Vertices of a Hyperbola

Introduction to vertices of a hyperbola:

A hyperbola is defined as a smooth curve which tends to lie down in the plane and it can be described by the geometric properties or by the type of equation for which it may be the solution set. A hyperbola can be defined in the form of a curve that is represented by the function f(x) = 1 / x in the Cartesian plane.

In this article we are going to see about the vertices of a hyperbola.

How to Find the Vertices of a Hyperbola:

In the case of ellipse, there exists a relationship between x, y, and z axes with ellipses, the computations are long and painful. So that, for hyperbolas (where x < z), the relationship is z2 – x2 = y2 or, which means the same thing, z2 = y2 + x2.  This has been proved using the Pythagorean Theorem.
When the transverse axis is in the horizontal plane (when the center, and vertices line up side by side which is parallel to the horizontal axis), then the x2 goes with the part of the hyperbola's equation, and the y part is subtracted.
((a – y) 2 / x2) – ((b – c) 2 / y2) = 1.

Example to Find the Vertices of a Hyperbola:

Example 1: Given a hyperbola (x2 / 16) – (y2 / 4) = 1. Find the vertices of the given hyperbola.

Solution:

The given hyperbola opens right or left because it is of the form x – y

Here a2 = 16, b2 = 4, c2 = 16+4 = 20.

Hence a = 4, b = 2 and c = 4.5

The vertices of a hyperbola are given by (a, 0), (-a, 0)

Thus the vertices are (4,0), (-4,0).

Example2: Given a hyperbola (y2 / 25) – (x2 / 9) = 1. Find the vertices and foci of the given hyperbola.

Solution:

The hyperbola opens up and down because it is of the form y – x.

Here a2 = 25, b2 = 9, c2 = 25 + 9 = 34.

a = 5, b = 3, c = 5.8

The vertices of a hyperbola are given by (0, a), (0,-a)

Vertices of a hyperbola: (0, 5) and (0,-5)

The foci of a hyperbola are given by (0, c), (0,-c)

Foci of the hyperbola: (0, 5.8), (0,-5.8)

Acute Triangle Calculator

Introduction about triangle:
A triangle is three side polygons. Triangle is one of the important shapes in geometry.

Types of triangle

Equilateral triangle
Scalene triangle
Right triangle
Isosceles triangle
Acute triangle
Obtuse triangle
In this article we shall discuss about the acute triangle. A triangle has three sides and three angles. A triangle angles are less than 90 degree, such an angle is called acute triangle.

Acute means the angle should be less than 90 degree.

To find area of the acute triangle = `1/2` * base * height

Acute Triangle Calculator:




In calculator you have to enter the value of height and base of the triangle after pressing the calculate button area of the triangle is calculated and display on answer box.

Problem 1:

Calculate the area of the acute triangle whose base is 10cm and height is 12 cm

Solution:

Area of the acute triangle = `1/2` * base * height

Base =10 cm Height = 12 cm

Area = `1/2` * 12 *10

= 60 cm2

The answer is 60 cm2

Problem 2:

Calculate the area of the acute triangle whose base is 8cm and height is 11 cm

Solution:

Area of the acute triangle = `1/2` * base * height

Base = 8 cm Height = 11 cm

Area = `1/2` * 8 *11

= 44 cm2

The answer is 44 cm2

Problem 3:

Calculate the base of the acute triangle whose area is 121cm2 and height is 6 cm

Solution:

Area of the acute triangle = `1/2` * base * height

Area =144 cm2 Height = 6 cm

Area = `1/2` * b *16

121 = `1/2` * b *16

B=121*`2/16`

B= 15.13 cm

The answer is 15.13 cm

Problem 4:

Calculate the height of the acute triangle whose base is 12cm and area is 196 cm2

Solution:

Area of the acute triangle = `1/2` * base * height

Base =12 cm Area = 196 cm2

Area = `1/2` * 12 *h

196 = `1/2` * 12 *h

h=196*`2/10`

B=39.2 cm

The answer is 39.2cm

Normal Sum Distribution

Introduction to normal sum distribution:

Let us see about normal sum distribution,

The normal distribution is also known as the normal probability distribution, happens to be the mainly helpful theoretical distribution for continuous variables. A lot of statistical data sum regarding business and economic problems are display in the type of normal sum distribution.In fact normal distribution is known as the ‘corner stone’ of  Modern statistics.Just like the Poisson distribution, the normal sum  distribution  may also be regarded as a limiting case of binomial distribution.

