Triangle Area Equation

Introduction to triangle area equation:
In this article we discuss about the triangle area equation. The triangle can be defined as three closed sided polygon. To find the area of triangle is multiplying by the height and base, and then dividing by two. Basically the triangle comes from parallelogram. The parallelograms can be divided into two triangles. The each triangle area is equal to the one-half the area of the parallelogram. The parallelogram and triangle figure is given below,

Triangle Area Equation:

The area of parallelogram equation is given below,

`Area=Base xx Height`

That is, `A=B xx H`

The area of the triangle equation is given below,

Area of triangle=`"parallelogram"/2` or

Area= `(Base xx Height)/2` or

Area= `1/2 xx (B xx H)`

Where

B is the base of the triangle

H is the height of the triangle

We know the length of all sides of the triangle; we can calculate the area by using the Heron’s equation. Please express your views of this topic how to find the volume of a triangular prism by commenting on blog.

In this case first we have to define a perimeter of a triangle.

Perimeter= (a side + b side + c side)

Semi-Perimeter= `(a side + b side + c side)/2`

Therefore the heron’s equation is given below,

Area=`sqrt(s.(s-a).(s-b).(s-c))`

The all sides of the triangle figure is given below,

Example Problems for Triangle Area Equation:
Triangle area equation problem 1:

To find the area of a triangle with base 40 inches and height 6 inches?

Solution:

Given data: base (B) =40 inches and height (H) =6 inches.

Area of a triangle=`1/2` `xx` (B `xx` H)

=`1/2` `xx` (40 `xx` 6)

=`1/2 xx` 240

`Area = 120 i n^2`

Therefore the area of the triangle value is `120 i n^2` .

Triangle area equation problem 2:

A triangle has side a =10 in, side b =14 in and side c =12 in. Find its area?

Solution:

Given data: side a=10 in, side b=14 in, side c=12 in

First we find the perimeter of the triangle.

The perimeter = 10+ 14 + 12 = 36.

Next we find the semi-perimeter of the triangle.

The semi-perimeter is one half of this or 18

Using Heron's formula,

area = `sqrt (s . (s-10). (s-14) .(s-12))`

= `sqrt (18 . (18-10) .(18-14) . (18-12))`

= `sqrt (18 .(8) . (4) . (6))`

= `sqrt (3456)`

=  `58.78 i n^2`

Event Space Probability

Introduction to event space probability:

The event space determines the output in the probability that refers the total number of possibilities for the experiment. The sample spaces of the each experiment are not same. For example, the sample space for tossing the coin and the sample space for rolling the die and the sample space for the getting the diamond card from the group of cards are different. The sample spaces of the each experiment determine the probability for that event.I like to share this Simple Events with you all through my article.

Terms Used in the Event Space Probability:

Trial is the process of performing any experiment is called as the trial.
Sample space represents a set of all necessary outcomes of the experiment.
Event is a subset of a sample space whose elements are the outcomes of a trial.
Equally likely events are the events that none of the events can be performed as like as the other event.
Mutually exclusive events are the events that two events cannot occur simultaneously.
Exhaustive event contains the set of all possible outcomes for the experiment.
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Example Problems for Event Space Probability:

Example 1 for event space probability:

A bag contains the 12 red balls, 3 green balls and 7 black balls. What is the sample space for getting the balls?

Solution:

The bag contains the 12 red balls, 3 green balls and 7 black balls. The total numbers of balls are 22 balls.

The sample space is 22.

Example 2 for event space probability:

Determine the sample space for rolling the single die.

Solution:

The die contains six faces.

The sample space for the die is 6.

It can be represented in the form of S= {1, 2, 3, 4, 5, 6}.

Example 3 for event space probability:

Find the sample space for tossing two coins.

Solution:

A single coin has two sides. For the single coin the sample space is 2.

For two coins the sample space is 4.

Construct a Perpendicular

Introduction to Construct a Perpendicular

Constructing a perpendicular means drawing a line at right angles to a given line from a given point. In case of curves, the perpendicular from a given point is to the tangent of the curve at the required point. Constructing perpendicular can be done in two ways. One method is by using a ruler and protractor and the other method is with the help of a compass and a ruler.

The second method is more accurate and let us study that.

Construct a Perpendicular – when the Point is on the Line

Look at the above diagram. To construct a perpendicular line l at point A in the line itself, the method is as follows.

