CV Therapeutics Wins FDA Approval

CV Therapeutics (CVTX NASDAQ) and Astellas Pharma Inc announced on Friday that the U.S. Food and Drug Administration has approved Lexiscan (regadenoson), a heart imaging agent.

Astellas signed a license agreement with CV Therapeutics for the development and sale of the agent in July 2007.

CV Therapeutics, Astellas Heart-Monitor Drug Approved

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A Million Dollars for Solving a Math Problem -- Will the Winner Show Up?

The Clay Mathematics Institute has announced the winner of the 1 million dollar prize for the resolution of the Poincare conjecture, which is a conjecture in a branch of mathematics known as topology. The announcement was made this week by Dr. James Carlson, the President of the Clay Institute, that Dr. Grigoriy Perelman of St.Petersburg, Russia is the winner of this prize.

Dr. Perelman in 2006 received the prestigious Fields Medal but never claimed it. The New York Times, in an article, is wondering whether he will (or will not) claim the million dollar prize for solving a longstanding mathematics problem that was one of seven selected for the Millenium Awards by the Clay Institute, which is located in Cambridge, Massachusetts.

Interestingly, Dr. Perelman, 7 years ago, in 3 papers p! osted on the Internet, provided the solution to this math problem, which was posed in 1904 by Poincare. The news of his results quickly spread (at least in math circles) and he embarked on a whirlwind series of speaking engagements, only to return back to Russia and then resign from his post at the Steklov Institute of Mathematics. He stopped answering email messages and, in a sense, disappeared professionally.

According to The New York Times, several teams of mathematicians, using Dr. Perelman’s papers as a guide, completed a full proof of the conjecture in manuscripts hundreds of pages long, showing that Dr. Perelman was right.

The Clay Institute plans to hold a conference to celebrate the solution of the Poincaré conjecture on June 8 and 9 in Paris, France. Dr. Carlson was quoted as saying that Dr. Perelman will let him know in due time whether he will accept this pri! ze.

There are 6 other math problems left as Millen! nium Pro blems, so for those who are interested, you may find the list here. The solution of any of these will garner you a million dollars.

solve math problem

How to find slope intercept

Introduction:

     An area of surface that tends evenly towards top or down is called as slope. The slope of a line in the plane containing the x and y axes is generally represented by the letter m, and is defined as the change in the y coordinate divided by the corresponding change in the x coordinate, between two distinct points on the line. For finding slope intercept we need to know the following things.

Finding slope intercept form:

Slope intercepts form:

The standard form of a line i! s,

ax+by+c=0    

We can express the line as slope intercept form. The slope intercept form of a line is,

y=mx+b, Where m is slope of a line and b is y-intercept of a line.

Finding slope Intercept:

When two points are given, we can use the following formula to find the slope intercept form of a line,

(y-y1)=m(x-x1), where m is slope,(x1,y1) is one point in the line

The slope m can be obtained by using the following formula,

m= (y2-y1)/! (x2-x1), here (x1, y1) and (x2, y2) are the

points in a line or segment.

Or

y-y1/y2-y1=x-x1/x2-x1

Hope you like the above explanation. Please leave your comments, if you have any doubts.

slope intercept form

Rating Climbs: a simple formula

Last post I reviewed some existing ratings for climbs. All fell short. With this in mind, I propose my own rating here.

First, a quick review of my rating philosophy:

1. For simplicity, the rating should be based on net climbing and distance.
2. For sufficiently gradual grades, the rider is assumed to be able to shift to remain in a comfort zone on the climb, and difficulty is then proportional to altitude gained.
3. Beyond a certain grade, climbing becomes much more difficult. It may be possible to construct bicycles which can climb steep grades relatively easily, but the rating is designed for a "typical" fit rider on a typical racing bicycle.

From the description, an obvious candidate for the rating is immediately self-evident for fans of polynomials:

rating = net climbing × (1 + [K × climbing / distance]^N)

The only question is then: what value of K, and what value of N?

First, K: K is the distance / climbing at which a climb is twice as steep as it would be were it more gradual, neglecting issues of added distance traveled (which we're not including in "climb difficulty"). When I'm riding, personally I find "really steep" kicks in at around 12%. Think Metcalf at this past Sunday's LiveStrong Challenge. But the key is that climbs with a high average grade typically vary a bit, or even more than a bit, about that average. So a rock-steady 12% average might be less difficult than the "typical" 12% average grade, which has sections at 15% or more, and others at a more modest (but still steep) 7%. So after playing around with ratings a bit and looking at climbs I know well, I determined a rate of climbing of 10% was a good number to use here. In other words, K = 10.

Next, there's the issue of N. N = 1 reduces to the formula used by Summerson (for infinite K). Honestly I don't think that does justice to really steep climbs like Filbert Street in San Francisco. Filbert gains only a bit over 20 meters, but it's a lot tougher than a 40 meter climb at 15%. So next I tried N = 2. That seemed to work well. N = 3 really kicks in the difficulty at steep grades, but didn't do enough for moderate grade roads.  To be honest, I was going to pick N = 3, but after experimenting with some local climbs, I went back to N = 2.

I plot the formula's result for grades up to 30%. Note this is difficulty per unit climbing, not unit distance traveled. To convert to grade you can use high school trigonometry:

grade = (climbing / distance) / √1 - (climbing / distance)²,

or equivalently

climbing / distance = grade / √1 + grade².

