Tuesday, August 31, 2010

Tutornext.com - Tutor Me


Online Tutoring specifically for K-12 and college tutoring has become very popular and is something that I certainly wish had been available several years ago when my daughter was going to school.  I could have been a lot smarter in helping with her home work if this type of assistance had been available,

In looking at this service today I find that a site such as  tutornext.com can help in providing online tutoring at the cheapest rate.  This looks like a great service.  It appears that they can help you with such various things as  Math answers , algebra 1, chemistry help, algebra 2, statistics help (that one would have really been nice to have), and much more.

So if you need help you can even get some  Free math tutoring online from tutornext.com.   They will provide a very high-quality one-on-one supplemental education for K-12 and college students.  Their assignment help and Online Tutoring services are convenient and affordable. These services allow students to connect to a tutor as often as they need help and just when they are ready to learn.

What more can you ask from a service especially as school gets started for 2010.


These are my thoughts on this.

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AP Calculus


Professor Posamentier, the dean of education at City College recently wrote an article in Newsday saying that AP calculus should not be taught in high school, that school's should emphasize algebra skills instead. He said kids are coming into the colleges with AP credit, yet their skills are severely lacking.

I know Professor Posamentier quite well. He was my main math education teacher as an under graduate, graduate and! post graduate student. He is a brilliant teacher and I credit him with my skills as a teacher. I completely and whole-heartedly disagree with the premise of this article. Kids might come into college ill prepared but, it is not the fault of AP calculus. Instead I would blame the poor curriculum of Math A and Math B. I would blame the early emphasis of calculator use. I would blame the use of manipulatives, instead of memorization (multiplication tables) in elementary school.

I have been teaching AP calculus for over ten years. The students I have must qualify for the class by taking pre-calculus and doing well in that class. By ta! king away AP calculus from these kids we are dummying down the curriculum and once again teaching to the lowest level possible. We should not assume that all kids are not capable. The kids in my school are. Sometimes I do end up with students who "sneak in." They don't have the prerequisites. When I find out, I try to "hide" them in the class. The five on the AP exam is not that important. The skills they learn in my class are. AP calculus is a class that teaches kids how to think, how to apply their knowledge to problem solving. They are able to solve problems numerically, analytically, algebraically, and graphically. They must be able to support or confirm answers through written exercises. They must understand that technology is used to support results. Studen! ts are taught that mathematics provides the foundation that allows technology to solve problems. No other math class does this.

Calculus taught on the college level does not delve as deeply into the subject as it is on the high school level. There simply is not enough time. Rather than remove AP calculus from the high school curriculum, we should be doing more to insure more students get to take it.

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Why Darwinius is not our ancestor

I have been blogging and tweeting the dead horse primate that is Ida Darwinius masillae for several days now, culminating in last night's twitter-fest during the BBC documentary Revealing Our Earliest Ancestor: The Link. During our lively little twitter event, which made the Times Online today, I was, apparently, pretty worked up: Andrew Maynard even wrote 'Warming my toes from the heat of your ire...'.

Of course, it's not only us twitterers bemoanin! g the Darwinius hype machine; there's going to be a special Darwinius blog carnival on Monday, and I'm pleased to see a growing number of mainstream outlets casting a critical eye over the story, or at least giving space to external critics. For example, today I found an excellent opinion piece by Chris Beard the NewScientist called Why Ida is not the missing link. In it, Beard dresses down not just the hype - "unbridled hoopla" as he called it - but also the science itself. His final paragraphs are particularly good:
!
'So, Ida is not ! a "missi ng link" – at least not between anthropoids and more primitive primates. Further study may reveal her to be a missing link between other species of Eocene adapiforms, but this hardly solidifies her status as the "eighth wonder of the world".

Instead, Ida is a remarkably complete specimen that promises to teach us a great deal about the biology of some of the earliest and least human-like of all known primates, the Eocene adapiforms. For this, we can all celebrate her discovery as a real advance for science.'

Great stuff. Great enough to make me shout "Yes!" out loud while alone in my flat. What Beard is saying here is that Darwinius is not the 'missing link' between anthropoid side of the primate family tree (includ! ing humans) and the lemur side because the authors of the paper in which Darwinius is described have not convincingly demonstrated that she belongs with the former and not the latter. Beard contends that the balance of evidence keeps Darwinius - and all the rest of the adapiforms by extention - anchored firmly on the lemur line. To explain this, he provides this useful diagram of primate evolutionary history:




Evolutionary tree diagram reproduced from
NewScientist.


The diagram shows the two competing hypotheses: the red dot in the diagram indicates where Chris Beard contends Darwinius belongs, on an early branch in the lemur (brown) lineage. The paler spot with the "?" indicates where the paper's authors claim Darwnius belongs, as an early member of the (blue) anthropoid lineage and, specifically, as they have contended in interviews and on their website if not in their paper, ON the line. In other words, they claim she is our direct ancestor.

I don't have any professional background in primate anatomy but I find Beard's argument - especially in combination with Laelaps' analysis - pretty compelling, certainly compelling enough to remain very skeptical of the authors' conclusions. We will have to wait for further analyses of the Darwinius specimen, however, before this controversy can be more soundly resolved.

But here's the thing: even if upon further analysis Beard is shown to be mistaken and the authors are right about Darwinius and her fellow adapids being on the anthropoid lineage, she will still not necessarily be the 'missing link', nor - perhaps more importantly because it! actually means something - 'our ancestor'. To explain my poin! t, I re- drew Beard's diagram:



My re-drawing of Chris Beard's diagram in which I shift his rep! resentation of the paper's conclusion about Darwinius' systematic position from ON the anthropoid line to a BRANCH off the anthropoid line.

I re-drew it this way for the simple reason that, considering the abundance of species upon the Earth at any one time, it would be very unusual for us to find a fossil on our direct ancestral line, rather than on a branch off of that line. As John Wilkins put it, "There is no missing link. Rather, there are an indefinite number of missing branches. [snip] We might have a species that is an ancestor of some other species, and yet not know enough to say that they are indeed the ancestor in question."