1)    Select two convenient points B and C on the line and on either side of A.

2)    Set the compass for a radius of approximately more than half of the length BC.

3)    Strike an arc from the point B..

4)    Without changing the compass setting strike an arc from the point C..

5)    Mark the point P where the arcs intersect.

6)    Draw a line m passing through P and A.

Line m is the perpendicular to the line l.

Construct a Perpendicular – from a Point not on the Line

In this diagram, the point A is not on the line. To construct a perpendicular in this case, the steps are as follows.

1)    With the help of the compass draw arcs from A to intersect the line at B and C.

2)    Set the compass for a radius of approximately more than half of the length BC.

3)    Strike an arc from the point B on the other side of the line.

4)    Without changing the compass setting strike an arc on the same side from the point C..

5)    Mark the point P where the arcs intersect.

6)    Draw a line m passing through A and P.

Line m is the perpendicular to the line l. Understanding 30 60 90 Triangles is always challenging for me but thanks to all math help websites to help me out.

Construct a Perpendicular – Proof

You will observe that, in both the cases the line m connects the points of intersections of circles of same radius. As per theorems on circles, the line m bisects the line l, joining their centers at right angle.

Variable Rate Calculator

Introduction to variable rate calculator:

The variable rate is defined as a rate of change in which it is expressed in the form of derivative. A calculator is a device that is used to perform arithmetic operations. The term rate can be calculated as the ratio of one thing or quantity to other. Also, the variable rate of any variables calculator with the help of a calculator. Here we discuss about the variable rate calculator.

Formula for Rate and Variable Rate of Change:

Rate:

It is a ratio in which the comparison of any two quantities can be done. It is calculated in a calculator in the form of proportion. In the calculator, it is calculated by the division operation. A rate can be found by dividing the distance with the time.

Variable rate of change:

The variable rate of change can be calculated as the change of rate value with any variables. The formula for variable rate is given as the change in variable of 'x' to the change in time.

Variable rate = `(dx)/(dt)`

Where,

dx is the change in variable

dt is the change in time.

Let us see about the calculator.

Calculator:

It is an important mathematical device which is used for calculation of all arithmetic operations.

In a calculator various symbols are there, they are used to do basic operations. For example, If we want to calculate the addition of 5 and 3 means, type the number 5 and then click the sign +. Now type the other number 3. Therefore the addition of two numbers is calculated as 8. I like to share this how to find the equation of a line with you all through my article.

Example Problem for Variable Rate Calculator:

We solve problem for calculating variable rate through calculator.

Problem for variable rate:

Calculate the variable rate of change with respect to variable x, for the given function f(x) = 6x2 + 4x.

Solution:

The given function is f(x) = 6x2 + 4x.

The formula for variable rate of change is (dx)/(dt).

So, we have to find the rate ofchange value of variable x in a calculator.

In calculator, the sign `d/dx` is used to find the rate of change.

Type the given function in calculator, then click the sign `d/dx`

Now the rate of change with respect to variable x is found or calculated as 12x + 4.

Hence the variable rate of change is calculated by a calculator.

These are about variable rate calculator.

Define Prime

Introduction to define prime

Normally the word prime is defined as “first, importance, greatest position, highest order, basics, and fundamentals”. In mathematics prime defines the prime number. Prime number is a whole number that is evenly divisible by 1 and the number itself. A natural number that has exactly two distinct divisors such as 1 and itself is called prime number (or simply a prime).There is infinity of prime numbers exist. The density of prime number compared to natural numbers is 0. The numbers 0 and 1 are not a prime number. Other than prime numbers are called composite numbers.

Calculation - Define Prime

How to calculate the prime numbers?

1 `->` The number is not a prime or composite.

2 `->` 1x2 = 2 and 2x1 =2. The number is divisible by 1 and itself (2), therefore this is even prime number.

3 `->` 1x3 = 3 and 3x1 = 3. The number is divisible by 1 and itself (3), therefore this is a prime number

4 `->` 1x4 = 4, 2x2 = 4 and 4x1 =4. The number is divisible by 1, 2 and itself (4), therefore this is a composite number.

5 `->` 1x5 = 5 and 5x1 =5. The number is divisible by 1 and itself (5), therefore this is a prime number.

6 `->` 1x6 =6, 2 x3=6, 3x2 =6 and 6 x 1 = 6. The number is divisible by 1, 2, 3 and itself (6). Therefore this is a composite number.

