Number Cube Probability

Number Cube Probability

Probability is the chance of the outcome of an event of a particular experiment. Probabilities are occurs always numbers between 0(impossible) and 1(possible). The set of all possible outcomes of a particular experiment is called as sample space. For example probability of getting a 3 when rolling a dice is 1/6. In this article we will discuss about probability problems using number cube.

Number Cube Probability - Example Problems

Example 1: If rolling a number cube, what is the probability of getting prime number?

Solution:

Let S be the sample space, n(S) = 6.

A be the event of getting prime number.

A = {2,3, 5}, n(A) = 3

P(A) = `(n(A))/(n(S))` = `3/6` = `1/2`

Example 2: If rolling two number cubes, what is the probability of getting a sum of odd?

Solution:

Let S be the sample space, n(S) = 6 * 6 = 36.

A be the event of getting a sum of odd.

n(A) = {{1, 2), (1, 4), (1, 6)....(6, 1), (6, 3), (6, 5)} = 18

P(A) = `(n(A))/(n(S))` = `18/36` = `1/2`

Therefore probability of getting a sum of odd is `1/2`

Example 3: If rolling two number cubes, what is the probability of getting a sum of 10 or 7?
Solution:

Let S be the sample space, n(S) = 6 * 6 = 36.

A be the event of getting 10.

n(A) = {{4, 6), (5, 5), (6, 4)} = 3

P(A) = `(n(A))/(n(S))` = `3/36` = `1/12`

Let B be the event of getting sum of 7.

n(B) = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} = 6

P(B) = `(n(B))/(n(S))` = `6/36` = `1/6`

P(A or B) = P(A) + P(B) = `1/12` + `1/6` = `1/4`

P(A or B) = `1/4`

Therefore probability of getting a sum of 10 or 7 is `1/4`


Number Cube Probability - Practice Problems

Problem 1: If rolling a number cube, what is the probability of getting composite number?

Problem 2: If rolling two number cubes, what is the probability of getting 9 or 6?

Answer: 1) `2/3` 2) `5/18`

Linear Approximation Table

Introduction to linear approximation table:

In mathematics, the procedure for determining the straight line is that closely fit curvature at various sites. Linear approximation equation is given as y = ax + b, the values of a and b are chosen that the line meet the curve at the preferred site, or value of x, and the evaluation of the line equals the rate of change of the curve at that site.

The majority curvature, linear approximation are first-rate only particularly close to the preferred x. The table consist both rows and columns. We can put the values in the table. It should be very easy for the student to study.

Formula-linear Approximation Table:

Linear approximation of the domain function represented as  f(x). Digression line to the table of f(x) at the point (`x_0, y_0` ) where `y_0` is given as `y_0` = f(`x_0` ). It is represented as

` y-y_0= f(x_0)(x-x_0)`

If x is closed the `x_0` , the formula can be written as `x_1 = x_0Deltax `

In mathematics, a linear approximation is an approximation of a universal function. The linear approximation uses a linear function. The linear function is more accurate and affine functions .They are widely used in the method of finite differences to create first order methods for solving or determining solutions to equations. It is given by,

f(x )= f(a) + f'(a)(x-a) + R2
f(x) = f(a) + f'(a)(x-a)

Linear approximation function:

A linear approximation functions to a function at a point can be computed by striking the primary expression in the Taylor series,

`F(x_0+Delta)`  = `f(x_0)+f(x_0)Delta+....`

The Newton’s method linear approximation can be estimated using linear approximation.

Example Problems- Linear Approximation Table:

Example 1- Using linear approximation table

Consider the linear function y = f(x) = `5x^2`

Solution:

Let` Deltax ` is an increment of x.

Therefore, `Deltay ` is also an increment of y.

Hence, we have

= f(x + `Deltax` )-f(x)

= 5(x + `Deltax` )`^2` – `5x^2`

= 25x( `Deltax` ) + 4 `(Deltax)^2`

The differential equation can be determined by using the given linear function.

Therefore `(dy)/(dx)` =25x

Hence, dy = 10x dx

Example 2- Using linear approximation table

Consider the linear function y = f(x) = `3x^2`

Solution:

Let` Deltax ` is an increment of x.

Therefore, `Deltay` is also an increment of y.

Hence, we have

= f(x + `Deltax` )-f(x)

= 3(x + `Deltax` )`^2` – `3x^2`

= 9x( `Deltax` ) + 2` (Deltax)^2`

The differential equation can be determined by using the given linear function.

Therefore `(dy)/(dx)` =9x

Hence, dy = 6x dx.

These all are the above explanation and examples make clear about this linear approximation.

Solve Math Fractions Problems

Solve math fractions problems:

In this article we are discussing  the solve math fractions problems. 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 were the common or "vulgar" fractions which are still used today (`1/2` , `5/8` , `3/4` etc.) and which consist of a numerator and a denominator. (Source – Wikipedia)

Solve math fractions problems is simple addition fraction, multiplication fraction, subtraction fraction and dividing fractions.

Solve Math Fractions Problems-example Problems:

Example 1:

Add the fractions for given two fraction, `3/9` + ` 3/9`

Solution:

The given two fractions are `3/9` + `3/9`

The same denominators of the two fractions, so

= `3/9 ` + `3/9`

Add the numerators the 3 and 3 = 3+3 = 6.

The same denominator is 9.

=` 6/9`

The addition fraction solution is `2/3.`

Example 2:

Subtract the fractions for given two fractions `4/6` - `6/5`

Solution:

The denominator is different so we take a (lcd) least common denominator

LCD = 6 x 5 = 30

So multiply and divide by 5 in first term we get

` (4 xx 5) / (6 xx 5)`

=`20/30`

Multiply and divide by 6 in second terms

= `(6 xx 6) / (5 xx 6)`

= `36/30`

The denominators are equals

So subtracting the numerator directly = `(20-36)/30`

Simplify the above equation we get = `-16/30`

Therefore the final answer is `-8/15`

Example 3:

Multiply the fractions for given two fractions, `4/5` x `5/4`

Solution:

The given two fractions are `4/5` x `5/4`

The multiply numerator and denominators of the two fractions, so

= `4/5` x `5/4`

Multiply the numerators the 4 and 5 = 4 x 5 = 20.

Multiply the denominators the 5 and 4 = 5 x 4= 20

= `20/20`

The multiply fraction solution is 1

Example 4:

Dividing fraction:

`4/3` divides `2/4`

Solution:

First we have to take the reciprocal of the second number

Reciprocal of `2/4 ` = `4/2`

Now we multiply with first term we get

`4/3` x `4/2`

Multiply the numerator and denominator

`(4 xx 4) / (3 xx 2)`

Simplify the above equation we get

= `16/6`

Therefore the final answer is `8/3`


Solve Math Fractions Problems-practice Problems:

Problem 1: Add the two fraction `2/8` +`2/8`

Solution: `1/2`

Problem 2: Subtract two fractions `10/10` – ` 6/10`

Solution: `2/5`

Problem 3: multiply two fractions `3/5` x  `5/5`

Solution: `3/5`

Problem 4: Dividing two fractions `6/3` and `2/4`

Solution: 4

Matrices Algebra 1

Introduction to matrices algebra 1.

An  rectangular arrangement of numbers in rows and columns within the paranthesis is called a matrix. If there are m rows and n columns in a matrix then m `xx` n is called order of the matrix. In this topic let us know three operations on the given matrices.

Two matrices A and B can be added or subtracted if their orders are same. If not matrix addition or subtraction is not defined.

Two matrices can be multiplied if the number of column in the first matrix is equal to the number of rows in the second matrix. Otherwise the matrix multiplication is not defined.

Now let us see few problems in this topic matrices algebra 1

Examples of Matrices Algebra 1

Ex 1: Find the sum of the given matrices A and B.

A = `[[1,2],[3,4]]`    and B = `[[5,7],[6,8]]`

Soln: Since A and B are having same orders 2 `xx` 2,

A + B =  `[[1,2],[3,4]]` + `[[5,7],[6,8]]`

= `[[1+5,2+7],[3+6,4+8]]`   [Add the corresponding values]

Therefore A + B =  `[[6,9],[9,12]]`

Ex 2: Find the value of the given equation 2A – B.

Here A = `[[4,6],[7,5]]` and B = `[[3,2],[1,4]]`

Soln: 2A – B  = 2 `[[4,6],[7,5]]` - `[[3,2],[1,4]]`

= `[[2xx4, 2xx6],[2xx7,2xx5]]` - `[[3,2],[1,4]]`

= `[[8,12],[14,10]]` `[[3,2],[1,4]]`

= `[[8**3, 12**2],[14**1,10**4]]` [Subtract the corresponding values]

= `[[5,10],[13,6]]`

Therefore 2A – B =` [[5,10],[13,6]]`


More Example Problems on Matrices Algebra 1

Ex 3: Multiply the following matrices if possible:

(i)  A = `[[2,3,4],[5,6,7]]` and B  = `[[2,7],[5,6]]`

(ii) A = `[[3,4],[2,8]] `  and  B = `[[7,2],[1,8]]`

Soln: (i) The order of matrix

A = `[[2,3,4],[5,6,7]]` is 2 `xx` 3and B  = `[[2,7],[5,6]]`

For B, it is 2 `xx` 2.

