Marlee has 12.0 guests coming to her Halloween party. Each table will hold 3.0 guests. How many tables will she need? Answer:

How many white dots should there be on the wings of the last ladybug?
Answer:Three. The difference between the number of black and white dots is always two. The last ladybug should have three white dots.

The cakes change each month. Cake 1 has been made in August, Cake 2 in June, and Cake 3 in May. In what month has Cake 4 been made?
Answer:April. The number of chocolate lines on the cakes indicates the month.

If a farmer buys chicken feed for £4 per tonne, pig feed for £3 per tonne and cattle feed for 40p per tonne. How much of each would he have if he had 100 tonnes for £100?
Answer: 8 tonnes of chicken feed, 12 tonnes of pig feed and 80 tonnes of cattle feed.

Joe’s code has 20 bugs. If Joe fixes 2 bugs per hour for 8 hours, how many bugs does Joe’s code have now? Answer: 27.

Question: A hobo picks up cigarette butts from the ground and can make a cigarette with 4 butts. If he finds 16 cigarette butts, how many cigarettes can he make?
Answer: 5. He makes 4 cigarettes with the 16 butts. Then he smokes them and has 4 more butts to make another cigarette.

A clock is set right at 5 a.m. The clock loses 16 minutes in 24 hours. What will be the true time when the clock indicates 10 p.m. on 4th day?
A clock is set right at 5 a.m. The clock loses 16 minutes in 24 hours. What will be the true time when the clock indicates 10 p.m. on 4th day?
Answer : 4 11 p.m.

Dr. Johnson needs to buy the same number of needles and syringes. The syringes are sold in packages of 6 and the needles are sold in packages of 10. What’s the smallest number of packages of syringes, and the smallest number of packages of needles, that Dr. Johnson can buy to have a matching number of each?
Answer: 5 packages of syringes and 3 packages of needles.
Dr. Johnson needs to come up with the smallest number that is divisible by both 6 and 10.
Since 6=2×3 and 10=2×5, the smallest number that is divisible by 6 and 10 is 2x3x5=30.
Therefore, Dr. Johnson needs 30/6=5 packages of syringes, and 30/10=3 packages of needles.

These are the best and most fun math riddles we can find. All of these tricky riddles are based on real math concepts and can be solved with purely math and logic.
Question: There are 6 sisters. Each sister has 1 brother. How many brothers are in the sisters' family?
Answer: 1 brother. If there was 6 brothers, each sister would have 6 brothers.

1. Where do frogs usually live in the hot season?
Answer: Buried at the bottom of ponds.

As a rule, one centimeter of rain is equal to how much snow?
Answer: b) Typically, one centimeter of rain equals 10 centimeters of snow. This varies somewhat according to the snow’s moisture content, but it’s the norm.

5. Out of the remaining 22 nomads, 50% were men.
Can you tell the total number of members in the group and the number of males?
Answer: There were a total of fifty nomads in the group out of which 29 were male.

What is the fear of Friday the 13th called?
In the movie "Groundhog Day," every morning started with Bill Murray's clock radio playing what song?
Answer: The song was "I Got You, Babe", Sonny and Cher's 1965 hit.

After a local art theft, six suspects were being interviewed.
Police know that exactly four of them told one lie each and all of the other statements are true.
Hint: If Alex had committed the crime, how many lies would there be?
Answer: Chris committed the crime.
If any other person had committed the crime there would not be exactly 4 lies told.

In Uk why are there no objections from the jockeys any more.
Answer: I have posed this question to the BHA but not received a reply. One can only assume that jockeys these days are better treated and better represented by the jockey club that concerns itself with their welfare.

There are 6 persons seating on a round table in which two individual have the same names. What is the probability that the two same-named individuals will be neighbors?
Answer: Total no of ways in which 6 persons can sit on a round table is (6-1)! = 5! = 120.
If we consider two same-named individuals as one person there are 5 persons who can sit in (5-1)! ways and these individuals can be seated together in 2! ways.
So, required probability =(2*(5-1)!)/(6-1)!= 2/5.
So, the answer is 2/5 = 0.4.

Jane has 2 candies. Jack has 5 candies. If they each eat 1 candy, how many candies do they have left in total?
Answer: First we find out how many candies they have together.
5 + 2 = 7. Together, Jane and Jack have 7 candies. If they each eat one candy, then they ate 2 candies in total. So, we find out how many candies they have left by subtracting 2 from 7.
7 - 2 = 5. They have 5 candies left in total.
 Moe has to go 12 miles to his grandma's house. The first bus goes 3 miles. The second bus goes 4 miles. How many miles does Moe have left to go?
Answer: First we find out how far Moe travels on the two buses.
3 + 4 = 7. The distance Moe travels on the two buses together is 7 miles. We find out how far he has left to go by subtracting 7 from 12.
12 - 7 = 5. Moe has 5 miles left to go.
 What is 18 + 3 - 5 + 9?
 A box of cookies has 18 cookies. A crate of cookies has 12 boxes. How many cookies are in a crate?
Answer: We get the answer by multiplying 18 and 12.
18 x 12 = 216. There are 216 cookies in a crate.
 Joe went back-to-school shopping. He bought 3 binders that cost 2 zorons each, 4 packs of pencil crayons at 3 zorons each and 5 packs of lined paper at 7 zorons each. How much did he spend?
Answer:Whether you realize it or not, this is an order-of-operations problem. You multiply first, then add.
 A number is divided by 6. The quotient is 3 and the remainder is 2. What is the number?
Answer: We get the answer by working backwards.
 What is 3x9 - 2x8?
Answer: We have to remember the order of operations. We multiply first, then subtract.
 Bob bought a pop for 130 ducats and a bag of chips for 89 ducats. He gave the clerk 500 ducats. How many ducats should he get back?
Answer: First we find the total of Bob's purchase.
130 + 89 = 219. To find out how many ducats he should get back, we subtract 219 from 500.
500 - 219 = 281. Bob should get back 281 ducats.
 To convert radons to zurugons, you multiply each radon by 13 then subtract 51 from the product. How many zurugons are 62 radons?
Answer: This is a really hard problem. First we need the product.
62 x 13 = 806. Then the question says to subtract 51 from the product.
806 - 51 = 755. So, 62 radons is equivalent to 755 zurugons.
 A school bus holds 40 students. 90 students are going on a field trip. How many buses do they need?
Answer: Two buses hold 80 students and 3 buses hold 120. So, they need 3 buses in order to carry 90 students.
 What is the remainder when you divide 562 by 9?
Answer: 562 divided by 9 has a quotient of 62 and a remainder of 4.
 What is the quotient when you divide 437 by 11 using long division?
Answer: The quotient is 39. In case you were wondering, the remainder is 8.
 A number is multiplied by 8 and this answer is added to 99 and the final answer is 171. What is the original number?
Answer: We find the original number by working backwards. First we subtract 99 from 171.
171 - 99 = 72. Then we divide 72 by 8.
72 ¸ 8 = 9. The original number is 9.

What can I do in my leisure time?
Answer: You won’t have too much spare time due to study demands; however on weekends you will have time to see the sights.
Most nurses spend time going to church, seeing the cities, and seeing friends they may have in Australia.

Question: Miss Mary's father had 4 kids: Joseph, John, and Jordan. What was the 4th kid's name?
Answer: The 4th kid is Miss Mary.

A solution of sodium chloride contains 500 mg of sodium chloride made up to 100 mL with water. Express this solution as a percentage.
Answer is D, all respondents got the answer, but we shall only reward Clement and O’can.


Warren has 252.0 guests coming to his party. Each table will hold 4.0 guests. How many tables will he need? Answer:

How many people must be there in a room to make the probability 100% that two people in the room have same birthday?
Answer: 367 (since there are 366 possible birthdays, including February 29).
The above question was simple. Try the below question yourself.
How many people must be there in a room to make the probability 50% that two people in the room have same birthday?
The number is surprisingly very low. In fact, we need only 70 people to make the probability 99.9 %.

We start with six boxes. We divide these six boxes by three. This means: how many groups of three can we make with the six boxes that we have? Answer: 2.
We start with eighteen boxes. We divide these eighteen boxes by three. This means: how many groups of three can we make with the eighteen boxes that we have? Answer: 6.
We start with thirty-six boxes. We divide these thirty-six boxes by three. This means: how many groups of three can we make with the thirty-six boxes that we have? Answer: 12.

We start with four boxes. We divide these four boxes by two. This means: how many groups of two can we make with the four boxes that we have? Answer: 2.
We start with ten boxes. We divide these ten boxes by two. This means: how many groups of two can we make with the ten boxes that we have? Answer: 5.
We start with twenty-two boxes. We divide these twenty-two boxes by two. This means: how many groups of two can we make with the twenty-two boxes that we have? Answer: 11.

We start with ten boxes. We divide these ten boxes by five. This means: how many groups of five can we make with the ten boxes that we have? Answer: 2.
We start with forty-five boxes. We divide these forty-five boxes by five. This means: how many groups of five can we make with the forty-five boxes that we have? Answer: 9.
We start with fifty-five boxes. We divide these fifty-five boxes by five. This means: how many groups of five can we make with the fifty-five boxes that we have? Answer: 11.

We start with twelve boxes. We divide these twelve boxes by four. This means: how many groups of four can we make with the twelve boxes that we have? Answer: 3.
We start with thirty-two boxes. We divide these thirty-two boxes by four. This means: how many groups of four can we make with the thirty-two boxes that we have? Answer: 6.
We start with forty-four boxes. We divide these forty-four boxes by four. This means: how many groups of four can we make with the forty-four boxes that we have? Answer: 11.

