algebra simple interest formula

Interest: Interest is the amount of money we pay for the use of some amount of money.

Types of interest: There are two types of interests,

a) simple interest

b) compound interest

Simple interest: Simple interest is the Interest paid / compensated only on the original principal, not on the interest accrued

Compound interest:Compound interest means that the interest Which  includes the  interest calculated on principal amount using the Compound Interest Formula .

Algebra Simple interest Formula :


Algebra Formula to find simple interest :

Simple interest I = PNR

Where,   P is the Principal amount, R is the  Rate of interest, N is Time duration.

When we knows interest I we can find p, n or r using the same formula ,

Different forms of algebra simple interest formula

P= `(I)/(NR)`

N=`(I)/(PR)`

R=`(I)/(PN)`


Algebra Problems on Simple Interest formula:


Ex 1:Interest Rate: 1% each year ,Starting Balance: $147 ,Time Passed: 6 year . Find the simple interest? What is the new total balance?

Solution:Algebra Formula for Simple Interest: I = PRT

P = principle = starting balance = $147

R = interest rate = 1%

T = time = 6 years

I = interest = principle × interest rate × time = 147 × `(1)/(100)` × 6 = $8.82

New Balance = starting balance + interest accrued = $147 + $8.82 = $155.82

I= $ 8.82, New Balance = $155.82

Ex 2:Interest Rate is 2% each year, Starting Balance is $184 ,Time Passed was 3 years .How much interest has ensued if we are

using simple interest? What is the new total balance?

Solution:algebra simple interest formula = PRT
P = principle = starting balance = $124

R = interest rate = 2 %

T = time = 9 year

I = interest = principle × interest rate × time = 124 × `(2)/(100)` × 9 = $22.32

New Balance = starting balance + interest accrued = $124 + $22.32 = $146.32

Interest = $22.32, New balance = $146.32

Ex 3 :John invests 1200 at the rate of 6.5 percent  per annum. How long it will take john earns 195 in interest ?

Solution : Algebra Simple interest formula = I = PTR/100

195 = (`(1200 * T*6.5)/(100)`) / 100

T = 2.5

John will earn $195 in interest in 2.5 years.

Area of Compound Figures

A compound figure is made up of two or more simple figures. The following figure is made up of a rectangle and a square. If we want to find the perimeter and area of a compound figure, then calculate the area of each simple figure and then find the sum of the area calculations.


Examples to Find the Area of Compound Figures:


Example 1: Find the area of the following compound figure.

Solution: Area of Rectangle Formula = l × b square units.

where, l is length,

b is breadth.

Rectangle 1 => 8 × 3 square units.

=> 24 square units.

Rectangle 2 => 6 × 3 square units.

=> 18 square units.

Rectangle 3 => 6 × 3 square units.

=> 18 square units.

Therefore the total area of compound figure = Area of Rectangle 1 + Area of Rectangle 2 + Area of Rectangle 3.

=> 24 + 18 + 18 square units.

=> 60 square units.

Hence the total area of compound figure is 60 square units.


Example 2: Find the area of the shaded part of the compound figure.

Solution: Area of Square = a2 square units.

where, a is length of the side of the square.

=> Area of Square = 82 square units.

=> 64 square units.

Area of Circle = ? r2 square units.

where, r is the radius of the circle.

=> Area of Circle = ? × 32 square units.

=> 3.14 × 9 square units.

=> 28.25 square units.

Therefore the area of the shaded part of the compound figure = Area of Square ? Area of Circle.

=> 64 ? 28.25 square units.

=> 35.75 square units.

Hence the area of the shaded part of the compound figure is 35.75 square units.


Practice Problems for Finding the Area of Compound Figures:


Questions: Find the area of the following compound figures.

Solutions: Compound Figure 1 = 28 square units.

Compound Figure 2 = 56 square units.

Solving Online Vertex of a Hyperbola

Hyperbola is a conic section.Hyperbola is all points found by keeping the whose difference from distances of two points (each of which is called a focus of the hyperbola) constant.The vertices of the hyperbola are located on the axis and are a unit from the center.Let us see some formulas for solving vertex of hyperbola online.