7 `->` 1x7 = 7 and 7x1= 7. The number is divisible by 1 and itself (7), therefore this is a prime number. I like to share this money math with you all through my article.

Example – Define Prime

Problem 1: Give the prime number from 10 to 20

Solution: Let us consider the values from 10 to 20 as

10 have divisors 1, 2, 5 and 10. This is composite number.

11 have divisors 1 and 11. This is prime number.

12 have divisors 1, 2,3,4,6 and 12. This is composite number.

13 have divisors 1 and 13. This is prime number.

14 have divisors 1, 2, 7 and 14. This is composite number.

15 have divisors 1, 3, 5 and 15. This is composite number.

16 have divisors 1,2,4,8 and 16. This is composite number.

17 have divisors 1 and 17. This is prime number.

18 have divisors 1, 2, 9 and 18. This is composite number.

19 have divisors 1 and 19. This is prime number.

20 have divisors 1, 2, 4,5,10 and 20. This is composite number

Venn Diagrams Finite Math Help

Introduction to venn diagram finite math help:-
Venn diagram  are diagrams that show all possible logical relations between a finite collection of sets. It is a collection of simple closed curves drawn on a plane. Sets are represented  by region.  There is a relation from one to the other.  Venn diagrams
comprise overlapping circles. The interior of the circle represent elements of the sets.  The exterior represent the elements that are not in the set.  The intersection of two circles contain the elements that belong to both the sets.

Venn Diagrams Finite Math Help with Images:

Venn diagram finite math help:-  Let us do a problem on the finite set A an B
Universal set ? = { 1,2,3,4,5,6,7,8}
Set A =  (1,2,5}
Set B = { 1,5, 8}
Represent these sets in a venn diagram This is a venn diagram where we can get finite math help.

We have represented the set A as yellow circle containing {1,2,5}  and the set B as green circle containing elements {1,5,8}
The elements {3,4,6,7} are outside the sets A and B.  The intersection of A and B contain elements {1,5}
Using the above venn diagram let us prove  ( A U B}'  =  A'nB'   This is De Morgan's law  for union of sets.
A U B =  {1,2,5,8}
(AUB)' =  ? - (AUB) =  {1,2,3,4,5,6,7,8} - {1,2,5,8} =  {3,4,6,7} ........ (1) ( Left hand side)
Now we must find A'   which is ? -  A = { 1,2,3,4,5,6,7,8} - {1,2,5} = {3,4,6,7,8}
Now let us find B'  which is ? - B = { 1,2,3,4,5,6,7,8} - {1,5,8} = {2,3,4,6,7}
Then we get A' n B' = { 3,4,6,7,8} n {2,3,4,6,7} = { 3,4,6,7} ............(2)(Right hand side)
Since (1) = (2) that is LHS = RHS we have proved that (AUB)'  = A' n B'. Is this topic free online math tutor hard for you? Watch out for my coming posts.

Venn Diagrams Finite Math Help with 3 Circles

Venn diagram finite math help can be used for a venn of three circles.
Now let us draw a 3 circle venn diagram.  Here we have A = {1,2,3,4,6} B = {2,3,4,7,8} C = {3,4,5,6,7} AnB = {2,3,4} BnC={3,4,7}  AnC = {3,4,6} and AnBnC = { 3,4}

This is a venn diagram with 3 circles,  Set A = { 1,2,3,4,6}   Set B = {2,3,4,7,8} and Set C = { 3,4,5,6,7}
Letus prove the De Morgan's law A - (BUC) = (A- B) n (A - C)
(BUC) = {2,3,4,5,6,7,8) {[ all the elements of B and C]
A - (BUC) =  { 1,2,3,4,6} - {2,3,4,5,6,7,8} = {1} .................(1)LHS
(A- B)  =  {1,2,3,4,6} - {2,3,4,7,8} = {1}
(A- C)  =  {1,2,3,4,6} - {3,4,5,6,7} = {1,2}
(A-B) n (A-C) = {1} n {1,2} = {1} .....................................(2) RHS
Since LHS = RHS we have proved the De Morgan's law
Venn diagram finite math help is very important because the diagrams help us to identify the  unions and intersections and
solve the problems very quickly.