Definition of Normal Sum Distribution:

Let we see about definition of normal sum distribution,

A variable X is told that a normal distribution with parameter µ and s (or µ and s2) if the probability function is,

Here, Z =(X – µ) s

Constants of Normal distribution:

Mean = µ

Variance = s2

Standard deviation = s

The graph sum of the normal curve is shown below,

The distribution with µ = 0 and s2 = 1 is known as the standard normal distribution.

Understanding problem solving with proportions is always challenging for me but thanks to all math help websites to help me out.

Normal Sum Distribution-example Problems:

1) If X is normally distributed having mean 7 and standard deviation 6 find. P(0 = X = 10).

Solution :

Given, µ = 7, s = 6

(i) P(0 = X = 10)

We know that Z =(X – µ) / s

When X = 0, Z =(0 – 7) / 6 = -1.16

When X = 10, Z =(10 – 6) / 6 = 4/6 = 0.67

? P(0 = X = 10) = P(-1.2 < Z < 0.67)

= P( 0 < Z < 1.2) + P( 0 < Z < 0 .67) (due to symmetry)

= 0.3849 + 0.258

= 0.516

2) If X is normally distributed having mean 10 and standard deviation 8 find. P(0 = X = 20).

Solution :

Given, µ = 10, s = 8

Find, P(0 = X = 20)

We know that Z =(X – µ) / s

When X = 0, Z =(0 – 10) / 8 = -1.3

When X = 10, Z =(20 – 10) / 8= 10/6 = 1.3

? P(0 = X = 20) = P(-1.3 < Z < 0.3)

= P( 0 < Z < 1.3) + P( 0 < Z < 0 .3) (due to symmetry)

= 0.4032 + 0.4032

= 0.8064

Constant Rate of Change Math

Introduction to constant rate of change math:

Rate on which reliant output modify by value to vary within independent input is describing the rate of change. Functional association be usually indicated through y = f(x). Alteration within y is indicated through ?y also alteration within x is indicated through ?x. Consequently, as of the function y = f(x), we contain y + ?y = f(x + ?x).  In addition to constant rates of alteration are known through method, ?y / ?x.

Constant Rate of Change in Math:

In math consider the function is the linear function that is graph of y also x is a line after that the function is known through y = mx + x. In linear function rates of change is a constant which is known through its slope, m. Slope of the line can be find out through the following

m = ?y / ?x =  (y2 – y1/(x2 – x1)       

In math consider the function is non-linear then its rates of alteration are dissimilar on dissimilar points. Therefore, for such functions derived are estimated through distinguish the function toward get the rates of modify. Derivative of y through value to x can be written as dy/dx.

Examples for Constant Rate of Change Math:

Example 1

Solve the rate of change for the line to exceeds through points (5, 4) and (7, 10)

Solution:

Here, x1= 5, x2=4, y1=7 and y2=10

so, slope m = (y2 -  y1)/(x2- x1)

m = (10 - 4)/(7 - 5) = 6/2 = 3

Consequently, rate of alteration otherwise slope of line is 3.

Example 2

Solve the rate of change of function, f(x) =4x2 + 10x at x = 6

Solution:

f(x) =4 x2 + 10x

Derivative f '(x) = 8x + 10

Substitute x = 6 in f '(x) which gives,

f '(6) = 8*6 + 10 = 48 + 10 = 58

Therefore, rate of change of function f(x) at x = 6 is 58.

