token_count
int32 48
481k
| text
stringlengths 154
1.01M
| score
float64 3.66
5.34
|
|---|---|---|
171 |
Last month saw unprecedented conditions in both Arctic and Antarctic. The animations below show the average extent of ice cover in September each year since 1979, when satellite observations began.
The US National Snow & Ice Data Center in Boulder, Colorado, divides the Arctic and Southern Oceans into a 25-kilometre grid, and considers cells covered if they have 15 per cent or more of ice. As the first animation reveals, this summer’s melt left the Arctic with only about half the ice cover it enjoyed in the 1980s.
Ice flowing from the continent of Antarctica extends into the ocean as floating ice shelves, which in winter are joined by an extended collar of sea ice. While this year’s winter maximum was a record, the increases in winter ice extent in the Southern Ocean are far less dramatic than the summer declines in the Arctic.<|endoftext|>
| 3.6875 |
2,403 |
# SET THEORY Set Notation Set is a collection
• Slides: 26
SET THEORY
Set Notation Set- is a collection or aggregate of definite, distinct objects. A well-defined set means that it is possible to determine whether an object belongs to a given set. Elements of a Set- the objects or members of a set. • Symbol: ϵ (epsilon)- use to denote the element of a set. Ex: r ϵ A ϵ – use to denote that an element is not an element of the given set. Ex: b ϵ A
Set Notation Ex. of well-defined sets (defined) 1. Set of ace cards 2. Colors of the rainbow 3. Days of the week Ex. of not well-defined sets (undefined) 1. Set of cards 2. Set of books 3. Set of beautiful women in Asia
Set Notation Different symbols are used when dealing with sets: 1. A pair of braces { } – is used to represent the idea of a set. 2. Capital letters of the English alphabet – are used to name sets. Example: A = { a, b, c, d, e } B = { 2, 4, 6, 8, 10 } Letters with subscripts can also be used, e. g. A 1, A 2, A 3, …etc.
Methods of Listing the Elements of a Set 1. Roster or Tabular Method – listing all the elements, enclosed it in braces and separated by comma. Ex: C = { a, b, c } B = { 1, 2, 3, 4, 5 } 2. Rule Method – a conditional way of listing method by writing and description using a particular variable. 3. Set Builder Notation – a modification of the rule method.
Methods of Listing the Elements of a Set Examples: Rule Set-builder 1. A = { counting numbers less than 5} A = { x/x is a counting number less than 5} 2. B = { days of the week that begin with letter S } B = { d/d is a day of the week beginning with letter S } v The symbol ( / ) means “wherein” or “such that. ”
Cardinality Refers to the number of elements contained in a set. Symbol: n(A) or |A| Examples: 1. A = {a, b, c, d} Ans: |A| = 4 2. Z = {x/x is a set of integers} Ans: n(Z) = ∞ or |Z| = ∞
Kinds of Sets
Subset – it is a part of a given set. Let A and B be sets. B is a subset of A if each element of B is an element of A. In symbols: B ⊂ A or B ⊆ A read as “B is a subset of A. ” Two kinds of subset: 1. Proper Subset – a part of a set, symbol ( ⊂ ) 2. Improper (Strict) Subset – the given set is equal to that set, symbol ( ⊆ )
Subset Example: Let P = {5, 6, 7, 8, 9, 10} Q = {5, 7, 9} R = {5, 9, 10} S = {1, 2} True or False 1. Q ⊆ P 2. R ⊂ P 3. P ⊂ Q 4. S ⊂ P 5. R ⊂ Q 6. Q ⊆ Q
Super set If B ⊆ A, then A is a super set of B. In symbols: A ⊇ B read as “A is a superset of B. ” Examples: A 1. A ⊇ B B 2. R ⊇ Q ⊇ Z
Kinds of Sets Empty or Null Set – sets having no elements. Symbol: { } or ∅ Note: An empty set is a subset of any set. Ø Universal Set (U) – also called the general set, is the sum of all sets or the totality of elements under consideration or a particular discussion. Example: U = {a, b, c, d, x, y, z} A = {a, b, c, d} B = {x, y, z} Ø
Kinds of Sets Unit Set – set having only one element. Ex: D = {y/y is a day of the week that begins with letter M} Ø Finite Set – sets having a limited or countable number of elements. Ex: set of counting numbers less than 5 set of colors of the rainbow Ø Infinite Set – sets having an unlimited or uncountable number of elements. Ex: set of counting numbers Ø
Kinds of Sets Equal Sets – sets having the same elements. Symbol: (=) Note: Two sets A and B are said to be equal if and only if A ⊆ B and B ⊆ A. Ex: A = {2, 4, 6} and B = {4, 6, 2} A = B and B = A Ø Equivalent Sets – sets having the same number of elements or cardinality, that is |A| = |B|. Symbol: (~) Ex: X = {1, 2, 3} Y = {a, b, c} |X| = |Y|, so X ~ Y and Y ~ X Ø
Kinds of Sets Joint Sets – sets that have elements in common. Ex: M = {5, 6, 7, 8} and N = {4, 6, 7, 9} are joint sets. Ø Disjoint Sets – sets that have no elements in common. They are mutually exclusive. Ex: C = {c, a, t} and D = {d, o, g} are disjoint sets or mutually exclusive. Ø
Power Set If A is a set, the power set of A is the set of all subsets of A denoted by: P(A) = {X/X ⊆ A} Number of Subsets of a Given Set: - If a set contains n number of elements then the number of subsets is 2ⁿ Examples: 1. What is the power set of O = {a, b, c}? 2. If B = {1, 2}, describe P(B). 3. What is P(ø)?
Venn Diagram It is a graphical representation, usually circular in nature. It is one way of showing the relationships of two or more sets by the use of pictures. This method was developed by John Venn (1834 -1923) thus, the name Venn Diagram. It consists of a rectangle representing the universal set and circles that represent the sets. Sometimes, circles can also represent the universal set.
Venn Diagram U A B a, b, c d, e, f
Set Operations 1. Union – it shows the unity of two or more sets. It is the joining of sets. ( ∪ ) In symbols: A ∪ B = {x/x ϵ A v x ϵ B} Examples: Find the union of sets and draw Venn diagrams a) A = {1, 2, 3} B = {4, 5, 6} b) S = {a, b, c, d} T = {c, d, m, n, o}
Set Operations 2. Intersection – it shows the intersection of the common elements of sets. ( ∩ ) In symbols: A ∩ B = {x/x ϵ A ʌ x ϵ B} Example: Find the intersection of sets and draw the Venn diagram. a) B = {m, o, p, q} C = {m, p, r, s} b) F = {1, 3, 5, 7} G = {2, 4, 6, 8}
Set Operations 3. Complement of a Set – it is the set whose elements are in the universal set but not in a set or a given set. ( ʼ ) In symbols: A’ = {x ϵ U / x ϵ A} Ex: Let U be the universal set. U = {2, 4, 6, 8, 10} A = {6, 8, 10} B = {2, 4, 6} Find: a) A’ b) B’
Set Operations Relative Difference – set of elements found in a set but not belong or found in other set. ( – ) In symbols: A – B = {x/x ϵ A ʌ x ϵ B} Ex: A = {1, 2, 3, 4, 5} B = {3, 4, 5, 6, 7} Find: a) A – B b) B – A 5. Symmetric Difference – the symmetric difference of A and B is denoted by: 4. A B = (A – B) U (B – A) Ex: A = {a, e, i, o, u} B = {e, o, n, s} Find: A B
Set Operations 6. Cartesian Product – For any sets A and B, the Cartesian product A x B is the set of all ordered pairs (a, b) where a ϵ A and b ϵ B. A x B is defined by: A x B = {(a, b) / a ϵ A ʌ b ϵ B} Examples: Let A = {0, 1} B = {x, y, z} Find: a) A x B b) B x A c) A 2
Exercises Given: Let U be the universal set U = {a, b, c, d, e, f, g, h} A = (a, b, c} C = {a, g, h} B = {d, e, f} D = { b, c, d, e, f} Find the ff. and draw Venn diagrams. 1) A’ ∩ B’ 6) P(A) 2) (A U B)’ 7) B x C 3) (C’ ∩ D) U A 8) A – C 4) (A ∩ B)’ 9) D – B 5) A’ U B’ 10) A D
Applications of Set Theory Examples 1. In a class, 15 are taking English, 20 are taking Filipino and 10 are taking both English and Filipino. How many students are there in all? 2. Of 1000 applicants for a mountain-climbing trip in the Himalayas, 450 get altitude sickness, 622 are not in good enough shape, and 30 have allergies. An applicant qualifies if and only if this applicant does not get altitude sickness, is in good shape, and does not have allergies. If there are 111 applicants who get altitude sickness and are not in good enough shape, 14 who get altitude sickness and have allergies, 18 who are not in good enough shape and have allergies, and 9 who get altitude sickness, are not in good enough shape, and have allergies, how many applicants qualify?
Do Worksheet 6<|endoftext|>
| 4.4375 |
1,885 |
## IMPORTANT QUESTION WITH TRICK
Published in: Mathematics
964 views
• ### Sardana Classes
• H No. 1367, Sector 28, Faridabad
• Area: Old Faridabad, Sector 15, Sector 16, Sector 16A, S...
• Courses: Mathematics, Physics, English, Chemistry, Biology,...
• Contact this Institute
IMPORTANT QUESTION WITH TRICK
• 1
ALGEBRA (SIMPLIFICATION)-I
• 2
Algebra is the part of mathematics in which letters and other general symbols are used to represent numbers and quantities in formulae and equations. Mostly equations are in linear & quadratic form i,e. x+7, x2- 3x+7, x2 -3 , x>3 etc. so basically here, we will discuss all algebraic formulas
• 3
(a + = a2 + b2 + 2ab Example : 78 x 78 + 22 x 22 + 2 x 22 Solution : Let 78 = a & 22 = b 78 x 78 + 22 x 22 +2 x 22 = (78+22)2 10000 Example: If x + y = 12 & xy = 32 then x2 + Y2 Solution : (x + Put x+y = 12 & xy =32 we will get 122 -I-Y2 +2 X 32 or 144 = + Y2 +64 or x2 + Y2 80
• 4
= a2 + b2 - 2ab Example : 63 x 63 + 33 x 33 -2 x 63 x 33 Solution : Let 63 = a & 33 = b 63 = (63 - - 302 900
• 5
Example : 161 x 161 -141 x 141 Solution : Let 161 a & 1 41 161 x 161 -141 x 141 —1612 -1412 Example : Solution : 6040 49 x 49 —13x13 72 49 x 49 —13x13 72 (49 + 72 - 13) b) 302 62x36 = 62/2=31 72
• 6
= a3 + b3 + 3ab(a + b) Example : If a + b — 10 & ab = 21 then a3 + b3 Solution : (a + a 3 + b 3 + 3 ab(a + b) so 103 000 or a3 + b3 - 630 = 370
• 7
= аз - b3 - 3ab(a - Example : The value of 3х0.6хо.4хо.2 Solution : . -0.4х0.4х0.4 - З 0.63 - 0.43 - 3х0.6хо.4(о.6 -0.4) = (0.6+0.4)3 0.23 0.008 The value of . -0,4х0.4х0.4 - з 0.008
• 8
- ab + b2) 13 x 13x 13+ 12x 12 Example : 13x 13 —12x 13 + 12x12 3 Solution : a 2 +122 —ab a2 +b2 —ab Example : Solution : 0.0347 x 0.0347 x 0.0347 +0.96538 0,03472 - 0,0347 x 0,9653 + 0,96532 3 a 2 +b2 —ab - 25 = 0.0347+0.9653 = a 2 +b2 —ab
• 9
(a3 - b3) = (a - b) (a2 + ab + b2) Example : Solution : Example : Solution : 13 x 13 x 13— 12 x 12 x 12 13 x 13 + 12 x 13 + 12 x 12 a 3 —b3 a 2 +b2 + ab —a— a 2 +b2 + ab 2.75 x 2.75 x 2.75 — 2.25 x 2.25 x 2.25 2.75 x 2.75 + 2.75 x 2.25 + 2.25 x 2.25 a3 —b3 -a- a 2 +b2 + ab b - 2.75 - 2.25 - 0.5
• 10
Example: If a + b + c Solution : Put a -k b -k C (9)2 81 _ 9 & a b + + c a = 40 then a 2 + b2 + c2 — 9 & ab -k bC + = 40 + c2 + 40 + c2 + 80
• 11
a3 + + c3 — 3abC = (a + b + c) (a 2 + b2 + c2 — ab - bc - ca) Example: Find a3 -k + c3 — 3abC, if a + b -k c = 9 and Solution : a3 + b3 + e — 3abC = (a + b + c) (a2 + b2 + c2 — ab — bc - ca) If a -k b -k c = 9 and ab -k bC -k — 11, here first of all we have to find out of the value of a2 + b2 + c2 By using this formulae put a + b + c 9 & a b + + c a (9)2 _ a2 + b2 + c2 + Il a2 + b2 + c2 + 22 81 -59 so, a 3 + 133 + c 3 — 3 abc = (9) {59 a b + bc+ c a)} = (9) {59 -l l} = =432 - ab - bc - ca)
• 12
If a + b c = O, then a3 + b3 + c3 = 3abC Example: If (x Then x Solution : let x- 9 = c Here, a 3 + b 3 + c 3 — 3abc so indirectly we can say that so finally, (x- 7 + x- 8 + x- 9) or 3x - 24 = 0 or 3x = 24 orx=8
• 13
Example: If x -y = 2 and x2 + Y2 Solution : — 20, (x + — 2(a2 + b2) Putting the value of x + y = 2 and x2 + Y2 - 20, we Will get (x + + (2)2 2(20) (x + y +4 = 40 so (x + = 40- 4 = 36 (2.697 + 0.498)2 + (2.697 - 0.498)2 Example: (2.6972 + 0.4982) Solution : consider a = 2.697 & b = 0.498 putting in question , we will get 2
• 14
= 4ab 1 Example: 5-+ 3 1 Solution : consider 5 3 136) (0.337 + o. 126)2 Example: 136) 3 16 (0.337 -o. 0.337 xo. 126 Solution : Considera = 0.337 & b = 0.126 Putting in question , we will get ab 4ab 4 ab
• 15
Some more important algebraic formulae: then x = 1 then x , then =N2-2 -2)2 -2 6 -2)3 -3(N2 -2)
• 16
4. If x + — then X x -1 66-1 1 then find the value of x 4 Example: If x + — x 1 , then we know that x6 Solution : x + — x So 48 42 36 -1 36
## Need a Tutor or Coaching Class?
Post an enquiry and get instant responses from qualified and experienced tutors.
Post Requirement
Mathematics
769 views
Mathematics
978 views
Mathematics
560 views
Mathematics
2,183 views
Mathematics
347 views
Mathematics
472 views
Mathematics
483 views
Mathematics
4,921 views
Mathematics
1,892 views
Mathematics
323 views
Drop Us a Query<|endoftext|>
| 4.46875 |
1,168 |
Rain gardens use run-off from roofs, driveways, other impervious surfaces. Rainwater is used by native plants rather than running to the storm sewer or ditch.
A rain garden is just that – a garden that utilizes the rain. It can be a personal contribution to cleaner water and an improved environment in a backyard and neighborhood.
It is a garden with a shallow depression to receive roof, yard, parking lot or street rain run-off. Rain gardens are being established in yards, boulevards, surrounding parking lots and more in urban and rural America. Many urban communities are working together to establish rain gardens in public and private areas.
Rain gardens increase the amount of water filtering into the ground to recharge the groundwater. They help reduce the amount of pollutants washing off imperious surfaces (roofs, sidewalks, pavement, street, parking lots) into lakes and streams via stormwater drains and ditches.
They are a beautiful way to showcase native plants, attracting birds and butterflies. Some shrubs are used, but the focus is on native perennial flower species.
A rain garden is not a pond. It is a depression that allows rapid infiltration into the soil, so there is not standing water for long periods of time. Therefore, they are not a breeding ground for mosquitoes. Mosquitoes need 7 to 12 days of standing water; it is only standing in the rain garden depression for a few hours at the most.
Why Plant a Rain Garden
- Rain gardens help prevent flooding and drainage problems.
- Rain gardens protect streams and lakes from contaminants, manage the amount of water reaching surface waters, and help reduce erosion of streambanks and lake shores.
- Rain gardens reduce the load on municipal storm water treatment structures.
- Rain gardens provide habitat for birds, butterflies and beneficial insects while enhancing the beauty of the area.
Considerations Before Creating a Rain Garden
- Locate the rain garden at least ten feet from the house foundation so infiltrating water will not seep into the foundation.
- Do not place over the septic system or a sewer pipe.
- Watch how rain drains off the roof; it may be most effective to select the side of the house with the greatest run-off.
- Integrate the rain garden into the full landscape of a home as would when considering any new flower bed.
- If establishing a rain garden to service a parking lot or boulevard, work with the community authorities before siting the garden.
Design the Rain Garden
Many resources exist to help design and plan a rain garden. Much research has been done by University or Wisconsin Extension Service and the Wisconsin Department of Natural Resources, which have resulted in some excellent resources for rain gardens.
Check with the Cooperative Extension Service, the USDA Natural Resource Conservation Service, county Soil and Water Conservation Service or greenhouses and nurseries for local resources.
Use these resources to accomplish the steps:
- Locate a proper site.
- Calculate square footage draining to the rain garden from the roof, yard, driveway, parking lot, etc.
- Use this information to size the rain garden. Generally, rain gardens are about 10% of the drainage area. For example, if the drainage area is 500 square feet, the rain garden will be roughly 50 square feet, although it can be any shape.
- Use information about soil types and the size needed to determine the depth and slope of the depression by evaluating soil compaction, soil type and texture, rate of water infiltration into the soil. In heavier soils, a deeper depression is often dug and sand is added to facilitate drainage.
- Add the appropriate spaces for surrounding upland (drier) plants to complement the plants in the depression.
- Use rope to mark the shape and size of the rain garden. The resources indicated above include some basic design plans as a good starting point.
- Many people have the resources to construct their own rain garden; others need to hire a landscape contractor.
- The first step is to dig the basin or depression, then create the berms, and prepare the entire bed for planting. Finally, the garden is planted with bedding plants (seeds may be used in the upland areas), mulch is added.
- A rain garden needs to be watered until it is established. If appropriate native plants have been used, additional water should only be needed in times of drought. Fertilizers should be used to get the garden established, and every spring thereafter.
Rain Garden Plant Selection
- Begin plant selection with native species of perennial flowers, grasses, shrubs or small trees that fit the USDA Hardiness Plant Zone for the location.
- Consider plants with deep root systems to encourage infiltration of run-off water.
- Native species are adapted to local conditions, so are more resistant to diseases and better adapted to the local climate.
- Consider the soil types in the rain garden and the amount of sunlight received each day.
- A diversity of plant types, colors, shapes and textures will provide interest and attract a variety of wildlife.
- The resources listed above contain sample plant lists; check catalogs and local plant nurseries for ideas.
Maintain the Rain Garden
Once the rain garden is established, very little additional water or weeding is needed. Rain gardens can be cared for similar to other perennial flower beds featuring native species.
Create a Rain Garden!
Consider building a rain garden to help conserve water, re-use rain run-off and create a natural environment for birds and beneficial insects in the yard.
A rain garden is a great community project to manage stormwater run-off from imperious surfaces in the area and limit pollutants entering streams and lakes in the area.<|endoftext|>
| 3.6875 |
943 |
Editor's Note:This post was co-authored by Joanna Maulbeck, a post-doctoral research associate and professor of education at Rutgers University.
An important component to supporting academic success is the relationships between home and school (Dodd & Konzal, 2002). Strained home-school relations may result in or contribute to low student performance. One way to foster these relations is by creating meaningful assignments and projects that connect life at home to life at school and vice versa, encouraging fluid, connected, and pertinent student experiences. When planning such assignments and projects, consider the questions and ideas below. Whether you use them or simply consider them as building blocks, they could inspire your own amazing ideas.
First, ask yourself three questions:
- How will the students benefit from this project?
- How will the parents benefit from this project?
- How will the teacher benefit from this project?
The best practices that link home and school will benefit all three stakeholders. Think about the needs of each stakeholder (students, parents, teachers) and then consider how you can meet those needs. Below are a few examples of projects and assignments that have potential to benefit all parties involved.
Support Bonding Between Children and Caregivers at Home
It's not uncommon for conversations between children and caregivers to be confined to "How was school today?" Providing an opportunity for children to have an in-depth conversation with adult caregivers could be powerful. One way to do this is through the use of interviews. Have children ask caregivers to share about their lives, perhaps about their best childhood friends or the first time they saw their children. Encourage students to ask caregivers about things they love or are scared of, earliest memories, or hopes for the future. In doing so, children are provided with an opportunity to connect with adult caregivers on a deeper level, as well practice interviewing, taking notes, and writing up results. When back in school, children can share findings, either in writing, through presentations, or both, further tapping into their literacy, as well as public speaking skills.
Homework is often seen as a solitary practice where students finish a few pages, read for a few minutes, and then get on with their life outside of school. However, homework could become more interactive between students and adults if the teacher scaffolds the experience for parents. Let them know exactly how you hope they'll be involved in their child's homework. This might be through an interactive notebook, calendar, or maybe even a web page. Also, provide tips for parents about how they can be helpful during homework time.
Encourage Family Outings
Create a list of historic sites, churches, and parks in your area. Almost every town has interesting places to visit that are free or very low cost. Share this list with children and caregivers as a way to encourage a family outing that is also a learning experience. You can ask families to share photos or just a few lines about what they liked or disliked about the site. The photographs and testimony could be saved in a book in the classroom or featured on a bulletin board so that students can share and learn from one another.
Invite Parents to the Classroom
Provide welcoming opportunities for parents to be involved in the classroom, where they can see what happens there and interact with children and teachers. There are various ways to involve parents. You can create a time in your classroom for "15-minute experts" where parents teach the children an interesting or useful skill -- cutting vegetables, changing a diaper, grooming a pet, folding towels, or any number of things. This activity has potential to reinforce students' listening skills. Teachers can also extend the activity by asking students to practice writing "how-tos." Parents can also be invited as guest readers, sharing their favorite stories with the class. Sharing stories as a community has great potential to foster your students' appreciation for literacy.
Teach Real-Life Lessons
It's not uncommon for students to ask, "Why are we learning this? Why do I need to know this?" One way to make learning relevant is to build a connection between home and school by teaching "real-life" lessons that expose students to content which can be used to help them live healthy lives. An example of such a lesson entails finding out how many hours of sleep the children in your classroom need based on their age.
Since effective education entails a partnership between home and school, we must creatively foster that partnership. This can be done through the projects and assignments described above, as well as many other ways. We hope that you will continue the conversation and share your ideas in the comments section below.<|endoftext|>
| 3.734375 |
914 |
You use operators to link variables, statements and functions together. You'll recognize some of these operators from arithmetic and math. We cover four types of operators here: arithmetic, comparison, Boolean and compound. All of these operators can be used in C++, too.
Arithmetic in code works exactly like arithmetic on paper, but you have more value-types available to you. If you want to divide a sensor reading by 4 and use that value in analogRead, for example, you can use:
and the arithmetic will happen using order of operations.
= assignment operator: value = value
Set two things equal to each other. This can be a variable equal to a number or a variable equal to a variable
+ addition: value + value
Add two things together. Variable + variable, number + number, variable + number, sensor reading + variable…
- subtraction: value - value
Subtract one value from another. Values can be variables, numbers, or sensor readings.
* multiplication: value * value
Multiply two values together. Values can be variables, numbers, sensor readings, results of other arithmetic.
/ division: value / value
Divide one value by another. You can use numbers, variables, other arithmetic, sensor readings, or pin numbers as your values.
% modulo: value % value
Modulo is a concept you don't use often in normal arithmetic. It divides two values, then returns to you the remainder of that division. You can use it with whole numbers or whole-number variables (ints, longs, bytes). Modulo is useful when you want to return, for example the 1's place of a temperature value. Assume you know it is in the 70s, but you don't know exactly what temperature it is. TempVariable % 10 divides your tempVariable by 10 and then returns the remainder of that division. If tempVariable was equal to 76, then tempVariable % 10 would return 6.
== equal to: value == value;
The 'equal to' comparator is two equal signs back-to-back. It is not setting two things equal, but is returning 'true' if two things are equal. You will often want to compare two things and do something if they are equal, like checking if the current time is equal to the alarm you have set.
< (less than): value < value; > (greater than): value > value;
The less than and greater than comparators return 'true' when their comparison is true. If the values are equal, these comparators will return 'false'.
<= (less than or equal to): value <= value; >= (greater than or equal to): value >= value;
The less-than-or-equal-to and greater-than-or-equal-to comparators return 'true' when the comparison is true. They also return true when the values are equal. To cover an entire range of numbers, you can combine a'greater than' and 'less-than-or-equal-to' comparators so that no matter what value you are comparing to, one statement will always return 'true'.
Boolean operators can be used to combine two comparisons to get a single result.
&& and: if (a == 0 && b == 1)
The && boolean operator combines two comparisons and returns a single result: 'true' or 'false'. 'True' is returned if both of the conditions you combined with && are true. 'False' is returned if either one or both of the conditions is false.
|| (or): if (a == 0 || b == 1)
The || (usually formed with the Shift + \ key) combines two comparisons and returns a single result: 'true' or 'false'. 'True' is returned if either of the conditions you combined with || are true or if both are true. If neither of the conditions is true, 'false' is returned.
! not: if (a != 0)
The 'not' operator can be added on to boolean variables and comparators to reverse whatever value they returned. For example, !(75 > 50) will return false.
x++ increment: x = x + 1 x-- decrement: x = x - 1
Compound operators simplify the job of adding one to a variable or subtracting one from a variable. They are simply a shortcut.<|endoftext|>
| 4.46875 |
2,176 |
Sherpa
Fractions, decimals and percentages are frequently used in everyday life. Knowing how to convert between them improves general number work and problem-solving skills. To convert a decimal to a fraction, use place value. The first number after the decimal place is worth tenths, the next is worth hundredths, the next thousandths and so on. Once a number is written as a decimal, it can be converted to a percentage. Remember that 'per cent' means 'per hundred', so converting from a decimal to a percentage can be done by multiplying by 100 (move the digits two places to the left).
Okay, we're looking at converting between fractions, decimals and percentages. So we're going to start off with the nicest of the six combinations. That's going to be decimals two percentages. So let's talk about pizza. So if I've got one whole pizza, what, I've got 1.0 pizzas, I've got one pizza.
This is the exact same to having as having 100% of my pizza. I bought 100% of a pizza. And so because these things are the exact same, they're just different ways of talking about the same thing, I know that 1.0 is the exact same as 100%. Well, how do I get from 1.0 to 100? What I do is I times by 100.
And in general, that's what we do. Imagine we've got no .12. I can times this by 100 and that should give me my percentage, right? I just move this whole thing up two spaces or the decimal point will move down two spaces and it'll end up there. So I get 12%.
So that's our decimal percentages, decimals, two fractions. Let's stick with this example. Let's stick with zero.
And if I want to convert this into a fraction, well, I could be super lazy and just write this as 0.12 over one, right? That's technically a fraction, but we don't tend to want to have decimal points in our fractions. So what we can do is we can make sure that we can get rid of this decimal point. How do I do that? In this case, I can times it by 100 or I can times the top and the bottom by 100.
I know that becomes twelve and the one becomes 100. And this is a much better fraction. Now we could simplify further. So I've got a common factor of two in the top and the bottom. I could divide them both by two and I get six over 50.
I can do the same thing again, divide by two, I get three over 25. And that would be my answer, converting between fractions and percentages. Next. So this can be nice or horrible, depending on what the question is. We'll do a nice one here and we'll do a horrible one here.
So these two methods are very similar. Let's start with maybe three eight and we want to convert this into a percentage. Well, even if we don't know what one 8th is, we're going to kind of work up from the bottom and kind of add our way to going three eight. We know what a half is, right? We know half is the exact same as 50%.
We know a quarter is the exact same as 25%. We're just halving both sides. I can have both sides again and again, 8.5%. And so I can think of three eight as just three lots of one eight, right? It's just three lots of one eight.
So I can think of this as three times an 8th. Well, I know what one eight is. I know one 8th is 12.5%. I just worked that out here. So I can think of this as three times twelve and that gives me 37.5%.
And that would be my answer, that we're kind of using our fundamental understanding of what a half is to kind of get to the stage that we can then kind of count up in eighths, converting fractions to decimals. So, yeah, very similar to fractions of percentages because we know decimals and percentages are very similar. They're just 100 times kind of separated. So fractions of decimals. Let's look at a more horrible question.
Let's go for, say, 1960. Well, this is just 19 divided by 60, right? A fraction is just the same as writing 19 divided by 60. And what do we do when we've got this? Well, normally when we do division, we take our number, we put it under the bus stop and we do the bus stop method.
And I'm going to give myself a few extra zeros because obviously 19 is smaller than 60. So we're going to continue on down into our decimal places. So we did 19 divided by 60. Well, first of all, we do the 60 going to one. No, it doesn't go.
We carry the one. The 60 going to 19. Nope, it doesn't go. We carry the 19 and then we do 60 into 190. Well, I know 60 times three gives me 180, right?
So another 360 can fit. In total, I've got a remainder of ten and now I do 60 into 100. Well, I know that 60 fits into 100 just once with a remains of 40. I now do 60 to 400. Well, because I know my six times tables.
I know six times 60 is 360. So, like, if it's 660 into 400 with the remainder of 40, and I just done this right, I know 60 fits into 406 whole times with the remainder of 40. And what we find is we're basically stuck in this neverending loop. We can keep basically doing this and adding on sixes. Well, we've basically solved it, right?
We know that our answer is therefore going to be 0.316 recurring for this one because we just get never running six s. Percentage to decimals. This is quite nice. It's just the opposite of this. From decimals to percentage, we times by 100.
From percentage to decimals, we just divide by 100.
For example, 12%, we just divide this by 100. Obviously, that decimal point next to the two is going to shift up two spaces. We're going to end up with 0.12. And obviously that's the same as what we started with here. So we're just kind of going back the other way.
Finally, percentages to fractions, this is very similar to percentages to decimals.
As an example, let's say we want 56% and we want this as a fraction. Well, per cent just means per 100, right? So I can think of this as 56 per 100 or 5600. And that's what this is, right? Obviously 100%, which is 100 over 100, which is just one or one whole one.
So we've got 56 over 100. We've got our fraction here, right? But we can probably simplify this down. We've got a common factor, I can see, of two in the top and the bottom. So I could write this as 28 over 50.
I could write this again, common factor of two. This is going to be 14 over 25. And I think that's as far as I can go. There's not any more common factors. All right, let's do a couple of exam style questions.
So we'll look at 54% as a decimal. Well, we know how to do this, right? We know that 100% is just equal to one. And so I know that to convert between the two, I want to divide by 100. But in this case, 54.4 divided by 100 is going to give me zero point 54.
And it's as simple as that. If you can read by force, 00:56 as a fraction, well, again, we can be really lazy and just write this as zero point 256 over one, that's ten to be a fraction. But we don't want to have any decimal places or decimal points in our numbers. So how can I get rid of this decimal point? Well, at the moment, if I look at this, I've got a digit in my thousandths column.
And so to get all three digits out of being on the right of the decimal point, I've got a times by 1000, times by 1000 on the top and the bottom. This is going to give me two, five, six over 1000. We can simplify further. Right, we've got a common factor of T. We could write this as one, two, eight over 500.
Again, I got a common factor of T. This is 64 over T, 50. And again, I've got a common factor of T. This is 32, one, two, five. And I don't think I can go any further than that.
So that's going to be my final answer, finally. 17 over 40 as a decimal. Well, just like we did before, if we think of this as basically 17 divided by 14, what all we're doing here is bus stop method, right? We've got the bus stop moving 17.00. We know we're going to probably need some decimal numbers.
Divided by 40 or 40 goes into 10 times, it goes into 170 times. So it carries 17, it goes into 174 times, right? Four times 40 gives me 160. I want 170. So I've got ten leftover.
40 goes into 102 times, right? Because two times 40 is 80. I've got another 20 left. Over 40 into 200 is going to go five times within remainders. So our answer, again, don't forget to point out our answer is going to be 00:45.
And that's the end of converting fractions, decimals and percentages.
# Book a lesson with this tutor
Oli W
A young, gifted tutor who uses his engineering experience to ignite your curiosity in Maths. A straight 'A' student, he knows exactly how you need to learn to help you ace your exams.<|endoftext|>
| 4.6875 |
1,475 |
Home » Cube Roots » Cube Root of 1000
# Cube Root of 1000
Table of Contents
The cube root of 1000 is the number, which multiplied by itself three times, is 1000. In other words, this number to the power of 3 equals 1000.
Besides the real value of
along with an explanation, on this page you can also find what the elements of the cube root of 1000 are called.
In addition to the terminology, we have a calculator you don’t want to miss:
## Calculator
Reset
³√1000 = 10
If you have been looking for the cube root of one thousand, then you are right here, too.
## Third Root of 1000
In this section we provide you with important additional information about the topic of this post:
The term can be written as ³√1000 or 1000^1/3.
As the index 3 is odd, 1000 has only one cube real root: ³√1000 sometimes called principal cube root of 1000.
If you want to know how to find the value of this root, then read our article Cube Root located in the header menu.
There, we also discuss the properties for index n = 3 by means of examples: multiplication, division, exponentiation etc.
Next, we have a look at the inverse function.
### Inverse of Cube Root of 1000
Extracting the cube root is the inverse operation of ^3:
In the following paragraph, we are going to name the elements of this ∛.
## What is the Cube Root of 1000?
You already have the answer to that question, and you also know about the inverse operation of 1000 cube root.
Keep reading to learn what the parts are called.
• ³√1000 is the cube root of 1000 symbol
• 3 is the index
• 1000 = radicand; the radicand is the number below the radical sign
• Cube root = 10
• √ is called radical symbol or radical only
Third root of 1000 = 10
As a sidenote: All values on this page have been rounded to ten decimal places.
Now you really know all about ³√1000, including its values, parts and the inverse.
If you need to extract the 3rd root of any other real number use our calculator above.
Simply insert the number of which you want to find the cube root (e.g. 1000); the calculation is done automatically.
If you like our information about ³√1000, then a similar cube root you may be interested in is, for example: cube root of 1002.
In the following table you can find the n-th root of 1000 for number n = 2,3,4,5,6,7,8,9,10.
## Table
The aim of this table is to provide you with an overview of the nth roots of 1000.
IndexRadicandRootSymbolValue
21000Square Root of 1000²√1000±31.6227766017
31000Cube Root of 1000³√100010
41000Forth Root of 1000⁴√1000±5.6234132519
51000Fifth Root of 1000⁵√10003.9810717055
61000Sixth Root of 1000⁶√1000±3.1622776602
71000Seventh Root of 1000⁷√10002.6826957953
81000Eight Root of 1000⁸√1000±2.3713737057
91000Nineth Root of 1000⁹√10002.15443469
101000Tenth Root of 1000¹⁰√1000±1.995262315
A few lines down from here we review the FAQs.
## Cube Root of One Thousand
If you have been searching for what’s the cube root of one thousand or 3rd root of 1000, then you are reading the right post as well.
The same is true if you typed 3 root of 1000 or 1000 3 root in the search engine of your preference, just to name a few similar terms.
Right below you can find the frequently asked questions in the context.
## FAQs About the Cube Root of 1000
Click on the question which is of interest to you to see the collapsible content answer.
### How Many Real Cube Roots Does 1000 Have?
1000 has one real cube root, because the index 3 is odd.
### What to the Third Power Equals 1000?
The cube root of 1000 to the power of 3 equals 1000.
### How Do You Find the Cube Root of 1000?
Start with an initial guess such that 3 times that value equals approximately 1000, then keep improving the guess until you have the required precision.
### What is 1000 to the Cube Root?
1000 to the cube root = 1000^1/3 = 10.
### What Number is the Cube Root of 1000?
The cube root of 1000 = 10.
### How Do You Solve the Cube Root of 1000?
To compute the 3rd root of 1000 use a calculator, and/or employ the Newton–Raphson method.
### What is the Value of 3 Root 1000?
The value of 3 root 1000 is 10.
If something remains unclear do not hesitate getting in touch with us.
We are constantly trying to improve our site, and truly appreciate your feedback.
Ahead is the wrap-up of our information.
## Summary
To sum up, the cube root of 1000 is 10.
Finding the third root of the number 1000 is the inverse operation of rising the ³√1000 to the power of 3. That is to say, (10)3 = 1000.
Further information related to ³√1000 can be found on our page n-th Root.
Note that you can also locate roots like ³√1000 by means of the search form in the menu and in the sidebar of this site.
If our article about the cube √ 1000 has been useful to you , then press some of the share buttons located at the bottom of this page.
BTW: A term closely related to cube roots is perfect cube. We tell you everything about perfect cubes in our article Cube Numbers.
If you have any questions about the 3rd root of 1000, fill in the comment form below.
Websites which are related to this one can be found in the “recommended sites” section in the sidebar.
Last, but not least, don’t forget to install our absolutely free PWA app (see menu or sidebar), or to bookmark us.
And come back soon!
Thanks for your visit.
Submitting...
Thank you, your sign-up request was successful! Please check your email inbox to confirm.
{{message}}<|endoftext|>
| 4.5 |
6,471 |
# 2.1: Real Numbers, Linear Inequalities, and Interval Notation
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
## Real Numbers
Algebra is often described as the generalization of arithmetic. The systematic use of variables, letters used to represent numbers, allows us to communicate and solve a wide variety of real-world problems. For this reason, we begin by reviewing real numbers and their operations.
A set is a collection of objects, typically grouped within braces, $$\{ \}$$, where each object is called an element. When studying mathematics, we focus on special sets of numbers.
\begin{align*} \mathbb { N } &= \underbrace{\{ 1,2,3,4,5 , \dots \}}_{\color{Cerulean}{Natural\: Numbers}} \\[4pt] W &= \underbrace{ \{ 0 , 1,2,3,4,5 , \dots \}}_{\color{Cerulean}{Whole\: Numbers}} \\[4pt] \mathbb{Z} &= \underbrace{ \{\dots ,-3,-2,-1,0,1,2,3,\dots\}}_{\color{Cerulean}{Integers}} \end{align*}
The three periods (…) are called an ellipsis and indicate that the numbers continue without bound. A subset, denoted $$\subseteq$$, is a set consisting of elements that belong to a given set. Notice that the sets of natural and whole numbers are both subsets of the set of integers and we can write:
$$\mathbb { N } \subseteq \mathbb{Z}$$ and $$W \subseteq \mathbb{Z}$$
A set with no elements is called the empty set and has its own special notation:
$$\{\:\:\:\}=\varnothing\: \color{Cerulean}{Empty\: Set}$$
Rational numbers, denoted $$\mathbb{Q}$$, are defined as any number of the form $$\frac { a } { b }$$ where a and b are integers and b is nonzero. We can describe this set using set notation:
$$\mathbb { Q } = \left\{ \frac { a } { b } | a , b \in \mathbb { Z } , b \neq 0 \right\} \color{Cerulean}{Rational\: Numbers}$$
The vertical line | inside the braces reads, “such that” and the symbol $$\in$$ indicates set membership and reads, “is an element of.” The notation above in its entirety reads, “the set of all numbers $$\frac{a}{b}$$ such that a and b are elements of the set of integers and b is not equal to zero.” Decimals that terminate or repeat are rational. For example,
$$0.05=\frac{5}{100}$$ and $$0.\overline{6}=0.6666…=\frac{2}{3}$$
The set of integers is a subset of the set of rational numbers, $$\mathbb{Z}\subseteq\mathbb{Q}$$, because every integer can be expressed as a ratio of the integer and 1. In other words, any integer can be written over 1 and can be considered a rational number. For example,
$$7=\frac{7}{1}$$
Irrational numbers are defined as any numbers that cannot be written as a ratio of two integers. Non-terminating decimals that do not repeat are irrational. For example,
$$π=3.14159…$$ and $$\sqrt{2}=1.41421…$$
Finally, the set of real numbers, denoted $$\mathbb{R}$$, is defined as the set of all rational numbers combined with the set of all irrational numbers. Therefore, all the numbers defined so far are subsets of the set of real numbers. In summary,
The set of even integers is the set of all integers that are evenly divisible by $$2$$. We can obtain the set of even integers by multiplying each integer by $$2$$.
$$\{\dots, −6,−4,−2, 0, 2, 4, 6,\dots\} \color{Cerulean}{Even\: Integers}$$
The set of odd integers is the set of all nonzero integers that are not evenly divisible by $$2$$.
$$\{\dots,−5,−3,−1, 1, 3, 5,\dots\} \color{Cerulean}{Odd\: Integers}$$
A prime number is an integer greater than $$1$$ that is divisible only by $$1$$ and itself. The smallest prime number is $$2$$ and the rest are necessarily odd.
$$\{2, 3, 5, 7, 11, 13, 17, 19, 23,\dots\} \color{Cerulean}{Prime\: Numbers}$$
Any integer greater than $$1$$ that is not prime is called a composite number and can be uniquely written as a product of primes. When a composite number, such as $$42$$, is written as a product, $$42=2⋅21$$, we say that $$2⋅21$$ is a factorization of $$42$$ and that $$2$$ and $$21$$ are factors. Note that factors divide the number evenly. We can continue to write composite factors as products until only a product of primes remains.
Therefore, the prime factorization of $$42$$ is $$2⋅3⋅7$$.
###### Example $$\PageIndex{1}$$
Determine the prime factorization of $$210$$.
Solution
Begin by writing $$210$$ as a product with $$10$$ as a factor. Then continue factoring until only a product of primes remains.
$$210=10⋅21$$
$$=2⋅5⋅3⋅7$$
$$=2⋅3⋅5⋅7$$
Since the prime factorization is unique, it does not matter how we choose to initially factor the number; the end result will be the same.
Answer: $$2⋅3⋅5⋅7$$
### The Number Line and Notation
A real number line, or simply number line, allows us to visually display real numbers and solution sets to inequalities.
Positive real numbers lie to the right of the origin and negative real numbers lie to the left. The number zero $$0$$ is neither positive nor negative. Typically, each tick mark represents one unit.
As illustrated below, the scale need not always be one unit. In the first number line, each tick mark represents two units. In the second, each tick mark represents $$\dfrac{1}{7}$$:
### Graph Inequalities on the Number Line
What number would make the inequality $$x>3$$ true? Are you thinking, "$$x$$ could be four"? That’s correct, but $$x$$ could be 6, too, or 37, or even 3.001. Any number greater than three is a solution to the inequality $$x>3$$. We show all the solutions to the inequality $$x>3$$ on the number line by shading in all the numbers to the right of three, to show that all numbers greater than three are solutions. Because the number three itself is not a solution, we put an open parenthesis at three.
We can represent inequalities using interval notation or set notation. In this text, we will use interval notation. There is no upper end to the solution to this inequality. In interval notation, we express $$x>3$$ as $$(3,\infty)$$. The symbol $$\infty$$ is read as “infinity.” It is not an actual number.
Before we begin, let us review the conventions of interval notation:
• The smallest term from the interval is written first.
• The largest term in the interval is written second, following a comma.
• Parentheses, $$($$ or $$)$$, are used to signify that an endpoint is not included, called exclusive.
• Brackets, $$[$$ or $$]$$, are used to indicate that an endpoint is included, called inclusive.
Some use parentheses and brackets when graphing solution sets on a number line. Others use open circles to denote when an endpoint is not included and a closed circle when an endpoint is included.
See Figure $$\PageIndex{5}$$ for a summary of interval notation.
###### Example $$\PageIndex{2}$$
Graph each inequality on the number line and write in interval notation.
1. $$x\geq −3$$
2. $$x\leq −\frac{3}{5}$$
Solution
1.
$$x \geq -3$$ Shade to the right of $$−3$$, and put a bracket at $$−3$$. Write in interval notation. $$[-3, \infty)$$
1.
$$x \leq -\dfrac{3}{5}$$ Shade to the left of $$−\frac{3}{5}$$, and put a bracket at $$−\frac{3}{5}$$. Write in interval notation. $$\bigg( -\infty, \dfrac{3}{5}\bigg]$$
###### You Try $$\PageIndex{1}$$
Graph each inequality on the number line and write in interval notation:
1. $$x\leq −4$$
2. $$0.5 \leq x$$
3. $$−\frac{2}{3} > x$$
1.
1.
1.
What numbers are greater than two but less than five? Are you thinking say, $$2.5,\space 3,\space 3\frac{2}{3},\space 4,\space 4,\space 99$$? We can represent all the numbers between two and five with the inequality $$2<x<5$$. We can show $$2<x<5$$ on the number line by shading all the numbers between two and five. Again, we use the parentheses to show the numbers two and five are not included. See Figure $$\PageIndex{6}$$.
###### You Try $$\PageIndex{2}$$
Graph each inequality on the number line and write in interval notation:
1. $$−2<x<1$$
2. $$−5\leq x<−4$$
3. $$1\leq x\leq 4.25$$
1. Interval notation: $$(-2,1)$$
1. Interval notation: $$[-5,4)$$
1. Interval notation: $$[1,4.25]$$
### Linear Inequalities
A linear inequality is a mathematical statement that relates a linear expression as either less than or greater than another. The following are some examples of linear inequalities, all of which are solved in this section:
$$5 x + 7 < 22$$ $$- 2 ( x + 8 ) + 6 \geq 20$$ $$- 2 ( 4 x - 5 ) < 9 - 2 ( x - 2 )$$
A solution to a linear inequality is any real number that will produce a true statement when substituted in for the variable. We graph the solution set on a number line and/or express the solution using interval notation. In the case that there is no solution to the inequality we indicate the solution using the empty set, $$\{\:\:\:\} \text{ or }\varnothing\:$$.
