 + (40 * 3)

= 4374 - 243 + 126 + 120

= 4377

Answer:

The final answer is 4377

Practice problems for calculus help with assignment

Calculus help with assignment problem 1:

Find the value f'(2). Given function f(x) = 12x3 - 4x2 + 9

Answer:

The final answer is 128

Calculus help with assignment problem 2:

Integrate the given function g(x) = 6x2 + 12. Find the value G(4).

Answer:

The final answer is 176

Percent Composition

Per cent composition of the element in the molecule is defined the ratio of the mass of the different elements present in the molecule expressed in term of percentage.    The easiest method for  determining the percentage mass of a particular element in the compound is to find out the molecular mass of the element.  The molecular mass of the element is calculated by adding all the atomic mass of the element in a compound  The find out the mass of the particular element that is represented in the molecule

The next step is to divide the mass of element upon the molar mass of the molecule and converting them in percentage.   For example for determining the % composition of the A in A3B4

% of A       =    (3 x atomic mass of A/(3 x atomic mass of A + 4 x atomic mass of B)) x 100

Example 1 for the determination of percent composition

Q. Calculate the % composition of  each element in NaOH?

Ans:  There are three element in NaOH.  Sodium, oxygen and hydrogen.

The molecular mass of NaOH = 23 + 16 +1 = 40g/mole

% composition of Na = (23/40) x 100 =57.5%

% composition of Oxygen = (16/40) x 100 = 40%

% composition of Hydrogen = (1/40)  x 100 =  2.5%

Example of percentage composition 2

Q.  Calculate the percentage composition of H2SO4?

Answer:  There are three element in H2SO4 they are hydrogen, sulfur and oxygen

The molecular mass of H2SO4 = 2x 1+32 + 4 x16 =98g/mole

% composition of Hydrogen = (2/98) x 100 = 2.04%

% composition of sulfur  = (32/98) x 100 = 32.65%

% composition of oxygen   = (64/98) x 100 = 65.31%

Excercise

Find the % composition of Carbon in carbon dioxide?
Find the % composition of chromium in potassium dichromate
Find the pecentage composition of water in copper(II) sulfate pentahydrate
Find the percentage composition of Mn in KMnO4

Expression Design Tutorial

An expression design tutorial is a finite combination of symbols that are well-formed according to the rules applicable in the context at hand. Symbols can designate values constants, variables, operations, relations, or can constitute punctuation or other syntactic entities. In algebra an expression may be used to designate a value, which value might depend on values assigned to variables occurring in the expression. Mathematical expressions include letters called variables.

(Source -Wikipedia)


Basic concepts of Expression design tutorial:

Basic concepts of Expression design tutorial:

Basic concepts of Expression:

Expression design Tutorial mean nothing but a to make the expression using constant, Operations, Variables, according to our given sign and operation of the expression.

Example:

A simple algebraic expressions like  6x + 8, 7y – 9. A variable can take various values. Its value is not fixed. On the other hand, a constant has a fixed value. Examples of constants are: 4, 100, and 17.

Based on the number of of terms expression should classified four types:

Monomials(having only one term)
Binomials(having two terms)
Trinomials(having three terms)
Polynomials(having many terms)

In word problems data’s are given in directed form when we solve the word problem first design the expression based on our given data and then solve the expression

Example problems in expression design tutorial:

Expression design tutorial:

Example problem 1:

16 years ago, Jenny age was half of the age his brother will be in 25 year.find the current age of jenny?

Here first we have to design the expression for solving jenny’s age

Solution:

Step 1:

Let us consider x is age of Jenny

14 years ago means that x-16

Step 2:

Half of the age of his brother will in 25years means =x+25

Step 3:

Now we have to write the equation or expression for solving age

x-16=1/2(x+25)(Designed expression)

Step 4:

2(x-16) =x+25(solving the expression)

2x-32=x+25

Step 5:

Subtract both side on x

2x-32-x=x-x+25

2x-x-32=25


Step 6:

Add both sides on 32

2x-x-32+32=25+32

Step 7:

X=25+32

X=57 year

Answer :jenny age was 57 year

Tangent Trigonometry

The word trigonometry is an origin of two Greek words “trigonon” meaning a triangle and “metron” meaning measurement. Metron means the science, which deals with the measurement of triangles. The familiar trigonometric terms are sine, cosine, tangent, cotangent, secant, cosecant. The tangent plane to a surface at a given point is the plane that "just touches" the surface at that point. The concept of a tangent is one of the most fundamental notions in differential geometry and has been extensively generalized; see Tangent space
                                                                                                                                                                                    - Source from Wikipedia

Looking out for more help on Trigonometry Formulas Sheet in algebra by visiting listed websites.