Example 3

Solve the rate of change of function, f(x) = 3x3 + 3x2 - 7x at x = 3

Solution:

f(x) =3x3 + 3x2 - 7x

Derivative f '(x) = 9x2+6x -7

Substitute x = 3 in f '(x) which gives,

f '(6) = 9(32)+6(3) - 7 = 81 + 18-7 = 92

Therefore, rate of change of function f(x) at x = 3 is 92.

Precalculus Math Final Help

Introduction to Precalculus Math Final Help:

The Precalculus is one of the foundation class of mathematics. The topics involved in Precalculus test are real  complex numbers, binomial theorem, composite & polynomial functions, vectors, parametric equations, polar coordinates, matrices, rational functions, solving inequalities & equations and trigonometry. In this article we shall discuss about the examples involved in Precalculus math final help. The following are the examples involved in Precalculus math final help.

Precalculus Math Final Help Problems:

Example 1:  Express the real and imaginary parts for 28 - i `sqrt(14) `

Solution:        Let z= 28 - i `"sqrt(14)`      

Re(z) = 28     Im(z) = `sqrt(14)`

Example 2: For what value of m the vectors `veca` and `vecb` are perpendicular to each other

`veca` =m`veci` + `vecj`` + ``veck`   `vecb`= 7`veci` - 9 `vecj` + 2 `vecj`

Solution: Given `veca` `_|_``vecb `

So, `veca` `vecb` = 0

=> (m`veci` +  `vecj`` + ``veck`) . (7`veci` - 9 `vecj``veck` )

=>  7m - 9 + 2

=>  7m  = 7

m =`7/7`  =1

Example 3: Solve for x,  ln(x-3) + ln(x-2) = ln(2x -6 )

Solution:   Simplify left side, So

ln(x-3) + ln(x-2)= ln(x - 3) (x - 2)

Therefore the given equation can be rewritten as

ln(x - 3)(x - 2) =  ln (2x - 6)

Taking exponent on above equation

eln(x - 3)(x- 2) = eln(2x - 6)

Simplifying the above equation

(x - 3) (x - 2)  = (2x - 6)

x2  - 5x + 6 = 2x - 6

x2  - 7x +12 = 0

Factoring the above equation, we get

(x - 4) (x - 3) = 0

When products of two factors are equal to 0, then one of the factors must be equal to 0.

So, x-4 = 0 => x=4

x-3 = 0 => x=3

Here x = 4 & 3 is the solution.

Example 4:   If f(x) = 14x-1, find f-1(y)?

Solution:Let y = 14x - 1

The given equation can be rewritten as

y + 1 =14x

=> x = y + 1
14

Therefore the f-1(y) = y + 1
14

Example 5: Write the given number in complex form `sqrt(-69)`

Solution: `sqrt(-69)` = `sqrt((-1)(69))`

= `sqrt(-1)` * `sqrt(69)`

= i `sqrt(69)`

Precalculus Math Final Help Practice Problems:

Problem 1:Write the given number in complex form `sqrt(-1415)`

Answer: = i `sqrt(1415)`

Problem 2:Write the real and imaginary parts for 89 - i `sqrt(75) `

Answer:   Re(z) = 89     Im(z) =`sqrt(75)`

Problem 3:   If f(x) = x-9, find f-1(y)?

Answer: y + 9

Radical Perfect Prime Numbers

Introduction to radical perfect prime numbers:
A radical perfect prime number is a symbol and it’s denoted as root of a number, the thought of a radical the root of a number can best be understood by first tackling the design of exponentiation, or raising a number to a given power. We point to a number raised to the nth power by writing xn. This expression denotes that we are multiplying x by itself n number of times. Example for radical perfect prime numbers, v5 =2.236, and v3 = 1.732. a radical perfect prime number is a number that has exactly 2 factors...1 and itself , and one thing is 1 is not a prime number ,so  that 2,3,5,7,11,13,17,19,23 etc  are all primes.

