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Science makes a difficult

Our “charticle” puts the current debate about
science and technology budgets into historical
context. A few things might surprise you. Click
here to download the PDF.
The White House Office of Science and
Technology Policy earlier this week unveiled its
plan for the fiscal year 2012 science and
technology budget in an event at the American
Association for the Advancement of Science in
Washington, D.C. The budget echoes the
priorities identified by the president in his State of
the Union address and in his “Strategy for
American Innovation,” which the White House
released last week. Here’s a look at how the two
documents match up the strategy with the
money.
A tough-love budget
In what OSTP Director John P. Holdren referred
to as a “tough-love budget” at the release event,
this year’s request to Congress keeps overall
nondefense spending flat while increasing the
crucial investments in science and technology
R&D , so-called STEM (science, technology,
engineering, and math) education, and 21st
century infrastructure that we need to “out-
innovate, out-educate, and out-build” our
economic competitors. Proposing $147.9 billion
for federal research and development overall, the
budget provides increases for areas identified by
the president as critically important to America’s
competitive future, such as sustainable energy,
information technology, advanced
manufacturing, and STEM education initiatives.
Programs that were uncompetitive with these
high-priority areas were reduced, including the
R&D programs at the Department of Veterans
Affairs (falling 12 percent from the 2010 enacted
level) and the Environmental Protection Agency,
which was cut by $11 million. Developmental
research overall saw a decrease in funding from
previous budgets with a total of $79.4 billion —a
decrease mostly due to cuts made in defense
research. The budget proposes a $4 billion
decrease for the Department of Defense R&D
budget from 2010 levels, putting it at $76.6
billion.
Holdren credited tactical, hard-nosed financial
discipline as the reason why the 2012 budget,
which begins in October this year, achieved what
most people thought wasn ’t possible. This year’s
budget calls for a total nondefense R&D budget of
$66.8 billion ($4.1 billion or 6.5 percent more
than the 2010 enacted budget).
Doubling path for key science agency
budgets
The FY 2012 budget proposed by the White
House does not lose focus on previous long-term
agency goals set by the Obama administration.
The three key agencies identified by the president
as crucial to national competitiveness —the
National Institute of Standards and Technology,
the National Science Foundation, and the
Department of Energy ’s Office of Science—
maintain financial momentum to reach their
budget-doubling goal by 2017. The budget
proposes an increase for these agencies of 12.2
percent from the 2010 enacted budget for a total
of $13.9 billion in funding. The NSF increases 13
percent from 2010 enacted levels to $7.8 billion,
the DOE Office of Science increases 10.7 percent
to $5.4 billion, and NIST intramural laboratories
increase to $764 million, a 15.1 percent increase
over 2010.
Clean energy innovation
To further advance American clean energy
innovation, $550 million was directed to continue
to fund DOE ’s Advanced Research Projects
Agency-Energy, or ARPA-E. Additionally, the
budget provides financial support to double the
number of Energy Innovation Hubs from three to
six in order to promote collaboration between
industry and academia.
The three new hubs will focus on rare earth
materials, advanced car batteries, and new
materials to advance the smart grid. Funding for
the existing hubs for building energy efficiency,
fuels from sunlight, and nuclear modeling
continue to receive funding.
Educating our children in science,
technology, engineering, and math
Recognizing that a workforce well trained in
science, technology, engineering, and math, or
STEM, is a critical building block for long-run,
innovation-driven economic success, the budget
also includes $3.4 billion across the federal
government for STEM education. New STEM
initiatives include a $100 million “down payment”
on preparing 100,000 new STEM teachers within
the upcoming decade with 80 percent going to
the Department of Education and 20 percent of
the down payment going to NSF.
Also included is $90 million for the Department of
Education to create a new agency called
Advanced Research Projects Agency-Education,
or ARPA-ED. This new agency mimics the
successful model used by DARPA and ARPA-E to
develop and commercialize game-changing and
transformational new technologies of national
importance. ARPA-ED, according to the OSTP,
will:
Push the field of education research,
development, and demonstration forward by:
sponsoring synthesis and vetting of public and
private R&D efforts; identifying breakthrough
development opportunities; shaping the next
wave of R&D; investing in the development of
new education technologies, learning systems,
and digital learning materials; and identifying and
transitioning the best and most relevant R&D
from other federal agencies.
Infrastructure to keep people, goods, and
information on the move
At the budget-release event at the AAAS on
Monday, U.S. Chief Technology Officer Aneesh
Chopra discussed how investments in 21st
century infrastructure will provide the
foundational capacity to foster the growth of new
jobs and industries. Specific infrastructure
investments that the budget makes include a one-
time $5 billion investment in the Universal Service
Fund to ensure all Americans have access to 4G
high-speed wireless, even those living in remote
areas. Some of this will be paid for by increasing
the wireless spectrum available for mobile
broadband. The auction of these new frequencies
to companies will cut the deficit by nearly $10
billion over the next decade.
Advancing manufacturing innovation
We’ve pointed out at Science Progress many
times that innovation does not just take place in
labs; it also happens on assembly lines. The
president ’s budget would increase funding for the
National Science Foundation, the National Institute
of Standards and Technology, the Department of
Energy, and the Defense Advanced Research
Projects Agency to support development of
advanced manufacturing technologies.
Investments in nanomanufacturing,
biomanufacturing, next-generation robotics, and
cyber-physical systems are important to keeping
American manufacturing on the cutting edge of
innovation. In addition, the budget proposes
reauthorizing the wildly successful and
oversubscribed section 48(c) clean energy
manufacturing tax credit for $5 billion.
Leveraging private-sector investment in
innovation
The FY 2012 budget also displays a solid
realization of the importance of private-sector
investment in innovation. The government
simply does not have the size, expertise, or
resources to directly develop the technologies of
the future. The research and experimentation tax
credit helps unlock private investment in research
and development by encouraging companies to
develop new technology.
But since 1981 the R&D tax credit has been
renewed by Congress on a temporary basis
every two to three years, creating considerable
uncertainty for businesses trying to make the
necessarily long-term investments in technology
research and development. The president ’s
budget proposes finally to expand, simplify, and
make permanent the R&D tax credit.
The budget also includes a new Innovation Fund
within the Small Business Investment Company,
or SBIC, program. The new fund will specifically
address the “valley of death” financing gap that
prevents promising technical ideas from
becoming job-creating business plans by making
$200 million in matching grant funds available to
augment private investments that support job-
creating and innovative startup technology
companies with high growth potential.
In addition, the budget includes $15 million for the
Small Business Administration ’s Emerging
Leaders initiative to “enhance small business
participation in regional economic clusters.” By
signaling to the private sector that investments in
research and development will carry tax benefits
long into the future, this administration hopes to
help kick start innovation across all of America ’s
industries.
Looming budget battle
Though these key investments in science and
innovation have made it into the president ’s
budget, they still face a long road ahead to secure
funding in the eventual FY 2012 budget that
Congress must pass and the president must sign.
President Obama ’s budget can be thought of as
an “opening bid,” as he characterized it earlier this
week, in a long process of haggling that will take
place between the Republican-controlled House
and the Democrat-controlled Senate and White
House.
Elaine Sedenberg is an Intern at Science Progress
and an undergraduate in honors biochemistry at
the University of Texas at Austin. Sean Pool is
Assistant Editor for Science Progress. Also see
our new “charticle” on U.S. Science R&D 101 (pdf).