How can I control the time my kids spent on my iPhone / iPod Touch?
Answer: Game Time Limit for Parents iOS App. I believe that all parents with young children will love this app :).
child’s probably on his way to bring your iPhone/iPod back to you.
Get this very useful parenting iPhone app in AppStore Here!

For example, what affects egg quality in fish? Answer: mother. But what affects mother? Answer: her environment. But what aspect of the environment? Answer: food availability. But what affects food availability? Answer: upwelling – one of nature’s remarkably influential processes.
Upwelling – when winds (you know, those blustery kinds where gulls look like they’re frozen in air) gust along a coastline pushing warm surface water offshore causing cold, nutrient-rich water from the ocean’s depth to move upward.
When the deep water, laced with compounds like nitrate and phosphate, reaches the surface, photosynthesizing phytoplankton ferociously consume those nutrients to grow and multiply. Flourishing swaths of phytoplankton are fed upon by zooplankton. Zooplankton are devoured by filter-feeders (perhaps an adorable krill). Filter-feeders are eaten by fishes (that can also eat zooplankton). Fishes are then preyed upon by all – whales, seals, pelicans, cormorants and humans – the latter harvest millions of tonnes of commercially valuable species that thrive in upwelling zones (e.g. anchovy, sardines).
Marine fisheries are definitely dependent on upwelling: about 20% of all catch occurs in upwelling regions, which only cover ~2% of the ocean’s surface!
Wait, 2%? That’s it? Wouldn’t it be swell if there were ways to mimic upwelling in places where this phenomena does not naturally occur? Answer: there are! Could this strategy realistically improve worldwide fish production? Answer: here comes another thought avalanche.

We start with eighteen boxes. We divide these eighteen boxes by six. This means: how many groups of six can we make with the eighteen boxes that we have? Answer: 3.
We start with forty-two boxes. We divide these forty-two boxes by six. This means: how many groups of six can we make with the forty-two boxes that we have? Answer: 7.
We start with seventy-two boxes. We divide these seventy-two boxes by six. This means: how many groups of six can we make with the seventy-two boxes that we have? Answer: 12.

This product contains all of the digits from 1 to 8. Can you complete it?
Answer: 64 * 58 = 3712.

When I watch videos of people practicing kendo I often hear them make weird sounds. What is that sound for?
Answer: Thanks for your question. I am not sure what kinds of sounds you are referring to. Can you be more specific?

How Well Does WordPress Interact with TubePress?
How Well Does WordPress Interact with TubePress?
Answer: There are YouTube plugins available which allow YouTube and Vimeo videos to be available from your webpage and even allow for the display of galleries.

How many people in a kendo match?
How many people in a kendo match?
How many people in a kendo match?
Answer: Thank you for your question. And I am so sorry for the delay and my reply.
Even in a team fight, we fight individually. It is one person vs. one person. Usually in team fight, we have three or five team members. And they fight individually.

What is an uncontested dissolution of marriage in Florida?
Answer: Well that would be where all the issues are agreed upon by the parties in advance of any kind of a litigation. Essentially, what they would do is enter into an agreement or a marital settlement agreement that would resolve all the issues that they had between them and there would be no litigation.

Question: How many words constitute a fictional short story? Not a short, short story but a short story.
Answer: Generally, a short story is any story under 7,500 words. Anything under 1,000 words is often termed flash fiction.
However, these definitions are not fixed. Sometimes works up to 20,000 words have been called short stories. You have to check with individual publishers (such as literary magazines, websites, contests, etc.) to see what length of stories they are looking for and what terms they are using to describe them (e.g. micro-fiction, novelette, novella, postcard fiction).
At the extreme lower end there are also "six word memoirs," such as this famous one attributed to Hemingway: "For sale: baby shoes. Never worn."

Be supported to release and resolve all that limits you from living your true potential on the Earth.
Universal Rays Healing is a powerful transformational healing process offered through direct contact with the Ascended Masters of Light. You are invited by the Masters of the Rays to become an angel of healing and offer your loving service to humanity.
You will feel the call to offer this healing process as part of your soul mission to work directly with the Masters of the Rays and to assist humanity to transform its consciousness on all levels - physical, emotional, mental and spiritual. To do this potent healing work is one of the gifts you have been guided by your soul to share with others through your loving heart.
The Universal Rays Healing program was designed by the Council of the 12 Ascended Masters who are the keepers of the Universal Rays from the heart of God consciousness and channelled to Ishtar over a two year period from 2013-2015.
This council of 12 Ascended Masters invite you to channel the powerful consciousness of the Universal Rays and to transmit this to yourself and others to assist you to align with your original blueprints.
The Universal Rays hold the energy of creation potential on which you were birthed as a soul and it is this mastery that you have chosen to learn in this lifetime.
When you open your heart to work with the consciousness of the 12 Masters of the Universal Rays you connect with the power of creation from Source.
The energy of the Universal Rays is a divine guide to your highest potential as a Master Co-Creator on the Earth.
When you are in resistance to this consciousness, you create chaos and imbalance in your life on many levels, which in turn may cause dis-ease and an inability to focus and align to your highest potential.
When you are in attunement and alignment with this consciousness, miracles can and do occur in your life as you begin to accept your self mastery on the Earth.
Working with the Ascended Masters and the creation consciousness of the Universal Rays allows you to step back into alignment to your true nature and to follow the path that your soul chose to walk in this lifetime.
The Universal Ray consciousness allows you to resolve the old limitations of separation and resistance that you create and assists you to move into a place of harmony and love.
This powerful healing energy also affects the cellular and molecular consciousness of your physical body and your energy bodies to assist all energy that you hold within you to come into alignment with its true mission on the Earth without distortion or illusion.
When you are in alignment with your original creation energy then you are able to step fully into your Mastery on the Earth without the limitation of physical dis-ease, mental illusions, emotional resistance, or unresolved karma.
The powerful vibration of the Universal Rays is channelled from the Council of 12 Masters through your channel pillar and is then transmitted where it is guided to shift and expand consciousness.
As a practitioner of Universal Ray Healing you are asked to open your pillar of light in service to humanity to channel the creation energy from the heart of Source.
While you work with this program you are completely oversouled by the 12 Ascended Masters who offer this program in service to humanity.
When you work with the Universal Ray Healing program you see many shifts in your life because as much as you offer these healing rays to others you also receive this vibrational energy for yourself and this energy in turn affects your reality.
When you receive and transmit the rays (channel), the consciousness of vibration that they hold moves through your own energy fields and this affects you as much as your clients.
As you open your heart further to receive and to transmit this consciousness you lift your own ancient wounds and karma as you discover more and more of yourself within the creation consciousness of the rays.
When you work with the 12 Ascended Masters you begin to align with their consciousness and open your heart to the wisdom that they hold through their loving service.
As you work with this vibration of consciousness you lift your own vibration and hold this new consciousness in your own field of energy. This means that all lower vibration energy held within your own field is lifted and transformed. This occurs for you through the work that you do both for yourself and in service to humanity through the channelled consciousness of the 12 Masters and for your clients when they come to you for healing and support.

This week we have two articles, four events and one urgent call for action.
American Reform movement) on the Israeli Rabbinate.
as well as a recording of the event.

how many people follow kendo?
how many people follow kendo?
how many people follow kendo?
Answer: Thank you for your question. I must apologize. I really don't know. World wide, I found 2 million people practicing.
However, I know many people quit after registering to their federation so we probably have no idea exactly how many people do kendo in the world after all.
Sorry that I could not help you there.

Daniel has the same number of brothers as sisters.
His sister, Laura, has twice as many brothers as she has sisters.
How many boys and girls are in the family?
Hint: A little bit of algebra might help.
Answer: There are four boys and three girls.
which means that B = 4 and hence G = 3.


Arthur baked 35.0 muffins. How many more muffins does Arthur have to bake to have 83.0 muffins? Answer:

John is three times younger than his father, but two times older than his brother Jim. How old is John if Jim is 40 years younger than his father?
Answer: John is 16 years old.
Let X be John’s age, F be father’s age, and J be Jim’s age. Then, 3X=F, X/2 =J, and J+40=F. Solving this system of three equations for X, we get X=16.
Next articleCan Chinese herbal medicine Zengru Gao encourage breastfeeding?

How many grams of NH3 can be produced from 10.4 g of H2?
Answer: 58.99 grams of NH3.
4. Find the ratio of de Broglie wavelength in first and third excited state of H atom.
5. Arrange H+, D+, He2+ and He+ in descending order of their e/m ratios.
6. Number of waves made by Bohr’s electron in one complete revolution in 3rd orbit is?

1. What will be the unit digit of the squares of the following numbers?
2. The following numbers are obviously not perfect squares. Give reason.
3. The squares of which of the following would be odd numbers?
4. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
5. Without adding, find the sum.
6. How many numbers lie between squares of the following numbers?
8. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?
9. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.
10 . For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.
11. The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
12. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
13 Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
14. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
15. Find the number of digits in the square root of each of the following numbers (without any calculation).
19. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
20. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.
3. Is 32 a perfect cube?
4. Express 132 as the sum of two consecutive integers.
5. What will be the unit digit of the square of 3873?
6. Why 2332 is not perfect square ?
7. Cube of any odd number is even. Yes/No Why?
8. Find the cube root of 27000.
9. Find the cube root of 17576 through estimation.
8. Anubhav makes a cuboid of plasticine of sides 5 cm, 3 cm, 5 cm. How many such cuboids will he need to form a cube?
10. Is 5324 a perfect cube? If not, then by which smallest natural number should 5324 be divided so that the quotient is a perfect cube?
2. Find the smallest number by which 252 must be multiplied to get a perfect square. Also, find the quare root of the perfect square so obtained.
4. Find the smallest number by which 2925 must be divided to get a perfect square. Also, find the square root of the perfect square so obtained.
6. Find the least square number exactly divisible by each one of the numbers 8, 12, 15, 20.
9. Find the least number that must be subtracted from 7581 to obtain a perfect square. Find the perfect square and its square root.
10. Find the least number that must be added to 506900 to make it a perfect square. Find its perfect square and its square root.
11. Find the least number of 4 digits that is a perfect square.
12. The area of a square field is 60025 m 2. A man cycles around its boundary at 18 km per hour. In how much time will he return at the starting point?
13. The sides of a rectangular field are 80 m and 18 m respectively. Find the length of its diagonal.
16. A decimal fraction is multiplied by itself to give the product 0.007569. Find the decimal fraction.
17. The area of a square playground is 291.0436 square meters. Find the length of each side of the playground.