Here the number of column in A is not equal to the number of rows in B (3≠ 2). Therefore AB is not possible.

(ii) Here in A and B, the number columns in A is equal number of rows in B (2=2). Therefore AB is possible

Therefore AB =  `[[3,4],[2,8]] ` `[[7,2],[1,8]]`

= `[[3 xx 7 + 4 xx 1,3 xx 2 + 4 xx 8],[2 xx 7 + 8 xx 1,2 xx 2 + 8 xx 8]]`

= `[[21 + 4,6 + 32],[14 + 8,4 + 64]]`

= `[[25,38],[22,68]]`

All Kinds of Math

Introduction to math:

Math is considered as the study of all quantity and all structure. Maths is considered as the tool in all fields. The math is mostly applicable in for the engineering and also for numerical analysis. The major kinds of the mathematics are algebra, trigonometry, statistics, probability and geometry. Here we are going to discuss about some kinds of math.

Kinds of Math

Kinds of math:

Algebra:

Algebra is considered as a kind of math which defines about the study of  all rules and also the operations. Algebra is also concerned with the equations, polynomials and all structure of algebra. This is considered as the main branch of mathematics. The algebra is classified into the various categories.

Statistics:

Statistics is a kind of math which makes use of a numerical data which is related to the group of individuals and experiments. Statistics deals with the collection, analysis and also the plan for the collection and analysis of data. Statistics includes the basic terms such as mean, median, Range, and mode.

Mean – process of finding the average of the numbers in the list.
Range – difference between the largest value and the smallest value in the list.
Median –finding the middle term from the list.
Mode – process of finding the frequently occurred term.
Trigonometry:

Trigonometry is the study of triangles which in particular defines the right triangles. This trigonometry takes place in both applied maths and also pure math which deals with the relations between the angle of the triangle and sides of the triangle. This includes three main functions such as sin, cos, tan.

Probability:

It is also a kind of math which defines the way for expressing the knowledge of events to occur. This probability is used in the areas such as statistics and finance.

Geometry:

Geometry is also a kind of math which defines the questions about the different kinds of shapes and also the study about the shapes. It is also related with the real life. Normal shapes used in geometry are triangle, circle, polygons, square and quadrilateral.

Example Problems

Example problem in algebra:

Solve: 2 (x+3) – (x+5) = 3(x+2) – (2x +5)

Solution:

2 (x+3) – (x+5) = 3(x+2) – (2x +5)

2x + (2*3) – (x+5) = 3x + (3 * 2) – 4x +7

2x + 6 – x + 5 = 3x + 6 -4x - 7

2x – x + 1 = 3x – 4x - 1

x + 1 = - x - 1

0 = 0

Hence the given equation is solved.

Example problems in statistics:

Find the mean, median, mode and range for the numbers 8, 6, 7, 4, 8, 5, 6, 8, 2.

Solution:

Mean:

Mean = `54/9` = 6.

Median:

Arrange the numbers in numerical order 2, 4, 5, 6, 6, 7, 8, 8, 8.

The term which is in middle is considered as the median. Therefore the answer is 6.

Range:

Range = maximum value – minmum value

Maximum value = 8.

Lowest value = 2.

Range = 8 – 2.

= 6.

Example in probability:

Three coins are tossed randomly. Find the sample space of getting 2 tails at the same time.

Solution:

Sample space = { HHH, HHT, HTT, HTH, THH, TTH, THT, TTT}

Probability of getting 2 tails = {TTH, THT, HTT, TTT}

= (Number of possible outcomes / Total number of outcomes}

= `4/8`

= `1/2`

Partial Products Math

Introduction of partial products math:

The partial products math is nothing but the numerical values and their operations in math. Partial products math topic involves the product of the natural numbers and the integers. Thus the partial products math not only deals with the products but also with the arithmetic operations like addition, subtraction and division. The whole numbers includes the natural numbers. Let capital N denotes the natural numbers.

Partial Products Math:

The partial products math is nothing but the summing of the two terms not only through the addition but also through the product method. First the terms are rounded and made to multiply with the left most term of the number. Then the second term is made to rounded and the values are made to multiplied with the another term of the left. Then the left most term can be made multiplied with left most term of another term. Then the right most term can be made to multiplied with the right most term of the another term.

At last the values are made to summing up and the total of the values gives the product of the both terms. This is how the product partial terms are executed. The partial math includes all the operations like addition and subtraction in the same manner. Then the second term is made to rounded and the values are made to multiplied with the another term of the left. Then the left most term can be made multiplied with left most term of another term.

Examples for the Partial Products Math:

Example1: Find the partial product of the term 83 x 27?



83
27
----
80*20 -> 1600
80* 7 ->  560
3*20 ->   60
3* 7 ->   21
----
2241
Answer: 2241

Example2: Find the partial product of the term 93 x 25?



93

25

--------

90*20 -> 1800

90*  5 ->   450

3*20 ->     60

3*  5 ->     15

---------

2325

Answer: 2325

Different Types of Math

Introduction mathematics:

The mathematic is deals with the logical calculation and quantitative calculation. The shape of the object is the arrangement, order, it has involved from the counting, measuring. The most significant branches of mathematics are algebra and analysis. A theoretical representative method used in the study of numbers, shapes, structure and change and the relationships. The different types of the math are the algebra, geometry, trigonometry, calculus, linear algebra, differential equations.

Different Type of the Math:

Algebra:

An algebra type is the branch of mathematics which treats of the associations and properties of measure by way of letters and other symbols. It is suitable to those associations that are correct of every class of magnitude.

Geometry:

Geometry types are the division of mathematics which investigate the relationships, property, and quantity of solids, surfaces, lines, and angles; the science which treat of the property and associations of magnitudes; the art of the relatives of space.

Trigonometry

A trigonometry types is the subdivision of the mathematics which is identified by the relative of the edges, and angle of the triangles. These kinds methods is give the assured position of the required position and also give the general relationship among the trigonometrically functions of arcs or angles.

Calculus:

The rate of probability is the calculus. The calculus can be divided into two type’s calculus; these are the differential calculus, and integral calculus. Differential calculus determines the rate of modify of a measure. Integral calculus is calculating the measuring rate of the change.

Additional Type of the Math:

Linear algebra:

The system of the solving is called as the linear equation. The linear conversion and vector spaces is the more frequent to the linear algebra. The equations are modifying as the function is the linear conversion. The system of the equation is the system of the transformation.

Differential equations:

A differential equation is a mathematical equation for an unidentified function of single or multiple variables that relate the ideals of the function itself and its derivatives of a variety of commands. The rate of the modification is used for the differential equation. They are most common a type of the differential equation is the ordinary differential equation and the partial differential equation.

Guided Reading Answers

Introduction to guided reading answers

The guided reading answers are nothing but getting help from others to reading the subjects.In online only we can get the help of tutor for reading any subjects.Initially the math problems can be solved by using some arithmetic operations  like addition,subtraction,division and multiplications and these can be denoted by (+ ,`xx` ,`-` ,÷ ).The following aticle shows some guided reading answers.

Solved Math Problems with some Guided Reading Answers

Problem 1:

Solve the 56x + 47y = 2632 given equation on the x and y intercepts.

Given:

56x + 47y = 2632

Solution:

56x + 47y = 2632

To find the x intercept of y = 0 and solve for x.

56x + 47(0) = 2632

Solve the value of x.

x = `2632/56`

x = 47

To find the y intercept of x=0 and solve for y.

56(0) + 47y = 2632

Solve the value of y

47y = 2632

y = `2632/47`

y = 56.

The equations of x intercept on (47,0) and y intercept on (0,56).

Problem 2:

Solve the problem 185 + 35 ( 40 + 27 ) ÷ 67 – 60 in method of order of operation
Solution:

Given:

`=>`  185 + 35 ( 40 + 27 ) ÷ 67 – 60

Step 1: we need to simplify the parentheses

`=>` 185 + 35 `xx` 67 ÷ 67 – 60

Step 2: We need to simplify the multiplication

`=>` 185 + 2345 ÷ 67 – 60

Step 3: We need to simplify the division

`=>` 185 + 35 – 60

Step 4: We need to simplify the addition

`=>`220 –  60

Step 5: We need to simplify the subtraction

`=>` 160

Answer: 185 + 35 ( 40 + 27 ) ÷ 67 – 60 = 160

Solved more Math Problems with some Guided Reading Answers

Problem 3:

Solve the given problem 11(s – 9) – 6s ` - ` 26 = 13(s + 33)
The Solutions follows below:

Step 1: Given expression is,

11(s – 9) – 6s ` - ` 26 = 13(s + 33)

Step 2:Multiplying the integer terms

11s – 99 – 6s – 26 = 13s + 429.