Taylor Alison Swift Profile

Taylor Swift yang lahir pada 13 Desember 1989 adalah
penyanyi sekaligus pencipta lagu musik country. Memulai
debutnya lewat jalur indie di tahun 2006, Swift memulai
perjalanan karir bermusiknya dengan merilis single Tim
McGraw dan langsung menempati posisi di tangga lagu U.S.
Billboard.
Swift lahir di Pennsylvania dan tinggal dengan orang tuanya
serta saudara lelakinya, Austin di Tennesse. Bakat
bermusiknya diperolehnya dari neneknya yang seorang
penyanyi opera serta dipengaruhi oleh penyanyi LeAnn
Rimes. Di usia 10 tahun, ia sudah mulai tampil di depan
umum dan mulai menulis lirik lagu.
Pada 24 Oktober 2006, Swift yang mempunyai nama asli
Taylor Alison Swift merilis albumnya yang berjudul
TAYLOR SWIFT. Hampir seluruh lagu di album ini ditulis
olehnya sendiri. Dan hasilnya, album penyanyi muda berbakat
ini dapat terjual hingga 61.000 kopi pada minggu pertamanya.
Tak berapa lama setelahnya, album ini menjadi juara di tangga
lagu Billboard Top Country.
Swift juga mengeluarkan album SOUNDS OF THE SEASON
yang dirilis pada 16 Oktober 2007. Sayangnya, album ini tidak
sesukses album perdananya.
Pada musim gugur 2008, pelantung tembang Love Story dan
White Horse ini merilis album keduanya, FEARLESS. Dalam
tiga minggu saja sudah terjual sebanyak 330 ribu keping.
Lewat album FEARLESS, Swift mendapatkan penghargaan di
ajang Academy of Country Music Awards Ke-44 dalam
kategori Album of The Year. Ia juga menerima trofi crystal
milestone sebagai artis musik country dengan penjualan
terbaik selama 2008.

Stay beautiful..

Drew looks at me, I fake a smile so he won't see
That I want and I'm needing everything that we should be
I'll bet she's beautiful, that girl he talks about
And she's got everything that I have to live without
Drew talks to me, I laugh cause it's so damn funny
That I can't even see anyone when he's with me
He says he's so in love, he's finally got it right,
I wonder if he knows he's all I think about at night
[Chorus:]
He's the reason for the teardrops on my guitar
The only thing that keeps me wishing on a wishing star
He's the song in the car I keep singing, don't know why I do
Drew walks by me, can he tell that I can't breathe?
And there he goes, so perfectly,
The kind of flawless I wish I could be
She'd better hold him tight, give him all her love
Look in those beautiful eyes and know she's lucky cause
[Repeat Chorus]
So I drive home alone, as I turn out the light
I'll put his picture down and maybe
Get some sleep tonight
He's the reason for the teardrops on my guitar
The only one who's got enough of me to break my heart
He's the song in the car I keep singing, don't know why I do
He's the time taken up, but there's never enough
And he's all that I need to fall into..
Drew looks at me, I fake a smile so he won't see.