Additionally, as Richard Carter notes, ! Ida herself can't be our direct ancestor because she died as a! juvenil e. But as I've just explained, even her whole species is very unlikely to include our ancestor.

Interestingly, as I mentioned in passing above, the hype machine (including the paper's authors themselves when interviewed) puts Darwinius directly on the line to us, while in Supplementary Figure 7 in the PLoS ONE paper, the authors put Darwinius on a early branch off the line. In other words, what they're saying in public isn't just hyped up, it's fundamentally different from what they're saying in the paper itself.

To summarize, if Darwinius is found, as the authors contend, to belong on 'our' side of the primate family tree (and even that! conclusion is shaky) she is not a 'missing link' (because there's no such thing), and she is very unlikley to be our ancestor.

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Homework homerun!!

Dear Parents,

We have a few more days left until the DC CAS state assessments.

During DC CAS testing in the spring, 6th to eighth grade students must not only know basic and advanced knowledge and skills, they must retain them, understand them, and use them in real life situations.

As parents you can help with improving your child/children's test scores by letting your child do their homework at home which I have linked from this class website. All of the links, the powerpoint and ! the practice probes and games here have been used presented in class and should be reviewed and practiced by your child. In the next few days, I will be uploading some videos that your child can watch at home related to the lessons that we are learning in the classroom from Discovery Education. Please ask your child to visit the class website and do the homework using the log-in codes listed.

As your child's teacher, I am working very hard to give him/ her the academic supports that he/ she needs even outside the classroom and even beyond the call of duty. I want to see your child successful in life as much as you do.

Please take a few minutes to review the class website and work with your child each night on some of the strategies offered. Even students not ! in my class and from other school districts are using our Prac! tice Pro bes and are having fun while learning from it.

Thank you for taking the time to always read my weekly emails. Your involvement is critical to your child's success.


Sincerely,
MARIA ANGALA
Room 118

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Free PMP Tests Resources

Free Test Websites:

PMP forums:
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Orkut (****) http://www.orkut.com/Main#Community?cmm=1280407

Yahoo Groups
http://finance.groups.yahoo.com/group/pmpbest/

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HeadFirst Labs (***) http://forums.oreilly.com/category/54/Head-First-PMP/

Google Groups (***) http://groups.google.com/group/pmp-! cert-online-study-group?lnk=srg



PMP Prep Materials:

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More issues surround DA?

From www.inlandpolitics.com
 
Several months ago a story exposing substantial political contributions to District Attorney Michael Ramos by two attorneys who control the San Bernardino County criminal defense conflict panel was published at www.iepolitics.com.
Now, more revelations are coming forward that may give a plausible explanation for $73,000 in political contributions given to Ramos from 01/01/2004 through 06/30/2009.

Sources who have had business dealings with the conflict panel tell of financial difficulties within the group plagued by late payments to employees and contractors. The issues have resulted in threats of going to state labor officials. It is being asserted that the funds from county conflict assignments now subsidize the private practice operations of the law firms involved.

One source also tells of the use of an investigator in the Victorville courthouse who has the primary task of meeting with pre-trial indigent defendants either in or out of custody with a so-called “plea sheet”. The investigator basically suggests to the defendants they should take a plea bargain. The investigator actually works out the plea arrangement and proposed sentence. Remember, this is a non-attorney.

Nearly all defendants agree.

The logic behind the process was explained this way. The use of the non-attorney investigator in working out plea deals saves money and only requires an actual panel attorney to spend very little time on each case thereby lowering the ratio of attorney man-hours per case. Since the contract between the conflict panel and the County of San Bernardino pays a flat rate per case, with the exception of complex defenses less attorney time per case, thus more profit.
County prosecutors apparently have no problem with the plea bargain milling of defendants. After all, a conviction is a conviction.

San Bernardino County Public Defender Doreen Boxer is apparently trying to cut back on the number of conflict cases being assigned to the conflict panel due to concerns over quality of representation. Boxer made a proposal to county supervisors in January 2009 to establish a separate alternate defense panel within her department. The structure would mirror that of Los Angeles County.
Currently, whenever the Public Defender or the court declares a conflict involving a defendant, the case is farmed out to the conflict panel.

The District Attorney lobbied against the Public Defender’s plan. County supervisors after one on one meetings with Boxer squelched the idea.

Here is a recap of political contributions by conflict panel attorneys James Spring and Earl Carter.

For the entire article please visit www.inlandpolitics.com.

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Solve your Algebra Problem in Tutornext.com

There is only little student have fun while studying algebra. I’m one of student that is really hated with algebra. I know algebra is very important, and will very useful in our live. That’s why I try hard to learn algebra.

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If you already in Algebra 2, this site will very helpful. They also offer Algebra 2 help, and you will get your Algebra 2 Answers. You can connect with the tutor whenever you want, while you have question for your algebra problem. With this online tutoring site, you can save your time (without having to come to the lessons).

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Slope calculator

In this blog we are going to learn about mathematics concept slope.
Slope: -It is the measure of a line which is inclined in relation to the horizontal.
Slope of any line in geometry means the ratio of vertical to the horizontal distance between any points on it.
Meaning of Slope of a line tangent to the graph in differential calculus is given by the particular functions derivatives & represents the rate of modify of the function with respect of modify in the independent variable.
In a graph of a position function, the slope of the tangent signifies an objects instantaneous velocity.
Point Slope Forms: -
Point Slope is the technique of representing or plotting an equation on a graph paper on an x-y axis. It is used to take a graph & find the equation in a specific contour.
The equation for Point Slope Form is given below: -
Y-y1=m(x-x1)
Here is the example of how we can find the slope.

Example:

Find the slope from the given equation, y = 2x + 4

Solution:

Here, the equation is given in slope intercept form,

y = mx + c

Where, m = slope

y = 2x + 4

m = 2.

The answer is 2.