###### Example $$\PageIndex{3}$$
Are $$x=−4$$ and $$x=6$$ solutions to $$5x+7<22$$?
Solution
Substitute the values in for $$x$$, simplify, and check to see if we obtain a true statement.
Check $$x=−4$$ Check $$x=6$$
$$\begin{array} { r } { 5 ( \color{Cerulean}{- 4}\color{Black}{ )} + 7 < 22 } \\ { - 20 + 7 < 22 } \\ { - 13 < 22 } \:\:\color{Cerulean}{✓} \end{array}$$ $$\begin{array} { c } { 5 ( \color{Cerulean}{6} \color{Black}{)} + 7 < 22 } \\ { 30 + 7 < 22 } \\ { 37 < 22 } \:\:\color{red}{✗} \end{array}$$
Table $$\PageIndex{2}$$
$$x=−4$$ is a solution and $$x=6$$ is not
To solve a linear inequality, one must isolate the variable on one side of the inequality symbol. If we add or subtract on both sides of an inequality, we will have an equivalent inequality.
###### Adding or Subtracting the Same Quantity from Both Sides of an Inequality
Let $$a$$ and $$b$$ be real numbers with $a < b \nonumber$
If $$c$$ is any real number, then $a + c<b+ c \nonumber$and$a−c<b−c \nonumber$That is, adding or subtracting the same amount from both sides of an inequality produces an equivalent inequality (does not change the solution).
There is more than one way to denote the solution to an inequality on a number line, as seen in the next example.
###### Example $$\PageIndex{4}$$
Solve for $$x: \quad x-2 ≤ 7$$. Sketch the solution on the real line, then use interval notation to describe your solution.
Solution
To “undo” subtracting $$2$$, we add $$2$$ to both sides of the inequality.
\begin{aligned} x-2 &\leq 7 & & \color {Red} \text { Original inequality. } \\ x-2+2 &\leq 7+2 & & \color {Red} \text { Add } 2 \text { to both sides. } \\ x &\leq 9 & & \color {Red} \text { Simplify both sides. } \end{aligned} \nonumber
To shade the real numbers less than or equal to $$9$$, we shade the number $$9$$ and all real numbers to the left of $$9$$.
Using interval notation, the solution is $$(-\infty, 9]$$.
When drawing the endpoints for a solution on a number line, one may use brackets or filled in circles which indicate that the number is a solution. A parenthesis or open circle indicates that the number is not a solution.
###### You Try $$\PageIndex{3}$$
Use interval notation to describe the solution of: $$x−7 < −8$$.
$$(-\infty,-1)$$
If we multiply or divide both sides of an inequality by a positive number, we have an equivalent inequality.
###### Multiplying or Dividing by a Positive Number on Both Sides of an Inequality
Let $$a$$ and $$b$$ be real numbers with $$a<b$$. If $$c$$ is a real positive number, then $a c<b c \nonumber$and$\dfrac{a}{c}<\dfrac{b}{c} \nonumber$
###### Example $$\PageIndex{5}$$
Solve for $$x : \quad 3 x \leq-9$$ Sketch the solution on the real line and state the solution in interval notation.
Solution
To “undo” multiplying by $$3$$, divide both sides of the inequality by $$3$$. Because we are dividing both sides by a positive number, we do not reverse the inequality sign.
\begin{aligned} 3x & \leq -9 & & \color {Red} \text { Original inequality. } \\ \dfrac{3x}{3} & \leq \dfrac{-9}{3} & & \color {Red} \text { Divide both sides by } 3. \\ x & \leq -3 & & \color {Red} \text { Simplify both sides. } \end{aligned} \nonumber
Shade the real numbers less than or equal to $$-3$$.
The solution in interval notaiton is $$(-\infty,-3]$$.
###### You Try $$\PageIndex{4}$$
Use interval notation to describe the solution of: $2 x>-8 \nonumber$
$$(-4, \infty)$$
When you multiply both sides of an inequality by a negative number, you must reverse the inequality sign to keep the statement true.
###### Multiplying or Dividing by a Negative Number
Let $$a$$ and $$b$$ be real numbers with $$a<b$$. If $$c$$ is a real negative number, then $a c>b c \nonumber$ and $\dfrac{a}{c}>\dfrac{b}{c} \nonumber$That is, when multiplying or dividing both sides of an inequality by a negative number, you must reverse the inequality sign.
###### Example $$\PageIndex{6}$$
Solve for $$x: \quad −2x<4$$. Sketch the solution on the real line, then use interval notation to describe your solution.
Solution
To “undo” multiplying by $$−2$$, divide both sides by $$−2$$. Because we are dividing both sides by a negative number, we reverse the inequality sign.
\begin{aligned} -2 x&< 4 & & \color {Red} \text { Original inequality. } \\ \dfrac{-2x}{-2}&> \dfrac{4}{-2} & & \color {Red} \text { Divide both sides by }-2 \\ x&> -2 & & \color {Red} \text { Reverse the inequality sign. } \\ x&> -2 & & \color {Red} \text { Simplify both sides. } \end{aligned} \nonumber
Shade the real numbers greater than $$−2$$.
Using interval notation, the solution is $$(-2, \infty)$$.
###### You Try $$\PageIndex{5}$$
Use interval notation to describe the solution of: $−3x≥− 6 \nonumber$
$$(-\infty, 2]$$
Sometimes you need to perform a sequence of steps to arrive at the solution.
###### Example $$\PageIndex{7}$$
Solve for $$x: \quad 2 x +5> −7$$. Sketch the solution on the real line, then use interval notation to describe your solution.
Solution
To “undo” adding $$5$$, subtract $$5$$ from both sides of the inequality.
\begin{aligned} 2x+5&> -7 & & \color {Red} \text { Original inequality. } \\ 2x+5-5&> -7-5 & & \color {Red} \text { Subtract } 5 \text { from both sides. } \\ 2x&> -12 & & \color {Red} \text { Simplify both sides. } \end{aligned} \nonumber
To “undo” multiplying by $$2$$, divide both sides by $$2$$. Because we are dividing both sides by a positive number, we do not reverse the inequality sign.
\begin{aligned} \dfrac{2x}{2}& >\dfrac{-12}{2} & & \color {Red} \text { Divide both sides by } 2 \\ x&> -6 & & \color {Red} \text { Simplify both sides. } \end{aligned} \nonumber
Shade the real numbers greater than $$−6$$.
Using interval notation, the solution is $$(-6, \infty)$$.
###### You Try $$\PageIndex{6}$$
Use interval notation to describe the solution of: $3x-2 ≤4 \nonumber$
$$(-\infty, 2]$$
###### Example $$\PageIndex{8}$$
Solve for $$x: \quad 3 −5x ≤ 2x + 17$$. Sketch the solution on the real line, then use interval notation to describe your solution.
Solution
We need to isolate terms containing $$x$$ on one side of the inequality. Start by subtracting $$2x$$ from both sides of the inequality.
\begin{aligned} 3-5x &\leq 2x+17 & & \color {Red} \text { Original inequality. } \\ 3-5x-2x &\leq 2x+17-2x & & \color {Red} \text { Subtract } 2x \text { from both sides. } \\ 3-7x &\leq 17 & & \color {Red} \text { Simplify both sides. } \end{aligned} \nonumber
We continue to isolate terms containing $$x$$ on one side of the inequality. Subtract $$3$$ from both sides.
\begin{aligned} 3-7x-3 &\leq 17-3 & & \color {Red} \text { Subtract } 3 \text { from both sides. } \\ -7x &\leq 14 & & \color {Red} \text { Simplify both sides. } \end{aligned} \nonumber
To “undo” multiplying by $$−7$$, divide both sides by $$−7$$. Because we are dividing both sides by a negative number, we reverse the inequality sign.
\begin{aligned} \dfrac{-7x}{-7} &\geq \dfrac{14}{-7} & & \color {Red} \text { Divide both sides by }-7 \\ x &\geq-2 & & \color {Red} \text { Simplify both sides. } \end{aligned} \nonumber
Using interval notation, the solution is $$[-2, \infty)$$.
###### You Try $$\PageIndex{7}$$
Use interval notation to describe the solution of: $4-x>2x +1 \nonumber$
$$(-\infty, 1)$$<|endoftext|>
| 4.59375 |
319 |
Part of Scholastic’s Journey Into Space project, the exciting “Moon Olympics” activity (grades 3–6) helps students understand the concept of gravity.
With fun animated graphics and text, this interactive activity first illustrates how athletes lift weights, dive, golf, skateboard a half-pipe, and parachute on Earth. Then, by answering multiple-choice questions, students can predict how each sport would change on the moon.
What would skateboarding be like on the moon?
- You’d be able to jump higher, and you’d glide back and forth much faster in the half-pipe.
- You’d be able to jump higher, but you’d glide back and forth more slowly in the half-pipe.
- When you jumped off the edge of the half-pipe, you’d said off into space!
More animation and text illustrate the correct answers, shedding light on how gravity and lack of atmosphere and air pressure would affect athletes on the moon.
- Knows that the Earth's gravity pulls any object toward it without touching it
- Knows that the Earth is one of several planets that orbit the Sun and that the Moon orbits the Earth
- Understands general concepts related to gravitational force (e.g., every object exerts gravitational force on every other object; this force depends on the mass of the objects and their distance from one another; gravitational force is hard to detect unless at least one of the objects, such as the Earth, has a lot of mass)<|endoftext|>
| 4.125 |
1,073 |
Do you often wonder if you’re using dashes properly? Dashes have distinct uses that often seem blurred in today’s society. Here’s what’s best for writers and those involved in business writing.
Q. When do I use a dash in text?
A. A dash usually replaces a comma, semicolon, colon, or parentheses. When used this way, it creates an EMPHATIC separation of words. Since a dash is versatile, people tend to use it to punctuate almost any break in a sentence. Don’t. It’s best used for EFFECT. And experts say we should never type a single hyphen to represent a “double” dash (em dash).
Author Amy Einsohn says, “The dash is best reserved for special effects: to prepare readers for a punchline or a U-turn.” The Chicago Manual of Style says, “A dash or a pair of dashes is used to denote a sudden break in thought that causes an abrupt change in sentence structure.” Professor Charles Darling suggests thinking of dashes as “super commas.”
There are four kinds of dashes: em, en, two-em, and three-em.
Generally, if you see the word “dash,” the writer means em dash. Most word processing programs give us access to em dashes (the width of a capital “m”). If you don’t have software that has this special character, type two hyphens with no spaces between the words on either side and the dashes, or do one of the following: Keyboard stroke: If you use Windows, hold down the Alt key and type 0151 on the keypad.
To make this autocorrect in Word for Windows:
Go to Menu bar / Tools / click AutoCorrect. In the “Replace” box, type two hyphens. In the “With” box, press Alt + 0151 to create an em-dash, and then click Add. Close.
If you use a Mac and Word, try typing two hyphens. Word should automatically convert them into an em-dash. If not, try this:
Go to Menu bar / Tools / click AutoCorrect. In the “Replace” box, type two hyphens. In the “With” box, press OPTION+SHIFT+- (hyphen) to create an em-dash, and then click Add. Close.
Regardless of whether you use an “em” dash or two hyphens, don’t leave any space before or after the dash.
Use an “en” dash to connect numbers in a range. It means “up to and including” when used like this: “During the years 1998-1999,” and “…people aged 55-63.”
If you don’t have access to an “en” dash, use a hyphen. If you use Windows, hold down the Alt key and type 0150 on the keypad.
Used to indicate that letters are missing from a word. If you don’t have access to this dash, type four consecutive hyphens with no spaces between. If letters are missing from within a word, leave no space before or after the two-em dash. If the letters are missing at the end of a word, leave no space before it, but leave one space after, unless punctuation is required. Examples:
* Mrs. Birming—- chose to remain anonymous.
* The diagnosis was made by Dr. Boy—-.
Note: You can type a two-em dash by using the keystrokes one after the other with no space between them.
Used to indicate that an entire word has been left out or needs to be provided. If you don’t have the character for the three-em dash, type six consecutive hyphens. You can type a three-em dash by using the keystrokes three times with no space between them.
Because this dash represents a complete word, leave a space before and after unless punctuation is needed. Three em-dashes are generally also used in bibliographies to represent an author’s name in subsequent entries, once the author’s full name has been given.
NOTE: These rules aren’t written on stone. Many documents you’ll type do not require using the various dashes. But when you’re typing manuscripts, print newsletters, special reports, or anything that requires good typography, use them as described above. If nothing else sticks, make sure you always use two hyphens when you can’t create an em dash. You’ll find more information on dashes at:
* The Chicago Manual of Style, 14th Edition ISBN 0-226-10389-7
* The Copyeditor’s Handbook, Amy Einsohn ISBN 0-520-21835-3
* The Gregg Reference Manual, Ninth Edition ISBN 0-02-804046-5<|endoftext|>
| 3.796875 |
324 |
Introduction to the Sirenia
Dugongs and Manatees
Sirenians are slow and passive mammals of tropical and sub-tropical waters. Their large thick bodies betray their heritage as relatives of elephants. There are only five living species of sirenians, known collectively as "sea-cows," including the dugong and the manatees. Until about 1770, an additional species known as the Stellar Sea-cow (Hydrodamalis) existed along the Eastern Pacific coast, but the last of these were hunted to extinction for their meat and fat in by American explorers. The West Indian Manatees are currently in danger of also becoming extinct.
The first Sirenians appeared in the early Eocene in Europe, but by the close of this epoch, they had spread to tropical Asia and North America. Dugongs were the prevalent group in the Caribbean and Mediterranean until the Late Miocene, when all but the Indo-Pacific species went extinct. Subsequently, the manatees of South America entered these areas in the Pliocene.
Sea-cows are herbivores that graze on sea-grasses and related plants, though some will eat algae and floating monocots, such as Pistia and Eichhornia, and they will occasionally eat shellfish and dead fish. Adults may eat as much as 30 pounds of food a day, and may reach more than 500 kg.
Sirenians are solitary creatures, who come together only to mate, or when favorable local conditions attract individuals for a short time. They may live more than 70 years.<|endoftext|>
| 3.734375 |
537 |
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
##### Thank you for watching the video.
To unlock all 5,300 videos, start your free trial.
# Mathematical Induction - Concept
Carl Horowitz
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Share
An important and fundamental tool used when doing proofs is mathematical induction. We can use mathematical induction to prove properties in math, or formulas. For example, we can prove that a formula works to compute the value of a series. Mathematical induction involves using a base case and an inductive step to prove that a property works for a general term.
So at some point in your Math career you will have or will have or have seen these formulas, okay? And what they are are basically sums of different numbers. You are looking at the just the sum of bunch of integers and we know that the sum is going to be n n+1 over 2. We also have a formula for the sum of squares and the sum of cubes. And on the surface these formulas look pretty complicated, okay. But actually how we can prove that they are true is by using what we call mathematical induction, okay? And what mathematical induction is is basically we prove it works for n=1. So we show that these equations all hold for n=1. We then assume that the equation holds for n=k some arbitrary k and then sorry, using that fact we show that it's true for n is equal to k+1. Okay? And basically how that works is if it works for 1, we can assume it works for any number. k could be any number. So k could be 1 as well. So if it works for 1, then this would show that it works for 2. If it works for 2 then we already know it works for the next one. It works for 3, it works for 4 [IB] 5 so on and so forth, okay?
So mathematical induction is basically a type of approach to proving a statement. You show it works for your first term, you assume it works for some arbitrary variable, typically k is used and then using the fact that it works you're assuming it works for k, you prove that the equation still holds for and is equal to k+1.<|endoftext|>
| 4.65625 |
344 |
The Earth’s oceans have been over-fished, leading to decreased biodiversity, with current practices being unsustainable.
Fish populations have been shown to rebound quickly following well enforced fishing bans. In fact, many experts believe fishing will only be sustainable if marine reserves, where fishing is banned, is expanded significantly.
That’s why some activists and scientists are discussing the idea of a fishing ban in international waters. International waters are approximately 58% of the ocean surface and include everything that is 200 miles or more from sovereign land. Fishing in this area is largely unregulated. Fishing boats use voluminous trawl nets, longlines miles in length, and other industrial gear to catch migrating fish and shark.
The environmental damage is tremendous, such as ancient corals being destroyed by deep trawl nets. All the while, the contribution to the global food supply is negligible.
Proponents say a fishing ban in this area will protect depleted species and create more fish in coastal waters.
University of British Columbia fisheries professor Daniel Pauly, has studied international water fisheries for years, and supports the idea of a ban. His research has shown that the global catch would be greater, with a fishing ban in international waters and the catch would be more evenly distributed. Currently, only a few nations catch most of the fish in these waters, including Japan, South Korea, Taiwan and Spain.
The discussion of a fishing ban in international waters is gaining rapid momentum as the United Nations convene in New York City to negotiate a treaty on protecting biodiversity in these waters from industrial activity, including fishing. The meeting could be the first step towards creating large marine reserves, if not a full closer to fishing in international waters.<|endoftext|>
| 3.75 |
691 |
Communication is so much more than talking. So as a speech-language pathologist I think a lot about gestures. The words “speech language” in my job title may have given you the impression that I solely focus on how children express themselves. However, a hugely important piece of communication development happens before your child says their first word.
We define a gesture as an action, or movement of part of the body, especially the hands or head, used with the intention to communicate an idea or meaning. Many gestures we do with our hands, like your child pointing to her cup on the counter to request it. But gestures can also be facial expressions such as your son smacking his lips indicating he wants to eat or your daughter arching her back to show she’s refusing something.
In case you were wondering, encouraging your child to use gestures will not hinder their speech and language development. Again, there is a strong link between gestures and communication development. For example, we know that a child who shows you or points to an object will likely learn the word for that object within three months. And children who produce more gestures early on have larger expressive vocabularies later in development. Additionally, research suggests that gestures pave the way for language development. In 2005, Iverson and Goldin-Meadow found that “children who first produced a gesture + word combination were also first in producing two-word combination,” an essential 2-year-old milestone.
Children will often begin using gestures as early as 8 or 9 months of age.
Here are some general guidelines for gesture development:
9-12 months: imitates clapping, shows and gives objects to adults, reaches to be picked up, waves bye, gains attention by making physical contact like grabbing your leg, requests objects by pointing, anticipates and initiates a social game like peek-a-boo (link) by covering their face with a blanket
12-15 months: hugs stuffed animals, claps to show excitement, dances to music, give objects to adults to ask for help, demonstrates the functional use of objects such as brushing hair with a comb
15-18 months: points to get you to do something, shakes head for no, points to objects on request, indicates all done by putting hands up or shaking hands, points to objects for adults to name or label
18-24 months: blows kisses, makes funny faces like sticking out their tongue to gain attention, slaps your palm for a “high five,” shrugs shoulds or puts hands up to indicate “what’s that?”
Gestures are crucial for language learning. One of the most valuable gestures for children to develop is pointing. I wrote about it, in depth, in a previous blog post. Pointing and other gestures allow a child to communicate their wants and needs months before they can produce verbal speech. Children learn communication by observing, listening, and imitating the world around them. So show your child what to do (i.e., demonstrate gesturing) when you play, talk, sing, and enjoy time together.
Ready to Learn More About Using Play to Boost Language Skills:
Start Playing With Purpose
Learn how to purposefully and intentionally interact with your child during play and help them increase opportunities for speech and language development with our Playing with Purpose eBook!<|endoftext|>
| 3.90625 |
1,003 |
Download This Sample
This sample is exclusively for KidsKonnect members!
To download this worksheet, click the button below to signup for free (it only takes a minute) and you'll be brought right back to this page to start the download!
Sign Me Up
The National Pearl Harbor Remembrance Day, or Pearl Harbor Remembrance Day, is the annual commemoration of the attack on Pearl Harbor, Hawaii on December 7, 1941. This attack led to the Americans joining World War II.
See the fact file below for more information about Pearl Harbor Remembrance Day or alternatively you can download our comprehensive worksheet pack to utilise within the classroom or home environment.
- On the morning of December 7, 1941, the Japanese Imperial Army attacked the American Army and Naval base in Pearl Harbor, Hawaii. The surprise attack from the Japanese resulted in more than 2000 deaths and 1000 injured American servicemen and civilians.
- Nearly 200 American aircrafts and numerous battleships stationed in the Pacific were also destroyed.
- The day after the attack, U.S President Franklin Roosevelt declared war against Japan and in doing so entered World War II. In his speech to Congress, President Roosevelt stated that the bombing of Pearl Harbor was “a date which will live in infamy.”
- The Second World War, or WWII, was the deadliest war in history. It involved more than 30 countries that were either members of the Allied or Axis Powers. WWII started when Nazi Germany, led by Adolf Hitler, invaded Poland in 1939.
- The chief members of the Allied Powers included Great Britain, France, the Soviet Union, the United States and China while the Axis Powers coalition was led by Germany, Italy and Japan.
- Shortly after, Germany declared war against the United States. American soldiers were guided by the slogan “Remember Pearl Harbor” through to the end of the war in 1945.
Pearl Harbor Remembrance Day Observations:
- In 1962, the USS Arizona Memorial was built in honor of the 102 sailors and marines who died on the USS Arizona during the attack.
- In 1972, a memorial on the northwest shore of Ford Island was built to honor the crew of USS Utah.
- By 1989, it was declared one of the National Historic Landmarks by the National Register of Historic Places. Furthermore, the USS Oklahoma Memorial, USS Missouri Memorial, USS Bowfin Submarine Museum and Park and Pacific Aviation Museum were also established.
- On November 5, 1990, the Pearl Harbor Commemorative Medal was released. In commemoration of the 50th year anniversary of the attack, US Armed Forces veterans and civilians who served during the attack were all eligible for medals.
- On August 23, 1994, the United States Congress named December 7 of each year as the National Pearl Harbor Remembrance Day.
- The Pearl Harbor Remembrance Day is not a federal holiday however organizations and civilians may hold events honoring those who were killed and injured during the attack.
- On the day of remembrance, all American flags should be displayed on all American homes and government offices at half-mast.
- Special services such as wreath laying ceremonies, keynote speeches, luncheons and school activities are held.
- A number of films were inspired by the attack on Pearl Harbor such as Pearl Harbor (2001), Saving Private Ryan (1998), Letters from Iwo Jima (2006), Flags of Our Fathers (2006), Empire of the Sun (1987), The Longest Day (1962) and the Pianist (2002).
Pearl Harbor Remembrance Day Worksheets
This bundle contains 11 ready-to-use Pearl Harbor Remembrance Day Worksheets that are perfect for students who want to learn more about The National Pearl Harbor Remembrance Day, or Pearl Harbor Remembrance Day, which is the annual commemoration of the attack on Pearl Harbor, Hawaii on December 7, 1941. This attack led to the Americans joining World War II.
Download includes the following worksheets:
- Pearl Harbor Remembrance Day Facts
- The Day of Infamy
- World War Powers
- War Leaders
- Mapping Pearl Harbor
- War Glossary
- Pearl Harbor Attack!
- Story of Survival
- Pearl Harbor in Numbers
- Photo Call
- In Memoriam
Link/cite this page
If you reference any of the content on this page on your own website, please use the code below to cite this page as the original source.
Link will appear as Pearl Harbor Remembrance Day Facts & Worksheets: https://kidskonnect.com - KidsKonnect, November 20, 2018
Use With Any Curriculum
These worksheets have been specifically designed for use with any international curriculum. You can use these worksheets as-is, or edit them using Google Slides to make them more specific to your own student ability levels and curriculum standards.<|endoftext|>
| 4.0625 |
930 |
The Pacific Ocean is the largest of the world's five oceans (followed by the Atlantic Ocean, Indian Ocean, Southern Ocean, and Arctic Ocean). Strategically important access waterways include the La Perouse, Tsugaru, Tsushima, Taiwan, Singapore, and Torres Straits. The decision by the International Hydrographic Organization in the spring of 2000 to delimit a fifth ocean, the Southern Ocean, removed the portion of the Pacific Ocean south of 60 degrees south.
note:includes Bali Sea, Bering Sea, Bering Strait, Coral Sea, East China Sea, Gulf of Alaska, Gulf of Tonkin, Philippine Sea, Sea of Japan, Sea of Okhotsk, South China Sea, Tasman Sea, and other tributary water bodies
planetary air pressure systems and resultant wind patterns exhibit remarkable uniformity in the south and east; trade winds and westerly winds are well-developed patterns, modified by seasonal fluctuations; tropical cyclones (hurricanes) may form south of Mexico from June to October and affect Mexico and Central America; continental influences cause climatic uniformity to be much less pronounced in the eastern and western regions at the same latitude in the North Pacific Ocean; the western Pacific is monsoonal - a rainy season occurs during the summer months, when moisture-laden winds blow from the ocean over the land, and a dry season during the winter months, when dry winds blow from the Asian landmass back to the ocean; tropical cyclones (typhoons) may strike southeast and east Asia from May to December
surface currents in the northern Pacific are dominated by a clockwise, warm-water gyre (broad circular system of currents) and in the southern Pacific by a counterclockwise, cool-water gyre; in the northern Pacific, sea ice forms in the Bering Sea and Sea of Okhotsk in winter; in the southern Pacific, sea ice from Antarctica reaches its northernmost extent in October; the ocean floor in the eastern Pacific is dominated by the East Pacific Rise, while the western Pacific is dissected by deep trenches, including the Mariana Trench, which is the world's deepest
surrounded by a zone of violent volcanic and earthquake activity sometimes referred to as the "Pacific Ring of Fire"; subject to tropical cyclones (typhoons) in southeast and east Asia from May to December (most frequent from July to October); tropical cyclones (hurricanes) may form south of Mexico and strike Central America and Mexico from June to October (most common in August and September); cyclical El Nino/La Nina phenomenon occurs in the equatorial Pacific, influencing weather in the Western Hemisphere and the western Pacific; ships subject to superstructure icing in extreme north from October to May; persistent fog in the northern Pacific can be a maritime hazard from June to December
the major chokepoints are the Bering Strait, Panama Canal, Luzon Strait, and the Singapore Strait; the Equator divides the Pacific Ocean into the North Pacific Ocean and the South Pacific Ocean; dotted with low coral islands and rugged volcanic islands in the southwestern Pacific Ocean
The Pacific Ocean is a major contributor to the world economy and particularly to those nations its waters directly touch. It provides low-cost sea transportation between East and West, extensive fishing grounds, offshore oil and gas fields, minerals, and sand and gravel for the construction industry. In 1996, over 60% of the world's fish catch came from the Pacific Ocean. Exploitation of offshore oil and gas reserves is playing an ever-increasing role in the energy supplies of the US, Australia, NZ, China, and Peru. The high cost of recovering offshore oil and gas, combined with the wide swings in world prices for oil since 1985, has led to fluctuations in new drillings.
Bangkok (Thailand), Hong Kong (China), Kao-hsiung (Taiwan), Los Angeles (US), Manila (Philippines), Pusan (South Korea), San Francisco (US), Seattle (US), Shanghai (China), Singapore, Sydney (Australia), Vladivostok (Russia), Wellington (NZ), Yokohama (Japan)
the Inside Passage offers protected waters from southeast Alaska to Puget Sound (Washington state); the International Maritime Bureau reports the territorial waters of littoral states and offshore waters in the South China Sea as high risk for piracy and armed robbery against ships; numerous commercial vessels have been attacked and hijacked both at anchor and while underway; hijacked vessels are often disguised and cargoes stolen; crew and passengers are often held for ransom, murdered, or cast adrift<|endoftext|>
| 3.734375 |
1,272 |
Bauxite mine is actually kind of ore minerals collectively…
ROYAL Machinery is professional manufacturer of calcite…
As micro ground calcium carbonate powder plays an important…
As kind of important raw mineral materials, clay mine’s…
Igneous rock, or magmatic rock, is one of the three main rock types. Igneous rock is formed through the cooling and solidification of magma or lava. The magma can be derived from partial melts of existing rocks in either a planet's mantle or crust.
Depending on the region of Ontario, there can be a variety of different minerals that one can discover in the field. Click on the links below to learn more about the areas in Ontario and the local rocks you can acquire there: Northern Ontario minerals Southern Ontario Paleozoic rocks East-Central Ontario minerals
Types of Rocks. These articles explore rocks of all types—from those you climb to those you collect—and explore how they formed and what they can teach us about the past.
There are so many types of rocks and minerals that grace the earth's surface. While rocks can be classified into three major types, minerals on the other hand have more than 1000 different types and more are still being discovered each day.
What is a Rock? Rocks ordinarily lie everywhere on the ground of the Earth. They constitute most of the landforms, as we often notice. For instance, rocks make up the mountains and most of the non-water portions of the earth's surface.
When most of us think about minerals, we relate to gold, silver and other elements like them. However, there are certain nutrients like iron and potassium which are also minerals. Here, we explain to you the various types of minerals.
The following is a list of rock types recognized by geologists. There is no agreed number of specific types of rocks. Any unique combination of chemical composition, mineralogy, grain size, texture, or other distinguishing characteristics can describe rock types.
Learning about rocks and minerals gives students a deeper appreciation of the story behind the scenery in our national parks. Rocks and minerals are all around us! They help us to develop new technologies and are used in our everyday lives. Our use of rocks and minerals includes as building material
Rocks are naturally occurring solid material consisting of various minerals. Earth's crust is largely made up of rocks. Rocks are classified by mineral composition, how they were formed and other physical attributes such as texture.
Rocks, Minerals & Soils (game) Rock Hounds : Layers of the Earth. Rocks in our World . The study of rocks and minerals introduces students to the science of geology. By examining different types of rocks and minerals found in the earth's crust, students will learn that the unique characteristics and properties of rocks and minerals are a result of how they were formed. Such properties ...
Minerals have distinct physical properties such as specific gravity, streak, and form which can easily distinguish the major rock-forming minerals. More detailed examination of minerals can be undertaken by examining a microscope thin section of a rock or mineral.
2016-06-17· Hey kids! What strikes your mind when you hear the word rock? Is that 'rock music?' Well not anymore, as Dr.Binocs is here to explain different types of rocks that exist around us. Tune into this ...
Home » Rocks. Rocks: Igneous, Metamorphic and Sedimentary Rocks hold the history of the earth and the materials that will be used to build its future.
Rocks and Minerals. Enjoy our wide range of fun facts and information about different types of rocks and minerals for kids. Learn what rocks and minerals are, what the difference is, examples of rocks and minerals, the difference between igneous, sedimentary and metamorphic rocks…
The Gallery of Minerals has examples of common minerals that are categorized by type. Each has a thumbnail picture of the specimen and a brief description of it.
2015-08-10· 3 Types of Rocks. On the bases of their formation, rocks are broadly divided into three types, namely igneous rocks, sedimentary rocks and metamorphic rocks. Igneous rocks are found in volcanic ...
Manitoba Mining Facts. The mining and petroleum industries make up Manitoba's largest primary resource industry. In 2017, the combined value of mineral production for metals ($1.37 billion), industrial minerals ($289.4 million) and petroleum ($877 million) totalled about $2.5 billion.
Common Rock-forming Minerals While rocks consist of aggregates of minerals, minerals themselves are made up of one or a number of chemical elements with a definite chemical composition. Minerals cannot be broken down into smaller units with different chemical compositions in the way that rocks can.
Rocks – the building blocks of everything Geology is the science that deals with the physical history and dynamics of the earth and the rocks it is composed of. Rocks are important – everything we have is derived from mining or grown (in soil which is mostly broken down rocks).
Rock type is usually defined as a particular kind of rock having a specific set of characteristics 1. Rock types are specific assemblages of minerals (most rocks are composed of minerals).
Mineral-rich metamorphic rock, baked or compressed igneous or sedimentary rock, shows up in the Shield and Cordillera. Synopsis This slide show consists of three images of different types of rocks: sedimentary, metamorphic and igneous, along with descriptions of each.
Sedimentary rocks . Sedimentary rock forms when layers of sand and pebbles are compressed enough to form rock. Fossils are mainly found in sedimentary rock, specifically limestone because limestone is formed in warm, shallow seas where organisms and shells get fossilized at the bottom.
Rock, in geology, naturally occurring and coherent aggregate of one or more minerals. Such aggregates constitute the basic unit of which the solid Earth is comprised and typically form recognizable and mappable volumes.
Minerals of Newfoundland and Labrador The following collection of mineral specimens were photographed as a project of the Mining Week Committee. They depict the quality of collectible mineral specimens available in Newfoundland and Labrador.<|endoftext|>
| 3.75 |
902 |
What are warbles?
Cuterebra is the genus or scientific family name of the North American botfly. Twenty-six species ofCuterebra are known to occur in the U.S. and Canada. They are also found in Mexico and the neo-tropical regions. Cuterebra larvae develop within the tissues of certain animal hosts, and during this phase of their life cycle, they are commonly referred to as 'warbles'.
What is the cuterebra or botfly lifecycle?
The adult botfly deposits its eggs near or in the opening of rodent and rabbit burrows. After hatching, the botfly larvae, which typically infect rodents and rabbits, enters the host's body through an opening such as the nose or mouth or through a skin wound. After several days, they migrate to the tissues beneath the skin where they encyst and continue their development.
Different species of Cuterebra flies have evolved to migrate to specific anatomical locations in different hosts. For example, Cuterebra horripilum tends to seek out the throat region in cottontail rabbits, and C. fontinella commonly selects the abdominal or caudal region in the deer mouse. The larvae encyst beneath the skin of the host and complete their development. Larval development within the host may last from 19 to 38 days in small rodents and from 55 to 60 days in jackrabbits. After leaving the host, the larva develops into a pupa in loose soil, debris or forest duff. The pupation period may be as long as 7 to 11 months or as short as 28 days, depending on the environmental temperature. Adult Cuterebra flies will mate within a few days after emergence and they seldom live more than two weeks.
How did my cat get warbles?
"Cats are accidental hosts ofCuterebra larvae."
Cats are accidental hosts of Cuterebra larvae. They are most commonly infected when they are hunting rodents or rabbits and encounter the botfly larvae near the entryway to a rodent's burrow. Most cases of warbles in cats occur around the head and neck.
How do I know if my cat has warbles?
The early stages of Cuterebra infection or warbles are rarely evident from external inspection of the skin. Most cases of warbles do not become noticeable until the larva enlarges and becomes a noticeable swelling that is seen or felt beneath the skin. A small "breathing" hole is often visible in the skin over the warble. The hole enlarges when the warble has fully matured and is about to leave the host. Sometimes, nothing abnormal is noticed until after the larva has left the host and the empty cyst becomes infected or develops into an abscess in the cat's skin.
In many cases, the secondary bacterial infection that develops in the empty cyst causes more damage to the host than the primary attack by the Cuterebra warbles. Most cats will develop a deep abscess or skin infection at the infection site after the warble has left the skin.
How is the condition treated?
Treatment depends on when the condition is discovered. If the condition is diagnosed before the warble leaves the skin, the warble will be removed and the injured tissues will be debrided or surgically removed. Antibiotics are usually prescribed to combat any secondary bacterial infection.
If the condition is noticed after the warble has left the skin, the infected area is cleaned and debrided and antibiotics prescribed.
What is the prognosis for my cat?
When only a few warbles are involved, the prognosis is very good for complete resolution and few, if any, permanent side effects occur. The prognosis is worse if a cat is infected with multiple warbles or if a warble migrates through or develops near a nerve or other sensitive tissue or organ.
How can I prevent my cat from getting warbles?
"Cuterebra is a common fly in North America."
Cuterebra is a common fly in North America. The best prevention is to keep your cat from hunting rodents. When this is impossible and if you live in an area with numerous rodents, rabbits or other small mammals, you should closely inspect your cat regularly for any signs of warbles. The earlier a warble is removed, the less likely the chance of permanent or serious damage to your cat.<|endoftext|>
| 3.65625 |
3,477 |
# Selina Solutions Concise Mathematics Class 6 Chapter 15: Decimal Fractions Exercise 15(C)
Selina Solutions Concise Mathematics Class 6 Chapter 15 Decimal Fractions Exercise 15(C) help students with the fundamental concepts of multiplication and division of decimal numbers. Students can clear their doubts instantly, by referring to the solutions PDF. The Solutions are prepared by subject experts, keeping in mind the important questions that would appear in the exams. Students can boost their logical thinking in solving complex problems with the help of solutions PDF. Selina Solutions Concise Mathematics Class 6 Chapter 15 Decimal Fractions Exercise 15(C), PDF links which are provided here
## Selina Solutions Concise Mathematics Class 6 Chapter 15: Decimal Fractions Exercise 15(C) Download PDF
### Access Selina Solutions Concise Mathematics Class 6 Chapter 15: Decimal Fractions Exercise 15(C)
Exercise 15(C)
1. Multiply:
(i) 5.6 and 8
(ii) 38.46 and 9
(iii) 0.943 and 62
(iv) 0.0453 and 35
(v) 7.5 and 2.5
Solution:
(i) 5.6 and 8
The multiplication of 5.6 and 8 is as follows
5.6 × 8 = 44.8
Hence, 5.6 × 8 = 44.8
(ii) 38.46 and 9
The multiplication of 38.46 and 9 is as follows
38.46 × 9 = 346.14
Hence, 38.46 × 9 = 346.14
(iii) 0.943 and 62
The multiplication of 0.943 and 62 is as follows
943
62 ×
________
1886
5658×
________
58466
________
We know that,
.943 × 62 = 58.466
Hence 0.943 × 62 = 58.466
(iv) 0.0453 and 35
The multiplication of 0.0453 and 35 is as follows
453
35×
_________
2265
1359×
__________
15855
___________
We know that,
453 × 35 = 15855
Hence 0.0453 × 35 = 1.5855
(v) 7.5 and 2.5
The multiplication of 7.5 and 2.5 is as follows
75
25 ×
________
375
150×
_________
1875
_________
We know that,
75 × 25 = 1875
Hence 7.5 × 2.5 = 18.75
2. Evaluate:
(i) 0.0008 × 26
(ii) 0.038 × 95
(iii) 1.2 × 2.4 × 3.6
(iv) 0.9 × 1.8 × 0.27
(v) 1.5 × 1.5 × 1.5
Solution:
(i) 0.0008 × 26
Since,
8 × 26 =208
0.0008 × 26 = 0.0208
∴ We get 0.0208 on multiplying 0.0008 × 26
(ii) 0.038 × 95
38
95×
_______
190
342×
________
3610
________
Since,
38 × 95 = 3610
.038 × 95 = 3.610
= 3.61
∴ We get 3.61 on multiplying 0.038 × 95
(iii) 1.2 × 2.4 × 3.6
12
24 ×
_______
48
24×
_________
288
36 ×
__________
1728
864×
_________
10368
_________
Since,
12 × 24 × 36 = 10368
1.2 × 2.4 × 3.6 = 10.368
∴ We get 10.368 on multiplying 1.2 × 2.4 × 3.6
(iv) 0.9 × 1.8 × 0.27
9
18 ×
_______
72
9×
_________
162
27 ×
________
1134
324 ×
_________
4374
_________
Since,
9 × 18 × 27 = 4374
0.9 × 1.8 × 0.27 = 0.4374
∴ We get 0.4374 on multiplying 0.9 × 1.8 × 0.27
(v) 1.5 × 1.5 × 1.5
15
15 ×
________
75
15 ×
_________
225
15 ×
__________
1125
225 ×
__________
3375
___________
Since,
15 × 15 × 15 = 3375
1.5 × 1.5 × 1.5 = 3.375
3. Multiply each of the following numbers by 10, 100 and 1000:
(i) 3.9
(ii) 2.89
(iii) 0.0829
(iv) 40.3
(v) 0.3725
Solution:
(i) 3.9
3.9 × 10 = 39
3.9 × 100 = 390
3.9 × 1000 = 3900
Hence, 39, 390 and 3900 are the required numbers
(ii) 2.89
2.89 × 10 = 28.9
2.89 × 100 = 289
2.89 × 1000 = 2890.00
= 2890
Hence, 28.9, 289 and 2890 are the required numbers
(iii) 0.0829
0.0829 × 10 = 0.829
0.0829 × 100 = 8.29
0.0829 × 1000 = 82.9
Hence, 0.829, 8.29 and 82.9 are the required numbers
(iv) 40.3
40.3 × 10 = 403
40.3 × 100 = 4030
40.3 × 1000 = 40300
Hence, 403, 4030 and 40300 are the required numbers
(v) 0.3725
0.3725 × 10 = 3.725
0.3725 × 100 = 37.25
0.3725 × 1000 = 372.5
4. Evaluate:
(i) 8.64 ÷ 8
(ii) 0.0072 ÷ 6
(iii) 20.64 ÷ 16
(iv) 1.602 ÷ 15
(v) 13.08 ÷ 4
Solution:
(i) 8.64 ÷ 8
8.64 ÷ 8 = 8.64 / 8
We get
= 1.08
Therefore, the value of 8.64 ÷ 8 = 1.08
(ii) 0.0072 ÷ 6
0.0072 ÷ 6 = (0.0072) / 6
We get
= 0.0012
Therefore, the value of 0.0072 ÷ 6 = 0.0012
(iii) 20.64 ÷ 16
20.64 ÷ 16 = (20.64) / 16
We get
= 1.29
Therefore, the value of 20.64 ÷ 16 = 1.29
(iv) 1.602 ÷ 15
1.602 ÷ 15 = (1.602) / 15
We get
1602 / (1000 × 15)
We get
= 106.8 / 1000
= 0.1068
Therefore, the value of 1.602 ÷ 15 = 0.1068
(v) 13.08 ÷ 4
13.08 ÷ 4 = 13.08 / 4
We get
= 3.27
Therefore, the value of 13.08 ÷ 4 = 3.27
5. Divide each of the following numbers by 10, 100 and 1000:
(i) 49.79
(ii) 0.923
(iii) 0.0704
Solution:
(i) 49.79
49.79 / 10 = 4.979
49.79 / 100 = 0.4979
49.79 / 1000 = 0.04979
Therefore, the required numbers are 4.979, 0.4979 and 0.04979
(ii) 0.923
0.923 / 10 = 0.0923
0.923 / 100 = 0.00923
0.923 / 1000 = 0.000923
Therefore, the required numbers are 0.0923, 0.00923 and 0.000923
(iii) 0.0704
0.0704 / 10 = 0.00704
0.0704 / 100 = 0.000704
0.0704 / 1000 = 0.0000704
Therefore, the required numbers are 0.00704, 0.000704 and 0.0000704
6. Evaluate:
(i) 9.4 ÷ 0.47
(ii) 6.3 ÷ 0.09
(iii) 2.88 ÷ 1.2
(iv) 8.64 ÷ 1.6
(v) 37.188 ÷ 3.6
Solutions:
(i) 9.4 ÷ 0.47
= 9.4 / 0.47
= (94 × 100) / (47 × 10)
On calculating further, we get
= 2 × 10
= 20
Hence, 9.4 ÷ 0.47 = 20
(ii) 6.3 ÷ 0.09
= 6.3 / 0.09
= (63 × 100) / (9 × 10)
We get
= 6300 / 90
= 630 / 9
= 70
Hence, 6.3 ÷ 0.09 = 70
(iii) 2.88 ÷ 1.2
= 2.88 / 1.2
= (288 × 10) / (12 × 100)
We get,
= 2880 / 1200
= 288 / 120
= 2.4
Hence, 2.88 ÷ 1.2 = 2.4
(iv) 8.64 ÷ 1.6
= 8.64 / 1.6
= (8.64 × 10) / (1.6 × 10)
We get,
= 86.4 / 16
= 5.4
Hence, 8.64 ÷ 1.6 = 5.4
(v) 37.188 ÷ 3.6
= 37.188 / 3.6
= (37188 × 10) / (36 × 1000)
We get,
= 371880 / 36000
= 2066 / 200
= 1033 / 100
= 10.33
7. Fill in the blanks with 10, 100, 1000, or 10000 etc:
(i) 7.85 × ……… = 78.5
(ii) 0.442 × ………. = 442
(iii) 0.0924 × ………. = 9.24
(iv) 0.00187 × ……… = 18.7
(v) 2.6 × …….. = 2600
Solution:
(i) 7.85 × 10 = 78.5
(ii) 0.442 × 1000 = 442
(iii) 0.0924 × 100 = 9.24
(iv) 0.00187 × 10000 = 18.7
(v) 2.6 × 1000 = 2600
8. Evaluate:
(i) 9.32 – 28.54 ÷ 10
(ii) 0.234 × 10 + 62.8
(iii) 3.06 × 100 – 889.4 ÷ 100
(iv) 2.86 × 7.5 + 45.4 ÷ 0.2
(v) 97. 82 × 0.03 – 0.54 ÷ 0.3
Solution:
(i) 9.32 – 28.54 ÷ 10
= 9.32 – 2.854
So, we get
= 9.320 – 2.854
= 6.466
Therefore, 9.32 – 28.54 ÷ 10 = 6.466
(ii) 0.234 × 10 + 62.8
Using BODMAS, we get
= 2.34 + 62.80
= 65.14
Therefore, 0.234 × 10 + 62.8 = 65.14
(iii) 3.06 × 100 – 889.4 ÷ 100
Using BODMAS, we get
= 3.06 × 100 – 8.894
= 306 – 8.894
= 306.000 – 8.894
= 297.106
(iv) 2.86 × 7.5 + 45.4 ÷ 0.2
Using BODMAS, we get
= 2.86 × 7.5 + 454 ÷ 2
On further calculation, we get
= 2.86 × 7.5 + 227.00
= (286 / 100) × (75 / 10) + 227.00
= (286 / 4) × (3 / 10) + 227.00
= (143 / 2) × (3 / 10) + 227.00
We get,
= 429 / 20 + 227.00
= 21.45 + 227.00
= 248.45
(v) 97. 82 × 0.03 – 0.54 ÷ 0.3
= 97.82 × 0.03 – 0.54 / 0.3
= 97.82 × 0.03 – (0.54 × 10) / (0.3 × 10)
On further calculation, we get
= 2.9346 – 5.4 / 3
= 2.9346 – 1.8
= 2.9346 – 1.8000
We get,
= 1.1346<|endoftext|>
| 4.875 |
524 |
Did you know that research has shown that developing early math skills can be a predictor of later learning? It may even be a greater predictor than basic reading skills1. Teachers were introduced to this information to help them focus on mathematizing everyday activities and to focus on more challenging and less basic math skill development. In order to support student learning beyond basic skills, the Learning Pathways to Numeracy document can be used to assist teachers in knowing what to do to help a child in their next steps along the mathematical progression.