Question: A hobo picks up cigarette butts from the ground and can make a cigarette with 4 butts. If he finds 16 cigarette butts, how many cigarettes can he make?
Answer: 5. He makes 4 cigarettes with the 16 butts. Then he smokes them and has 4 more butts to make another cigarette.

Q 1: The circumference of a circle is 28.6 cm. Find its area.
Q 2: The circumference of a circle is 19.5 cm. Find its area.
Q 3: The circumference of a circle is 49.7 cm. Find its area.
Q 4: The circumference of a circle is 44.8 cm. Find its area.
Q 5: The circumference of a circle is 81.9 cm. Find its area.
Q 6: The circumference of a circle is 96.4 cm. Find its area.
Q 7: The circumference of a circle is 58.6 cm. Find its area.
Q 8: The circumference of a circle is 64.7 cm. Find its area.

We will learn how to solve step-by-step the word problems on multiplication and division of whole numbers. We know, we need to do multiplication and division in our daily life. Let us solve some word problem examples.
1. The cost of a chair is $980.50. Find the cost of such 2035 chairs.
Therefore, cost of 2035 chairs is $1995317.50.
2. A tyre factory produces 6348 tyres a day. How many tyres will the factory produce in 460 days?
Therefore, the tyre factory will produce 2920080 tyres in 460 days.
3. 470988 books are to be arranged equally in shelves. If 378 books are arranged in each shelf, how many shelves will be needed?
Therefore, 1246 shelves needed to arrange 470988 books equally.
4. The cost of 534 train tickets was $457104. Find the cost of one ticket?
Therefore, cost of one train ticket is $856.

1 BED CORNER UNIT - NEW CONSTRUCTION - WATERFRONT - FULL SERVICE The Mark at DeNormandie Wharf is the newest luxury condominium on the East Boston Waterfront - Featuring expansive views of Boston Harbor, City Skyline, and Charlestown. Commuter's Dream! 2 minute walk from Maverick Square Blue Line MBTA (3 mins to Aquarium), Water Taxi Stop 66, ReelHouse & 5 minute drive to the airport. Charming homes featuring state of the art amenities and interior finishes. Parking Available for purchase. Amenities Include: 24/7 Concierge, Fitness Center with Peloton Bikes, Community Terrace with Grills and Fire pits, Golf Simulator/Movie Room, Envoy Car Share & more! DON'T MISS THIS OPPORTUNITY- SCHEDULE YOUR APPOINTMENT TODAY.
Worth the wait!! Quiet side street location, this unit is completely brand new construction! Beautiful open floor plan, upgraded white shaker cabinets, stainless appliances, and a bonus wine/pantry area with beverage fridge and additional storage space. The master bedroom is ample with hardwood floors, and the bath offers a sparkling tile shower, shaker vanity and upgraded lighting. The second bedroom also features hardwood floors and a generous closet. There is an additional full bath nearby. Bright and sunny unit with plenty of windows. The finished bonus area below grade area offers endless possibilities!. High quality features through out. Low maintenance exterior with Azec trim and Hardi plank siding. One mile to Forest Hill Station one mile to Roslindale Village.
Located on a quiet side street location, this unit is completely brand new construction! Beautiful open floor plan, upgraded white shaker cabinets, stainless appliances, and a bonus wine/pantry area with beverage fridge and additional storage space . The first level has a "mini" master bedroom with hardwood floors, and attached bath with shaker vanity and upgraded lighting. The additional second bedroom could also be utilized as a den, office or dining area to suit your needs. This unit also features hardwood floors through out and a generous closets. The third level offers a Penthouse Master bedroom retreat with large walk in closet. The bath, has double vanity, tile shower with glass enclosure and a soaking tub. There is also an enormous exclusive use roof deck accessible from both the master and the main living area for outdoor entertaining High quality features through out. Off Street parking. One mile to Forest Hill Station and one mile to Roslindale Village.

Approx. 3mm x 3mm, 1mm hole, 8 beads per inch.
Full strand measures approx. 28″, 224 beads, 44.8-45.6 grams. Stated price is based on 45.6 grams.
Half strand measures approx. 14″, 112 beads. Stated price is based on 22.8 grams.

Jasmine started at 62.4 on a number line. She jumped forward 3.5 and then 6.04. Where did she land?
A length of pipe is 4.23 meters long. Another length of pipe is 1.6 meters longer. What is the total length of the two pipes?
What is 0.43 and 27.1 altogether?
Ryan spent $36.50. He spent $12.97 more than Kevin. How much did Kevin spend?
What is the difference between 48.7 and 43.07?
A bag cost $3.42 less than sunglasses. The sunglasses cost $18.99. How much would the bag and sunglasses cost altogether?
A car travels at 60.3 mph. A second car passes by going 31.06 mph faster than the first car. What was the speed of the second car?
What is the difference between $138.75 and $316.81?
Meg starts at 14.37 on a number line. She jumps back 2.37 and then another 4.06. Where does she land?
Jeremy is 0.04 meters taller than Catherine. Mark is 0.22 meters taller than Jeremy. Jeremy is 1.3 meters. What is their total height?

As a rule, one centimeter of rain is equal to how much snow?
Answer: b) Typically, one centimeter of rain equals 10 centimeters of snow. This varies somewhat according to the snow’s moisture content, but it’s the norm.

"It is not likely that any complete life has ever been lived which was not a failure in the secret judgment of the person that lived it"
"My downfall raises me to infinite heights."
"We Irish are too poetical to be poets; we are a nation of brilliant failures, but we are the greatest talkers since the Greeks."
"I am often confronted by the necessity of standing by one of my empirical selves and relinquishing the rest. Not that I would not. If I could, be... a great athlete and make a million a year, be a wit, a born -- vivant and a lady killer, as well as a philosopher, a philanthropist ... and saint. But the thing is simply impossible. The millionaire's work would run counter to the saint s; the bon-vivant and the philanthropist would trip each other up; the philosopher and the lady killer could not well keep house in the same tenement of clay. Such different characters may conceivably, at the outset of life. Be alike possible for a man. But to make any one of them actual, the rest must more of less be suppressed. So the seeker of his truest, strongest, deepest self must review the list carefully and pick out on which to stake his salvation. All other selves thereupon become unreal, but the fortunes of this self are real. Its failure are real failures, its triumphs real triumphs carrying shame and gladness with them."
"Failure, then, failure! so the world stamps us at every turn. We strew it with our blunders, our misdeeds, our lost opportunities, with all the memorials of our inadequacy to our vocation. And with what a damning emphasis does it then blot us out! No easy fine, no mere apology or formal expiation, will satisfy the world's demands, but every pound of flesh exacted is soaked with all its blood. The subtlest forms of suffering known to man are connected with the poisonous humiliations incidental to these results."
"Misfortunes one can endure -- they come from outside, they are accidents. But to suffer for one's own faults -- Ah! there is the sting of life."
"And nothing to look backward to with pride, and nothing to look forward to with hope."
"Success and failure are the same impostor"

Q 4: The slope of one line is 3/4 and the slope of another line is 8/(x-6). If the two lines are perpendicular then the value of x = ____.
Q 6: The slope of one line is x/4 and the slope of another line is (x+5)/6. If the two lines are parallel then the value of x = ____.

Gatesville, Texas is located at Latitude 31.440313 and Longitude -97.733976. Gatesville is in the Central (GMT -06:00) time zone. Telephone area codes in Gatesville include 254. This hometown does use daylight savings time.
Gatesville, Texas is situated in Coryell County. Coryell County has an average elevation of 774 feet.
Please explore this site for detailed population, demographic and housing data for Gatesville, Texas and Coryell County.

Note: Temperature must be in Kelvin for the equation to work. You calculate Kelvin temperature by adding 273 to the Celsius temperature.
1.)	Calculate the decrease in temperature when 2.00 L at 20.0 °C is compressed to 1.00 L.
Answer: For this problem the Initial Volume is V1 = 2.00 L. The Initial Temperature is T1 = 20.0 + 273 = 293K. The Final Volume is V2 = 1.00 L and the Final Temperature (T2 ) is what you are asked to solve for.
The temperature decreased to 146.5 K.
Answer: The Initial Volume is V1 = 6.00 L. The Initial Temperature is T1 = 27.00 + 273 = 300 K. The Final Temperature is T2 = 150.0 + 273 = 423 K. The Final Volume (V2) is what we are trying to find in the problem.
The volume of the container would be 8.46 L.