Formulas for Solving Online Vertex of a Hyperbola

Standard Formulas Of Hyperbola:

Equation of hyperbola  standard  formulas



1)The hyperbola  opens left  and right(horizontal axis) if the term `(x-h)^2`  is positive.

2)The hyperbola opens up and down(vertical axis) if the term is `(y-k)^2`

3)c= 

is the distance from the center (h,k) to each focus

4)asymptote have slopes given by.

5) The equation of asymptotes  are  given by since the asymptote contain center(h,k)

Example of Solving Online Vertex of a Hyperbola

Example 1:Find the vertex and center  for the given hyperbola.

9x^2-16y^2+18x+160y-247=0

Solution:

First put the equation into standard form.

9x² - 16y² + 18x + 160y - 247 = 0

Now complete the square.

9(x² + 2x + 1) - 16(y² - 10y + 25) = 247 + 9 - 400
9(x + 1)² - 16(y - 5)² = -144

Multiply thru by -1 since the right hand side is negative.

16(y - 5)² - 9(x + 1)² = 144

Set equal to one.

(y - 5)²/9 - (x + 1)²/16 = 1

Since y² is the positive squared term, the pair of hyperbolas open vertically up and down.

The center (h,k) = (-1,5).

a² = 9 and b² = 16
a = 3 and b = 4

The vertices are (h,k-a) and (h,k+a) or
(-1,5-3) and (-1,5+3) which is (-1,2) and (-1,8).

c² = a² + b² = 9 + 16 = 25
c = 5

The foci are (h,k-c) and (h,k+c) or
(-1,5-5) and (-1,5+5) which is (-1,0) and (-1,10).

Example 2:

Find the vertices of the given hyperbolic equation:y^2/16-x^2/9=1

Step 1:
center (0, 0)

Step 2:
a2 = 9; a= 3
b2 = 16; b = 4

Step 3:
Vertices = (0, 3)
= (0, -3)

Step 4:
Find c value
c2=a2+b2
c2 = 9 + 16
c2 = 25
c = 5

Focus = (0, 5)
= (0, -5)

The vertical hyperbola of the given equation is ((3, 0) (-3, 0)

Three Distance Theorem

The three-distance theorem states that there are at most three distinct gaps between consecutive elements in the set of fractional parts of the first n multiples of any real number. In a real world of three - dimensions, we must extend our knowledge of geometry ti three dimensional space.

To find the distance between two points whose co-ordinates are given. Let P( x1 , y1 , z1 ) and Q ( a2, b2, c2 ) be the two given points. Through P and Q draw planes parallel to the co-ordinate planes to form a rectangular box whose one diagonal is PQ. Geometrically to find the distance PQ is the computing of the length of the diagonal PQ of the box ( as shown in the figure) by means of Pythagoras theorem.



Since MQ is perpendicular to the plane PAMB and PM lies in the plane so MQ is perpendicular to PM `=>` `angle` PMQ = 900 .

in triangle PMQ, `angle` PMQ = 900 , therefore by pythagoras theorem we get,

PQ2 = PM2 + MQ2 ........(1)

since, AM is perpendicular to AP, so `angle` MAP = 900

In triangle AMP, `angle` MAP = 900 , therefore, by pythagoras theorem

PM2 = AP2 + AM2 .....(2)

From (1) and (2), we get

PQ2 = AP2+ AM2+ MQ2 ......(3)

Clearly from the figure

AP = x2 - x1 , AM = y2 -y1 , MQ = z2-z1

Substituting these values in (3) we get

PQ2 = (x2 - x1 ) + ( y2 - y1 ) + (z 2- z1)

PQ = `sqrt(( x_2-x_1)+ ( y_2-y_1) + (z_2-z_1))`

Distance from origin: The distance of P(x,1y,1z1) from the origin o(0,0,0) is `sqrt((x_1-0)^2 +( y_2 - 0)^2 + (z_2 - 0)^2)` = `sqrt((x_1)^2 + (y_1)^2 + (z_1)^2)`



Example Problems on three Distance Theorem:

Problem1: Find ' a' so that the distance between ( -2,1,-3) and ( 1, 3, -6) be 7 units?