Step 3:Grouping the above terms

5s –125 = 13s + 429

Step 4: Add 125 on both sides

5s –125 + 125 = 13s + 429 + 125

Step 5:Grouping the above terms

5s = 13s + 554

Step 6:Subtract 13s by on both sides

5s `-` 13s = 13s `-` 13s + 554

Step 7:Grouping the above terms

–8s = 554

S = `- 554/8`

The required answers is

S = `- 554/8`

Problem 4:

Solve the given problem (156x2 – 141x – 91) + (213x2 – 181x – 144) `-` (–916x2 +   41x + 20)
Solution:

The problem can be solved in simplifying method .

Step 1:(156x2 – 141x – 91) + (213x2 – 181x – 144) `-` (–916x2 +   41x + 20)

Step 2: 156x2 – 141x – 91 + 213x2 – 181x – 144 + 916x2 `-`    41x `-` 20

Step 3:  1285x2 – 363x – 255

The required answer is

(156x2 – 141x – 91) + (213x2 – 181x – 144) `-` (–916x2 +   41x + 20) = 1285x2 – 363x – 255

Trigonometry Test Answers

Introduction to trigonometry test answers:

Trigonometry is an important branch of Mathematics. The word Trigonometry has been consequent from three Greek words Tri (Three), Goni (Angles), Metron (Measurement). Literally it means “measurement of triangles”. The fundamental trigonometric functions are Sine, Cosine and Tangent functions of a triangle takes an angle and give the sides of the triangle. The ordinary trigonometric terms are Sine, Cosine and Tangent.

Solving Trigonometric Functions on Trigonometry Test Answers:

Trigonometry test answers 1:

Solve the trigonometric equation.

(Find all solutions) 2 Cos x + 2 = 3

Solution:

First we have to solve for Cos x

2Cos x + 2 = 3

2Cos x =3 – 2

2Cos x = 1

Cos x = 1 / 2

X = Cos-1x (1/2)

X = 60

Trigonometry test answers 2:

Find the solutions for the trigonometric function:

-5 Cos 2x + 9 Sin x = -3

Solution:

-5 Cos2x + 9 Sin x = -3

-5(1- Sin2x) + 9Sinx = -3

-5 -5n2x + 9Sinx = -3

-5 -5Sin2x + 9Sinx +3 = 0

-5Sin2x + 9Sinx -2 = 0

Let us take y = Sin x

-5y2+ 9y – 2 = 0

Y = -2                                                     Y =-.2

Now plug y =sin x

Sin x = -2                                              Sin x = .2

X = Sin-1(-2)                                             X = Sin2(- .2)

X = - Sin-1(2)                                            X =-Sin-1(.2)

Trigonometry Test Answers 3 on Trigonometry Test Answers:

Prove (tanx+secx)/(cosx-tanx-secx)=-cscx

(tanx+secx) / (cosx-tanx-secx)
[(sinx+1)/cosx] / [(cos²(x)/cosx - sinx/cosx - 1/cosx]
[(sinx+1)/cosx] / [-(1-cos²x)/cosx - sinx/cosx]
[(sinx+1)/cosx] / [-sin²(x)/cosx - sinx/cosx]
[(sinx+1)/cosx] / [(-sin²(x)-sinx)/cosx]
[(sinx+1)/cosx]*[cosx/(-sin²(x)-sinx)]
(sinx+1)/-sin²(x)-sinx
sinx+1 / -sinx(sinx+1)
1/-sinx = -cscx

Trigonometry test answers 4:

Problem for trigonometric cosine function:

In a triangle adjacent side is 5 and hypotenuse is 25 then finds the angle A of the triangle?

Cos A= (adjacent) / (hypotenuse)

Cos A = 5/25

A = cos-1(5/25)

A = 78.46.

Trigonometry test answers 5:

Problem for trigonometric cosine function:

In a triangle adjacent side is 7 and hypotenuse is 2 then finds the angle A of the triangle?

Cos A= (adjacent) / (hypotenuse)

Cos A = 7/2

A = cos-1(7/2)

A = 0.998

How to Learn Fractions Exercises

In this article we shall discuss about how to learn fractions exercises. Here, fractions are also denoted as division of a whole. A fraction is can be creation over to a decimal all through dividing the upper digit, or numerator, during the lower digit, or denominator. Fractions are as an alternative of as ratios, and significance for fraction which is one of the main math processes. Thus the fractions `5/7` are also used to point out the ratio 5:7 and the fractions 5 ÷ 7 as well.



Example Problems Based on How to Learn Fractions Exercises:

The example problems based on how to learn fractions exercises are given below that,

Example 1:

How to learn fractions exercises of 129 divide by 5?

Solution:

Step 1:

Here, 129 divide by 5 is meant by `129/5` .

Step 2:

Now, 129 divide by 5 is given below that,

Here, using long division method. So, long division method is shown given below that,

25.8

5)129             [129 > 5, now divide `129/5` ]

125             [(hint: 25 × 5 = 125)]

40           [4 < 5, so add one zero]

40           [(hint: 8 × 5 = 40)]

0

Step 3:

The final answer is 25.8

Example 2:

How to learn fractions exercises of 467 divide by 13?

Solution:

Step 1:

Here, 467 divide by 13 is meant by `467/13.`

Step 2:

Now, 467 divide by 13 is given below that,

Here, using long division method. So, long division method is shown given below that,

35.92

13)467            [467 > 13, now divide `467/13`]

455            [(hint: 35 × 13 = 455)]

120          [12 < 13, so add one zero]

117          [(hint: 9 × 13 = 117)]

30        [3 < 13, so add one zero]

24        [(hint: 2 × 13 = 26)]

6

Step 3:

The final answer is 35.92





Practice Problems Based on How to Learn Fractions Exercises:

The practice problems based on how to learn fractions exercises are given below that,

Problem 1:

How to learn fractions exercises of 383 divide by 5?

Answer: The final answer is 76.6

Problem 2:

How to learn fractions exercises of 461 divide by 8?

Answer: The final answer is 57.625

Problem 3:

How to learn fractions exercises of 567 divide by 9?

Answer: The final answer is 63

Math and Multplication

Introduction to math multiplication:

Multiplication (symbol "×") is the mathematical operation of scaling one number by another. It is one of the four basic operations in elementary arithmetic (the others being addition, subtraction and division). (Source from Wikipedia)

For example:

2*4 = 2*2*2*2 = 8

Here we are going to study about different method s of multiplication in math and its example problems.

Example Problems for Math Multiplication:

Example: 1

Multiply the following integer

23*46

Solution:

Here we have to perform simple normal math multiplication

2 3

4 6 *

-----------------

1 3 8          (6 multiply with 23 we get 138)

9 2             (4 multiply with 23 we get 92)

-----------------

1 0 5 8         (Add both the numbers we get final answer)

-----------------

There fore the final answer is 1058.

Example: 2

Multiplication of binomial

(2x+3) (3x+3)

Solution:

Here we have two binomial terms

First we take the first term 2x multiply with other terms

6x2 + 6x

Next term 3

9x+ 9

Combine the like terms we get

= 6x2 + 6x+ 9x+ 9

= 6x2+15x+9

Therefore the final answer is 6x2+15x+9

Example: 3

Multiplication of exponential

Solve 2x3y4.3x4.2y3

Solution:

We know the multiplication property of exponents

am.an = a ( m+n)

Combine the like terms first

2x3 3x4 y4 2y3

2*3 x (3+4) .2y(4+3)

Simplify the above expression we get

6x7.2y7

Therefore the final answer is 6x7.2y7

Example: 4

Fraction multiplication

`(3/4)` * `(2/4)`

First we have to multiply the numerator

3*2 = 6

Next multiply the denominator we get

4*4 = 16

= `6/16`

The simplest fraction is `3/8`

Therefore the final answer is `3/8`

Example for Math Multiplication: 5

Find the area of rectangle with base 8 meter and height 6 meter.

Solution:

We know that area of rectangle is multiplication base and height

Area = base * height

Substitute base and height value in the above formula we get

Area = 8 * 6 =48 meter square

Therefore area of given rectangle is 48 meter square.

Whole Number Word Problems

Introduction to whole number word problems

Whole number word problems are about the word problems that we work out with under many chapters in mathematics. Some of the chapters to solve whole number word problems are arithmetic operations, number system and many more. Whole numbers are the numbers that do not contains any decimal numbers fractional part or rational numbers. Working on whole number word problems will be simple and easier one.

Example of Whole Number Word Problems

A school consists of 4320 boys and 3840 girls. There are 983 teachers working in the school. What is the total strength of the school including the teachers?