This is just an example on find slope.Next time we will see slope calculator.Before knowing slope calculator,going through slope worksheet you should know about slope.
Also you should know the concept of ogive,as its a equally important concept in math.

slope intercept calculator

Vedic Mathematics Lesson 43: Polynomial Division 1

Most of us are familiar with the multiplication of polynomials. It is one of the first things taught in algebra. Polynomial multiplication, as you are well aware, involves the application of the distributive property of multiplication over addition, the addition of exponents, collecting of like terms, etc. to get a new polynomial from two or more other polynomials. However, the division of one polynomial by another is not familiar to most because it is seldom taught or even mentioned.

You can find all my previous posts about Vedic Mathematics below:

Introduction to Vedic Mathematics
A Spectacular Illustration of Vedic Mathematics
10's Complements
Multiplication Part 1
Multiplication Part 2
Multiplication Part 3
Multiplication Part 4
Multiplication Part 5
Multiplication Special Case 1
Multiplication Special Case 2
Multipli! cation S pecial Case 3
Vertically And Crosswise I
Vertically And Crosswise II
Squaring, Cubing, Etc.
Subtraction
Division By The Nikhilam Method I
Division By The Nikhilam Method II
Division By The Nikhilam Method III
Division By The Paravartya Method
Digital Roots
Straight Division I
Straight Division II
Vinculums
Divisibility Rules
Simple Osculation
Multiplex Osculation
Solving Equations 1
Solving Equations 2
Solving Equations 3
Solving Equations 4
Mergers 1
Mergers 2
Mergers 3
Multiple Mergers
Complex Mergers
Simultaneous Equations 1
Simultaneous Equations 2
Quadratic Equations 1
Quadratic Equations 2
Quadratic Equations 3
Quadratic Equations 4
Cubic Equations
Quartic Equations

In the previous lesson, we identified one of the solutions to a quartic equation based on the values of the coefficients of the different powers of the unknown variable, x. Once the solution is identified, we needed to perform a polynomial division to get the residual cubic equation from the given quartic equation and one of its solutions. We touched upon the procedure for performing this tangentially, but we did not take it up formally. In this lesson, we will formalize the procedure for performing simple polynomial divisions. In particular, we will deal with the div! ision of higher-order polynomials by polynomials of the first degree (linear polynomials) in this lesson. In future lessons, we will deal with divisions where the divisor is a polynomial of higher degree.

In the most general terms, in this lesson, we are dealing with divisions of the following form:

Divide ax^n + bx^(n-1) + ... + constant by (cx + d)

In the previous lesson, (cx + d) was either (x + 1) or (x - 1). But the method we used in that lesson is applicable to other divisors also. We already know how to do some basic divisions without even realizing we are doing polynomial division. As in any other division, the answer will have a quotient and a remainder. In some cases, the remainder may be zero. In some cases, the quotient may be zero too.

Consider the following example of division by a linear polynomial, wh! ich practically anyone can actually perform without any hesita! tion:

Divide 4x^2 + 6x + 7 by 2x.

In this case, c = 2 and d = 0. Just by inspection, we can say that the quotient of this division is 2x + 3, and the remainder is 7. The way to verify whether we obtained the correct answer, just as in the case of arithmetic division is to check whether dividend = quotient * divisor + remainder.

In the above case, we can indeed verify that 4x^2 + 6x + 7 = 2x*(2x + 3) + 7. Thus, we can be confident that our division was correct.

Now, consider another example of polynomial division which may not be readily apparent as polynomial division. Divide 4x^2 + 6x + 7 by 2. In this case c = 0 and d = 2. Once again, just be inspection, we can say that the quotient is 2x^2 + 3x + 3, and the remainder is 1. We can verify that this is the correct answer by seeing that 4x^2 + 6x + 7 = 2*(2x^2 + 3x + 3) + 1.

However, when neither c nor d is zero, the situation becomes trickier. That is the kind of divisio! n that is not dealt with in school in any detail. It is the kind of division that is not taught to students as a normal part of algebra education for whatever reason. It is actually very useful to teach this kind of division, as we will see later in this lesson!

Consider the division of 4x^2 + 6x + 7 by x + 2. Essentially, we have to find (4x^2 + 6x + 7)/(x + 2). How exactly do we perform this division? The answer turns out to be quite simple actually. We have already seen some examples of how we do such a division in the previous lesson. Let us formalize the method in this lesson.

Consider the division of a polynomial ax^n + bx^(n-1) + ... + constant by the first degree expression cx + d. Consider the ratio c/d. The secret to polynomial division is to rewrite the numerator using this ratio just like we did in our lessons on cubic equations and quartic equations.

We have to rewrite the dividend in the form below:

ax^n + e! x^(n-1) + fx^(n-1) + ... + yx + zx + constant1 + constant2

And the coefficients have to satisfy the conditions below:

c/d = a/e = f/g = ... = z/constant1
e + f = b
...
y + z = coefficient of x in the given dividend
constant1 + constant2 = constant

In some cases, constant2 could be zero. When that is the case, the division does not have a remainder. When constant2 is not zero, that is the remainder (at least an intermediate remainder) of the division.

Let us see how these rules are applied in the case of (4x^2 + 6x + 7)/(x + 2). In this case, c/d = 1/2. Thus, we have to rewrite the dividend using the ration 1/2.

4x^2 + 6x + 7 = 4x^2 + ax + bx + constant1 + constant2

where

1/2 = 4/a = b/constant1
a + b = 6
constant1 + constant2 = 7

We see that we can accomplish this by writing the dividend as below:

4x^2 + 8x - 2x - 4 + 11

This can then be factorized as below:

4x(x + 2) - 2(x + 2) + 11
This then tells that the quotient of the division is (4x - 2), and the remainder is 11. As you can see, the method itself is quite simple and straightforward. If you can work with ratios, then you can perform polynomial division! We will spend some time on a couple more examples, then we will move on to some special cases, and then conclude with a very useful practical application of the technique.

Consider (4x^3 + 6x^2 - 8x + 15)/(2x - 1). In this case, c/d = -2/1. Thus, we rewrite the dividend using this ratio as below:

4x^3 -2x^2 + 8x^2 - 4x - 4x + 2 + 13

We can then factorize the expression above as below:

2x^2(2x - 1) + 4x(2x - 1) - 2(2x - 1) + 13

Thus, the quotient of the division is 2x^2 + 4x - 2, and the remainder is 13.