Here are some highlights of the learning:
- Realizing some areas that pose stumbling blocks in learning like language. In English, we use words that don’t make sense or lack connection like, eleven versus twenty-one where the term connects to the number value.
- Enhancing learning with complexity using questioning techniques. Asking questions like:
- How do you know?
- Can you show me another way?
- Who has the same number as you?
- How many do you and your neighbor have together?
- Math talk is very important. Learning to replace statements like:
- put on your shoes, parents and teachers would say put on your two shoes and replace general terms like few, some, or lots by saying things like put the four dishes on the table. This helps build conceptual knowledge of numbers that go beyond memorization of numbers and basic counting.
- Mathematize reading. Move beyond counting books and mathematize stories. After reading a story for enjoyment, such as Five Creatures by Emily Jenkins, teachers can incorporate activities. Children can draw the number of creatures in their house, stack cubes for each number and share which is most or least and how they know. This creates a more complex learning task.
- Making dot plates to develop the capacity to subitize (determining the number of objects without counting) which leads to composing and decomposing skills.
- Using whole-body activities like forming lines based on teachers’ questions to develop skills around more, less, or the same.
- Using dot cards to explore more, less, the same and extending their use by composing and decomposing numbers, building operation and algebraic thinking.
- Playing games – Shake & Spill, Five Octopi, Concentration, Cross Out, and more.
- Using 5-frames and 10-frames to make multiple representations of numerical values (5 and 2 more is 7, 3 and 4 more is 7…)
Help kids be school ready and prepare them for future academic success. Mathematize your world!<|endoftext|>
| 4.3125 |
553 |
# Climbing Alpine Peaks
## OBJECTIVE: To learn how to graph a line on the Cartesian plane using a point and the slope
In a previous lesson, you learned how to graph a line on the Cartesian plane using the slope and the y-intercept. Riding the Berms: https://www.geogebra.org/m/nkwt8g3y In this lesson, you'll be introduced to yet another form of the equation of a line—the POINT-SLOPE FORM, (y - y₁) = m(x - x₁) where (x₁, y₁) are the coordinates of a point and m is the slope. The process of graphing a line using a point and the slope is very similar to graphing a line using the slope and the y-intercept. In fact, the y-intercept is just another point, but it is a special point in the sense that it always lies on the y-axis. Interact with the applet below to learn how to graph a line using a point (any point) and the slope.
## INSTRUCTIONS:
1. Move point A to the position indicated by (x₁, y₁) in the linear equation. 2. Use rise/run to locate point B on the line. 3. If your graph is correct, the expression will appear. Otherwise, you'll have to try again. 4. Click "Generate New Line" to come up with another problem. 5. Repeat as many times as needed to master the concept.
The applet below shows a variation of the POINT-SLOPE FORM, y = m(x - x₁) + y₁ where (x₁, y₁) are the coordinates of a point and m is the slope. Interact with the applet to learn how to graph a line using a point (any point) and the slope.
## INSTRUCTIONS:
1. Move point A to the position indicated by (x₁, y₁) in the linear equation. 2. Use rise/run to locate point B on the line. 3. If your graph is correct, the expression will appear. Otherwise, you'll have to try again. 4. Click "Generate New Line" to come up with another problem. 5. Repeat as many times as needed to master the concept.
QUESTION: In getting the coordinates (x₁, y₁) in the second formula, what do you observe about the sign of y₁?
## TODAY you learned how to graph a line on the Cartesian plane using a point and the slope.
In future lessons, you'll learn how to work out the equations of a line from the graphs. Did you ENJOY your lesson today?<|endoftext|>
| 4.5 |
129 |
Is There a Central Brain Area for Hearing Melodies and Speech Cues?Released November 28, 2011
- The perceptual feature of sound known as pitch is fundamental to human hearing, allowing us to enjoy the melodies and harmonies of music and recognize the inflection of speech. Previous studies have suggested that a particular hotspot in the brain might be responsible for perceiving pitch. However, auditory neuroscientists are still hotly debating whether this “pitch center” actually exists. A new review article discusses a recent study claiming that this pitch center may not exist after all, or alternatively, may not be located where previous research has suggested.<|endoftext|>
| 3.796875 |
1,365 |
15
Q:
# A room contains 3 brown, 5 black and 4 white chairs. Two chairs are picked and are put in the lawn. What is the probability that none of the chairs picked is white ?
A) 14/33 B) 14/55 C) 12/55 D) 13/33
Explanation:
Total number of chairs = (3 + 5 + 4) = 12.
Let S be the sample space.
Then, n(s)= Number of ways of picking 2 chairs out of 12
12×11/2×66
Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.
=> 8C8×7/2×28
Therefore required probability = 28/66 = 14/33.
Q:
When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12?
A) 35/36 B) 17/36 C) 15/36 D) 1/36
Explanation:
When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36
Required, the sum of the two numbers that turn up is less than 12
That can be done as n(E)
= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5) }
= 35
Hence, required probability = n(E)/n(S) = 35/36.
3 629
Q:
In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random and are placed in a box. If a coin is now picked from the box, find the probability of it being a rupee coin?
A) 4/7 B) 2/3 C) 1/2 D) 5/6
Explanation:
Total coins 30
In that,
1 rupee coins 20
50 paise coins 10
Probability of total 1 rupee coins = 20C11
Probability that 11 coins are picked = 30C11
Required probability of a coin now picked from the box is 1 rupee = 20C11/30C11 = 2/3.
7 1024
Q:
In a box, there are 9 blue, 6 white and some black stones. A stone is randomly selected and the probability that the stone is black is ¼. Find the total number of stones in the box?
A) 15 B) 18 C) 20 D) 24
Explanation:
We know that, Total probability = 1
Given probability of black stones = 1/4
=> Probability of blue and white stones = 1 - 1/4 = 3/4
But, given blue + white stones = 9 + 6 = 15
Hence,
3/4 ----- 15
1 ----- ?
=> 15 x 4/3 = 20.
Hence, total number of stones in the box = 20.
10 1073
Q:
What is the probability of an impossible event?
A) 0 B) -1 C) 0.1 D) 1
Explanation:
The probability of an impossible event is 0.
The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.
The probability of a certain event is 1.
9 1487
Q:
In a box, there are four marbles of white color and five marbles of black color. Two marbles are chosen randomly. What is the probability that both are of the same color?
A) 2/9 B) 5/9 C) 4/9 D) 0
Explanation:
Number of white marbles = 4
Number of Black marbles = 5
Total number of marbles = 9
Number of ways, two marbles picked randomly = 9C2
Now, the required probability of picked marbles are to be of same color = 4C2/9C2 + 5C2/9C2
= 1/6 + 5/18
= 4/9.
9 1813
Q:
A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that it is either pink or red?
A) 2/3 B) 1/8 C) 3/8 D) 3/4
Explanation:
Given number of balls = 3 + 5 + 7 = 15
One ball is drawn randomly = 15C1
probability that it is either pink or red =
14 1673
Q:
Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M?
A) 1/4 B) 1/6 C) 1/8 D) 4
Explanation:
Required probability is given by P(E) =
19 2389
Q:
14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.
A) 11/379 B) 21/628 C) 24/625 D) 26/247
Explanation:
Total no of ways = (14 – 1)! = 13!
Number of favorable ways = (12 – 1)! = 11!
So, required probability = $11!×3!13!$ = $39916800×66227020800$ = $24625$<|endoftext|>
| 4.5625 |
3,680 |
It is this type of recurrence relation that we will learn to solve today, starting from the simplest ones: linear recurrence relations of first order. 2 Nonhomogeneous linear recurrence relations When f(n) 6= 0, we will search for a particular solution apn which is similar to f(n). Therefore, the same recurrence relation can have (and usually has) multiple solutions If both the initial conditions and the recurrence relation are specified, then the sequence is uniquely determined. If there are 3 constants then we need 3 equations. Answer: Extending the example of ROTOR, it is a 5 letter palindrome. Now plug back in. Hence the solution is the sequence {a n} with a n = 3.2 n - 5n (c) a n = 6 a n-1 -8 a n-2, a 0 = 4, a 1 = 10 The characteristic equation of the recurrence relation is r2 -6r +8 = 0 Its roots are r= 2 and r= 4. recurrence relations is to look for solutions of the form a n = rn, where ris a constant. Multiply by the power of z corresponding to the left-hand side subscript Multiply both sides of the relation by zn+2 A solution of a recurrence relation in any function which satisfies the given equation . For any , this defines a unique sequence with as .
We will still solve the homogeneous recurrence relation setting f(n) temporarily to 0 and the solution of this homogeneous recurrence relation will be ah nand a n= a p n+ah n. The following table provides a good . 2 n + n + , where , , and are suitable real numbers that can be found by taking. A recurrence relation is an equation that expresses each element of a sequence as a function of the preceding ones. There are four methods for solving Recurrence: Substitution Method. In this case the solution could be expressed in the same way as in the case of distinct real roots, but in order to avoid the use of complex numbers we writer1=rei, r2=rei,k1=c1+c2,k2= (c1c2)i, which yields:1 ( 2) n + n 5 n + 1 Putting values of F 0 = 4 and F 1 = 3, in the above equation, we get a = 2 and b = 6 Hence, the solution is F n = n 5 n + 1 + 6. What is the solution to the recurrence t'n't n 2 )+ n? Note: c is a constant. The simplest form of a recurrence relation is the case where the next term depends only on the immediately previous term. The solution of the recurrence relation can be written as F n = a h + a t = a .5 n + b. Other recurrence relations may be more complicated, for example, f(N) = 2f(N - 1) + 3f(N - 2) And so a particular solution is to plus three times negative one to the end The value of these recurrence relations is to illustrate the basic idea of recurrence relations with examples that can be easily verified with only a small effort Using the . 2) Case 2 can be extended for f (n) = (n c Log k n) For example consider the recurrence T (n) = 2T (n/2) + n We guess the solution as T (n) = O (nLogn). The solution of the following recurrence relation with the given initial condition is. It is lower bounded by (x+y) QUESTION: 4. Use the substitution method to identify the big-Oh of the represented by the following recurrence relation. Iteration Method for Solving Recurrences. S(1) = 2 S(n) = 2S(n-1) for n 2 Let r 1,r 2 be the roots of C 0r2 +C 1r +C 2 = 0. Eg. Let a 99 = k x 10 4. Let us find the solution of the recurrence relation a_n = a_{n-1} + 2a_{n-2} a n = a n 1 + 2 a n 2 , with a_0 = 2 a 0 = 2 and a_1 = 7 a 1 = 7. Definition: A second order linear homogeneous recurrence with constant coefficients is a recurrence relation of the form Squaring yields i, and squaring two High School Math Solutions - Algebra Calculator, Sequences When formulated as an equation to be solved, recurrence relations are known as recurrence equations, or sometimes difference . Solving for k, we get k = n - 1. Search: Recurrence Relation Solver Calculator. Let us solve the characteristic equation k^2=k+2k2=k+2 which is equivalent to k^2-k-2=0k2k2=0, and hence by Vieta's formulas has the solutions k_1=-1k1 =1 and k_2=2.k2 =2. Note, you likely need to rewrite the . 6.1.1 De nition. Edit: I think the hint is given due to the fact that $2^n + 3n$ looks nothing like linear homogenous recurrence nor does it even look like a "typical" linear nonhomogenous recurrence relation, since $2^n + 3n$. General solutions to recurrence relations For the purpose of these notes, a sequence is a sequence of complex numbers (although our results should hold if we replace C by any algebraically closed eld). Introduction to Recurrence Relations The numbers in the list are the terms of the sequence T(n) = 5 if n More precisely: If the sequence can be defined by a linear recurrence relation with finite memory, then there is a closed form solution for it but this is not a barrier to building useful PRNGs So far, all I've learnt is, whenever you . So our solution to the recurrence relation is a n = 32n. For , the recurrence relation of Theorem thmtype:7 Use the method of Frobenius to obtain two linearly independent series solutions about x 0 The additional solution to the complementary function is the particular integral, denoted here by y p Recurrence Equations aka Recurrence and Recurrence Relations That kind of formula is called a closed .
Characteristic equation: r 1 = 0 Characteristic root: r= 1 Use Theorem 3 with k= 1 like before, a n = 1n for some constant . TROTORT There are 26 possible options for adding a letter. + a 0 f ( n d) = 0 for all n 0. Q&A Forum for Sage The idea is simple 5 dn 2+ (t 1- 0 What is the general form of the particular solution guaranteed to exist by Theorem 6 of the linear nonhomogeneous recurrence relation an = 8an2 16an4 + F (n) if The base cases in the recursive denition are A linear homogeneous recurrence relation with constant coecients is a . Correct answer: Find the solution of the recurrence relation an = 4an1 3an2 + 2 n + n + 3 , a0 = 1 and a1 = 4 Sikademy Practice Recurrence Relation - Algorithm previous year question of gate cse. . Suppose we have been given a sequence; a n = 2a n-1 - 3a n-2 Now the first step will be to check if initial conditions a 0 = 1, a 1 = 2, gives a closed pattern for this sequence. C 0crn +C 1crn1 +C 2crn2 = 0. There are mainly three ways for solving recurrences.
We can say that we have a solution to the recurrence relation if we have a non-recursive way to express the terms. Recurrence relations and their closed-form solutions 6.1. abstract = "Many linear recurrence relations for combinatorial numbers depending on two indices - like, e.g. So to get the next palindrome using this word we will have to prefix and suffix this word with the same letter . 12.Argue that the solution to the recurrence T(n) = T(n=3) + T(2n=3) + an, where a>0 is a constant, is T(n) = ( nlogn), by using an appropriate recursion tree. Big-O, small-o, and the \other" This notation is due to the mathematician E. Landau and is in wide use in num-ber theory, but also in computer science in the context of measuring (bounding above) computational complexity of algorithms for all \very large inputs". Solution The above example shows a way to solve recurrence relations of the form an = an 1 + f(n) where n k = 1f(k) has a known closed formula. The initial conditions give the first term (s) of the sequence, before the recurrence part can take over. b a n = a n 1 for n 1;a 0 = 2 Same as problem (a). Method 1 Arithmetic Download Article 1 Consider an arithmetic sequence such as 5, 8, 11, 14, 17, 20, .. [1] 2 Since each term is 3 larger than the previous, it can be expressed as a recurrence as shown. 1. ( 2) n 2.5 n Generating Functions Definition. Solution The above example shows a way to solve recurrence relations of the form an =an1+f(n) a n = a n 1 + f ( n) where n k=1f(k) k = 1 n f ( k) has a known closed formula. However, relations such as x n =(x n-1) 2 + (x n-2) 5 or x n = x n-1 x n-4 + x n-2 are not. Now we use induction to prove our guess. Find the value of constants c 1, c 2, , c k by using the boundary conditions. You might want to comment about that, but I think this book will cover this in future chapters. We look for a solution of form a n = crn, c 6= 0 ,r 6= 0. Write down the general form of the solution for this recurrence (i (2 . T (n) = 2T (n/2) + cn T (n) = 2T (n/2) + n These types of recurrence relations can be easily solved using Master Method. But we can simplify this since 1n = 1 for any n, so our solution . Ifr1=r2=r, the general solution of the recurrence relation is xn=c1r n+c 2nr n, wherec1,c2are arbitrary constants. The textbook only briefly touches on it, and most sites I've searched seem to assume I already know how. In other words, kerf() is the solution set of (2). Question: Show that the sequence {an} is a solution of the recurrence relation an = -3an-1 + 4an-2 if an = 0. an = 1. an = (-4)n. an = 2(-4)n + 3. We want T(1). Search: Closed Form Solution Recurrence Relation Calculator. Recurrence Relations Many algo rithm s pa rticula rly divide and conquer al go rithm s have time complexities which a re naturally m odel ed b yr . Iteration Method. The given three cases have some gaps between them. The first and third algorithm are new and the second algorithm is an improvement over prior algorithms for the second order case. Recurrence Relations Example: Consider the recurrence relation a n = 2a n-1 - a n-2 for n = 2, 3, 4, Is the sequence {a n } with a n limited to the solutions of linear recurrence relations; the provided references contain a little more information about the power of these techniques. The solution of this recurrence relation, if the roots are distinct, is T ( n) = i = 1 k c i r i n Where c 1, c 2, , c k are constants. Download these Free Solution of Recurrence Relations MCQ Quiz Pdf and prepare for your upcoming exams Like Banking, SSC, Railway, UPSC, State PSC. If the roots are not distinct then the solution becomes The characteristic equation of the recurrence is r2 r 2=0. It is often useful to have a solution to a recurrence relation 2 Closed-Form Solutions and Induction 3 Vizio Tv Hack Recurrence Relations Annual Report on Form 10-K for the fiscal year ended December 31, 2019, filed with the SEC on April 13, 2020, and our Quarterly Report on Form 10-Q for the quarter ended September 30, 2020, filed Annual . Generalized recurrence relation at the kth step of the recursion: T(n) = T(n-k) + 2*k . The most common recurrence relation we will encounter in this course is the uniform divide-and-conquer recurrence relation, or uniform recurrence for short. We obtain C 0r2 +C 1r +C 2 = 0 which is called the characteristic equation. For eg. Solve the recurrence system a n= a n1+2a n2 with initial conditions a 0= 2 and a 1= 7. Chapter 3.2 Recurrence Relations Reading: 3.2 Next Class: 4.1 Motivation Solving recurrence relations using an iterative or recursive algorithm can be a very complex and time consuming operation.
Show transcribed image text Expert Answer. 2 Homogeneous Recurrence Relations Any recurrence relation of the form xn=axn1+bxn2(2) is called a second order homogeneous linear recurrence relation. So we let n-k = 1. The first algorithm `Find 2F1' finds a gt-transformation to a recurrence relation satisfied by a hypergeometric series u (n) = hypergeom ( [a+n, b], [c],z), if such a transformation exists. Find . Remark 1. Recurrence Relations and Generating Functions Ngy 8 thng 12 nm 2010 Recurrence Relations and Generating Functions. Hence, (a n ) is a solution of the recurrence i a n= 1 2 n+ 2 (1)n for some constants 1and 2 From the initial con- ditions, we get a 0=2= Consider the recurrence relation a 1 = 8, a n = 6n 2 + 2n + a n-1. 3 Inductively Verifying a Solution Exercises 3 The argument of the functional symbol may be a non negative integer, an expression of the form n-k where k is a (possibly negative) integer, or of the The above example shows a way to solve recurrence relations of the form an =an1+f(n) a n = a n 1 + f (n) where n k=1f(k) k = 1 n f (k . Recursive Problem Solving Question Certain bacteria divide into two bacteria every second. 2 = 01 2 = So the solution is a n = 2 1n. For example, the recurrence T (n) = 2T (n/2) + n/Logn cannot be solved using master method.
Solve the recurrence relation an = an1+n a n = a n 1 + n with initial term a0 = 4. a 0 = 4. The above expression forms a geometric series with ratio as 2 and starting element as (x+y)/2 T (x, y) is upper bounded by (x+y) as sum of infinite series is 2 (x+y). A sequence (x n) for which the equation is true for any n 0 is considered a "solution". Remark 1. The use of the word linear refers to the fact that previous terms are arranged as a 1st degree polynomial in the recurrence relation Recurrence Relations in A level In Mathematics: -Numerical Methods (fixed point iteration and Newton-Raphson) Call this the homogeneous solution, S (h) (k) Solve the recurrence relation and answer the . Solve the recurrence with a1 = 1. arrow_forward. So the solution is (Logn) Notes: 1) It is not necessary that a recurrence of the form T (n) = aT (n/b) + f (n) can be solved using Master Theorem. If you rewrite the recurrence relation as an an 1 = f(n), and then add up all the different equations with n ranging between 1 and n, the left-hand side will always give you an a0. Search: Recurrence Relation Solver. Consider the recurrence relation an = an1 + An + B where A and B are constants. How many ways the recurrence relations can be solved? T (n) = cn + T (n/2) + T (n/2), T (1) = c. arrow_forward. A closed form solution is an expression for an exact solution given with a nite amount of data Try to join/form a study group with members from class and get help from the tutors in the Math Gym (JB - 391) Find the solution to each of these recurrence relations with the given initial conditions Equations like Equations \ref{eq:7 Equations .<|endoftext|>
| 4.8125 |
943 |
Otger Campàs, an assistant professor in UC Santa Barbara’s Department of Mechanical Engineering, helped develop a compelling new technique which will allow scientists to study the mechanical environment of cells for the first time in their natural, 3D conditions: the living tissue.
Based on the biochemical and mechanical cues they sense in their environment, cells are able to take important actions, such as deciding what type of cell to develop into. Factors known to affect cell behavior include forces applied to cells (whether neighboring cells are pushing or pulling on them) and mechanical properties like rigidity or fluidity of their environment.
Campàs, who holds the Mellichamp Chair in Systems Biology and Bioengineering at UCSB, said, “For instance, stem cells that feel a soft environment like Jell-O, decide to become neurons because the compliance of brain tissue is similar to Jell-O. Cells can also decide to divide or not depending on the compliance of the environment. Most important decisions that cells make are influenced by the mechanical properties they feel.”
Previously, scientists were limited to observing how the mechanical environment of cells influenced their behavior in synthetic laboratory dishes. While permitting researchers to gain understanding of the traction forces of cells, in vitro studies were insufficient for learning about the forces cells produce within forming tissues and organs and for measuring mechanical properties within three-dimensional developing tissues and organs. But presently, the technique provides a way to study these inquiries in the 3D, native habitat of cells.
The procedure is carried out by injecting magnetically responsive droplets – each just slightly larger than a single cell – inside tissues and in between cells, and then applying a magnetic field to deform the droplets. When they deform, the droplets generate forces onto the surrounding cells.
“[This] allows us to literally poke individual cells inside developing organs,” Campàs said. “Moreover, we monitor the shape of the droplet over time when a magnetic field is applied on it which allows us to probe the compliance or fluidity of the cell’s environment inside living tissues. In other words, we can reveal what is the mechanical environment of cells and poke them in their natural environment, which allows us to know what mechanical cues they are exposed to as tissues develop.”
Using the technique, Campàs and his research team explored how vertebrate embryos, specifically zebrafish embryos, shaped their body axis.
“We mapped how the mechanical properties of the growing [zebrafish] embryo changes along its axis,” said Campàs. “We found that the tissues are fluid, like very viscous honey, but the viscosity is lowest at the end of the body axis, exactly where it needs to grow. This shows that the physical principles behind sculpting embryos are similar to shaping molten glass in glassblowing.”
The technique will enable Campàs and other researchers to study how organs and tissues are created in 3D settings, which is not yet understood. Indeed, it holds great potential for numerous implementations. By helping scientists understand how tissues behave mechanically, this could be significantly conducive in tissue engineering applications. It has opened the possibility of examining how the mechanical cues cells respond to affect cellular decisions in living tissues and grasping a better understanding of mechanical signaling – the “sense of touch,” according to Campàs – of cells.
Furthermore, as many diseases, including cancerous tumor growth, are linked to abnormal changes in the mechanical environment of cells, scientists now have a means to investigate how these atypical mechanical conditions contribute to the progression of disease.
“We will be able to study how mechanical cues affect disease progression using living tissues, without having to take the cells [from] their environment and study them on synthetic dishes,” Campàs said. “This is important because lab dishes provide a very different mechanical environment than the native 3D tissue. We will also be able to understand how the genetic code and the physical world interplay to build these complex structures.”
The latter application of the technique is what Campàs and his colleagues are currently focusing on.
“We are now using this technique to understand the role of mechanical cues in shaping embryonic tissues. Beyond the formation of 3D tissues and organs, we are studying mechanical cues in tumor progression compared to [those in] healthy tissues,” he said.
Campàs’s pioneering research has been published in the journal Nature Methods.
Jacqueline serves as the Science and Tech Editor. She enjoys watching movies and learning about science and tech.<|endoftext|>
| 3.953125 |
293 |
# 1969 Canadian MO Problems/Problem 8
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Let $\displaystyle f$ be a function with the following properties:
1) $\displaystyle f(n)$ is defined for every positive integer $\displaystyle n$;
2) $\displaystyle f(n)$ is an integer;
3) $\displaystyle f(2)=2$;
4) $\displaystyle f(mn)=f(m)f(n)$ for all $\displaystyle m$ and $\displaystyle n$;
5) $\displaystyle f(m)>f(n)$ whenever $m>n$.
Prove that $\displaystyle f(n)=n$.
## Solution
It's easily shown that $\displaystyle f(1)=1$ and $\displaystyle f(4)=4$. Since $\displaystyle f(2) $\displaystyle f(3) = 3.$
Now, assume that $\displaystyle f(2n+2)=f(2(n+1))=f(2)f(n+1)=2n+2$ is true for all $\displaystyle f(k)$ where $\displaystyle k\leq 2n.$
It follows that $\displaystyle 2n Hence, $\displaystyle f(2n+1)=2n+1$, and by induction $\displaystyle f(n) = n.$<|endoftext|>
| 4.5 |
1,069 |
Student: What’s the answer to 0/0?
Teacher: It’s undefined.
Student: Then why can’t we define it?
Teacher: Because that’s just how division works.
The exchange above is one that happens often when students start learning division. It’s a simple question, arising since we can divide by any other number, so why not this one? Unfortunately, the answers given to this question don’t attack the consequences of defining 0/0, but explains it as something given by authority.
I often like to remind students of a very important lesson when they start wondering about the definitions and restrictions that come up. We get to make up the definitions used in mathematics. We aren’t forced to use a certain definition for a concept if we don’t want to! Mathematics is about defining concepts and building from them, but we get to choose those starting points.
One specific example where things can get controversial is when we start considering an expression like 0/0. In this case, what is the answer?
As the teacher said above, we leave this expression as undefined. In other words, we say that the statement just doesn’t make sense, and we move on. However, if the definitions are up to us, why can’t we define these to take on certain values? What’s the harm in that?
This is a good question, and answering it will give us insight into the notion of divisibility. What does it mean for a number to “divide” another? If you ponder this question, you will find that we can divide b by a if we can write b=ac, where c is another element. Note that I’m using the word “element” here because we don’t necessarily have to be working with integers, though that is the setting which is most familiar.
There’s another property that we would like, even though you may have implicity assumed this. It’s that if we can indeed write b=ac, the element c is unique. If we take the example of 20/5, we know that the only way to write this is 20=(5)(4). Four is the unique number that we get when performing the division.
So what about zero? If we want to find an answer to 0/0, we need to find a number such that 0=0a. What kind of numbers fit the bill? One works, since (0)(1)=0, but two works as well. Actually, you will quite quickly realize that all numbers work. That’s because you’re multiplying by zero, which “collapses” every single number to zero when you multiply them together.
Now we’re faced with a bit of a dilemma. Which number should we choose to be the answer? Remember, we have all the freedom in the world to create our own definition of things! Let’s say we choose 0/0=5. Then, what happens if we consider (0/0)(0/0)? On the one hand, we know that each term in the parentheses is 5, so we should get a result of 25. However, if we do the multiplication in our usual way, we also get that (0/0)(0/0)=(0/0)=5. As such, we would conclude that 5=25. This is clearly not a very good number system, since any time I owe you twenty-five dollars, I’ll only give you back five. We know that those two numbers should be different, so it’s a bit of a concern when we manage to say that they are equal to each other.
The lesson we learn from this is that, while we do have the freedom to define concepts however we wish, we also want to be consistent. If we can transform five into twenty-five, it turns out that we can make basically any number we want equal to five. This is not a system in which the usual arithmetic rules apply. That’s not inherently bad, but we need to ask ourselves if the tradeoffs are worth it. Is it a good idea to have 0/0 being equal to any number we want as soon as we say it’s equal to a specific number, or is it better to leave it as undefined? As a community, mathematicians have obviously chosen the path of not defining terms like 0/0, and it’s precisely for this reason. Our common sense goes out the window once we start allowing these kinds of expressions.
In fact, you may have seen “pseudo-proofs” that 1=2, and these proofs rely on the fact that they are sneaking in a “divide by zero” operation at some point. Of course, these “proofs” won’t mention that, but that is the trick that is being played.
While it is easy to define these expressions to take on a certain value, there’s a reason why this isn’t done. We aren’t just lazy and don’t want to divide by zero. It’s that zero is a fundamentally different sort of number, and dividing by it doesn’t give us a unique answer.
However, it is much more fascinating to delve into why we don’t divide by zero, rather than simply forcing you to memorize this in class, don’t you think?<|endoftext|>
| 4.53125 |
508 |
# BerkswichCE Primary School
## Achieve, Believe and Care
Translate Translate
# Understanding times tables
Understanding Multiplication Tables
Times tables is more than memorising facts: in maths lessons, children will develop their understanding. Using apparatus, models, images, play, investigations and more, the children will gain a deep understanding of multiplications. This includes:
#### 1. Multiplication is repeated addition
In other words, 5 x 5 is the same as 5 + 5 + 5 + 5 + 5.
Children need experience of using apparatus and equipment, such as counters or multilink cubes and pictorial representations of objects which show this repeated addition.
#### 2. Multiplication is commutative
In other words, 4 x 5 is the same as 5 x 4.
Children build on their existing understanding using arrays such as the one above, turning the arrays around to show that you now have 5 groups of 4 and they will still total 20. This can then be linked to recalling multiplication facts; i.e., if they know their 5 times table as facts but not their 4 times table, they can use 4 x 5 to work out 5 x 4.
#### 4. Number families
In other words, there are related facts connected to each times table fact:
4 x 5 = 20, 5 x 4 = 20, 20 ÷ 5 = 4, 20 ÷ 4 = 5
Due to their commutative understanding, children will develop their understanding of whole number families. For many children this will need to be pointed out and discussed. Many children will be able to explore this in its abstract form (like in the list above), but this is usually investigated through arrays like the grid above.
From here it is only a short jump to understanding that any missing number can be worked out through knowledge of number families; for example, 4 x [ ] = 20 or [ ] ÷ 4 = 5.
There are other methods children can use to work out missing numbers, but our goal is to increase working memory in order to increase instant recall from long term memory. Being able to bounce around a number family will achieve that.
Please return to the menu page for more information about the stage your child may be at in their understanding. Teachers are always happy to discuss your child's stage and needs.
Top<|endoftext|>
| 4.53125 |
3,593 |
# 2.3: Subtracting Integers
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
In Section 1.2, we stated that “Subtraction is the opposite of addition.” Thus, to subtract 4 from 7, we walked seven units to the right on the number line, but then walked 4 units in the opposite direction (to the left), as shown in Figure $$\PageIndex{1}$$.
Thus, 7 − 4 = 3. The key phrase is “add the opposite.” Thus, the subtraction 7 − 4 becomes the addition 7 + (−4), which we would picture on the number line as shown in Figure $$\PageIndex{2}$$.
Figure $$\PageIndex{1}$$ and Figure $$\PageIndex{2}$$ provide ample evidence that the subtraction 7−4 is identical to the addition 7+(−4). Again, subtraction means “add the opposite.” That is, 7 − 4=7+(−4).
Defining Subtraction
Subtraction means “add the opposite.” That is, if a and b are any integers, then
$a − b = a + (−b).\nonumber$
Thus, for example, −123−150 = −123+(−150) and −57−(−91) = −57+91. In each case, subtraction means “add the opposite.” In the first case, subtracting 150 is the same as adding −150. In the second case, subtracting−91 is the same as adding 91.
Example 1
Find the differences: (a) 4 − 8, (b) −15 − 13, and (c) −117 − (−115).
Solution
In each case, subtraction means “add the opposite.”
a) Change the subtraction to addition with the phrase “subtraction means add the opposite.” That is, 4−8 = 4+(−8). We can now perform this addition on the number line.
Thus, 4 − 8=4+(−8) = −4.
b) First change the subtraction into addition by “adding the opposite.” That is, −15 − 13 = −15 + (−13). We can now use physical intuition to perform the addition. Start at the origin (zero), walk 15 units to the left, then an additional 13 units to the left, arriving at the answer −28. That is,
\begin{aligned} −15 − 13 & = −15 + (−13) \\ ~ & = −28. \end{aligned}\nonumber
c) First change the subtraction into addition by “adding the opposite.” That is, −117 − (−115) = −117 + 115. Using “Adding Two Integers with Unlike Signs” from Section 2.2, first subtract the smaller magnitude from the larger magnitude; that is, 117 − 115 = 2. Because −117 has the larger magnitude and its sign is negative, prefix a negative sign to the difference in magnitudes. Thus,
\begin{aligned} −117 − (−115) & = −117 + 115 \\ & = −2. \end{aligned}\nonumber
Exercise
Use each of the techniques in parts (a), (b), and (c) of Example 1 to evaluate the difference −11 − (−9).
−2
## Order of Operations
We will now apply the “Rules Guiding Order of Operations” from Section 1.5 to a number of example exercises.
Example 2
Simplify −5 − (−8) − 7.
Solution
We work from left to right, changing each subtraction by “adding the opposite.”
\begin{aligned} -5-(-8) -7=-5+8+(-7) ~ & \textcolor{red}{ \text{ Add the opposite of } -8, \text{ which is 8.}} \\ ~ & \textcolor{red}{ \text{ Add the opposite of 7, which is } -7.} \\ =3 +(-7) & \textcolor{red}{ \text{ Working left to right, } -5+8=3.} \\ =-4 ~ & \textcolor{red}{3 +(-7) = -4.} \end{aligned}\nonumber
Exercise
Simplify: −3 − (−9) − 11.
−5
Grouping symbols say “do me first.”
Example $$\PageIndex{1}$$
Simplify −2 − (−2 − 4).
Solution
Parenthetical expressions must be evaluated first.
\begin{aligned} -2(-2-4)=-2-(-2+(-4)) ~ & \textcolor{red}{ \text{ Simplify the parenthetical expression}} \\ ~ & \textcolor{red}{ \text{ first. Add the opposite of 4, which is } -4.} \\ = -2 -(-6) ~ & \textcolor{red}{ \text{ Inside the parentheses, } -2 + (-4) = -6.} \\ =-2 + 6 ~ & \textcolor{red}{ \text{ Subtracting a } -6 \text{ is the same as adding a 6.}} \\ =4 ~ & ~ \textcolor{red}{ \text{ Add: } -2 + 6 = 4.} \end{aligned}\nonumber
Exercise
Simplify: −3 − (−3 − 3).
3
## Change as a Difference
Suppose that when I leave my house in the early morning, the temperature outside is 40 Fahrenheit. Later in the day, the temperature measures 60◦ Fahrenheit. How do I measure the change in the temperature?
The Change in a Quantity
To measure the change in a quantity, always subtract the former measurement from the latter measurement. That is:
$\colorbox{cyan}{Change in a Quantity} = \colorbox{cyan}{Latter Measurement} - \colorbox{cyan}{Former Measurement}\nonumber$
Thus, to measure the change in temperature, I perform a subtraction as follows:
\begin{aligned} \colorbox{cyan}{Change in Temperature} & = \colorbox{cyan}{Latter Measurement} & - & \colorbox{cyan}{Former Measurement} \\ ~ & = 60^{ \circ} \text{F} & - & 40^{ \circ} \text{F} \\ ~ & = 20^{ \circ} \text{F} \end{aligned}\nonumber
Note that the positive answer is in accord with the fact that the temperature has increased.
Example 4
Suppose that in the afternoon, the temperature measures 65Fahrenheit, then late evening the temperature drops to 44 Fahrenheit. Find the change in temperature.
Solution
To measure the change in temperature, we must subtract the former measurement from the latter measurement.
\begin{aligned} \colorbox{cyan}{Change in Temperature} & = \colorbox{cyan}{Latter Measurement} & - & \colorbox{cyan}{Former Measurement} \\ ~ & = 44^{ \circ} \text{F} & - & 65^{ \circ} \text{F} \\ ~ & = -11^{ \circ} \text{F} \end{aligned}\nonumber
Note that the negative answer is in accord with the fact that the temperature has decreased. There has been a “change” of −11 Fahrenheit.
Exercise
Marianne awakes to a morning temperature of 54 Fahrenheit. A storm hits, dropping the temperature to 43 Fahrenheit. Find the change in temperature.
−11◦ Fahrenheit
Example 5
Sometimes a bar graph is not the most appropriate visualization for your data. For example, consider the bar graph in Figure $$\PageIndex{3}$$ depicting the Dow Industrial Average for seven consecutive days in March of 2009. Because the bars are of almost equal height, it is difficult to detect fluctuation or change in the Dow Industrial Average.
Let’s determine the change in the Dow Industrial average on a day-to-day basis. Remember to subtract the latter measurement minus the former (current day minus former day). This gives us the following changes.
Consecutive Days Change in Dow Industrial Average Sun-Mon 6900 - 7000 = -100 Mon-Tues 6800 - 6900 = -100 Tues-Wed 6800 - 6800 = 0 Wed-Thu 7000 - 6800 = 200 Thu-Fri 7100 - 7000 = 100 Fri-Sat 7200 - 7100 = 100
We will use the data in the table to construct a line graph. On the horizontal axis, we place the pairs of consecutive days (see Figure $$\PageIndex{4}$$). On the vertical axis we place the Change in the Industrial Dow Average. At each pair of days we plot a point at a height equal to the change in Dow Industrial Average as calculated in our table.
Note that the data as displayed by Figure $$\PageIndex{4}$$ more readily shows the changes in the Dow Industrial Average on a day-to-day basis. For example, it is now easy to pick the day that saw the greatest increase in the Dow (from Wednesday to Thursday, the Dow rose 200 points).
## Exercises
In Exercises 1-24, find the difference.
1. 16 − 20
2. 17 − 2
3. 10 − 12
4. 16 − 8
5. 14 − 11
6. 5 − 8
7. 7 − (−16)
8. 20 − (−10)
9. −4 − (−9)
10. −13 − (−3)
11. 8 − (−3)
12. 14 − (−20)
13. 2 − 11
14. 16 − 2
15. −8 − (−10)
16. −14 − (−2)
17. 13 − (−1)
18. 12 − (−13)
19. −4 − (−2)
20. −6 − (−8)
21. 7 − (−8)
22. 13 − (−14)
23. −3 − (−10)
24. −13 − (−9)
In Exercises 25-34, simplify the given expression.
25. 14 − 12 − 2
26. −19 − (−7) − 11
27. −20 − 11 − 18
28. 7 − (−13) − (−1)
29. 5 − (−10) − 20
30. −19 − 12 − (−8)
31. −14 − 12 − 19
32. −15 − 4 − (−6)
33. −11 − (−7) − (−6)
34. 5 − (−5) − (−14)
In Exercises 35-50, simplify the given expression.
35. −2 − (−6 − (−5))
36. 6 − (−14 − 9)
37. (−5 − (−8)) − (−3 − (−2))
38. (−6 − (−8)) − (−9 − 3)
39. (6 − (−9)) − (3 − (−6))
40. (−2 − (−3)) − (3 − (−6))
41. −1 − (10 − (−9))
42. 7 − (14 − (−8))
43. 3 − (−8 − 17)
44. 1 − (−1 − 4)
45. 13 − (16 − (−1))
46. −7 − (−3 − (−8))
47. (7 − (−8)) − (5 − (−2))
48. (6 − 5) − (7 − 3)
49. (6 − 4) − (−8 − 2)
50. (2 − (−6)) − (−9 − (−3))
51. The first recorded temperature is 42F. Four hours later, the second temperature is 65F. What is the change in temperature?
52. The first recorded temperature is 79F. Four hours later, the second temperature is 46F. What is the change in temperature?
53. The first recorded temperature is 30F. Four hours later, the second temperature is 51F. What is the change in temperature?
54. The first recorded temperature is 109F. Four hours later, the second temperature is 58F. What is the change in temperature?
55. Typical temperatures in Fairbanks, Alaska in January are −2 degrees Fahrenheit in the daytime and −19 degrees Fahrenheit at night. What is the change in temperature from day to night?
56. Typical summertime temperatures in Fairbanks, Alaska in July are 79 degrees Fahrenheit in the daytime and 53 degrees Fahrenheit at night. What is the change in temperature from day to night?
57. Communication. A submarine 1600 feet below sea level communicates with a pilot flying 22,500 feet in the air directly above the submarine. How far is the communique traveling?
58. Highest to Lowest. The highest spot on earth is on Mount Everest in Nepal-Tibet at 8,848 meters. The lowest point on the earth’s crust is the Mariana’s Trench in the North Pacific Ocean at 10,923 meters below sea level. What is the distance between the highest and the lowest points on earth? Wikipedia http://en.Wikipedia.org/wiki/Extremes_on_Earth
59. Lowest Elevation. The lowest point in North America is Death Valley, California at -282 feet. The lowest point on the entire earth’s landmass is on the shores of the Dead Sea along the Israel-Jordan border with an elevation of -1,371 feet. How much lower is the Dead Sea shore from Death Valley?
60. Exam Scores. Freida’s scores on her first seven mathematics exams are shown in the following bar chart. Calculate the differences between consecutive exams, then create a line graph of the differences on each pair of consecutive exams. Between which two pairs of consecutive exams did Freida show the most improvement?
1. −4
3. −2
5. 3
7. 23
9. 5 11. 11 13. −9 15. 2 17. 14 19. −2 21. 15 23. 7 25. 0 27. −49 29. −5 31. −45 33. 2 35. −1 37. 4 39. 6 41. −20 43. 28 45. −4 47. 8 49. 12 51. 23◦ F 53. 21◦ F 55. −17 degrees Fahrenheit 57. 24,100 feet 59. 1,089 feet lower
This page titled 2.3: Subtracting Integers is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Arnold.<|endoftext|>
| 4.40625 |
1,244 |
3
Q:
# What is $\frac{1}{5}$ as a decimal?
A) 0.333 B) 0.666 C) 0.200 D) 0.452
Explanation:
Here I converted the original fraction to a fraction with the denominator being 10, because decimals can be written in the tenths place.
Q:
2/3 as a decimal
A) 0.47 B) 6.77 C) 0.67 D) 0.07
Explanation:
To convert a fraction into decimal, just take the number on top which we called the numerator and divide it by the number at the bottom which we called the denominator.
2/3
How it comes :
÷ 0 bring down the decimal point.
20 ÷ 6 and carry the remainder 2 to make 20
20 ÷ 6 and carry the remainder 2 to make 20
20 ÷ 6 and carry the remainder 2 to make 20
etc...
This gives the recurring decimal 0.6666...
If we round off to two decimals , then it becomes 0.67.
Hence, decimal fraction of 2/3 = 0.67.
0 66
Q:
3/8 as a decimal
A) 1.375 B) 2.375 C) 0.375 D) 3.75
Explanation:
Expressing 3/8 as a decimal fraction:
We can convert 3/8 into a decimal fraction by just dividing 3 by 8 or making 3 into 8 parts.
Then, it is 3/8 = 0.375 as a decimal fraction.
Other Method::
3/8 as a Decimal :
In this we should convert the denominator into 10, 100, 1000, 10000, 100000,... to make it simple.
Now to make 8 into 1000, we multiply numerator and denominator with 125
=> (3 x 125)/(8 x 125) = 375/1000 = 0.375.
2 48
Q:
3/4 divided by 2 is
A) 3/2 B) 2/3 C) 8/3 D) 3/8
Explanation:
3/4 divided by 2
= 3/4/2
= 3/4x2
= 3/8.
Hence, 3/4 divided by 2 is equal to 3/8.
1 59
Q:
1/8 as a decimal?
A) 0.125 B) 0.225 C) 0.325 D) 0.425
Explanation:
1/8 to convert into decimal,
Find an equivalent fraction with a denominator of a power of 10.
In this case, we will use 1000.
This can now be written as a decimal with 3 decimal places (because we have thousandths.
i.e, 0.125
Hence,
2 239
Q:
125 over 1000 in Simplest Form?
A) 2/9 B) 1/8 C) 3/8 D) 4/7
Explanation:
125 over 1000 in Simplest Form means $\frac{\mathbf{125}}{\mathbf{1000}}$ in its simple fraction form.
Now, to get the simplest form of 125/1000, find the HCF or GCD of both numerator and denominator i.e, 125 and 1000.
HCF of 125, 1000 = 125
Then, divide both numerator and denominator by 125
i.e,
Hence, $\frac{\mathbf{1}}{\mathbf{8}}$ is the simplest form of 125 over 1000.
3 252
Q:
$\frac{\mathbf{1}}{\mathbf{8}}$ as a decimal?
A) 0.375 B) 0.247 C) 0.214 D) 0.125
Explanation:
Here $\frac{\mathbf{1}}{\mathbf{8}}$ in decimal means 8th part in 1.
We know that $\frac{\mathbf{1}}{\mathbf{10}}$ = 0.1
Then,
=>
3 335
Q:
Express 0.875 and 0.375 as a Fraction.
A) 3/11 & 7/11 B) 7/9 & 9/13 C) 7/8 & 3/8 D) 7/9 & 9/11
Explanation:
Given 0.875 and 0.375
0.875 ::
After decimal point there are 3 digits 875
Now, mutiply and divide 0.875 with 1000
Similarly,
0.375 ::
After decimal point there are 3 digits 375
Now, mutiply and divide 0.375 with 1000
4 546
Q:
Express 3/8 and 1/10 as a Decimals?
