Math Homework Help
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Math Homework Help

Finding The Way To Freedom

published on November 14th, 2006 . by Vanaja

Today let’s relax a bit.

A peasant is convicted. He gets the death penalty. The judge allows him to say a last sentence in order to determine the way the penalty will be carried out. If the peasant lies, he will be hanged, if he speaks the truth he will be beheaded. The peasant speaks a last sentence and to everybody surprise some minutes later he is set free because the judge cannot determine his penalty.

What did the peasant said?

The peasant said: “I shall be hanged!”

If the peasant was lying, he would be hanged. But that’s what the peasant was saying. So he speaks the truth. But if he speaks the truth, he would be beheaded, so then he was not speaking the truth. So it is impossible for the judge to determine whether the peasant speaks the truth or not. So therefore the judge cannot determine the penalty and sets the peasant free.

Don’t go away……. I need your help…….

My 2 year old son put a coin in an empty bottle and insert a cork into the neck of the bottle.
The bottle has a very nice shape. So I don’t want to break it.
How could I remove the coin without taking the cork out ?

Answer

Don’t worry for a solution. He himself found a way and took it out. He pulled the cork into the bottle and the coin came out.

Rational Root(Zero) Theorem

published on November 9th, 2006 . by Vanaja

Roots of a Polynomial

A root or zero of a function is a number that, when plugged in for the variable, makes the function equal to zero. Thus, the roots of a polynomial P(x) are values of x such that P(x) = 0.

The Rational Zeros (Roots) Theorem

We can use the Rational Zeros Theorem to find all the rational zeros of a polynomial.

  1. Arrange the polynomial in descending order
  2. Write down all the factors of the constant term. These are all the possible values of p.
  3. Write down all the factors of the leading coefficient (The coefficient of the first term of a polynomialwhen writing in descending order.)
    These are all the possible values of q.
  4. Write down all the possible values of p/q . Remember that since factors can be negative p/q, and - (p/q)must both be included. Simplify each value and cross out any duplicates.
  5. Use synthetic division or remainder theorem to determine the values of p/q for which P(p/q) = 0. These are all the rational roots of P(x).

Example:

Find all the possible rational roots of

Mysterious Number 1089

published on November 8th, 2006 . by Vanaja

  1. Think of a 3 digit number where all the digits are different.
  2. Reverse the digits
  3. Subtract the smaller number from the larger number
  4. Reverse the digits again
  5. Add the numbers in steps 3 and 4

Whatever 3 digit number you choose the number will always be 1089!

Here is an example

  1. 543
  2. 345
  3. 543 - 345 = 198
  4. 891
  5. 891 +198 =1089

Online Mathematics Textbooks

published on November 6th, 2006 . by Vanaja

I have found a site that offers many free math ebooks. Most of them are in PDF format. If you don’t have a PDF reader download it here.
Here is the link to free Online Mathematics Textbooks.

Some useful formulea from Analytical Geometry

published on November 6th, 2006 . by Vanaja

Approach To School Affects How Girls Compare With Boys In Math

published on November 2nd, 2006 . by Vanaja

More women are pursuing higher education and doctoral degrees than ever before, but women still are rare in the math-oriented professions. Yet, researchers say, girls perform just as well as boys on achievement tests and tend to earn better grades in math than do boys during the earlier school years.

A new study in the journal Developmental Psychology indicates that how girls and boys approach their schooling underlies the differences in math grades. It also suggests that although the girls’ approach to school may give them an edge in the grades they earn in math, it may not buy them much when it comes to math scores on achievement tests because girls are not more confident than the boys about their skills in math.

The study examined 518 boys and girls as they went through fifth and seventh grades in three primarily white, middle- to upper-class school districts in Illinois. Using children’s reports, researchers looked at how the children approached their schoolwork, including their goals and in-class behavior. The children also reported on how confident they were about their ability to do well in math. Researchers also reviewed the young students’ grades and achievement test scores in math.

In the classroom, girls outperformed boys at both time points of the study, with the girls’ grades rising over time, while the boys’ grades remained the same, said Eva Pomerantz, a professor of psychology at the University of Illinois at Urbana-Champaign. The study was part of the doctoral work done by Gwen A. Kenney-Benson, who now is at Allegheny College in Pennsylvania.

Concerns with outperforming others and engaging in disruptive behaviors while in the classroom, both of which characterized boys more than girls, were tied to lower grades in math by the researchers.

“This was due in large part to the fact that such competitive and disruptive leanings were associated with decrements in learning strategies such as preparing for tests, seeking help, and persisting even when things were challenging that led to higher grades,” Pomerantz said.

Girls consistently used these learning strategies more than the boys did, the researchers found. It appears that, in contrast to boys, girls are more concerned with learning than with outperforming their classmates. They also engaged in less disruptive classroom behavior. As a consequence, girls used more focused learning strategies, giving them an edge over boys in terms of grades, Pomerantz said.

The researchers noted that the differences in grades between girls and boys disappeared once children’s concerns with learning versus outperforming others, engagement in disruptive behavior and learning strategies were taken into account.

At achievement test time, however, girls’ lost their advantage in math; their scores were the same as those of boys. After examining various factors, what stood out, Pomerantz said, was children’s confidence in their ability to do well in math.

In the classroom, she said, children may be less likely to feel that they will be judged based on their gender, believing instead that their own behavior, knowledge and effort will determine their grades. Thus, she added, the girls’ approach to schoolwork will pay off in the classroom, while the boys’ approach will not. It also could be, the researchers theorized, that higher grades given to the girls reflect rewards from their teachers for better behavior.