Fraction Math Word Problems

Introduction to fraction math word problems:

Numbers can be represented in different forms like decimals, fractions in the mathematics. Fraction is used to represent the values accurately. Fractions consist of two parts, the numerator and denominator values and it can undergo all the basic mathematical operations. Here we will prepare word problems for fractions. In this article we will illustrate about fraction math word problems.

Fraction Math Word Problems:

Fraction math word problems simply involves in preparing word problems for fractions. Some sample problems are given as follows,

Example problem 1- Fraction math word problems

In the car showroom we have 50 cars; one- fifth of the car is Lenovo. Find out how many Lenovo cars present in the car showroom?

Solution:

`1 / 5` of 50 cars are Lenovo.

To find the number of Lenovo,

Divide 50 ÷ 5 = 10

10 Lenovo are` 1 / 5` of 50.

10 Lenovo are in the car showroom.

Example problem 2:

Of the soft drinks sold yesterday at Kevin’s supermarket, `2/4` was Pepsi and another `3 /4` was coke. Find out at what fraction of the soft drinks sold was either Pepsi or coke?

Solution:

Fraction of Pepsi sold = `2/4`

Fraction of coke sold = `3/4`

Fraction of soft drinks sold was either Pepsi or coke =?

Add the Fraction of Pepsi and coke to get the fractions for either Pepsi or coke

= `2 / 4` + `3 / 4`

Here the denominators are same, so add the numerators.

= `5 / 4`

Fraction of soft drinks sold was either Pepsi or coke is` 5/4` .

Example problem 3:

Quen prepared some cookies in the morning. She used` 5 / 6 ` of a cup of flour, and `3 / 6 ` cup of sugar for the preparation of cookie. How much more flour did Quen use than sugar?

Solution:

Amount or quantity of flour used = `5/ 6`

Amount or quantity of sugar used = `3 / 6`

Amount or quantity of flour used than sugar =?

By subtracting the amount of sugar from the flour, we can get the amount of flour used.

=  ` 5/ 6` –` 3 / 6`

=` 2 / 6`

Amount or quantity of flour used than sugar = `1 / 3.`

Example problem 4:

Ben owns 20 acres of farmland where he planted flowers on `1/6` of the land. On how many acres of land does Ben plant flowers?

Solution:

Ben grows flowers on `1/6` of 20 acres of land.

Acres of land that Ben grow flowers =?

=      20  ` xx`   `1 / 6`

=   `10 / 3`

Ben plants flowers on `10 /3 ` of an acre of land.I have recently faced lot of problem while learning how to calculate simple interest, But thank to online resources of math which helped me to learn myself easily on net.

Practice Problem: Fraction Math Word Problems

Practice problem 1 - Fraction math word problems

Quen prepared some cookies in the morning. She used `8 / 6` of a cup of flour, and `7 / 6` cup of sugar for the preparation of cookie. How much more flour did Quen use than sugar?

Answer: `1/ 6`

Practice problem 2 - Fraction math word problems

Of the soft drinks sold yesterday at Kevin’s supermarket, `7/10` was Pepsi and another `4 /10` was coke. Find out at what fraction of the soft drinks sold was either Pepsi or coke?

Answer:   `11/10`

Twice Differentiable Function

Introduction for twice differentiable function:

The twice differentiable function explains the process which takes double time differentiation on the operations of differentiation equations on the different functions like algebraic function, trigonometric functions, logarithmic functions, exponential functions etc..   In this article we are going to deal with the different functions with different variables to differentiate for the complex equations. If we have the constant in the second time differentiation then its results in the zero value.

Examples to Explain Twice Differentiable Function

Review on the twice differentiable function for the equation in the form of function f(x) = `x^2 + 3x + 10`
Solution:

The given function is f(x) = `x^2 + 3x + 10`

Differentiate the above function

`f(x)` = `2x + 3 + 0`

`f(x)` = `2x + 3`

Again differentiate (twice differentiation)  the above function f(x),

`f'(x)` = `2 + 0`

`f'(x)` = `2 `    is the required solution obtained after twice differentiation.

Review on the twice differentiable function for the equation in the form of function `f(x)` = `x^3 - x^2 + 3x + 10`
Solution:

The given function is` f(x) ` =  `x^3 - x^2 + 3x + 10`

Differentiate the above function

`f(x)` = `3x^2 - 2x + 3 + 0`

`f(x)` = `3x^2 - 2x + 3 `

Again differentiate (twice differentiation)  the above function f(x),

`f'(x)` = `6x + 2 + 0`

`f'(x)` = `6x + 2 `    is the required solution obtained after twice differentiation.