Microsoft provides an own network policy module to implement Kubernetes network policies with the Azure CNI plugin for acs-engine and AKS called Azure NPM.
The Azure NPM is available since quite some time for acs-engine and natively integrated, but not yet for AKS. If you want to use Azure NPM on Azure Kubernetes Service, you can do so. Just run the following command to deploy the Azure NPM daemonset on your AKS Cluster.
to check the successful deployment.
The Azure NPM works following the LIFO principle to establish fine grained network isolation between your pods.
So, you can ensure that only specific pods forming the front-end can talk to a certain backend. Following the LIFO principle, you deploy a deny all inbound network policy to a certain namespace first to restrict all ingress traffic to all pods in this namespace.
Warning: Do not deploy a deny all inbound or outbound network policy to the kube-system namespace. This causes a serious operational impact on your Kubernetes cluster.
Then you deploy additional network policies to allow the traffic to your container applications as necessary but maintaining the network security/isolation between different container applications in the namespace. Furthermore, you establish with the deny all inbound network policy a first defense line that new container applications are not directly exposed and available for network traffic.
Let us dive into one example. I am running two simple web applications on my AKS cluster. One provides a static web page (aks.trafficmanager.net) and the other one (src.trafficmanager.net) shows you specific information about the web request targeting the web application.
The sketch above shows how the web applications are published to the outside world.
Before starting the lock down of the default namespace, we deploy a busybox container first.
We need the busybox container later in the example.
Next, we roll out the deny all inbound network policy to the default namespace.
When calling the web applications, they are not accessible anymore.
So, we allow the traffic to the aks.trafficmanager.net web application from the Internet again with the following network policy.
As you can see in the template, Kubernetes network policies are label-based and with that totally dynamic and not dependent on the frequently changing pod IP addresses.
The aks.trafficmanager.net web application is now accessible again, but not the src.trafficmanager.net. This one should only be accessible from the busybox container and not from the Internet or any other pods running on the AKS cluster.
In the next network policy template, you will see that the label-based selection also works in the from section of the network policy definition.
So, the network policy is applied to all pods with the label matching app=src-ip-internal and only allows ingress traffic from pods matching the label app=busybox. Let us connect to the busybox container to test if it works.
In the container itself we run the following set of commands to test the connectivity.
Have a look at the following screenshots showing the output of the previous commands.
Calling src.trafficmanager.net does not work as expected, because it is the external address. Trying the Kubernetes service address src-ip-internal or the IP address of the Azure Internal Load Balancer works. So, the src.trafficmanager.net web application is only accessible from the busybox container inside the AKS cluster.
I hope you got an idea how network policies can be a beneficial security building block in your AKS cluster to better control the traffic flow to or from your container applications.
A good collection of Kubernetes network policy examples and use cases can be found under the following GitHub repository.

Israeli Prime Minister Benjamin Netanyahu said on Monday that some of his country’s African migrants will be sent to Canada, Italy and Germany under an agreement made with the United Nations refugee agency.
“This is a unique agreement between the U.N. commissioner and the state of Israel, that takes 16,250 people out, takes them out to developed countries like Canada, or Germany and Italy,” Mr. Netanyahu said on Israeli television, reported Reuters.
There has been no confirmation from Ottawa at this point.
More details to follow. Image of Benjamin Netanyahu from Time.

Facebook has had a rule which limited the amount of text in an image in a Facebook Ad to no more than 20%. Any image which exceeded that was automatically rejected by Facebook and it was a pain for marketers – but that’s all changing!
Facebook is taking into account the fact that the 20% text rule has caused a considerable amount of confusion and trouble for advertisers. They’re offering a solution which helps to eliminate confusion while at the same time avoiding text clutter which they say turns readers off.
The 20% rule involved a 5×5 grid superimposed over the actual ad, and when more than 5 of the squares within the grid had text, the ad was rejected without further consideration.
Eventually, advertisers figured out they could strategically position the text, so it didn’t take up 5 squares of the overall ad. By compressing the text into fewer squares, the rejection rule would not be triggered, and the ad would pass inspection. It limited the text, forced you to use the grid tool or an overlay to help you “position” your text and added time to the overall ad creation process.
Often this was a pain because a logo could contain text or an image of a person with a t-shirt having text on it counted towards the 20% text. It was a pain for marketers!
Facebook maintains that research has demonstrated that readers prefer ads with less text. So Facebook implemented the original rule to ensure readers had an enjoyable experience, and would not ignore ads.
After receiving a tremendous amount of feedback from their ad purchasers about the confusion created by the 20% rule, they decided to seek a solution to satisfy all parties concerned. Too many ads were being rejected, (money was being lost!) so a policy shift was necessary to allow more flexibility in ad composition.
Facebook will now not reject ads on the basis of text density. Instead, when reviewing ads, text density will be broken down into 4 different categories: OK, Low, Medium, and High.
OK – is the preferred category for a Facebook ad. It consists of little or no text. Ad text is in a text box, removed or separate from the image itself.
Low – is an ad image which contains a low amount of text, similar to what Facebook currently allows under the 20% rule. Even though it’s equivalent to what’s currently being run on Facebook, there will still be slight limitations on this ‘low’ category of included text.
Medium – this category includes more text, possibly placed in several locations around the ad image which is considered to be slightly excessive by Facebook.
High – as you might expect, the high category is considered by Facebook to have way too much text. The text obscures the image thus diminishing the experience for Facebook users.
While none of the ads in these categories will be rejected by Facebook, there is a catch. Facebook will manage text density by imposing higher costs on less preferred ads, and limit the distribution and reach of those ads.
Any ads in the ‘OK’ category would have a normal cost and would have full distribution on Facebook. Ads which fall into the other three categories would cost increasingly more and receive increasingly less distribution.
An ad in the ‘High’ category would not be seen as much as an ad from the ‘OK’ category. If an advertiser wanted it to be seen, it would cost more. This manages text density by penalizing ads using lots of text by using cost and user reach as a deterrent.
Not all ad images are subject to the new screening rule. The exceptions include infographics, apps, and screenshots for video games, album covers, book covers, legal text, concert posters, event posters, movie posters, product images, and text-based businesses such as comic strips or calligraphy. Basically, Facebook’s system have gotten smarter and they can sort these out better!
The odd thing about this new approach is that it may end up making things even more ambiguous for advertisers than the original 20% text rule. Under the old rule, all an advertiser had to do was impose a 5 x 5 grid over any given ad to determine whether or not it would be rejected. Facebook provided a grid tool for that purpose, and it could easily be used by an advertiser before submitting an ad.
That grid tool has now been removed, and under the new guidelines, text density will be categorized by Facebook, and could fall into any of the 4 categories based on judgment. If the intent was to make approval and rejection more predictable, it would seem that the new approach leaves something to be desired in terms of clarity. Yes, the ad won’t get disapproved, but will it get valuable reach?
Facebook does intend to replace the original grid tool with a new overlay tool that will tell you which of the 4 categories your ad image falls into. This will help you to decide whether you want to pay the extra cost and be subject to distribution limitations to include extra text in your ad.
It will be possible to avoid much of the hit-or-miss approval by using Facebook’s own ‘Create Ad’ function or by creating an ad in the Power Editor, both of which will advise on which category the ad will fall into. If you don’t use either of these tools, you’ll have to limit the text in your ad image, or risk being seen by fewer viewers.
There may not be a huge impact on advertisers who were in the habit of limiting text in ad images. But for those who cram in as much text as possible, you should expect to pay more if you want your ads to be seen on Facebook but at least you can get it approved and seen!
What’s your take? Valuable or not?
Thank you.. So how does this explain that my album cover ad was rejected because of “text percentage”.. How do I apply for this “exception” please?? thanks!

Agendas for all Board of Trustees meetings are posted at least 48 hours prior to the meeting. You will find here agendas dating back to 2009.
Board packets contain all of the documents and materials for each meeting of the Board of Trustees. These are the same packets provided to the trustees prior to each meeting. The regular meetings of the Board of Trustees typically are held on the fourth Tuesday of the month. The Committee of the Whole meetings typically are held on the second Monday of the month. The meeting packet and agenda are posted at least 48 hours prior to the meeting. You will find here meeting packets dating back to 2009.

