Intercept Theorem

Introduction:

                  Intercept theorem plays a vital role in elementary geometry. It deals with the ratios of various line segments, which are created if two intersecting lines are intercepted by a pair of parallels. This theorem is equivalent to ratios of similar triangles. Traditionally it is attributed. The intercept theorem is  also called as Thale's Theorem. If a transversal line makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts.

Concept of Intercept Theorem:

                  If a transversal line makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts. AP || BQ || CR.

Types of intercept Theorem

                  If a transversal line makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts. AP || BQ || CR.

What Does Transversal Mean in Math

Introduction to Transversal in Math:

Definition:

A line that cuts (passes through) across two or more (usually parallel) lines then it is said to be a transversal.
It can also be defined as a line that intersects two or more co-planar lines each at a different point is also termed as a Transversal.

Transversal Postulates:
If two lines are parallel to each other and intersected by a transversal then the corresponding angles are congruent.

Some important points:

• Supplementary angles are the angles whose sum is equal to 180⁰.
• Vertical angles are the angles that are not adjacent angles and are formed by two intersecting lines and are always equal to each other.

Properties of transversal of parallel lines:

• If two parallel lines are cut by a straight line (transversal), then the corresponding angles around each intersection are equal in measure or we can say mathematically as these angles are congruent.
• If two parallel lines are cut by a straight line (transversal) then the alternate interior angles are congruent.
• If two parallel lines are cut by a straight line (transversal) then the interior angles on the same side of the transversal are supplementary.

Proofs for transversal by solved examples

Solved Problems to prove the above given points:
Ex 1: In the given figure, If the angles 2 and 3 are congruent then prove that r and s are parallel.

parallel lines 

Sol: Step 1: Given in the problem that angle 2 = angle 3
       Step 2: From the data given above angle 1 and angle 2 are vertically opposite angles and are congruent.
                               therefore, angle 1 = angle 2
       Step 3:The transitive property says that if a=b and b=c then a=c
 Using this property we have,
          angle 2 = angle 3    (given in the problem)
          angle 1 = angle 2    (vertically opposite angles)
so ,    angle 1 = angle 3   
As the angles made by the transversal with the two lines are equal the two lines are parallel.
Ex 2: Show that the transversal shown in the figure cuts the parallel lines.

Sol: In the above figure the angles 4 and 3 are known as alternate interior angles.Accoding to above discusssed points the alternate interior angles are congruent.
           `=>`    angle 4 = angle 3
So, the two lines are parallel lines.
Ex 3: Prove that the lines a and b shown in the figure below are parallel.

Sol: Clearly it is shown that the angles 1 and 2 are equal to` 90@` . That means that they are supplementary angles based upon the above discussion.
As the  angles made by the transversal with the two straigth are supplementary(equal) the two lines are parallel.Hence, proved.

Practice Problem on traversal line

Pro: Find the measures of the unknown angles in the following figure.Given r is parallel to s and angle 1 = `60@.`

Ans:  angle 2 =`120@.`
angle 3 = `60@` .
angle 4 = `120@`
angle 5 = angle 1 = 60⁰
angle 6 = angle 2 = 120⁰
angle 7 = angle 3 = 60⁰
angle 8 = angle 4 = 120⁰

Solving Genetics Probability Problems

In probability method, we have to decide the basic terms of probability.. The main aim of genetics probability subject is to give the fundamental properties the importance of their use with some basic examples of genetics. In a genetics probability, the basic term and properties is not easier to understand but work with sums we will start to get comfortable with in genetics probability.


Basic Definitions of genetics probability:

In a genetic probability method, an testing method is continuously repeated for infinite numbers of time and it happened by the expected probability in the genetics

Genetics probability solving problems:

Problem 1:

To solve consider a locus with two chromosome, C and c. If the frequency of CC is 0.64, what is the frequency of A under Hardy-Weinberg?

Solution:

Solving methods of genetic probability,

Under H-W, if q = freq(C) , then q2 = freq(CC), hence q2 =0.64 or q = 0.8.

Problem 2:

To solve if the genotypes CC, Cc, and cc have frequencies 0.7, 0.49, and 0.49 (respectively), what are q = freq (C)? r = freq (c)? After a single generation of random mating, what is the expected frequency of CC? Of Cc? Of cc?

Solution:

Solving methods of genetic probability,

p = freq (CC) + (1/2) freq (Cc) = 0.7 + (1/2)(0.49) = 0.945

q = 1-p = 0.945

freq (CC) = p2 = 0.9452 = 0.8930

freq (Cc) = 2qr = 2*0.945*0.8930 =1.6877

freq (cc) = r2= 0.89302= 0.797449

Problem 3:

To solve the suppose 60 out of 850 women’s are green eyes. What is frequency of green eyes? If a random women’s is chosen, what is the probability they are a green eyes?