During achievement tests, the researchers suggest, the environment changes. Removed for girls is the familiarity of the classroom, which is replaced with uncertainty and increased stress. In such a situation, confidence mattered more than in the classroom.

Because confidence was found to be a predictor of scores on math achievement tests, Pomerantz said, girls may not have kept the edge they had while in the classroom because confidence levels did not differ along gender lines.

It may be that while many girls are going on into higher education, they continue to steer away from “stereotypically masculine fields, such as science and engineering” because the “more competitive environment of these fields is not a good fit with how girls approach school,” the researchers wrote.

“Consequently, even if the topic is of interest,” Pomerantz said, “the girls’ more learning-oriented approach may not match the work environment, where the atmosphere in these fields may provide a better fit to boys’ more competitive approach.”

Courtesy: Science News

Distance Formula

published on November 1st, 2006 . by Vanaja

Distance formula is useful to find the distance between any two given points in a cartetian plane.

Remainder Theorem and Factor Theorem

published on October 30th, 2006 . by Vanaja

Remainder Theorem
Let p(x) be any polynomial of degree n>0 ,and a any real number. If p(x) is divided by ( x-a), then the remainder is p(a).

The Remainder Theorem can be proved as follows.

Proof
Let us suppose when p(x) is divided by ( x-a), the quotient is q(x) and remainder is r(x).

So we have,
p(x)=(x-a) q(x)+r(x), where r(x)=0 or degree of r(x)< degree of x-a.

Since degree of ( x-a) is1, either r(x)=0 or degree of r(x)=0
So r(x) must be a constant,say r.

Thus for all values of x,
p(x)=(x-a) q(x)+r ……..(1), where r is a constant.

In particular, when x=a,
p(a)=0.q(x)+r
=0+r
=r
Hence the theorem.

Factor Theorem
Let p(x) be a polynomial of degree n>0. If p(a)=0 for a real number a, then (x-a) is a factor of
p(x). Conversely, if (x-a) is factor of p(x), then p(a)=0.

Proof

First part:
Let p(a)=0
Then by remander theorem, r=0
So equation (1) becomes
p(x)=(x-a) q(x)

==>(x-a) is a factor of p(x).

Second Part:
By remainder theorem,

p(x)=(x-a) q(x)+r
ie. p(x)=(x-a) q(x)+p(a)
Since (x-a) is a factor, p(a) must be zero.
This proves the theorem.

Example
Find the remainder when p(x)= x^2 +3x+1 is divided by x+1.

Determine whether( x-2) is a factor of p(x) or not.

Solution
x+1= x-(-1)
[Always check whether the divisor is in the form of (x-a)or not. Otherwise rewrite that in the form of(x-a)]

So, here a=-1
There fore by remainder theorem, the required remainder is
p(a)= p(-1)
=(-1)^2+3(-1)+1
=1-3+1
=-1

We know, by factor theorem,

if (x-2) is a factor of p(x), then p(2) must be zero.

Here p(2)=2^2+3(2)+1
=4+6+1
=11 which is not equal to zero.

So (x-2) is not a factor of p(x).

An Interesting E-Mail

published on October 29th, 2006 . by Vanaja

Today I received an interesting E-Mail. Here is the Mail and the reply I have sent.

Plot the point with coordinates (3, –5).

Graph 2x – y = 4.

Graph using the intercept method: 2x + y = 4.

Write the equation of the line with slope 4 and y-intercept (0, –5). Then
graph the line.

Math For You to P

Hi P
Thanks for writing to me
I think you have four questions. So my fee willbe $6. If that is O.K for you, I will send the detailed solutions for you.

P to me
You don’t have to send me a solution for my problems because the site said
that it was free.

Math For You to P
No. Where did you see that? I have clearly mentioned about my service at the home page.Here also you can see more details. http://math-homeworkhelp.blogspot.com/2006/07/i-dont-do-math-homework-for-them.html.Why did you send the questions then?

P to me
Like I mentioned earlier the site said something about getting free help
witrh your math. I got the site from another math student so maybe I
clicked on the wrong site. That is why I posted the questions. Once again
you can disregard the questions and I will found help elsewhere. Sorry if
I wasted your time.

Math For You to P
No P,
You came from MSN for the keyword ‘Math Genius’.(http://search.msn.com/results.aspx?q=math ) Not even for math help.So you are not a targeted visitor for me.So it do not worry me.(otherwise also no problem.I have been getting enough clients.) You are from Bastrop,Louisiana, U.S.You go to anywhere for help.(are you really a student?)That is not my topic. But I have a humble request : Don’t lie.I have the detailed track record of your visit.
Don’t worry.Take a deep breath and Relax.

P to me
I think I misread the article it said free lessons. Like I say disregard
the questions because I already have the solution.

Math For You to P
That is O.K. No problem.

Train Problem

published on October 27th, 2006 . by Vanaja

Two trains start from two different stations A and B. One is going from station A to station B and another from station B to station A. If first train takes 8 hours to complete the journey and the second one takes 12 hours, when do the two trains cross each other?

Answer
The first and second train travel 1/8 and 1/12 respectively of the total distance of AB in one hour . So they cover 1/8 +1/12=5/24 of the distance AB in one hour. That means they will cross each other after 24/5 (4 hours 48 minutes) hours.

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