ABRAHAM LINCOLN, SLAVERY, THE CIVIL WAR, THE SENATE
Washington, D.C., Jan. 28, 1861
His speech was a memorable one — that was the only thing that
everyone could agree upon. The anti-slavery New-York Tribune
called it “pathetic.” A pro-Southern newspaper in Washington
found it “solemn and impressive.” Today, the words have even
more resonance than in 1861. For they may have been the very
first eerie echo of what would become a famous American
rallying cry: “The South will rise again.” They also marked a
moment in national politics whose repercussions are still being
felt.
Library of Congress
Sen. Alfred Iverson of Georgia
After six years in the Senate, Alfred Iverson of Georgia had come
to the Capitol that morning to bid farewell to his fellow legislators,
to Washington — and to the United States of America. He waited
impatiently through the chaplain’s morning prayer, a report from
the secretary of the Navy, and multiple petitions from citizens
begging their lawmakers to forge a Union-saving compromise. At
last he could sit quiet no longer and interrupted one of his
colleagues, begging leave to speak.
Iverson began by having the Senate secretary read Georgia’s
ordinance of secession, passed little more than a week earlier.
Unlike other Southern politicians, he had never claimed that his
state ’s withdrawal was constitutional. It was, he frankly admitted,
simply a revolution. And now he challenged the Northern
legislators:
You may acquiesce in the revolution, and acknowledge the
independence of the new confederacy, or you may make war
on the seceding States, and attempt to force them back into a
Union with you. If you acknowledge our independence, and
treat us as one of the nations of the earth, you can have
friendly intercourse with us; you can have an equitable
division of the public property and of the existing public debt
of the United States. If you make war upon us, we will seize
and hold all the public property within our borders or within
our reach.
As he went on to imagine this war that might well lie ahead, the
senator ’s rhetoric grew more and more heated:
You boast of your superior numbers and strength.
Remember that “the race is not always to the swift, nor the
battle to the strong.” You have your hundreds of thousands of
fighting men. So have we; and, fighting upon our own soil, to
preserve our rights, vindicate our honor and defend our
homes and firesides, our wives and children from the invader,
we shall not be easily conquered.
You may possibly overrun us, desolate our fields, burn our
dwellings, lay our cities in ruins, murder our people and
reduce us to beggary, but you cannot subdue or subjugate us
to your government or your will. Your conquest, if you gain
one, will cost you a hundred thousand lives, and more than a
hundred million dollars. Nay more, it will take a standing army
of 100,000 men, and millions of money annually, to keep us
in subjection.
You may whip us, but we will not stay whipped. We will rise
again and again to vindicate our right to liberty, and throw off
your oppressive and accursed yoke, and never cease the
mortal strife until our whole white race is extinguished and our
fair land given over to desolation.
“The Rubicon is passed,” Iverson concluded, “and it shall never,
with my consent, be recrossed.” There was still a chance, he
conceded, that other Southerners might agree to a compromise.
“ I may safely say, however, that nothing will satisfy them, or
bring them back, short of a full and explicit recognition of the
guarantee of the safety of their institution of domestic slavery and
the protection of the constitutional rights for which in the Union
they have so long been contending, and a denial of which, by
their Northern confederates, has forced them into their present
attitude of separate independence. ”
Library of Congress
Members of the Senate, Thirty-Sixth Congress (1859-1861), in a
composite photograph by Mathew Brady. Jefferson Davis is in the
lower left-hand corner, Robert Toombs in the lower right. David
Levy Yulee is the third senator directly above Davis. Stephen
Mallory is just above Toombs, diagonally to the left. CLICK TO
ENLARGE
Not all of Iverson’s colleagues listened to him with the respect that
he might have hoped for. According to one Washington
newspaper, some of the Republican senators interrupted several
times with “derisive cheers and laughter.” (Its editor expressed his
“deep sense of shame and disgust that men occupying such high
and dignified positions could be guilty of such unseemly levity.”)
But the legislators were growing weary of operatic farewells.
Iverson ’s swan song was just the latest of many throughout the
preceding month. As they joined the new Confederacy, one
Southern senator after another had risen to declaim his valedictory
address.
Some left bitter recriminations as their last entries in the
Congressional annals, others gave polite and regretful farewells.
On Jan. 7th Iverson ’s fellow Georgian, Robert Toombs, had used
his departure speech to fire parting shots at “Black Republicans”
and abolitionists: “We want no negro equality, no negro
citizenship; we want no negro race to degrade our own; and as
one man [we] would meet you upon the border with the sword
in one hand and the torch in the other. ”
Library of Congress
The Senate Chamber, 1859
On Jan. 21, no fewer than five senators had departed in quick
succession. Stephen Mallory of Florida blasted the North with
brimstone: “You cannot conquer us. Imbrue your hands in our
blood and the rains of a century will not wipe from them the stain,
while coming generations will weep for your wickedness and
folly. ” His fellow Floridian, David Levy Yulee, was somewhat more
genteel, “acknowledging, with grateful emotions, my obligations
for the many courtesies I have enjoyed [from] the gentlemen of
this body, and with most cordial good wishes for their personal
welfare. ” Alabama’s Clement Clay and Benjamin Fitzpatrick likewise
departed.
Finally it was the turn of the man whom many already suspected
would be the first president of the new Confederacy: Jefferson
Davis of Mississippi. In a low, hoarse voice, weakened by recent
illness and by the emotion of the moment, he explained why his
state had seceded:
It has been a conviction of pressing necessity — it has been a
belief that we are to be deprived in the Union of the rights
which our fathers bequeathed to us — which has brought
Mississippi to her present decision. She has heard proclaimed
the theory that all men are created free and equal, and this
made the basis of an attack upon her social institutions; and
the sacred Declaration of Independence has been invoked to
maintain the position of the equality of the races.
Then Davis bade a gracious farewell to his longtime colleagues: “I
carry with me no hostile remembrance. Whatever offense I have
given which has not been redressed, or for which satisfaction has
not been demanded, I have, Senators, in this hour of parting, to
offer my apology. … Senators, having made the announcement
which the occasion seemed to me to require, it only remains to
me to bid you a final adieu. ”
At these words, Davis and his four fellow Southerners turned to
make their way slowly up the aisle toward the door. It is said that
spectators sobbed in the gallery, as stern legislators choked back
tears. The Union seemed truly — perhaps irrevocably —
dissolved. Democrats and a few moderate Republicans crowded
around to shake the five men ’s hands and wish them well. The
rest of the Northerners sat, hands folded, at their desks.
RELATED
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An unfolding history of the Civil War with photos and articles
from the Times archive and ongoing commentary from Disunion
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That dramatic exit overshadowed another piece of business
transacted in the Senate that day, one that may, however, have
been just as momentous. After the Southerners ’ departure, their
remaining colleagues passed a bill that had languished for almost a
year: one that admitted Kansas to the Union as a free state. The
newly powerful Republican majority thus settled peacefully an
issue over which so much American blood had already been
spilled.
A week later, on Jan. 28, as Iverson was speaking in the Senate,
the House of Representatives made Kansas ’ statehood official. The
balance of power had shifted for good that week, in the nation and
in Congress. Throughout the Civil War and Reconstruction years,
a Congress dominated by Northern Republicans would pass
legislation that, collectively, would change America forever. And
the South might “rise again and again” over the following century,
but it would never regain the political leverage that it had just
willingly abdicated.
Join Disunion on Facebook »
Sources: Congressional Globe, 36th Congress, 2nd Session; New
York Times, Jan. 29, 1861; New-York Tribune, Jan. 29, 1861; The
Constitution (Washington, D.C.), Jan. 30, 1861; Thomas Ricaud
Martin, ed., “The Great Parliamentary Battle and Farewell
Addresses of the Southern Senators on the Eve of the Civil War.”
Note: The exact text of Iverson’s address varies a bit from source
to source, due to the vagaries of period newspaper reports. I have
relied on the official record, the Congressional Globe.
Adam Goodheart is the author of the forthcoming book “1861: The
Civil War Awakening.” He lives in Washington, D.C., and on the
Eastern Shore of Maryland, where he is the Hodson Trust-
Griswold Director of Washington College ’s C.V. Starr Center for
the Study of the American Experience.
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7 Corp Circle terbesar didunia

7 crop circle terbesar di dunia

Oleh Agung Dwi Cahyadi  | Senin, 24 Januari 2011 | alam

Wikipedia

 

Fenomena crop circle di Sleman, Daerah Istimewa Yogyakarta boleh jadi yang pertama kali di Indonesia. Namun sebenarnya, ada beberapa pola serupa dengan ukuran jauh lebih besar di beberapa negara di dunia.

Crop circle  adalah suatu pola teratur yang terbentuk karena adanya perebahan sebagian tanaman, biasanya ditemukan di ladang pertanian, yang terkadang membentuk pola-pola tertentu. Pola tersebut seringkali terbentuk hanya dalam waktu semalam. Karena pola yang ditemukan tidak selalu berbentuk lingkaran, crop circle juga sering disebut crop formation.

Sepanjang paruh ketiga abad 20, sekira 10 ribu crop formation dilaporkan ditemui di 26 negara (27 jika termasuk fenomena di Sleman, Yogyakarta) di seluruh dunia dan 90% di antaranya ada di selatan Inggris. Dari sekian banyak fenomena crop formation di seluruh dunia, berikut ini beberapa di antaranya yang dilaporkan sebagai crop formation terbesar. Namun, tidak semua crop formation ini misterius. Beberapa di antaranya merupakan buatan manusia.

1. The Human Butterfly. Sebuah crop formation berbentuk manusia bersayap kupu-kupu ditemukan di selatan Belanda, dekat kota Goes pada bulan Agustus 2009. Ini adalah crop formation terbesar di dunia yang pernah ditemui dengan ukuran 530 x 450 meter. Sekelompok orang dari Project Atlas membentuk crop formation ini sebagai simbol keindahan sekaligus kerapuhan manusia.

2. Dalian Cornfiled Maze. Seperti namanya, crop formation ini memang sebuah maze raksasa berukuran 22,7 hektare yang terletak di dekat jalan raya Dandong-Dalian, China. Rute terpendek maze ini sepanjang 3.800 meter dan perlu waktu sekira satu jam untuk menyelesaikan maze ini. crop formation ini sengaja dibuat pemerintah setempat sebagai sarana promosi untuk menarik investor mengembangkan area pertanian di daerah Dalian.

3. Pada tahun 2005 sebuah ladang mint di Dalponte Farms, Richland, New Jersey, AS tiba-tiba ditemukan bergambar kelelawar seluas lebih dari 6.070 meter persegi. Namun setelah diperhatikan, gambar kelelawar tersebut merupakan logo dari Bacardi, sebuah perusahaan yang memproduksi minuman beralkohol.

4. Crop formation juga ditemui di Italia. Salah satu yang terbesar dilaporkan terlihat di Torino pada 13 Juni 2010. Pola yang ditemui di negeri ini berbentuk bunga dengan enam kelopak. Sebagian pengamat menafsirkan pola crop formation itu dengan menggunakan perpektif galaksi.