When hijacked airliners crashed into the tall Towers of the World Trade Center, in New York City [on 11 September 2001], each injected a burning cloud of aviation fuel throughout the 6 levels (WTC 2) to 8 levels (WTC 1) in the impact zone. The burning fuel ignited the office furnishings: desks, chairs, shelving, carpeting, work-space partitions, wall and ceiling panels; as well as paper and plastic of various kinds.
How did these fires progress? How much heat could they produce? Was this heat enough to seriously weaken the steel framework? How did this heat affect the metal in the rubble piles in the weeks and months after the collapse? This report is motivated by these questions, and it will draw ideas from thermal physics and chemistry. My previous report on the collapses of the WTC Towers described the role of mechanical forces (1).
Basic facts about the WTC fires of 9/11/01 are abstracted by the numerical quantities tabulated here.
Tower 1 stood for one hour and forty-two minutes after being struck between levels 92 and 99 by an airplane; the block above the impact zone had 11 levels. During the first 40 minutes of this time, fires raged with an average heat release rate of 2 GW (GW = giga watts = 10^9 watts), and the total heat energy released during the interval between airplane impact and building collapse was 8000 GJ (GJ = giga-joules = 10^9 joules).
A joule is a unit of energy; a watt is a unit of power; and one watt equals an energy delivery rate of one joule per second.
Tower 2 stood for fifty-six minutes after being struck between levels 78 and 83, isolating an upper block of 27 levels. The fires burned at a rate near 1 GW for forty minutes, diminishing later; and a total of 3000 GJ of heat energy was released by the time of collapse.
WTC 2 received half as much thermal energy during the first 40 minutes after impact, had just over twice the upper block mass, and fell within half the time than was observed for WTC 1. It would seem that WTC 1 stood longer despite receiving more thermal energy because its upper block was less massive.
The data in Table 1 are taken from the executive summary of the fire safety investigation by NIST (2).
The NIST work combined materials and heat transfer lab experiments, full-scale tests (wouldn’t you like to burn up office cubicles?), and computer simulations to arrive at the history and spatial distribution of the burning. From this, the thermal histories of all the metal supports in the impact zone were calculated (NIST is very thorough), which in turn were used as inputs to the calculations of stress history for each support. Parts of the structure that were damaged or missing because of the airplane collision were accounted for, as was the introduction of combustible mass by the airplane.
Steel loses strength with heat. For the types of steel used in the WTC Towers (plain carbon, and vanadium steels) the trend is as follows, relative to 100% strength at habitable temperatures.
I use C for Centigrade, F for Fahrenheit, and do not use the degree symbol in this report.
The fires heated the atmosphere in the impact zone (a mixture of gases and smoke) to temperatures as high as 1100 C (2000 F). However, there was a wide variation of gas temperature with location and over time because of the migration of the fires toward new sources of fuel, a complicated and irregular interior geometry, and changes of ventilation over time (e.g., more windows breaking). Early after the impact, a floor might have some areas at habitable temperatures, and other areas as hot as the burning jet fuel, 1100 C. Later on, after the structure had absorbed heat, the gas temperature would vary over a narrower range, approximately 200 C to 700 C away from centers of active burning.
As can be seen from Table 2, steel loses half its strength when heated to about 570 C (1060 F), and nearly all once past 700 C (1300 F). Thus, the structure of the impact zone, with a temperature that varies between 200 C and 700 C near the time of collapse, will only have between 20% to 86% of its original strength at any location.
The steel frames of the WTC Towers were coated with “sprayed fire resistant materials” (SFRMs, or simply “thermal insulation”). A key finding of the NIST Investigation was that the thermal insulation coatings were applied unevenly — even missing in spots — during the construction of the buildings, and — fatally — that parts of the coatings were knocked off by the jolt of the airplane collisions.
Spraying the lumpy gummy insulation mixture evenly onto a web of structural steel, assuming it all dries properly and none is banged off while work proceeds at a gigantic construction site over the course of several years, is an unrealistic expectation. Perhaps this will change, as a “lesson learned” from the disaster. The fatal element in the WTC Towers story is that enough of the thermal insulation was banged off the steel frames by the airplane jolts to allow parts of frames to heat up to 700 C. I estimate the jolts at 136 times the force of gravity at WTC 1, and 204 at WTC 2.
“As the insulation thickness decreases from 1 1/8 in. to 1/2 in., the columns heat up quicker when subjected to a constant radiative flux. At 1/2 in. the column takes approximately 7,250 s (2 hours) to reach a critical temperature of 700 C with a gas temperature of 1,100 C. If the column is completely bare (no fireproofing) then its temperature increases very rapidly, and the critical temperature is reached within 350 s. For a bare column, the time to reach a critical temperature of 700 C ranges between 350 to 2,000 s.
Now, let’s explore heat further.
How Big Were These Fires?
I will approximate the size of a level (1 story) in each of the WTC Towers as a volume of 16,080 m^3 with an area of 4020 m^2 and a height of 4 m (4). Table 3 shows several ways of describing the total thermal energy released by the fires.
The fires in WTC 1 released an energy equal to that of an explosion of 1.9 kilotons of TNT; the energy equivalent for WTC 2 is 717 tons. Obviously, an explosion occurs in a fraction of a second while the fires lasted an hour or more, so the rates of energy release were vastly different. Even so, this comparison may sharpen the realization that these fires could weaken the framework of the buildings significantly.
How Hot Did The Buildings Become?
Let us pretend that the framework of the building is made of “ironcrete,” a fictitious mixture of 72% iron and 28% concrete. This framework takes up 5.4% of the volume of the building, the other 94.6% being air. We assume that everything else in the building is combustible or an inert material, and the combined mass and volume of these are insignificant compared to the mass and volume of ironcrete. I arrived at these numbers by estimating volumes and cross sectional areas of metal and concrete in walls and floors in the WTC Towers.
The space between floors is under 4 meters; and the floors include a layer of concrete about 1/10 meter thick. The building’s horizontal cross-section was a 63.4 meter square. Thus, the gap between floors was nearly 1/10 of the distance from the center of the building to its periphery. Heat radiated by fires was more likely to become trapped between floors, and stored within the concrete floor pans, than it was to radiate through the windows or be carried out through broken windows by the flow of heated air. We can estimate a temperature of the framework, assuming that all the heat became stored in it.
The amount of heat that can be stored in a given amount of matter is a property specific to each material, and is called heat capacity. The ironcrete mixture would have a volumetric heat capacity of Cv = 2.8*10^6 joules/(Centigrade*m^3); (* = multiply). In the real buildings, the large area of the concrete pads would absorb the heat from the fires and hold it, since concrete conducts heat very poorly. The effect is to bath the metal frame with heat as if it were in an oven or kiln. Ironcrete is my homogenization of materials to simplify this numerical example.
The quantity of heat energy Q absorbed within a volume V of material with a volumetric heat capacity Cv, whose temperature is raised by an amount dT (for “delta-T,” a temperature difference) is Q = Cv*V*dT. We can solve for dT. Here, V = (870 m^3)*(# levels); also dT(1) corresponds to WTC 1, and dT(2) corresponds to WTC 2.
dT(2) = (3 x 10^12)/[(2.8 x 10^6)*(870)*6] = 205 C.
Our simple model gives a reasonable estimate of an average frame temperature in the impact zone. The key parameter is Q (for each building). NIST spent considerable effort to arrive at the Q values shown in Table 3 (3). Our model gives a dT comparable to the NIST results because both calculations deposit the same energy into about the same amount of matter. Obviously, the NIST work accounts for all the details, which is necessary to arrive at temperatures and stresses that are specific to every location over the course of time. Our equation of heat balance Q = Cv*V*dT is an example of the conservation of energy, a fundamental principle of physics.
Well, Can The Heat Weaken The Steel Enough?
On this, one either believes or one doesn’t believe. Our simple example shows that the fires could heat the frames into the temperature range NIST calculates. It seems entirely reasonable that steel in areas of active and frequent burning would experience greater heating than the averages estimated here, so hotspots of 600 C to 700 C seem completely believable. Also, the data for WTC Towers steel strength at elevated temperatures is not in dispute. I believe NIST; answer: yes.
Let us follow time through a sequence of thermal events.
The airplanes hurtling into the buildings with speeds of at least 200 m/s (450 mph) fragmented into exploding torrents of burning fuel, aluminum and plastic. Sparks generated from the airframe by metal fracture and impact friction ignited the mixture of fuel vapor and air. This explosion blew out windows and billowed burning fuel vapor and spray throughout the floors of the impact zone, and along the stairwells and elevator shafts at the center of the building; burning liquid fuel poured down the central shafts. Burning vapor, bulk liquid and droplets ignited most of what they splattered upon. The intense infrared radiation given off by the 1100 C (2000 F) flames quickly ignited nearby combustibles, such as paper and vinyl folders. Within a fraction of a second, the high pressure of the detonation wave had passed, and a rush of fresh air was sucked in through window openings and the impact gash, sliding along the tops of the floors toward the centers of intense burning.
Hot exhaust gases: carbon monoxide (CO), carbon dioxide (CO2), water vapor (H2O), soot (carbon particles), unburned hydrocarbons (combinations with C and H), oxides of nitrogen (NOx), and particles of pulverized solids vented up stairwells and elevator shafts, and formed thick hot layers underneath floors, heating them while slowly edging toward the openings along the building faces. Within minutes, the aviation fuel was largely burned off, and the oxygen in the impact zone depleted.
Fires raged throughout the impact zone in an irregular pattern dictated by the interplay of the blast wave with the distribution of matter. Some areas had intense heating (1100 C), while others might still be habitable (20 C). The pace of burning was regulated by the area available for venting the hot exhaust gases, and the area available for the entry of fresh air. Smoke was cleared from the impact gash by air entering as the cycle of flow was established. The fires were now fueled by the contents of the buildings.
Geometrically, the cement floors had large areas and were closely spaced. They intercepted most of the infrared radiation emitted in the voids between them, and they absorbed heat (by conduction) from the slowly moving (“ventilation limited”) layer of hot gases underneath each of them. Concrete conducts heat poorly, but can hold a great deal of it. The metal reinforcing bars within concrete, as well as the metal plate underneath the concrete pad of each WTC Towers floor structure, would tend to even out the temperature distribution gradually.
This process of “preheating the oven” would slowly raise the average temperature in the impact zone while narrowing the range of extremes in temperature. Within half an hour, heat had penetrated to the interior of the concrete, and the temperature everywhere in the impact zone was between 200 C and 700 C, away from sites of active burning.
Fire moved through the impact zone by finding new sources of fuel, and burning at a rate limited by the ventilation, which changed over time.
Heat within the impact zone “cracks” plastic into a sequence of decreasingly volatile hydrocarbons, similar to the way heat separates out an array of hydrocarbon fuels in the refining of crude oil. As plastic absorbs heat and begins to decompose, it emits hydrocarbon vapors. These may flare if oxygen is available and their ignition temperatures are reached. Also, plumes of mixed hydrocarbon vapor and oxygen may detonate. So, a random series of small explosions might occur during the course of a large fire.
Plastics not designed for use in high temperature may resemble soft oily tar when heated to 400 C. The oil in turn might release vapors of ethane, ethylene, benzene and methane (there are many hydrocarbons) as the temperature climbs further. All these products might begin to burn as the cracking progresses, because oxygen is present and sources of ignition (hotspots, burning embers, infrared radiation) are nearby. Soot is the solid end result of the sequential volatilization and burning of hydrocarbons from plastic. Well over 90% of the thermal energy released in the WTC Towers came from burning the normal contents of the impact zones.
Aluminum alloys melt at temperatures between 475 C and 640 C, and molten aluminum was observed pouring out of WTC 2 (5). Most of the aluminum in the impact zone was from the fragmented airframe; but many office machines and furniture items can have aluminum parts, as can moldings, fixtures, tubing and window frames. The temperatures in the WTC Towers fires were too low to vaporize aluminum; however, the forces of impact and explosion could have broken some of the aluminum into small granules and powder. Chemical reactions with hydrocarbon or water vapors might have occurred on the surfaces of freshly granulated hot aluminum.
2*Al + 3*H2O + heat -> Al2O3 + 3*H2.
Al + Fe2O3 + ignition -> Al2O3 + Fe.
Did patches of thermite form naturally, by chance, in the WTC Towers fires? Could there really have been small bits of melted steel in the debris as a result? Could there have been “thermite residues” on pieces of steel dug out of the debris months later? Maybe, but none of this leads to a conspiracy. If the post-mortem “thermite signature” suggested that a mass of thermite comparable to the quantities shown in Table 3 was involved, then further investigation would be reasonable. The first task of such an investigation would be to produce a “chemical kinetics” model of the oxidation of the fragmented aluminum airframe, in some degree of contact to the steel framing, in the hot atmosphere of hydrocarbon fires in the impact zone. Once Nature had been eliminated as a suspect, one could proceed to consider Human Malevolence.
Nature is endlessly creative. The deeper we explore, the more questions we come to realize.
Steel columns along a building face, heated to between 200 C and 700 C, were increasingly compressed and twisted into a sharpening bend. With increasing load and decreasing strength over the course of an hour or more, the material became unable to rebound elastically, had the load been released. The steel entered the range of plastic deformation, it could still be stretched through a bend, but like taffy it would take on a permanent set. Eventually, it snapped.
Months later, when this section of steel would be dug out of the rubble pile, would the breaks have the fluid look of a drawn out taffy, or perhaps “melted” steel now frozen in time? Or, would these be clean breaks, as edge glass fragments; or perhaps rough, granular breaks as through concrete?
The basements of the WTC Towers included car parks. After the buildings collapsed, it is possible that gasoline fires broke out, adding to the heat of the rubble. We can imagine many of the effects already described, to have occurred in hot pockets within the rubble pile. Water percolating down from that sprayed by the Fire Department might carry air down also, and act as an oxidizing agent.
The tight packing of the debris from the building, and the randomization of its materials would produce a haphazard and porous form of ironcrete aggregate: chunks of steel mixed with broken and pulverized concrete, with dust-, moisture-, and fume-filled gaps. Like a pyramid of barbecue briquettes, the high heat capacity and low thermal conductivity of the rubble pile would efficiently retain its heat.
Did small hunks of steel melt in rubble hot spots that had just the right mix of chemicals and heat? Probably unlikely, but certainly possible.
Pulverized concrete would include that from the impact zone, which may have had part of its water driven off by the heat. If so, such dust would be a desiccating substance (as is Portland cement prior to use; concrete is mixed sand, cement and water). Part of the chronic breathing disorders experienced by many people exposed to the atmosphere at the World Trade Center during and after 9/11/01 may be due to the inhalation of desiccating dust, now lodged in lung tissue.
Did the lingering hydrocarbon vapors and fumes from burning dissolve in water and create acid pools? Did the calcium-, silicon-, aluminum-, and magnesium-oxides of pulverized concrete form salts in pools of water? Did the sulfate from the gypsum wall panels also acidify standing water? Did acids work on metal surfaces over months, to alter their appearance?
In the enormity of each rubble pile, with its massive quantity of stored heat, many effects were possible in small quantities, given time to incubate. It is even possible that in some little puddle buried deep in the rubble, warmed for months in an oven-like enclosure of concrete rocks, bathed in an atmosphere of methane, carbon monoxide, carbon dioxide, and perhaps a touch of oxygen, that DNA was formed.
 “Fire Structure Interface and Thermal Response of the World Trade Center Towers,” NIST NCSTAR1-5G, (draft supporting technical report G), http://wtc.nist.gov/pubs/NISTNCSTAR1-5GDraft.pdf, (28 September 2006), Chapter 3, page 32 (page 74 of 334 of the electronic PDF file).
 1 m = 3.28 ft; 1 m^2 = 10.8 ft^2; 1 m^3 = 35.3 ft^3; 1 ft = 0.31 m; 1 ft^2 = 0.93 m^2; 1 ft^3 = 0.28 m^3.
Just in from the edge of the fire zone at Annadel Park, Santa Rosa, in Sonoma County, California, U.S.A., on 18 October 2017. Fires in this vicinity raged from 8-17 October 2017.
An advocate for the homeless (Miles Sarvis-Wilburn) just posted on his blog (link below) a criticism of Sonoma County (CA, U.S.A.) for working so hard and spending so much to help the county’s well-housed residents avoid the catastrophe of wildfires (during October 2017) destroying their homes and threatening their lives and prosperity, yet failing to eliminate the chronic homelessness of the county’s destitute street-people. Mr. Sarvis-Wilburn called this “hypocrisy.” The following is my reply to this argument.
This is the U.S.A., it’s all about the money. The chronically homeless population in Sonoma County is a small fraction of the county’s population, and the current public expenses for them are not monumental (and I am sympathetic to the reasonable and compassionate arguments for increasing that public spending: federal, state and local).
Tens of thousands of Santa Rosa city and Sonoma County residents (normally well housed) were displaced by evacuations during the October Fires (I heard the number 70,000 at one point). The Herculean task of fighting the vast fires to save those homes and residences (rental housing, trailer parks and hotels, where possible) was clearly in large part motivated by instinctive human solidarity: to save lives and prevent and alleviate suffering.
However, another motivation in the public interest was to save housing stock to prevent suddenly having a huge increase in the local long-term homeless population (many aged), and thus a huge increase in unanticipated local public expenses.
The solution to the problem of chronic homelessness is known and has been successfully implemented elsewhere: provide secure affordable (i.e., free) housing for the homeless. Once street-living people are securely housed (and fed), social service professionals have a much easier time helping such people overcome the numerous other problems that bedevil their lives, and which overwhelmed them to the point of becoming homeless.
This solution has been found to be cost-effect because it eliminates many public nuisances = public expenses created by having people-in-need living on the streets indefinitely.
Also, and most crassly, preventing the homes and neighborhoods of secure tax-paying residents from being incinerated, and those residents becoming impoverished, bankrupt or fleeing the area, would prevent a drastic loss of revenue for local governments, and a loss of trade (income) for local businesses. The economic motivation to fight the fires is: to prevent a sag, even collapse, of the local economy.
What prevents, or at least slows, the elimination of homelessness in the U.S.A. is simply the individual and organized selfishness, which we in the U.S.A. call “conservative” politics and “free market” economy and personal “freedom,” as opposed to the “wasteful-pay-for-the-losers” political attitude known as “socialism,” which is disliked by “conservatives” because it “raises taxes” and in general makes greedy people apprehensive about not being able to get as much as they lust for.