A) 1.375 & 1 B) 0.375 & 1 C) 13.75 & 1.1 D) 0.375 & 0.1
Explanation:
3/8 as a Decimal :
In this we should convert the denominator into 10, 100, 1000, 10000, 100000,... to make it simple.
Now to make 8 into 1000, we multiply numerator and denominator with 125
=>
1/10 as a Decimal :
1 divided into 10 parts of each 0.1.
Therefore, 3/8 = 0.375 and 1/10 = 0.1 in decimals.<|endoftext|>
| 4.5625 |
365 |
In Common is a quick game which can be used as an ice breaker or warmer or perhaps at the end of a lesson to fill 5 minutes. It’s easy to set up, requires little or no preparation and is suitable for all ages and levels of class.
Running the Game
Simply divide the class into groups of 4 or 5. Then ask each group to find 3 things which all the members of the group have in common.
This will involve each member of the group asking and answering questions about themself and the others. Perhaps they all have a younger brother or they have all been on holiday to Malaga – who knows but it’s all about finding out about each other.
After a few minutes get each group to tell the rest of the class what they all have in common. The most interesting and inventive group wins. (This stops them from coming up with simplistic and obvious things like all having two legs or learning English, etc).
Now rearrange the students into new groups and ask them to find another thing they all have in common!
Further work with Venn Diagrams
The example here shows three students: Karla, Kyriaki, and Klaus and the subject was restricted to activities they enjoy. They have worked together to find one activity they all have in common (playing the bagpipes); activities only two of them have in common (e.g. Kyriaki and Karla enjoy playing football but Klaus doesn’t) and then activities only one person likes (e.g. Karla likes rock climbing but the others don’t).
It’s easy to prepare a blank version of this Venn diagram (or describe it to your class so they have to draw it) and have them complete it like this.<|endoftext|>
| 3.921875 |
1,165 |
1. ## integral
evalute: $\displaystyle \int \frac{x + 3}{x^2 + 6 x}dx$
2. Originally Posted by viet
evalute: $\displaystyle \int \frac{x + 3}{x^2 + 6 x}dx$
use substitution. u = x^2 + 6x
3. Originally Posted by viet
evalute: $\displaystyle \int \frac{x + 3}{x^2 + 6 x}dx$
Or you can use partial fractions:
$\displaystyle x^2 + 6x = x(x + 6)$
So let's find an A and B such that
$\displaystyle \frac{x + 3}{x^2 + 6x} = \frac{A}{x} + \frac{B}{x + 6}$
Well...
$\displaystyle \frac{A}{x} + \frac{B}{x + 6} = \frac{A(x + 6) + Bx}{x^2 + 6x} = \frac{(A + B)x + 6A}{x^2 + 6x}$
Comparing the two expressions gives me
A + B = 1
6A = 3
Giving me that $\displaystyle A = B = \frac{1}{2}$.
So
$\displaystyle \int dx \frac{x + 3}{x^2 + 6x} = \int dx \frac{1/2}{x} + \int dx \frac{1/2}{x + 6}$
Can you finish this?
-Dan
4. heres what i've done:
$\displaystyle \int \frac{x + 3}{x^2 + 6 x}dx =$
$\displaystyle \int \frac{x}{x^2 + 6 x}dx + \int \frac{3}{x^2 + 6 x}dx$
$\displaystyle u = x^2+6x, du = 2x+6$
$\displaystyle ln(x+6)+ \int \frac{3}{x^2 + 6 x}dx$
is this right sofar?
5. Originally Posted by topsquark
Or you can use partial fractions:
$\displaystyle x^2 + 6x = x(x + 6)$
So let's find an A and B such that
$\displaystyle \frac{x + 3}{x^2 + 6x} = \frac{A}{x} + \frac{B}{x + 6}$
Well...
$\displaystyle \frac{A}{x} + \frac{B}{x + 6} = \frac{A(x + 6) + Bx}{x^2 + 6x} = \frac{(A + B)x + 6A}{x^2 + 6x}$
Comparing the two expressions gives me
A + B = 1
6A = 3
Giving me that $\displaystyle A = B = \frac{1}{2}$.
So
$\displaystyle \int dx \frac{x + 3}{x^2 + 6x} = \int dx \frac{1/2}{x} + \int dx \frac{1/2}{x + 6}$
Can you finish this?
-Dan
That's nice! never thought of doing partial fractions. substitution was the way i immediately saw to do it. and since it's one of the easiest integration techniques, i didn't bother looking for another
$\displaystyle \int \frac {x + 3}{x^2 + 6x}dx$
Let $\displaystyle u = x^2 + 6x$
$\displaystyle \Rightarrow du = 2x + 6dx = 2(x + 3) dx$
$\displaystyle \Rightarrow \frac {1}{2}du = (x + 3)dx$
So our integral becomes:
$\displaystyle \frac {1}{2} \int \frac {1}{u}du$
6. Originally Posted by viet
heres what i've done:
$\displaystyle \int \frac{x + 3}{x^2 + 6 x}dx =$
$\displaystyle \int \frac{x}{x^2 + 6 x}dx + \int \frac{3}{x^2 + 6 x}dx$
$\displaystyle u = x^2+6x, du = 2x+6$
$\displaystyle ln(x+6)+ \int \frac{3}{x^2 + 6 x}dx$
is this right sofar?
dont think so. how did you get $\displaystyle \ln ( x + 6)$?
you didnt split the integral incorrectly, but i think you're getting yourself into more trouble than it's worth. use topsquark's or my method. do you understand them?
7. yeah i understand them, needed more practice on it. thanks for the help topsquark & Jhevon
8. $\displaystyle \int {\frac{{x + 3}} {{x^2 + 6x}}~dx} = \frac{1} {2}\int {\frac{{\left( {x^2 + 6x} \right)^\prime }} {{x^2 + 6x}}~dx} = \frac{1} {2}\ln (x^2 + 6x) + k$<|endoftext|>
| 4.40625 |
1,390 |
# Cross-multiplication
Wikipedia view on Wikipedia
In mathematics, specifically in elementary arithmetic and elementary algebra, given an equation between two fractions or rational expressions, one can cross-multiply to simplify the equation or determine the value of a variable.
The method is also occasionally known as the "cross your heart" method because a heart can be drawn to remember which things to multiply together and the lines resemble a heart outline.
Given an equation like:
${\displaystyle {\frac }={\frac }}$
(where b and d are not zero), one can cross-multiply to get:
${\displaystyle ad=bc\qquad \mathrm \qquad a={\frac }.}$
In Euclidean geometry the same calculation can be achieved by considering the ratios as those of similar triangles.
## Procedure
In practice, the method of cross-multiplying means that we multiply the numerator of each (or one) side by the denominator of the other side, effectively crossing the terms over.
${\displaystyle {\frac }\nwarrow {\frac }\quad {\frac }\nearrow {\frac }.}$
The mathematical justification for the method is from the following longer mathematical procedure. If we start with the basic equation:
${\displaystyle {\frac }={\frac }}$
we can multiply the terms on each side by the same number and the terms will remain equal. Therefore, if we multiply the fraction on each side by the product of the denominators of both sides—bd—we get:
${\displaystyle {\frac }\times bd={\frac }\times bd.}$
We can reduce the fractions to lowest terms by noting that the two occurrences of ${\displaystyle b}$ on the left-hand side cancel, as do the two occurrences of d on the right-hand side, leaving:
${\displaystyle ad=bc}$
and we can divide both sides of the equation by any of the elements—in this case we will use d—getting:
${\displaystyle a={\frac }.}$
Another justification of cross-multiplication is as follows. Starting with the given equation:
${\displaystyle {\frac }={\frac }}$
multiply by d/d = 1 on the left and by b/b = 1 on the right, getting:
${\displaystyle {\frac }\times {\frac }={\frac }\times {\frac }}$
and so:
${\displaystyle {\frac }={\frac }.}$
Cancel the common denominator bd = db, leaving:
${\displaystyle ad=cb.}$
Each step in these procedures is based on a single, fundamental property of equations. Cross-multiplication is a shortcut, an easily understandable procedure that can be taught to students.
## Use
This is a common procedure in mathematics, used to reduce fractions or calculate a value for a given variable in a fraction. If we have an equation like this, where x is a variable we are interested in solving for:
${\displaystyle {\frac }={\frac }}$
we can use cross multiplication to determine that:
${\displaystyle x={\frac }.}$
For example, suppose we want to know how far a car will travel in 7 hours, if we know that its speed is constant and that it already travelled 90 miles in the last 3 hours. Converting the word problem into ratios we get
${\displaystyle {\frac }}={\frac } }}.}$
Cross-multiplying yields:
${\displaystyle x={\frac \times 90\ \mathrm } }}}$
and so:
${\displaystyle x=210\ \mathrm .}$
Note that even simple equations like this:
${\displaystyle a={\frac }}$
are solved using cross multiplication, since the missing b term is implicitly equal to 1:
${\displaystyle {\frac }={\frac }.}$
Any equation containing fractions or rational expressions can be simplified by multiplying both sides by the least common denominator. This step is called clearing fractions.
## Rule of Three
The Rule of Three[1] was an historical shorthand version for a particular form of cross-multiplication that could be taught to students by rote. It was considered the height of Colonial math education[2] and still figures in the French national curriculum for secondary education.[3]
For an equation of the form:
${\displaystyle {\frac }={\frac }}$
where the variable to be evaluated is in the right-hand denominator, the Rule of Three states that:
${\displaystyle x={\frac }.}$
In this context, a is referred to as the extreme of the proportion, and b and c are called the means.
This rule was already known to Chinese mathematicians prior to the 7th century CE,[4] though it was not used in Europe until much later.
The Rule of Three gained notoriety[citation needed] for being particularly difficult to explain. Cocker's Arithmetick, the premier textbook in the 17th century, introduces its discussion of the Rule of Three[5] with the problem, "If 4 Yards of Cloth cost 12 Shillings, what will 6 Yards cost at that Rate?" The Rule of Three gives the answer to this problem directly; whereas in modern arithmetic, we would solve it by introducing a variable x to stand for the cost of 6 yards of cloth, writing down the equation:
${\displaystyle {\frac } }}={\frac }}}$
and then using cross-multiplication to calculate x:
${\displaystyle x={\frac \times 6\ \mathrm } }}=18\ \mathrm .}$
An anonymous manuscript dated 1570[6] said: "Multiplication is vexation, / Division is as bad; / The Rule of three doth puzzle me, / And Practice drives me mad."
### Double Rule of Three
An extension to the Rule of Three was the Double Rule of Three, which involved finding an unknown value where five rather than three other values are known.
An example of such a problem might be If 6 builders can build 8 houses in 100 days, how many days would it take 10 builders to build 20 houses at the same rate? and this can be set up as
${\displaystyle {\frac {\frac } }} }}={\frac {\frac }} }}}$
which, with cross-multiplication twice, gives
${\displaystyle x={\frac \times 100\ \mathrm \times 6\ \mathrm } \times 10\ \mathrm }}=150\ \mathrm }$
Lewis Carroll's The Mad Gardener's Song includes the lines "He thought he saw a Garden-Door / That opened with a key: / He looked again, and found it was / A double Rule of Three"[7]<|endoftext|>
| 4.8125 |
1,880 |
# How to Calculate the Area of a Rectangle
Author Info
Updated: October 22, 2019
A rectangle is a quadrilateral[1] with two sides of equal length and two sides of equal width that contains four right angles. To find the area of a rectangle, all you have to do is multiply its length with its width. If you want to know how to find the area of a rectangle, just follow these easy steps.
### Method 1 of 3: Understand the Basics of a Rectangle
1. 1
Understand the rectangle. The rectangle is a quadrilateral, which means it has four sides.[2] Its opposite sides are equal in length, so the sides along its length are equal, and the sides along its width are equal as well. If one side of the rectangle is 10, for example, then the opposite side's length will also be 10.
• Also, every square is a rectangle, but not all rectangles are squares. So treat squares like rectangles in terms of finding its area.
2. 2
Learn the equation for finding the area of a rectangle. The equation for finding the area of a rectangle is simply A = L * W. This means that the area is equal to the length of the rectangle times its width.[3]
### Method 2 of 3: Find the Area of a Rectangle
1. 1
Find the length of the rectangle. In most cases, you will be given the length, but if not, you can find it using a ruler.[4]
• Note that the double hash marks on the long sides of the rectangle mean that the lengths of the two sides are the same.
2. 2
Find the width of the rectangle. Use the same methods to find it.
• Note that the single hash marks on the wide sides of the rectangle mean that the two widths have equal length.
3. 3
Write the length and width next to each other. In this example, the length is 5 cm and the width is 4 cm.
4. 4
Multiply the length times the width. Your length is 5 cm and your width is 4 cm, so you should plug them into the equation A = L * W to find the area.[5]
• A = 4 cm * 5 cm
• A = 20 cm^2
5. 5
• You can write your final answer in one of two ways: either 20 cm.sq. or 20 cm^2.
### Method 3 of 3: Find the Area if You Only Know the Length of One Side and the Diagonal
1. 1
Understand the Pythagorean theorem. The Pythagorean theorem is a formula for finding the third side of a right triangle if you know the value of two of the sides. You can use it to find the hypotenuse of a triangle, which is its longest side, or its length or width, which meet at a right angle.[7]
• Since a rectangle is comprised of four right angles, the diagonal that cuts through the shape will create a right triangle, so you can apply the Pythagorean theorem.
• The theorem is: a^2 + b^2 = c^2, where a and b are sides of the triangle and c is the hypotenuse, or longest side.[8]
2. 2
Use the Pythagorean theorem to solve for the other side of the triangle. Let's say that you have a rectangle with a side of 6 cm and a diagonal of 10 cm. Use 6 cm for one side, use b for the other side, and take 10 cm as your hypotenuse. Now just substitute your known quantities into the Pythagorean theorem and solve. Here's how to do it:[9]
• Ex: 6^2 + b^2 = 10^2
• 36 + b^2 = 100
• b^2 = 100 - 36
• b^2 = 64
• square root (b) = square root (64)
• b = 8
• The length of the other side of the triangle, which is also the other side of the rectangle, is 8 cm.
3. 3
Multiply the length times the width. Now that you've used the Pythagorean theorem to find the length and width of the rectangle, all you have to do is multiply them.[10]
• Ex: 6 cm * 8 cm = 48 cm^2
4. 4
## Community Q&A
Search
• Question
How do I find the area in meters?
You cannot find an area in meters, because area is expressed in square meters. If the values of your rectangle sides are not in meters, then you must first convert those units to meters, and then multiply the converted values to get the area in square meters.
• Question
If the problem says length is 105 breadth is 81 and other said of length is 103 and breadth is 53, how do I find the area of the rectangle?
A rectangle has 2 sides of equal length and 2 sides of equal width (breadth). By definition, the object in which you are describing is not a rectangle. The object you have is an irregular quadrilateral.
• Question
How do I find the area of a rectangle when each side is a different length?
It is not a rectangle if each side is a different length. It is either an irregular shape or a trapezoid. There is no formula for finding the area of an irregular shape. The area of a trapezoid is found by multiplying its height by the average of its bases.
• Question
How can I find the area of a rectangle with only one number?
Atheia
If both the sides are labeled with that same number, then it is a square. Square the number to get the area. If it's not a square, you can't, unless there's more information available than you're letting on (like an equation to calculate the length of the other side).
• Question
How can I find the area of a rectangle if I only know the diagonal?
The area of a rectangle is calculated by multiplying its width times its length. You'll need this info, not the diagonal, to solve the problem.
• Question
How can I find the length of a rectangle if given only the area and width?
The area of a rectangle is equal to its length multiplied by its width. This equation is represented by A=L*W. If you have only the area and width, you can use the same equation to solve for the area. For example, if the area is 60 and the width is 5, your equation will look like this: 60 = x*5. Divide 60 by 5 to find x, or the length. The answer here is 12.
• Question
How do I calculate the area of an irregular rectangle?
Onebluethinker
There is no such thing as an irregular rectangle. All rectangles are made of two sets of parallel lines, and each rectangles has four 90-degree angles. The area is calculated by multiplying the length of the rectangle by its width.
• Question
If I am only given the diagonal of a rectangle how can I find the area?.
You can't with just that information.
• Question
How do I find the area of a quadrilateral?
Donagan
There is no area formula for a non-specific quadrilateral. There are commonly available area formulas for a square, rectangle, parallelogram, rhombus, and trapezoid.
• Question
How can I find the length and width of a rectangle with only the perimeter and area?
Donagan
This is a "two equations in two unknowns" situation. Write two equations, one for the area and one for the perimeter, both in terms of length and width. Solve either equation for one of the unknowns. Plug that value back into either equation, which will give you the value of the other unknown.
200 characters left
## Tips
• All squares are rectangles. However, not all rectangles are squares.
Thanks!
Thanks!
wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 26 people, some anonymous, worked to edit and improve it over time. Together, they cited 10 references. This article has also been viewed 433,678 times.
Co-authors: 26
Updated: October 22, 2019
Views: 433,678
Article SummaryX
To quickly calculate the area of a rectangle, find the length of the base. Then, multiply the base by the height of the rectangle to get the area. For example, a rectangle with a base of 6 and a height of 9 has an area of 54. Be sure to include the units of the measurements in your answer. If you need to find the area if you only know the area or the length of 1 side and a diagonal, keep reading the article!<|endoftext|>
| 4.78125 |
175 |
How do you solve 6z - 10- z + 5= 0?
Jun 24, 2017
z = 1
Explanation:
1. Combine like terms.
$\textcolor{red}{6 z} \textcolor{b l u e}{- 10} \textcolor{red}{- z} \textcolor{b l u e}{+ 5} = 0$
$5 z - 5 = 0$
2. Add 5 to both sides to isolate the variable ($z$).
$5 z - 5 + 5 = 0 + 5$
$5 z = 5$
3. Divide both sides by 5 to find $z$.
$\frac{5 z}{5} = \frac{5}{5}$
$z = 1$<|endoftext|>
| 4.4375 |
305 |
Perimeter of a trapezoid
("trapezium" in British usage)
The total distance around the outside of a
Drag any orange dot to resize the trapezoid. The perimeter is calculated as you drag.
How to find the perimeter of a trapezoid
Like any polygon, the perimeter is the total distance around the outside,
which can be found by adding together the length of each side.
Or as a formula:
perimeter = a+b+c+d
a,b,c,d are the lengths of each side
In the figure above, drag any orange dot to resize the trapezoid. From the side lengths shown, calculate the perimeter
and verify your result matches the formula at the top of the diagram.
In coordinate geometry, if you know the coordinates of the four vertices,
you can calculate various properties of it, including the area and perimeter.
For more on this, see Trapezoid area and perimeter (Coordinate Geometry)
Things to try
- In the figure above, click on "hide details".
- Drag the four orange dots to reshape the trapezoid.
- Calculate the perimeter.
- Click "Show details" to check your answer.
Other polygon topics
Types of polygon
Area of various polygon types
Perimeter of various polygon types
Angles associated with polygons
(C) 2011 Copyright Math Open Reference. All rights reserved<|endoftext|>
| 3.921875 |
1,974 |
# Open Library - открытая библиотека учебной информации
## Категории
### Образование Chapter 04.06 просмотров - 249
GaussianElimination
After reading this chapter, you should be able to:
1. Solve a set of simultaneous linear equations using Naïve Gauss Elimination,
1. Learn the pitfalls of Naïve Gauss Elimination Method,
2. Understand the effect of round off error on a solving set of linear equation by Naïve Gauss Elimination Method,
3. Learn how to modify Naïve Gauss Elimination method to Gaussian Elimination with Partial Pivoting Method to avoid pitfalls of the former method,
4. Find the determinant of a square matrix using Gaussian Elimination,
5. Understand the relationship between determinant of co-efficient matrix and the solution of simultaneous linear equations.
How are a set of equations solved numerically?
One of the most popular techniques for solving simultaneous linear equations is the Gaussianelimination method. The approach is designed to solve a general set of n equations and n unknowns
. .
. .
. .
Gaussianelimination consists of two steps
1. Forward Elimination of Unknowns: In this step, the unknown is eliminated in each equation starting with the first equation. This way, the equations are “reduced” to one equation and one unknown in each equation.
2. Back Substitution: In this step, starting from the last equation, each of the unknowns is found.
Forward Elimination of Unknowns:
In the first step of forward elimination, the first unknown, x1 is eliminated from all rows below the first row. The first equation is selected as the pivot equation to eliminate x1. So, to eliminate x1 in the second equation, one divides the first equation by a11 (hence called the pivot element) and then multiply it by a21. That is, same as multiplying the first equation by a21/ a11 to give
Now, this equation can be subtracted from the second equation to give
or
where
This procedure of eliminating , is now repeated for the third equation to the nth equation to reduce the set of equations as
. . .
. . .
. . .
This is the end of the first step of forward elimination. Now for the second step of forward elimination, we start with the second equation as the pivot equation and as the pivot element. So, to eliminate x2 in the third equation, one divides the second equation by (the pivot element) and then multiply it by . That is, same as multiplying the second equation by and subtracting from the third equation. This makes the coefficient of x2 zero in the third equation. The same procedure is now repeated for the fourth equation till the nth equation to give
. .
. .
. .
The next steps of forward elimination are conducted by using the third equation as a pivot equation and so on. That is, there will be a total of (n-1) steps of forward elimination. At the end of (n-1) steps of forward elimination, we get a set of equations that look like
. .
. .
. .
Back Substitution:
Now the equations are solved starting from the last equation as it has only one unknown.
Then the second last equation, that is the (n-1)th equation, has two unknowns - xn and xn-1, but xn is already known. This reduces the (n-1)th equation also to one unknown. Back substitution hence can be represented for all equations by the formula
for i = n – 1, n – 2,…,
and
## Читайте также
• - Using the words from task I try to make up the game described in the chapter.
TASKS CAMP LAURENCE CHAPTER 12 Translate the following passage into English. Compare it with the original. Dwell upon the results of the experiment. - Я поняла, Марми! - воскликнула Мег. - Ведь ты нарочно ушла из дома, чтобы посмотреть, как мы управимся без тебя и Ханны. - Ну да! Я хотела, чтобы вы поняли, насколько уют в нашем... [читать подробенее]
• - Reread the description of the girls in the chapter and present each of them
Explain the words in bold type. Find in the text and translate the following passages: 1. “We shouldn't enjoy ourselves half so much as we do now.” - So Meg went down, wearing an injured look, and wasn't at all agreeable at breakfast time. 2. Jo gave her sister an encouraging pat on the shoulder… - Other friends told the Marches that they had lost all chance of being remembered in the rich old lady's will, but the unworldly Marches only said… 3. The old lady wouldn't speak to... [читать подробенее]
• - Pick up some more examples from Chapters 17 and 18 to prove that critical situations change us all out of recognition and help to distinguish between real and false values.
Tasks Find in the text and translate into Russian the following words and expressions: to do one’s best to devote oneself to… to feel very anxious to keep house to miss smb. irksome to keep the watch despondently as the winter night began to wane. Recall the situations from the story suggested by the following sentences: Meg stayed at home, lest she should infect the Kings. … Beth was very patient, and bore her pain uncomplainingly as long as she could... [читать подробенее]
• - Chapter III
The Financier, by Theodore Dreiser It was in his thirteenth year that young Cowperwood entered into his first business venture. Walking along Front Street one day, a street of importing and wholesale establishments, he saw an auctioneer’s flag hanging out before a wholesale grocery and from the interior came the auctioneer’s voice: “What am I bid for this exceptional lot of Java coffee, twenty-two bags all told, which is now selling in the market for seven dollars and thirty-two... [читать подробенее]
• - Chapter XII THE COMPOSITE SENTENCE
І Incombination of sentences into larger units we may observe two different types of grammatical relationship based upon relative position and interaction of sentences. These are co-ordination and subordination. This classification remains the prevalent scheme of the structural classification of sentences in the grammars of all types in various languages. A very important syntactic concept developed along with this classification is the concept of syndeton and asyndeton. Sentences... [читать подробенее]
• - Chapter VIII THE ADVERB
Adverbs make up a rather complicated group of words varying widely in form and distribution. Considered in their morphemic structure, adverbs may be classified in eight groups. 1—2. The two largest groups are those formed from derived and base adjectives by adding the suffix -ly, e. g.: hopefully, physically, strangely, falsely, etc. 3. The third group consists of those that are formed by means of the derivational prefix -a (phonemically [э]) to nouns, adjectives or verbs. Of about... [читать подробенее]
• - Chapter VII THE PRONOUN
PERSONAL PRONOUNS In the category of person English makes distinction between three classes of personal pronouns denoting respectively the person (s) speaking (first person); the person (s) spoken to (second person) another person(s) or thing(s) — third person. Person distinctions are naturally closely related to the category of number. There is no formal distinction of persons in plural, e. g.: we speak, you speak, they speak. There is no distinction of number in the 1st and 2nd persons... [читать подробенее]
• - Chapter V THE VERB
Revision Material 1. Comment on analytical and inflected form of comparison. 2. Give comments on the classification of adjectives in terms of meaning. 3. Comment on the distinction between base adjectives and derived adjectives. 4. Illustrate the statement that relative adjectives can develop qualitative meanings. Give examples of such metaphoric extension. 5. Comment on the noticeable change going on in present-day in the formation of the comparative and superlative of dissyllabic... [читать подробенее]
• - Chapter IV THE ADJECTIVE
An adjectiveis a word which expresses the attributes of substances (good, young, easy, soft, loud, hard, wooden, flaxen). As a class of lexical words adjectives are identified by their ability to fill the position between noun-determiner and noun and the position after a copula-verb and a qualifier. Considered in meaning, adjectives fall into two large groups: a) qualitative adjectives, b) relative adjectives. Qualitative adjectivesdenote qualities of size, shape, colour, etc. which an... [читать подробенее]
• - Chapter HI THE NOUN
The lexico-grammatical meaning of the noun is denoting "substance" In Modern English the marked formal characteristics of nouns are as follows: nouns have inflections for number and case, they may be associated with the definite or indefinite article. There is no grammatical gender in Modern English 1. The noun does not possess any special gender forms, neither does the accompanying adjective, pronoun or article indicate any gender agreement with the head noun. Unlike many... [читать подробенее]<|endoftext|>
| 4.46875 |
713 |
# Domain of a Function
The functions introduction page explains how the domain of a function, is the values that go in to the function.
Sometimes, there needs to be restrictions on the numbers that can be in the domain of a specific function.
Often, this means that some values will have to be excluded from a function domain.
Though before looking at some basic domain of a function examples though, it helps to get familiar with some Interval Notation first.
## Domain of a Function, Examples
(1.1)
f(x) = 3x + 1
This function is fine to work for any value of x.
So for the domain we can say x or (-∞ , ∞).
(1.2)
f(x) = x²
Again like (1.1), this function is fine to work for any x values.
The domain is x or (-∞ , ∞).
(1.3)
Now here we can’t have values of x = 3 in the domain of f(x).
This would result in a division of 0. Which can't be done.
We can say x ∈ , x 3. Meaning x is any real number, but not 3.
In the alternative notation, we have 2 cases here.
(- , 3) and ( 3 , ∞).
We can combine them to specify the domain of a function.
In Set Notation, there is the union symbol U.
This symbol is used to refer to one group OR another group.
For example to represent group A or B, this would be A U B.
This U can be used for interval notation also.
So (- , 3) and ( 3 , ∞) as a domain can be put together as:
(- , 3) ( 3 , ∞).
Which in plain English, means that the domain can be a number less than 3, OR larger than 3, but NOT 3 itself.
(1.4)
Again like with (1.3), we can’t have the divisor equal to 0.
x² - 5x + 6 ≠ 0
Factor:
( x - 3 )( x - 2 ) 0
x ≠ 3 , x ≠ 2
Domain: x x ≠ 2, x ≠ 3
Or (- , 2) U (2 , 3) U ( 3 , ∞)
(1.5)
When dealing with the real numbers. We can’t have a negative number inside a square root.
x 4 will have to be larger or equal to 0.
x 4 > 0 => x > 4
Domain: x > 4 or [4 , ∞)
(1.6)
Here, not only does the number inside the square root have to not be negative.
As it’s the denominator, it also can’t equal 0.
So x 4 has to be greater than 0.
x 4 > 0 => x > 4
Domain: x > 4 or (4 , ∞)
› › Domain of a Function<|endoftext|>
| 4.6875 |
2,376 |
As is generally known, both the velocity of rotation and the centrifugal force at the various points of a rotating object are proportional to the distance from its center of rotation, the axis (Figure 83); i.e., the velocity is that much greater, the further the point in question is distant from the axis and that much less, the closer it is to the axis; it is equal to zero on the theoretical axis of rotation itself.
Figure 83. Velocity of rotation and centrifugal force on a rotating object.
Key: 1. Centrifugal force; 2. Velocity of rotation; 3. Distance from the center of rotation; 4. Center of rotation; 5. Rotating object; 6. Axis of rotation.
Accordingly, the rotating part of the space station must be structured in such a manner that its air lock and the cable connections in the center of the entire structure are in the axis of rotation because the least motion exists at that point, and that those parts, in which a gravitational effect is to be produced by centrifugal force, are distant from the axis on the perimeter because the centrifugal force is the strongest at that point.
Figure 84. The Habitat Wheel. Left: Axial cross section. Right: View of the side constantly facing the sun, without a concave mirror, partially in cross section.
Key: 1. Wheel rim; 2. Well of the staircase; 3. Elevator shaft; 4. Axial segment; 5. Circular corridor; 6. Turnable air lock; 7. Elevator; 8. Bull's eyes with mirrors; 9. Condenser pipes; 10. Evaporation tube; 11. Bull's eye (window); 12. Cable connection.
These conditions are, however, best fulfilled when the station is laid out in the shape of a large wheel as previously indicated (Figures 84, 89 and 90): the rim of the wheel is composed out of cells and has the form of a ring braced by wire spokes towards the axis. Its interior is separated into individual rooms by partitions; all rooms are accessible from a wide corridor going around the entire station. There are individual rooms, larger sleeping bays, work and study areas, mess hall, laboratory, workshop, dark room, etc., as well as the usual utility areas, such as a kitchen, bath room, laundry room and similar areas. All rooms are furnished with modern day comforts; even cold and warm water lines are available. In general, the rooms are similar to those of a modern ship. They can be furnished just like on Earth because an almost normal, terrestrial gravitational state exists in these rooms.
Figure 85. Directional relationships in the habitat wheel.
Key: 1. Direction of the centrifugal force, that is, of apparent gravity; 2. Everything vertical is tilted instead of parallel; 3. "Lowest" region; 4. Partition; 5. Down; 6. Up; 7. Bathtub; 8. The water level is curved instead of straight (flat); 9. Vertical direction; 10. Vertical; 11. Axis of rotation (center) of the habitat wheel; 12. "Highest" point.
However, to create this gravitational state, the entire station, assuming a diameter of 30 meters, for example, must rotate in such a manner that it performs a complete rotation in about 8 seconds, thus producing a centrifugal force in the rim of the wheel that is just as large as the gravitational force on the Earth's surface.
While the force of gravity is directed towards the center of mass, the centrifugal force, on the other hand, is directed away from the center. Therefore, "vertical" in the habitat wheel means the reverse of on Earth: the radial direction from the center (from the axis of rotation) directed outward (Figure 85). Accordingly, "down" now points towards the perimeter and at the same time to the "lowest" part, while "up" now points towards the axis and at the same time to the "highest" point of this manmade celestial body. Taking its smallness into account, the radial orientation of the vertical direction, which in most cases is irrelevant on the Earth due to its size, now clearly becomes evident in the space station. The consequence of this is that all "vertical" directions (such as those for human beings standing erect, the partitions of the rooms, etc.) are now convergent instead of parallel to one another, and everything "horizontal" (e.g., water surface of the bathtub) appears curved instead of flat (see Figure 85).
A further peculiarity is the fact that both the velocity of rotation and the centrifugal force, as a result of their decrease towards the center of rotation, are somewhat less at the head of a person standing in the habitat wheel than at his feet (by approximately 1/9 for a wheel diameter of 30 meters) (Figure 83). The difference in the centrifugal forces should hardly be noticeable, while that of the velocities of rotation should be noticeable to some degree, especially when performing up and down (i.e., radial) movements, such as lifting a hand, sitting down, etc.
Figure 86. a) Top view onto the external door of the rotating air lock of the habitat wheel. b) Axial cross section through the rotating air lock of the habitat wheel.
(See Figure 84 and the text.)
The ball bearings are designed in such a manner that they allow movement in the direction of the axis through which closing and/or releasing is possible of the external air seal which connects the air lock airtight to the inside of the habitat wheel when the inside door is open.
Key: 1. Rotation of the habitat wheel; 2. Rotation of the air lock; 3. Axial segment; 4. Inside door; 5. To the air pump; 6. Air intake valve; 7. External air seal; 8. Motor pinion gear; 9. Gear on the rotor of the lock; 10. Outside door; 11. Ball bearing; 12. Rotating air lock, movement in the axial direction.
However, all of these phenomena make themselves felt that much less, the larger the diameter of the wheel. In the previously selected case (30 meters in diameter), only a slight effect would be perceptible.
Since the equipment for connecting to the outside is installed in near the axis (because at that point the least motion exists!), the axial segment forms a kind of "entrance hall" of the entire station. This segment has a cylindrical shape. At its ends (near those points where it is penetrated by the theoretical axis of rotation), the air lock is positioned on one side and the cable connection on the other side (Figure 84, S and K).
The air lock is made rotatable in order to ease the transition between the rotational movement of the habitat wheel and the state of rest of outer space (Figure 86). When "outgoing," the air lock is stationary with respect to the habitat wheel (thus, it is rotating with respect to outer space). People can, therefore, move easily out of the habitat wheel into the air lock. Then, the latter begins to rotate by electrical power opposite to the direction of rotation of the habitat wheel until it reaches the same rotational speed as the habitat wheel. As a result, the air lock is stationary in relation to outer space and can now be departed just as if the habitat wheel were not even rotating. The process is reversed for "incoming." With some training, rotating the air lock can, however, be dispensed with because the habitat wheel rotates only relatively slowly at any rate (one complete revolution in approximately 8 seconds in the previously assumed case with a 30 meter diameter of the wheel).
Even the cable connection at the other side of the axle segment is designed in a basically similar manner in order to prevent the cable from becoming twisted by the rotation of the habitat wheel. For this reason, the cable extends out from the end of a shaft (Figure 87), which is positioned on the theoretical axis of rotation of the habitat wheel and is continually driven by an electrical motor in such a manner that it rotates at exactly the same speed as the habitat wheel, but in the opposite direction. As a result, the shaft is continually stationary in relation to outer space. The cable extending from the shaft cannot, in fact, be affected by the rotation of the habitat wheel.
Figure 87. A. Top view onto the cable connection of the habitat wheel. B. Axial cross section through the cable connection of the habitat wheel.
(See Figures 84, K, and the text.)
Key: 1. Rotation of the habitat wheel; 2. Rotation of the shaft; 3. Cable; 4. Shaft; 5. Compound cable; 6. Ball bearings; 7. Passageways sealed airtight; 8. Vacuum;
9. Sliding contact rings; 10. Pressurized; 11. High and low voltage lines on the inside of the habitat wheel;
12. Heating tube.
Stairs and electrical elevators installed in tubular shafts connect the axial segment and the rim of the wheel. These shafts run "vertically" for the elevators, i.e., radially (Figure 84, A). On the other hand, for the stairs, which must be inclined, the shafts are taking the divergence of the vertical direction into account curved along logarithmic spirals that gradually become steeper towards "up" (towards the center) (Figures 88 and 84, T) because the gravitational effect (centrifugal force) decreases more and more towards that point. By using the stairs and/or elevators in an appropriately slow manner, the transition can be performed gradually and arbitrarily between the gravitational state existing in the rim of the wheel and the absence of gravity in outer space.
Figure 88. Well of the living wheel staircase.
Key: 1. Axial beam; 2. Elevator shaft; 3. Rim of the wheel; 4. Well of the staircase; 5. Railing; 6. Logarithmic spiral with a slope of 30°.
Supplying the habitat wheel with light, heat, air and water takes place in the fashion previously specified in general for the space station by employing the engineering equipment described there. The only difference being that the wall of the wheel rim always facing the sun also acts to heat the habitat wheel; for this reason, this wall is colored dull black (Figures 89 and 84), in contrast to the otherwise completely highly polished external surfaces of the station. A small solar power plant sufficient for emergency needs of the habitat wheel is also available.
Figure 89. Total view of the side of the habitat wheel facing the sun. The center concave mirror could be done away with and replaced by appropriately enlarging the external mirror.
All storage rooms and tanks for adequate supplies of air, water, food and other materials, as well as all mechanical equipment are in the wheel rim. The concave mirrors associated with this equipment and the dull black colored steam generator and condenser pipes are attached to the habitat wheel on the outside in an appropriate manner and are rotating with the habitat wheel (Figures 84, 89 and 90).
Figure 90. Total view of the shadow side of the habitat wheel.
Finally, attitude control motors and thrusters are provided; besides the purposes previously indicated, they will also generate the rotational motion of the habitat wheel and stop it again; they can also control the rate of rotation.<|endoftext|>
| 3.78125 |
584 |
Definition - What does Polyethylene (PE) mean?
Polyethylene (PE) is an organic polymer made by the polymerization of monomer subunits. The chemical formula of polyethylene is (C2H4)n. Polyethylene is a combination of similar polymers of ethylene with different values of n. A typical polyethylene molecule can contain more than 500 ethylene subunits.
Polyethylene has shown good mechanical, thermal, chemical, electrical and optical properties. It is cheap, flexible, and electrically and chemically resistant.
Polyethylene is the most widely used plastic in the world. It can be processed into any shape for flexible or hard and strong products. It is used for lining tubes and tanks, and to wrap pipes to protect against corrosive materials.
Corrosionpedia explains Polyethylene (PE)
Polyethylene (PE) is a light, versatile synthetic resin produced from the polymerization of ethylene. It is a member of the important family of polyolefin resins. Polyethylene is a thermoplastic polymer. Polyethylene has very distinct properties:
- Light weight
- Long lasting
- Low friction
- Low cost
- Electrically resistant
- Sun resistant
- Corrosion resistant
Polyethylene doesn't biodegrade easily. It can remain in a landfill for hundreds of years. It is easily recycled and polyethylene scrap can be melted down and reused.
Polyethylene is a good insulator and resists caustic materials. It is almost unbreakable. It is reliable and usable under any environmental conditions from extreme hot to extreme cold.
Polyethylene is classified according to its density and branching. The three main types are:
- High-density polyethylene (HDPE). HDPE has a low degree of branching. It is the sturdiest and most inflexible type. It has high tensile strength and is used to produce milk jugs, detergent bottles, butter tubs, garbage containers and water pipes.
- Low-density polyethylene (LDPE). LDPE has a high degree of short- and long-chain branching, which give it a lower tensile strength and increased ductility. This gives molten LDPE unique and desirable flow properties. It is used for both rigid containers and plastic film applications, such as plastic bags and film wrap.
- Linear low-density polyethylene (LLDPE). LLDPE has a substantially linear polymer with significant numbers of short branches. It has higher tensile strength than LDPE, and it exhibits higher impact and puncture resistance than LDPE. LLDPE is extremely tough and inflexible. These features are suitable for larger items, such as covers, storage bins and some types of containers.<|endoftext|>
| 3.859375 |
1,006 |
Language can reveal the invisible, study shows
It is natural to imagine that the sense of sight takes in the world as it is — simply passing on what the eyes collect from light reflected by the objects around us.
But the eyes do not work alone. What we see is a function not only of incoming visual information, but also how that information is interpreted in light of other visual experiences, and may even be influenced by language.
Words can play a powerful role in what we see, according to a study published this month by UW–Madison cognitive scientist and psychology professor Gary Lupyan, and Emily Ward, a Yale University graduate student, in the journal Proceedings of the National Academy of Sciences.
“Perceptual systems do the best they can with inherently ambiguous inputs by putting them in context of what we know, what we expect,” Lupyan says. “Studies like this are helping us show that language is a powerful tool for shaping perceptual systems, acting as a top-down signal to perceptual processes. In the case of vision, what we consciously perceive seems to be deeply shaped by our knowledge and expectations.”
And those expectations can be altered with a single word.
To show how deeply words can influence perception, Lupyan and Ward used a technique called continuous flash suppression to render a series of objects invisible for a group of volunteers.
Each person was shown a picture of a familiar object — such as a chair, a pumpkin or a kangaroo — in one eye. At the same time, their other eye saw a series of flashing, “squiggly” lines.
“Essentially, it’s visual noise,” Lupyan says. “Because the noise patterns are high-contrast and constantly moving, they dominate, and the input from the other eye is suppressed.”
Immediately before looking at the combination of the flashing lines and suppressed object, the study participants heard one of three things: the word for the suppressed object (“pumpkin,” when the object was a pumpkin), the word for a different object (“kangaroo,” when the object was actually a pumpkin), or just static.
Then researchers asked the participants to indicate whether they saw something or not. When the word they heard matched the object that was being wiped out by the visual noise, the subjects were more likely to report that they did indeed see something than in cases where the wrong word or no word at all was paired with the image.
“If language affects performance on a test like this, it indicates that language is influencing vision at a pretty early stage. It’s getting really deep into the visual system.”
“Hearing the word for the object that was being suppressed boosted that object into their vision,” Lupyan says.
And hearing an unmatched word actually hurt study subjects’ chances of seeing an object.
“With the label, you’re expecting pumpkin-shaped things,” Lupyan says. “When you get a visual input consistent with that expectation, it boosts it into perception. When you get an incorrect label, it further suppresses that.”
Experiments have shown that continuous flash suppression interrupts sight so thoroughly that there are no signals in the brain to suggest the invisible objects are perceived, even implicitly.
“Unless they can tell us they saw it, there’s nothing to suggest the brain was taking it in at all,” Lupyan says. “If language affects performance on a test like this, it indicates that language is influencing vision at a pretty early stage. It’s getting really deep into the visual system.”
The study demonstrates a deeper connection between language and simple sensory perception than previously thought, and one that makes Lupyan wonder about the extent of language’s power. The influence of language may extend to other senses as well.
“A lot of previous work has focused on vision, and we have neglected to examine the role of knowledge and expectations on other modalities, especially smell and taste,” Lupyan says. “What I want to see is whether we can really alter threshold abilities,” he says. “Does expecting a particular taste for example, allow you to detect a substance at a lower concentration?”
If you’re drinking a glass of milk, but thinking about orange juice, he says, that may change the way you experience the milk.
“There’s no point in figuring out what some objective taste is,” Lupyan says. “What’s important is whether the milk is spoiled or not. If you expect it to be orange juice, and it tastes like orange juice, it’s fine. But if you expected it to be milk, you’d think something was wrong.”<|endoftext|>
| 3.90625 |
486 |
|Nepenthes bicalcarata (by David Sucianto, via Wikimedia Commons)|
Why would a photosynthetic organism need to trap animals? After all, they still have green leaves and chlorophyll, like any other plant. But other nutrients, especially nitrogen, are important to plant growth as well. Pitcher plants can grow in nutrient poor soil because they can supplement their intake of nitrogen and other nutrients by trapping and digesting animals, which are especially nitrogen-rich.
Not all animals will die in pitchers, however. Some insects, including the larvae of several dipteran (fly) species, can live in the pitchers and feed on the organic matter found there. The pitcher species Nepenthes bicalcarata also plays host to an ant species, Camponotus schmitzi, that is found only with N. bicalcarata. The ants are somehow able to walk on the slippery inner surface of the pitcher, and predate upon the fly larvae and other organic material, and also feeds on nectar from the plant.
It would seem at first that this is a lousy deal for the plant. The flies and ants are stealing its food right from its mouth! A new research paper published in PLoS ONE shows how the ants and pitcher plants actually derive mutual benefit.
By looking at the nitrogen isotope ratios in the plant tissue, and using isotope labeling experiments, the researchers showed that nitrogen is being transferred from the ants to the nutrients. They also observed how the ants predate upon the fly larvae that live and mature inside the pitchers. Left to their own devices, these larvae would consume the pitcher's nutrient supply, and then literally fly away with the stolen nutrients when they metamorphose into adults. For this they are (harshly) called kleptoparasites, or "thief-parasites". By capturing and eating the flies while they are still larvae or pupae, the ants put a stop to this thievery. The plant itself then recovers these nutrients in the form of the ant colony's waste products.
The ants are hence not only improving the pitcher's prey-capture efficiency, by keeping the slippery pitcher walls clean, but also prevent the nutrients from escaping with the insects. A fascinating story of symbiosis, that reveals just how dynamic and interconnected all these nutritional and behavioral relationships are in Nature.<|endoftext|>
| 3.96875 |
388 |
DC Electricity is "direct current" and is the electricity generated by your solar panels and stored in your batteries.
AC Electricity is "alternating current" as connected to most houses and buildings in most areas of the world. This is the electricity your inverter will produce to run your appliances like the TV stereo and blender in a desirable manner.
A Volt is the unit of measure for electrical potential. It is also considered to be like water pressure in a pipe carrying a certain volume of water (current)
Commonly referred to as Current - It is the amount of electricity flowing (volume) and is proportional to the amount of power being used. A definition would be that an item that uses a large amount of electricity, say a heater or an iron will draw more current than a little electrical appliance like a light globe. Current is measured in amps or more correctly amperes. While voltage could be called the potential of electricity, current is the amount of power that comes out of the wire.
A WATT is the unit for measuring electrical power, i.e., the rate of doing work, in moving electrons by, or against, an electrical potential. Formula: Watts = Amperes x Volts.
A WATT-HOUR is the unit of measure for electrical energy expressed as Watts x Hours.