Definition of tangent trigonometry:

                A function of an acute angle in a right-angled triangle can be defined as the ratio of the opposite side of length and the adjacent side of the length is called tangent. It is expressed by the gradient of a line. Find either sides or angles in a right-angled triangle by using this function.
      
  In Diagram (A) represents the Thangent to curve
 In Diagram (B) represents the tangent of an angle
 In Diagram (c) represents the tangent to a circle.
                       In Right angle triangle Tangent  can be expressed in mathe matical form. If an angle is` theta`
               ` tan theta ` = `(Opposite side)/(adjacent side)`

Tangent trigonometry problems:

Tangent trigonometry problem 1:
          Find the height of the given right-angle triangle.
           
         Solution:
                               tan 30° = `(opposite)/(adjacent)`
                                             = `h/5`                                                                                 we know, tan 30°= `1/ sqrt3 `

                       height  (h) = 5 × tan 30°
                                           = 5 × `1/ sqrt3 `
                                          =  2.886 (approximately)
      So, The height of the right-angle triangle (h) =2.886 cm


Tangent trigonometry problem 2:
           Find the hypotenuse value of the given right-angle triangle. length and height are in centimeter.

                
      Solution:
                                           Tan` theta`   = `(opposite side)/(adjacent side)`
                                                       = `4/3`
                                   Tan `theta` = 1.333
                                            `theta` = tan^-11.333
                                            theta = 53.12°
                  Now the trigonometry relation,  sin` theta` = `(opposite side) / (hypotenuse)`
                                                               sin 53.12° =` 4 / (hypotenuse)`
                                                     hypotenuse (X) =` 4 / sin 53.12 `                                we know sin 53.12° = 0.7999
                                                     hypotenuse (X) = `4 / 0.7999`
                                                     hypotenuse (X) = 5.000
                  So, the hypotenuse value of right-angle triangle is 5 cm.

Descriptive Geometry

The descriptive geometry is a branch of geometry, which deals three dimensional objects in two dimensions by using the lines, curves, solids, surfaces and points in space. Geo means “earth” and metron  means “measurement”. ”Euclid, is a Greek mathematician, called the father of geometry. A point is used to represent a position in space. A plane to be a surface extending infinitely in every directions such that all points lying on the line joining any two points on the surface. The descriptive geometry example problems and practice problems are given below. 

Example problems for descriptive geometry:

Example problem 1:
          Show that the straight lines 2x + y − 9 = 0 and 2x + y − 10 = 0 are parallel.
Solution:
           Slope of the straight line 2x + y − 9 = 0 is m₁ = − 2
           Slope of the straight line 2x + y − 10 = 0 is m₂ = − 2 ∴ m₁ = m₂
           The given straight lines are parallel.

Example problem 2:
      Find the co-ordinates of orthocentre of the triangle formed by the straight lines x − y − 5 = 0, 2x − y − 8 = 0 and 3x − y − 9 = 0
Solution:
      Let the equations of sides AB, BC and CA of a ΔABC be represented by
                x − y − 5 = 0 … (1)
               2x − y − 8 = 0 … (2)
               3x − y − 9 = 0 … (3)
      Solving (1) and (3), we get A as (2, − 3)
       The equation of the straight line BC is 2x − y − 8 = 0. The straight line perpendicular to it is of the form
             x + 2y + k = 0
       A(2, − 3) satisfies the equation (4) ∴ 2 − 6 + k = 0 ⇒ k = 4
       The equation of AD is x + 2y = − 4 … (5)
       Solving the equations (1) and (2), we get B as (3, − 2)
       The straight line perpendicular to 3x − y − 9 = 0 is of the form x + 3y + k = 0
       But B(3, − 2) lies on this straight line ∴ 3 − 6 + k = 0 ⇒ k = 3
       The equation of BE is x + 3y = − 3 … (6)
       Solving (5) and (6), we get the orthocentre O as (− 6, 1).