Solution

Number of boys in the school        = 4 3 2 0
Number of girls in the school         = 3 8 4 0
Number of teachers in the school =     9 8 3 +
-------------
Total strength of the school    =  9 1 4 3
-------------
2.  The sum of the ages of Daniel and his brother David is 46 years. If the age of the David is 24 years what will be the age of the Daniel?

Solution:

Let the age of Daniel be x years.

The age of his brother David = 24 years.

Sum of both of their ages =(x+ 24) years.

But this sum is given as 46years.

So x + 24 = 46.

x =46 - 24

x =22

Thus the age of the Daniel is 22 years.

3. Jack plays basketball and can sink the ball in the basket 80% of the time. If he takes 120 shots, how many did Jack will sink?

Solution

Jack can sink = 80%                                   

Number of shots attempted = 120

Total sink Jack made = 80% of 120

= `80 / 100` * 120

= 96 sinks

Practice Problems of Whole Number Word Problems

1. For a foot ball tournament 32 teams participated. Each team has 16 players and 2 coaches. Find the total number of players and coaches participated in the tournament.

2. The sum of the ages of Nancy and his sister Mary is 42 years. If the age of Mary is 19 what will be the age of the Nancy?

Answers for practice problem of whole number word problems.

Total members = 576
Nancy age = 23

Degree of Polynomial Calculator

INTRODUCTION ABOUT DEGREE OF POLYNOMIAL CALCULATOR:       

In arithmetic, a polynomial is an expression of finite length constructed from variables and constants, using only the addition, subtraction, multiplication operations, and non-negative, whole-number exponents. The maximum power of the variable in a polynomial is called the degree of the polynomial. A term with no variables is called a constant term and its degree is zero.I like to share this Polynomial Solver with you all through my article.

Find the Degree of Polynomial:

CONDITION FOR POLYNOMIAL:

In polynomial the variable should not involve in division operation.

Ex: `x^2 -4x+7, x^3 +5x^2-x+6, ` it is a polynomial.

Ex: `x^2 -4/x+3/2,` not a polynomial.

General form for degree of Polynomial Calculator:

A polynomial function is a function that can be defined as,

`f(x)=a_nx^n+a_(n-1)x^(n-1)+...+a_2x^2+a_1x+a_0`

Where n is a non-negative integer and `a_0, a_1,a_2,` ...,` a_n ` is constant coefficients.

First Degree of polynomial calculator:  

The first degree polynomial is the polynomial that has only one variable. The variable may be x, y or z and it is easy to find the value for the variable. The combinations of two first degree polynomials are called as second degree polynomial.

Second degree polynomial calculator:

In this equation 6x2+5x+2  the power of x is 2. so it is a second-degree polynomial or a polynomial of degree 2.

POLYNOMIAL CALCULATOR:

To solve the polynomial in calculator, first we have to enter the two polynomials then click anyone of the button like add, subtract, multiply, division and factor and the equivalent operation will be performed and answer will be displayed. Thus the polynomial calculator is shown below.



Example and Practice Problem for first and Second Degree Polynomial Calculator:

First degree polynomial calculator:

Example 1:

Determine the value for the first degree polynomial 2x+10 = 12.

Solution:

The given first degree polynomial is 2x+10 = 12.

For the above polynomial, subtract 10 on both sides of the equation.

2x +10-10 =12-10

2x = 2

Divide by 2 on both sides of the above equation.

`(2x)/2 = 2/2`

x = 1

The value for the first degree polynomial 2x+10 = 12 is 1.

Second degree polynomial calculator:

Example 1

Find whether the given expression is a second degree polynomial or not.

(x - 1) (x - 2) = 0

Solution:

Multiply two expressions in the left hand side,

(x - 1) (x - 2) = 0

x(x-2) -1(x-2) =0

x2 -2x -x + 1(2) = 0

x2 -3x +2 = 0

So, the highest exponent of variable x is 2.

So, the given polynomial equation is second degree polynomial equation.

Practice Problem for first Degree Polynomial:

1.Find the value for the first degree polynomial of 2x+ 5 = 6.

Answer: x= 1/2.

2.Compute the value for the first degree polynomial 4x+12 = 24.

Answer: x=3.


Practice Problem for Second Degree Polynomial:

1.    Find whether the given expression is a second degree polynomial or not.

(x - 10) (x - 5) = 0

Answer: The polynomial equation is second degree polynomial equation.

Lower Quartile

Introduction to lower quartile:

In descriptive statistics, a quartile is any of the three values which divide the sorted data set into four equal parts, so that each part represents one fourth of the sampled population. Lower Quartile is one type of quantile.In that quartile, the 25 % of data which lie in the lower half of the data set is called as lower quartile. We can also define it as the middle value of the lower half.

For example,

21, 5, 22, 33, 17, 36, 14, 42, 15, 35, 27


Arrange the data in ascending order,

5, 14, 15, 17, 21, 22, 27, 33, 35, 36, 42

In the above Data set,

15 is lower quartile.

Calculation of Lower Quartile:

Lower Quartile for Odd data set:

Let us see how to calculate the lower quartile of the data set that contains odd numbers.

Step1:  First find the median of the data set. The median divide the data set into upper and lower part.

Step2:  Find the median for lower part of the data set .That median is called as lower quartile.                    

Ex: Data Set: 8, 49, 53, 19, 45, 44, 11, 41, 47, 51, 39

Ordered Data set: 8, 11, 19, 39, 41, 44, 45, 47,49,51,53

Median (Mid value) =44

Lower part = 8, 11, 19, 39, 41

Upper part = 45, 47,49,51,53

Lower quartile (Median of lower part) = 19

Lower Quartile for Even Data set:

Let us see how to calculate the lower quartile of the data set that contains even numbers.

Step 1: Find the Median of the data set by calculating the average of the data set.

Step 2: Next in the lower part, find the median by calculating the average. That value is known as lower quartile.

Ex: Data set: 13, 6, 24, 33, 9, 38, 16, 46, 19, 53, 26, 65

Ordered data set: 6,9,13,16,19,24,26,33,38,46,53,65

Median: (24+26)/2= 25

Lower quartile:  (24+33)/2= 28.5



Practice Problems on Lower Quartile

Find the lower Quartile for the following;

1.   17,11,47,27,69,20,6,34,12,93,33

2.   7,4,18,23,30,20,14,36,41,11,48,64

Answer Key:

1.12

2.12.5

Compound Dependent Events

This page is based on compound dependent events which is a study under dependent events in statistics. In probability, an event is a one or more possible outcomes from an experiment. An event consisting of one or more simple events is called compound event. An event is called dependent event, if an event does affect the other event.

Compound Events Dependent - Examples

Ex 1: A bag contains 20 bulbs, in which 5 bulbs are defective bulbs. If 2 bulbs are taken at one by one without replacement, what is the probability of getting good and defective bulbs?

Sol:

Let S be the sample space, n(S) = 20

Number of good bulbs = 20 - 5 = 15

A be the event of getting a good bulb, n(A) = 15

B be the event of getting a defective bulb, n(B) = 5

P(A) = `(n(A))/(n(S))` = `15/20`= `3/4`

Now 19 balls will be remaining in the box, so

P(B) = `(n(B))/(n(S))` = `5/19`

P(A and B) = `3/4` · `5/19` = `15/76`.

P(Good and Defective bulb) = `15/76` .

Ex 2: A jar contains 7 yellow candies, 5 green candies, and 8 blue candies. If 2 candies are taken one by one without replacement, what is the probability of getting yellow and green candy?

Sol:

Let S be the sample space, n(S) = 7 + 5 + 8 = 20

A be the event of getting a yellow candy, n(A) = 7

B be the event of getting a green candy, n(B) = 5

P(A) = `(n(A))/(n(S))` = `7/20`

Now 19 candies will be remaining in a jar.

P(B) = `(n(B))/(n(S))` = `5/19`

P(A and B) =P(A) · P(B) = `7/20` · `5/19` = `7/76`

P(Yellow and Green) = `7/76`

Practice Problems

Pr 1: A bag contains 15 bulbs, in which 7 bulbs are defective bulbs. If 2 bulbs are taken at one by one without replacement, what is the probability of getting good and defective bulbs?

Pr 2: A jar contains 10 yellow candies, 6 green candies, and 4 blue candies. If 2 candies are taken one by one without replacement, what is the probability of getting yellow and green candy?

Ans: 1) ` 4/15` 2) `3/19`

Poisson Normal Distribution

Poisson normal distribution an Introduction:

Normal distributon:

In probability theory and statistics, the normal distribution or Gaussian distribution is an absolutely continuous probability distribution with zero cumulants of all orders higher than two.

Poisson distributon:

In probability theory and statistics, the Poisson distribution  (or Poisson law of large numbers) is a discrete probability distribution that expresses the probability of a number of events occurring in a fixed period of time if these events occur with a known average rate and independently of the time since the last event.


Poisson Distributon in Poisson and Normal Distribution:

The three basic assumptions on poisson distribution are as follows

The probability of one photon arriving in ?t is proportional to ?t when ?t is very small.