When you buy a microwave oven you want to know how intense the microwave field is, not how much the oven draws from the wall. So a microwave oven that boasts 600 watts on the box, will often have 1200 watts on the boilerplate in the back.
Stereo manufacturers are bigger liars than politicians. Sometimes they use peak output power (milliseconds), sometimes they use power drawn from the wall, but often they just look at the competition's carton front and add 10%. However the truth is available: look at the boilerplate sticker, which has been evaluated by standards organization.<|endoftext|>
| 3.90625 |
739 |
Scientists recently unveiled the tiniest electric motor ever built. You could stuff hundreds of them into the period at the end of this sentence.
One day a similar engine might power a tiny mechanical doctor that would travel through your body in the ultimate house call.
The motor works by shuffling atoms between two molten metal droplets in a carbon nanotube [watch it run].
One droplet is even smaller than the other. When a small electric current is applied to the droplets, atoms slowly eek off the larger droplet and join the smaller one. The small droplet grows - but never gets as big as the other droplet - and eventually bumps into the large droplet. As they touch, the large droplet rapidly sops up the atoms it had previously sloughed off. This quick shift in energy produces a power stroke.
The technique exploits the fact that surface tension -- the tendency of atoms or molecules to resist separating -- becomes more important at small scales. Surface tension is the same thing that allows some insects to walk on water.
The motor, a surface-tension-driven nanoelectromechanical relaxation oscillator, was built by a team of researchers led by Alex Zettl at the University of California, Berkeley.
Although the amount of energy produced is small -- 20 microwatts -- it is quite impressive in relation to the tiny scale of the motor. The whole setup is less than 200 nanometers on a side, or hundreds of times smaller than the width of a human hair. If it could be scaled up to the size of an automobile engine, it would be 100 million times more powerful than a Toyota Camry's 225 horsepower V6 engine, the researchers say.
In 1988, Berkeley electrical engineering professor Richard Muller and colleagues made the first operating micromotor, which was 100 microns across, or about the thickness of a human hair. In 2003, Zettl's group created the first nanoscale motor. Last year they built a nanoconveyor, which moves tiny particles along like cars in a factory.
A paper discussing the latest invention was published in the March 21 issue of Applied Physics Letters.
Nanotechnology engineers try to mimic nature, building things atom-by-atom. Among other things, nanomotors could be used in optical circuits to redirect light, a process called optical switching. Futurists envision a day when nanomachines, powered by nanomotors, roam inside your body to find disease and repair damaged cells.
"This oscillator is ideal for locomotive applications, since it is so powerful for its size," Chris Regan, a Zettl group researcher, told LiveScience. "For instance, in a nanobot it could be used as the motor that drives crawling, walking, swimming, jumping, or flying."
More immediately, the new motor could be used within two years in existing Micro-Electro-Mechanical Systems (MEMS) technology, Regan said.
MEMS are tiny components made by etching away parts of a silicon wafer or adding tiny layers. As small as a grain of pollen, MEMS are larger than the nanoworld. They are employed in such things as microaccelerometers, the devices that activate automotive airbags. Optical MEMS are used in many home theater systems.
Regan said the shift to NEMS, the nano-equivalent of MEMS, "will take longer."
- Self-Assembling Nanomachines
- A Step Toward Single-Molecule Computers
- Tiny Paddle Built to Ply Your Body<|endoftext|>
| 3.765625 |
514 |
## Ms. Kylie
### Target 1
###### Lesson Type:
Continuation
Number Operation
:
Computation
Solve multi-step word problems within 1,000,000 (addition, subtraction, multiplication, and division).
###### 1:
Identify the correct arithmetic processes based on the information presented in word problems.
###### 2:
Find the key words in word problems that indicate the correct arithmetic process to use: Addition: add, altogether, both, combined, in all, plus, sum, total; Subtraction: difference, fewer, how many/much more, left, less, minus, remains; Multiplication: times, product, each, twice; and Division: how many in each, divided by, quotient, goes into, split evenly.
###### 3:
Find and use the needed information in a word problem to solve.
4th
###### Vocabulary:
CUBES
Activities:
-Students were given multi-step word problems to solve. They used the CUBES strategy to help solve the problem and solve the problem in steps.
-Then the students explained how they solved their word problems to each other.
## Absent Students:
### Target 2
:
###### 1:
Understand that fractions represent quantities that have a portion that is less than a whole.
###### 2:
Understand that as the number of equal parts of the whole increases, the size of those pieces’ decreases.
###### 3:
Convert improper fractions to mixed numbers (and vice versa), understanding that both have “wholes” as a part of their quantity.
4th
###### Vocabulary:
Denominator, Numerator, Improper fraction, Mixed number
Activities:
-Students reviewed converting improper fractions to mixed numbers and vice versa by drawing a picture of the fraction
-Students learned how to convert a mixed number to an improper fraction by multiplying the whole number with the denominator and adding the numerator to obtain the numerator to the improper fraction. The denominator stays the same because the mumber parts don't change.
### Target 3
:
###### 1:
Understand that geometric figures can be drawn on the coordinate plane.
###### 2:
Draw quadrilaterals on the coordinate plane.
###### 3:
Become familiar with how the coordinate plane can be used to determine distance or length of geometric shapes.
5th
###### Vocabulary:
X-axis, Y-axis, Coordinate pair, Coordinate plane
Activities:
-Students drew quadrilateral shapes on the coordinate plane and then located and wrote each point of the quadrilateral on the coordinate plane.<|endoftext|>
| 4.5625 |
7,565 |
Droughts in the United States
Drought in the United States is similar to that of other portions of the globe. Below normal precipitation leads to drought, which is caused by an above average persistence of high pressure over the drought area. Changes in the track of extratropical cyclones, which can occur during climate cycles such as the El Niño-Southern Oscillation, or ENSO, as well as the North Atlantic Oscillation, Pacific Decadal Oscillation, and Atlantic multidecadal oscillation, modulates which areas would be more prone to drought and when drought develops. Increased drought frequency is expected to be one of the effects of global warming. In dry areas, removing grass cover and going with a more natural vegetation for the area can reduce the impact of drought, since a significant amount of fresh water is used to keep lawns green. Droughts are periodic, alternating with floods over a series of years.
The worst droughts in the history of the United States occurred during the 1930s and 1950s, periods of time known as 'Dust Bowl' years in which droughts lead to significant economic damages and social changes. In particular, relief and health agencies became overburdened and many local community banks had to close.
The U.S. Drought Monitor provides national database to track the duration and severity of droughts in the United States, hosted by the University of Nebraska-Lincoln with assistance from the United States Department of Agriculture and the National Oceanic and Atmospheric Administration. Their standardized measurements track droughts on a severity scale from "Abnormally Dry" (D0) to "Exceptional" (D4).
Mechanisms of producing precipitation include convective, stratiform, and orographic rainfall. Convective processes involve strong vertical motions that can cause the overturning of the atmosphere in that location within an hour and cause heavy precipitation, while stratiform processes involve weaker upward motions and less intense precipitation over a longer duration. Precipitation can be divided into three categories, based on whether it falls as liquid water, liquid water that freezes on contact with the surface, or ice.
If these factors do not support precipitation volumes sufficient to reach the surface over a sufficient period of time, the result is a drought. Drought can be triggered by a high level of reflected sunlight and above average prevalence of high pressure systems, winds carrying continental, rather than oceanic air masses, and ridges of high pressure areas aloft can prevent or restrict the developing of thunderstorm activity or rainfall over one certain region. Once a region is within drought, feedback mechanisms such as local arid air, hot conditions which can promote warm core ridging, and minimal evapotranspiration can worsen drought conditions. Winters during El Niño are warmer and drier than average in the Northwest, northern Midwest, and northern Mideast United States, so those regions experience reduced snowfalls.
Activities resulting in global climate change are expected to trigger droughts with a substantial impact on agriculture throughout the world, and especially in developing nations. Overall, global warming will result in increased world rainfall. Along with drought in some areas, flooding and erosion will increase in others. Paradoxically, some proposed solutions to global warming that focus on more active techniques, solar radiation management through the use of a space sunshade for one, may also carry with them increased chances of drought.
Certain regions within the United States are more susceptible to droughts than others. Droughts can be more damaging than tornadoes, tropical cyclones, winter storms and flooding combined. Unlike a hurricane, tornado or flooding, the onset of droughts happen gradually over a long period of time.
In the Nevada "cash for grass" program, the people are paid to remove grass and put in desert landscaping. Xeriscaping calls for the planting of vegetation which is local in origin and more resistant to drought.
When California suffered a severe drought from 1985 to 1991, a California company, Sun Belt Water Inc. was established for the purpose importing water from Canada in marine transport vessels formerly used for oil transport and converted to water carriers. The idea was commercially viable and Sun Belt Water Inc., was selected by the Goleta Water District to enter a long term contract. When the government of British Columbia reversed its existing bulk water export policy, the change in government policy led to a claim by Sun Belt Water Inc. against Canada under the provisions of Chapter 11 of the North American Free Trade Agreement (NAFTA).
The National Integrated Drought Information System (NIDIS) Act was signed into law in 2006 (Public Law 109-430). The Western Governors' Association described the need for NIDIS in a 2004 report, Creating a Drought Early Warning System for the 21st Century: The National Integrated Drought Information System. The NIDIS Act calls for an interagency, multi-partner approach to drought monitoring, forecasting, and early warning, led by the National Oceanic and Atmospheric Administration (NOAA). NIDIS is being developed to consolidate data on drought's physical, hydrological and socio-economic impacts on an ongoing basis, to develop drought decision support and simulation tools for critical, drought-sensitive areas, and to enable proactive planning by those affected by drought. NIDIS (www.drought.gov) draws on the personnel, experience, and networks of the National Drought Mitigation Center, the NOAA Regional Climate Centers, and the Regional Integrated Sciences and Assessments (RISAs), among others. Federal agencies and departments partnering in NIDIS include the U.S. Army Corps of Engineers, the Bureau of Reclamation, the U.S. Geological Survey, NASA, the U.S. Department of Energy, the U.S. Environmental Protection Agency, the National Science Foundation, and the Natural Resources Conservation Service.
This section needs to be updated.July 2018)(
This section appears to be slanted towards recent events. (July 2012) (Learn how and when to remove this template message)
There are indications for megadroughts between 900 and 1300. Drought apparently struck what is now the American Southwest back in the 13th century, which probably affected the Pueblo cities, and tree rings also document drought in the lower and central Mississippi River basin between the 14th and 16th century. The droughts of that period might have contributed to the decline and fall of the Mississippian cultures.
The 18th century seems to have been a relatively wet century in North America, but there were apparently droughts in Iowa in 1721, 1736, and from 1771 to 1773.
There were at least three major droughts in 19th-century North America: one from the mid-1850s to the mid-1860s, one in the 1870s, and one in the 1890s. There was also a drought around 1820; the periods from 1816 to 1844 and from 1849 to 1880 were rather dry, and the 19th century overall was a dry century for the Great Plains. While there was little rain-gauge data from the mid-19th century in the middle of the US, there were plenty of trees, and tree-ring data showed evidence of a major drought from around 1856 to around 1865. Native Americans were hard hit, as the bison they depended upon on the Plains moved to river valleys in search of water, and those valleys were full of natives and settlers alike. The river valleys were also home to domestic livestock, which competed against the bison for food. The result was starvation for many of the bison.
The 1870–1877 drought brought with it a major swarm of Rocky Mountain Locusts, as droughts benefit locusts, making plants more nutritious and edible to locusts and reducing diseases that harm locusts. Locusts also grow more quickly during a drought and gather in small spots of lush vegetation, enabling them to swarm, facts which contributed to the ruin of much of the farmland in the American West. The evidence for this drought is also primarily in tree-ring, rather than rain gauge, data.
The 1890s drought, between 1890 and 1896, was the first to be widely and adequately recorded by rain gauges, with much of the American West having been settled. Railroads promised land to people willing to settle it, and the period between 1877 and 1890 was wetter than usual, leading to unrealistic expectations of land productivity. The amount of land required to support a family in more arid regions was already larger than the amount that could realistically be irrigated by a family, but this fact was made more obvious by the drought, leading to emigration from recently settled lands. The Federal government started to assist with irrigation with the 1902 Reclamation Act.
The Dust Bowl or the Dirty Thirties was a period of severe dust storms causing major ecological and agricultural damage to American and Canadian prairie lands from 1930 to 1936 (in some areas until 1940). The phenomenon was caused by severe drought coupled with decades of extensive farming without crop rotation, fallow fields, cover crops or other techniques to prevent erosion. Deep plowing of the virgin topsoil of the Great Plains had displaced the natural grasses that normally kept the soil in place and trapped moisture even during periods of drought and high winds.
During the drought of the 1930s, without natural anchors to keep the soil in place, it dried, turned to dust, and blew away eastward and southward in large dark clouds. At times the clouds blackened the sky reaching all the way to East Coast cities such as New York and Washington, D.C. Much of the soil ended up deposited in the Atlantic Ocean, carried by prevailing winds which were in part created by the dry and bare soil conditions itself. These immense dust storms—given names such as "Black Blizzards" and "Black Rollers"—often reduced visibility to a few feet (around a meter). The Dust Bowl affected 100,000,000 acres (400,000 km2), centered on the panhandles of Texas and Oklahoma, and adjacent parts of New Mexico, Colorado, and Kansas.
Millions of acres of farmland became useless, and hundreds of thousands of people were forced to leave their homes; many of these families (often known as "Okies", since so many of them came from Oklahoma) traveled to California and other states, where they found economic conditions little better than those they had left. Owning no land, many traveled from farm to farm picking fruit and other crops at starvation wages. Author John Steinbeck later wrote The Grapes of Wrath, which won the Pulitzer Prize, and Of Mice and Men about such people.
Negative effects included bank closures and overburdened relief and health agencies. Economic migrants also had mixed success as native workers in many areas resented the intense competition for dwindling jobs. The National Drought Mitigation Center has reported that financial assistance from the government alone may have been as high as $1 billion (in 1930s dollars) by the end of the drought.
Drought began in the Southwestern United States in 1944 and continued through the entire rest of the decade; one of the longest recorded droughts observed there. This drought continued into the 1950s.
Other severe drought years in the United States happened through the 1950s. These droughts continued from the 1940s drought in the Southwestern United States, New Mexico and Texas during 1950 and 1951; the drought was widespread through the Central Plains, Midwest and certain Rocky Mountain States, particularly between the years 1953 and 1957, and by 1956 parts of central Nebraska reached a drought index of −7, three points below the extreme drought index. From 1950 to 1957, Texas experienced the most severe drought in recorded history. By the time the drought ended, 244 of Texas's 254 counties had been declared federal disaster areas. Drought became particularly severe in California, with some natural lakes drying up completely in 1953. Southern California was hit hard by drought in 1958–59, badly straining water resources. A widespread, 1930s-style dust storm affected the Plains and beyond on 19 February 1954 driven by winds of up to 100 mph/161 km/h, drifting soil to 3 feet/a metre deep in some areas.
The Northeastern United States were hit with devastating drought which lasted almost four to five years in the 1960s. The drought affected multiple regional cities from Virginia into Pennsylvania, New Jersey and New York; the drought also affected certain Midwest States. Drought continued in parts of California in the early 1960s. Southern California recorded their worst drought of the 20th century in 1961.
Short term droughts hit particular spots of the United States during 1976 and 1977. California's statewide snowpack reached an all-time low in 1977. Water resources and agriculture (especially livestock) suffered; negatively impacting the nation's economy. This drought reversed itself completely the following year.
Droughts also affected the Northeast US, Corn Belt and Midwest States during 1980 and 1983. The 1983 Midwestern States Drought was associated with very dry conditions, severe heat and substandard crop growth which affected prices and caused hardship for farmers. Multiple disaster declarations went out in Indiana and neighboring states because of the 1983 drought. Readings of 100 °F (38 °C) or higher became prevalent in 1983 during these dry spells across the Midwest, Ohio Valley Regions and Great Lakes. Kentucky declared the 1983 drought their second-worst in the 20th century; 1983 was Ohio's driest calendar year. Los Angeles received more rainfall than Cleveland that year. The drought forced many trees and shrubs into dormancy and created water shortages in many towns. The associating heat waves killed between 500–700 people in the United States. Similar spells during 1980 caused between 4000 and 12000 deaths in the United States along with $24 billion in damage 1980 USD.
A severe drought struck the Southeast from 1985 through 1987. It began in 1985 from the Carolinas west-southwest into Alabama, when annual rainfall was reduced by 5 to 35 percent below what was normal. Light precipitation continued into the spring of 1986, with Atlanta, Georgia recording their driest first six months on record. High amounts of precipitation during the winter of 1987 ended the drought.
The Western United States experienced a lengthy drought in the late 1980s. California went through one of its longest observed droughts, from late 1986 through early 1991. Drought worsened in 1988-89, as much of the United States also suffered from severe drought. In California, the five-year drought ended in late 1991 as a result of unusual persistent heavy rains, most likely caused by a significant El Niño event in the Pacific Ocean and the eruption of Mount Pinatubo in June 1991.
Another significant drought in the United States occurred during 1988 and 1989. Following a milder drought in the Southeastern United States the year before, this drought spread from the Mid-Atlantic, Southeast, Midwest, Northern Great Plains and Western United States. This drought was widespread, unusually intense and accompanied by heat waves which killed around 4,800 to 17,000 people across the United States and also killed livestock across the United States. One particular reason that the drought of 1988 became very damaging was farmers might have farmed on land which was marginally arable. Another reason was pumping groundwater near the depletion mark. The drought of 1988 destroyed crops almost nationwide, residents' lawns went brown and water restrictions were declared many cities. The Yellowstone National Park fell victim to wildfires that burned many trees and created exceptional destruction in the area. This drought was very catastrophic for multiple reasons; it continued across the Upper Midwest States and North Plains States during 1989, not officially ending until 1990.
The conditions continued into 1989 and 1990, although the drought had ended in some states thanks to normal rainfalls returning to some portions of the United States. Dry conditions, however, increased again during 1989, affecting Iowa, Illinois, Missouri, eastern Nebraska, Kansas and certain portions of Colorado. The drought also affected Canada in certain divisions. The drought of 1988 became the worst drought since the Dust Bowl 50 years before in the United States; 2008 estimates put damages from the drought somewhere between $80 billion and almost $120 billion in damage (2008 USD). The drought of 1988 was so devastating that in later years it was compared against Hurricane Andrew in 1992 and against Hurricane Katrina; in addition, it would be the costliest of the three events: Hurricane Katrina comes second with $81 billion (2005 United States dollars), Hurricane Andrew coming in third. The drought of 1988 qualifies being the costliest natural disaster in the history of the United States.
During 1993 the Southeastern United States experienced high temperatures and conditions of drought for extended periods. The heat waves associated caused the deaths of seventeen people and overall damage from the Southeastern-state drought of 1993 was somewhere between $1 billion and $3 billion in damage (1993 U.S. dollars).
Similar drought conditions hit the Northeast United States during 1999 – the Northeast, including Kentucky, New York, New Jersey, Pennsylvania and Maryland were pummeled by extensive heat waves which killed almost 700 people across the Northeastern U.S. and unusually dry conditions caused billions of dollars in destruction during 1999. This unusually damaging drought was reminiscent of the Northeast United States drought of the 1960s considering it affected similar states within the Northeast United States and New England.
The Midwest and Rocky Mountain regions had a drought during 2002, which was accompanied by dry conditions, wildfires and hot temperatures over the Western US and Midwest areas. The U.S. drought of 2002 turned a "normal" fire season into a very dangerous, treacherous and violent season. Denver was forced to impose mandatory limits regarding water for the first time in twenty-one years, as Colorado and other states in the Southwest were hit particularly hard by the severe drought conditions in 2002. The Quad Cities had around 8 inches (200 mm) below average rainfall during 2002 (normal precipitation is 38.06 inches (967 mm) every year, during 2002 30.00 inches (762 mm) were recorded). The 2001–02 rain season in Southern California was the driest since records began in 1877. San Diego recorded only 2.99 inches (76 mm), compared to the annual average of 10.34 inches (263 mm). Records were broken in an even worse drought just five years later, during the 2006–07 rain season in Los Angeles (3.21 inches (82 mm) compared to the annual average of 15.14 inches (385 mm)).
The U.S. drought of 2002 was reminiscent of the 1988 drought and was compared to the droughts of the 1930s, the 1983 drought and the dry spells of the 1950s. The drought also affected Saskatchewan, Manitoba and Alberta, in Canada.
Although the Western United States and Southwestern U.S. are most likely to be hit, droughts can also happen over the Upper Midwestern States, the Central Great Plains, Southeast United States, the Middle Atlantic, the Great Lakes Region, the Ohio River Valley, Northeastern United States and even New England. Droughts vary in severity and have potential for causing elevated to exceptional damage wherever they focus their area toward.
There were extensive droughts through the 2000s (decade) all over the Southeastern United States, continuing as far westward as Texas. The Southeastern United States were affected by heavy droughts extending from the Carolinas toward Mississippi and even into Tennessee and Kentucky. Droughts affecting Florida were so severe that lakes were actually drying out. Wildfires, forest fires, and brush fires were very prevalent in association with the 2000s (decade) drought in the Southeastern United States. The drought of 2006–07 in California contributed to the extreme severity of the 2007 California wildfires.
Missouri, Arkansas, (portions of) Louisiana, Tennessee, southeast Iowa and northern Illinois were hit with severe droughts and heat during 2005. The conditions caused $1 billion in overall damage, there were no deaths attributed to the drought and associated heat spells. The Quad Cities themselves received only 17.88 inches (454 mm) of precipitation during 2005.
In 2008 and 2009, much of south and south-central Texas were in a state of exceptional drought.
California also experienced a multiyear drought, peaking in 2007–2009, when a statewide drought emergency was issued. Although reports of widespread agricultural losses were reduced in later analysis, large decreases were seen in many fish populations in the region, and additional reliance on groundwater in farming may have set the precedent for further damages in the 2012–2015 California drought.
The California drought continued through 2010 and did not end until March 2011. The drought shifted east during the summer of 2011 to affect a large portion of the Southwest and Texas. See above for additional information on this drought. In 2013 and early 2014, the California drought returned and intensified, expanding to much of the western US. In 2013, many places in California set all-time low precipitation records, with very little measurable rain falling across much of the state from January 2013 into mid-February 2014. San Francisco nearly halved its previous annual record low in 2013, receiving only 5.59 inches compared to a normal of 23.65. The 2012–13 and 2013–14 winter snowpacks were among the lowest recorded in the last 100 years. In January 2014, the state cut allocations from its State Water Project to zero percent (revised upwards to five percent in April), a record low, as reservoirs dropped to critical levels. Municipal districts in the northern and central parts of the state, including the capital, Sacramento, enacted water rationing while over half a million acres (2000 km2) of Central Valley farmland were fallowed. In 2015, wildfires burned over 7 million acres, primarily in the western U.S. and Alaska, which is approaching the all-time national record. Although the drought continues, a major El Niño episode forecast to develop late in 2015 may bring some relief to drought areas.
In 2011 intense drought struck much of Texas, New Mexico and a large portion of the Southwest bringing much of the region its worst drought seen since the Dust Bowl years of the 1930s. Most of the drought in Texas ended or had it impacts ease by spring and summer 2012 as precipitation returned to the region, while the New Mexican drought continued unbroken into 2014. The Texas and Southwest U.S. drought was also accompanied by a severe heat wave that brought record setting heat to much of Texas, including but not limited to bringing a 40-day stretch of temperatures at or above 100 °F (38 °C) to Dallas, Texas. Drought of severe magnitude also affected a large portion of the Southeastern US, especially Georgia and South Carolina. It is believed that a combination of La Niña and climate change had contributed to the intense drought.
In 2012, much of the U.S. had drought conditions develop through the late winter and spring months and lasting into the summer, creating the 2012 North American drought. Meanwhile, severe to extreme drought developed in the lower Midwest and Ohio Valley as well as the southern and central Rockies. This led to large wildfires in Colorado including the record setting Waldo Canyon fire, the most destructive in Colorado history. Drought conditions have led to numerous firework show cancellations and voluntary water restrictions in much of the Ozarks, Mid-Mississippi and Ohio River Valleys. Lagging effects of La Niña, climate change, and also a large persistent upper level ridge of high pressure present over much of North America since the late winter have all contributed to the drought and above average temperatures since February 2012. This further lead to the vicious cycle of reduced evaporation and decreased rainfall all through the spring of 2012. While the summer of 2011 was the second-warmest (74.5 °F (23.6 °C)) in U.S. history after the Dust Bowl era of 1936 74.6 °F (23.7 °C) the summer of 2012 was the third-warmest at (74.4 °F (23.6 °C)). This intense heat wave contributed to the intensification of the drought particularly over the Midwest and the Northern Plains. Because the drought conditions were forcing American farmers to sell off livestock, the Department of Defense sought to buy up meat at "fire sale" prices in order to stockpile meals for the lean times ahead.
High wheat prices caused by the drought have discouraged farmers from investing in alternative drought-tolerant crops.
The United States Drought Monitor observed “extreme drought” conditions in much of the eastern half of Massachusetts, southeastern New Hampshire and the southern part of Maine in September 2016.
In summer 2016, severe drought affected the temperate New England and New York area, including a Massachusetts and New York drought that persisted into the fall. While not as severe as other major, more well-known droughts in the south and west where the climate is semi-arid, it was among the most severe for the northeastern region.
Hydrologist Royce Fontenot said the impact of drought will linger according to the La Niña pattern.
- "Drought in the Dust Bowl Years". Drought.unl.edu. 1935-05-06. Retrieved 2015-08-14.
- "U.S. Drought Monitor Classification Scheme". University of Nebraska-Lincoln. Retrieved 22 August 2014.
- Emmanouil N. Anagnostou (2004). "A convective/stratiform precipitation classification algorithm for volume scanning weather radar observations". Meteorological Applications. Cambridge University Press. 11 (4): 291–300. Bibcode:2004MeApp..11..291A. doi:10.1017/S1350482704001409.
- A.J. Dore; M. Mousavi-Baygi; R.I. Smith; J. Hall; D. Fowler; T.W. Choularton (June 2006). "A model of annual orographic precipitation and acid deposition and its application to Snowdonia". Atmospheric Environment. 40 (18): 3316–3326. Bibcode:2006AtmEn..40.3316D. doi:10.1016/j.atmosenv.2006.01.043.
- Robert Penrose Pearce (2002). Meteorology at the Millennium. Academic Press. p. 66. ISBN 978-0-12-548035-2. Retrieved 2009-01-02.
- Robert A. Houze, Jr. (1994). Cloud Dynamics. Academic Press. p. 348. ISBN 0080502105.
- Roland Paepe; Rhodes Whitmore Fairbridge; Saskia Jelgersma (1990). Greenhouse Effect, Sea Level and Drought. Springer Science & Business Media. p. 22. ISBN 0792310179.
- Joseph S. D'Aleo; Pamela G. Grube (2002). The Oryx Resource Guide to El Niño and La Niña. Greenwood Publishing Group. pp. 48–49. ISBN 1573563781.
- https://web.archive.org/web/20080625001539/http://www.economics.noaa.gov/library/documents/benefits_of_weather_and_climate_forecasts/drought-climate_change-implications_for_west.doc. Archived from the original on June 25, 2008. Retrieved February 6, 2016. Missing or empty
- "Irish business, finance news on economics". Finfacts. Retrieved 2015-08-14.
- https://archive.is/20120913201531/http://www.timesdaily.com/article/20070914/NEWS/709140329/1011/RSS&source=RSS. Archived from the original on September 13, 2012. Retrieved February 6, 2016. Missing or empty
- "Nigerian Scholar Links Drought, Climate Change to Conflict in Africa – US Department of State". Web.archive.org. Archived from the original on 2005-10-28. Retrieved 2015-08-14.
- "Is Water the New Oil? | Common Dreams | Breaking News & Views for the Progressive Community". Common Dreams. 2008-11-02. Retrieved 2015-08-14.
- Brahic, Catherine (2007-08-02). "'Sunshade' for global warming could cause drought". New Scientist. Retrieved 2015-08-14.
- "Sunbeltwater.com". Sunbeltwater.com. Retrieved 2015-08-14.
- Peter Folger. "Drought in the United States: Causes and Issues for Congress" (PDF). Digitalcommons.unl.edu. Retrieved 2015-08-14.
- Cleaveland, M.K., Duvick, D.N., 1992. Iowa climate reconstructed from tree rings, 1640–1982. Water Resources Research 28, 2607–2615.
- "Causes and consequences of 19th-century droughts in North America". Ldeo.columbia.edu. Retrieved 2015-08-14.
- "Agricultural and Forest Meteorology" (PDF). Web.missoui.edu. Retrieved 2015-08-14.
- "Drought: A Paleo Perspective – 20th Century Drought". National Climatic Data Center. Retrieved 2009-04-05.
- Hakim, Joy (1995). A History of Us: War, Peace and all that Jazz. New York: Oxford University Press. ISBN 0-19-509514-6.
- Climatestations, Charles Fisk, retrieved 2013-06-11
- Water Issues during the 1950s, Living History Farm, retrieved 2009-04-19
- "Texas State Parks in the 1950s – Texas State Library | TSLAC". Tsl.state.tx.us. 2014-02-28. Retrieved 2015-08-14.
- Hurt 1956 Ch 9
- Plight of Farmers in State Deepening in Worst Drought (Report). The New York Times. Oct 1, 1964. Retrieved Jun 22, 2016.
- "California Drought of 1976 and 1977—Extent, Damage, and Governmental Response". Retrieved 2013-06-11.
- "Oilseeds: situation and outlook 1984". Bureau of Agricultural Economics. Archived from the original on 2011-07-25. Retrieved 2009-04-17.
- Malcolm, Andrew H. (1983-09-03). "U.S. Drought Disaster Declared by U.S". The New York Times. Retrieved 2009-04-13.
- "The Top Ten Heat Events". NOAA. Retrieved 2009-04-20.
- American Society of Civil Engineers. Task Committee on Sustainability Criteria, Unesco/IHP-IV Project M-4.3 (1998). Sustainability criteria for water resource systems. ASCE Publications. p. 204. ISBN 978-0-7844-0331-0.
- Water Resources Support Center, Institute For Water Resources, U.S. Army Corps of Engineers (1992). Lessons learned from the California Drought (1987–1992). ASCE Publications. p. 122.CS1 maint: Multiple names: authors list (link)
- Billion Dollar Disasters (Northern Plains Drought in Summer 1989) (Report). Live Science. Retrieved 2009-04-18.
- Robbins, William (1989-09-16). "Drought Stricken Areas Find Relief after Rains". The New York Times. Retrieved 2009-04-13.
- Improving Drought Management (PDF) (Report). The University of Nebraska. Archived from the original (PDF) on 2010-06-15. Retrieved 2009-04-19.
- Spotty Rains Define 1989 Drought (Report). Christian Science Monitor. Retrieved Apr 6, 2015.
- Drought of 1989 (Report). NOAA. Retrieved November 10, 2017.
- Drought Condition in Utah, Utah Division of Water Resources, archived from the original on 2009-03-31, retrieved 2009-04-19
- Earth Observatory/Droughts, NASA, archived from the original on September 16, 2008, retrieved 2009-04-20
- Billion Dollar Disasters (Report). Live Science. Retrieved 2009-04-18.
- The Climate of 1999 (Report). NCDC. Retrieved 2009-04-17.
- "The US Drought of 2002". NOAA. Retrieved 2009-04-17.
- "NOAA Drought Monitor". NOAA. Archived from the original on 2008-09-05. Retrieved 2009-04-17.
- The Drought of 2002 in Colorado (PDF) (Report). The Colorado State University. Retrieved 2009-04-17.
- The Drought of 2002 (PDF) (Report). Retrieved January 22, 2010.
- Los Angeles Climate Data/Records (Report). National Weather Service. Retrieved June 11, 2013.
- San Diego Climate Data/Records (Report). National Weather Service. Retrieved June 11, 2013.
- Canada Drought of 2002 (Report). CDC News. Retrieved January 23, 2010.
- "U.S. drought conditions worsen". The Brownfield Network. Archived from the original on 2009-05-06. Retrieved 2009-04-19.
- "The Quad Cities is under Extreme Drought". WHBF-TV4 Rock Island, Illinois. Retrieved 2009-04-18.
- Iowa Annual Weather Summary 2005 (PDF) (Report). Iowa Agriculture. Retrieved November 10, 2017.
- Ron Smith (2009-08-03). ""Exceptional drought" covers 32 million acres in south Texas | Management content from". Southwest Farm Press. Archived from the original on 2012-07-19. Retrieved 2015-08-14.
- Gleick, Peter (2011-06-16). "The California Drought (2007–2009): Myth versus Reality". The San Francisco Chronicle.
- Smith, Scott (2014-04-19). "California farmers to get more water". Washington Post.
- Mooney, Chris (2015-08-19). "With a stunning 7 million acres burned so far, the U.S. wildfire situation is looking dire". The Washington Post.
- "Archived copy". Archived from the original on 2015-09-12. Retrieved 2015-08-25.CS1 maint: Archived copy as title (link)
- Worland, Justin (2015-07-09). "Strong El Niño Set to Bring California Drought Relief". Time.
- http://web.b.ebscohost.com/ehost/pdfviewer/pdfviewer?sid=f2f628dd-a305-4320-8319-2bd701634078%40sessionmgr114&vid=5&hid=118[dead link]
- John Pike (2012-08-13). "DOD to Review Meat Purchases During Drought". Globalsecurity.org. Retrieved 2015-08-14.
- "Researchers Seek New Strains for Biofuels | Defense News". defensenews.com. Retrieved 2015-08-14.
- Jess Bidgood (September 26, 2016). "Scenes From New England's Drought: Dry Wells, Dead Fish and Ailing Farms". The New York Times. Retrieved September 27, 2016.
- Dylan McGuinness (November 17, 2016). "Despite all the rain, the drought got much worse". Boston Globe. Retrieved November 18, 2016.
- Grover, Hannah (2.4.2019). "Hydrologist says drought impacts will linger". Farmington Daily Times. Archived from the original on 3.4.2019. Check date values in:
- National Integrated Drought Information System
- GIDMaPS Global Integrated Drought Monitoring and Prediction System, University of California, Irvine
- Social & Economic Costs of Drought from "NOAA Socioeconomics" website initiative
- National Oceanic and Atmospheric Administration – National Weather Service
- National Drought Mitigation Center
- 1980s Drought and Subsequent Food Crisis from the Dean Peter Krogh Foreign Affairs Digital Archives
- Study predicts future megadroughts for western U.S. "The chance of a megadrought hitting the western United States rises to 80 percent if emissions fail to slow down in the coming decades." (February 2015), United Press International
- US 'at risk of mega-drought future' (February 2015), British Broadcasting Corporation
- Drought in the United States: Causes and Current Understanding Congressional Research Service<|endoftext|>
| 3.8125 |
634 |
Even though there are various methods for generating hydrogen from water such as electrolysis and photocatalytic methods, the primary challenge in the whole process remains safe and economical storage of hydrogen. Using aluminum, however, seems to be a solution to this challenge. Chemical and physical properties of aluminum favor production of hydrogen gas especially on alloys of aluminum. When aluminum is modified through plasma activation process, the hydrogen production process becomes more efficient. This is because the activation process induces the effects of hydrophilicity which enables the metal to sink in water. It aids polar molecules formation leading to high surface tension and formation of hydrogen bonds hence allowing the metal sink.
Researchers from Center for Hydrogen Energy Technologies at Lithuanian Energy Institute and led by Dr. Darius Milcius investigated the effect of plasma-modified aluminum on hydrogen production. Their experiments involved plasma treatment and analysis using XRD and XPS analysis methods as well as measurements of the hydrogen volumes produced. Due to its effectiveness and efficiency, the work has been published in Energy Technology journal.
The research team was able to measure the amount of hydrogen produced. Inert gas fusion was used to facilitate the measurement due to high reactive nature of hydrogen. From the tests, it only required one minute from the start of the reaction to start generating hydrogen. The process is, however, longer and could continue for approximately 20 minutes before the reaction completes.
Aluminum was chosen by the authors due to its advantages. First, it is cheap and readily available as compared to other metals which could have however been used as well. Aluminum oxide is a highly reactive metal quickly oxides to form a protective layer. The layer prevents any reaction between aluminum and water to take place. As a result, aluminum was modified to gain hydrophilic property that enables it to sink in water and hence inducing the required reaction. The best practice is to perform the plasma-activation process under low temperatures.
The reaction between aluminum and water involves three phases. According to the authors, the first phase otherwise known as induction time refers to the period from which the metal is immersed in water up to the time when hydrogen generation begins. First reaction stage follows. Most of the reactions take place since the oxide layer preventing reaction no longer exists. In the final step, all the reactions are completed and required end products formed. A modified aluminum plasma performs very well and produces a significant amount of end
Darius Milcius and his research team are focused on revolutionizing hydrogen production. The method is inexpensive, provides a better way of harnessing and storing produced hydrogen or oxygen and hence should be adopted by technologists, chemists and other scientists and researchers. The reaction between water and plasma-modified aluminum is environment-friendly due to limited pollution effects as compared to other available methods. Another added advantage is that other products such as aluminum can also be produced through this kind of reaction.
Urbonavicius, M., Varnagiris, S., & Milcius, D. (2017). Generation of Hydrogen through the Reaction between Plasma-Modified Aluminum and Water. Energy Technology, 5(12), 2300-2308.Go To Energy Technology<|endoftext|>
| 3.8125 |
953 |
Listeria monocytogenes is a Gram-positive, rod-shaped, bacterium – miscroscopic in appearance, but looming large among public health officials for its ability to result in fatalities.
First observed in scientific literature in the 1920s, it’s been widely known among microbiologists since the 1960s. This stealthy organism is ubiquitous in the environment, often found in moist conditions, including soil, surface water and decaying vegetation. Salt-tolerant and able to both withstand and grow in temperatures below 1°C, L. monocytogenes is a hardy pathogenic organism. Its flagella – hair-like appendages protruding out of its cell structure – allow it to swim through a variety of hosts, though at a warmer and narrower temperature band.
Listeria monocytogenes is the most infamous of the seventeen known species in the Listeria genus. Only one other member of its tribe (Listeria ivanovii) is considered pathogenic, but it mainly resides in animals, not humans.
“L. mono” is not a leading cause of foodborne illness, however, it is among the leading causes of death from foodborne illness. The U.S. Centers for Disease Control and Prevention (CDC) estimates that, in the U.S., domestically acquired foodborne L. monocytogenes results in 255 fatalities per year. These numbers are down from figures seen prior to the turn of the century, when the CDC estimated approximately 500 such deaths in the United States annually.
Persons most affected by L. monocytogenes’ bacterial infection, listeriosis, are those with immune system deficiencies, caused by anticancer drugs, AIDS and other reasons. Pregnant women are disproportionately infected with L. monocytogenes, but they usually only experience mild, flu-like symptoms. Sadly, the babies they’re carrying don’t always fare as well – passing away in utero or born stillborn at concerning rates.
Overall, the most severe form of the infection has a case-fatality rate of 15 to 30 percent of the population. When listerial meningitis occurs, the case-fatality rate may be as high as 70 percent. In perinatal/neonatal infections, it’s more than 80 percent.
Many foods have been associated with L. monocytogenes, especially when they’re in raw form. Prime examples include ready-to-eat processed products, poultry and meats, vegetables, seafood and improperly or unpasteurized milk, cheeses (particularly soft varieties) and ice cream.
Contamination can come from these raw materials, from the air and from food workers and food- processing environments. Most experts agree that post-processing contamination from food-contact surfaces represent the clearest and most present danger, and surfaces can harbor the organism for years. For example, the identical strain of L. monocytogenes collected from a human listeriosis case in 1989 was isolated more than a decade later from a fresh slice of turkey produced by the very same processing plant.
Several high-profile L. monocytogenes outbreaks over the past decade have plagued food companies and their consumers, drawing more concern and attention from inspectors, the media and the general public. Some have dubbed the swelling interest in this pathogen as a “Listeria hysteria.”
Certainly, control of Listeria species, including Listeria monocytogenes, is vital. Food processors mustn’t let their guard down to test for this persistent pathogen, methodically sampling end product and key areas within food production environments as well as other food contact and non-food contact surfaces in plants that could be reservoirs for Listeria. They also must be mindful of the fact that L. monocytogenes can grow in refrigerated temperatures.
To assist these critical efforts, food producers count on technology to potentially grow selective organisms like Listeria monocytogenes to detectable levels while simultaneously preventing less harmful microorganisms from also growing and interfering with accurate testing. Then they must evaluate its presence and prevalence.
Rapid-test kits for Listeria species as well as the Listeria monocytogenes pathogen have been among the fastest growing tests in the food industry. As global regulatory requirements have changed and become more stringent, the demand for Listeria testing has dramatically increased. In addition, as the food industry has become more global and increasingly competitive, processors want results that can be delivered faster and with more details.<|endoftext|>
| 3.828125 |
698 |
# If a magnet of pole strenth m is divided into four parts such that the length and width of each part is half that of initial one, then the pole strength of each part will be
A
m/4
B
m/2
C
m/8
D
4m
Video Solution
Text Solution
Generated By DoubtnutGPT
## To solve the problem, we need to analyze how the pole strength of a magnet changes when it is divided into smaller parts. Here’s a step-by-step breakdown of the solution:Step 1: Understand the Initial Conditions- We have a bar magnet with an initial pole strength denoted as m.- The dimensions of the magnet are given as length l and width b.Step 2: Determine the Cross-Sectional Area- The cross-sectional area A of the magnet can be calculated as: A=l×bStep 3: Analyze the Division of the Magnet- The magnet is divided into four parts, and each part has its length and width halved.- Therefore, the new dimensions for each part will be: - New Length = l2 - New Width = b2Step 4: Calculate the New Cross-Sectional Area- The new cross-sectional area A′ for each part will be: A′=(l2)×(b2)=l×b4=A4Step 5: Understand the Relationship Between Pole Strength and Cross-Sectional Area- The pole strength m of a magnet is proportional to its cross-sectional area. Thus, if the area changes, the pole strength will also change accordingly.- Since the new area A′ is A4, we can express the new pole strength m′ as: m′=k×A′=k×A4 where k is a constant of proportionality.Step 6: Relate New Pole Strength to Initial Pole Strength- Since the initial pole strength m is given by m=k×A, we can substitute this into our equation for m′: m′=m4Conclusion- Therefore, the pole strength of each part after dividing the magnet will be: m′=m4Final AnswerThe pole strength of each part will be m4.---
|
Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc
NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams.
Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation<|endoftext|>
| 4.625 |
269 |
# Chapter 3 - Equations and Problem Solving - Chapter 3 Test - Page 139: 15
The area of the rectangle is 576 square inches.
#### Work Step by Step
To find the area of the rectangle, we need to know both the length and the width of the rectangle. To find the width, we use the perimeter equation. The formula for the perimeter of the rectangle is width + width + length + length. Let w represent the width and l represent the length. Recall: Perimeter = 2w + 2l Since the perimeter is 100 inches and the length is 32 inches, we can write the equation as: 100 = 2w + 2$\times$ 32 100 = 2w + 64 36 = 2w Divide both sides by 2. w = 18 Now that we know both width and length, we can find the area. Area of a rectangle is equal to width $\times$ length. Area = w $\times$ l Area = 18 $\times$ 32 Area = 576 The area of the rectangle is 576 square inches.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.<|endoftext|>
| 4.5625 |
2,918 |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# Tan Theta Formula
Last updated date: 29th May 2024
Total views: 350.4k
Views today: 4.50k
## Tan Theta
The tangent is defined in right triangle trigonometry as the ratio of the opposite side to the adjacent side (It is applicable for acute angles only because it's only defined this way for right triangles). To find values of the tangent function at different angles when evaluating the tangent function, we first define the reference angle created by the terminal side and the x-axis. Then we calculate the tangent of this reference angle and determine whether it is positive or negative based on which quadrant the terminal side is in. In the first and third quadrants, the tangent is positive. In the second and fourth quadrants, the tangent is negative. The slope of the terminal side is also equal to the tangent.
Let us discuss an introduction to Trigonometry in detail before looking at the formula. Trigonometry is a branch of mathematics concerned with the application of specific functions of angles to calculations. In trigonometry, there are six functions of an angle that are widely used. Sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (csc) are their names and abbreviations. In relation to a right triangle, these six trigonometric functions. The sine of A, or sin A, is defined as the ratio of the side opposite to A and the side opposite to the right angle (the hypotenuse) in a triangle. The other trigonometric functions are defined similarly. These functions are properties of the angle that are independent of the triangle's size, and measured values for several angles were tabulated before computers made trigonometry tables outdated. In geometric figures, trigonometric functions are used to calculate unknown angles and distances from known or measured angles. Trigonometry has a wide range of applications, from specific fields such as oceanography, where it is used to measure the height of tides in oceans, to the backyard of our home, where it can be used to roof a building, make the roof inclined in the case of single independent bungalows, and calculate the height of the roof etc. Here, we will discuss the tan theta formula in detail.
### How to Find the Tangent?
You must first locate the hypotenuse to find the tangent. The hypotenuse is typically the right triangle's longest side. The next task is to decide the angle. There are only two angles to choose from. You cannot choose the right angle. After you've chosen an angle, you will mark the sides. The side opposite to this angle will be the opposite side and the side next to the angle is the adjacent side. After labeling the sides, you can take the required ratio. Let’s discuss ratios, what is tan theta and it’s practical applications?
### What is Tan Theta?
The length of the opposite side to the length of the adjacent side of a right-angled triangle is known as the tangent function or tangent ratio of the angle between the hypotenuse and the base.