Now, consider (3x^4 + 4x^3 - 11x^2 + 6x - 5)/(2x + 1). In this case c/d = 2/1. Thus, we rewrite the dividend using this ratio as below:

3x^4 + 1.5x^3 + 2.5x^3 + ! 1.25x^2 - 12.25x^2 - 6.125x + 12.125x + 6.0625 - 11.0625
!
We can then factorize the expression as below:

1.5x^3(2x + 1) + 1.25x^2(2x + 1) - 6.125x(2x + 1) + 6.0625(2x + 1) - 11.0625

The quotient is then (1.5x^3 + 1.25x^2 - 6.125x + 6.0625), and the remainder is -11.0625. We can rewrite the quotient as 24x^3/16 + 20x^2/16 - 98x/16 + 97/16, and the remainder as -177/16. Thus, the quotient can be rewritten as (24x^3 + 20x^2 - 98x + 97)/16, and the remainder is -177/16.

Thus, we see that the quotient and remainder can contain fractional terms. And, in this case, the remainder was negative too! Thus, polynomial division can result in some results that can cause us to scratch our heads. But, we can verify that the answer we got is correct by simply multiplying the quotient with the divisor, and adding the remainder to the result to see whether we get back our original dividend.

Now consider a division as below:

(2x^3 + 5x^2 - 6x + 8)/(2x + 2)

We see that the ratio c/d ! in this case is 2/2 = 1. Thus, we could rewrite the dividend expression as below:

2x^3 + 2x^2 + 3x^2 + 3x - 9x - 9 + 17

We would then factorize it as below:

2x^2(x + 1) + 3x(x + 1) - 9(x + 1) + 17

We immediately see that the expressions in the parentheses are x + 1, not our divisor, 2x + 2. Thus, if we decide, based on the factorization, that our quotient is 2x^2 + 3x - 9, with a remainder of 17, we will find that we are wrong. In fact, (2x^2 + 3x - 9)(2x + 2) + 17 = 4x^3 + 10x^2 - 12x - 1, which is not what we started with as our original dividend. This, then tells us that the actual factorization of the rewritten dividend is as below:

x^2(2x + 2) + (3x/2)(2x + 2) - (9/2)(2x + 2) + 17

Thus, the quotient would be (x^2 + 3x/2 - 9/2) and the remainder would be 17. This can be verified as the correct answer. Thus, we need to pay extra attention when the ratio c/d can be simplified by the presence of common! factors. The common factor, in this case, prevented us from ! getting the divisor as a common factor during the factorization when we factorized the rewritten dividend. We then had to introduce fractions during the factorization to get the divisor as the common factor during the factorization.

A simpler solution to the above problem would be to take the common factor out of the divisor and divide the dividend by this common factor in advance. In this case, the common factor in the divisor is 2. So, we can convert the dividend to x^3 + (5/2)x^2 - 3x + 4 before we attempt the division. Now, our division problem is reduced to (x^3 + (5/2)x^2 - 3x + 4)/(x + 1). The ratio of terms in the divisor is 1/1 = 1. Thus, we can perform the division by rewriting the dividend as below:

x^3 + x^2 + (3/2)x^2 + (3/2)x - (9/2)x - 9/2 + 17/2

This will then enable us to perform the factorization as below:

x^2(x + 1) + (3x/2)(x + 1) - (9/2)(x + 1) + 17/2

The quotient is then x^2 + 3x/2 + 9/2. But since w! e divided the numerator and denominator by the common factor, 2, remember that the remainder is now to be multiplied by this common factor to get the final remainder. Thus, the remainder is not 17/2, but is instead 17!

Having fractional terms as part of the dividend may appear a little intimidating. So, another solution would be to take the common factor out of the divisor and keep it aside. Perform the division as usual, then divide the resulting quotient by the common factor.

In this case, our initial division resulted in a quotient of 2x^2 + 3x - 9. We see that we can then divide this by the common factor, 2, to get the correct quotient. Remember not to divide the remainder by the common factor!

Now, let us consider the last special case. Consider the division of 6x + 7 by x + 1. In this case, the dividend is also a linear polynomial. But the method is still the same as before. The ratio, c/d, in this case is 1/1 = 1. Thus, we r! ewrite the dividend expression as below in preparation for fac! torizati on:

6x + 6 + 1

This then leads to the factorization as below:

6(x + 1) + 1

We then conclude that the quotient is 6 and the remainder is 1.

Let us now record our observations about polynomial division, with a first-degree polynomial as the divisor, below:

  • The quotient will be a polynomial of one degree lower than the original dividend (thus, if the dividend is also a linear expression, as in our last example, then the quotient will be just a constant with no x term)
  • The remainder will be a constant, but can be either positive or negative
  • If the divisor has a constant as a common factor, we can set the constant aside, divide by the divisor reduced to lowest terms, then divide the quotient by the common factor to get the correct quotient (no change needs to be made to the remainder)
  • Alternatively, we can divide the dividend by the common factor, then perform the division by the div! isor reduced to lowest terms to get the quotient, then multiply the remainder by the common factor to get the correct remainder

As you can see, the division of a polynomial by a linear polynomial is quite easy once the method of ratios is laid out. This is an application of the Madhyamadhyena Adhyamanthyena sutra. Now, as promised, let us examine why this technique is useful. The secret to its usefulness lies in the insight that practically every real number is a polynomial. Consider the number 324, for instance.

324 = 3*10^2 + 2*10 + 4

This is the same as 3x^2 + 2x + 4, where x = 10! In fact, we can even express it is 3x^2 + 3x - 6, where x = 10. In fact there are several ways to express a given number as a polynomial, even using just x = 10. Using a different value of x enables us to rewrite the number in other bases (such as octal, where x = 8, hexadecimal, where x = 16 and binary, where x = 2). Because of this, it is easy to ! see that we can use our insights from polynomial division to p! erform r egular arithmetic division with just as much ease! In fact, this is the practical application of polynomial division that I have been hinting at since the beginning of this lesson!!