Solution:

Let A, B be ( -2,1,3) ,( 1, 3, -6) respectively then

`=>` `sqrt((a-2)^2 + (3-1)^2 + (-6 + 3 )^2)` = 7

`=>` a2 + a + 4 + 4 + 9 = 49 `=>` a2 + 4a - 32 = 0

`=>` (a-4) (a+ 8) = 0 `=>` a= 4 , -8

Problem2: Find the distance between the points A(1,2,3) from he point B(3, 6, 8)?

Solution: The distance

`sqrt((3-1)^2 +(6-3)^2 + ( 8-3)^2)`

AB = `sqrt(2^2 + 3 ^2 + 5^2)`

AB = `sqrt(4 + 9 + 25)`

AB = `sqrt(38)` <br>

Practice Problems on three Distance Theorem

Problem2: Find the distance of the point P( 4, 6, 8) from the point Q ( 9,5,2)? ( Answer: 3`sqrt(7)` )

problem3: By using the distance formula prove that the points A( 3, -5, 1), B ( -1, 0 , 8) and C( 7, -10, -6) are collinear ("Answer : The sum of two side is equal to the third side so the points are collinear)

Dihedral Angle Calculator

Dihedral angle is the one of the special kind of angles in mathematical geometry.  Dihedral angle is an angle that forms between two plans.  Usually a plane is the three dimensional flat surface.  The points in the plane take the form of (x, y, z).  In this topic we are going to study about how to calculate the dihedral angle when the points of the plane are given through the dihedral angle calculator.

Explanation about Dihedral Angle through Calculator

Important Guidelines:

The following steps are the important guide lines to calculate the dihedral angle.

• First we have to choose the three points on the plane 1 and then choose the three points on the plane 2

• Enter the (x, y, Z) values for each point on the dihedral angle calculator we have to get the equation of the plane 1 and plane 2 separately

• The equation of the plane 1 is taking the form of A1x + B1y + C1z + D1 =0 and then the equation of the plane 2 is taking the form of A2x + B2y + C2z + D2 =0

Now we have to calculate the dihedral angle through the following formula,

cos`alpha` = `((A_1)(A_2) + (B_1)(B_2) + (C_1) (C_2))/ (sqrt((A_1)^2 +(B_1)^2 +(C_1)^2) * sqrt((A_2)^2 +(B_2)^2 +(C_2)^2))`

This formula is used for our manual calculation.

Example Problem on Calculate Dihedral Angle via Calculator

Calculate the dihedral angle for the plane 1 and plane 2.  The points on the plane 1 are (1, 2, 3) and (3, 2, 1) and then (2, 1, 3) and the points on the plane 2 are (4, 1, 2) and (1, 4, 4) and then (2, 4, 2)

Solution:

Now we have to substitute the values of the points in the following dihedral Angle Calculator,



This calculator gives the distance between each points and equation of the each plane and then the angle in between the planes in both degrees and radians.  These are the main usage of the dihedral angle calculator.

Distributive Law Definition

The distributive law is placed in set theory. The distributive law is the piece of sets that used to learn about the sets. Group of items is called the sets. Set theory have some operations union, intersection, difference, complement and performs some law.The set theory is used to select the number of objects simultaneously. Here we will see about the distributive law definition with examples.

Examples - Distributive Law Definition:

Now we are going to solve the examples for distributive law definition.

AU (B∩C) = (AUB) ∩ (AUC)

A∩ (BUC) = (A∩B) U (A∩C)

Example 1

Get the solution of the given sets by using distributive law. A={ a,b,c,d } B={ b,c,g } and C={ b,c,k,m }.

Solution:

The given sets are A={a,b,c,d} B={b,c,g} and C={b,c,k,m}.

Distributive law for set is as follows,

i)AU (B∩C)=(AUB) ∩(AUC)

ii)A∩(BUC)=(A∩B) U (A∩C)

i)AU (B∩C)=(AUB) ∩(AUC)

First find the solution left hand side.

AU (B∩C)

First we can solve the inner bracket set.

B∩C

Intersection means we choose the common elements from the set B and C.

So B∩C =   B= {b, c, g} ∩ C= {b, c, k, m}.

B∩C = {b, c}

Now find the AU (B∩C)

Grouping the values of A and B∩C

AU (B∩C) = {a, b, c, d} U (b, c)

= {a, b, c, d}

So AU (B∩C) = {a, b, c, d} -------------- (1)

(AUB) ∩ (AUC)

Groping the values of A and B sets.