Example problem 3:
         Find the values of a and b if the equation (a − 4)x² + by² + (b − 3)xy + 4x + 4y − 1 = 0 represents a circle.
Solution:
   The given equation is (a − 4)x² + by² + (b − 3)xy + 4x + 4y − 1 = 0
            (i) coefficient of xy = 0 ⇒ b − 3 = 0 ∴ b = 3
            (ii) coefficient of x² = co-efficient of y² ⇒ a − 4 = b
                        a = 7
                Thus a = 7, b = 3

Practice problems for descriptive geometry:

Practice problem 1:
          If the equation 12x² − 10xy + 2y² + 14x − 5y + c = 0 represents a pair of straight lines, find the value of c. Find the separate equations of the straight lines and also the angle between them.
     Answer: C = 2 ; 6x − 2y + 1 = 0 and 2x − y + 2 = 0 ; tan⁻¹ (1/7)
Practice problem 2:
          Find the equation of the straight line which passes through the given intersection of the straight lines 2x + y = 8 and 3x − 2y + 7 = 0 and is parallel to the straight line 4x + y − 11 = 0
      Answer: 28x + 7y − 74 = 0

Algebra is a subdivision in math, which comprises of infinite number of operations on equations, polynomials, inequalities, radicals, rational numbers, logarithms, etc. Graphing algebra equations or function is also a part of algebra. Graphing is nothing but the pictorial view of the given function or equation, it may be a line, parabola, hyperbola, curve, circle, etc.The most commonly used graphing functions with their graph is given in the following sections.

Common graph functions:

The most commonly used graph functions are,
Linear function which gives a straight line
                f(x) = ax +b
Functions with degree 2
               f(x) = ax² +b
Functions with degree 3,
               f(x) = ax³ + bx
Functions of square root,
              f(x) =` sqrt(x)`

Graph for the common functions:

The graphical view of common functions is shown below,
1. Linear function.

Linear functions are in  the format where the function has degree 1. So that the function varies linearly. The general format for linear functions is shown below,

 f(x) = ax +b

where, x power is 1. b is a constant.

The graph for the linear function is a line and the graph is shown below,

2. Function with degree 2,

The function for degree 2 is in the format,

f(x) = ax² . where, a is a constant

The graph for this function looks as shown below,

3. Function with degree 3 is as follows,

f(x) = ax³

where, a is a constant.

The graph looks as shown below,

4. Function with degree 4 is in th format,

f(x) = ax⁴

where, a is a constant.
The graph is as follows,

5. The function with square root is as follows,

f(x) = sqrt (x)

The graph is as follows,

The above shown functions are functions of x, if it is changed to function of y, then the graph changes to the y axis.

Non Linear Data

Non linear equation solver involves solving non linear equation with the help of formulas and it is mainly used to find the unknown variable value. The quadratic equation is one of the non linear equation in which it transforms the polynomial function into normal linear function. Non linear equation is also related with the families of vectors called vector spaces or non linear spaces. The following are the example problem deals with non linear equation solved with detailed solution.

Non linear equation solver example problems:

Example 1:

Solve the non linear quadratic equation.

√ (x ² – 11x+28) = 2

Solution:

Given function is

√ (x ² – 11x+28) = 2

Squaring on both sides, then the above equation becomes

[√ (x ² – 11x+28)] ² = (2) ²

And simplify.

x ² – 11x+28= 4

Make the above equation in factor form.

x ² – 11x + 24 =0

The above equation is in quadratic form with two solutions.

x = 3 and x = 8

Conclusion:

The given equation has two real values as solution x = 3 and x = 8.

Example 2:

Solve the non linear quadratic equation.

         y ² – 4y + 13 = 0
Solution:

Given equation is

y ² – 4y + 13 = 0

The discriminant is given as

D = b² - 4ac

= (-4)2 - 4(1)(13) = -36

Since the discriminant results negative, the square root of the discriminant value is a pure imaginary number.

√(D) = √(-36) = √(-1)√(36) = 6i

where i is the imaginary part defined as i = √(-1).

To find the solution we use b / 2a formula.

y₁ = (4 + 6i)) / (2*1) = 2 + 3i

y₂ = (4 - 6i) / (2*1) = 2 - 3i

Conclusion:

So the given equation has two imaginary solutions 2 + 3i and 2 – 3i.

Non linear equation solver practice problems:

1) Solve the non linear quadratic equation.
                      y ² + 9 = 0
Answer: (y – 3i) (y + 3i) is the solution.
2) Solve the non linear quadratic equation.
                         -y ² + 2y = -3
Answer: (y + 1) (y – 3) is the solution.