Review on the twice differentiable function for the equation in the form of function `f(x)` = `5x^2 + x^3 - x^2 + 3x + 10`
Solution:

The given function is` f(x) ` =  `5x^2 + x^3 - x^2 + 3x + 10`

This equation is rewritten as

` f(x) ` =  `x^3 + 3x^2 + 3x + 10`

Differentiate the above function

`f(x)` = `3x^2 + 6x + 3 + 0`

`f(x)` = `3x^2 + 6x + 3 `

Again differentiate (twice differentiation)  the above function f(x),

`f'(x)` = `6x + 6 + 0 `

`f'(x)` = `6x + 6 `    is the required solution obtained after twice differentiation.Understanding statistics help is always challenging for me but thanks to all math help websites to help me out.

Problems to Explain Twice Differentiable Function

Review on the twice differentiable function for the equation in the form of function f(x) = `x^2 + 3x + sin x`
Solution:

The given function is f(x) = `x^2 + 3x + sin x`

Differentiate the above function

`f(x)` = `2x + 3 + cos x`

Again differentiate (twice differentiation)  the above function f(x),

`f'(x)` = `2 + 0 - sin x`

`f'(x)` = `2 - sin x`        is the required solution obtained after twice differentiation.

Review on the twice differentiable function for the equation in the form of function f(x) = `x^2 + 3x + sin x + cos x`
Solution:

The given function is f(x) = `x^2 + 3x + sin x + cos x`

Differentiate the above function

`f(x)` = `2x + 3 + cos x - sin x`

Again differentiate (twice differentiation)  the above function f(x),

`f'(x)` = `2 + 0 - sin x - cos x`

`f'(x)` = `2 - sin x - cos x`        is the required solution obtained after twice differentiation.

Solving Making Box and Whisker Plots

Introduction to making box and whisker plots:

Box and whisker plots is one of the important concept in math.  This method is used to split the data into different number of quartiles.  For this we can calculate the values of median and upper quartile and lower quartile range values.  In this topic we are going to seen about how to solving and making a box and whisker plots problems with suitable diagrams.

Description about Making a Box and Whisker Plots

Median:

First we can solve the median for the given set of values.  So we can arrange the numbers in to the format of ascending or else descending order. Then count the number of values if it is odd middle value is median or else find the mean for the middle two values.

Lower quartile part:

The group of values before the median is noted as Lower quartile part.  Again we can measure the median for this LQR values.

Upper quartile part:

The group of values after the median is noted as upper quartile part.  Again we can find the median for this UQR values.

Number line:

Draw a number line with particular scale level.  Then mark the given values and shade the LQR and UQR and median values.  These gives the following four quartiles.

Lowest value to LQR value
LQR value to Median value
Median value to UQR value
UQR value to greatest value

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Problems on Making a Box and Whisker Plots

Solving and making a box and whisker plots for the first 10 multiples of 20

Solution:

The first 10 multiples of 20 are 20, 40, 60, 80, 100, 120, 140, 160, 180, 200

Solving for Median:

Ascending order: 20, 40, 60, 80, 100, 120, 140, 160, 180, 200

Number of values: 10

10 is even number so median is a mean of 100 and 120.  That is 110

Median = 110

Solving for LQR:

Lower quartile Part: 20, 40, 60, 80, 100

Number of values: 5

5 is a odd number so median is center number 60

LQR = 60

Solving for UQR:

Upper quartile Part: 120, 140, 160, 180, 200

Number of values: 5

5 is a odd number so median is center number 160

UQR = 160

Number line:



Quartiles:

20 to 60
60 to 110
110 to 160
160 to 200

Practice Prime Numbers

Introduction – Practice prime numbers:

A prime number is a natural number. This has exactly divisible by one and itself only. If any one number divisible by more than these two factors that is not a prime number. First twenty-five prime numbers are listed below up to 100.

First twenty-prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.The only one even prime number is 2. Because other than 2 all even numbers will divisible by two. 

Finding Prime Numbers – Practice Prime Numbers:

Finding prime number less than two digits are very easy. If it is three digits we have to follow some rules.