Take the case of 324/11 for instance. This can be rewritten as (3x^2 + 2x + 4)/(x + 1), where x = 10. We already know how to do this with no problems. We rewrite the dividend expression as below:

3x^2 + 3x - x - 1 + 5

We then factorize it to get:

3x(x + 1) -1(x + 1) + 5

We then conclude that the quotient is 3x - 1 and the remainder is 5. 3x - 1 when x = 10 is 29. Thus, we have just performed the division, 324/11 and found that the quotient is 29 with a remainder of 5. I encourage you to verify that this is indeed correct! Let us now take on a few more examples to see just how easy the method is.

Take 372/12 for instance. This can be rewritten as (3x^2 + 7x + 2)/(x + 2) where x = 10. The ratio, c/d, is then 1/2. We can th! en rewrite the dividend expression as below, and factorize it:

3x^2 + 6x + x + 2, which can be factorized as
3x(x + 2) + 1(x + 2)

Thus, we see that the quotient is 3x + 1 (or 31 with 10 as the value of x), and the remainder is 0!

Now, let us take a few slightly trickier cases. Consider 412/12, for instance. This can be rewritten as (4x^2 + x + 2)/(x + 2), where x = 10. The ratio, c/d is once again 1/2. Then, the dividend can be rewritten and factorized as below:

4x^2 + 8x - 7x - 14 + 16 which can be factorized as
4x(x + 2) - 7(x + 2) + 16

This then tells us that the quotient is 4x - 7 (or 33, since x = 10), and the remainder is 16. This can not be correct. We can verify using a calculator that the quotient in this case should be 34 and the remainder should be 4. It turns out that our solution is correct, but it does not meet the standards for normal division. In particular, in this case, the remainder ! turned out to be bigger than the divisor. To correct this, ad! d one to the quotient and reduce the remainder by the divisor. If necessary, repeat the operation until the remainder becomes less than the divisor. Following this procedure, we can then adjust our solution to a quotient of 34 (add one to 33), and a remainder of 4 (subtract the divisor, 12, from the original remainder of 16). Now, we can say that the answer is in truly correct form!

Now consider the case of 271/13. This can be rewritten as (2x^2 + 7x + 1)/(x + 3), where x = 10. The ratio, c/d is 1/3 in this case. We can then rewrite the dividend expression, and do the factorization as below:

2x^2 + 6x + x + 3 - 2 which can be factorized as
2x(x + 3) + 1(x + 3) - 2

We can then interpret this as saying that the quotient is (2x + 1), or 21 since x = 10, and the remainder is -2. Once again, we are left with an answer that seems wrong. We are not used to dealing with negative remainders when we do divisions normally. But the correction for t! his problem, once again, is very simple. We simply subtract one from the quotient and add the divisor to the remainder. Repeat this until the remainder is positive. Thus, we can correct our answer to a quotient of 20 (subtract 1 from 21), and a remainder of 11 (add the divisor, 13, to -2). It is easy to verify that a quotient of 20 and a remainder of 11 is indeed correct.

Next, let us consider 4878/18. We can write it as (4x^3 + 8x^2 + 7x + 8)/(x + 8), where x = 10. But the ratio, c/d now becomes 1/8. Using that ratio, we would be forced to rewrite the dividend expression as below:

4x^3 + 32x^2 - 24x^2 - 192x + 199x + 1592 - 1584

Obviously, this is correct, and does give us the following factorization:

4x^2(x + 8) - 24x(x + 8) + 199(x + 8) - 1584

This then translates to a quotient of 4x^2 - 24x + 199, which equals 400 - 240 + 199 = 359, and a remainder of -1584. To get this answer to standard form would require ! a lot of subtractions of 1 from the quotient and additions of ! 18 to th e remainder!

Instead, let us consider a different approach to this problem. We can rewrite the given problems as (4x^3 + 8x^2 + 7x + 8)/(2x - 2), where x = 10. Now, we see that c/d = -1 and the divisor has a common factor of 2. Let us take the common factor out initially and do the rewriting and factorization as below:

4x^3 - 4x^2 + 12x^2 - 12x + 19x - 19 + 27 which can be factorized as
4x^2(x - 1) + 12x(x - 1) + 19(x - 1) + 27

This translates to a quotient of 4x^2 + 12x + 19, which can be translated as 539 by substituting x = 10. Now, remember to divide the quotient by 2 (the common factor in the divisor), and we get 269.50. Since the remainder is greater than 18, we also have to subtract 18 from it and add 1 to the quotient. This gives us a quotient of 270.50 and a remainder of 9. Since 0.50*18 is 9, we can further simplify it to a quotient of 270 with a remainder of 18 (subtract the 0.50 from the quotient and add 0.50*18 to the! remainder), which once again can be reduced to a quotient of 271 with no remainder. We can verify that this is indeed the correct answer.

The approach we took the second time around is reminiscent of the use of vinculums in various arithmetic operations. Vinculums are dealt with in great detail in a lesson dedicated to them. It may be useful to review that lesson once more for more insights into the usefulness of the concept of vinculums.

Now, let us consider a more tricky case. We will try 3421/77. This can be rewritten as (3x^3 + 4x^2 + 2x + 1)/(7x + 7), where x = 10. We see that there is a common factor of 7 in the divisor, and a c/d ratio of 1. Using the ratio, we can rewrite the dividend and factorize it as below:

3x^3 + 3x^2 + x^2 + x + x + 1, which can be factorized as
3x^2(x + 1) + x(x + 1) + 1(x + 1)

This directly translates to a quotient of 311 with a remaind! er of 0. We now have to remember to divide the quotient by the common factor, 7. This presents us problems since we are left with another division problem almost as difficult as the original problem. However, we have reduced the problem by at least an order of magnitude, and in fact, we can perform the division by 7 in our heads to get a final quotient of 44 and a remainder of 3. Notice that this remainder is from a division by 7 while the real remainder needs to be from a division by 77. So, to get the true remainder to the problem, we now multiply this remainder of 3 from this step by 77/7 (original divisor/common factor), which is 11. Thus, the final answer is 44 with a remainder of 33.