5. Crop fromations terpanjang terlihat di Etchilhampton, Wiltshire, Inggris pada tahun 1996. Bentuknya berupa lingkaran dan jalan kecil yang saling bertautan sepanjang kurang lebih 1250 meter dari satu ladang ke ladang lainnya.

6. Sebuah crop formation berbentuk mandala dengan tujuh kelopak ditemukan pada tahun 1998 di Alton Barnes, Wiltshire, Inggris. Area yang membentuk pola tersebut membentang seluas 6 ribu meter persegi.

7. Rekor untuk crop formation dengan desain terbesar dan lingkaran terbanyak dalam sebuah formasi masih dipegang sebuah motif yang ditemukan di Milk Hill, Wiltshire, Inggris. Pada 12 agustus 2001, pola dengan 409 lingkaran kecil membentuk sebuah desain berlengan enam berdiameter sekira 243 meter.

Crop Circle Di Sleman Yogyakarta

Pola geometris mendadak muncul di Sleman

Oleh Administrator  | Minggu, 23 Januari 2011 | alam

Hendi Kurniawan/Tribunjogja.com

 

Pola geometris yang tiba-tiba muncul di areal persawahan Desa Rejosari, Sleman.

Fenomena unik untuk pertama kalinya muncul di wilayah Yogyakarta, bahkan kemungkinan pertama di Indonesia. Fenomena yang juga pernah muncul di berbagai negara ini sering dikaitkan dengan UFO meskipun kesahihannya belum dapat dipastikan. 

Jauhari, warga Kebondalem, Sleman, mengaku takjub menyaksikan peristiwa langka di persawahan yang terletak tak jauh dari Bandara Internasional Adisucipto ini. Jauhari dan warga lain mendaki Gunung Suru yang jalannya licin untuk melihat pola itu secara keseluruhan. "Ini seperti fenomena pendaratan UFO yang sering dibahas di televisi, tapi saya tidak tahu apakah benar adanya. Hanya Allah yang tahu sebabnya," ucapnya kepada Tribun di lokasi, Minggu (23/1).

Sebatas yang ia lihat, Jauhari tak mendengar keterangan apa pun terkait terbentuknya pola misterius itu, termasuk suara gemuruh seperti yang didengar oleh warga lain. Yudi, pemuda Desa Rejosari yang pertama kali melihat pola itu di sawah, juga mengaku tak mendengar suara apa-apa. Padahal, Yudi nongkrong di depan rumah Basori sampai pukul 03.00. "Bahkan semalam tidak ada hujan atau angin. Tahu-tahu tadi pagi sudah terbentuk pola ini (lingkaran) di tengah sawah," imbuh Yudi.

Beberapa warga lain mengaku mendengar suara gemuruh selama sekitar 30 menit. "Tetapi saya tidak gubris suara itu. Saya pikir itu suara helikopter lewat," kata Basori, warga yang rumahnya berada di sebelah utara sawah tersebut. Pengakuan Basori itu juga didukung oleh pernyataan istrinya.

Pola geometris di beberapa negara dihasilkan oleh sekelompok seniman. Para seniman bisa membuat hasil karya itu dalam waktu semalam. Beberapa pola geometris seperti ini bahkan digunakan oleh perusahaan untuk beriklan. John Lundberg adalah salah satu contoh seniman yang telah membuat pola geometris di Inggris dan beberapa negara lain dalam rangka karya seni dan komersial. (Hendi Kurniawan/Tribunjogja.com, Wikipedia)

HP shakes up board, adds Meg Whitman and 4 others

By Gabriel Madway

SAN FRANCISCO | Fri Jan 21, 2011 3:45am EST

 

SAN FRANCISCO (Reuters) - Hewlett-Packard Co is shaking up a board criticized by many as dysfunctional, bringing in five new directors including former eBay chief Meg Whitman, as new CEO Leo Apotheker remakes the company.
HP's board had for years come under fire from shareholders and business leaders such as Oracle's Larry Ellison, most recently after it forced out Mark Hurd as CEO in controversial fashion.
The new directors will bring fresh thinking to the world's largest technology company by revenue, including much-needed expertise in areas such as telecommunications and international experience, the company said.
In addition to Whitman, HP named as directors Shumeet Banerji, CEO of Booz & Co; Gary Reiner, former chief information officer of General Electric Co; Patricia Russo, former CEO of Alcatel-Lucent; and Dominique Senequier, CEO of AXA Private Equity.
The technology giant said directors Joel Hyatt, John Joyce, Robert Ryan and Lucille Salhany will not stand for reelection by shareholders.
"This change was two things -- the handling of the Mark Hurd situation, which was very controversial, and that with a new CEO it makes sense to have someone new," said Kaufman Bros analyst Shaw Wu.
Hurd was accused last June of sexual harassment by a female contractor, but a board investigation found no evidence to back that up. Hurd resigned August 6 after the board said he filed inaccurate expense reports to conceal a "close personal relationship" with that contractor, something Hurd's representatives have disputed.
The four directors leaving are doing so voluntarily, HP said. Hyatt and Joyce have been directors since 2007, Ryan since 2004, and Salhany since 2002. The changes leave HP with 13 board members.
According to a source familiar with the matter, Salhany and Ryan pushed hard for Hurd's ouster from the outset. Hyatt and Joyce were more supportive of Hurd at first but eventually went along with the unanimous decision to seek his resignation.
NEW BLOOD
Whitman is arguably the best-known of HP's new directors. At eBay, she oversaw a period of robust growth but was criticized for her acquisition of Web telephone company Skype. She recently lost out in a bid to become governor of California.
"With Leo coming in as CEO, we both thought it was appropriate to look at the board," HP Chairman Ray Lane in an interview.
Lane said Russo will bring a strong track record in the telecommunications industry, called Reiner an "iconic CIO," and lauded Banerji's and Senequier's international experience.
Gleacher & Co analyst Brian Marshall said investors should welcome the move.
"To shake it up a little bit and get some of the old-school guys out of there and get some new-school blood in there, highlighted by Meg Whitman, this is a positive development," he said.

Sample questions for those of you who are interested in algebra.