Examples for Radical Perfect Prime Numbers:

Addition for radical perfect prime number:

Example 1: Find the value of 2v3 +3 v3

Solution:

=v3 (2+ 3)
=5v3

Here the root is square root, only take the base are same so that we can add the roots easily.

So the result is =5v3

Example 2: Find the value of v11 +6 v23+5v11 +7 v23

Solution:

= v11 (1+5)+ v23 (6+7)

= v11 (6)+ v23 (13)

= 6 v11 + 13 v23

Here the root is square root, the base are same in different terms so that we can add the roots easily.

So the result is 6 v11 + 13 v23

Examples for different radical perfect prime numbers:

Example for square root:

v3 = 1.732

v5 = 2.23

v7 =  2.64

v11 = 3.31                              

Some other Examples for Radical Perfect Prime Numbers:

Multiplication for radical perfect prime numbers:

Example 1:

Solution:

To find multiplication of v3x . v1y    

= v(3 x y)            

=  v3x y   

The result is =  v3x y    

Both the radicals have an same index of two, so we are multiply mutually under one radical keeping the 2 as its index number.

Example 2:

Solution:

To find multiplication of v(23xy2). vy    

= v(23 x y3)            

= v(23 x y3)

The result is v(23 x y3)

Both the radicals have an same index of two, so we are multiply mutually under one radical keeping the 2 as its index number.


Factorization for Radical Perfect Prime Numbers Examples:

Example 1:

To find the value of v75

v75 = v5*3*3

= 3v5

Radical perfect prime factors for v75 = 3v5

So the result is in the form of prime numbers and radical.

Integral Trapezoidal Rule

Introduction to integral trapezoidal rule :

The integral trapezoidal rule is a method which has been used to calculate the definite integral that has been given by

a
∫ f(x) dx
b

A definite integral is defined as the integral which has both the upper and the lower limits. In this topic we are going to see about the integral trapezoidal rule. The integral trapezoidal rule is also called as the trapezium rule or the trapezoid rule.

Integral Trapezoidal Rule:

The trapezoidal rule has been used by considering the region that has been under the graph of the given function f(x) as a trapezoid and then finding its area.

a
∫ f(x) dx ≈ (b-a) f(a)+f(b)
b                        2

The next step is to split the interval of the integration (a, b) into a number of smaller intervals and then apply the trapezoidal rule on every subintervals. This is given by

a                                            n-1
∫ f(x) dx ≈ (b-a)   (f(a)+f(b))/2   + ∑ f(a+k((b-a)/n)
b                 n                         k=1

This integration can also be written as

a
∫ f(x) dx ≈ ((b-a)/2n)(f(x0)+2f(x1)+2f(x2)+…+2f(xn-1)+f(xn))
b

Where the value of xk = a + k(b-a)/n, for the values of k = 0, 1, …. N

Between, if you have problem on these topics Derivative of Parametric Equations, please browse expert math related websites for more help on Integration by Parts Table.

Error Analysis of the Integral Trapezoidal Rule:
The error of the integral trapezoidal rule is given by the difference of the integral and the result.

b                                                n-1     
Error = ∫ f(x) dx -  b-a   (f(a)+f(b)/2) + 2   + ∑ f(a+k((b-a)/n)
a                 n                             k=1

This can be simply written as

Error = - (b-a)3    f′′(ξ)
12n2

Integral trapezoidal rule for non uniform intervals:

Let us consider the given values are x1 ….xn and y1,…..yn, then according to the integral trapezoidal rule

n
∫ydx ≈ ½ ∑ (xi – xi-1)(yi+yi-1)
i=2

The integral trapezoidal rules are said to be exact when the periodic functions are said to be integrated with their periods and the trapezoidal rule converges very rapidly for the periodic functions. When the function is said to be periodic and it integrates over one complete period then there exists a number of graphs which makes the error of the trapezoidal rule to be cancelled.