An easier way to remember this might be to express the answer as 44 and 3/7 with a remainder of 0. Now, we convert the 3/7 to the final remainder by multiplying by the divisor, 77, to get 33 as the final remainder. We can verify that 44*77 + 33 = 8843, so we know our answer is c! orrect.

To obviate the need for a manual division, we can once again the use the concept of vinculums to rewrite the given problem as (3x^3 + 4x^2 + 2x + 1)/(8x - 3). This then gives us a c/d ratio of -8/3. But this leads to new problems as we see below:

3x^3 + 4x^2 + 2x + 1 can be rewritten with a ratio of -8/3 as
3x^3 - (9/8) x^2 + (41/8)x^2 - (123/64)x + (251/64)x - 753/512 + 1264/512 which can be factorized as
(3/8)x^2(8x - 3) + (64/41)x(8x - 3) + (512/251)(8x - 3) + 1264/512

This then gives us a quotient of 300/8 + 640/41 + 512/251 (after substituting x = 10 in the expression above) with a remainder of 1264/512. As you can see, this is not a very convenient expression to work with and convert into a standard form for presentation as a normal quotient and remainder. So, the lesson to take away from this is to be careful and not carried away by vinculums too much!

The concept of polynomial division is useful to le! arn since it can translate directly into an easy method for ar! ithmetic division also. But, sometimes, it does not produce results any faster or easier than other methods of division. This is important to recognize as a shortcoming of the method. Different methods have their own strengths and weaknesses. If the ratio we need to work with is inconvenient and can not be converted into a more convenient ratio by the use of vinculums or other tricks, we have to recognize that polynomial division may not be the best approach to the division problem.

Moreover, in this lesson, we have dealt with only linear divisors. This can limit the technique to just 2-digit divisors under most conditions. We will leave this lesson with an example of a division by a 3-digit divisor, but in general, we will deal with division by more digits in the next lesson, when we will deal with polynomial division where the divisor is not linear.

Consider the division problem 8843/126. This can be expressed as (8x^3 + 8x^2 + 4x + 3)/(x^2 + 2x + 6), ! where x = 10, but this does not leave us with a linear divisor. Instead, let us rewrite this as (8x^3 + 8x^2 + 4x + 3)/(12x + 6), where x = 10. Now, we have a linear divisor, so we can proceed as before. We see that the ratio c/d = 2, and 6 is a common factor in the divisor. We will set aside the common factor for now, and rewrite and factorize the dividend as below:

8x^3 + 4x^2 + 4x^2 + 2x + 2x + 1 + 2 which can be factorized as
4x^2(2x + 1) + 2x(2x + 1) + 1(2x + 1) + 2

This then translates to a quotient of 421, and a remainder of 2. Since we took out a common factor of 6 from the divisor, we now divide the quotient by 6. Dividing 421 by 6 gives us the new quotient of 70 and a remainder of 1. To convert this remainder to a remainder that would be obtained by division by 126 rather than by 6, we need to multiply it by 126/6, which is 21. Now, we add this remainder to the remainder from the factorization to get a final remainder of 23.

An easier way to remember this might be to express the ! result o f the division as 70 and 1/6. To move the 1/6 to the remainder, we need to multiply it by the divisor, 126. This gives us 21, which is then added to the already existing remainder of 2 to give us a final remainder of 23. We can verify that 70*126 + 23 = 8443, so we know that our answer is correct.

As you can see, polynomial division has several uses, chief among them, the factorization of polynomial expressions. But, as we saw in this lesson, it can also translate into an easier way to do some arithmetic divisions also. Hope you will take the time to practice some of the techniques, both with polynomial expressions as well as with arithmetic expressions, so that you can be confident about their application when it is appropriate. Good luck, and happy computing!

simplify polynomial calculator

WHAT CLOTHES SHOULD I WEAR FOR MY BODY SHAPE?

Types of Body Shapes Everyone’s body is made differently so instead of trying to look like someone else start to feel confident with yourself and find a style that will complement your shape. To help you we have put together a useful guide to buying clothes for your body shape so you can figure out what suits you.
Athletic – An athletic build is an ideal shape and you can just about wear anything. Be careful though, this does not mean that you go overboard and try and mix too many trends all at once.
Hourglass - This is where the bust and hips are in proportion, and there is a defined waistline. However, if your waist is small and your hips curvy then it is important not to draw too much attention to the waist area as it will make your hips look even larger. Cropped jackets look great or slightly looser tops that don’t cling tightly to the waist.
Big hips – If you have big hips then flatter the figure with wide leg trousers in dark colo! urs this will make you look taller and will take the attention away from the hips. Tailored pieces with a great cut are ideal for minimising the appearance of big hips.
Top heavy – Minimise you assets by wearing scooped or V-necks. Avoid high necklines and polo necks this will simply accentuate your 'top half'.
Pear-shaped – A Pear shaped body is where your hips are wider than your shoulders. In order to achieve a proportionate balance you should invest in a pair of straight legged trousers or A-line skirts teamed with loose fitting tops. Dresses with an empire line are great for the pear shaped body. Accentuate your bust area with low necklines or accessorise with a scarf and beads as this will draw the attention away from the hip area.Apple-shaped - Apple-shaped bodies are larger on top and smaller below. People with this shape often have a short neck and it is important to make the neck look longer by wearing tops with a v-neckline. Avoid anything f! igure hugging as this will only draw attention to your waist. ! Wide leg ged trousers or flared skirts look great on Apple Shaped bodies.
Straight body - This body shape has no defined waist so avoid high waisted clothes and belts. Instead, clothes with a dropped waist are ideal or a wrap dress. Accentuate other areas of your body such as the bust and legs as this will draw the attention away from the waistline.
Arms and legs – Arm and legs can be troublesome areas when buying clothes. Take a long hard look at your upper arms. If your upper arms are flabby then wear three-quarter or half-length sleeve shirts and tops this will stop you feeling self-conscious. Long lean legs will suit any style of skirt but if you are unsure what skirt will suit just remember that a skirt should end just above the slimmest part of your leg.
Must-have’s for your shape Flatter your figure with our guide to the ultimate must-have’s for your body shape.
Curves - Curves need to be show! n off so don’t hide underneath baggy clothes, you need to flaunt it: Invest in a pair of tailored trousers that will elongate you legs teamed with a fitted jacket which will polish off the look perfectly. Corsets and pencil skirts are a great way to accentuate the waist area.Tall and slim – Most trends will look great on this type of figure but there are a few fashion faux pas that you have to stay away from. For instance, if you are flat chested then stay away from low cut tops. Skinny jeans are ideal for this shape teamed with a pair of heels for the night or ballet shoes for the day. Most tops will look great on this shape and you can even wear horizontal stripes as this will help to minimise your torso area.Petite – It is important to find the right fit. Don’t be afraid to mix and match from the petite and standard range to get your ideal look. Dark colours are great for the petite frame as this helps to make you look taller. Fitted tops are grea! t to create a slim silhouette teamed with a pair of tailored t! rousers and kitten heel boots the look is simply great. Above-the-knee skirts and dresses suit the petite frame and are a great way to show off those legs.
Pear shaped – A great shape for showing off your bust area. Try corset style tops, tailored jackets, a halterneck dress and A-line skirts. The wrap dress is also key for this shape as it is a great solution for different sized tops and bottoms.