Bisector in Geometry

In geometry, bisector is a line dividing something into two equal or congruent parts. The most often considered types of bisectors are the segment bisector (a line that passes through the midpoint of a given segment) and the angle bisector (a line that passes through the apex of an angle, that divides it into two equal angles). (Source: From Wikipedia).

A plane is used as a bisector in bisecting three dimensional shapes. It is called as bisector plane. Bisectors are used to divide line segments, shapes, and solids in to two equal, similar and congruent parts.

In this article we are going to learn, drawing bisector to a line segment.

Drawing bisector to a line segment

Here we are going to learn how to draw bisector of a line segment step by step.

Step 1

Draw a line segment PQ

Step 2

Fix a compass in point P, and expand it over half the length of the line segment.

Step 3

Now, draw two arcs, on each side of the line segment.

Step 4

Without changing or adjusting the measure of the compass, draw two another arcs on each side of the line segment from the point Q.

Step 5

Now we have drawn two set of arcs on each side of the line segment PQ, from points P and Q. And we have two points on either side of the equations formed by the arcs drawn from the points P and Q.

Now join those two points of intersection between the arcs either sides of the line segment by a ruler.


Step 6

The line joined by the points of intersection between arcs is called as the bisector of the line segment PQ. The bisector is perpendicular to the line segment PQ. The point, where the bisector cuts the line segment PQ, is the mid point of the line segment PQ. So, the line segments PJ and JQ are equal in length. PJ = JQ.

Trigonometric Functions of Sum

trigonometric functions of sum and difference of two angles:

The trigonometric functions of sum and difference of two angles A are B are listed below:
Sin (A+B) = Sin (A) Cos (B) + Cos (A) Sin (B) .................. (1)
Sin (A-B) = Sin (A) Cos (B) - Cos (A) Sin (B) .................... (2)
Cos (A+B) = Cos (A) Cos (B) - Sin (A) Sin (B) ................... (3)
Cos (A-B) = Cos (A) Cos (B) + Sin (A) Sin (B) ................... (4)
Adding (1) and (2) we get, 
Sin (A+B) + Sin (A-B) = 2 Sin (A) Cos (B)
Subtracting (1) and (2), we get,
Sin (A+B) - Sin (A-B) = 2 Cos (A) Sin (B)
Similarly, adding (3) and (4), we get,
Cos (A+B) + Cos (A-B) = 2 Cos (A) Cos (B)
Subtracting (3) and (4), we get,
Cos (A+B) - Cos (A-B) = - 2 sin (A) sin (B)
These can be written in the forms
Sin (A+B) + Sin (A-B) = 2 Sin (A) Cos (B) ............................ (5)
Sin (A+B) - Sin (A-B) = 2 Cos (A) Sin (B) ............................. (6)
Cos (A+B) + Cos (A-B) = 2 Cos (A) Cos (B) ........................ (7)
Cos (A - B) - Cos (A+B) = 2 sin (A) sin (B) ........................... (8)
Note: The order on the right-hand side of (8) must be carefully noted
The sum and difference of two angles A and B of a Tangent function are given below:
Tan (A+B) = [Tan (A) + Tan (B)] / [1 - Tan (A) Tan (B)] .................. (9)
Tan (A-B) = [Tan (A) - Tan (B)] / [1 + Tan (A) Tan (B)] ................... (10)

Example for trigonometric functions of sum and difference of two angles

 Express as the sum of two trigonometrical ratios 
sin (5θ) cos (3θ)
Sol : Using the Identity:  2 Sin (A) Cos (B) = Sin (A+B) + Sin (A-B)  
On substitution, we get,
sin (5θ) cos (3θ) = (1/2) {Sin (5θ+3θ) + Sin (5θ-3θ)  
= (1/2) {Sin (8θ) + Sin (2θ) 

More examples for trigonometric functions of sum and difference of two angles

Change in to sum of trigonometric functions: Sin (70^o) Sin (20^o)
Sol: Using,   2 sin (A) sin (B) = Cos (A - B) - Cos (A+B)
Sin (70^o) Sin (20^o) = (1/2) { Cos (70^o - 20^o) - Cos (70^o+20^o) }
                            = (1/2) { Cos (50^o) - Cos (90^o) }
= (1/2) {Cos (50^o) }           Since Cos(90^o) = 0

Square Pyramid

Pyramid is one where the surface of outer area looks triangular and it converges at that point. They are congruent .Square pyramid is a pyramid in which the base of pyramid looks to be square. The most popular form of square pyramid is Johnson square pyramid. If the apex is perpendicularly on top of the center of the square, it will have C4v symmetry.