As discussed, the tangent function is one of the three most common trigonometric functions, along with sine and cosine. The tangent of an angle in a right triangle is equal to the length of the opposite side (O) divided by the length of the adjacent side (A). It is written simply as 'tan' in a formula.
⇒ tan x = O/A
tan(x) is the symbol for the tangent function which is also called the tan x formula. It is one of the six trigonometric functions that are commonly used. Sine and cosine are most often associated with the tangent. In trigonometry, the tangent function is a periodic function that is very useful.
The tan formula is as follows:
What is tan theta in terms of sine and cos?
⇒ tan x = sin x/cos x
or, tan theta = sin theta/cos theta (here, theta is an angle)
The sine of an angle is equal to the length of the opposite side divided by the length of the hypotenuse side, while the cosine of an angle is the ratio of the adjacent side to the hypotenuse side.
Hence, sin x = Opposite Side/Hypotenuse Side
cos x = Adjacent Side/Hypotenuse Side
Therefore, (tan formula) tan x = Opposite Side/Adjacent Side
### Finding the Tangent of the Triangle
Angles A and B are the two angles we will deal with in this triangle. To find our tangent, we must first find out the hypotenuse. Our right angle is clear to me. Can you see it? That means our hypotenuse is directly across it, and the side that measures 5 is the hypotenuse. Can you see it? Okay, as we have our hypotenuse, let's choose an angle to work with. We'll choose angle B. As B is our angle, our opposite side is the side that measures 3. Our adjacent side is the one that measures 4 because it is the only side next to angle B which is not the hypotenuse.
This means that our tangent of angle B will be the ratio of the opposite side over the adjacent side or we can write it as 3/4 which will be equal to 0.75. Similarly, If we choose angle A, our sides will change and the tangent will be 4/3 which will be equal to 1.33.
### Tangent to Find the Missing Side
In a few problems, we will have to find the missing side of the triangle. This particularly has real-life applications like when construction companies are building on hills. In these kinds of problems, an angle is given to us. To solve these problems, we first have to locate the side that is missing. In the triangle given above, the side missed is the adjacent side to the angle given. Right? What we have to do now is just to write the equation from the definition of the tangent. Once we have the equation, we can move ahead, use algebra to solve for the variable which will be missing.
By multiplying both sides by x and then dividing both sides by tan 66, we've identified our variable. And by using the calculator to measure tan 66, we get the answer as 2.22.
### Arctan - Inverse Tangent Function
There is an inverse function for any trigonometry function, for example, tan has arctan, which works in a reverse manner. These inverse functions have the same name as the originals, but with the word 'arc' added to the start. So, arctan is the opposite of tan. If we know the tangent of an angle and want to know the actual angle, we use the inverse function.
### Large & Negative Angles
The two variable angles in a right triangle are always less than 90 degrees. However, we can find the tangent of any angle, irrespective of its height, as well as the tangent of negative angles. We can also graph the tangent function.
### Tangents will be Used to Calculate the Height of a Building or a Mountain:
You can easily find the height of a building if you know the distance from where you observe it and also the angle of elevation. Similarly, you can find another side of the triangle if you know the value of one side and the angle of depression from the top of the house. All you need to know is a side and an angle of the triangle.
### Conclusion
In fields like astronomy, mapmaking, surveying, and artillery range finding, trigonometry emerged from the need to compute angles and distances. Plane trigonometry deals with issues involving angles and distances in a single plane. Spherical trigonometry considers applications of related problems in more than one plane of three-dimensional space.
### All You Need to know About Tan theta
Students in the regular classes learn about the Triangles and various geometric measurements and operations performed over them. Trigonometry is another branch of mathematics concerned with the measurements of angles and sides corresponding to it in any right-angle triangle. Most commonly heard of functions in introductory chapters of Trigonometry are Sine theta (sin), Cosine theta (cos), tangent theta (tan), cotangent theta (cot), secant theta (sec), and cosecant theta (codec). As we all know a right angle triangle has three sides namely base, height and hypotenuse. These 6 functions of Trigonometry are nothing more than the different combinations of the three sides of the triangle in pairs concerning their proportions with each other. The functions are independent of the size or length of sides but are determined by the angle produced by the two corresponding sides.
Tangent theta or Tan theta is the ratio of the height of a right-angled triangle over the length of the base. Locating the hypotenuse of any triangle other than the right triangles sometimes becomes the difficult part of this approach. Tan theta is also used for obtaining the length of the missing side after measuring the distance between the point of observation and the origin of that side or video versa. But in this case, the value of the angle formed by the base and the line projecting to the top of the other side is required to be known before calculation.
## FAQs on Tan Theta Formula
1. What are the other functions of Trigonometry and their values?
Theta is the term used for the angle of the triangle. It can be anything between 0 to 360 degrees. Likewise, the Trigonometric function values correspond to each value of theta. For example, the value of Sin 30° is ½ while the value of Sin 45° is 1/√2. Similarly, another example is that the value of Tan 30° is 1/√3, Tan 60° is √3 and Tan 45° is 1. Students can get the values of all the functions from any standard Book or reference book with an explanation. Vedantu also provides study materials for the same which you can download and read by yourself.
2. When is the right to learn about the concepts of Trigonometry?
Students learn about various types of triangles such as Acute angled Triangle, Obtuse angled Triangle, Isosceles Triangle, Equilateral Triangle or Right-angled Triangle very early in the classes for secondary education. But Trigonometry as a chapter is introduced from class 9 or 10 in the CBSE or ICSE pattern of the curriculum. Students can learn about triangles before arriving in the 9 class but do have to be conceptually clear about the different types of triangles and the knowledge about the concept and use of angles.
3. Do I need to Practice questions on the Tan Theta chapter?
Trigonometry Is essentially a chapter of Mathematics. So mere reading of the subject matter will not be sufficient to get a good grasp of the concepts. Every student. Mathematics needs to work on the exercise questions provided after each topic. It not only clears the concept of the subject read but gives you an understanding of how to apply the concept to solve the problems in real life. You can also arrange for extra questions to practice by following any reference or complement book publications available in the market and online.
4. Do Vedantu Online courses make available Practice Questions for its Students?
Vedantu Online Course has been started with the motto of helping out the students of both CBSE and ICSE preparing for their term exams, board exams of any class or any other competitive exams after 12th board. As most of the students in India can't afford the expensive coaching classes or special books in the market. So Vedantu provides for them all types of courses to learn the subject in just one click. We also have compiled question sets other than those in the standard Book of every topic. Any student with a smartphone and internet connection can avail of the facilities.
5. Where Can I obtain all the study materials and practice questions set available on Vedantu? And what is the total cost?
The materials available on the Vedantu have been arranged according to the proper course structure and relevant web pages for the subject. The students are required to get enrolled to avail the study materials and other products available on this website. Registration is very easy and free with some basic data to be filled in. Then students can get access to all the pages they want. Students can also download the same materials available in PDF format by clicking on the link provided and without paying any cost to do so.
6. What is Tan Theta Equal to?
In the right triangle, the tangent is defined as the ratio of the opposite side to the adjacent side. In other words, the length of the opposite side to the length of the adjacent side will be known as the tangent function or tangent ratio of an angle.
7. Can Trigonometry Be Used in Everyday Life?
While trigonometry does not have clear applications in solving practical problems, it is used in a variety of things that we appreciate. For example, sound travels in waves in music, and this pattern, although not as normal as a sine or cosine function, is still useful in computer music creation. Since a machine can't listen to or comprehend music as we can, it's represented mathematically by its constituent sound waves. As a result, sound engineers must have a basic understanding of trigonometry. And, thanks to trigonometry, the good music that these sound engineers make is used to help us recover from our hectic, stressful lives.
8. What is the Use of the Tan Theta Formula?
The most efficient and progressive application of trigonometry is the analysis and simplification of equations using various trigonometric functions such as sine, cosine, and tangent, among others. In engineering fields such as mechanical engineering, electronics, and mechatronics, the analytical use of trigonometry is important. So, with the basics of the tan theta formula we can easily solve and simplify higher-order problems such as 1 tan theta, tan 3 theta, tan 3 theta, tan 3 theta 1 tan 2 theta, tan3x formula and so on.<|endoftext|>
| 4.75 |
606 |
# A non-trivial problem related to surface area and volume of
Find the breadth and the height of a cuboid with square base, being given the volume 63 $\mathrm{in}^3$ and surface area 102 $\mathrm{in}^2$.
Discussion:This problem is from the book "How to Solve It" by George Polya. Let us reproduce Polya's words.
What are the unknowns? The side of the base, say $x$, and the height of the cuboid, say $y$.
What are the data? The volume, 63, and the area, 102.
What is the condition? The cuboid whose base is a square with side $x$ and whose height is $y$ must have the volume 63 and the area 102.
Separate the various parts of the condition. There are two parts, one concerned with the volume, the other with the area.
We can scarcely hesitate in dividing the whole condition just in these two parts; but we cannot write down these parts "immediately". We must know how to calculate the volume and the various parts of the area. Yet, if we know that much geometry, we can easily restate both parts of the condition so that the translation into equations is feasible. We write on the left hand side an essentially rearranged and expanded statement of the problem, ready for translation into algebraic language.
.
Of a cuboid with square base find the side of the base $x$ and the height $y$ First. The volume is given. 63 The area of the base which is square with side $x$ $x^2$ and the height $y$ determine the volume which is their product. $x^2y=63$ Second. The area of the surface area is given. 102 The surface consists of two squares with side $x$ $2x^2$ and of four rectangles, each with base $x$ and height $y$, $6xy$ whose sum is the area. $2x^2+4xy=102$
Translation to the algebraic expression set up following equations
\begin{align} x^2y=63 \\ 2x^2+4xy=102. \end{align}
The first equation gives $y=63/x^2$. Substitute this into the second equation to get,
\begin{align} x^3-51x+126=0. \end{align}
This is a cubic equation. One factor of $x^3-51x+126$ is $(x-3)$ and the other is $(x^2+3x-42)$. Thus, one solution is $x=3$. Substitute $x=3$ in $y=63/x^2$ to get $y=7$. We encourage you to solve $x^2+3x-42=0$ to get the other solution(s).<|endoftext|>
| 4.5625 |
526 |
# One container contains 6 red and 4 white balls, while a second container contains 7 red and 3 white balls
One container contains 6 red and 4 white balls, while a second container contains 7 red and 3 white balls. A ball is chosen at random from the first container and placed in the second. Then a ball is chosen at random from the second container and placed in first.
a. What is the probability to choose a red ball of the second container?
b. What is the probability the ball was red chosen from the first container if know that from the second selected white?
Solution for question 1:
Probability that a red ball will be chosen on first step: $\frac{6}{10}$
In this case, we will have 8 red and 3 white balls in the second container, and the probability to chose a red ball on second step is $\frac{8}{11}$.
Overall probability in this case gives $\frac{6}{10}*\frac{8}{11}$ = $\frac{48}{110}$. Let's mark it as p$_1$.
Probability that a white ball will be chosen on first step: $\frac{4}{10}$
Similarly to the previous case, we will get probability to choose a red ball from the second container is p$_2$=$\frac{4}{10}*\frac{7}{11}=\frac{28}{110}$
Because our cases do not intersect, we have total probability p=p$_1$+p$_2$=$\frac{76}{110}$=$\frac{38}{55}$.
Solutin for question 2:
Here we shall use Bayes's theorem: $$P(A|B)=\frac{P(B|A)*P(A)}{P(B)}$$
Where event A means that a red ball was chosen from the first container, B means that a white ball was chosen from the second container.
What we know:
$P(A)=\frac{6}{10}$
$P(B)=1-\frac{38}{55}=\frac{17}{55}$ (1-probability that a red ball will be chosen from the second container) $P(B|A)=\frac{6}{10}*\frac{3}{11}=\frac{9}{55}$
And here we are:
$$P(A|B)=\frac{\frac{9}{55}*\frac{6}{10}}{\frac{17}{55}}=\frac{27}{85}$$<|endoftext|>
| 4.75 |
1,585 |
What is multiple myeloma?
Multiple myeloma and other plasma cell neoplasms (cancers) are diseases in which the body makes too many plasma cells.
Plasma cells develop from B lymphocytes (B cells), a type of white blood cell that is made in the bone marrow. Normally, when bacteria or viruses enter the body, some of the B cells will change into plasma cells. The plasma cells make a different antibody to fight each type of bacteria or virus that enters the body, to stop infection and disease.
Plasma cell neoplasms are diseases in which there are too many plasma cells, or myeloma cells, that are unable to do their usual work in the bone marrow. When this happens, there is less room for healthy red blood cells, white blood cells and platelets. This condition may cause anemia or easy bleeding, or make it easier to get an infection. The abnormal plasma cells often form tumors in bones or soft tissues of the body. The plasma cells also make an antibody protein, called M protein9, which is not needed by the body and does not help fight infection. These antibody proteins build up in the bone marrow and can cause the blood to thicken or can damage the kidneys.
In multiple myeloma, abnormal plasma cells (myeloma cells) build up in the bone marrow, forming tumors in many bones of the body. These tumors may prevent the bone marrow from making enough healthy blood cells. Normally, the bone marrow produces stem cells (immature cells) that develop into three types of mature blood cells:
- Red blood cells that carry oxygen and other materials to all tissues of the body.
- White blood cells that fight infection and disease.
- Platelets that help prevent bleeding by causing blood clots to form.
As the number of myeloma cells increases, fewer red blood cells, white blood cells and platelets are made. The myeloma cells also damage and weaken the hard parts of the bones.
A tumor can damage the bone and cause hypercalcemia (a condition in which there is too much calcium in the blood). This damage can affect many organs in the body, including the kidneys, nerves, heart, muscles and digestive tract, and cause serious health problems.
In rare cases, multiple myeloma can cause organs to fail, which may be caused by a condition called amyloidosis. Antibody proteins build up and may bind together and collect in organs — such as the kidney and heart —causing the organs to become stiff and unable to function.
What are the risk factors for multiple myeloma?
Age can affect the risk of developing plasma cell neoplasms. Plasma cell neoplasms are found most often in people who are middle aged or older.
Other risk factors include the following:
- Being black.
- Being male.
- Having a brother or sister who has multiple myeloma.
- Exposure to petroleum and other chemicals.
- Exposure to high amounts of radiation.
What are the symptoms of multiple myeloma?
Sometimes multiple myeloma does not cause any symptoms. The following symptoms may be caused by multiple myeloma or other conditions. A doctor should be consulted if any of the following problems occur:
- Bone pain, often in the back or ribs.
- Bones that break easily.
- Fever for no known reason or frequent infections.
- Easy bruising or bleeding.
- Trouble breathing.
- Weakness of the arms or legs.
- Feeling very tired.
How is multiple myeloma diagnosed?
In addition to a complete medical history and physical examination, diagnostic procedures for myeloma bone disease may include the following:
- X-ray – a diagnostic test that uses invisible electromagnetic energy beams to produce images of internal tissues, bones and organs onto film.
- Blood and urine tests.
- Bone marrow aspiration and/or biopsy – a procedure that involves taking a small amount of bone marrow fluid (aspiration) and/or solid bone marrow tissue (called a core biopsy), usually from the hip bones, to be examined for the number, size and maturity of blood cells and/or abnormal cells.
- Magnetic resonance imaging (MRI) – a diagnostic procedure that uses a combination of large magnets, radio frequencies and a computer to produce detailed images of organs and structures within the body.
How are prognosis and staging determined?
Certain factors affect prognosis (chance of recovery) and treatment options. The prognosis depends on the following:
- The type of plasma cell neoplasm.
- The stage of the disease.
- Whether a certain immunoglobulin (antibody) is present.
- Whether the kidney is damaged.
- Whether the cancer responds to initial treatment or recurs (comes back).
After multiple myeloma and other plasma cell neoplasms have been diagnosed, tests are done to find out the amount of cancer in the body. This process is called staging. It is important to know the stage in order to plan treatment.
There are three stages for multiple myeloma. The number of myeloma cells in the body is determined by the following:
- Level of hemoglobin in the blood.
- Levels of calcium and creatinine in the blood.
- Amount of bone damage.
- Amount of antibody M protein in the blood and/or urine, and how well the kidneys are working.
The following stages are used for multiple myeloma:
- Stage I – In stage I multiple myeloma, there is a low number of myeloma cells in the body.
- Stage II – In stage II multiple myeloma, there is a moderate number of myeloma cells in the body.
- Stage III – In stage III multiple myeloma, there is a large number of myeloma cells in the body.
Treatment for multiple myeloma
Specific treatment will be determined by your physician based on:
- Your age, overall health and medical history.
- Extent of the disease.
- Your tolerance for specific medications, procedures or therapies.
- Expectations for the course of the disease.
- Your opinion or preference.
Patients without symptoms may not need treatment. When symptoms appear, treatment for any stage of multiple myeloma may include the following:
- Bone marrow or stem cell transplantation.
- High-dose corticosteroid therapy.
- Thalidomide therapy.
- Combination chemotherapy.
- Biologic therapy with monoclonal antibodies.
- Radiation therapy for tumors of the spine.
- Alpha interferon – a biological response modifier (a substance that stimulates or improves the ability of the body’s immune system to fight disease) that interferes with the division of cancer cells, therefore slowing tumor growth. Interferons are substances normally produced by the body but that also can by produced in the laboratory.
Clinical trials available
The VCU Massey Cancer Center has a number of Phase I and Phase II clinical trials for multiple myeloma. Please click to learn about the benefits of enrolling in a clinical trial.
To locate a trial at Massey, please visit “Find a clinical trial” on this Web site.
Attribution: Sections of this article have been reprinted from the article “Multiple Myeloma and Other Plasma Cell Neoplasms” prepared by the National Cancer Institute. For more information from the NCI, please write to this address:
NCI Public Inquiries Office
6116 Executive Boulevard, MSC8322
Bethesda, Maryland 20892-8322<|endoftext|>
| 3.90625 |
157 |
the newer branches of engineering, and dates back to the late 19th century. It is the branch of engineering that deals with the technology of electricity. Electrical engineers work on a wide range of components, devices and systems, from tiny microchips to huge power station generators.
Early experiments with electricity included primitive batteries and static charges. However, the actual design, construction and manufacturing of useful devices and systems began with the implementation of Michael Faraday’s Law of Induction, which essentially states that the voltage in a circuit is proportional to the rate of change in the magnetic field through the circuit.
This course will help you become familiar with engineering numbers and notation, and learn about the two most important electrical quantities: current and voltage.
No Reviews found for this course.<|endoftext|>
| 4.09375 |
1,524 |
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
The relationship between the length and resistance of a wire.
Extracts from this document...
Introduction
The relationship between the length and resistance of a wire
Aim
For this experiment, I will be finding out the resistance of a wire in which its length is the variable. I will be investigating the process of how much the resistance increases due to the increase in the size of the resistor (the wire).
Investigation
Resistance is the property of an electrical conductor, to work against the flow of the current and change some of the electrical energy into heat (Taken from “The Oxford mini reference for science”) The amount of resistance in an electric circuit determines the amount of current flowing in the circuit for any given voltage applied to the circuit, according to Ohm's law. The unit of resistance is an ohm and resistance is measured in amperes. The abbreviation for electronic resistance is “R”, and the symbol for ohms in electronic circuits is Ω.
To find the value of resistance, we use the equation:
Resistance = Voltage / Current
The resistance of an object is determined by the substance of the material, the amount of particles it has to pass through. The temperature can also change the resistance of the resistor; the temperature affects the amount of energy in particles of materials.
Middle
Length of wire (mm)
Voltage across wire (volts)
Current through wire (amps)
Resistance (Ω)
200mm
3.00 V
4.63 A
0.65 Ω
300mm
3.00 V
3.04 A
0.99 Ω
400mm
3.00 V
2.36 A
1.27 Ω
500mm
3.00 V
1.86 A
1.61 Ω
600mm
3.00 V
1.58A
1.90 Ω
700mm
3.00 V
1.37 A
2.19 Ω
800mm
3.00 V
1.18 A
2.54 Ω
2nd set of results
Length of wire (mm) Voltage across wire (volts) Current through wire (amps) Resistance (Ω) 200mm 3.00 V 4.67 A 0.64 Ω 300mm 3.00 V 3.03 A 0.99 Ω 400mm 3.00 V 2.36 A 1.27 Ω 500mm 3.00 V 1.86 A 1.61 Ω 600mm 3.00 V 1.57 A 1.91 Ω 700mm 3.00 V 1.37 A 2.19 Ω 800mm 3.00 V 1.18 A 2.54 Ω
3rd set of results
Length of wire (mm)
Conclusion
If I were to make improvements on the experiment, I would have tried to make the wire lengths as accurate as possible, although in this experiment there were no extremely odd results.
An experiment that I could have done to make sure that the experiment fairer would have been to keep the resistor under a constant temperature. The circuit would be the same, but the resistor would be under a temperature so that the resistivity of the wire was kept the same. Possibly under water the temperature would have been the same.
The reason this would make the experiment fairer would be because there would be no slant at the end of the graph. Without a constant temperature, there is more resistance if the temperature rises, because to atoms with kinetic energy.
I have made my experiment accurate, by turning the power pack off while changing the length of the resistor, to stop the resistor heating up and causing more resistance.
I have made my experiment reliable, by doing the experiment several times, and averaging my results emitting any anomalies.
For my conclusion, I have found out that the resistance and the voltage are the same if the current is kept the same and that the resistance and the voltage are in proportion with the length of the wire.
Jack Cummins 10LB
This student written piece of work is one of many that can be found in our GCSE Electricity and Magnetism section.
Found what you're looking for?
• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month
Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Related GCSE Electricity and Magnetism essays
1. Investigate the relationship between the length of wire and its resistance when a voltage ...
The resistance of a piece of wire depends on the affect of the 4 factors: Length, Cross sectional area, Material and Temperature. In this investigation we will be looking at the length. > Length - The length of the metal conductor can affect the resistance.
2. For this experiment, I will be finding out the resistance of a wire in ...
If this experiment was to be done under perfect conditions, this is what the graph should look like if the graph was plotted with the resistance against the length of the wire. [2] Voltage (V) Current (I) Resistance (?) Length of wire (cm)
1. To investigate how the length (mm) and the cross-sectional (mm2) area of a wire ...
in this experiment, so it is important that I include it in my procedure to gain extensive results. As stated above for length, the procedure is the same, except that I will repeat the procedure a number of times for the different types of fuse wires that have different cross-sectional areas.
2. For this experiment, I will be finding out the resistance of a wire in ...
able to pass through it, as there is more room within the wire. The material the resistor is made out of will also make a difference as if the wire has a high density, there will be more atoms in the wire causing high resistance, but if the wire has
1. I will be finding out the resistance of a wire in which its length ...
plenty of kinetic energy in the resistor and the atoms will be moving about. The current will find it hard to pass through the wire with atoms moving about so there will be more resistance. For the experiment, the variables except the length will be kept the same, as it will make it a fair test.
2. To find a relationship between the molecular size of different alcohols and the associated ...
+ C - O + O - H 3(435) + 358 + 464 = 2127 kJ/mol Ethanol: 5 (C - H) + C - C + C - O + O - H 5(435) + 347 + 358 + 464 = 3344 kJ/mol Propanol: 7 (C - H) + 2 (C - C)
• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to<|endoftext|>
| 4.40625 |
575 |
Reading : Along with the new year, comes a shift in the reading focus in 1-Bu. The main focus for all of the reading units from here on out will be on comprehension. Of course, we won't leave decoding and fluency in the dust, but now we're going to be looking very closely at many different kinds of texts and working on extracting meaning from those texts.
Our current reading unit is focusing on schema and using our schema to better understand the text. One's schema is information that he/she knows about a topic. (I'll get much more into it as we go through our unit.) To kick off the unit, we'll be talking about 'real reading' and 'fake reading' next week. The goal is for all of the students to be doing 'real reading,' which is reading the words AND thinking about and understanding what the words and story mean. 'Fake reading' is just reading the words, but not thinking about the text.
Author Study : Last week we started reading books by Ezra Jack Keats. The students are enjoying his books. We spent some time discussing Ezra Jack Keats' illustrations, as he tends to use collages in many of his books. As your child to explain what a collage is.
Writing : We finally wrapped up our Small Moments writing unit last week, and we will be moving on to a 'How To' unit this week. Ask you child what a 'how to' book is. The students know that 'how to' texts are everywhere, from recipes, to game directions, to a book that teaches you how to draw. I have told the students that if they have examples of 'how to' writing at home, to please bring them in. These could be directions for how to put together a lego set, how to make macaroni and cheese, or even how to play a board game. It's always great for the students to see many examples of what 'how to' writing looks like in the real world.
Math : We've been doing a great job with our measuring unit. Before break, the students learned the importance of using standard units of measure. We've learned about measuring temperature in degrees Fahrenheit and length in feet and inches. The students loved using tape measures to measure various body parts last week. (See pictures of that below.)
Phonics : Now that we have learned about all of the consonants and short vowels, we're taking it to the next step. Recently, we have been focusing on blends. Blends are when two consonants blend their sounds together. Last week we focused on s blends (i.e. sc, sn, sw) and l blends (i.e. pl, sl, bl, fl).
Here are some pictures of us in the classroom last week :<|endoftext|>
| 4.0625 |
4,354 |
Home » Math Theory » Numbers » Place Value of Tenths, Hundredths, and Thousandths
# Place Value of Tenths, Hundredths, and Thousandths
## Introduction
As a basic number system, we use decimals. The number 10 is the basis of the decimal system. It is sometimes referred to as a base-10 number system.
We will cover in this article more about the decimal place value and place value of tenths, hundredths, and thousandths.
## What is a place value?
### Definition
A digit’s position within a number determines its place value. It sets the value the number represents.
Let us say, for example; we have 1, 10, 100, and 1000.
1: ones place
10: tens place
100: hundreds place
1000: thousands place
The place value of 1 represents the value it holds. As the place value moves to the left, the value of the number becomes greater by 10 times.
The image below shows the place value from millions to ones.
## Decimal Place Value
The decimal place value shows the parts of a number less than 1. They appear to the right of the decimal point. Each place gets ten times smaller moving from left to right. After the decimal point, the decimal place values are tenths, hundredths, thousandths, and so on.
From the decimal point, the first digit is the tenths place.
From the decimal point, the second digit is the hundredths place.
From the decimal point, the third digit is the thousandths place.
Let us say, for example; we have the number 2.345.
2 is the digit which is a whole number.
3 is 3 tenths and represents $\frac{3}{10}$.
4 is 4 hundredths and represents $\frac{4}{100}$.
5 is 5 thousandths and represents $\frac{5}{1000}$.
For another example, let us have the number 7.654.
7 is the digit which is a whole number.
6 is 6 tenths and represents $\frac{6}{10}$.
5 is 5 hundredths and represents $\frac{5}{100}$.
4 is 4 thousandths and represents $\frac{4}{1000}$.
## Decimal Point
### Definition
The decimal point, which separates a number’s whole number part from its fractional part, is a dot. The decimal numbers to the right of the decimal point have values that are less than one.
Let us say, for example; we have 0.1, 0.01, and 0.001.
0.1: tenths
0.01: hundredths
0.001: thousandths
Since the decimal point is like a fraction, 0.1 is $\frac{1}{10}$, 0.01 is $\frac{1}{100}$, and 0.001 is $\frac{1}{1000}$.
The image below shows the decimal place values tenths, hundredths, thousandths, ten thousandths, hundred thousandths, and millionths.
Thousandths is the place value that appears as the third digit right after the decimal point. Its value is less than one or represents a fractional number. It is 1/1000 position of the digit. Therefore, it is 1 ÷ 1000 = 0.001, which is 1 thousandth.
## Tenths
Tenths is the place value that appears right after the decimal point. Its value is less than one or represents a fractional number. It is 1/10 position of the digit. Therefore, it is 1 ÷ 10 = 0.1, which is 1 tenth.
As shown below, from the decimal point, the first digit is the tenths place.
The digits in the tenths place can be written as a fraction out of 10. Hence, 0.2 is 2 tenths which is $\frac{2}{10}$ as a fractional number.
As for more examples, let us write the following numbers as fractions.
( a ) 0.4
( b ) 0.6
( c ) 0.9
( d ) 0.7
( e ) 0.5
( a ) 0.4 is read as 4 tenths. It is equal to $\frac{4}{10}$.
( b ) 0.6 is read as 6 tenths. It is equal to $\frac{6}{10}$.
( c ) 0.9 is read as 9 tenths. It is equal to $\frac{9}{10}$.
( d ) 0.7 is read as 7 tenths. It is equal to $\frac{7}{10}$.
( e ) 0.5 is read as 5 tenths. It is equal to $\frac{5}{10}$.
## Hundredths
Hundredths is the place value that appears as the second digit right after the decimal point. Its value is less than one or represents a fractional number. It is 1/100 position of the digit. Therefore, it is 1 ÷ 100 = 0.01, which is 1 hundredth.
Starting from the decimal point, the second column to the right is the hundredths place, as shown in the table below.
The digits in the hundredths place can be written as a fraction out of 100. Hence, 0.02 is 2 hundredths which is $\frac{2}{100}$ as a fractional number.
As for more examples, let us write the following numbers as fractions.
( a ) 0.05
( b ) 0.08
( c ) 0.03
( d ) 0.07
( e ) 0.04
( a ) 0.05 is read as 5 hundredths. It is equal to $\frac{5}{100}$.
( b ) 0.08 is read as 8 hundredths. It is equal to $\frac{8}{100}$.
( c ) 0.03 is read as 3 hundredths. It is equal to $\frac{3}{100}$.
( d ) 0.07 is read as 7 hundredths. It is equal to $\frac{7}{100}$.
( e ) 0.04 is read as 4 hundredths. It is equal to $\frac{4}{100}$.
## Thousandths
Thousandths is the place value that appears as the third digit right after the decimal point. Its value is less than 1 or represents a fractional number. It is 1/1000 position of the digit. Therefore, it is 1 ÷ 1000 = 0.001, which is 1 thousandth.
As shown in the table below, from the decimal point, the thousandths place is the third digit.
The digits in the thousandths place can be written as a fraction out of 1000. Hence, 0.002 is 2 thousandths which is $\frac{2}{1000}$ as a fractional number.
As for more examples, let us write the following numbers as fractions.
( a ) 0.007
( b ) 0.009
( c ) 0.005
( d ) 0.003
( e ) 0.006
( a ) 0.007 is read as 7 thousandths. It is equal to $\frac{7}{1000}$.
( b ) 0.009 is read as 9 thousandths. It is equal to $\frac{9}{1000}$.
( c ) 0.005 is read as 5 thousandths. It is equal to $\frac{5}{1000}$.
( d ) 0.003 is read as 3 thousandths. It is equal to $\frac{3}{1000}$.
( e ) 0.006 is read as 6 thousandths. It is equal to $\frac{6}{1000}$.
## Place Value of Tenths, Hundredths, and Thousandths
The figure shows the place value table for tenths, hundredths, and thousandths.
Tenths is the place value that appears right after the decimal point. Its value is less than 1 or represents a fractional number. It is 1/10 position of the digit.
Hundredths is the place value that appears as the second digit right after the decimal point. Its value is less than 1 or represents a fractional number. It is 1/100 position of the digit.
Thousandths is the place value that appears as the third digit right after the decimal point. Its value is less than 1 or represents a fractional number. It is 1/1000 position of the digit.
For example, let us write each number in a place value chart:
( a ) 3.7
( b ) 12. 6
( c ) 75. 87
( d ) 938. 047
( e ) 867.076
## More Examples
Example 1
Write the place value of each digit.
( a ) 56.389
( b ) 217.936
( c ) 6.143
( d ) 837.124
Solution
( a ) 56.389
In 56.389, the place value of 5 holds 5 tens or 50.
The place value of 6 is 6.
The place value of 3 is 3 tenths or $\frac{3}{10}$.
The place value of 8 is 8 hundredths or $\frac{8}{100}$.
The place value of 9 is 9 thousandths or $\frac{9}{1000}$.
( b ) 217.936
In 217.936, the place value of 2 holds 2 hundreds or 200.
The place value of 1 is 1 ten or 10.
The place value of 7 is 7.
The place value of 9 holds 9 tenths or $\frac{3}{10}$}.
The place value of 3 holds 3 hundredths or $\frac{3}{100}$.
The place value of 6 holds 6 thousandths or $\frac{6}{1000}$.
( c ) 6.143
The place value of 6 is 6.
The place value of 1 holds 1 tenth or $\frac{9}{10}$.
The place value of 4 is 4 hundredths or $\frac{3}{100}$.
The place value of 3 holds 3 thousandths or $\frac{3}{1000}$.
( d ) 837.124
The place value of the digit 8 holds 8 hundreds or 800.
The place value of the digit 3 holds 3 tens or 30.
The place value of 7 is 7.
The place value of 1 is 1 tenth or $\frac{1}{10}$.
The place value of 2 holds 2 hundredths or $\frac{2}{100}$.
The place value of 4 is 4 thousandths or $\frac{4}{1000}$.
Example 2
Write each number in the place value table.
( a ) 0.968
( b ) 12.859
( c ) 47.028
( d ) 348.956
( e ) 274.018
Solution
Remember that the digits of the whole number part start from the right to left before the decimal point. On the other hand, the place value of the fractional part appears after the decimal point.
Example 3
Identify the place value of the digit 5 in each decimal.
( a ) 487.509
( b ) 62.957
( c ) 1.295
( d ) 76.3859
( e ) 298.1905
Solution
( a ) 487.509
The place value of 5 in 487.509 is 5 tenths or $\frac{5}{10}$.
( b ) 62.957
The place value of 5 in 62.957 is 5 hundredths or $\frac{5}{100}$.
( c ) 1.295
The place value of 5 in 1.295 is 5 thousandths or $\frac{5}{1000}$.
( d ) 76.3859
The place value of 5 in 76.3859 is 5 thousandths or $\frac{5}{1000}$.
( e ) 298.1905
The place value of 5 in 298.1905 is 5 ten thousandths or $\frac{5}{10000}$.
Example 4
Identify the place value of the underlined digit in each number below.
( a ) 283.058
( b ) 12.938
( c ) 2. 3896
( d ) 23.4987
( e ) 384.9271
Solution
( a ) 283.058
The place value of 5 in 283.058 is 5 hundredths or $\frac{5}{100}$.
( b ) 12.938
The place value of 8 in 12.938 is 8 thousandths or $\frac{8}{1000}$.
( c ) 2. 3896
The place value of 3 in 2.3896 is 3 tenths or $\frac{3}{10}$.
( d ) 23.4987
The place value of 9 in 23.4987 is 9 hundredths or $\frac{9}{100}$.
( e ) 384.9271
The place value of 1 in 384.9271 is 1 ten thousandth or $\frac{1}{10000}$.
Example 5
In the number 298.734, write the decimal value of each digit.
Solution
The place value of 2 holds 2 hundreds or 200.
The place value of 9 is 9 tens or 90.
The place value of 8 is 8.
The place value of 7 holds 7 tenths (0.7) or $\frac{1}{10}$.
The place value of 3 holds 3 hundredths (0.03) or $\frac{3}{100}4. The place value of 4 is 4 thousandths (0.004) or$\frac{4}{1000}\$.
## Summary
A digit’s position within a number determines its place value. It sets the value the number represents.
The decimal point, which separates a number’s whole number part from its fractional part, is a dot. The digits to the right of the decimal point have values that are less than one.
Tenths is the place value that appears right after the decimal point. Its value is less than 1 or represents a fractional number. It is 1/10 position of the digit.
Hundredths is the place value that appears as the second digit right after the decimal point. Its value is less than 1 or represents a fractional number. It is 1/100 position of the digit.
Thousandths is the place value that appears as the third digit right after the decimal point. Its value is less than 1 or represents a fractional number. It is 1/1000 position of the digit.
The table below shows the place value table for tenths, hundredths, and thousandths.
## Frequently Asked Questions on the Place Value of Tenths, Hundredths, and Thousandths
### What is a place value chart?
A table that helps in determining each digit’s place value based on where it appears in the number is known as a place value chart.
The place value chart shows the whole numbers and the fractional part of the number.
The decimal point ( . ) is the symbol that separates the whole numbers and those less than 1 in value.
Whole numbers
The ones place, starting from the decimal point, is the first digit to the left.
The tens place, starting from the decimal point, is the second digit to the left.
The hundreds place, starting from the decimal point, is the third digit to the left.
The thousands place, starting from the decimal point, is the fourth digit to the left.
The ten thousands place is the fifth digit to the left of the decimal point.
The hundred thousands place, starting from the decimal point, is the sixth digit to the left and so on.
Fractional Part
From the decimal point, the first digit to the right is the tenths place. It is 1/10 position of the digit.
From the decimal point, the second digit to the right is the hundredths place. It is 1/100 position of the digit.
From the decimal point, the third digit to the right is the thousandths place. It is 1/1000 position of the digit.
From the decimal point, the fourth digit to the right is the ten thousandths place. It is 1/10000 position of the digit.
From the decimal point, the fifth digit to the right is the hundred thousandths place. It is 1/100000 position of the digit.
### How do tens place and tenths place differ from one another?
From the decimal point, the first digit to its right is the tenths place while the second digit to its left is the tens place.
For example, for the number 123.456, digit 2 holds the tens place while digit 4 holds the tenths place.
### What is the difference between hundreds place and hundredths place?
From the decimal point, the second digit to its right is the hundredths place while the third digit to its left is the hundreds place.
For example, for the number 123.456, the digit 1 holds the hundreds place while the digit 5 holds the hundredths place.
### What is the difference between thousands place and thousandths place?
From the decimal point, the third digit to its right is the thousandths place while the fourth digit to its left is the thousands place.
For example, for the number 7123.456, the digit 7 holds the thousands place while the digit 6 holds the thousandths place.
### How do you write 1 tenth, 1 hundredth, and 1 thousandth?
Tenths is the place value that appears right after the decimal point. Its value is less than 1 or represents a fractional number. It is 1/10 position of the digit. Therefore, it is 1 ÷ 10 = 0.1, which is 1 tenth.
Hundredths is the place value that appears as the second digit right after the decimal point. Its value is less than 1 or represents a fractional number. It is 1/100 position of the digit. Therefore, it is 1 ÷ 100 = 0.01, which is 1 hundredth.
Thousandths is the place value that appears as the third digit right after the decimal point. Its value is less than 1 or represents a fractional number. It is 1/1000 position of the digit. Therefore, it is 1 ÷ 1000 = 0.001, which is 1 thousandth.
### What distinguishes the fractional and whole number parts of the place value chart?
Starting from the decimal point, the digits to the left stand in for the whole number part, while the digits to the right represent the fractional part.
Whole numbers
The ones place, starting from the decimal point, is the first digit to the left.
The tens place, starting from the decimal point, is the second digit to the left.
The hundreds place, starting from the decimal point, is the third digit to the left.
The thousands place is the fourth digit to the left of the decimal point.
The ten thousands place is the fifth digit to the left of the decimal point.
The hundred thousand place, starting from the decimal point, is the sixth digit to the left.
Fractional Part
The tenths place, starting from the decimal point, is the first digit to the right. It is 1/10 position of the digit.
The hundredths place, starting from the decimal point, is the second digit to the right. It is 1/100 position of the digit.
The thousandths place, starting from the decimal point, is the third digit to the right. It is 1/1000 position of the digit.
The ten thousandths place, starting from the decimal point, is the fourth digit to the right. It is 1/10000 position of the digit.
The hundred thousandths place is the fifth digit to the right of the decimal point. It is 1/100000 position of the digit.<|endoftext|>
| 4.9375 |
647 |
LR Circuits Video Lessons
Concept
# Problem: To understand the mathematics of current decay in an L-R circuit.A DC voltage source is connected to a resistor of resistance R and an inductor with inductance L, forming the circuit shown in the figure. For a long time before t = 0, the switch has been in the position shown, so that a current I0 has been built up in the circuit by the voltage source. At t = 0 the switch is thrown to remove the voltage source from the circuit. This problem concerns the behavior of the current from the circuit. This problem concerns the behavior of the current I (t) through the inductor and the voltage V (t) across the indicator at time t after t = 0.Part A. What is the differential equation satisfied by the current I (t) after time t = 0? Express dI(t)/dt in terms of I (t), R, and L.Part B. What is the expression for I(t) obtained by solving the differential equation that I(t) satisfied after t = 0? Express your answer in terms f the initial current I0 as well as L, R, and t.
###### FREE Expert Solution
As electric current flows in an inductor, energy is stored in the inductor.
This energy is expressed by:
$\overline{){\mathbf{I}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}{\mathbf{R}}{\mathbf{+}}{\mathbf{L}}\frac{\mathbf{d}\mathbf{I}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}{\mathbf{0}}}$
Part A
From the equation of the energy stored in a conductor, we can solve for the differential, dI(t)/dt as:
84% (3 ratings)
###### Problem Details
To understand the mathematics of current decay in an L-R circuit.
A DC voltage source is connected to a resistor of resistance R and an inductor with inductance L, forming the circuit shown in the figure. For a long time before t = 0, the switch has been in the position shown, so that a current I0 has been built up in the circuit by the voltage source. At t = 0 the switch is thrown to remove the voltage source from the circuit. This problem concerns the behavior of the current from the circuit. This problem concerns the behavior of the current I (t) through the inductor and the voltage V (t) across the indicator at time t after t = 0.
Part A. What is the differential equation satisfied by the current I (t) after time t = 0? Express dI(t)/dt in terms of I (t), R, and L.
Part B. What is the expression for I(t) obtained by solving the differential equation that I(t) satisfied after t = 0? Express your answer in terms f the initial current I0 as well as L, R, and t.<|endoftext|>
| 4.5 |
744 |
1793–1863, American frontier hero and statesman of Texas, b. near Lexington, Va.
He moved (c.1806) with his family to Tennessee and lived much of his youth with the Cherokee, by whom he was adopted. Serving (1814) in the Creek campaign under Andrew Jackson, he was seriously wounded (1814) while fighting bravely at the battle of Horseshoe Bend. He returned to Tennessee, was admitted (1818) to the bar, practiced law in Lebanon, Tenn., and held many state offices.
Tall, vigorous, and dramatic in speech and in action, Houston, like Jackson, captured the popular imagination. He was sent (1823, 1825) to the U.S. Congress as a Democrat. Elected (1827) governor of Tennessee, Houston seemed in 1829 to have a bright political future, with his reelection almost assured and the Democrats strengthening themselves in national politics. Suddenly, however, his wife, Eliza Allen Houston, left him, and he immediately resigned (1829) his governorship. He rejoined the Cherokee in what is now Oklahoma. There he lived with them as government post trader and as adviser, drinking heavily during much of this period.
In 1833 Houston moved on through Arkansas to Texas. He had little to do with the preliminaries of the Texas Revolution, although he watched the struggle closely. He was a member of the convention that set up a provisional government in Texas and of the convention (1836) that declared Texas independent. He was made commander in chief of the revolutionary troops. After the surrender of the Alamo (Mar., 1836), Houston's army persistently retreated before the numerically superior forces of Santa Anna, and there was panic among Texas settlers and much criticism of Houston. He brilliantly redeemed himself at the battle of San Jacinto (Apr. 21, 1836), when by a surprise attack he decisively defeated the Mexicans and captured Santa Anna himself.
In 1836 Houston was elected the first president of the new Republic of Texas. The independence of Texas was recognized by the United States and other countries. Replaced (1838) by Mirabeau Lamar, Houston served as president again from 1841 to 1844, but during these years his government was perplexed by financial problems and by border troubles.
Texas was admitted to the Union in 1845, and Houston was one of the first to represent his state in the U.S. Senate. After serving 14 years in the Senate, he was defeated because of his uncompromising Unionism. Challenging his opponents and drawing upon his popularity, Houston was elected (1859) governor of Texas. The aged statesman preached preservation of the U.S. Constitution in the face of secession, but the tide was against him. After the people of Texas voted (Feb., 1861) to secede from the Union, Houston refused to join the Confederacy and was removed (Mar., 1861) from the governorship. He accepted the verdict, refused help from the North to defend his prerogative, and retired.
Sam Houston was an American politician and military leader who played a vital role in the founding of Texas during the early 19th century. He was al
Sam Houston, photograph by Mathew Brady Credit:Library of Congress, Washington, D.C. (born March 2, 1793, Rockbridge county, Va., U.S.—died July
1798–1858, last president of the Texas republic (1844–46), b. Seekonk section of Great Barrington, Mass. He studied medicine and after an itinerant<|endoftext|>
| 3.75 |
276 |
The rapid pace of species extinctions globally has been called one of the most pressing environmental issues facing humanity. Canada has made national and international commitments to protect biodiversity, particularly endangered species. Canada’s response to these commitments is complicated by a divided constitutional jurisdiction over the environment. The federal, provincial, and territorial governments have endeavoured to find a solution to this shared responsibility by executing the National Accord for the Protection of Species at Risk (National Accord.) In it, each jurisdiction commits to take specific steps to implement legal protection for endangered species. Alberta’s answer to endangered species protection has been to enact amendments to the Wildlife Act. Although passed before the National Accord was signed, these amendments are all Alberta intends to do to fulfil its responsibilities...
Francis, Wendy. 1999. Endangered Species Protection in Alberta: "Where's the Beef?" (in Proc. Conference Biology & Management of Species and Habitats at Risk). Ministry of Environment, Lands and Parks. Proceedings - Conference Biology & Management. Vol. 1
Topic: Species and Ecosystems at Risk
Keywords: Alberta, legislation, National Accord
Other Identifier: University College of the Cariboo
To copy the URL of a document, Right Click on the document title, select "Copy Shortcut/Copy Link", then paste as needed. Only documents available to the public have this feature enabled.<|endoftext|>
| 3.6875 |
1,588 |
In telecommunications, frequency-division multiplexing (FDM) is a technique by which the total bandwidth available in a communication medium is divided into a series of non-overlapping frequency sub-bands, each of which is used to carry a separate signal. This allows a single transmission medium such as the radio spectrum, a cable or optical fiber to be shared by multiple independent signals. Another use is to carry separate serial bits or segments of a higher rate signal in parallel.