P(1;?t) = a?t for small ?t

where a is a constant whose value is not yet determined.

The probability that more than one photon arrives in ?t is negligible when ?t is very small.

P(0;?t)+P(1;?t) = 1 for small ?t

The number of photons that arrive in one interval is independent of the number of photons that arrive in any other non-overlapping interval.

Variance of the Poisson Distribution:

var(k) = E[k2] - E2[k] = at

Note that for a poisson distribution mean and the variance are equal .

Algorithm for poisson distribution:

Algorithm :Poisson random number:

init:

Let L ? e-?, k ? 0 and p ? 1.

do:

Generate uniform random number u in [0,1] and let p ? p × u.

while p > L.

return k - 1.



Normal Distribution in Poisson and Normal Distribution:

Description of normal distribution:

For any of situation in which the absolute value of a continuous variable is changed randomly from trial to trial.

The random improbability or random error.

Some properties of normal distribution Bell curve:

The curve will be Symmetric,

It will be unimodal,

The value will be extends to +/- infinity,

And the area under the curve is always one (it will be the fixed value).

The normal distribution bell curve is fully based on the mean and standard deviation functions.

These are some of the important points of  poisson and normal distribution.

Numerator of Fractions

Introduction to fractions

A fraction is a number that can represent part of a whole. A fraction consists two parts, a numerator and a denominator, the numerator represents the number of equal parts and the denominator represents number of equal parts make up a whole.

For example, `2/3` is a fraction. Here, the number above the fraction bar is called as the numerator (2) and number below the fraction bar is called as the denominator (3) of the fraction.

Here we are going to study about fractions

Sample Problems on Fractions

Here we are going to study about adding and subtracting fractions with same denominators and different numerators.

Example 1

`2/11` + `5/11`

Solution

Here the given problem is to add the fractions `2/11` and `5/11`

Notice that the denominators of both fractions are same, so we can just add the numerators.

So,

`2/11 + 5/11` = `(2 + 5)/11`

= `7/11`

So, the sum of `2/11` and `5/11` is `7/11`

Example 2

`6/14 - 3/14`

Solution

Here the given problem is to subtract `3/14` from `6/14`

Here the denominators are same, so we can just subtract the numerators.

`6/14 - 3/14` = `(6 - 3)/14`

= `3/14`

Few more Sample Problems on Fractions

Here we are going to study about adding and subtracting fractions with different numerators and denominators.

Example 1

`2/5 + 1/4`

Solution

Here the given problem is to add `2/5` and `1/4`

Notice that the denominators are different in the given fractions and we cannot add the fractions directly. The given fractions can be added with the help of least common denominator.

The least common denominator of `2/5` and `1/4` is 20. Because, 20 is the least common multiple of the denominators 5 and 4.

So, the fractions can be re-written as,

`2/5` = `2/5` * `4/4` = `8/20`

`1/4` = `1/4` * `5/5` = `5/20`

Now the problem becomes,

`8/20 + 5/20` = `(8 + 5)/20`

= `13/20`

Example 2

`3/7 - 1/3`

Solution

Here the given problem is to subtract `1/3` from `3/7`

Notice that the denominators are different in the given fractions and we cannot subtract the fractions directly. The given fractions can be subtracted with the help of least common denominator.

The least common denominator of `1/3` and `3/7` is 21. Because, 21 is the least common multiple of the denominators 3 and 7.

So, the fractions can be re-written as,

`3/7` = `3/7` * `3/3` = `9/21`

`1/3` = `1/3` * `7/7` = `7/21`

Now the problem becomes,

`9/21 - 7/21` = `(9 - 7)/21`

= `2/21`

Antiderivative of Log X

Introduction to anti-derivative of log x:

The natural logarithm is the logarithm to the base e, where e is an irrational constant approximately equal to 2.718. The natural logarithm is generally written as ln(x), loge(x) or sometimes, if the base of e is implicit, as simply log(x). Formally, ln(a) may be defined as the area under the graph of `1/x ` from 1 to a, that is as the anti-derivatives or integral,

ln a  = `int_1^a(1/x)dx`

This defines a logarithm because it satisfies the fundamental property of a logarithm:

ln (ab) = ln a + ln b

Source Wikipedia.

Anti-derivative Logarithmic Formulas:

1. `int` `(1/x)` dx = log x + c

2.  `int` e x dx = e x + c

3. `int (dx) / (a^2 - x^2) ` = `(1/(2a)) log [(a + x) / (a - x)] + c`

4. `int (dx) / (x^2 - a^2)` = `(1/(2a)) log [(x - a) / (x + a)] + c`

5. `int (dx) / sqrt(a^2 - x^2)` = `sin^-1(x / a) + c`

6. `int (dx) / sqrt(x^2 - a^2) ` = `log [(x + sqrt(x^2 - a^2)] + c`

Anti-derivative Logarithmic Problems:

Anti-derivative logarithmic problem 1:

Find the anti-derivative of given logarithmic function, log x  with respect to x

Solution:

Given logarithmic function, ` int ` log x. dx

Let,    u = log x                            dv = dx.

`(du)/(dx) ` = `1/x`                                    v = x

du = `1/x` dx

We know anti-derivative parts formula,  `int ` u dv = uv - `int ` v du

`int ` log x. dx  = log x . x - `int`` x ((dx) /x) `

= x. log x - `int` dx

= x. log x - x + c

= x( log x - 1) + c

Answer: Anti-derivative of log x is    x( log x - 1) + c

Anti-derivative logarithmic problem 2:

Find the anti-derivative of given logarithmic function, `(1 + 25x)/x^2`   with respect to x

Solution:

Given function, ` int` `(1 + 25x)/x^2` . dx

`int``(1 + 25x)/x^2`. dx  =` int` `dx/x^2` + `int` ` (25x)/x^2 ` dx            

= `int` `x^(-2) ` dx + ` int `` 25x^(-1)` dx               

= `x^(-1)` +  25 log x + c

Answer:  Anti-derivative of  `(1 + 25x)/x^2`  is x-1 + 25 log x + c

Anti-derivative logarithmic problem 3:

Find the anti-derivative of given logarithmic function, `e^(3x)/(1- e^(3x))` with respect to x

Solution:

Let u = 1- e3x           du = - `3` `e^(3x)` dx      

So, substitute the u and du

`int `  `e^(3x)/(1- e^(3x))` dx = `int`` (-1/3)(du)/u`                                 

= `(-1/3)` ` int` `1/u` du

= `(-1)/3` ln u + c                                                      we know u = 1- e3x

=` (- ln (1-e^(2x)))/3` +c

= `((-1)/3)` ln(e3x -1) + c

Answer:  Anti-derivative of `e^(3x)/(1- e^(3x))` is  `((-1)/3)` ln(e3x -1) + c

Subtracting Mixed Fractions

Introduction subtracting mixed numbers:

Natural number and a fraction is called mixed number. If numerator is less than denominator then it is said to be proper fraction. Improper fraction is formed when numerator is greater than denominator. For subtracting mixed numbers we have to first convert it in to proper fractions and then subtract the numbers so obtained.

Problem Subtracting Mixed Numbers with Different Denominators:

Pro 1 : 2 `3/5` - 3` 5/7`

Sol : Here the both are mixed fraction,

Step 1:  convert into the proper fraction and subtracting mixed numbers

2 `3/5` first we have to change into the proper numbers `(5xx2+3)/5` , we get

`13/5` is the proper numbers

3 `5/7=` `[(7xx3)+5]/7` = `26/7 ` is the proper number

Here we round mixed numbers and adding both functions

`13/5` -` 26/7`

In this step subtracting mixed numbers taking the Least Common Multiplier of 5 and 7 is 35. `[(13 xx 7) - (26 xx 5)]/(35)`  = `(91-130)/35 = (-39)/(35) `= -1.1142

Problem Subtracting Mixed Numbers with same Denominator:

Ex 1:      2 `3/4` - 3 `5/4`

Sol:  Here the fraction 4 is the equivalent fraction mixed number, subtracting both mixed numbers

2 `3/4` first we have to change into the proper numbers `(4xx2+3)/4` we get

`11/4 ` is the proper numbers

3 `5/4 = [(4xx3)+5]/(4) = (17)/(4)`

Here we round mixed numbers and adding both functions

`11/4 ` - `17/4 = (11-17)/(4) = (-6)/(4) = (-3/2) = -1.5`

Tangent Function Graph

Introduction to tangent function graph:
In geometry, the tangent line (or simply the tangent) to a curve at a given point is the straight line that "just touches" the curve at that point (in the sense explained more precisely below). As it passes through the point where the tangent line and the curve meet, or the point of tangency, the tangent line is "going in the same direction" as the curve, and in this sense it is the best straight-line approximation to the curve at that point. The same definition applies to space curves and curves in n-dimensional Euclidean space.

Source: Wikipedia.