what is square pyramid

Formulae for square pyramid:

Volume of a square pyramid V = `1/3` (Ah)

= `1/3` a2 h     (Area of square = a2)

Height of square pyramid h = (3 * V) / a2

Where, A -----> area of base of a pyramid

V -----> Volume of a pyramid

a ----->  Side length

what is square pyramid

Surface area of a square pyramid = side2 + 2 x side x length

what is square pyramid

Example problems for square pyramid:

ExampleExample 1:

Find the surface area of a square based pyramid with the given sides 6 cm, height 7 cm, and the slant height 8 cm.

Solution:

Formula:

Surface area of a square pyramid = side2 + 2 x side x length

Given:

Side = 6 cm, length = slant height = 8cm

Surface area             = (6)2 + 2 x 6 x 8

= 36 + 96

= 132

The Surface area of square pyramid is 132 square cm.

ExampleExample 2:

Find the volume of square pyramid for a given dimensions base = 5cm, height = 6cm.

Solution:

Formula:

Volume of the pyramid = `1/3` * Area of base * height.

Given:

Base = 5cm, height = 6cm.

Area of base = b2

Volume = `1/3` * 52 * 6 cm3

= `1/3` * (150) cm3

= 50 cm3

The Final answer is 50 cm3.

ExampleExample 3:

Find the volume of the square pyramid whose base area is 16 cm and height is 12 cm.

Solution:

Formula:

Volume of the pyramid = `1/3` * B*h

Given:

Base area is 16 cm and height is 12 cm.

Volume of the pyramid = `1/3` * 16 * 12

= `1/3` * 192

= `192/3`

= 64 cm3

Practice problems:

Problem 1:

Find the volume of square pyramid for a given dimensions base = 2cm, height =3cm.

Solution:

Volume of a square pyramid is 4cm3.

Problem 2:

Find the volume of the square pyramid whose base area is 15 cm and height is 13 cm.

Solution:

Volume of a square pyramid is 65cm3.

Decimal in Words

In mathematics, a decimal in words is nothing but, it is only represent the decimal numbers. The heart of our number system is considering by place value. Many ways can be represent the decimal numbers. But the important one is place value. The numbers in a base-10 numeral system is specified as decimal notation. A dot with a decimal number, like to present in 41.602. Decimal powers in decimals, (1, 10, 100, and 1000) and secondary symbols for half these values (5, 50, and 500) are contained as roman numerals.

How to write Decimal System in words

Below are examples which will show you how to write decimals in words:

1. Write the decimals in words for given decimal number, 5.32

Solution:

Given the decimal number is 5.32,

The (1.32) decimals in words is write,

Five and thirty two hundredths.

2. Write the decimals in words for given decimal number, 65.44

Solution:

Given the decimal number is 65.44,

The (65.44) decimals in words is write,

Sixty five and forty four hundredths.

3. Write the decimals in words for given decimal number, 24.65

Solution:

Given the decimal number is 24.65,

The (24.65) decimals in words is write,

Twenty four and sixty five hundredths.

4. Write the decimals in words for given decimal number, 36.875

Solution:

Given the decimal number is 36.875,

The (36.875) decimals in words is write,

Thirty six and eight hundred seventy five thousandths.

5. Write the decimals in words for given decimal number, 0.02

Solution:

Given the decimal number is 0.02,

The (0.02) decimals in words is write,

Two hundredths.

Solved Examples

1. Write the decimals in words for given decimal number, 6000.32

Solution:

Given the decimal number is 6000.32,

The (6000.32) decimals in words is write,

Six thousand and thirty two hundredths.

2. Write the decimals in words for given decimal number, 475.00055

Solution:

Given the decimal number is 475.00055,

The (475.00055) decimals in words is write,

Four hundred seventy five and fifty five hundred thousandths.

3. Write the decimals in words for given decimal number, 98.84

Solution:

Given the decimal number is 98.84,

The (98.84) decimals in words is write,

Ninety eight and eighty four hundredths.

4. Write the decimals in words for given decimal number, 637.674.

Solution:

Given the decimal number is 637.674,

The (637.674) decimals in words is write,

Six hundred thirty seven and six hundred seventy four thousandths.

5. Write the decimals in words for given decimal number, 9.69

Solution:

Given the decimal number is 9.69,

The (9.69) decimals in words is write,

Nine and sixty nine hundredths.