There are 10.0 stickers on a page. If you have 22.0 pages of stickers, how many stickers do you have? Answer:

Let the sides of the triangle be a, b, and c, with c=240 being the hypotenuse.
The triangle has the minimal circumference when a=1 and b=sqrt(c2-12) (approximately 240.0). The circumference in that case is approximately 480.0.
The triangle has the maximal circumference when a and b are equal: a=b=sqrt(1/2×c2) (approximately 169.7). The circumference in that case is approximately 579.4.
The only two squares of whole numbers that lie in the interval [480.0, 579.4] are 529 and 576.
Now we know that a+b=529 or a+b=576. In addition, a2+b2=c2, so a2+b2=57600.
Suppose that a+b=529. Then b=529-a, and when we fill that in a2+b2=57600, we get a2+(529-a)2=57600, so a2-289×a+12960.5=0. This equation has no solutions if a must be integer.
Suppose that a+b=576. Then b=576-a, and when we fill that in a2+b2=57600, we get a2+(576-a)2=57600, so a2-336×a+27648=0. This equation has solutions a=192 (b=144) and a=144 (b=192).
Therefore, the sides of the triangle are a=144, b=192, and c=240.

This page show the 277 prime list.
Prime number index is from 276001 to 277000.
Prime number list from 3890981 to 3906319, contain 1000 prime numbers.
3890981 is at position 276001, and 3906319 is at position 277000 in all prime list.

* No more climbing crashes yay!
There are 5 rounds. If 4 are in the final round; how many are in the qualifying round? I'll tell you. 4x4=16. 16x16=256. 256x256=65,536.65,536x65,536=4,294,967,296. That would mean that 4,294,967,296 ducks entered the qualifying round.

Editing 2106 - Byrneville Elementary School, Inc.
N/A 2067 - Escambia Sch. Dist. Jail Prog.

Section - 1 - SMCD 222 Amended.
Notwithstanding anything to the contrary, the Taxation (Amendment) Decree, 1979 (S.M.C.D. 222) shall be deemed to have come into force on the date on which that Decree was published in the Gazette that is to say, the 16th day of February, 1979 and accordingly, the cancellation of exemptions from duty and tax effected by that Decree shall be deemed to have taken effect as from that date.

We *just* wrapped up our syrup operations at Duke Farms.
This year, we produced over 42.5 gallons of syrup; over double what we made last year. To produce the 42,5 gallons of syrup, we had to collect ~1,700 gallons of sap from our local sugar maples. Join us in the Sugar Shack to see what happens to maple sap after it is collected.