Unit 1
1. Multiply and simplify by factoring: a) √( 15x2y) √
( 6y2) ( √ is square root)
Step1: Multiply the two radicands together. √
( 15x2y) √( 6y2) = √(90 x2y3)
Step2: Write √(90 x2y3) as the product of a
perfect square = √(9x 2y2 )√(10y)
Step3: Take the square root of the radicand that is
a perfect square = 3xy √(10y)
________________________________________
2. State whether each number is rational,
irrational, or complex.
a) -5/6 rational b) 8.93 rational (repeating) rational
d) √ 6 irrational e) 5.2323 rational f) 3 + 4i complex
Note: Although not usually done so, a) b) c) d)
and e) can be thought of as a complex number
a + bi where b = 0.
________________________________________
3. Multiply: (2 – 3i)(2 + 3i)
Step1: 4 -6i + 6i – 9i2
Step2: Remember that i2 = -1
Step3: 4 -6i + 6i – 9i2 = 4 – 9 (-1) = 4 + 9 = 13
________________________________________
4. Find the conjugate of 3 - 7i. The conjugate of a
number a + bi is a – bi. Here, a = 3, b = -7 so the
conjugate is 3 – (-7)I = 3 + 7i
________________________________________
5. Divide: (-5 + 9i) by (1 – 2i)
Step1: Write as a fraction (-5 + 9i)/(1-2i)
Step2: Multiply numerator and denominator by (1
+ 2i)
Step3: The numerator becomes (-5 + 9i)(1 + 2i) =
-23 - i
Step4: The denominator becomes (1 -2i)(1 + 2i) =
5
Step5: = (-23/5) – (1/5)i
________________________________________
6. Expand (3x – 2)6
(x+t)n = nC0xn+nC1xn−1t+nC2xn−2t2+• • •+nCn
−2x2tn−2+nCn−1xtn−1+nCntn
Step1: (3x – 2)6 = 6C0(3x)6+6C1(3x)5(-2)+6C2(3x)
4(-2)2+• • •+6C4(3x)2(-2)4+6C5(3x)(-2)5+6C6(-2)6
Step2: nCr = n!/(r!(n-r)!)
Step3: 6C0 = 1, 6C1 = 6 , 6C2 = 15 , 6C3 = 20 , 6C4
= 15 , 6C5 = 6, 6C6 = 1
Step4: (3x – 2)6 = (3x)6+(6)(3x)5(-2)+(15)(3x)4(-2)2
+(20)(3x)3(-2)3+15(3x)2(-2)4+6(3x)(-2)5+(-2)6
Step5: (3x – 2)6 = 729x6+(6)(243x5)(-2)+(15)(81 x4)
(4)+(20)(27x3)(-8)+15(9x2)(16)+6(3x)(-32)+64
Step6: (3x – 2)6 = 729x6 - 2816x5 + 4860 x4 -
4320 x3 + 2160x2 - 576x + 64
________________________________________
7. Complete the square: Let q(x) = 3x2 - 24x + 50
Step1: a = 3, b = -24, c = 50
Step2: q[u - b/(2a)] = au2 - (b2 – 4ac)/4a
Step3: b/2a = -24/2(3) = -24/6 = -4
Step4: (b2 – 4ac)/4a = [(-24)(-24) – [4(3)(50]/4(3) =
(576 – 600)/12 = -24/12 = -2
Step5: q(u – b/2a) = q[u – (-4 )] q(u +4) = 3u2 –
(-2) = 3u2 +2 q(u+4) = 3u2 +2
Step6: x = u – (b/2a), therefore u = x + (b/2a) u = x
– 4
Step7: q(u + 4) = 3u2 +2 q(x – 4 + 4) = 3u2 +2 q(x)
= 3u2 +2 q(x) = 3 (x – 4)2 + 2
Unit 2
1. Are the following systems of equations
consistent, inconsistent, or dependent (infinitely
many solutions)?
a. 2x – y = 4, 5x – y = 13 b. 6x – 2y = 2, 9x – 3y =
3
Solution:
a) 2x – y = 4 can be written as y = 2x – 4, 5x – y =
13 can be written as y = 5x – 13
2x – 4 = 5x – 13 9 = 3x x = 3
(Another way to solve this is to write the first
equation as y = 2x – 4 and substitute in the
second equation 5x – (2x – 4) =13 when you solve
this you also obtain x = 3)
Substituting x = 3 in the first equation we get, 2(3)
– y = 4 6 – y = 4 y = 2
But when we substitute in the second equation,
5(3) – y =13 y = 2
These equations are consistent (and independent).
Answer: consistent
b) If you divide 6x – 2y = 2 by 2 3x – y = 1
If you divide 9x – 3y = 3 by 3 3x – y = 1
Since these equations reduce to the same
equation, they are dependent.
Answer: dependent
________________________________________
2. Solve the set of equations for x and y:
2x – y = -1
3x – 3y = -2
You can use the substitution method.
y = 2x + 1 and substitute in 3x – 3y = -2 3x - 3(2x +
1) = -2
3x -6x -3 = -2 -3x -3 =-2 -3x = 1, x = -1/3
Substituting in y = 2x + 1 y = -2 (1/3) + 1 y = -2/3 +
1 y = 1/3
Answer: x = -1/3, y = 1/3
________________________________________
3. If
write a formula for x in terms of A, B, C, D, E, and
F.
Solution: G(Bx + C) = A(Ex – F) è GBx + GC = AEx –
AF è GC + AF = AEx – GBx
GC + AF = (AE – GB)x è Divide each side of the
equation by (GC + AF)
Answer: x = (GC + AF)/( AE – GB)
________________________________________
4. A third-order determinant can be defined by:
a1 b1 c1
a2 b2 c2
a3 b3 c3
= a1 times
b2 c2
b3 c3
- a2 times
b1 c1
b3 c3
+ a3 times
b1 c1
b2 c2
Find the value of the determinant:
-1 -2 -3
3 4 2
0 1 2
=
(-1) times
4 2
1 2
-(3) times
-2 -3
1 2
0
Unit 3
1. If f(x) = (125y3 – 8) and g(x) = (5y – 2), find f(x)/
g(x)
Answer: 25y2 + 10y + 4
________________________________________
2. When (x3 +9x2 - 5) is divided by (x2 – 1), find
the value of the quotient and the remainder.
Answer: quotient is x + 9 and remainder is x + 4
________________________________________
3. When (x2 – 3x +2k) is divided by x + 2, the
remainder is 7. Find the value of k.
Solution:
Quotient: x - 5
Divisor: x + 2
Dividend: x2 – 3x +2k
x2 +2x
-5x + 2k
-5x – 10
Remainder: 2k + 10
2k + 10 = 7 2k = -3, k = -3/2
Answer: k = -3/2
________________________________________
4. Show that the triangle whose vertices are A =
(-1, 2), B = (4, -3), and C = (5, 3) is an isosceles
triangle.
Solution:
Use the distance formula: d = √[(x2 – x1)2 + (y2 –
y1)2]
AB = √50, BC = √37, AC = √37
Since 2 sides are equal, the triangle is isosceles.
________________________________________
5. Quadrilateral PQRS has vertices P = (-1, 3), Q =
(2, 3), R = (3, -4), and S = (-3, -3). What are the
coordinates of the vertices if it is translated by the
equations:
x’ = x – 1, y’ = y + 2?
Solution: This is demonstrated for 1 translation: P:
x = -1, y = 3
x ’ = x – 1 x’ = -1 – 1 = -2
y’ = y + 2 y’ = 3 + 2 = 5
P’: (x’, y’) = (-2, 5)
Answer: P’ = (-2, 5), Q’ = (1, 5) , R’ = (2, -2), S’ =
(-4, -1)
________________________________________
6. The x, y coordinate system is translated in
such a way that the origin Q of the new system
is (-3, -1).
a. What are the transformation equations?
Answer: x’ = x + 3, y’ = y + 1
b. Point A (3,4) in the original system is equivalent
to what coordinates in the new system?
Solution:
Use the same method as in example 5.
Answer: (6, 5)
c. Point B (3, -2) in the new system is equivalent
to what coordinates in the original system?
Solution:
x ’ = x + 3 3 = x + 3 è x = 0
y’ = y + 1 -2 = y + 1 è y = -3
Answer: (0, -3)
________________________________________
7. Town A is in a French-speaking country and
town B is in an English speaking country. They
are 100 miles apart and connected by a straight
road. When giving directions to a point on the
road between them the people in town A will say
that it "x km from here "(i.e. from town A) while
the people in town B will say that it is "x ' miles
from here" (i.e. from town B). Thus every point
on the road has an x- coordinate given it by town
A and an x' coordinate given it by town B.
(For this problem 8 km = 5 miles)
a. If the border is 30 miles from town A the
people in town B will say that it is "___________
miles from here".
b. Give a general formula for calculating the x
' (miles from B) coordinate in terms of the x (km
from A) coordinate. x ' = ______________ * x + _______
____________
Solution:
Assume that the relation between the two
coordinate systems (x for the A-people and x ′
for the B-people) is x ′ = px + q.
Note the fact that the A-town itself would have
coordinates x = 0 and x ′ = 100. Thus, we have
100 = p(0) + q q = 100.
Similarly, for the B-town, we have x ′ = 0 and x
= 160 (100 miles = 160 km). So, 0 = p(160) + 100
where we have used the known value of q. Thus
p = -100/160 = -5/8 and the formula is:
Answer: x ′ = (-5/8)x + 100