Definitions of a Trigonometry Tangent Function:

Definition of tangent function is defined by using the right angle triangle

Tangent of an angle is the ratio of length of the adjacent and the opposite side.
tan x = opposite / adjacent

The tangent is the inverse of the cotangent and is given by
tan x = 1 / cot x

Quotient of sine and cosine functions is called as tan or tangent function.
tan x = sin x / cos x

Properties of a Tangent Function Using the Graph



Example Problem for Tangent of a Graph:

Example 1:

Find the function value of tan 35 o.

Solution:       

Use the tangent's co function identity to solve the problem.

tan x = cot (90o – x)
tan 35o = cot (90o – 35o)
= cot (55o)
tan 35o = 0.70021
The tangent of 60o is 0.70021.

Example 2:

Find the hypotenuse of the right angle triangle using the Pythagorean Theorem and find the tangent of an angle value.

Solution:

Use the Pythagorean Theorem, for finding the hypotenuse

In the given right angle triangle

AC2 = AB2 + BC2

Here,

AB = Opposite side

BC = Adjacent side

AC = Hypotenuse

AC2 = AB2 + BC2

= 42 + 92

= 16 + 81

AC2 = 97

AC = 9.84

Hypotenuse for the given right angle triangle is 9.84

Tangent function:

Tan ? = Sin ?/ Cos ?

= adj / opp

= 4 / 9

Tan ? = 4 / 9

? = tan -1(4/9)

? = 23.96 °

Solving Distance Formula

Introduction to solving distance formula:
Distance formula is a formula which is used to find the distance between two points (x1, y1) and (x2, y2). This formula is based on the Pythagorean Theorem.Solving problems using distance formula is one of the easier method to find the distance between two given points.

The distance d between the points (x1, y1) and (x 2, y2) is given as,

d = v ((x2 - x1)2 + (y2 - y1)2)

Example Problems on Solving Distances Using Distance Formula:


1) Find the distance between the points (-2, -3) and (-4, 4).

Solution:

Let (x1, y1) = (-2, -3)

(x2, y2) = (-4, 4)

By distance formula, d = v ((x2 - x1)2 + (y2 - y1)2)

=  v (((-4 - (-2))2 + (4 - (-3))2)

=  v ((-2)2  + (7)2)

=  v (4 + 49)

= v53

Solving for d, we get d  ˜ 7.28

2) Find the distance between the points (2, 4) and (5, -1).

Solution:

Let (x1, y1) = (2, 4)

(x2, y2) = (5, -1)

By distance formula, d = v ((x2 - x1)2 + (y2 - y1)2)

=  v ((5 - 2)2 + (-1-4)2)

=  v (32 + (-5)2)

=  v (9 + 25)

= v 34

Solving for d, we get d ˜ 5.83

3) Find the distance between the points (-2, 1) and (1, 5).

Solution:

Let (x1, y1) = (-2, 1)

(x2, y2) = (1, 5)

By distance formula, d = v ((x2 - x1)2 + (y2 - y1)2)

=  v ((1 - (-2))2 + (5 - 1)2

=  v (32 + 42)

=  v (9 + 16)

d  = v 25

Solving for d, we get   d = 5

4) If the distance from the point is (1, 2) to the point (3,y) is v8.  Find the value of  y.

Solution:

Let (x1, y1) = (1, 2)

(x2, y2) = (3, y)

d = v8

By distance formula, d = v ((x2 - x1)2 + (y2 - y1)2)

v8 = v ((3 - 1)2 + (y - 2)2

8 = 22 + (y - 2)2

8 = 4 + (y - 2)2

8 - 4 = (y 2 - 4y + 4)

4 = y 2 - 4y + 4

0 =  y 2 - 4y

0 = y (y - 4)

Solving for y. we get

y = 0 ;   y = 4
   


Solve the Problems Using Distance Formula:

Find the distance between the following points,

(1, 4) and (4, 0)

(2, 8) and  (16, 4)

(8, 5) and (9, 6)

(5, 6) and (-12, 40)

Solutions:

d = 5
d = 14.56
d = 1.41
d = 38.01

Tangent and Secant of a Circle

Introduction to tangent and secant of a circle:

A tangent of a circle is a line drawn from a point passing through the circle just at one point and secant is a line passing through two points of a circle. In case, a tangent and a secant are drawn from the same point outside the circle, we can find an interesting relation. Similar is the case when two secants are drawn from the same point outside the circle.

Let us study these two situations.

Tangent and Secant of a Circle – a Tangent and a Secant

Look at the above diagram.

A tangent OT is drawn from O touching the circle at P. From the same point O, a secant OS is drawn passing through the points Q and R on the circle. Join PR and PQ.

The angle OPQ and ORP are congruent as both of them are subtended by the intercepted arc PQ.

Angle POR is subtended both by the chords PQ and PR at O. Therefore the triangles OPQ and ORP are similar.

Applying the rule of similarity,

OP/OR = OQ/OP

or,  OP2 = OQ*OR



Tangent and Secant of a Circle – Two Secants

Look at the above diagram.

Two secants are drawn from the same point O to the circle. One secant passes through the points A and B on the circle and the other passes through the points C and D on the circle. Join BC and DA.

The angles BAD and BCD are subtended by the same arc BD on the circumference on the circle. Hence these two angles are congruent. The angle at O is common to the triangle AOD and BOC. Therefore, the triangles AOD and BOC are similar.

Applying the rule of similarity,

OC/OA = OB/OD

or,   OA*OB = OC*OD 

The two properties derived are very useful in solving problems related circles with tangents and secants.

Adding Negative Exponents

Introduction :

The term exponent in math is used to find the exponential value of the particular value it may be integer  or fraction . We can easy to get  the power of a  number using online calculator ,Consider  the unknown number , Here x is base and y is the power  of x

Adding Exponents means how many times to divide the number .

that is , `a ^(-n) = 1/a^n`

Steps for Adding Negative Exponents :

Negative exponents are added in the same way as the exponents are added with just a negative sign.

The given terms exponents are combined in a way such that the negative exponents are added or combined in case of same base.
In case of different bases , we have to simplify by convert it to positive exponent ,
Then we have to simplify the exponent value . Make the negative exponent to the positive one .
Now simplify further to get the results.
Ex  :   `5^-2 + 5^-4 = 5^((-2-4))`

`=5^-6`

=`1/5^6`

=`1/15625`

= 0.000064



Examples to Add Negative Exponents:

Ex 1:  A number  with negative exponents `8^(-3)`

Sol :     `8^(-3)` = `1/ (8^3)`

=`1/ (8 *8*8)`

= `1/ 512`

=0.00195

Ex 2:    Add the neagtive exponents` 7^-3`    + `5^-2`

Sol :    `=1/7^3 + 1/5^2`

`=1/343 +1/25`

=0.0029 +0.04

=0.0429

Ex 3:   Add the negative exponents `6^-2 + 6^-3`

Sol :  Using the rule , we have to add the exponents as 6 is the common base

= `6^(-2 + -3)`

` 6^(-2-3)`

=`6^-5`



= `1/(6^5)`

= 0.00012

Example 4:

Add the negative exponents  of `5^-3 +5^-5 +5^-2`

Solution:

Use the property to add the negative exponents , we have

=`5^-3 +5^-5 +5^-2`

=`5^(-3 + -5 + -2)`

=`5^(-3 -5 -2)`

= `5^-10`

= `1/(5^10)`

=0.04832

Example 5:

Add the negative exponents  of` 3^-3 + 3^-6`

Solution:

Use `a^-n = 1/a^n`   rule

We  get ,

=`1/3^3 +1/3^6`

=0.03703 +0.0013

=0.03833

Example 6:

Add the negative exponents  of `5^-2 + 5^-3`

Solution:

Use a ^-n = 1/a^n

Then we  get ,

=`1/5^2 +1/5^3` 

=0.04+0.008

=0.048

Practice Problem to Add Negative Exponents:

Pro1 : Add the negative exponents  of` 2^-2 +2^-4`

Ans : 0.3125

Pro2 :Add the negative exponents  of` 3^-3 +3^-2`

Ans : 0.1481

Selecting on the Dependent Variable

Introduction about selecting on the dependent variable:

Independent variable is a variable which does not depends on any other variable. But the dependent variable should depend only on independent variable. The dependent variable’s value depends on direct or indirect variation. In direct variation if the value of independent variable increases then the value of dependent variable also increases and if independent variable decreases then the value of dependent variable also decreases. Here we are going to learn about some example problems of selecting on the dependent variable.

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Simple Example Problems of Selecting on the Dependent Variable.

Example 1:

What is the dependent variable in the function f(x) =4-x?

Solution:

In the function f(x) = 4 - x, the value of f(x) depends on the value of x. So, f(x) is dependent variable.

Example 2:

What is the dependent variable in the function f(x) =10-x?

Solution:

In the function f(x) = 10- x, the value of f(x) depends on the value of x. So, f(x) is dependent variable.

