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.

Looking out for more help on Confounding Variable in algebra by visiting listed websites.

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.