Now the given border is 30 miles from A, so it is
70 miles from B.
Thus the border has x ′ = 70. The question (a)
was actually asking only this much!
We do more. The x-coordinate for the border
(distance in km from A) is found by solving:
70 = ( -5/8)x + 100 x = 48 The border is 48 km
from A
Unit 4
1. Write parametric equations for 2x + 4y = 11.
Solution:
Write 2x + 4y = 11 as 4y = -2x + 11 y = (-1/2)x +
11/4
Let x = t and substitute in y y = (-1/2)x + 11/4 y =
(-1/2)(t) + 11/4 y = (11 -2t)/4
Answer: x = t and y = (11- 2t)/4
________________________________________
2. Find the x- and y-intercepts of the equation: y =
(2/5) x - 2
Solution:
The equation is already in the slope-intercept
form so the y-intercept is -2
To find the x-intercept let y=0 and solve for x 0 =
(2/5) x – 2 2 = (2/5) x x = 5
Answer: x-intercept is 5 and y-intercept is -2
________________________________________
3. Write an equation in slope-intercept form of the
line whose parametric equations are x = 3t - 4
and y = -t + 8.
Solution:
Write y = -t + 8 as t = 8 - y and substitute in x =
3t – 4 x = 3(8 – y) - 4 x = 24 – 3y -4
x = 20 – 3y 3y = -x + 20 y = (-1/3)x +20/3 in slope-
intercept form
Answer: y = (-1/3)x + 20/3
________________________________________
4. The midpoint of a line-segment is (-1, 2). One
end of the segment is (-5,-3). Find the coordinates
of the other end of the segment.
Solution:
Use the midpoint formula xM = (1/2)(x1 + x2), yM
= (1/2)(y1 + y2)
The coordinates of the other endpoint are (x2,y2)
-1 = ½(-5 + x2) -2 = -5 + x2 x2 = 3
2 = ½(-3 + y2) 4 = -3 + y2 y2 = 7
Answer: (3, 7)
________________________________________
5. Find the coordinates of the point which is 3/4
of the distance from the point (2, 3) to the point
(6, -5).
Solution:
x = a1 + t(a2-a1), y=b1 + t(b2-b1) and use t=3/4
x = 2 + (3/4)(6-2) x = 2 + (3/4)(4) = 5
y = 3 + (3/4)(-5-3) = 3 + (3/4)(-8) = -3
The point is (5,-3)
To check, find the distance between (2,3) and
(5-3) and the distance between (2,3) and (6,-5).
When you simplify the radicals, the distance
between (2,3) and (5-3) is 3 √5 while the distance
between (2,3) and (6,-5) is 4√5.
Answer: (5,-3)
________________________________________
6. What is the slope-intercept equation of the line
that contains the point (3, 4) and is perpendicular
to the line y = 1/3x – 6?
Solution:
The slope of the line y = 1/3x – 6 is 1/3 so the
slope of the perpendicular line will be -3.
Use the point slope formula, where (3,4) is on the
new line.
y – y1 = m(x – x1) y – 4 = -3(x - 3) y – 4 = -3x +9
y = -3x + 15
Answer: y = -3x + 15
________________________________________
7. Find the equation of the line whose x-intercept
is -3 and y-intercept is 2.
Solution:
Use the intercepts formula: x/p + y/q = 1 where
(p,0) and (0, q) are the intercepts.
(x/-3) + y/2 = 1 -2x + 3y = 6
Answer: -2x + 3y = 6
________________________________________
8. Find the equation of a line through the points
(5,3) and (1,7) in the slope-intercept form.
Solution:
Use the equation y2-y1=m(x2-x1)
7-3 = m(1-5) 4 = -4m m = -1
Substituting the point (5,3) in the formula y - y1=
m(x-x1) y-3=-1(x-5) y-3=-x+5
y = -x + 8
Answer: y = -x + 8
________________________________________
9. Consider the parametric Line 1: x = t+1, y = −3t
+1 and another Line 2: x = −2s + 4, y = −s + 1.
Find their common point.
Solution:
Write the two implied equations for the common
point:
t + 1 = −2s + 4 and − 3t + 1 = −s + 1.
Solve the two equations for s and t. We get t =
3/7 and s = 9/7.
Substitute t = 3/7 in the parametric equations of
Line 1: x = t + 1, y = −3t + 1
x = (3/7) + 1 x = 10/7
y = -3(3/7) + 1 = -9/7 + 1 = -2/7
Answer: The point is (10/7, -2/7)
________________________________________
10. Consider the parametric line: x=3 t+3, y=t+2.
Determine the slope-intercept equation for the
same line.
Solution:
Make a suitable combination of the two equations
to eliminate the parameter:
x = 3t + 3 x = 3t + 3
-3(y = t + 2) -3y = -3t - 6
x – 3y = -3 x + 3 = 3y y = (1/3)x + 1
Answer: y = (1/3)x + 1
Unit 5
1. If f(x) = 2x3 -3x2 + 1 find:
a. f(-1)
Solution:
f(-1) = 2(-1) 3 – 3(-1) 2 + 1 = 2(-1) – 3(1) + 1 = -2 – 3
+ 1 = -4
b. f(3)
Solution:
f(3) = 2(3) 3 – 3(3) 2 + 1 = 2(27) – 3(9) + 1 = 28
c. f(0)
Solution:
f(0 = 2(0) 3 – 3(0) 2 + 1 = 0 – 0 + 1 = 1
________________________________________
2. What is the value of ceil (3.7)? ceil (-3.7)?
Solution:
ceil(x) = n if n − 1 < x ≤ n for integer n
ceil(3.7) = 4
ceil(-3.7) = -3
________________________________________
3. A rental car company charges $32 a day plus $
0.20 a mile.
a. Write an equation that describes the function
for this situation.
Solution:
f(x) = 32 + .20x
b. Find the charge when the returned car was
driven 200 miles in one day.
Solution:
f(x) = 32 + .20x = 32 + .20 (200) = 32 + 40
Answer: $72
________________________________________
4. Find the inverse function.
a. y = f(x) = 5x + 1
Solution:
Solve the equation y = f(x) for x and then replace
y by x.
y = 5x + 1 x = (y – 1)/5
Replace y by x y = (x – 1)/5
Answer: y = (x – 1)/5
b. y = f(x) = (x-2)2
Solution:
y = (x - 2)2 ± √y = x – 2 x = ±√y + 2
Replace y by x
Answer: y = ± √x + 2 or y = 2 ± √x
________________________________________
5. Find the equation of a circle where P (1,3) and
Q (4,7) are the endpoints of a diameter.
Solution:
Method A:
The center must be the midpoint of PQ. Midpoint
is (5/2, 5)
The radius must be half the diameter. Calculate
the distance from (5/2, 5) to one of the points.
The distance is 5/2
Equation is (x – (5/2))2 + (y – 5)2 = 25/4
x2 – 5x + 25/4 + y2 - 10y + 25 = 25/4 x2 – 5x + y2
- 10y + 25 = 0
Answer: x2 – 5x + y2 - 10y + 25 = 0
Method B:
Use the diameter form of circle: (x − a1)(x − a2) +
(y − b1)(y − b2) = 0,
where P = (a1, b1), Q= (a2, b2)
(x − a1)(x − a2) + (y − b1)(y − b2) = 0
(x − 1)(x − 4) + (y − 3)(y − 7) = 0 x2 -5x + 4 + y2
-10y +21 = 0
x2 - 5x + y2 - 10y + 25 = 0
________________________________________
6. Find the center and radius of circle x2 +y2 + 4x
– 6y - 3 = 0
Solution:
Complete the square.
x2 +y2 + 4x – 6y - 3 = 0 x2 + 4x + y2 – 6y = 3 ( x2
+ 4x + 4) + (y2 – 6y +9) = 3 + 4 + 9 (x + 2)2 + (y -
3)2 = 16
Answer: Center is (-2, 3) with radius 4
________________________________________
7. A Pythagorean triple is a sequence of three
integers (a,b,c) such that a2+b2=c2. So each
Pythagorean triple corresponds to a right triangle
whose sides are all of integer length. For example
(3,4,5) is a Pythagorean triple since 32+42=52. It is
easy to check that if s and t are integers that are
both odd or both even then (s2-t2)/2, s t, (s 2 +t
2)/2 is a Pythagorean triple. Thus s=3 and t=1
gives the Pythagorean triple (4, 3, 5). T he triple
obtained from s=13 and t=1 is (84, 13, 85).
a. Find the Pythagorean triple that corresponds to
s=37, t=1
Solution:
Substitute s=37, t=1 in the formula for the triple:
(s2-t2)/2, s t, (s 2 +t 2)/2
Answer: The numbers are 37, 684, 685.
You can check that these are correct. 372 + 6842
= 6852
b. Find the values of s and t corresponding the
Pythagorean triple (84, 13, 85).
Solution:
Solving (s2-t2)/2 = 84 and (s 2 +t 2)/2 = 85 2s2 =
338 s2 = 169 s = 13
Substitute in s t = 13 (the middle number) t = 1
Answer: s = 13, t = 1
________________________________________
8. Find the equation of a parabola through the
points (1, 0), (2,-3), and (0,5).
Solution:
The formula for a parabola is y = ax2 + bx + c
Substituting the coordinates of the 3 points, you