Useful tips
Look in the mirror and get to know your body. Everyone is different so feel confident with your body shape and the right clothes will enhance your confidence.

shape areas

Teaching negative exponents

Continuing on with the exponents, I have another video about negative exponents. Check also the "prelude" to this, my video on zero exponent.


negative expon! ents: Learn Them with a Pattern!


In books we usually find the definition that x−n = 1/xn or perhaps x−n = (1/x)n (you take the reciprocal of x and use the opposite of the exponent).

But again, I want to highlight that we DON'T have to just "announce" to students this definition or how negative exponents are done. We can justify it. In the video, you can see this done first with the help of a pattern, where we go from the positive exponents to zero exponent to negative ones.

I also show another way of showing your students why negative exponents have been defined the way they are. If you look at the shortcut for dividing powers with the same base (I used 63/65 in the vide! o), we can simplify as usual and arrive at 1/62. Ho! wever, t he shortcut says we can just subtract the exponents, so 63/65 also equals 6−2. The same reasoning applies for any non-zero x and any positive whole number exponent.

This way, we teach our students not only the "how" but also the "why" of mathematics, and help them become better thinkers. Mathematics is not just about memorizing things and then learning to "spit" them out at tests - it also has a LOT to do with proof and logical thinking. And surely our kids need that today, more than ever.


negative exponents

The LA TIMES Cracks Out of Turn When It Doesn't Know The Shot


David Mamet*


I'm not sure what I was doing or where I was the day the LOS ANGELES TIMES went from being a newspaper to being a national leader in evaluating teacher quality. Perhaps it was supposed to be kept secret, like what was discussed and said at the meeting Dick Cheney had with Big Oil executives a decade ago. If so, several  TIMES reporters have blown it with a recent article, WHO'S TEACHING LA'S KIDS?

In it, three reporters, Jason Felch, Jason Song and Doug Smith, present ratings of "teacher effectiveness." In particular, they single out one particular fifth grade teacher, John Smith, and claim he is the least effective teacher for his grade level in his school. Mr. Smith's photo is at the beginning of the article that purports to know, based on kids' scores on administrations of a single standardized test, which teachers are helping their students, teaching effectively, and making a positive difference, and, of course, which, like Mr. Smith, are allegedly failing to move their students ahead. 


I wrote the following to these reporters today and will be fascinated to see if any of them respond. I know that were I John Smith, I'd be speaking to my attorney and considering lawsuits against several parties, not the least of whom are Jason Felch, Jason Song, and Doug Smith. As I am not, the best I can do is try to point out how wrongheaded, how irresponsible, and how ultimately counterproductive is both their article and the methods they employ to smear the professional integrity of many fine teachers who for any number of reasons may not "measure up." The professional integrity I call into question, however, is that of these reporters, their editors, and others who profit from the publication of this sort of cheap-shot, ignorant journalism. 



If you start with the absurd assumption that multiple-guess
standardized test scores tell us anything (let alone EVERYTHING) we
need to know about teacher effectiveness or student learning of
subject matter or all the other things that teachers and schools are
about (not all of which are good, but that's another debate entirely),
then it follows that the LA TIMES is as qualified as anyone else with
no expertise whatsoever in psychometrics to determine which teachers are "most
effective" and which are "least effective." Further, with the same
starting assumption, there's nothing unconscionable about reporters
and editors  from that noble publication choosing to print a photo of
a so-called "ineffective" 5th grade teacher and include the following
in the article:

Yet year after year, one fifth-grade class learns far more than the
other down the hall. The difference has almost nothing to do with the
size of the class, the students or their parents.

It's their teachers.

With Miguel Aguilar, students consistently have made striking gains on
state standardized tests, many of them vaulting from the bottom third
of students in Los Angeles schools to well above average, according to
a Times analysis. John Smith's pupils next door have started out
slightly ahead of Aguilar's but by the end of the year have been far
behind."

But if the assumption is false, then what the TIMES and its reporters have done is to pillory one 5th grade teacher on the wheel of meaningless test scores. They have, in fact, violated two  fundamental principles of psychometrics: never use a test designed to measure one thing (e.g., student achievement) to measure something it was not designed to measure (e.g., teacher effectiveness), and never use a single test score or measurement type to draw definitive conclusions (particularly not in the social sciences). Further, they have made the fundamental error of assuming that correlation (Teacher A's kids scores are higher than Teacher B's scores) equates with causation (Scores rose primarily BECAUSE of the superior teaching skills and methods of Teacher A).