For example:
Determine the number 281 is prime number or not?
Step 1:

Find the square root for the given number. That is 281 = 16.76
Then we have to round- up the decimal value16.76 as 17.
Step 2:

Mention all the prime number up to 17. That is 2, 3, 5, 7, 11, 13 and 17.
Step 3:
Now divide the given number by all obtained prime numbers. The number is not divisible by any number. So the given number is a prime number.

Example Problems – Practice Prime Numbers:

Problem 1:

Determine the number 150 to 155 is prime number or not?

Solution:

150 – Factors for 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75 and 150. So 150 is not a prime number.

151 – Factors for 151 are 1 and 151 only. So 151 is a prime number.

152 – Factors for 152 are 1, 2, 4, 8, 19, 38, 76 and 152. So 152 is not a prime number.

153 – Factors for 153 are 1, 3, 9, 17, 51 and 153. So 153 is not a prime number.

154 – Factors for 154 are 1, 2, 7, 11, 14, 22, 77 and 154. So 154 is not a prime number.

155 – Factors for 155 are 1, 5, 31 and 155. So 155 is not a prime number. I like to share this prime factorization of 72 with you all through my article.

Problem 2:

Determine the number 307 is prime number or not?

Solution:

The given number 307 has two factors only. Those factors 307 are 1 and 307 only. So the given number 307 is considered as a prime number.           

Problem 3:

Determine the number 330 is prime number or not?

Solution:

The number 330 has more than two factors. Those factors are 1, 2, 3, 5, 6, 10, 11, 15, 22, 30, 33, 55, 66, 110, 165 and 330. Hence the given number 330 is not a prime number.

Practice problems – practice prime numbers:

Problem 1:

Determine the number 593 is prime number or not?

Solution:

The given number 593 is a prime number.

Problem 2

Determine the number 641 is prime number or not?

Solution:

The given number 641 is a prime number.

Problem 3

Determine the number 432 is prime number or not?

Solution:

The given number 432 not a prime number.

Methods of Analyzing Data

Introduction to methods of analyzing data:

Any collection of information in the form of giving the required information is called data. Methods of analyzing data are used to compare the collection of data. Data analysis method Graphs mostly help to analysis the various types of database. For example methods are statistics, bar graphs, histogram graphs, pie charts, and line graphs are used to easily analysis the data. In this article, we are going to see about methods of analyzing data.

Example Problem for Methods of Analyzing Data:

1. Solve the statistics data analyzing using the given data set, find the mean, mode, median, and range of the given data set.

59, 57, 56, 59, 54, 55, and 66

Solution:

Rearrange the data set for ascending order.

54, 55, 56, 57, 59, 59, 66

Mean:

Mean is the sum of data divided by the number of data in the given data set.

Mean = (sum of data)/ (number of data)

Mean = `406/7`

= 58

Mode:

Mode is the most common value in given data set.

Most common value is 59.

Therefore, mode is 59.

Median:

Median is a middle number else we have two middle value means; we find the average of their two values is median. We find it can be arranged in ascending order.  

Median:

In this problem we have a middle number.

Therefore, Median is the middle number 57.

Range:

Range is difference between greatest and least values in the given data set.

Range = maximum value – minimum value

= 66 - 54.

Range = 12.

Please express your views of this topic descriptive statistics definition by commenting on blog.

More Example Problem for Methods of Analyzing Data:

1. Solve and create a bar chart for data analyzing using the given data set, worker name with their salary, given below.

Worker name

Salary in dollar($)

Aldo

4500

Dillan

6000

Joan

5500

Philip

7000

Alec

5000

Steps to draw bar charts and learn examples                                        

Step 1:

First, we need to take a paper and draw the horizontal line and vertical line. In bar chart, horizontal line called as x-axis and another line that vertical line called as y-axis.

Step 2:                                                                                

The horizontal axis mention the information that ‘worker name’ and in the vertical axis, we take the other information namely ‘salary in dollar ($)’.

Bar chart:



The above bar graph analyzed the given data set that which one is high range and low range. This is the needed bar chart for given data set.

Pre Calculus Problems Exam

Introduction to pre calculus problems exam:

Pre calculus is the advanced form of school algebra and its exam. Actually, pre algebra gives us a brief introduction to algebra while Pre Calculus exam explores some of the important topic about calculus. Pre calculus includes sets, real numbers, complex numbers solving inequalities and equations, properties of functions, composite functions, polynomial functions, rational functions, etc.