Solution:

Solving methods of genetic probability,

Freq (Redheads) = 60/850 = 0.070 or 7.05 percent

Probability of a Redhead = 30/850, or 7.05 percent

Problem 4:

Solve the genotype cc is lethal and yet population has an equilibrium frequency for c of 30.  Here the Cc is the fitness and its value is one.  Find the CC genotype fitness?

Recall if the genotypes CC: Cc: cc have fitness 1-s : 1 : 1-t, then the equilibrium frequency of A is s/(r+s).

Here, s= 1 (as cc is lethal), so that freq(C) = 1- freq(c) =0.6 = 1/(1+s)

Solving gives 1+s = 1/0.8, or s = 1/0.8 -1 = 3.5

Hence fitness of CC = 1-s = 1-3.5=-2.5.

Learn Derivative at a Point

Learn derivative at a point involves the process solving derivative problems at an exam point view. The rate of change of the given function is determined with the help of calculus. To find the rate of change the calculus is divided into two types such as differential calculus and integral calculus. At exam point of view the derivative is easily carried out by differentiating the given function with respect to input variable. The following are the solved example problems in derivatives at exam point of view for learn.

Learn derivative at a point example problems:


Example 1:

Find the derivative for the given differential function.

f(e) = 3e 3 -4 e 4  - 5e

Solution:

The given function is

f(e) = 3e 3 -4 e 4  - 5e

The above function is differentiated with respect to e to find the derivative

f '(e) = 3(3e 2 )-4(4e 3  ) - 5

By solving above terms

f '(e) = 9e 2 -  8e 3 - 5

Example 2:

Compute the derivative for the given differential function.

f(e) = 6e6 - 5 e5 - 4 e4 - e

Solution:

The given equation is

f(e) = 6e6 - 5 e5 - 4 e4 - e

The above function is differentiated with respect to e to find the derivative

f '(e) =  6(6e 5)  -5 (5 e4 ) -4(4 e3) - 1

By solving above terms

f '(e) =  36e 5  - 25 e4  - 16 e3 - 1

Example 3:

Compute the derivative for the given differential function.

f(e) = 2e 2 -4 e 4  - 15

Solution:

The given function is

f(e) = 2e 2 -4e 4  - 15

The above function is differentiated with respect to e to find the derivative

f '(e) = 2(2e  )-4(4 e 3 ) - 0

By solving above terms

f '(e) = 4e -16e3

Example 4:

Compute the derivative for the given differential function.

f(e) = 5e5 -4e 4 -3e 3  - 2

Solution:

The given function is

f(e) = 5e5 -4e 4 -3e 3  - 2

The above function is differentiated with respect to e to find the derivative

f '(e) = 5(5e 4 )-4(4e 3 ) -3( 3e 2) -0

By solving above terms

f '(e) = 25e 4 -16e 3  -9 e 2


Learn derivative at a point practice problems:


1) Compute the derivative for the given differential function.

f(e) = 2e 3 -3 e 4  - 4 e 5

Answer: f '(e) = 6e 2 -12 e3 - 20 e 4

2) Compute the derivative for the given differential function.

f(e) = e 3-e5 - 4 e 6

Answer: f '(e) = 3e2 - 5e4 - 24 e 5

Linear Algebra Test

Linear function is a function with polynomial degree 1. This linear function matches a dependent variable with independent variable and maintains a relation in a simpler way. The linear algebra is the important method which is used in the linear function, variable function, linear equation, vector, matrix etc. This linear function is used in the high school and college students are using the linear function by using the linear algebra.

Sample Problem for linear algebra:


f is a linear function. Value of x and f(x) are given in the table below; complete the table using linear algebra

X                           f(x)

-1                          11

1                           –

–                           1

6                          -12

7                         --

–                          -35

Solution:

f is a linear function in linear algebra whose formula has the form to test is

f(x) = a x + b

where a and b are constants to be test and found. Note that 2 ordered pairs (-2,18) and (5,-19) are given in the table. These are two ordered pairs to test are used to write a system of linear equations as follows

11 = - 1 a + b and -12 = 6 a + b

Solve the above system to obtain a = - 3 and b = 2 and write the formula for function f as follows

f(x) = - 3 x + 2

We now use the formula for the f to find f(x) given x or find x given f(x).

for x = 0 , f(0) = -3(0) + 2 = 2

for f(x) = 1 , 1 = -3 x + 2 which gives x = 1/3

for x = 7 , f(7) = -3(7) + 2 = - 19

for f(x) = - 30 , -30 = -3 x + 2 which gives x = 32/3

We now put the values of the calculated above in the answer.