The most natural example of frequency-division multiplexing is radio and television broadcasting, in which multiple radio signals at different frequencies pass through the air at the same time. Another example is cable television, in which many television channels are carried simultaneously on a single cable. FDM is also used by telephone systems to transmit multiple telephone calls through high capacity trunklines, communications satellites to transmit multiple channels of data on uplink and downlink radio beams, and broadband DSL modems to transmit large amounts of computer data through twisted pair telephone lines, among many other uses.
An analogous technique called wavelength division multiplexing is used in fiber optic communication, in which multiple channels of data are transmitted over a single optical fiber using different wavelengths (frequencies) of light.
How it works
The multiple separate information signals that are sent over a FDM system, such as the video signals of the television channels that are sent over a cable TV system, are called baseband signals. At the source end, for each frequency channel, an electronic oscillator generates a carrier signal, a steady oscillating waveform at a single frequency that serves to "carry" information. The carrier is much higher in frequency than the baseband signal. The carrier signal and the baseband signal are applied to a modulator circuit. The modulator alters some aspect of the carrier signal, such as its amplitude, frequency, or phase, with the baseband signal, "piggybacking" the data onto the carrier.
The result of modulating (mixing) the carrier with the baseband signal is to generate sub-frequencies near the carrier frequency, at the sum (fC + fB) and difference (fC − fB) of the frequencies. The information from the modulated signal is carried in sidebands on each side of the carrier frequency. Therefore, all the information carried by the channel is in a narrow band of frequencies clustered around the carrier frequency, this is called the passband of the channel.
Similarly, additional baseband signals are used to modulate carriers at other frequencies, creating other channels of information. The carriers are spaced far enough apart in frequency that the band of frequencies occupied by each channel, the passbands of the separate channels, do not overlap. All the channels are sent through the transmission medium, such as a coaxial cable, optical fiber, or through the air using a radio transmitter. As long as the channel frequencies are spaced far enough apart that none of the passbands overlap, the separate channels will not interfere with each another. Thus the available bandwidth is divided into "slots" or channels, each of which can carry a separate modulated signal.
For example, the coaxial cable used by cable television systems has a bandwidth of about 1000 MHz, but the passband of each television channel is only 6 MHz wide, so there is room for many channels on the cable (in modern digital cable systems each channel in turn is subdivided into subchannels and can carry up to 10 digital television channels).
At the destination end of the cable or fiber, or the radio receiver, for each channel a local oscillator produces a signal at the carrier frequency of that channel, that is mixed with the incoming modulated signal. The frequencies subtract, producing the baseband signal for that channel again. This is called demodulation. The resulting baseband signal is filtered out of the other frequencies and output to the user.
For long distance telephone connections, 20th century telephone companies used L-carrier and similar co-axial cable systems carrying thousands of voice circuits multiplexed in multiple stages by channel banks.
For shorter distances, cheaper balanced pair cables were used for various systems including Bell System K- and N-Carrier. Those cables didn't allow such large bandwidths, so only 12 voice channels (Double Sideband) and later 24 (Single Sideband) were multiplexed into four wires, one pair for each direction with repeaters every several miles, approximately 10 km. See 12-channel carrier system. By the end of the 20th Century, FDM voice circuits had become rare. Modern telephone systems employ digital transmission, in which time-division multiplexing (TDM) is used instead of FDM.
The concept corresponding to frequency-division multiplexing in the optical domain is known as wavelength-division multiplexing.
Group and supergroup
A once commonplace FDM system, used for example in L-carrier, uses crystal filters which operate at the 8 MHz range to form a Channel Group of 12 channels, 48 kHz bandwidth in the range 8140 to 8188 kHz by selecting carriers in the range 8140 to 8184 kHz selecting upper sideband this group can then be translated to the standard range 60 to 108 kHz by a carrier of 8248 kHz. Such systems are used in DTL (Direct To Line) and DFSG (Directly formed super group).
132 voice channels (2SG + 1G) can be formed using DTL plane the modulation and frequency plan are given in FIG1 and FIG2 use of DTL technique allows the formation of a maximum of 132 voice channels that can be placed direct to line. DTL eliminates group and super group equipment.
DFSG can take similar steps where a direct formation of a number of super groups can be obtained in the 8 kHz the DFSG also eliminates group equipment and can offer:
- Reduction in cost 7% to 13%
- Less equipment to install and maintain
- Increased reliability due to less equipment
Both DTL and DFSG can fit the requirement of low density system (using DTL) and higher density system (using DFSG). The DFSG terminal is similar to DTL terminal except instead of two super groups many super groups are combined. A Mastergroup of 600 channels (10 super-groups) is an example based on DFSG.
FDM can also be used to combine signals before final modulation onto a carrier wave. In this case the carrier signals are referred to as subcarriers: an example is stereo FM transmission, where a 38 kHz subcarrier is used to separate the left-right difference signal from the central left-right sum channel, prior to the frequency modulation of the composite signal. A television channel is divided into subcarrier frequencies for video, color, and audio. DSL uses different frequencies for voice and for upstream and downstream data transmission on the same conductors, which is also an example of frequency duplex.
FDMA is the traditional way of separating radio signals from different transmitters.
In the 1860s and 70s, several inventors attempted FDM under the names of Acoustic telegraphy and Harmonic telegraphy. Practical FDM was only achieved in the electronic age. Meanwhile, their efforts led to an elementary understanding of electroacoustic technology, resulting in the invention of the telephone.
- Duplex (telecommunications)
- Single-sideband modulation
- Orthogonal frequency-division multiplexing (OFDM),
- Time-division multiplexing (TDM)
- Harold P.E. Stern, Samy A. Mahmoud (2006). "Communication Systems: Analysis and Design", Prentice Hall. ISBN 0-13-040268-0.<|endoftext|>
| 3.96875 |
339 |
FREE 5-7 DAY SHIPPING for all orders over $25 shipped within the USA
Computational Thinking with Cubelets
What’s the best way to reason through complex problems? How can we be better prepared for the technology we’ll be using to solve our biggest challenges? Computational thinking!
Computational thinking is a problem-solving framework that turns complex, many-layered problems into solutions a computer could execute. Cubelets help transform the abstract concepts of computational thinking into easy-to-practice skills, making it simple to integrate into your classroom.
When looking at a big, complex problem, it can be hard to figure out where to start untangling the issues. That’s why computational thinking is so important. The first step, decomposition, is when we zoom in to notice the smaller, more manageable parts within the problem.
Finding patterns in the small parts of a problem can reveal how it all fits together. It allows us to zoom out from the details, and determine what those parts have in common. Grouping pieces by similarities helps us create better hypotheses about how a system works.
Abstracting is something our brains do naturally when we notice patterns or choose to focus only on one part of the system. Think of abstracting as filtering the world around us so we can pay better attention to the most important or useful information.
One last step in computational thinking involves solving our problem in a logical, repeatable, step-by-step manner. This solution is called an algorithm, and it’s a set of instructions that will help a person — or a program — arrive at the correct answer.<|endoftext|>
| 3.9375 |
1,357 |
Home Math Including Polynomials with Algebra Tiles, Vertically, and Horizontally
### Including Polynomials with Algebra Tiles, Vertically, and Horizontally
Including polynomials by combining like phrases collectively and utilizing algebra tiles is the purpose of this lesson. We’ll present you the next 3 ways so as to add polynomials.
• Utilizing algebra tiles
• Including vertically
• Including horizontally
## Including polynomials with algebra tiles
We’ll begin with algebra tiles because the course of is much more easy and concrete with tiles. To this finish, research the mannequin under with nice care.
Instance #1:
Add 2x2 + 3x + 4 and 3x2 + x + 1
Step 1
Mannequin each polynomials with tiles.
Step 2
Mix all tiles which can be alike and depend them.
• You bought a complete of 5 mild blue sq. tiles, so 5x2
• You bought a complete of 4 inexperienced rectangle tiles, so 4x
• You bought a complete of 5 blue small sq. tiles, so 5
Placing all of it collectively, we get 5x2 + 4x + 5
I hope from the above modeling, it’s clear that we are able to solely mix tiles of the identical kind. For instance, you might not add mild blue sq. tiles to inexperienced rectangle tiles identical to it will not make sense so as to add 5 potatoes to five apples. Attempt including 5 potatoes to five apples and inform me when you bought 10 apples or 10 potatoes. It simply doesn’t make sense!
Instance #2:
Add 2x2 + -3x + -4 and -3x2 + -x + 1
Step 1
Mannequin each polynomials with tiles
Step 2
Mix all tiles which can be alike and depend them. Tiles which can be alike, however have totally different colours will cancel one another out. We present every cancellation with a inexperienced line.
• You might be left with of 1 purple sq. tile, so -x2
• You bought a complete of 4 mild purple rectangle tiles, so -4x
• You might be left with 3 robust pink small sq. tiles, so -3
Placing all of it collectively, we get –x2 + -4x + -3
## Maintain these essential info in thoughts when including polynomials
• We name tiles which can be alike or are the identical kind “like phrases”
• Like phrases are phrases with the identical variable and the identical exponent. For instance, 2x2 and 3x2 in instance #1 are like phrases as a result of they’ve the identical variable, x and the identical exponent, 2.
• So as to add like time period, simply add the coefficients, or the numbers hooked up to the time period, or the quantity on the left aspect of the time period. 2x2 + 3x2 = (2 + 3)x2 = 5x2
## Including polynomials vertically
Instance #3:
Add 6x2 + 8x + 9 and 2x2 + -13x + 2
Line up like phrases. Then add the coefficients.
6x2 + 8x + 9
+ 2x2 + -13x + 2
——————————
8x2 + -5x + 11
Instance #4:
Add -5x3 + 4x2 + 6x + -8 and 3x3 + -2x2 + 4x + 12
Line up like phrases. Then add the coefficients.
-5x3 + 4x2 + 6x + -8
+ 3x3 + -2x2 + 4x + 12
————————————–
-2x3 + 2x2 + 10x + 4
## Including polynomials horizontally
Instance #5:
Add 6x2 + 2x + 4 and 10x2 + 5x + 6
Mix or group all like phrases. You may use parentheses to maintain issues organized.
(6x2 + 2x + 4) + (10x2 + 5x + 6) = (6x2 + 10x2) + (2x + 5x) + (4 + 6)
(6x2 + 2x + 4) + (10x2 + 5x + 6) = (6 + 10)x2 + (2 + 5)x + (4 + 6)
(6x2 + 2x + 4) + (10x2 + 5x + 6) = 16x2 + 7x + 10
Instance #6:
Add 2x4 + 5x3 + -x2 + 9x + -6 and 10x4 + -5x3 + 3x2 + 6x + 7
(2x4 + 5x3 + –x2 + 9x + -6) + (10x4 + -5x3 + 3x2 + 6x + 7)
= (2x4 + 10x4) + (5x3 + -5x3) + (-x2 + 3x2) + (9x + 6x) + (-6 + 7)
= (2 + 10)x4 + (5 + -5)x3 + (-1 + 3)x2 + (9 + 6)x + (-6 + 7)
= (12)x4 + (0)x3 + (2)x2 + (15)x + (1)
= 12x4 + 2x2 + 15x + 1
Discover that if the time period is x2, you possibly can rewrite it as 1x2, so your coefficient is 1. We did this in instance #6, third line with (-x2 + 3x2)
(-x2 + 3x2) = (-1x2 + 3x2)<|endoftext|>
| 4.875 |
673 |
Most of us use PowerPoint for creating presentations, a tool launched in 1990… For over 20 years we have been using the metaphor of slides introduced by Microsoft. It is quite challenging for us to start thinking differently, from a new perspective. Sometimes it is very helpful to think outside the box.
Is our thinking linear at all? Learning has more to do with a network than a line. It consists of nodes and connections between them. Second language acquisition is considered by some researches as a chaotic/complex system. That’s the way we learn. To see why it is so, you as a teacher and your learner need to be able to zoom in and zoom out.
A great example of non-linear and linear thinking in action is Prezi, a new form of digital presentation. Why? Because Prezi uses a map-like metaphor.“See the big picture” takes on a whole new meaning. Once you created your Prezi you can give your viewers control over it: they can zoom in and out, drag the page around. Besides, Prezi is a very eye-catching tool. You can always use it to wow your audience and surprise your learners.
If you are new to Prezi you’ll need to explore some tutorials and have some subsequent “play-time”. Once you get used to this style of navigation many interesting ways to structure and organize information become possible.
When we talk about the EAL/ESL context Prezi can be used in more ways than just a presentation tool.
You can use it to introduce the major concepts on your program syllabus to your learners for them to be able to explore all the details on their own. You can easily embed links, videos, even your learners’ testimonials and student-created materials. Make your Prezi a landing page for your learners!
Create a prezi with some visual information (graph/chart/diagram/process) and ask your EAP learners to describe some parts of the process in the context of a big picture.
Now you have an easy way to focus on key details. So, use it to create speaking activities. Gradually reveal the wider context of a photo by first zooming in on a small detail and then gradually zooming out. Use this activity to practice asking question, describing, analyzing, making predictions, expressing opinion, asking for clarification and so on.
Prezi is an infinite canvas, so now you and your learners can create detailed timelines to provide comprehension support for reading and listening activities.
And yes! Your learners can work on Prezi collaboratively! You can start creating a new prezi from scratch together. Or you can start it and let your learners contribute. Watch how your prezi grows in real time. Your learners can drag and drop objects to sort them into groups while having pro/con arguments, or they can brainstorm problem solutions, or you prezi can become an invaluable repository of pieces of knowledge collected collaboratively.
Be creative and remember you are no longer limited by step1, step 2, step 3. Not only details, but the whole canvas can convey meaning.
Watch the webinar recording or look though the handout to get links to the examples and much more.
Thanks and see you online!<|endoftext|>
| 3.828125 |
115 |
The amplitude of a variable is a measure of its change over a single period. The maximum height observed in the wave is called as Amplitude. It is denoted by A and is given in decibels (dB). The sine wave is represented as
Where A is amplitude and $\omega$ is angular frequency. The Amplitude Formula is given by
Where D is the distance traveled by the wave and
F is the frequency of the wave.
Amplitude Formula is used to the amplitude of the given wave if its distance and frequency are known.<|endoftext|>
| 4.15625 |
1,487 |
Breaking News
Home / Civil Engineering / Bar Bending Schedule / How to calculate the cutting length of reinforcement bars in circular R.C.C slab
# How to calculate the cutting length of reinforcement bars in circular R.C.C slab
### How to calculate the cutting length of reinforcement bars in circular R.C.C slab
Now, let us go through the procedure to find out the cutting length of the bars in a circular slab as shown below.
### Given data :
Diameter of the circular slab = 1600mm = 1.6m.
Clear cover for the reinforcement = 25mm.
Rebar diameter = 10mm.
Rebar spacing = 150mm c/c.
The formula for finding the length of the bar
=2ār2 – d2
### Where,
r = [diameter of the slab – (2nos.Ć clear cover )]Ā Ć· 2
d = c/c distance of the individual bars from the central bar of the circle.
As you can observe in the above-given drawing, I have drawn a red-colored circle by deducting the clear cover. Radius r in the above formula will be the radius of this red circle.
i.e. r = [1600mm – ( 2nos.Ć 25mm)] Ć· 2
Ā Ā Ā Ā Ā = [1550mm ] Ć· 2
### Ā Ā Ā Ā =Ā 775mm.
TheĀ value of r will be the same for the cutting length calculation of every individual bar of the slab.
### Note: All the rebars of the circular slab act as a chord of the red circle. To gain a basic understanding of this theory, you have to go through the article,
The length of the individual bars in the upper half part of the circular slab will be different.
The cutting length of the bars in the lower half portion will be equal to the upper half part bars.
I have named the bars for the half portion of the Circular slab, and let us find out the cutting length of each of these bars.
### 2. Cutting length of bar DD1
=2ār2 – d2
Here, d = 150mm.
= 2 Ć ā7752 – 1502
= 2 ĆĀ ā 600625 – 22500
= 2 ĆĀ ā 578125
= 2 Ć 760.345
= 1520.69mm =Ā 1.521m.
### 3. Cutting length of the bar FF1
Here, d = 300mm.
=2ār2 – d2
= 2 Ć ā7752 – 3002
= 2 ĆĀ ā 600625 – 90000
= 2 ĆĀ ā 510625
= 2 Ć 714.58
= 1429.16mm =Ā 1.429m.
### 4. Cutting length of the bar GG1
Here, d = 450mm.
=2ār2 – d2
= 2 Ć ā7752 – 4502
= 2 ĆĀ ā 600625 – 202500
= 2 ĆĀ ā 398125
= 2 Ć 630.971
= 1261.94mm =Ā 1.261m.
### 5.Cutting length of the bar HH1
Here, d = 600mm.
=2ār2 – d2
= 2 Ć ā7752 – 6002
= 2 ĆĀ ā 600625 – 360000
= 2 ĆĀ ā 240625
= 2 Ć 490.535
= 981.07mm =Ā 0.981m.
### 6. Cutting length of the bar JJ1
Here, d = 750mm.
=2ār2 – d2
= 2 Ć ā7752 – 7502
= 2 ĆĀ ā 600625 – 562500
= 2 ĆĀ ā 38125
= 2 Ć 195.256
= 390.51mm =Ā 0.39 m
3. ### Basic Components of Building Structure | Building Elements
The calculation should be done until the cumulative value of d is less than r. So, we have calculated the cutting length ofĀ 5 barsĀ on the upper side of the central bar CC1.
### No. of bars in a circular slabĀ
The no. of top bars inĀ the circular slab
=[( diameter of the slab) Ć· c/c bar spacing ] +1
= [ (Ā 1600mm ) Ć·Ā 150mm] +1
=Ā 11 nos.
That means 5 bars on either side of the central bar CC1
### Total no. of barsĀ
= Top bars + Bottom bars
= 11nos + 11nos.=22nos.
### Estimating Of Steel Bar Quantity for Construction
As the upper half part of the circle is identical to the bottom half part, you will have 2nos of central common bars and 4 nos of the chord bars having the same cutting length.
1. Cutting length of central bar CC1 = 1.55mĀ šĀ 2nos.
2. Cutting length of bar DD1Ā Ā Ā Ā Ā Ā = 1.521mĀ šĀ 4nos.
3. Cutting length of bar FF1Ā Ā Ā Ā Ā Ā = 1.429mĀ šĀ 4nos.
4. Cutting length of bar GG1Ā Ā Ā Ā Ā Ā = 1.261mĀ šĀ 4nos.
5. Cutting length of bar HH1Ā Ā Ā Ā Ā Ā = 0.981mĀ šĀ 4nos.
6. Cutting length of bar JJ1Ā Ā Ā Ā Ā Ā Ā = 0.390mĀ šĀ 4nos.
————–<|endoftext|>
| 4.4375 |
4,474 |
Bergen-Belsen was liberated by the British troops on 15 April 1945. It was not the first camp to be liberated, but it was most likely the most infamous camp at that time. The gruesome scenes of Bergen-Belsen which were captured by British army photographers, soon went round the world and pictured the cruel Nazi regime. These pictures finally revealed the true nature of the concentration camps to the world. Bergen-Belsen would soon serve as an example of Nazi camps. But this is not entirely in keeping with the truth.
The town Bergen-Belsen does in fact not exist. Belsen is a small village, while Bergen was a larger town with 13,000 residents. Both are located about sixty kilometres northeast of Hannover. In 1935 the German Wehrmacht built a barracks with practice ground between both locations. It was a modern barracks for the Panzertruppenschule with well-equipped sleeping-places, canteens, kitchens, etc. To provide housing for the workers who were building the barracks, a camp with approximately thirty barracks was built.
During the May days of 1940, the former practice ground was used as a prisoner-of-war camp, where about 600 Belgian and French soldiers were accommodated. A year later it received its first official name Stalag 311 (XI C). A few weeks after the start of operation Barbarossa, the German attack on the Soviet Union, the first prisoner transports arrived at Bergen-Belsen. By July 1941, around 21,000 Russian prisoners of war were locked up in Bergen-Belsen. Their living conditions were miserable: in open air surrounded by barbed wire, without any sanitary provisions. By February 1942, more than 18,000 prisoners had died. Many died as a consequence of dysentery, but especially the typhoid epidemic of November 1941 caused most victims to die. At the same time, hundreds of Soviet commissioners were deported to Sachenhausen to be executed in concordance with Hitler’s guidelines.
The SS took over command of Bergen-Belsen in April 1943.
The SS had different intentions with the camp. Reichsführer-SS Heinrich Himmler had adopted a plan from the Ministry of Foreign Affairs, which argued that Germans who were incarcerated abroad would be exchanged for Jews. The guidelines of the Reichssicherheitshauptamt (RSHA) of 31 August 1943 made clear who would be eligible for detention in this camp: "1. Juden, die verwandtschaftliche oder sonstige Beziehungen zu einflußreichen Personen im feindlichen Ausland haben; 2. Juden, die unter Zugrundelegung eines günstigen Schlüssels für einen Austausch gegen im feindlichen Ausland internierte oder gefangene Reichsangehörige in Frage kommen; 3. Juden, die als Geiseln und als politische oder wirtschaftliche Druckmittel brauchbar sein können; 4. Jüdische Spitzenfunktionäre." So therefore only Jewish top officials, Jews with important connections abroad or those with a British or American passport, with a passport from one of the neutral countries (such as Sweden, Spain) or with a Palestine certificate (authorization to immigrate to Palestine) could be accommodated here. From now on, Bergen-Belsen was a Zivilinternierungslager. But when it became clear that camps with such a designation could be inspected by the Red Cross or international committees, the official name was changed into "Aufenthaltslager Bergen-Belsen" in June 1943.
Part of the camp was still being used by the Wehrmacht, mainly as military hospital. To prepare the camp for the arrival of thousands of Jews, 500 prisoners from Natzweiler and Buchenwald were transported to Bergen-Belsen. Here they were part of the Aufbaukommando that had to extend the camp. They were also responsible for the improvement of the sanitary provisions. A small crematorium was also built. In July 1943 the first Jewish prisoners arrived at the camp. But the prisoners were not put up all together: there were separate compartments. These compartments were separated from each other by barbed wire fences, making contact between the different compartments impossible.
There was the ordinary prisoners’ camp, including the Aufbaukommando. Because most prisoners had died due to the terrible living conditions, this commando was shut down in February 1944. From now on, many weakened prisoners, who were not capable of performing labour, were sent to this camp section from other camps. Bergen-Belsen was considered an "unproductive" camp (there was no general work duty) by the SS-Wirtschafts- und Verwaltungshauptamt (SS-WVHA) so it seemed a good idea to send those that were no longer "economically useful" to this section. But those that ended up in this section of the camp were subjected however to a form of work duty. Officially Bergen-Belsen was never a concentration camp, but this Häftlingslager [prisoner camp] was managed as such: the prisoners wore striped outfits and were being tortured and abused by the Kapos. On 27 March 1944 the first transport arrived: 1,000 prisoners from the underground V1-weapon factory in Nordhausen.
There was also the Sternlager, with 4,000 Jews eligible for exchange (so-called Austauschjuden). A mixed group of men and women were accommodated here. They could keep their civilian clothes, but wearing the Star of David was obligatory. In the Sternlager the largest group was made up of Dutch prisoners from camp Westerbork. Life was evidently better here: families could stay together and they did not have to work. Shmuel Huppert, who stayed with his mother as Austauschjude in Bergen-Belsen, later recalled: "From a certain point of view, life was quite reasonable. Reasonable in the sense that we received three blankets so that we wouldn’t be cold and that we had some food. It was not much, but we could survive. We didn’t work. I learned how to play chess in Bergen-Belsen and I still play chess. But the most important thing was that we were together, that I was never separated from my mother."
In the Neutralenlager several hundreds of Jews from neutral countries were accommodated. Here too the living conditions were bearable; the prisoners were not assigned to the work commands and usually there was enough food. The section was governed by a Jewish council chaired by a Jewish chairman.
Polish Jews with Latin-American passports or Palestine certificates were put up in the Sonderlager. They were separated from the other prisoners, because they could report about the SS crimes in the east. Most of them were eventually deported to Auschwitz.
In July 1944 the Hungarian camp was established. In it were 1,683 Hungarian Jews that Himmler wanted to exchange for money and goods. This group also had its own management, was not part of the work groups and did not have to take part in roll calls.
In early August 1944, a Sternlager was set up in the open terrain behind the encampment where thousands of women were accommodated. Initially the women transports mainly came from Poland (Warsaw). Later thousands of women arrived from Auschwitz and Birkenau. Amongst them was Anne Frank, who would not survive the camp. After a violent autumn storm had destroyed most tents in November 1944, the women were put up in newly built barracks. Most women would however not stay very long in Bergen-Belsen. The SS sent them to other places to work. They mostly ended up in the satellite camps Buchenwald and Flossenbürg.
|Date:||Camp section:||Definition:||1944||Häftlingslager I||Section where prisoners from other camps who were too weak to work would be accommodated.|
|Sternlager||Section where the so-called Austauschjuden – the Jews eligible for exchange – would stay.|
|Neutralenlager||Here the Jews from neutral countries were put up.|
|Sonderlager||tdSection for Polish Jews with Latin-American passports or Palestine certificates.|
|Ungarnlager||Established in July 1944 for Hungarian Jews that Himmler wanted to exchange for money and goods.|
|Zeltlager||Established in August 1944 to receive the increasing number of transports of women from Poland and Auschwitz|
|Kriegsgefangenenlazarett||Originally established to accommodate severely ill patients. Discontinued in January 1945 and replaced with a large Frauenlager [women’s camp].|
|1945||Häftlingslager II||Because Häftlingslager I was becoming too small, as of 1945 part of the Sternlager was used to receive the large stream of prisoners.|
With the appointment of Josef Kramer as camp commander, the transition to an ordinary concentration camp was completed, although that was never confirmed officially by a name change. At that time, approx. 15,000 prisoners were locked up here. With the arrival of Kramer the hell of Bergen-Belsen truly began. It’s due to this hellish period that the camp "thanks" its reputation. The appointment of Kramer to camp commander – together with the abolishment of the privileges for the Austauschjuden and the flood of new prisoners – was the start of what later would be called the "hell of Bergen-Belsen".
At the West front the Americans and British advanced further, while at the same time the march of the Soviets in the East could not be stopped. Despite this, the SS did not set free the KZ-prisoners. The threatened camps were evacuated by means of the infamous dead marches. Tens of thousands of people were sent on their way by foot. For many of them Bergen-Belsen was the final destination. The number of people in the camp quickly rose and was no longer in proportion to the capacity of the camp. At the beginning of January 1945, 15,000 prisoners were housed in Belsen. At the end of the month the number had risen to 22,000. Late February 41,000 and 60,000 when the camp was liberated. In this period more than 35,000 people died.
Only a few of them died as a consequence of cruelties by the guards. It was also not necessary to gas or execute prisoners. Anita Lasker Walfisch, who ended up in Bergen-Belsen after Auschwitz, explained: "In Auschwitz people were murdered; in Bergen-Belsen they would just let them die." The circumstances in the camp were terrible. The main cause of death was mainly lack of food and epidemics. The guards made no effort to improve the living conditions. Many barracks were not completed. Those that were fully built, quickly fell apart. Because of the overpopulation there were hardly any wooden beds. There were no sanitary provisions at all: no heating, no water tabs, no toilets… The prisoners were only given a minimum of food: two slices of bread and half a litre water soup a day. As the war progressed, in many parts of the camp no food was supplied at all. Supposedly two hundred cases of cannibalism took place.
Because of the lack of sanitary provisions and streaming water, the already weakened prisoners in the crowded camp were prone to diseases. The many rodents and bodies that the small crematorium could not process, were a feeding ground for diseases such as (spot)typhoid, tuberculosis, dysentery and cholera. Most prisoners suffered from digestive disorders and couldn’t control their bowel movements. Those who could not stand up anymore, relieved themselves on their wooden beds. In February 1945, the worst epidemic broke out. Typhoid killed thousands of people in the last months before the liberation. The camp was littered with piles of bodies.
In April 1945, another 7,000 Austauschjuden from Bergen-Belsen were transported to Theresienstadt, but soon their place was taken by thousands of new prisoners from Nordhausen. To improve the lack of space by any means possible, the nearby Wehrmacht barracks was seized. But this initiative proved just a drop in the ocean.
In the last days before the liberation, the SS tried to do something about the ever growing piles of bodies. Kramer ordered two thousand prisoners to drag the bodies in the streets of the camp to mass graves – huge open pits. For hours at a time, forced by whips from the SSers and the Kapos, the tired prisoners dragged thousands of bodies. It is estimated that 17,000 bodies were thrown into those open or shallow pits. At the background the prisoners’ band played dance music, 'to improve the atmosphere’, according to Kramer’s ordinance.
The end of Bergen-Belsen is a bizarre chapter in the history of the Second World War.
On 11 April 1945, Himmler gave his recently appointed Reich under Commissioner of the camps, SS-Standartenführer Kurt Becher, permission to transfer Bergen-Belsen to the British. One day later, two German officers, Colonel Hanns Schmidt and lieutenant-colonel Bohnekamp reached the allied borders to announce the surrender of the camp. Avoiding fights on the camp site was crucial to avoid the spreading of the typhoid epidemic amongst the fighting parties.The talks led to a local cease-fire. An area of around 48 square metres was declared neutral territory.The Germans would place warning signs with "Danger – Typhoid" on all access roads to the camp and transfer the Wehrmacht barrack to the British. 800 German and 1500 Hungarian soldiers were stationed in this barrack. They stayed put to avoid an outbreak of the epidemic. In the coming days, the Hungarians would serve the British, while the Germans were sent back to their own lines. Most SS guards, who were considered prisoners of war, took off on 13 April.
At first, the British did not really realise what they had stumbled upon. Initially they thought that it was just an "ordinary" internment camp. They were much more interested in the Wehrmacht barrack. In the afternoon of Sunday 15 April, a group of British officers, led by lieutenant-colonel Robert Daniell, commander of the 13th Regiment Royal Horse Artillery, entered the camp. They were awaited by camp commander Kramer and a small group of SSers. It was only after they visited the camp themselves, when they became aware of the true horrific nature of the camp. The British locked up Kramer in an underground basement beneath the officers’ accommodation. Here he had to sleep on the bare floor with no blankets. Additionally 25 female SS camp guards and 28 male SSers were imprisoned. On 18 April, Kramer was transferred to the prisoner-of-war camp at Celle.
Captain William Roach described the camp as follows: "There was no sound or sign of life. The prisoners didn’t even smile. They just stood there and gazed at us. And most of us, just gazed back at them." Often, people imagine that freeing a concentration camp was a joyful occasion. But the truth is clearly different, as was evident from this testimony, explained Zdenka Ehrlich. Together with 300 women she occupied a barrack in Bergen-Belsen. Here place was in the back of the barrack, so she could not see the camp site. Someone said: "Are they here yet?" "Who?", asked someone else. "The English." "Oh," said Zdenka.' That was her reaction. Such indifferent reactions were common for the starved and exhausted prisoners. Many of them hardly realised that they had been liberated.
The prisoners that had any strength left in them, slowly started to realise that their suffering was over and sought contact with the British liberators. They begged for water and food. Russian and Polish prisoners took it upon themselves and broke into SS rooms to search for food. This meant that they finally could have a warm meal again, albeit between thousands of bodies that had not yet been buried.
The situation in Bergen-Belsen was so severe that lieutenant-colonel Richard Taylor, who had been given the task to take over the camp, had to call in the help of other units. He sent a liaison-officer to the headquarters of the VIII Army Corps asking them to urgently send water, food and other aid. The N° 76 Field Hygiene Section was the first to respond to the call for help and helped to improve the hygienic circumstances in the camp. One day after the discovery of the camp the first vehicles with food and water arrived. And the day after that, 27 more water tanks arrived. When the British discovered a stream in the area they built a water pipe with materials found in the camp to one of the water reservoirs. Other units, including the 11th Light Field Ambulance, took care of medical provisions. They tried to take care of the sick and disinfect them. A field hospital was set up in the Wehrmacht barrack which was close by. Here the severely sick were put up. Even German doctors and nurses were deployed to treat the great number of sick prisoners. The evacuation of camp prisoners with a better condition started on 24 April. A month later the barracks were burned down to exterminate the typhoid germs. This is why hardly anything of Bergen-Belsen remained.
To avoid the spreading of diseases, it was hugely important to bury the bodies. The arrested SSers had to help removing the bodies from the camp streets and to throw them into large mass graves. Because that took too long, a bulldozer was deployed, which could shove the many bodies into the pits quickly. It was impossible to keep track of the number of dead. It’s estimated that around 15 to 20,000 dead were buried in the last two weeks of April 1945. Each of the mass graves was given a number. In total there were 11 mass graves.
Despite all these efforts, another 14,000 people would die in the weeks after. Most people fell victim to the typhoid epidemic, but there were also people who died because they ate too much too quickly. The intestines of the starved prisoners could not digest large quantities of food. The total amount of victims in Bergen-Belsen in the period 1943-1945 is estimated to be 50,000.
On 25 September 1945, following a congress of liberated Jews in the British occupation zone, a provisional wooden commemoration cross was placed between the mass graves. On 15 April 1946, the Belsener Jüdische Komitee revealed a stone commemoration monument with Hebrew and English inscription: a large square stone with Jewish symbols on a triangular base. This monument is the centre of Jewish commemorative ceremonies each year. The English inscription reads:Israel and the world shall remember
The British military government ordered the erection of a large, worthy monument in October 1945. Early 1947, the build of the central monument started – a 24 metre high obelisk and a 50 metre long wall with inscriptions. Each country which mourned the loss of victims in Bergen-Belsen received a spot on the wall to place a text. For the Netherlands, the text below was used:HET ZWAARST GESCHUT EN 'T BEST GEWEER
In 1952, Lower Saxony was given the responsibility of taking care of the Bergen-Belsen monument (Gedenkstätte), while on 30 November of that same year, the official inauguration of the Mahnmal took place. In total 14 mass graves are located on the site of the former concentration camp. Each mass grave has a memorial stone with the number of dead. Spread across the site, a number of single graves and memorial stones can be found, for instance for Anne Frank. The last mass grave was constructed in 1964, when bones and ashes were discovered in the former crematorium. Not all victims of Bergen-Belsen are buried at the site of the Gedenkstätte. Numerous victims were buried at the Großen Friedhof für die Opfer der nationalsozialistischen Gewaltherrschaft in Bergen-Hohne. As of 1987, the Kriegsgräberstätte Hörsten, where a monument for the thousands of Soviet victims was erected, is also part of the Gedenkstätte Bergen-Belsen.
In 1966, the first small documentation centre was opened, that was replaced by a bigger centre in 1990. The new documentation centre included, amongst others, a library and archive. Information can be found on the history of Bergen-Belsen. A special exhibit focuses on the National Socialist system of prosecution.
As of 1991 several youth organisations from Lower Saxony started with a kind of renovation project. During project weeks and work camps they try to excavate the foundations of the former concentration camp. The YMCA is still undertaking these excavation works.<|endoftext|>
| 3.734375 |
289 |
The United States president’s judicial powers include nominating judges to the Supreme Court and granting pardons. The president can also shorten prison terms and grant amnesty.
The president of the United States is responsible, under Article II of the Constitution, to execute and enforce the laws that are created by Congress. To facilitate this, the president is responsible for appointing 15 cabinet members. Each cabinet member is the head of a department, such as the Department of Labor or the Department of Energy, which enforces these laws.
The president also nominates judges to the Supreme Court, which is the highest court in the country. The president's nominations must first be confirmed by the Senate.
The president can grant pardons, which legally forgive crimes and cancel penalties as well as reprieves. Whereas a pardon eliminates the legal consequences of a crime, a reprieve only postpones the sentence.
The president can also commute sentences that are already in effect, by reducing prison terms or lowering penalties. The only crime for which the president cannot grant a pardon is impeachment.
The president can grant amnesty, which is a legal pardon for a group that has committed treason, or a similar crime. The first presidential amnesty in the U.S. was granted by George Washington to the members of the Whiskey Rebellion.
The president's judicial powers are closely tied to his or her legislative powers. These include signing legislation into law and vetoing bills.<|endoftext|>
| 3.921875 |
701 |
· How does your understanding of differentiation compare to what you have observed in your practicum?
· How will strategies for differentiation and engaging all learners be reflected in your practice?
One of the greatest ways we can help our students is through differentiation. What is differentiation? Basically, it’s the way we should have been teaching all along. Every student learns a different way; students have their own needs. As a teacher, we must teach in a way that encompasses most, if not all, of these different styles and needs. This can be exceptionally difficult to do, especially in a large class with many different learning styles. However, we can try our best to use a variety in our classroom. In my current practicum placement, I see differentiation all the time. One student in particular has a very specific learning style, and very specific needs. His needs are consistently met through assignments (given more difficult problems in and out of the classroom), seating arrangement (can sit in a certain spot that will help him learn), and through communication (letting the student know what will be coming up next, in addition to other techniques). These are only a few to name for this specific student, but there are many strategies for the other students as well. One other example is for a student who has difficulty with spelling. This technique works with others as well (universal design), however it especially works with student A. During times in which students must journal, or create text-to-text/self connections, the spelling is not something that is always looked at. We encourage students during these times to simply write, and not think about the spelling. Content is just as important as spelling. For student A, he becomes more motivated to write, knowing that this bit of differentiation is being used.
I strive to use differentiation in my practice. In the past, I have been in a split grade (one and two). Certain students needed the opposite grade’s worksheet, and I was aware of which students needed this. In this case, I would tell all my students that they could start with either their grade worksheet, or the other one, and then once they were finished with that, move on to the next one. This allowed for all Grade 2 students to use the Grade 1 worksheet as a refresher, and the Grade 1 students to see what they could do from a Grade 2 worksheet. Students that needed this differentiation did not feel out of place using another grade’s worksheet.
During this year’s practicum, whenever there is a group discussion, I make sure to write down all points on an anchor chart. Students that need to look back at what we spoke about have this opportunity. The anchor chart will remain up for the duration of their activity. I also put schedules on the blackboard of the tasks they need to complete, and in what order. Lastly, I make sure to have visuals. If the students are creating something, most of the time, I have an example of the creation (one case I did not give an example was when they were building structures. I wanted to see what they could create on their own!).
These are, of course, only a few examples of differentiation that either I do, or am apart of. I feel that differentiation is something all teachers and educators should strive to do in their classrooms. It is beneficial for all students, and will make school a safer and more trusting place.<|endoftext|>
| 4.0625 |
815 |
The art to developing science inquiry skills.
Timms’ paper “Measuring learning in complex learning environments” explores the impact of technology on student learning, specifically focusing on the impact of virtual scientific environments in the development of inquiry skills. This concurrent paper presented at the 2013 ACER Research Conference discusses how “in science, students often have difficulty connecting concepts to real world phenomena and in understanding how to use scientific practices in investigating those phenomena.” Timms suggests that online interactive learning environments provide students with a richer learning experience, beyond the scope of a textbook. Through engaging with an online environment and manipulating scientific variables, students are able to ‘engage in the construction of knowledge, rather than just receive the facts.’ Timms further suggests that such learning experiences aid the contentious debate of how much guidance is enough guidance, as these elearning resources allow students to experiment with their own learning in a controlled environment.
Extract: Measuring learning in complex learning environments
Technology offers the opportunity to enhance the learning experience through providing students with learning environments that bring to them other worlds outside the classroom. The use of animations, simulations and augmented reality can help to show dynamic processes such as geological events over time, virtual chemistry laboratories or events from history in deeper and richer ways than are possible in textbooks. These technological tools also offer the chance to allow students to explore and manipulate the virtual environments that are created, bringing opportunities for learners to engage in the construction of knowledge rather than just receiving facts.
How interactive learning environments can assist student learning
Interactive learning environments hold a lot of promise for assisting learners in ways that are tailored to the needs of each learner. Well designed interactive learning environments combine pedagogical approaches that are based on cognitive theory of learning in interactive ways in electronic environments with methods of measuring the progress of learners and techniques for providing assistance at key moments.
The challenge of providing assistance in inquiry science instruction
The goal of inquiry learning is to allow students to induce the characteristics of a domain through their own experiments and exploration (de Jong, 2006). But, even in curricula with hands-on laboratories and the opportunity to engage in inquiry learning, students are typically asked to replicate standard experiments rather than perform their own inquiries.
…Research on scaffolded inquiry learning suggests that teaching the critically important skills associated with scientific inquiry can be greatly improved if supported by the right kind of guidance (Linn & Hsi, 2000; Sandoval & Reiser, 2004; Slotta, 2004; van Joolingen, de Jong, Lazonder, Savelsbergh & Manlove, 2005; White & Frederiksen, 1998).
But what exactly is the right amount and type of guidance? While past work with inquiry learning environments makes clear that some guidance is necessary, it doesn’t fully answer this question, which in the learning sciences more generally has been variously investigated under the guise of ‘desirable difficulty’ (Schmidt & Bjork, 1992), the ‘assistance dilemma’ (Koedinger & Aleven, 2007) and ‘productive failure’ (Kapur, 2009). Essentially the issue is to find the right balance between, on the one hand, full support and, on the other hand, allowing students to make their own decisions and, at times, mistakes. There are cost benefits associated with each end of this spectrum.
Assistance giving allows students to move forwards when they are struggling and to experience success, yet can lead to shallow learning, non-activation of long-term memory and the lack of motivation to learn on their own. On the other hand, assistance withholding encourages students to think and learn for themselves, yet can lead to “floundering, frustration and wasted time when students are unsure of what to do.
ImpactResults from trials of the SimScientists and ChemVlab+ modules indicate that the kinds of feedback built into the systems are producing learning gains and, more interestingly, that they might benefit particular students.<|endoftext|>
| 3.671875 |
370 |
# Point A is at (3 ,2 ) and point B is at (-7 ,3 ). Point A is rotated pi/2 clockwise about the origin. What are the new coordinates of point A and by how much has the distance between points A and B changed?
Mar 21, 2016
Rotating Point A $\left(3 , 2\right)$ clockwise through $\frac{\pi}{2}$ yields Point A' $\left(2 , - 3\right)$. We can then calculate the distance between Point A' and Point B as approximately $10.8$ $u n i t s$.
#### Explanation:
Now, $\frac{\pi}{2}$ is $\frac{1}{4}$ of a full rotation ($2 \pi$) or ${90}^{o}$. If we rotate the point $\left(3 , 2\right)$ through this angle clockwise, we move from the first to the fourth quadrant, so the $x$ value will still be positive and the $y$ value will be negative.
You should probably draw a diagram, but it turns out that the new point, which we can call Point A', has the coordinates $\left(2 , - 3\right)$.
We can calculate the distance between Point A' and Point B:
$r = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}} = \sqrt{{\left(\left(- 7\right) - 2\right)}^{2} + {\left(3 - \left(- 3\right)\right)}^{2}} = \sqrt{81 + 36} \approx 10.8$<|endoftext|>
| 4.53125 |
223 |
# How do you simplify square root of 70 / square root of 40?
Sep 19, 2015
$\frac{\sqrt{70}}{\sqrt{40}} = \frac{\sqrt{7}}{2}$
#### Explanation:
Using prime factorization, $\frac{\sqrt{70}}{\sqrt{40}}$ can be rewritten as sqrt(7*5*2) / sqrt(5*2^3
$\sqrt{5 \cdot 2}$ is found in both the numerator and denominator and cancel each other out:
$\frac{\sqrt{7 \cdot 5 \cdot 2}}{\sqrt{5 \cdot {2}^{3}}} = \frac{\sqrt{7}}{\sqrt{{2}^{2}}}$
The denominator $\sqrt{{2}^{2}}$, or $\sqrt{4}$, is a perfect square which simplifies to $2$:
$\frac{\sqrt{7}}{\sqrt{{2}^{2}}} = \frac{\sqrt{7}}{2}$
This is the simplified expression.<|endoftext|>
| 4.46875 |
1,265 |
500 Years Later, Martin Luther's Legacy Lives On
On this day in 1517, a young Catholic monk rejected corruption and so changed the world.
Today marks the 500th anniversary of one of the most transformational and contentious days in all of Christendom. It’s the day that a lowly German monk, Martin Luther, nailed his 95 theses to the door of Wittenberg Castle Church, challenging many of the teachings of the Catholic Church and creating a schism in the church that would launch the Protestant Reformation, reverberating throughout the centuries.
To Catholics, the teachings of Luther were and are a heresy that challenged the power of the universal church, sowing confusion and disunity, leaving factions where there was once a unified body of Christ. To Protestants, it was a needed challenge to a Catholic hierarchy that had grown decadent and abused its power, corrupting the pure doctrines of Christ as revealed through Holy Scripture, growing wealthy and powerful by making its priestly class the sole conduit through which the will of God was revealed to mankind.
Martin Luther was an unlikely candidate to be the catalyst for such a monumental change. Born in 1483 in Eisleben, Germany, Luther was the son of a businessman who sent his son to law school so young Martin could return to assist in the family’s business endeavors. As the story goes, the boy Martin was one day trapped in a terrible storm and prayed that if God would spare him, he would dedicate his life to serving God by joining the priesthood. Martin kept his word and became a monk.
As Luther studied the Bible with deepening intensity, he struggled more and more to reconcile what he read there with the teachings of the Catholic Church. Especially egregious to him was the Catholic practice of issuing indulgences; essentially purchasing forgiveness for sins, whether one’s own or one’s already deceased relatives. It was the zealousness with which a local priest, Johann Tetzel, sold indulgences to the poor in exchange for promises of salvation of the souls of their families from purgatory that led Luther to write his 95 theses. He declared the supremacy of the Bible and expounded his views of salvation through grace alone.