Climbing difficulty per altitude gained
This seems to capture things fairly well. It says Filbert Street, the stiff portion of which gains around 18 meters @ 32% actual grade, corresponding to climbing / height = 30.5%, is equivalent to a 5% climb gaining around 150 meters. Well, difficulty on such a steep climb is hard to assess (for example, I might personally find 35% unclimbable), but for a quick-and-dirty rating, not too bad.

So there it is: my proposed formula for ranking the diffulty of climbs:

rating = net climbing × (1 + [10 × climbing / distance]²)

There's still one open question, however: how do I define a climb? Answer: a "climb" is a segment of road which has a higher climbing rating than any overlapping segment of road. What this implies is no "climb" should be rated lower than any subset of that climb. So suppose a climb starts gradually,then is steep for awhile, then levels out, like Metcalf Road.    One should be careful to consider using the steep portion rather than the absolute endpoints of the altitude-gaining segment for the rating.  More on that when I show some examples, but this is a bit of a point of weakness in the rating scheme versus one which used a detailed profile of the climb.  I'll make an attempt at such a rating system in a future post.

simple discount formula

Goodbye

I'm closing Restraint of the Heartless now.

I think now you deserve to know the meaning of "Nulono".

In Esperanto:

• "nul" is the word for zero.
• "-on-" is the suffix for creating a fraction ("kvar", meaning four, plus "-on", plus "-o", makes "kvarono", a fourth or a quarter)
• "-o" is the standard grammatical ending for nouns

Thus a "nulono" would be a "zero-th", or one divided by zero.

Now, division by zero is impossible (as far as we know in 02009), and would have some very serious implications if it was possible. Consider the expression a=b, where a and b can be any numbers. Let's say, for the sake of discussion, that a=2 and b=3002.

• 0*2 = 0
• 0*3002 = 0

By the reflexive property, 0=0.

Because 0*2 and 0*3002 are both equal to zero, we can substitute them in as such:

• 0*2 = 0*3002

So, by the multiplicative property of equality, if divide both sides by zero (multiply them by one "nulono"), we will arrive at an equation that still holds true.

Thus:

• (0*2)*(1/0) = (0*3002)*(1/0)

or

• (0*2)/0 = (0*3002)/0

Simplifying, we get:

• 2 = 3002

This exact same procedure works with any two numbers. Thus, the existence of 1/0 would make any number equal to every other number. Numbers, thus, would be completely meaningless, and the entire field of mathematics would come crumbling down.

What doe! s this h ave to do with me, you ask?

Yes, I am an atheist (in fact, I am an antitheist). I am also a liberal (in fact, I'm a communist). I am also a feminist (In fact, I support the Equal Rights Amendment*). I am also pro-gay. I am! also in favor of helping those in need. I am also pro-science.

Yet I am also pro-life.

I make many people's stereotyping world-views crumble to the ground, just like 1/0 would do to mathematics. I chose "Nulono" because I pretty quickly decided "Cde. Oxymoron" was setting me up for trouble.

*, though equal prote! ction is already in Amendment XIV, so it's technically redunda! nt.

reflexive property

Quadrilateral Practice

Test your knowledge of quadrilaterals!




Homework:
18.3 Quadrilaterals- both sides all
Reading- Read In the Days of King Adobe and complete Think About It questions. Complete workbook pages 126 & 127 in reading workbook.
Complete Blue Math sheet for tomorrow.
Phy. Ed. tomorrow.

quadrilaterals pictures

Finding better explanations

Another advantage for individual students taking math tutoring sessions

Recently, one of my students told me:

“I like your lecture style better than my professor’s. You have a much better way of explaining the subject. He just starts doing problems, and that’s it. Last time he was having a hard time explaining how he was using the absolute value to solve a problem. We were all confused about it, nobody was understanding what he was doing.”

Sometimes when you are studying math, and you see a topic for the first time, you struggle to understand it, and you work out examples until you find a way to get it. Then if you are a teacher, and you only have that one way of understanding the subject, you go out teaching it that way, and som! etimes you confuse all of your students.

Some teachers care a lot about their students understanding their lectures but some other teachers do not care that much. Sometimes they think: “Well, if they don’t get it, though luck.” However, teachers who care about their students understanding the subject, they spend a lot of time thinking up alternative explanations, or better examples, or better ways to illustrate what’s happening.

I remember a few times (years back when I was a teacher) I felt kind of depressed, disappointed, or frustrated at the end of a lecture because I couldn’t find a way for my students to understand what I was trying to explain. Then, afterwards, I would spend hours, days, even weeks sometimes looking for better ways to explain a particular topic, and the next time I taught that course I was able to explain those topics much better.

I noticed when I started private tutoring, that really sped ! up the process for me, of finding better explanations, because! sitting with students one-on-one, and taking the time to go in depth and in detail with them over their doubts and questions, many times I was able to discover exactly how my students in class were looking at specific problems.

That allowed me to discover faster the reasons why they were not understanding a subject, or why some of my explanations were not working. By tutoring individual students, I was able to find a lot faster a lot more alternative ways of explaining subjects when my students in a large class felt the need for those better explanations.

Tutoring individual students has helped me to focus on finding the best way for each student to understand a given subject, rather than focusing only on covering the whole subject fast in front of a big class.

online math tutoring