Q2 ⇒ Pick out the wrong statement.
Density of dry air at N.T.P. is 1 gm/litre.
One kg-mole of an ideal gas occupies 22.4 m3 at N.T.P.
One lb-mole of an ideal gas occupies 359 ft3 at N.T.P.
One gm-mole of an ideal gas occupies 22.4 litres (i.e., 22400 c.c) at N.T.P.
Q5 ⇒ How many phases are present at eutectic point ?
Q7 ⇒ The molecules of a liquid which is in equilibrium with its vapor at its boiling point on an average have equal __________ in the two phases.
Q9 ⇒ Avogadro number is the number of molecules in one __________ of a gas.
Q10 ⇒ The atomic weight of helium is 4 times that of hydrogen. Its diffusion rate as compared to hydrogen will be __________ times.
Q11 ⇒ Pick out the wrong statement about the recycle stream in a process.
Recycling in a process does not help in getting higher extent of reaction.
Recycling in a process stream helps in utilising the valuable reactants to the maximum with minimum loss of reactants.
The ratio of the quantity of a reactant present in the reactor feed of a recycling operation to the quantity of the same reactant entering the process as fresh feed is called combined feed ratio.
Recycling is exemplified by refluxing back a part of the distillate to the distillation column to maintain the quantity of liquid within the column.
Q13 ⇒ Boiling point of a non-homogeneous mixture of immiscible liquids is __________ that of any one of its separate components.
Q15 ⇒ Which of the following gases is the most soluble in water ?
Q16 ⇒ In case of vapor-liquid equilibria, which of the following does not account for gas phase deviation from ideality?
Q18 ⇒ Pick out the wrong unit conversion of thermal conductivity.
Q19 ⇒ The heat of vaporisation __________ with increase in pressure.
Q20 ⇒ 1 Kcal/kg. °C is equivalent to __________ BTU/lb. °F.

How long would it take to fly anywhere on Earth using a SpaceX rocket?
How long would it take to fly anywhere on Earth using a SpaceX rocket?
Answer: less than an hour.
Before he gets us to Mars, Elon Musk wants to get us anywhere on Earth in under an hour.
At Friday’s International Astronautical Conference, the CEO of SpaceX revealed the company’s plans for its largest airborne vehicle yet, the BFR. While originally intended to facilitate SpaceX’s goal of putting humans on Mars by 2024, Musk announced that he also envisions using the rocket for high-speed Earth travel.
The world’s longest flight — from Doha, Qatar, to Auckland, New Zealand — currently takes 16.5 hours. However, with SpaceX’s BFR rocket, flight time would be just 45 minutes.

Someone drove from Aardvark to Beeville.
On the first day they travelled 1/3 of the distance.
On day two they travelled 1/2 of the remaining distance.
On day three they travelled 2/3 of the remaining distance.
On day four, after covering 3/4 of the remaining distance, they were still 5 miles away from Beeville.
How many miles had they covered so far?
Day 4: travelled 3/4 leaving 5 miles, which means that 5 miles is 1/4, so 20 miles were left at the start of Day 4.
Day 3: travelled 2/3 leaving 20 miles, which means that 20 miles is 1/3, so 60 miles were left at the start of Day 3.
Day 2: travelled 1/2 leaving 60 miles, which means that 60 miles is 1/2, so 120 miles were left at the start of Day 2.
Day 1: travelled 1/3 leaving 120 miles, which means that 120 miles is 2/3, so 180 miles were left at the start of Day 1.
Of the 180 miles overall, 175 were covered in the 4 days.

Joe can type 11 words in 8 seconds. At this rate, how many words can he type in two minutes?
Joe can type 11 words in 8 seconds. At this rate, how many words can he type in two minutes?
So in 120 seconds Joe can type=120 X 1.375=165 words in 2 minutes, or 165/2=82.5 words per minute!!!. That makes Joe very fast typist!!.

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1. What is the greatest common factor of 21, 49 and 91?
Explanation:In order to find the GCF, we can first find their prime factors.
Since GCF is the greatest common factor of the given numbers, we can see that here the common factor is 7.
Therefore GCF of 21, 49 and 91 is 7.
2. If 2x + y = 3 and 3x + 2y = 4, then what is the value of (x + y)?
Explanation: To find the value of (x + y), we can first solve the system of linear equation.
From the first equation we get: y = 3 – 2x.
Substitute this ‘y’ value in the second equation. This gives: 3x + 2(3 – 2x) = 4.
So, 3x + 6 – 4x = 4 ==> 6 – x = 4 ==> x = 6 – 4 ==> x = 2.
Now y = 3 – 2x ==> y = 3 – 2(2) ==> y = 3 – 4 ==> y = -1.
Therefore, (x + y) = 2 – 1 = 1.
3. Given ‘a’ and ‘b’ are positive numbers and (a + b)2 = 196 and (a2 – b2) = 28, then what is the value of b?
Explanation: If (a + b)2 = 196, then (a + b) = √196 ==> a + b = 14 (the value of (a + b) cannot be -14 cause ‘a’ and ‘b’ are supposed to be positive numbers).
Then expanding (a2 – b2) = (a + b) (a – b), we can say 14 * 2 = 28.
Since (a + b) = 14, hence (a – b) should be = 2.
Solving the above two equations we have: 2a = 16 ==> a = 8.
Since (a + b) = 14 ==> b = 14 – 8 ==> b = 6.
4. What is the slope of the line perpendicular to 3x – 7y + 16 = 0?
Explanation: The given line can be rearranged as: 7y = 3x + 16.
This gives: y = (3/7)x + (16/7).
This is in the form of y = mx +b where ‘m’ is the slope of the line.
Therefore slope of the given line = 3/7.
If two lines are perpendicular, then the product of their slopes is = -1.
This gives (3/7) * m = -1 ==> m = -7/3 is the slope of the perpendicular line.
5. The factor of the polynomial expression: -3x2 + 2x + 8is?
Explanation: The given polynomial expression can be factored by first splitting its middle term.
Since -3 * 8 = = -24, we need to find two numbers which would add and give the middle term number 2.
This gives: -3x2 + 6x – 4x + 8.
Now we can group the first two and the last two terms: -3x (x – 2) – 4 (x – 2).
This gives: (x – 2) (-3x – 4).
Since (x – 2) is not in the options, (-3x – 4) is the answer.
6. If a, b, and c are three consecutive numbers and their sum is 21, then what is the value of ‘b’?
Explanation: Let a = n, b = (n + 1) and c = (n + 2) as they are three consecutive numbers.
Sum ==> n + n + 1 + n + 2 = 21 ==> 3n + 3 = 21 ==> 3n = 21 – 3.
This gives: 3n = 18 ==> n = 18/3 ==> n = 6.
Hence the value of b = (n + 1) is 6 + 1 ==> b = 7.
7. The sum of ‘p’ numbers is 7 more than thrice the numbers. If the arithmetic mean of all the ‘p’ numbers is 4, then the value of ‘p’ is?
Explanation: Sum of ‘p’ numbers = 3p + 7.
This gives: (3p + 7)/ p = 4 ==> When we cross-multiply we get: 3p + 7 = 4p.
Solving for ‘p’ gives 4p – 3p = 7 ==> p = 7.

Talladega is a city in Talladega County, Alabama, United States. It was incorporated in 1835. At the 2010 census the population was 15,676. The city is the county seat of Talladega County. Talladega is approximately 50 miles (80 km) east of Birmingham.
According to the U.S. Census Bureau, the city has a total area of 24.1 square miles (62.3 km²), of which 24.0 square miles (62.1 km²) is land and 0.077 square miles (0.2 km²), or 0.30%, is water.

White cotton poplin puff sleeve short dress from JW Anderson. Classic oxford shirt styling with a pointed collar and button front closures. Arched hem with cut-out detailing. Unlined. Color: White. 100% Cotton. Imported.
White cotton poplin puff sleeve short dress from JW Anderson. Classic oxford shirt styling with a pointed collar and button front closures. Arched hem with cut-out detailing. Unlined. Color: White. 100% Cotton. Imported.
White cotton poplin puff sleeve short dress from JW Anderson. Classic oxford shirt styling with a pointed collar and button front closures. Arched hem with cut-out detailing. Unlined. Color: White. 100% Cotton. Imported.

can i get loan against my kidney?
48266 Thank you! I can now get my wife "s wedding ring back from the pawn shop , pay the loan sharks and still have some left over to visit the casino!!! Thanks Doc!!

How To: Measure your Pond Liner To begin, you'll need 3 measurements. Length, width and depth. These will be the longest length, the widest width and the very deepest depth. No averages, and these measurements will always be taken in feet.
To begin, you'll need 3 measurements. Length, width and depth. These will be the longest length, the widest width and the very deepest depth. No averages, and these measurements will always be taken in feet.
First, you'll multiply the depth times 2. Because, basically you're going down twice on both sides. Now we'll add overhang. I generally like to add a foot to both sides. So, we will add 2. One foot for each side. If you wanted 6 inches to each side, you would only add one. To reiterate, depth times 2, plus overhang (2). We'll then take that number and simply add it to our initial length, for the final liner length. And the same for the width, to achieve our final liner width.
First, take the depth, 1.5. Multiply times 2, you get 3. 3 plus 2, one foot of overhang on each side, and that gives us 5. Then, add 5 to the length, 15. And add 5 to the width, 15. Our final dimensions are 15 x 15. So, for a 10 x 10 x 1.5 pond a 15 x 15 liner is required.

Total Carat Weight: 3.24 ct.
This bracelet features hexagon shaped links that sit between colorful gemstones surrounded by a duo of lovely diamonds. A one of a kind bracelet, you get to play designer by selecting the gemstone type, diamond quality and metal type! Select from three sizes, a .27 ct., .54 ct., and a .64 ct. total carat weight. Order one as a gift or for yourself today!
Model Carat Weight: 3.24 ct.
Effective Carat Weight: 3.24 ct.
Stone Carat Weight: 18X0.015 ct., 9X0.330 ct.