X                           f(x)

-1                          11

1                           2

1/3                         1

6                         -12

7                         -19

32/3                      -35


Practice problem:


Problem 1:
f is a linear function in linear algebra. Value of x and f(x) to test are given in the table below; complete the table.

X                              f(x)

-5                             15

1                               –

–                              1

10                         -14

8                             --

–                           -35


Problem 2:
f is a linear function. Value of x and f(x) to test are given in the table below; complete the table.

X                               f(x)

-6                             18

1                              –

–                             1

9                           -20

10                             --

–                             -40

Exponentiation Tutoring

This article is a online exponentiation tutoring.

Generally the exponentiation is used in all sciences including the mathematical, physical, chemical or even in biological sciences but it is assumed or preferably called a mathematical operation which generally written as an it is clear from this notation that the exponentiation involves two numbers one is called the base i.e. a and other is called the exponent i.e. n. When n is any positive integer then the exponentiation simply means the repeated multiplication of the number ‘a’ to itself n times which are written as:

 

Generally, exponent is the superscript to the base. Exponentiation above will be read as a raised to the power n or exponent of n, or a to the n. A few exponents have their pronunciation: for instance, a2 is read as a square and a3 as a cube.

Exponentiation comes to use in economics or biology or chemistry or physics or computer science to calculate compound interest or population growth.


Example problems on exponentiation tutoring:


Positive integer exponents:

a2 is pronounced as a square because any square with side a has area equal to a2. The expression a3 is referred to as the cube of a since the volume of a cube with side length a is a3.

So 32 is called as "three squared", and 23 is "two cubed".

The numbers with positive integer exponents may be defined by the initial condition

a1 = a

thus in general the recurrence relation

an+1 = a·an.


Exponents of one and zero:


If a is not zero and n an m are two positive integer exponents such that the exponent (n – m) is also positive then we can say that



If we consider a special case in which the exponent’s n and m are equal then we can wrote the equality as

 

Thus we lead to the following rule: Any number with raised to the power 1 is the number itself.

Solving Quadratic Formula

Quadratic Equation:

An equation which has one or more terms are squared but no higher power in terms, having the syntax, ax2+bx+c =0 where a represents the numerical coefficient of x2, b represents the numerical coefficient of x, and c represents the constant numerical term.

Types of quadratic equation

Pure quadratic equation:

The numerical coefficient cannot be zero. If b=0 then the quadratic equation is called as a ‘pure’ quadratic equation

Complete quadratic equation:

If the equation having x and x2 terms such an equation is called a ‘complete’ quadratic equation. The constant numerical term ‘c’ may or may not be zero in a complete quadratic equation. Example, x2 + 5x + 6 = 0 and 2x2 - 5x = 0 are complete quadratic equations.


Quadratic Equation Formula


The quadratic equation has the solutions ax2+bx+c =0

x =√(b2-4ac)/2a

Consider the general quadratic equation

ax2+bx+c =0

With a`!=` 0. First divide both sides of the equation by a to get

x2+b/a x + c/a =0

which leads to

x2+ b/a x = - c/a

Next complete the square by adding ((b)/(2a) )2to both sides

X2+ ((b)/(a) )x+((b)/(2a) )2 = -((c)/(a) )+ ((b)/(2a) )2

(x+(b)/(2a) )2=-((c)/(a) ) + ((b^2)/(4a^2) )

(x+(b)/(2a) )2 = (b^2-4ac)/(4a^2)

Finally we take the square root of both sides:

x+(b)/(2a) = +-(sqrt(b^2-4ac))/(2a)

or

x =-(b)/(2a)  +-(sqrt(b^2-4ac))/(2a)

The final form of Quadratic Formula is

x =-b+-sqrt(b^2-4ac)/(2a)

The two roots of the equation is

``          -b-(sqrt(b^2-4ac))/(2a)

-b+(sqrt(b^2-4ac))/(2a)


Example Problem on solving Quadratic formula


Example:

Find the roots of the equation by quadratic formula method, x2-10x+25=0

Solution:

Step 1:  From the equation, a = 1, b = - 10 and c = 25.

Step 2:  To Find X:
plug-in the values in the formula below
x = ``

Step 3:  We get the roots, x = ``
x = 5 and x = 5
which means x1 = 5 and x2 = 5.

Here x = 5 is root of the equation.