A= {a, b, c, d} B= {b, c, g}

AUB = A= {a, b, c, d} U B= {b, c, g}

= {a, b, c, d, g}

Joining the values of A and C sets.

AUC:

A= {a, b, c, d} U C= {b, c, k, m}.

= {a, b, c, d, k, m}

(AUB) ∩ (AUC):

AUB ∩AUC = {a, b, c, d, g} ∩ {a, b, c, d, k, m}

(AUB) ∩ (AUC) = {a, b, c, d} ------------- (2)

So AU (B∩C) = (AUB) ∩ (AUC)

ii) A∩ (BUC) = (A∩B) U (A∩C)

A={ a,b,c,d } B={ b,c,g } and C={ b,c,k,m }.

Now solve the left side.

A∩ (BUC)

BUC:

= {b, c, g} U {b, c, k, m}

BUC = {b, c, g, k, m}

A∩ (B U C):

A= {a, b, c, d} ∩ {b, c, g, k, m}

= {b, c} -------------- (1)

(A∩B) U (A∩C)

A∩B= {a, b, c, d} ∩ B= {b, c, g}

= {b, c}

A∩C= A={ a,b,c,d } ∩ C={ b,c,k,m }.

= {b, c}

(A∩B) U (A∩C) = {b, c}

A∩ (BUC) = (A∩B) U (A∩C)


Example 2 – Distributive Law Definition:

A= {2, 7, 9} B= {5, 10, 11} C= {5, 8, 10}

What is AU (B∩C)

(B∩C):

= {5, 10, 11} ∩ {5, 8, 10}

= {5, 10}

AU (B∩C):

= {2, 7, 9} U {5, 10}

= {2, 7, 9, 5, 10}

These are examples for distributive law definition.

That’s all about distributive law definition.

Resultant of three Vectors

This article is about resultant of three vectors. Resultant of three vectors is vectors that results from adding two or more vectors together. There are two ways to calculate the resultant vector. There are many different websites to help the students on obtaining the resultant of three vectors. Tutor vista is the famous website to provide help on finding the resultant of three vectors with the help of highly qualified tutors. Below are some of the problems regarding resultant of three vectors.

Methods - Resultant of three Vectors:

There are two different methods to find the resultant vector.



The head to tail method:

The head to tail method is represented with the arrow mark. The arrow mark is the head and the tail is with out the arrow mark. Place the two vectors such that the head of the vectors joins the tail of another vector. Draw the resultant vector as shown in the figure above. To find the resultant vectors we use the Pythagorean theorem. There is also another method to calculate the resultant vector which is studied in college grade.

Parallelogram Method:

The another method is parallelogram method. To use this parallelogram method it is very important to be well know with trigonometry basics.


Example Problems - Resultant of three Vectors

vector addition is also the resultant of 3 vectors.

Example 1: The vector a = 3 and vector b = 4. Find the resultant vector using Pythagorean theorem.

Solution

We know that Pythagorean theorem is a2+b2 = c2

So given a =3 and b = 4

So the resultant vector c2 = a2 + b2

c =` sqrt (a^2 + b^2)`

= `sqrt (3^2 + 4^2)`

=` sqrt (9+16)`

= `sqrt(25)`

c = 5

So the resultant vector is 5


Example 2: Add 5`veci` +4`vecj`+3`veck` with 2`veci`+1`vecj`+0`veck`

Solution

Here two set of vectors are given. we have to perform the sum of two vectors by adding the magnitudes alone

5`veci` + 4`vecj` +3`veck`
2`veci` + 1`vecj` +0`veck`
----------------------
7`veci` + 5`vecj` + 3`veck`
-----------------------
Example 3: `veca =4vecp + 3vecq+2vecr` , `vec b = 5vecp +2vecq+vecr` and `vecc = 2 vecp + vecq + 4 vecr`
Solution
Resultant vector is `veca + vecb + vecc`
`4 vecp + 3vecq+2 vecr`
`5 vecp + 2vecq+ vecr`
`2 vecp + vecq + 4 vecr`
---------------------------------
`11 vecp +6 vecq + 7 vecr`
---------------------------------