obtain 3 simultaneous equations.
0 = a + b + c
-3 = 4a + 2b + c
5 = c
You can use Cramer ’s Rule to solve these
equations. a = 1, b = -6, c = 5
Answer: The equation is y = x2 – 6x + 5
________________________________________
9. Is the point P (3, 4) inside, outside or on the
circle with equation (x + 2)2 + (y - 3)2 = 9?
Solution:
First find the distance from the center of the circle
to point P.
From the equation of the circle, the center C is at
(-2, 3), and the radius r = 3.
Find the distance from C to P.
d = √([3 - (-2)] 2 + [4 - 3]2) = √(52 +12) = √(26)
which is approximately equal to 5.1.
Since this distance is greater than the radius 3, the
point is outside the circle.
Answer: outside the circle
________________________________________
10. Find the line joining the intersection points of
the two circles:
x2 + y2 -7x - 2y +7 = 0 and 3x2 +3y2 -7x + y = 0
Solution:
Method A:
Rewrite the first equation as:
3x2 + 3y2 -21x - 6y + 21 = 0
3x2 +3y2 -7x + y = 0
Subtracting, you get: -14x – 7y = -21 y = -2x + 3
Answer: y = -2x + 3
Method B:
To check, you can substitute this in one of the
equations:
x2 + y2 -7x - 2y +7 = 0 x2 + (-2x + 3)2 -7x – 2(-2x
+ 3) +7 = 0
x2 + 4x2 -12x + 9 -7x + 4x -6 + 7 = 0 x2 – 3x + 2 =
0
Solving this equation: x = 2, x = 1
Substitute these values and solve for y.
The coordinates of the two intersection points are
(2, -1) and (1,1).
Now you can write the equation of the line
through these points.
m = (y2 – y1)/(x2 – x1) = (-1 -1)/(2-1) = -2
The equation of a line through a point is y – y1 =
m (x2 –x1)
Use the slope m= -2 and one of the points, i.e. (1,
1)
y – 1 = -2(x – 1) y -1 = -2x + 2 y = -2x + 3
Unit 6
1. Write the equation of circle passing through
points (7, 1), (-1,-5), and (6,2).
Solution:
The equation of a circle is in the form x2 + y2 + ax
+ by + c =0
Substituting the values of the coordinates we get
3 simultaneous equations:
-a -5b +c = -26
7a + b + c = -50
6a + 2b +c = -40
You can subtract the second equation from the
first and eliminate variable c.
You can subtract the third equation from the
second and eliminate variable c.
-8a - 6b = 24
a – b = -10
-8a – 6b = 24
6a – 6b = -60
Subtracting the second equation -14a = 84 a = -6
Substituting for b: a – b = -10 (-6) – b = -10 b = 4
Substituting for a and b: -a -5b +c = -26 6 -20 + c
= -26 c = -12
The equation of the circle is x2 + y2 + ax + by + c
= 0 x2 + y2 - 6x + 4y - 12 = 0
Complete the square for the equation to be in
circle/radius form.
(x2 - 6x) + (y2 + 4y) = 12 (x2 - 6x + 9) + (y2 + 4y +
4) = 12 + 9 + 4
(x- 3)2 + (y + 2)2 = 25
Answer: (x- 3)2 + (y + 2)2 = 25
________________________________________
2. Find the equation of the circle which is centered
at (3,5) and which has the line x + y = 2 as a
tangent.
Solution:
Find the slope of the line L: x + y = 2 y = -x + 2 m
= -1
The slope of L ’ is 1 (since L and L’ are
perpendicular).
The equation will be y = x + b
Since the point (3, 5) will be on this line, substitute
the values 5 = 3 + b b = 2
The equation of line L ’ is y = x + 2
To find where L and L’ meet on the circle, solve
the simultaneous equations
x + y = 2 and y = x + 2 The point of tangency is at
(0 , 2).
The distance between the center of the circle and
the point of tangency is equal to the radius r of
the circle and is given by the distance from (3, 5)
to (0, 2).
d = √(3 – 0)2 + (5 - 2)2 = √(9 + 9) = √18
Equation of the circle is (x - 3) 2 + (y - 5) 2 = 18
Answer: (x - 3) 2 + (y - 5) 2 = 18
________________________________________
3. If L is the line given by the equation y = -2 -2x,
are A(-2,-5) and B(4,-3) on the same side of the
line? or are they on opposite sides of the line?
Solution:
Write the equation of the line y = -2 -2x in
standard form ax + by + c = 0 2x + y + 2 = 0
For any point P(x, y) we can evaluate the
expression f(x,y) = 2x + y + 2. Obviously, it
evaluates to 0 on the line itself. f(x,y) keeps a
constant sign on one side of the line.
Evaluating the point A(-2,-5), f(x,y) = 2(-2) + (-5) +
2 = -7
Evaluating the point B(4,-3), f(x,y) = 2(4) + (-3) + 2
= 7
Answer: A and B are on opposite side of the line
________________________________________
4. If L is the line given by the equation y = -2 -2x,
are C(4, -17) and D(5, -19) on the same side of the
line? or are they on opposite sides of the line?
Evaluating the point C(4, -17), f(,y) = 2(4) + (-17)
+2 = -7
Evaluating the point D(5, -19), f(x,y) = 2(5) + (-19)
+ 2 = -7
Answer: C and D are on opposite sides of the line
________________________________________
5. The line L is given parametrically by the
equations x(t) = 3t + 4, y(t) = 4t + 7 and C is the
circle of radius 10 with center at (4,7).
a. What are the values of the parameter t which
correspond to the points on L at which L meets
C?
Solution:
The equation of the circle is (x – 4)2 + (y – 7)2 =
100 x2 -8x + 16 + y2 – 14y + 49 = 100
x2 - 8x + y2 – 14y = 35 Since L meets the circle,
you can substitute x(t) = 3t + 4, y(t) = 4t + 7
(3t + 4)2 – 8(3t + 4) + (4t + 7)2 – 14(4t + 7) = 35
25t2 = 100 t2 = 4 t = 2, t = -2
Answer: t = 2, t = -2
b. What are the points of intersection of the line L
and the circle C? Be sure to give both coordinates
for each.
Solution:
When t = 2 x(t) = 3(2) + 4, y(t) = 4(2) + 7 x = 10, y
= 15
When t = -2 x(t) = 3(-2) + 4, y(t) = 4(-2) + 7 x = -2,
y = -1
Answer: (10, 15) and (-2, -1)
________________________________________
6. Find the distance from the point (1, 1) to the line
3x – 3y + 1 = 0 .
Solution:
The distance from (p, q) to ax + by + c = 0 is (|ap
+ bq + c|)/ √(a2 + b2) for point (p, q)
For the line 3x – 3y + 1, a = 3, b = -3, c = 1, p =1,
q=1
Substituting in the formula (|(3)(1) + (-3)(1) + 1|)/√
(32 + (-32)
1/ √18 √18/18 = 3√2/18 = √2/6
Answer: √2/6
________________________________________
7. Simplify the following expression. [1 - sin 4x] /
[1 + sin 2x]
Solution:
[1 - sin 4x] / [1 + sin 2x] [(1 - sin2x) (1 + sin2x)]/ (1 +
sin2x) 1 - sin2x = cos2x
Answer: cos2x
________________________________________
8. A bicycle traveled a distance of 100 meters.
The diameter of the wheel of this bicycle is 40
cm. Find the number of rotations of the wheel (to
the nearest whole number).
Solution:
For every one rotation of the wheel, the bicycle
moves a distance equal to the circumference of
the wheel. The circumference C of the wheel is
given by the formula C = πd C = 40π cm
The number of rotations N of the wheel is
obtained by dividing the total distance traveled,
100 meters = 10000 cm, by the circumference.
N = 10000 cm/40 π cm = 80 rotations (rounded
to the nearest unit)
Answer: 80 rotations
________________________________________
9. Simplify the following trigonometric
expression: [sec(x) sin 2x] / [1 + sec(x)]
Solution:
Substitute sec (x) that is in the numerator by 1/
cos (x) and simplify.
[sec(x) sin 2x] / [1 + sec(x)] = sin 2x / [cos x (1 +
sec (x)] = sin 2x / [cos x + 1]
Substitute sin 2x by 1 - cos 2x, factor and
simplify.
= [1 - cos 2x] / [cos x + 1] = [(1 - cos x)(1 + cos x)]
/ [cos x + 1] = 1 - cos x
Answer: 1 - cos
________________________________________
10. From a distance of 1753 feet the top of a
tower forms an angle of 40 degrees with the
ground. What is the height of the tower (to the
nearest foot)?
(sin 40 degrees = .6428, cos 40 degrees = .7660,
tan 40 degrees = .8391 ).
Solution:
tan 40° = x/1753 .8391 = x/1753 x = (.8391)( 1753)
= 1471 feet
Answer: 1471 feet