Sample Pre Calculus Problems Exam:

Pre calculus problem 1:

Find the perpendicular bisector equation of the line which passes through (4, -2) and (6, 6).

Solution;

The perpendicular bisector is given by

P = (x1 + x2) / 2, (y1 + y2) / 2

P = ((4 + 6) / 2, (-2 + 6) / 2).

= (5, 2)

The slope of the line segment (m1) is given by

m1 = (- 2 - 6) / (4 - 6)

= 4

So the slope of the perpendicular bisector m2 will be

m2 = -1 / m1

= - 1/4.

Therefore, the equation for perpendicular bisector is,

y – 2 = - 1 / 4(x - 5)

y = (- 1 / 4) x + (5 / 4) + 2

y = (-1 / 4) x + 13 / 2

Pre calculus problem 2:

Find an equation of the line which passes through (2, 6) and is parallel to the line 4x - 3y = 24

Solution:

4x - 3y= 24,

Solve the above last equation for y

we get,                    y = (4/3) x - 8

So this line has the slope m = 4/ 3.

The parallel in has its slope is also 4 / 3,

So the equation becomes

y- (-6) = (4/3) (x - (-2))

y+6 = (4/3) (x+2)

y = (4/3) x + 8/3 - 6

y = (4/3) x - 10/3

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Pre Calculus Problems Exam-test Problems:

Exam problem 1:

Find the perpendicular bisector equation of the line which passing through the points (6, -3) and (2, 5).

Answer: y = 1 / 2x -1

Exam problem 2:

Find an equation of the line which passes through (4, -3) and is perpendicular to the line 2x – 5x = 20.

Answer: y = - 5 / 2x +7

Exam problem 3:

Find an equation of the circle with center C (7, -5) which is tangent to the x-axis.

Answer: (x-7)2 + (y +5)2 = 25

Solving One and Two Step Equations

Introduction to solving one and two step equations:
Solving one and two step equations is the problems for systems of linear equations. Solving one step equation is the problem that involves the single step of solving the equation and finding the values of the variables. Solving two step equations is the problem that involves the two step of solving the equation and finding the values of the variables.Let us see sample problems for solving one and two step equations.

Solving One and Two Step Equations:

Solving one step equations:

To solve a one-step equation, we must get the variable alone on one side of the equation. This can be done by using inverse operations. Addition and Subtraction (or adding the opposite) are inverse operations. Multiplication and Division are example of inverse operations. Whatever you do to one side of an equation, you must do to the other side (like a balance scale).

Example1:

1) Simplify  `X/5` = 15

`X/5` ·5 = 15 · 5

X = 75

2) Simplify  X – 12 = 18

X – 12 + 12 = 18 +12

X – 12 + 12 = 18 +12

X =30

3)  Simplify   3X =9

`(3X) / X` = 9 /3

`(3X) / X ` = 9 /3

X =3

Example 2:

x +1 = 7  , 1 is being added to the variable, so we must subtract 1 (or add -1) from (to) both sides

x = 6

Example 3:       

6 = y - 5 , 5 is being subtracted from the variable, so we must add 5 to both sides

y =11

Example 4:

y - 1/5 = 1/10 , 1/5 is being subtracted from the variable, so we must add 1/5 to both sides

y = 3/10

Example 5:

-20.2+ 4.1= n - 38.1

n = 22

First we must add -20.2 and 4.1, 38.1 is being subtracted from the variable, so we must add 38.1 to both sides

Example 6:

5 - x =12 , because the 5 is positive, we must subtract 5 (or add -5) to both sides, then the opposite of x is 7, so x is -7

-x = 7

x = -7

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Solving One and Two Step Equations:

Solving Two-Step Equations - Getting variable alone

1. Remove constant by adding opposite to both sides.

2. If variable is being divided by a number, multiply both sides by the number. If variable is being multiplied by a number, divide both sides by the number (remember dividing by a fraction is the same as multiplying by the reciprocal).

Examples:

1) Simplify     `X/3` + 3 = 12

`X/3` + 3 – 3 = 12 – 3

`X/3` + 3 – 3 = 12 – 3

`X/3` = 9

`X/3` = 9

`X/3` · 3 = 9 ·3

X =  `9*3`

X = 27

2) Simplify     4X – 5 = 6

4X – 5 + 5 = 6 + 5

4X – 5 + 5 = 11

4X = 11                  

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

X = `11/4`

X = 2.75.