These are the examples of selecting on the dependent variable.



Some more Examples of Selecting on the Dependent Variable:

Example 3:

Check whether the d variable dependent or not in the equation d=8b

Solution:

Step 1: The given equation is d=8b.Here =d is dependent variable and b is independent variable.

Step 2: Now plug different values for b and check the value of d for each value of b.

Step 3: When b = 1 =the value of d is 8.

Step 4: When b = 2 =the value of d is 16.

Step 5: When b = 3 =the value of d is 24.

Step 6: When b = 4 =the value of d is 32.

Step 7: When b = 5 =the value of d is 40.

Step 8: If the value of b increases then the value of d is also an increase.

Step 9: So, d is a dependent variable.

These are the example problems selecting on the dependent variable.

Triangle Pyramid Net

Introduction to triangle pyramid net:

In this article we see about definition of triangular pyramid. Pyramid is a polyhedron 3 – dimensional geometric shape. The pyramid has 4 – vertices. Out of them 3 are base of the pyramid and one is top of the pyramid.

Types of pyramid:

Square pyramid
Rectangular pyramid
Triangular pyramid
Pentagonal pyramid
These are some of the types of pyramid. In this section we see about definition of triangular pyramid.

Definition – Triangle Pyramid Net:

The net definition for triangular pyramid is given below. The base of the pyramid is triangle in shape is said to be triangle pyramid.

Types of triangular pyramid:

Equilateral triangular pyramid – Pyramid base is in the shape of equilateral triangle
Isosceles triangular pyramid – Pyramid base is in the shape of isosceles triangle
Scalene triangular pyramid – Pyramid base is in the shape of scalene triangle
Volume of a triangular pyramid = `1/3` area of the triangle x length

Volume = `1/3` x `1/2 ` x base x height x length

That is volume = `1/6` x b x h x l

Let we workout some of the example problems for triangle pyramid net.



Example Problems – Triangle Pyramid Net:

Example problems 1 – triangle pyramid net:

Find out the volume of triangular pyramid where the base is 3.5 m, height is 11.5 m and length is 12.2m.

Solution:

Given:

Base b = 3.5 m

Height h = 11.5 m

Length l = 12.2 m

Volume of triangular pyramid = `1/6` x b x h x l

= `1/6` x 3.5 x 11.5 x 12.2

= `1/6` 491.05

= 81.84

Answer: Volume of a triangular pyramid = 81.84 cubic meter.

Example problem 2 – triangle pyramid net::

Find out the volume of triangular pyramid where the base is 4 m, height is 7 m and length is 5 m.

Solution:

Given:

Base b = 4 m

Height h = 7 m

Length l = 5 m

Volume of triangular pyramid = `1/6` x b x h x l

= `1/6` x 4 x 7 x 5

= `1/6` 140

= 23.3

Answer: volume of a triangular pyramid = 23.3 cubic meter.

Prime Factorization Chart Fractions

Introduction to Prime Factorization chart fractions:

A Prime Number is a complete number, larger than 1, so as to can be evenly divided just by 1 otherwise itself. "Prime Factorization" is established which prime numbers require to multiply as one to obtain the original number.

A few of the prime numbers are:  1,7,13,19 etc...

The prime factorization of a digit is multiplying prime factors of a digit.

For example

38  = 2x19

Basic Prime Factorization Chart Fractions:

Factors:

"Factors" are the facts you multiply mutually to obtain another number: 45 = 3x3x5. In prime factorization, every factor will be prime numbers. 

Example:

39 = 3 x 13    (use chart to verify the answer)

Here multiplication of 13 x 3 is called as prime factorization.
                
Example Problems for Prime Factorization Chart Fractions:

Problem 1:

What are the prime factorization chart fractions of `1/16` ?

Solution:

It is best to create work with the least prime number, which is `1/2` , so let check:

`1/16`  ÷ `1/2` = `1/8`

But `1/8` is not a prime number, so we need to factor it further:

`1/8` ÷ `1/2` = `1/4`

But `1/4` is not a prime number, so we need to factor it further:

`1/4`  ÷ `1/2` = `1/2`

`1/2`  ÷ `1/2` = `1/1`

And 1 is a prime number,

`1/16`  = `1/2` × `1/2` × `1/2` × `1/2`

All factors are a prime number, so the answer should be correct.

The prime factorization of `1/16` is `1/2` × `1/2` × `1/2` × `1/2`


Problem 2:

What are the prime factorization chart fractions of `1/150` ?

Solution:

It is best to create work with the least prime number, which is `1/2` , so let check:

`1/150` ÷ `1/2` = `1/75`

But `1/75` is not a prime number, so we need to factor it further:

`1/75`  ÷ `1/5` = `1/15`

But `1/15` is not a prime number, so we need to factor it further:

`1/15`  ÷ `1/3` = `1/5`

`1/5`  ÷ `1/5` = `1/1`

And `1/1` is a prime number,

`1/150`  = `1/5` × `1/5` × `1/3` × `1/2`

All factors are a prime number, so the answer should be correct.

The prime factorization of fractions `1/150` is `1/5` × `1/5` × `1/3` × `1/2`

Algebra Foil Method

Introduction 

The algebra is a basic topic in mathematics and it is related with binomial expansion. The binomial expansion is done by foil method. In algebra, the foil method is multiplying the binomials. The foil method is considered as rule and the expansion of foil is first – outer – inner – last. Now we are going to see about algebra foil method.

Explanation for Algebra Foil Method
Algebra in math:

The algebra is a simple topic in math and it is defining the relations, rules and so on. The algebra problems are based on variables. In foil method also the variables are used.

Algebra foil method steps:

Multiplication of binomials process is called as foil method and it is done with variables. Steps for foil method:

Binomial’s first terms are multiplied.
Binomial’s outer terms are multiplied.
Binomial’s inner terms are multiplied.
Binomial’s last terms are multiplied.
The terms are combined with each other.




More about Algebra Foil Method

Example problems for algebra foil method:

Problem 1: Use the foil method and determine the binomial expansion of (x + 11) and (x + 5).

Solution:

The binomials are given as (x + 11) and (x + 5).

We can expand the binomials as.

Binomial’s first terms are multiplied as x^2.
Binomial’s outer terms are multiplied as 5x.
Binomial’s inner terms are multiplied as 11x.
Binomial’s last terms are multiplied as 55.
Combine the terms as x^2 + 5x + 11x + 55.
The result is x^2 + 16x + 55.
Problem 2: Use the foil method and determine the binomial expansion of (x + 8) and (x + 1).

Solution:

The binomials are given as (x + 8) and (x + 1).

We can expand the binomials as.

Binomial’s first terms are multiplied as x^2.
Binomial’s outer terms are multiplied as x.
Binomial’s inner terms are multiplied as 8x.
Binomial’s last terms are multiplied as 8.
Combine the terms as x^2 + x + 8x + 8.
The result is x^2 + 9x + 8.
Exercise problems for algebra foil method:

1. Use the foil method for binomial expansion.

(x – 2) (2x + 3)

Solution: The binomial expansion is 2x^2 – x – 6.

2. Use the foil method for binomial expansion.

(3x + 1) (x + 1)

Solution: The binomial expansion is 3x^2 + 4x + 1.

Opposite Rays Geometry

Introduction to opposite rays:

In geometry ray is a line but it has starting point with no ending point. Two opposite rays join at a single vertex and forms straight angles.  Straight angles are one of the angles in mathematical geometry.Straight angle is a straight line angle it measures 180 degree angle. Therefore we can say that an angle formed by opposite rays is the straight angle.  In this article we are going to learn about how two opposite rays make a straight angle in geometry.

Explanation about Opposite Rays in Geometry

Ray:                      

Ray is the important concept in mathematical geometry. A ray is the one of the part of the line. All angles are lies between two rays. Generally a ray as a starting point but is has no ending point it ends up to infinity level.  Starting point of the ray is mentioned by the dot and then ending point of the ray is mentioned by the arrow.

Opposite rays:
Two rays that share the starting point and it goes up to infinity level on both directions is known as opposite rays.  It looks like a straight line so it is forms an straight angle.

The above diagram AB and BC are two opposite rays they share a common point B and it forms 180 degree straight angle.  Here B is known as the vertex of the straight angle 180 degree.



Description about Pictures and Points of the Opposite Rays

Different pictures of the opposite rays:

Collinear points:

If the point are in the straight line that has been similar to two opposite then that points are called as the collinear points. Simply known as a set of points in the straight line then those points is called as the collinear points.  In the above diagram 1 A, B and then C are the collinear points because they lie in the same straight line and formed by two opposite rays.

Complex matrix inverse

Complex matrix inverse - Introduction

          In algebra, the determinant is a particular number related with some square matrix. The essential geometric significance of a determinant is a scale factor for compute when the matrix is considered as a linear transformation. Thus a 3× 3 matrix with determinant 3 when useful to a place of points with fixed area will convert those points into a place with double the area. Determinants are significant together in calculus, where they go into the replacement rule for various variables, and in multi linear algebra.