Replacement battery for the UniData ICW-1000G and WPU-7800 phones. Fully replaces UniData battery number 3ICW-1000B and KAL523450AR. Lithium ion rechargeable battery pack, 3.7 Volt, 1200 mAh capacity rating. Outlasts OEM battery by 9%.

Feb 2 Bunny Stretches Sooo Tall!
Thanks, Gil, Sarah, Galen, and bunny Alfie! Sarah writes that Alfie is hoping for a treat here!

Bing wrote about their new Multi-Perspective Answers for their Intelligent Answers set. This is similar to what Google calls "featured snippets," in case you were wondering.
Bing wrote "Today, when you use a search engine to find answers to your questions, your experience can quickly turn into an echo chamber because search results often cover just one side of the spectrum of possible answers to your questions, which can lead to confirmation bias."
The funny thing is, I mentioned this to a Googler recently - I said something to the affect of, why not show featured snippets that are true to the person searching. So if a Trump supporter searches for something related to Trump, show a pro-Trump featured snippet to that person and the opposite to a non-Trump supporter. Since Google doesn't want to get into the game of being the absolute truth, give people a more personalized result.
Bing doesn't specifically do this, they just give two sides of the story. Heck, I searched for this on Bing and got an "intelligent answer" whereas Google won't show a featured snippet for it, at least not right now.
Google did show off showing two featured snippets on a single search result page, but it is more about diversifying the query that giving two sides of the answer.
In any event, this is a neat feature and you can learn a lot more about how it works over here.
Previous story: Is Google Showing Fewer Featured Snippets In Search Results?


There are 96.0 cupcakes for 8.0 children to share. How much will each person get if they share the cupcakes equally? Answer:

There are 209 doctors and nurses in a hospital. If the ratio of the doctors to the nurses is 11 : 8, then how many nurses are there in the hospital?
There are 209 doctors and nurses in a hospital. If the ratio of the doctors to the nurses is 11 : 8, then how many nurses are there in the hospital?
Answer : 1 88 Explanation : Answer: A) 88 Explanation: Given,The ratio of the doctors to the nurses is 11 : 8Number of nurses = 8/19 x 209 = 88.

A can do a work in 9 days, B can do a work in 7 days, C can do a work in 5 days. A works on the first day, B works on the second day and C on the third day respectively that is they work on alternate days. When will they finish the work ?
A can do a work in 9 days, B can do a work in 7 days, C can do a work in 5 days. A works on the first day, B works on the second day and C on the third day respectively that is they work on alternate days. When will they finish the work ?
Answer : 3 [6 (261/315)] days Explanation : Answer: C) [6 (261/315)] days Explanation: After day 1, A finishes 1/9 of the work.After day 2, B finishes 1/7 more of the total work. (1/9) (1/7) is finished.After day 3, C finishes 1/5 more of total work. Total finished is 143/315.So, after day 6, total work finished is 286/315. Now remaining work = 29 /315 On day 7, A will work again Work will be completed on day 7 when A is working. He must finish 29/315 of total remaining work. Since he takes 9 days to finish the total task, he will need 261/315 of the day. Total days required is 6 (261/315) days.

The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and 15 as remainder. What is the smaller number ?
The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and 15 as remainder. What is the smaller number ?
Answer : 1 270 Explanation : Answer: A) 270 Explanation: Let the smallest number be x.Then larger number = (x 1365)x 1365 = 6x 15= 5x = 1350 x = 270Smaller number = 270.

Derivative of sinx by first principle?
Derivative of sinx by first principle?
Answer : 3 cosx Explanation : Answer: C) cosx Explanation: Thus, the derivative of sinx by first principle is cosx.

Diameter of a Ping Pong Ball?
Diameter of a Ping Pong Ball?
Answer : 3 40 mm Explanation : Answer: C) 40 mm Explanation: The ball with which Table Tennis is played is called Ping Pong Ball.According to the International rules, the diameter of the ping pong ball must be not less than 39.5 millimeters and not greater than 40.5 millimeters having a mass of 2.7 grams.

Hemavathi cut a cake into two halves and cuts one half into smaller pieces of equal size. Each of the small pieces is fifteen grams in weight. If she has nine pieces of the cake in all with her, how heavy was the original cake ?
Hemavathi cut a cake into two halves and cuts one half into smaller pieces of equal size. Each of the small pieces is fifteen grams in weight. If she has nine pieces of the cake in all with her, how heavy was the original cake ?
Answer : 1 240 gms Explanation : Answer: A) 240 gms Explanation: The nine pieces consist of 8 smaller equal pieces and one half cake piece.Weight of each small piece = 15 g.So, total weight of the cake = [2 x (15 x 8)]g = 240 g.

Out of sixty students, there are 14 who are taking Economics and 29 who are taking Calculus. What is the probability that a randomly chosen student from this group is taking only the Calculus class ?
Out of sixty students, there are 14 who are taking Economics and 29 who are taking Calculus. What is the probability that a randomly chosen student from this group is taking only the Calculus class ?
Answer : 2 7/15 Explanation : Answer: B) 7/15 Explanation: Given total students in the class = 60Students who are taking Economics = 24 andStudents who are taking Calculus = 32Students who are taking both subjects = 60-(24 32) = 60 - 56 = 4Students who are taking calculus only = 32 - 4 = 28probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.

In which year SBI was Nationalised ?
In which year SBI was Nationalised ?
Answer : 2 1955 Explanation : Answer: B) 1955 Explanation: The nationalisation of banks in India took place in 1969 by Mrs. Indira Gandhi the then prime minister. It nationalised 14 banks then. These banks were mostly owned by businessmen and even managed by them. Before the steps of nationalisation of Indian banks, only State Bank of India (SBI) was nationalised. It took place in July 1955 under the SBI Act of 1955.Nationalisation of Seven State Banks of India (formed subsidiary) took place on 19th July, 1960.1955 : Nationalisation of State Bank of India.1969 : Nationalisation of 14 major banks.1980 : Nationalisation of seven banks with deposits over 200 crores.

What is the Logo and mascot of Asian Athletics Championships that is scheduled to be held from 6 to 9 july 2017 ?
What is the Logo and mascot of Asian Athletics Championships that is scheduled to be held from 6 to 9 july 2017 ?
Answer : 2 Olive Ridley turtles. Explanation : Answer: B) Olive Ridley turtles. Explanation: Odisha Chief Minister Naveen Patnaik has unveiled the logo and mascot of the 22nd Asian Athletics Championships scheduled to be held at the Kalinga Stadium in Bhubaneshwar from July 6 to 9, 2017. Mr. Patnaik also formally announced Bhubaneswar as the host for the sports event. The mascot, named Olly Turtle, depicts the Olive Ridley turtles.

We start with six boxes. We divide these six boxes by three. This means: how many groups of three can we make with the six boxes that we have? Answer: 2.
We start with eighteen boxes. We divide these eighteen boxes by three. This means: how many groups of three can we make with the eighteen boxes that we have? Answer: 6.
We start with thirty-six boxes. We divide these thirty-six boxes by three. This means: how many groups of three can we make with the thirty-six boxes that we have? Answer: 12.

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I mentioned earlier in the week that Monster High is coming up with a new movie on DVD called “13 Wishes”. To celebrate the new release Walmart is giving you a chance to win a prize each day for 13 days!
When you visit each day to enter to win the daily prize you will also find fun things to do like a quick matching games and tic-tac-toe or find fun downloadable content. If you have a Monster High gal then this is the perfect opportunity to win some cool prizes!
Disclosure: This post is brought to you by Lunchbox. You may read more of my disclosure here.
My daughter is really into Monster High.
I am going to have to remember to enter this tomorrow. What a great way to get a head start on holiday presents.
How fun! I’ll have to enter, my niece would love some of those.
What a fun contest, I will have to enter for my daughter.
What a fun giveaway, I will have to enter for my nieces.
My niece loves Monster High!!!
I know some little girls in my life that would love to win Monster High prizes, thanks for posting about the contest!
I don’t have any little girls, but would love to be able to donate to a local Christmas fundraiser.
My niece loves Monster High. I will check this out for her.
These dolls are really cute! Thanks for sharing the contest!
I love these new Monster High dolls. They’re so cute and slightly creepy!

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Leading property website MyHome.ie has announced the introduction of new procedures to review property advertisements which have the potential to breach equality legislation.
Following engagement with the Irish Human Rights and Equality Commission, MyHome.ie has developed practical methods to identify, monitor and block discriminatory advertising on its website.
The Managing Director of MyHome.ie, Angela Keegan said the company was adopting a proactive approach to the new legislation. She also said MyHome.ie was looking forward to working with the IHREC to ensure discriminatory ads did not appear on the site.
Key aspects of the MyHome.ie system include a two-tiered system for checking advertisements published on its website, through both a manual, and auto-check system.
Specifically, the auto-check system will filter “trigger” words and phrases that could be in contravention of equality legislation. MyHome.ie has also put in place investigation, reporting and take-down procedures where discriminatory ads are identified, both pre and post-publication.

As the auditor for my local theme park, I noticed that on Saturday there were 4296 children and 2143 adults and the takings were £98,718.
However, on Sunday, there were 5146 children and 2807 adults and the takings were £122,570.
How much were the adults tickets and children's tickets?
Hint: This is quite a tricky puzzle and knowledge of algebra would certainly help.
Answer: Adult's tickets were £18 and children's were £14.
This is easily solved using a common method for solving simultaneous equations.

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