Apple iPad 2

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11/01/2011 14:38

New York: Apple akan segera merilis generasi terbaru komputer tablet populer iPad 2 dalam waktu 3-4 pekan ke depan. Demikian dikatakan pendiri situs content-sharing Digg Kevin Rose, seperti dilansir situs Xinhua, Selasa (11/1).

"Saya memiliki otoritas untuk mengumumkan rilis ini yang  kemungkinan akan jatuh pada Selasa, 1 Oktober 2011," kata Rose dalamblog resminya. "iPad 2 itu akan menampilkan fitur layar retina dan kamera depan/belakang. Jika Anda ingin membeli iPad diharapkan bersabar dan lebih baik membeli versi terbaru ini saja."

Selain kamera, menurut laporan dari PC World, pada generasi terbaru iPad itu juga terdapat fitur kamera, video FaceTime chat, dan port mini-USB. Rencana sebelumnya, Apple ingin memproduksi secara massal iPad 2 pada Januari lalui, namun ditunda sampai Februari karena hardware dan spesifikasi tablet belum diselesaikan. Hingga saat ini, pihak Apple belum mengonfirmasi secara resmi perilisan iPad 2.(SHA)

The Curious Case Of Julian Almighty

My name is Julian Setyanto, born on July 6, 1988. My father named Hardjo Oetomo and my mother named Ngatiyah. Born from a pair of simple peasant, I began to pioneer life. I graduated from high school Negeri 3 Purworejo in 2006, now studying at prestigious universities in the city of Bogor, Bogor Agricultural University majoring in Information Management. I began to pioneer life in Bogor with minimum investment. I worked at a leading private company in Bogor. The money from work I put aside for college. Hopefully everything goes according to plan.

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