In fact, the above-cited article is so fraught with error and leaps of logic (and bad faith) as to be utterly, irredeemably worthless, not unlike the test scores upon which its false (and probably libelous) conclusions are based. But then, the article's authors began with a patently incorrect assumption, and  they very likely had its conclusions well in mind to begin with.

So I am moved to ask: may we expect in the near future an article by the same reporters on which LA TIMES journalists are "most effective" and "least effective" based on how sales of the paper are impacted by their articles and reportage?

May we expect that the reporters will be taking, say, the tests given to high school kids in LAUSD (I assume all these journalists graduated from college) or perhaps the SAT or ACT (or, Darwin forbid! the GRE) and publishing the results in the paper? How about the politicians who pushed and voted for using these tests as fair measures of a host of things they were never designed to assess? (And I include in that list not only state and local officials, but every US senator, congressperson, US Department of Education secretary, and every US president from William Jefferson Clinton to George W. Bush to Barack Hussein Obama who has supported these tests as the measure of all things.)

Let's shed more sunshine on the test-based competence of our reporters and politicians. Publish and publicize their scores. Threaten, meaningfully, to hoist these folks by their own petard and we'll see some critical examination of the assumption that the tests are valid and reliable, as well as adequate measures alone of effectiveness or lack thereof. My suggestion is no worse than what politicians and reporters are doing now with kids, administrators, schools, districts, and, of course, everyone's favorite scapegoat, public school teachers.

Until such time, shame on Messrs. Felch, Song, Smith; shame on the LA TIMES. Shame on anyone and everyone who buys into the ridiculous test-mad nonsense that has this country by the throat.

Michael Paul Goldenberg

p.s.: Lest someone suggest otherwise, my last official GRE scores, taken in October 1991, are Verbal 800; Mathematics 780; and Logical Reasoning 720. I feel safe in putting those scores up against those on any comparable test of the three reporters, any member of the LA TIMES staff, any current legislator in California or the United States Congress. I've spent over 30 years preparing students for various standardized tests and debunking many of the myths surrounding them. I'll happily meet anyone on the standardized test battle ground, No. 2 pencils aready at dawn or high noon.


*For those wondering what David Mamet's photo is doing at the beginning of this blog entry, it has to do with the title of my post. Mamet is very fond of the language of con artists. Apparently, our intrepid LA TIMES reporters are not unfamiliar with both the short and long cons. Or perhaps it's just their editors, the publisher, and others with vested interests in destroying US public education.

measurement math problems

Grammar-Land Worksheets Part 1

I've mentioned one of the free resources we are using for grammar in a previous post, but I made some worksheets today to go along with the book and I wanted to share them with you all.  Grammar-Land by M.L. Nesbitt is a wonderful book, and a perfect choice for Charlotte Mason inspired families.  You can view it online or download the pdf file, or buy the book.  We read it on my Sony Reader (the best homeschooling investment I have made!)

At the end of each chapter are some exercises woven into the story.  I've taken those exercises and turned them into worksheets.  Without something printed out and ready to go we tend to forget to do the exercises and I really think they are helpful in cementing concepts.  So far I have 6 worksheets completed.  I believe there are 17 chapters in the book, and I hope to have them all completed by the end of the week.

Worksheet 1 for Grammar-Land or, Grammar in Fun for the Children of Schoolroom-Shire by M.L. Nesbitt (by Je...                  &n! bsp;                                                                                                                         

Worksheet 2 for Grammar-Land or, Grammar in Fun for the Children of Schoolroom-Shire by M.L. Nesbitt (by Je...                  &n! bsp;                                                                                                                           

Worksheet 3 for Grammar-Land or, Grammar in Fun for the Children of Schoolroom-Shire by M.L. Nesbitt (by Je...

Worksheet 4 for Grammar-Land or, Grammar in Fun for the Children of Schoolroom-Shire by M.L. Nesbitt (by Je...                  &n! bsp;                                                                                                                         

Worksheet 5 for Grammar-Land or, Grammar in Fun for the Children of Schoolroom-Shire by M.L. Nesbitt (by Je...                  &n! bsp;                                                                                                                         

Worksheet 6 for Grammar-Land or, Grammar in Fun for the Children of Schoolroom-Shire by M.L. Nesbitt (by Je...                  &n! bsp;                                                                                                                           

math 4 today worksheets

brains in a vat


"I want now to ask a question which will seem very silly and obvious .., but which will take us to real philosophical depths rather quickly. [..]
The argument I am going to present is an unusual one, and it took me several years to convince myself that it is really right. [..]
It first occurred to me when I was thinking about a theorem in modern logic, the Skolem-Löwenheim Theorem ..."

Hilary Putnam about brains in a vat and related thoughts about the mind-body problem.


kind of sets

If you are looking for help or directions at NYSML


Here's how to find the people who can help you.

Our small army of volunteers will be wearing distinctive badges like the ones on the table above, made by our awesome sponsor Tough Traveler, in their Schenectady factory.



Some of them, like our Lead Zone Proctor Alexandra Schmidt, a math teacher from Hebrew Academy of the Capital District, also like to wear math t-shirts, so they'll be wearing the same light blue Albany Area Math Circle T-shirts that our students wear to help them stand out doubly from the crowd.



Our UAlbany student guides will carry pennants that look like this:



If you are looking for definitive answers to those all important questions like "Where is our team's classroom?" or "Where is the gym?" or "Where is the restroom?" or "Where am I supposed to be right now?" or "The light went out in our classroom? How do we get it turned back on" or "How long until lunch?" or "Why is there no proctor in our classroom yet?" or "I am an alternate, but I don't remember what team I'm supposed to be working with?" these folks can help you.

If your question relates to the UAlbany campus (like location of buildings and restrooms), your campus guides with the purple and gold Athena pennants are the best go-to p! eople. If you question relates to the NYSML contest (like "Where am I supposed to be right now?" the folks with the Tough Traveler badges are the best bet.) For the most challenging questions, look for volunteers like Alexandra who are wearing light blue math circle shirts AND Tough Traveler badges.



These awesome volunteers will either know the answer to your questions OR they will know how to contact someone who can answer them!

Thanks to all our great volunteers.

help with math