          `A = [[A,B],[C,D]] If AD - BD != 0` , then A has an inverse, Denoted` A^-1`

          `A^-1 = 1/(AD-BC) [[D,-B],[-C,A]]`

Complex Matrix Inverse - Examples:

Complex matrix inverse - Example 1:

Find the Complex inverse matrix of the matrix below?

`A = [[4,2],[3, 8]]`

Solution:

Inverse A =` 1/((8*4)-(3*2)) [[8,-3],[-2,4]]`

           = `1/(32-6) [[8,-3],[-2,4]]`

           =` 1/26[[8,-3],[-2,4]]`

          = `1/26 * [[8,-3],[-2,4]] = [[8/26, -3/26],[-2/26,4/26]]`

          = `[[4/13, -3/26], [-1/13, 2/13]]`

Complex matrix inverse - Example 2:

What is the inverse of the matrix below?

          `[[9, 1], [6, 3]]`

Solution:

Inverse of the matrix is the

Inverse A =` 1/((9*3)-(6*1)) [[3,-1],[-6,9]]`

           = `1/(27-6) [[3,-1],[-6,9]]`

           = `1/21[[3,-1],[-6,9]]`

          = `1/21 * [[3,-1],[-6,9]] = [[3/21, -1/21],[-6/21,9/21]]`

         ` [[3/21, -1/21],[-6/21,9/21]] = [[1/7, -1/21],[-2/7,3/7]]`

Complex matrix inverse - Example 3:

What is the inverse of the matrix below?

  A =  ` [[7, 1], [5, 3]]`

Solution:

Inverse of the matrix is the

Inverse A =` 1/((7*3)-(5*1)) [[3,-1],[-5,7]]`

           = `1/(21-5) [[3,-1],[-5,7]]`

           = `1/16[[3,-1],[-5,7]]`

          = `1/16 * [[3,-1],[-5,7]]`

          = `[[3/16, -1/16],[-6/16,9/16]]`

`[[3/16, -1/16],[-6/16,9/16]] = [[3/16, -1/16],[-3/8,9/16]]`




Complex Matrix Inverse - more Examples:

Complex matrix inverse - Example 1:

What is the inverse of the matrix below?

         ` [[8, 2], [6, 4]]`

Solution:

Inverse of the matrix is the

Inverse A = `1/((8*4)-(6*2)) [[4,-2],[-6,8]]`

           `= 1/(32-12) [[4,-2],[-6,8]]`

           `= 1/20[[4,-2],[-6,8]]`

         ` = 1/20* [[4,-2],[-6,8]]`

          `= [[4/20, -2/20],[-6/20,8/20]]`

`[[4/20, -2/20],[-6/20,8/20]]= [[1/5, -1/10],[-3/10,2/5]]`

Complex matrix inverse - Example 2:

Find the inverse Complex matrix of the matrix below?

          `[[9, 3], [7, 5]]`

Solution:

Inverse of the matrix is the

Inverse `A = 1/(9*5)-(7*3) [[5,-3],[-7,9]]`

          `= 1/(45-21) [[5,-3],[-7,9]]`

          `= 1/24[[5,-3],[-7,9]]`

          `= 1/24* [[5,-3],[-7,9]]`

          `= [[5/24, -3/24],[-7/24,9/24]]`

`[[5/24, -3/24],[-7/24,9/24]] = [[5/24, -1/8],[-7/24,9/24]]`

Complex matrix inverse - Example 3:

What is the inverse of the matrix below?

          `[[10, 4], [8, 6]]`

Solution:

Inverse of the matrix is the

Inverse `A = 1/(10*6)-(8*4) [[6,-4],[-8,10]]`

          `= 1/(60-32) [[6,-4],[-8,10]]`

           `= 1/28 [[6,-4],[-8,10]]`

          `= 1/28* [[6,-4],[-8,10]]`

          `= [[6/28, -4/28],[-8/28,10/28]]`

`[[6/28, -4/28],[-8/28,10/28]] = [[3/14, -1/7],[-2/7,5/14]]`

Nonnegative Integers

Introduction for non negative integers:

                      Non negative integers are defined as a numbers greater than zero, numbers at the right side of the zero. Thus the non negative integers are all the integers from zero on upwards, and the non negative real are all the real numbers from zero on upwards. The set of all non-negative integers forms a commutative mono id under addition. Non negative integers always proceed by a non negative sign (+). An integer with or without the non negative integers is always non negative.

Operations and Examples of Non negative Integers:

Operation for non negative integers:

          1. Reduce anything in the parentheses for solving

          2. Reduce the exponents for solve non negative integer.

          3. Multiplication or division for solve non negative integer.

          4. Finally, Addition or Subtraction for solve non negative integer.

Examples of non negative integers:

                   Integers are similar to whole numbers, but they also consist of negative numbers but still fractions are not allowed. So integers can be negative {-1, -2,-3, … }, positive {1, 2, 3, … }, or zero {0}.

Operations on Non negative Integers:

1. Adding non negative integers:

           Non negative + Non negative = Positive: 8 + 4 = 12

           Negative + Negative = Negative: (- 6) + (- 4) = - 10

           Addition of a negative and a non negative integer: Use the sign of the larger number and subtract, examples are,

           (- 5) + 3 = - 2

           7 + (-4) = 3

2. Subtracting non negative integers:

           Negative - Non negative = Negative: (- 4) - 2 = -4 + (-2) = -6

           Non negative - Negative = Non negative + Non negative = Non negative: 6 - (-2) = 6 + 2 = 8

           Negative - Negative = Negative + Non negative = Use the sign of the larger number and subtract (Change double negatives to a non negative)

          (-8) - (-3) = (-8) + 3 = -5

          (-3) - (-8) = (-3) + 8 = 5

3. Multiplying non negative integers:

           Non negative x Non negative = Non negative: 4 x 3 = 12

           Negative x Negative = Non negative: (-3) x (-2) = 6

           Negative x Non negative = Negative: (-3) x 3 = -9

           Non negative x Negative = Negative: 2 x (-5) = -10

4. Dividing non negative integers:

           Non negative ÷ Non negative = Positive: 14 ÷ 7 = 2

           Negative ÷ Negative = Non negative: (-12) ÷ (-4) = 3

           Negative ÷ Non negative = Negative: (-18) ÷ 3 = -6

           Non negative ÷ Negative = Negative: 15 ÷ (-3) = -5

Introduction of Decimal and Fraction

Introduction of Decimal and Fraction:

            Decimal is also termed as the fraction whose denominator of the fraction is 10 to the power. Example: `(3)/(10)` = 0.3 decimal. Fraction: It is part of the entire object. We can make chart from decimal to the fraction by following certain procedure. In this article, we see how to make decimal to fraction chart.

Decimal to Fraction Chart:

Procedure - Decimal to Fraction Chart:

Step 1: check the how many digit present after the decimal point.

Step 2: Multiply the 10 to power depending up on the digit present after the decimal point.

Step 3: Now we get the fraction, and simplify as much as possible.

Now decimal get convert to the fraction.

Writing decimal numbers:

We know the place value of the decimal numbers the following diagram shows the place value of the decimal numbers.

Example Problem - Decimal to Fraction Chart:

Example 1:

Make the decimal to fraction chart of the following decimal.

a)     1.35

b)    0.78

c)     2.8

Solution:

a) Converting the decimal 1.35 to fraction

Solution:

Step 1: Check the digit after the decimal point. Here there are two digits.

Step 2: Hence make the decimal as the whole and it divided by the 10 to the power of 2.

`(135)/(100)`

Step 3: Simplify the fraction `(135)/(100)`

1.35 = `(27)/(20)`

b) Converting the decimal 0.78 to fraction.

Solution:

Step 1: Check the digit after the decimal point. Here there are two digits.

Step 2: Hence make the decimal as the whole and divided it by the 10 to the power of 2.

`(78)/(100)`

Step 3: Simplify the fraction`(78)/(100)`,  we get

0.78 = `(39)/(50)`

c) Converting the decimal 2.8 to fraction.

Solution:

Step 1: Check the digit after the decimal point. Here there is one digit.

Step 2: Hence make the decimals as the whole and divided it by the 10 to the power of 1.

`(28)/(10)`

Step 3: Simplify the fraction `(28)/(10)`,  we get

`(7)/(25)`

Make the chart:

Decimal	Fraction
1.35	`(27)/(20)`
0.78	`(39)/(50)`
2.8	`(7)/(25)`

Example 2:

Make the chart of the following decimal to fraction.

0.45, 0.6, 0.75, 0.125

Solution:

Decimal	Fraction
0.45	`(45)/(100)`
0.6	`(6)/(10)`
0.75	`(75)/(100)`
0.125	`(125)/(1000)`