The Protestant Reformation rejected many of the teachings of the Catholic Church, including the claim of the inerrancy of the papacy. To Catholics, this led to theological and moral chaos because the infallibility of papal proclamations was replaced with an individual search for truth through the study of the Bible and prayer.
Luther’s influence extended far beyond these 95 theses, however. While in hiding from Catholic authorities, he accomplished his greatest work — translating the Bible into German for the first time. In doing so, he unified the various German dialects into a common language, making the Word of God accessible for the first time to the common man, and in doing so stripping the Catholic priesthood of its aura of primacy and mystique as intermediaries with God. The invention of the printing press in 1453 helped Luther to spread his message throughout Germany, and eventually throughout Europe.
The Catholic Church, which dominated Europe in almost every way, did not take lightly Luther’s teachings, which undermined its authority and therefore its power. It persecuted Luther for the rest of his life, along with other reformers like William Tyndale, a British priest and scholar who translated the Bible into English after having been denied permission to do so by the Bishop of London (who claimed such translations were illegal). For this disobedience Tyndale would pay with his life, strangled and then burned at the stake in 1536, even in death praying for enlightenment for his executioner.
The schism between the Catholics and the Protestants would rend Europe with violence for centuries. In 2016, Pope Francis asked for forgiveness from Protestants and other Christian churches for the Catholic Church’s persecutions.
Ironically, in Western Europe, the home of the Protestant Reformation, today the schism between Protestant and Catholic has given way to apathy, or even antagonism, toward Christianity. While the majority of Europeans still identify as Christian, it’s only nominally so. Ever more they reject Christianity as the moral foundation of their society. Whereas faith was once the dominant force in German life, and European life more broadly, that has now been replaced with an almost totally secularist view. This trend is evident, for example, in the policies of German Chancellor Angela Merkel, a self-professed Lutheran who allowed in roughly a million refugees from the Middle East on the basis of her Christian faith, who nevertheless tends toward secularism in policy matters.
Interestingly, it is immigrants from North Africa and the Middle East who are providing resurgence in the faith. In Amsterdam, for example, the majority of the 350 churches in the city are led by immigrants who, having fled persecution in their native lands, now embrace Protestant Christianity in their adopted homelands.
This illuminates an interesting parallel with the birth of Christianity itself. The gospel of Christ was first brought to the Jews, who rejected it and persecuted its followers, and was then taken to the Gentiles. Likewise, in our day, even as Europeans, heirs to the Protestant Reformation, have largely turned their backs on their inheritance, immigrants to their lands are adopting the Christian faith as their own, and in doing so, transforming one soul at a time.
The Bible oft reveals that the Lord works in mysterious ways. The history of the world was transformed by the teachings of a humble Nazarene carpenter’s son. Even for those who doubt the divinity of Christ, His impact on the world is beyond dispute.
Fifteen hundred years later, Martin Luther, a humble monk, sought to better understand the true nature of the teachings of Christ, rejecting what he saw as errors in Catholic teachings. In publishing his thoughts, he transformed the world yet again, even laying much of the intellectual foundation for the American embrace of individual liberty. Though many may dispute whether his legacy is one of good or ill, none can dispute that, like his Master, Jesus Christ, Martin Luther’s influence has been felt long after his death.<|endoftext|>
| 3.9375 |
631 |
Financial analysis is defined as an assessment of the going concern, leverage and profitability of a business or a joint venture. It is often referred to as an accounting or financial statement analysis. To perform financial analysis, many analysts use models, and these models can be either univariate or multivariate.
A univariate analysis is the most basic type of quantitative analysis. It is carried out using a single variable. For example, if the variable revenue per reporting period was the subject of the analysis, then the researcher would look at how many reporting periods showed a high, average or low revenue figure. The main quantitative method for univariate analysis is the median, mean or average. It also describes the dispersion of data: the range, maximum and minimums, variance with average and standard deviations, as well as frequency distributions.
A multivariate analysis is based on two or more variables within a financial model. This is a far more sophisticated financial analysis as it can take into account relationships, correlations and interdependencies. Multivariate analyses describes causes and its major use is to explain, compare and highlight relationships.
Differences between univariate and multivariate analysis
Univariate analysis is primarily used at the initial stages, by analyzing data that is already available. Multivariate analysis is used for inferential research, as two or more variables can be unknown or approximated. Univariate analysis is usually utilized for descriptive purposes, whereas multivariate analysis is aimed towards explanations.
In financial analysis, the Univariate Model usually used is the one proposed by W. Beaver, where a company is classified as a failure if any of the three events occurred: bankruptcy, bond default, or an overdrawn account. To forecast financial failure, analysts focused on three financial ratios:
Cash Flow / Total Debt Net Income / Total Assets Total Debt / Total Assets
The univariate model assumes a business fails when any one of these financial ratios indicate financial difficulty; this is a simplification as start-ups may have low cash flow and large debt balances but cannot be classified as failures. The Univariate Model does not capture all scenarios of possibilities given its limited variable input.
The Multivariate analysis used in financial circles is known as the Altman Z-Score, a multi-variable analysis which uses five financial ratios, weighted according to importance to help increase the accuracy of the model’s prediction. Using these five variables, the model produces an overall discriminate score, a Z-score or zeta number. This multivariate analysis uses different indicators of risk and profitability, resulting in a model that showed a company’s risk of failure relative to a standard.
This allows for a more comprehensive view of the business’ current financial standing, and also standardizes different sized companies so that a meaningful comparison can be conducted.
However, a multivariate financial analysis can be difficult and costly to run. Furthermore, some financial information may be deemed confidential and be unavailable to analysts. It also takes far more time to be set up and tested successfully, whereas a univariate model is a quick, easy way of getting a feel for a company’s financial position.<|endoftext|>
| 3.71875 |
554 |
# What is the shortest path between the three points (4, 7), (2, 3) and (0, 1)?
hala718 | High School Teacher | (Level 1) Educator Emeritus
Posted on
To find the shortest path,we will calculate the length of the path between all three points.
The distance Between (4,7) and (2,3)
==> D1 = sqrt( 4-2)^2 + (7-3)^2 = sqrt(4+16) = sqrt20
The distance between (4,7) and ( 0,1)
==> D2 = sqrt(4^2 + (7-1)^2 = sqrt(16+36) = sqrt(52)
The distance between ( 2,3) and (0,1)
==> D3 = sqrt(2^2 + (3-1)^2 = sqrt(4+4) = sqrt8
Now we can conclude that the shortest distance is sqrts20.
Then we should go from (0,1) to (2,3) to (4,7).
The shortest path is sqrt8+sqrt20.
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
You want the length of the shortest path which has all the three points (4, 7), (2, 3) and (0, 1) on it.
Let us first find the distance between each of the points. We use the relation for the distance between tow points (x1, y1) and (x2, y2) as sqrt [(x1 – x2) ^2 + (y1 – y2) ^2].
(4, 7) and (2, 3): sqrt [(4 – 2) ^2 + (7 – 3) ^2] = sqrt [4 + 16] = sqrt 20
(4, 7) and (0, 1): sqrt [(4 – 0) ^2 + (7 – 1) ^2] = sqrt [16 + 36] = sqrt 52
(2, 3) and (0, 1): sqrt [(2 – 0) ^2 + (3 – 1) ^2] = sqrt [4 + 4] = sqrt 8
Therefore the shortest path would be if you go from (0, 1) to (2, 3) and then to (4, 7) and the total distance to be covered would be sqrt 8 + sqrt 20
The required result is sqrt 8 + sqrt 20<|endoftext|>
| 4.4375 |
1,084 |
WESTERN & CENTRAL AFRICA
A new IUCN report released today evaluates the state of West and Central Africa’s terrestrial and freshwater fauna and highlights the inadequacy of responses to rapid wildlife decline in the region. Improved legislation and much more effective protection is urgently required to meet international targets for protected areas and halting biodiversity loss.
The report attributes the erosion of West and Central Africa’s biodiversity to habitat loss and degradation due to rapid urbanization, agricultural expansion and unsustainable resource exploitation, as well as hunting for bushmeat and the illegal wildlife trade.
The 22 countries of West and Central Africa are home to a rich tapestry of species, habitats and ecosystems. Many areas, such as the Upper Guinea forests, the Afromontane forests of the Nigeria and Cameroon border and the Albertine Rift, and the Congo Basin, have long been considered conservation priorities. However, the rapidly growing human population is projected to rise to over 600 million in little over a decade, placing tremendous pressure on the region’s natural heritage.
10% of the 2,471 amphibian, bird and mammal species native to West and Central Africa are threatened with extinction, as well as 17% of the more than 1,600 freshwater fish species. In the last decade, both African rhinos, the Black Rhinoceros (Diceros bicornis) and White Rhinoceros (Ceratotherium simum), have been extirpated from the region, while the Scimitar-horned Oryx (Oryx dammah) became Extinct in the Wild in the 1980s. Gambia, Mauritania, Senegal, and Mali have lost five or more of their historically native large mammal species. A few species, including Dama Gazelle (Nanger dama) and Dryad Monkey (Cercopithecus dryas) have global populations now down to only a few 100 individuals in the wild, while regional subpopulations of African Wild Dog (Lycaon pictus), Lion (Panthera leo), Cheetah (Acinonyx jubatus), Giraffe (Giraffa camelopardalis), and Giant Eland (Tragelaphus derbianus) are all Critically Endangered.
The decline of wildlife in West Africa in particular, can be attributed to extensive deforestation and forest fragmentation, primarily via wide-scale clear-cutting to replace forests with agricultural land and commercial plantations. Central African forests remain relatively intact, but roughly one-third of remaining forests are in logging concessions, suggesting that pressures are increasing. The region as a whole is also subject to extensive and increasing exploitation of its mineral and oil reserves, involving both large commercial, open-cast operations and artisanal activities. Mining operations have already led to the downsizing and degazettement of protected areas, including one World Heritage site. Even where forests remain intact, bushmeat hunting, especially for ungulates, is prevalent and off-take rates are often not sustainable. The black market demand for ivory and, more recently, pangolin scales, is further driving wildlife declines.
The authors point out the inadequacy of national legislation in the region, especially in meeting global targets for biodiversity and protected areas. Furthermore, many sites important for biodiversity remain unprotected in the region, including more than one-third of sites known to hold the last remaining population of a highly threatened species. Nonetheless, despite considerable pressures, tremendous complexity, and regional instability, there is an excellent track record of civil society organizations supporting, and assuming a mandate for, wildlife conservation interests in the region.
The information in the report can be used to guide urgently needed improvements to legislation and enforcement in order to fulfil obligations to international agreements and to allow scarce resources to be targeted with much greater efficiency. The findings will inform further targeted conservation action and expand IUCN’s existing work through its SOS – Save Our Species initiative. To date, this activity has helped improve management in eight protected areas in the region, such as Bouba-Njiidda National Park in Cameroon and Conkouati-Douli National Park in Congo, and at least 25 threatened vertebrates, among them the Slender-snouted Crocodile (Mecistops cataphractus), Sawfish (Pristis spp.), and thePygmy Hippopotamus (Choeropsis liberiensis). Similarly these findings will guide national and regional protected area capacity development and policy support activities through the EU-funded BIOPAMA (Biodiversity and Protected Areas Management) programme. It will also add important context to the forthcoming EU strategic approach for African wildlife conservation.
This Situation Analysis was undertaken to inform responses to several resolutions made at the 5th World Conservation Congress in 2012 about the plight of large vertebrates in West and Central Africa. The study was produced with support from the 10th European Development Fund through the BIOPAMA programme, jointly implemented by IUCN, the European Commission’s Joint Research Centre (JRC)and the Deutsche Gesellschaft für Internationale Zusammenarbeit (GIZ).
The full report, An IUCN situation analysis of terrestrial and freshwater fauna in West and Central Africa, is available for download here.<|endoftext|>
| 3.71875 |
293 |
About The Course
The discovery of a new fundamental particle at the Large Hadron Collider (LHC), CERN is the latest step in a long quest seeking to answer one of physics’ most enduring questions: why do particles have mass? The experiments’ much anticipated success confirms predictions made decades earlier by Peter Higgs and others, and offers a glimpse into a universe of physics beyond the Standard Model.
As Professor Peter Higgs continues his inspiring role at Edinburgh University’s School of Physics & Astronomy, the experiments at the LHC continue.
This free online course introduces the theoretical tools needed to appreciate the discovery, and presents the elementary particles that have been discovered at the tiniest scales ever explored. Beginning with basic concepts in classical mechanics, the story unfolds through relativity and quantum mechanics, describing forces, matter and the unification of theories with an understanding driven by the tools of mathematics.
Narrating the journey through experimental results which led to the discovery in 2012, the course invites you to learn from a team of world-class physicists at Edinburgh University. Learners participate in discussion of the consequences of the Higgs boson, to physics and cosmology, and towards a stronger understanding and new description of the universe.
This course will also give you the opportunity to purchase a Statement of Participation. If you want to demonstrate your knowledge further, you can register for an end-of-course exam to earn a Statement of Attainment.<|endoftext|>
| 4 |
201 |
# How do you solve 1/5 ( 10x - 15 ) = 6 ( 1/3x - 1/2 ) + 10?
May 9, 2018
No solution.
#### Explanation:
$\frac{1}{5} \left(10 x - 15\right) = 6 \left(\frac{1}{3} x - \frac{1}{2}\right) + 10$
$\frac{1}{5} \times 10 x - \frac{1}{5} \times \left(- 15\right) = 6 \times \frac{1}{3} x - 6 \times \frac{1}{2} + 10$
$2 x + 3 = 2 x - 3 + 10$
$2 x - 2 x = - 3 + 10 - 3$
$0 \ne 4 \to$ no solution<|endoftext|>
| 4.4375 |
2,059 |
The black grouse has experienced a serious decline over the last century. Restoring a healthy mosaic of habitats is crucial for the recovery of this attractive bird.
The black grouse is found throughout northern Eurasia, with a continuous distribution from Great Britain to south-eastern Siberia. The most southerly populations are found in Kyrgyzstan and North Korea, with the northern extent of its range being northern Norway. In western and central Europe its population is highly fragmented: populations are very small and isolated in a number of countries and it no longer breeds in Denmark. Its overall population has declined, particularly in Europe, although its population is more stable in its Russian stronghold. Black grouse usually require a mosaic of habitats, ideally including heath and bog along with open woodland and scrub for cover. There are thought to be seven or eight subspecies of black grouse, the only geographically isolated one being the most westerly of all, Britain's Tetrao tetrix britannicus.
This grouse was once distributed through most of Great Britain. However, its numbers have declined drastically over the last century, and its range has retreated rapidly northwards. A 2005 survey coordinated by the Royal Society for the Protection of Birds reported an alarming 22% decline in the preceding ten years. Its core population is now in Scotland, where there are an estimated 3,344 of the UK's 5,078 displaying males. There are smaller populations in northern England and Wales, although the decline has actually been reversed in the latter.
The reasons for this decline are varied and complex. Changes in agricultural practices, and in particular the increase in sheep numbers, have led to herb-rich and wet pasture areas becoming overgrazed, with a resulting decline in the insects essential for the survival of young grouse. In addition, overgrazing by sheep and deer has damaged areas of heather moorland; heather (Calluna vulgaris) being an important winter food source for black grouse. Excessive grazing has also suppressed native woodland regeneration, depriving the birds of some valuable food sources as well as refuge from predators. Drainage of bogs (which are important foraging grounds for chicks) has also contributed to the decline, while commercial afforestation has shaded out important berry plants such as blaeberry (Vaccinium myrtillus). High tensile deer fences pose a serious hazard to black grouse: when alarmed, these birds tend to fly fairly low, and seek refuge in nearby trees. They often do not see fences in time and are killed as they hit them. Furthermore, illegal shooting during the breeding season can affect both population and breeding success. Climate further complicates the picture, as cold, wet weather in June and July has increased mortality among newly-hatched chicks, which are unable to regulate their own body temperature and are therefore vulnerable to poor weather.
A number of measures are being used to increase the black grouse population, including restoration of preferred habitats, as well as predator control. Redundant fences are being removed, while those essential for woodland regeneration are marked in various ways to make them more visible, and are strategically positioned so as to pose less of a hazard. Otherwise, alternative methods such as deer culling are preferred ways of encouraging regeneration. Mature, open pine and birch woodland, bogs and heath were once much more widespread in Scotland, and restoration of these habitats is playing a key role in black grouse recovery.
Locally extinct in many regions, the black grouse is the fastest declining bird in the UK and is on the Red List of Birds of Conservation concern. It is protected under the Game Acts (closed season 11 December-19 August); while it may be legally hunted in the open season, many estates choose not to do so. It is classed as Vulnerable in Europe, and is in Annex II/2 of the EC Birds Directive and Appendix III of the Bern Convention. In Britain, the black grouse is a Priority Species, and therefore the subject of a Biodiversity Action Plan (BAP), under the UK government's response to the Convention on Biological Diversity signed at the Earth Summit in 1992. However, its global population is not thought to be endangered, and it is classified as being of Least Concern.
The black grouse (also known as 'blackgame') is a member of the grouse family Tetraonidae. Roughly the size of a domestic hen, the male, or 'blackcock', is 49-58 cm. in length with a tail measuring around 15 cm.; he weighs approximately 1.25 kg. The female or 'greyhen' is slightly smaller at 40-45 cm., and weighs 950 g. The male has black plumage with a bluish sheen. He has white wing-bars, which are only visible in flight, white under-tail coverts, and a distinctive lyre-shaped tail. The female has grey-brown plumage, (providing excellent camouflage when nesting on the ground) with pale wing bars, and her tail has a slight notch at the end. Both sexes have a red patch called a 'wattle' above each eye. Among its calls are a rapid cackling and a harsh hissing sound. When displaying, the males produce a pigeon-like, bubbling call, interspersed with a 'sneezing' noise.
The diet varies considerably through the year, and also in different parts of the bird's range. Birch buds (Betula spp.) along with heather and blaeberry shoots are important winter food sources. Cotton grass (Eriophorum spp.) and larch buds (Larix spp.) are eaten in spring, pine pollen (Pinus sylvestris) in early summer, blaeberries in late summer, and rowan berries (Sorbus aucuparia) in autumn and winter. For the first three weeks of their lives, chicks feed on protein-rich invertebrates, particularly moth caterpillars and sawfly larvae (suborder: Symphyta).
Blackcocks participate in a fascinating territorial and breeding display known as a lek (from the Swedish leka - play). They gather on an open area of ground where they strut, fanning out their tails, spreading their wings slightly and inflating the wattles above their eyes. The dominant males gain the favoured position at the centre of the lek, increasing their chances of mating. The females, who only attend in late March to mid-May, are positioned around the edge of the lek, watching and selecting the most impressive males to mate with. Lekking takes place almost year-round, except for July and August when the males moult. Outside of the breeding season they lek to establish hierarchy and defend territories.
From late April to early June, greyhens typically lay a brood of 6-10 eggs, which are buff with brown spots. They are normally laid in a simple hollow in the ground, with little lining, although they are usually concealed by surrounding tall vegetation such as heather or rushes (Juncus spp.). The eggs are incubated for about 24 days; the chicks hatch in mid to late June and fledge after about four weeks. Black grouse do not form pair bonds, and the male plays no part in rearing the chicks. The average lifespan of a black grouse is around five years.
As with other herbivorous animals, the black grouse plays a role in keeping the vegetation structure in an area more varied than if it were completely ungrazed. During their insectivorous phase as fledglings they also help to regulate numbers of the caterpillars and other invertebrates they eat.
The black grouse is prey for numerous predators, including raptors such as golden eagles (Aquila chrysaetos) and hen harriers (Circus cyaneus). Pine martens (Martes martes), foxes (Vulpes vulpes), crows (Corvus corone), and stoats (Mustela erminea) eat the birds as well as their eggs.
Black grouse are host to a variety of parasites. These include feather mites, for example Tetraolichus gaudi, tapeworms such as Hymenolepis spp., and blood parasites such as the protozoan Leucocytozoon lovati. On an evolutionary timescale, these parasites improve the health and vigour of the species overall, by selecting out weaker individuals.
The black grouse depends on other animals to provide some of the habitat it requires. In some areas conservationists use cattle to maintain the open habitats the grouse need for their leks. It is likely that in former times, herbivores such as the extinct wild cattle, or auroch (Bos primigenius) would have played a role in keeping patches of the forest relatively open.
Human management for the black grouse also has a knock-on effect for other wildlife. As this bird requires a variety of habitats, management for the recovery of black grouse benefits a wide range of other species. Conservationists are working hard to aid the recovery of this attractive bird, and there is still hope that the overall downward trend can be reversed.
Mullarney, K., Svensson, L., Zetterstrom, D., Grant, P.J. (1999). Collins Bird Guide. Collins: London.
Dan Puplett https://www.danpuplett.net/
Sign up to our mailing list to receive our monthly ‘Tree News’ e-newsletter and other occasional emails about volunteering, events, appeals and fundraising. It’s the perfect way to stay up to date with the latest news about the wild forest and it’s wonderful wildlife.
Live locally and want to know more about volunteering with us? Get updates about our Conservation Days straight to your inbox.
Support Trees for Life and receive our exclusive magazine!<|endoftext|>
| 4.03125 |
1,100 |
More is more – nowhere is that truer than at the world’s most powerful atom smasher, the Large Hadron Collider in Switzerland, where scientists last week concluded a six-month series of experiments where they forced infinitesimally tiny particles to smash against each other at double the energy level ever recorded.
The higher energy level – 13 trillion electronvolts – will increase physicists’ chances of answering some of the most daunting questions in science. Through their work, researchers hope to find out if there are extra dimensions in the universe other than the three we’re familiar with. They also hope to elucidate what dark matter might be – that’s the “stuff” that makes up about a quarter of the universe.
And there might even be surprises along the way, said physicist Michael Barnett, of the Lawrence Berkeley National Laboratory.
“We don’t know what we don’t know,” said Barnett, who recently spent a week at the Large Hadron Collider, in Geneva. “All we do is collide protons.”
The collider smashes tiny constituents of matter called protons against other protons inside a 17-mile ring so long that it straddles the border of Switzerland and France.
The giant accelerator’s first run started in 2010 and culminated two years later with the discovery of the Higgs boson, also known as the “God particle” because it has the god-like ability to confer mass to other particles. Scientists like Barnett hope that it will take two more years to find clues about extra dimensions and dark matter.
The process involves looking for phenomena that can only be created inside a particle accelerator, such as microscopic black holes that disappear in less than a millionth of a second, leaving only traces to be pored over by scientists.
“It’s like fireworks,” said Barnett, “with tails that become more and more elaborate.”
Many of the technologies that made the Large Hadron Collider possible were pioneered in the Bay Area.
Physicists on the University of California, Berkeley, campus in the 1930s and at the Stanford Linear Accelerator Center, in Menlo Park, in the 1970s, created precursors to the Large Hadron Collider that led to key discoveries about the tiny constituents of the atom – from the nucleus all the way down to quarks.
In its first iteration, the cyclotron created by UC Berkeley physicist Ernest Lawrence in 1930 fit in the palm of his hand. It was a breakthrough because, without requiring much energy, it could produce very energetic particles in a small space. This allowed physicists to readily investigate the atom’s nucleus by creating elements with large nuclei.
The resulting new field of nuclear science has a complicated legacy, said Lawrence Berkeley Lab nuclear physicist Larry Phair. Nuclear physics were used to build the atomic bomb, as well as to create the medical accelerators that are now commonly used to fight cancer.
Subsequent versions of the cyclotron were so big that they were housed in their own buildings. The Lawrence Berkeley Lab started out as the facility that Ernest Lawrence built above the UC Berkeley campus to house his ever-bigger cyclotrons.
When it opened in Menlo Park in 1966, the Stanford Linear Accelerator Center, now the SLAC National Accelerator Laboratory, was the longest particle accelerator in the world. The linear accelerator sent electron beams traveling down a two-mile row of microwave-oven-like devices and smashed them against a stationary target. Physicists used these accelerated electrons to investigate what was inside the protons and neutrons, and in 1968 they found that they were made up of minuscule constituents they called quarks.
A few years later, SLAC physicist Burton Richter built a collider – a type of particle accelerator in which particle beams are smashed against each other to reach high energy levels.
“All the energy of those two beams could get transformed into new kinds of particles,” said Richter.
The so-called SPEAR collider that Richter built led him and his team to discover a more massive quark called the charm quark, and won him the Nobel Prize in physics.
“It was a revolutionary idea, to collide two beams against each other,” said Barnett. The SPEAR collider became a precursor to the Large Hadron Collider.
Today, dozens of physicists and graduate students at the Lawrence Berkeley Lab and SLAC are working at the Large Hadron Collider, making regular trips to Geneva and crunching data back home in their labs.
The particle accelerators at both facilities have been given new uses.
The cyclotron at the Lawrence Berkeley Lab is used to test computer chips that go into satellites, by exposing them to high-radiation conditions similar to those they’ll encounter in space.
And the X-rays emitted by accelerated particles at SLAC are being used to study the impact of climate change on coral reefs.
For Richter, the Large Hadron Collider offers the tantalizing possibility of answering fundamental questions about the universe, one by one.
“The blackboard is covered with Post-it notes now,” said Richter.
He looks forward to “going down the line and removing them all.”<|endoftext|>
| 3.71875 |
227 |
## Algebra: A Combined Approach (4th Edition)
$$x=22$$ $$y=-6$$
Let $x$ be the larger number and $y$ be the smaller number. Equation 1: $x\:+\:y = 16$ Equation 2: $3x-y=72$ We can solve for the numbers using the addition method. Adding equations & 2: $$x\:+\:y = 16$$ $$+$$ $$3x-y=72$$ $$=$$ $$4x=88$$ Divide both sides by $4$: $$\frac{4x}{4}=\frac{88}{4}$$ $$x=22$$ Substitute this value of $x$ to equation 1: $$x\:+\:y = 16$$ $$22+y=16$$ Subtract $22$ from both sides: $$22-22+y=16-22$$ $$y=-6$$ Use equation 2 to check: $$3x-y=72$$ $$3(22)-(-6)=72$$ $$66+6=72$$ $$72=72$$<|endoftext|>
| 4.46875 |
159 |
The resource has been added to your collection
The idea with the learning object is to practice the different forms of comparison. Since the student is to solve the exercise through clues, there will also be game-like features in the learning scenario. By clicking the pictures of different persons, the student will hear descriptions of these people. The student task is to place the persons in correct sequence (for example, youngest-oldest) based on the information he or she hears. The student can listen to the descriptions as many times as needed. To check the exercise, you click the word "Control". If the student has placed persons correctly, the description of the persons will be heard again and their names are visible on the screen.
This resource has not yet been reviewed.
Not Rated Yet.<|endoftext|>
| 3.9375 |
1,686 |
# TCS previous placement paper - 22
1. X takes 4 days to complete one-third of a job,Y takes 3 days to complete one-sixth of the same work and Z takes 5 days to complete half the job.If all of them work together for 3 days and X and Z quit, how long will it take for Y to complete the remaining work done.
a. 6 days b. 7 days c. 5.1 days d. 8.1 days
Answer: c
Explanation:
X takes 12 days to complete the full work. Y take 18 days, Z takes 10 days.
3 days work = $3\left( {\dfrac{1}{{12}} + \dfrac{1}{{18}} + \dfrac{1}{{10}}} \right)$ = $\dfrac{{43}}{{60}}$
Remaining work = $1 - \dfrac{{43}}{{60}} = \dfrac{{17}}{{60}}$
This work should be completed by Y in $\dfrac{{17}}{{60}} \times 18$ = 5.1 days
2. Thomas takes 7 days to paint a house completely whereas Raj would require 9 days to paint the same house completely. How many days will take to paint the house if both them work together.(give answers to the nearest integer)?
a. 4 b. 2 c. 5 d. 3
Answer: a
Explanation:
Simple formula = $\dfrac{{xy}}{{x + y}}$ = $\dfrac{{7 \times 9}}{{7 + 9}} \simeq 4$
3. One day, Eesha started 30 minutes late from home and reached her office 50 minutes late, while driving 25% slower than her usual speed. How much time in minutes does Eesha usually take to reach her office from home?
a. 20 b. 40 c. 60 d. 80
Answer: c
Explanation:
She got late to the office 20 minutes late as she drove at 3/4 th of the speed.
Given, $\dfrac{d}{{\dfrac{3}{4}s}} - \dfrac{d}{s} = 20$
⇒ $\dfrac{d}{s}\left( {\dfrac{4}{3} - 1} \right) = 20$
⇒ $Time = \dfrac{d}{s} = 60$
4. Curious Elva asked her father what he would gift for her nineteenth birthday. Father replied that it would depend on the day of the week and be one of SUNglasses, MONeybag, ..., FRIedcake, and SATchel. Please help Elva find the day of the week on 08-Jan-2029
a. Monday b. Tuesday c. Thursday d. Saturday
Answer: a
Explanation:
Number of odd days upto 2000 = 0
From 2001 to 2028 = 28 + 7 = 35 = 0 ($\because$ 35/7 remainder zero)
From 2019 January 1 to 7 = 7 = 0
So 08 - Jan - 2029 falls on the same week day as 1-1-1 which is Monday.
5. All even numbers from 2 to 98 inclusive, except those ending 0, are multiplied together. What is the rightmost digit (the units digit) of the product?
a. 6 b. 2 c. 0 d. 4
Answer: a
Explanation:
2 × 4 × 6 × 8 × 12 × 14 × . . . . . × 98
Now units digit of 2 × 4 × 6 × 8 = 4
Also 12 × 14 × 16 × 18 also 4. So on
Total 10 times 4 occurs in the units digit = ${4^{10}} = 6$
6. In 2003, there are 28 days in February and there are 365 days in the year. In 2004, there are 29 days in February and there are 366 days in the year. If the date March 11, 2003 is Tuesday, then which on of the following would the date March 11, 2004 be?
a. Monday b. Thursday c. Wednesday d. Tuesday
Answer: b
Explanation:
March 11, 2003 is Tuesday. So March 11, 2004 weekday will be 2 days after Tuesday. i.e., Thursday.
7. 8 year old Eesha visited her grandpa. He gave her this riddle.
I started working at 13. I spent 1/6 of my working life in a factory. I spent 1/4 of my working life in an office, and I spent 1/4 of my working life as a school caretaker. For the last 32 years of my working life I've been doing social service. How old am I?
a. 109 b. 102 c. 105 d. 113
Answer: a
Explanation:
Let $x$ be the number of years he worked.
$\Rightarrow \dfrac{x}{6} + \dfrac{x}{4} + \dfrac{x}{4} + 32 = x$
$\Rightarrow x = 96$
His age = 96 + 13 = 109
8. 100 students appeared for two examinations. 60 passed the first, 50 passed the second and 30 passed both. Find the probability that a student selected at random has failed in both the examinations?
a. 1/5 b. 5/6 c. 1/7 d. 5/7
Answer: a
Explanation:
$n(A \cup B)$ = $n(A)$ + $n(B)$ - $n(A \cap B)$
$n(A \cup B)$ = 60 + 50 - 30 = 80
So 80 passed in atleast one of the exams. 100 - 80 = 20 failed in both.
Probability = 20/100 = 1/5
9. What is the greatest power of 143 which can divide 125! exactly
a. 12 b. 11 c. 8 d. 9
Answer: d
Explanation:
143 = 11 × 13. So highest power of 13 should be considered in 125!.
Highest power of 11 in 125! is 12 but highest power of 13 is only 9. That means, 125!=$11^{12}\times13^{9}\times....$
So only nine 13's are available. So we can form only nine 143's in 125!. So maximum power of 143 is 9.
10. Three containers A, B and C are having mixtures of milk and water in the ratio of 1:5, 3:5, 5:7 respectively. If the capacities of the containers are in the ratio 5:4:5, find the ratio of milk to water, if all the three containers are mixed together.
a. 53:115 b. 53:113 c. 54:115 d. 54:113
Answer: a
Explanation:
Weighted average rule can be applied = $\dfrac{{5 \times \frac{1}{6} + 4 \times \dfrac{3}{8} + 5 \times \dfrac{5}{{12}}}}{{5 + 4 + 5}}$ = $\dfrac{{53}}{{168}}$
So milk and water concentration = 53 : (168 - 53) = 53 : 115<|endoftext|>
| 4.59375 |
416 |
# How many joules does it take to melt 35 g of ice at 0° C?
Apr 29, 2016
It takes 12,000 Joules of energy to melt 35 grams of ice at 0 °C
#### Explanation:
The enthalpy change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol. This means for every mole of ice we melt we must apply 6.02 kj of heat. We can calculate the heat needed with the following equation:
$q = n \times \Delta H$
where:
$q$ = heat
$n$ = moles
$\Delta H$ = enthalpy
In this problem we would like to calculate the heat needed to melt 35 grams of ice at 0 °C. This problem can be broken into three steps:
1. Calculate moles of water
2. multiply by the enthalpy of fusion
3. Convert kJ to J
Step 1: Calculating moles of water
$35 g \times \left(\frac{1 m o l}{18.02 g}\right) = 1.94 m o l s$
Step 2: Multiply by enthalpy of fusion
$q = n \times \Delta H = 1.94 \times 6.02 = 11.678 k J$
Step 3: Convert kJ to J
$11.678 k J \times \left(\frac{1000 J}{1 k J}\right) = 11 , 678 J$
Finally rounding to 2 sig figs (since 34°C has two sig figs) we get
$q = 12 , 000 J$
For more examples on phase changes and enthalpy, see the video below:
One last note, if the temperature were not 0 °C then the ice would have to be heated in addition to melted. This would be a phase change problem combined with a heat capacity problem.<|endoftext|>
| 4.5 |
1,277 |
The Homestake Mine started yielding gold in 1876. If you had asked George Hearst, the operator at the time, if the mine would someday yield the secrets of the universe I bet he would have laughed you out of the room. But sure enough, by 1960 a laboratory deep in the mine started doing just that. Many experiments have been conducted there in the five and a half decades since. The Large Underground Xenon (LUX) experiment is one of them, and has been running is what is now called the Sanford Underground Research Facility (SURF) for about four years. LUX’s first round of data was collected in 2013, with the experiment and the rest of the data slated to conclude in 2016. The method, hardware, and results wrapped up in LUX are utterly fascinating.
It’s All About Noise Filtering
Detecting dark matter is exceedingly hard. So much so that we’ve never actually done it. A sufficiently large detector like the LUX can hope to catch just a few interactions per year. To make sure that these aren’t missed, it’s important to filter out as many non-events as possible. This is the reason behind the underground location; 4,850 feet of earth stand between the detector and open air to reduce cosmic rays by a factor of 1 million. That doesn’t get rid of everything but it becomes possible to discern false readings.
The ultra pure Xenon further filters this noise in the system. What is left is a very dark environment waiting for Weakly Interactive Massive Particles (WIMPs) to pass through it. Theory tells us that somewhere between millions and billions of the WIMP dark matter particles move through one square centimeter of space every second. But one square centimeter on a subatomic scale is a vast and empty space. With the noise filtered out, researchers are just waiting for a WIMP to collide with a Xenon nucleus.
Xenon is a scintillator; when the nucleus is struck by a fast-moving subatomic particle, it gives off a photon. Ionizing electrons are also a result of the interaction and they in turn create secondary scintillation. The pattern of primary and secondary light is specific to the type of interaction and, if measured properly, can be used to distinguish a dark matter interaction from other events. The good news is that we’re really good at measuring light. In fact, you can probably already guess what mechanism is used in the measurement: a Photomultiplier Tube (PMT).
PMTs are used in all kinds of scientific measurement equipment, and also appear often in medical devices and photography equipment. We seen this last example as a source for a PMT that Kerry Wong used to demonstrate the speed of light.
LUX has 122 PMTs in its sensor array. A fair amount of signal conditioning sets the levels before being fed into the custom triggering system. Then things really start to get interesting. Readings are summed into 16 groups of PMTs which are then processed by the triggering system to establish if the pattern is that of a possible dark matter interaction. If you need to do a lot of very fast, parallel processing, what kind of hardware do you choose? You’re right, you reach for an FPGA and build up a system around it.
If you’re on the edge of your seat for details, the research team has come through in a big way. In November they published a paper entitled FPGA-based Trigger System for the LUX Dark Matter Experiment. And even if you don’t want to dig that deep, Professor Frank L.H. Wolfs at the University of Rochester has a concise set of pages dedicated to the triggering hardware which was designed by Wojtek Skulski.
The current iteration of the trigger pulse digitizer board goes by the part number DDC-8DSP. Two of these boards use a Xilinx Spartan-3A to capture the signals using 14-bit resolution at 64 MHz. Each of these boards collects and processes the signals, then send the data to a Trigger Builder board via HDMI cables. This board gets its name because it is responsible for inspecting for a characteristic pattern of primary and secondary scintillation that indicates a WIMP interaction (and differentiates from other interactions). The diagram on the left is from the published paper and illustrates the signal pattern that the hardware is looking for.
Did it Work?
Ah, be careful what you ask. Yes, LUX worked, delivering the first round of data in 2013 as anticipated. The real question is did it detect dark matter? Evidence of dark matter has not been proven in that data. But this, the most sensitive detector ever build for this purpose, hasn’t gone to waste. The research team has been able to further establish what properties WIMPs do not possess.
The experiment is currently running another round, having been further configured based on the 2013 data. This data set is expected to be ready in June of this year and will be the last run for this iteration of LUX. In the works is the LUX-ZEPLIN project which will use a much larger detector apparatus with vastly more liquid xenon, 488 PMTs in the sensor array, and several other improvements.
Experiments that Deserve Celebrity Status
When first hearing about LUX, you might be reminded of similar experiments to detect neutrinos because both experiments need to be located underground to filter out cosmic noise. Neutrino experiments have a higher public profile, although they’re probably not at celebrity status like the Large Hadron Collider. And there will be an exciting neutrino detection experiment sharing he SURF facility with LUX before too long.
With one billion dollars committed to build the new Deep Underground Neutrino Experiment (DUNE), planned to start searching for neutrinos in 2022, it is easy to look at these types of experiments as the new space race. Surely, investment in experimentation will yield technological advancements similarly transformative as those which can be attributed to mankind making our way into space. The missing piece of the puzzle is widespread public awareness and excitement for scientific discovery.<|endoftext|>
| 3.734375 |
1,487 |
In one of my weekly visits to the library, I came across a new book called Noah Webster & His Words by Jeri Chase Ferris (Houghton Mifflin 2012). This book, a biography of the creator of the first American dictionary, inspired me to write a post about teaching dictionary skills. The book itself is an adorable picture book that takes readers through Webster’s entire life while exploring his motivations for creating his famous dictionary. My favorite thing about the book is that there are “dictionary entries” peppered throughout. The author immediately follows each challenge word like “unite” with the word’s definition in brackets. What a great way to add a little vocabulary lesson to the reader’s experience and to honor Webster’s legacy!
I think that third graders are at the perfect stage in their educational development to benefit from a unit about dictionaries. Third graders are starting to read more, write more, and research more than the K-2 set, so the dictionary is an important reference with which they should become acquainted.
You should plan to spend a block of about 5-7 library sessions on dictionary skills (see the Lesson Plan Calendar in The Tibrarian Handbook for an outline of what to cover each session). By the end of your unit on dictionaries, students should be able to:
- List the main functions of a dictionary (word definitions, spelling, parts of speech, sample usage, origin, pronunciation)
- Identify the section of the dictionary in which a specific word can be located (ex. The word “nullify” would be found in the third quarter of the dictionary)
- Use guide words to locate a specific word in the dictionary
- Effectively use the dictionary to find out information about a word (ex. “What is the definition of abet?'” “What part of speech is the word lovely?”)
Some ideas for how to help students learn to use guide words:
- Copy a page from the dictionary onto an overhead transparency, scan a page, save on your computer and project onto an interactive whiteboard, scan a dictionary page and blow it up into poster size (if your system is lucky enough to have a poster printer). In other words, use the resources available to you to present a page of the dictionary to your students so that they can all see it and interact with it. The best methods will allow you to mark on the page and then reuse it for your next class. In this manor, you can show your students what guide words look like, and where to find them, as well as get them started using those guide words to find words on the page.
- Choose a page from a dictionary that you have in your library. Write the two guide words from the top of one page on two separate note cards/sentence strips/pieces of paper. Tape the words onto the ends of two of your shelves, or on the wall, making sure to leave significant space between the words. You can then give each student a word written on a note card. Students should look at their words and place themselves either in-between, before, or after the guide words to show where their word would be found in the dictionary. Getting students up and moving helps them to visualize this concept more clearly. You can also end the activity by asking students to put their words in alphabetical order.
- After students have become a bit more comfortable using guide words to locate words in the dictionary, you will be ready to challenge them with a game. Dictionary games are a lot easier if you have a class set of dictionaries. If you don’t have enough dictionaries available in the library, try to supplement your supply by borrowing dictionaries from the classroom teachers. With a bunch of dictionaries in hand (or on a cart–those suckers are heavy!), you can play several games with your students that will give them a chance to use their searching skills. Group games are always fun and are also a good way to support those students who are still struggling with the concept. Separate your students into groups of four or five and give each group a stack of dictionaries. Alternate between displaying a word for groups to locate and saying a word aloud (so that students can practice finding words that they don’t know how to spell). Set a timer for two minutes, and give two points/tokens to the first group to locate the word and one point/token to any group that finds the word before the timer runs out. If you want students to compete on their own, you can give each student a dictionary, a list of definitions, and a pencil. Alternate between displaying and calling out the words whose definitions are on the handout. Give students two minutes to look up each word and then write that word next to its definition on the handout. Students who correctly match all words and definitions can be awarded a prize.
An idea for an end-of-unit project:
Once you have spent several class periods working with your students on the functions of the dictionary and how to efficiently locate words in the dictionary, you will be ready to see what they have learned. A great way to do this is to assign a project that will get your students using the dictionary. Start by assigning a “challenge” word to each student. Ideally, these should be words that are not already in the students’ everyday vocabularies. Your choice of words will allow you an easy way to differentiate your assignment based on students’ ability levels (i.e. assign simpler nouns to your struggling students, as those types of words will be easier for students to illustrate and use in a sentence/advanced students will have fun writing creative sentences using challenging adjectives). You can also allow your lowest functioning students to work in groups so that they can support each other.
Once each student has a word, outline the expectations of the project. Feel free to use my Dictionary Assignment Handout (pdf) so that students will know what you want them to accomplish. The goal of the project is for students to create informative posters about their challenge words. The posters should include definitions, parts of speech, pronunciation, a sample sentence that uses the word correctly, and an illustration that will help others to understand the meaning of the word. Make sure to give students access to as many different dictionaries as possible while they are working on this project. You should plan to give students about two class periods in which to complete this project.
After students have completed their projects, you can use the challenge word posters that your students create to decorate a dictionary-themed bulletin board or simply to adorn your library walls (think about hanging them where students line up to be dismissed so that they will have something interesting and educational to look at as they wait!).
Noah Webster & His Words would certainly be a great book to read to students to introduce your dictionary unit. Another book that I love to share with students at the end of a dictionary unit is The Book Who Loved Words by Roni Schotter. This homage to the powers of language is book that will capture your students’ attention and make them excited to expand their vocabularies.<|endoftext|>
| 4 |
386 |
The Microbiome Could Be the New Fingerprint
The unique culture of bacteria living inside you may one day identify you as easily as a fingerprint does. The human microbiome—meaning the genetic material of all the organisms that call our bodies home, including bacteria, fungi, and viruses—is relatively unique across populations, and might be capable of identifying individuals within a population.
In a new study published in the journal PNAS, microbiologists and biostatisticians developed an algorithm to match the genetic coding found in microbiome samples collected as part of the Human Microbiome Project. They found strong variations between the types of organisms that live inside different individuals, and those organisms stayed in the body for a long time. Up to a year after the first sample, 80 percent of individuals from a group of 242 people could still be traced back to their microbial features.
It seems that poop is particularly revealing of an individual's unique microbiome. Using genetic codes derived from stool samples (representing organisms in the gut), the researchers were able to uniquely match 86 percent of the microbiome samples to their hosts after a period of 30 to 300 days. About a third of the individuals surveyed could be later identified using samples taken from the elsewhere on body. The results tended to match either the owner of the sample or no one at all, meaning there were few false positives in the test.
This brings up important questions about privacy protection in microbiome studies, since microbial samples from participants in such studies were thought to be biologically anonymous. If people can be linked back to their microbiome data, studies of the microbiome and associated health outcomes or particular environments—an especially popular area of scientific inquiry right now—need to be designed to make participants' data more anonymous. You're probably not handing it out to just anybody, but still: that stool sample could be as unique to you as any other genetic material, so be careful who you give it to.<|endoftext|>
| 3.84375 |
736 |
As summer temperatures rise, residents of New Jersey and other parts of the Northeast will head outside to enjoy the warm weather with family and friends. Unfortunately, the summer’s arrival means that tick-borne disease season is also here.
Lyme disease is one of the most serious tick-borne illnesses and typically arises in the Northeast, mid-Atlantic region. Named after an outbreak in Lyme, Conn., in the 1970s, Lyme disease is spread by the blacklegged tick. It is the most common tick-borne disease in America, and its prevalence is growing. The current number of cases Tick per year, which is around 300,000, has tripled in the past 20 years. In fact, a recent article published on Time Magazine says that Lyme disease is becoming more prevalent in states outside of the Northeast. The Center for Disease Control and Prevention (CDC) warns that climate change is aiding in the spread of this potentially fatal disease. Ticks thrive in densely forested areas and their populations are usually kept in balance by predatory white mice who feed on them. Recent increased deforestation efforts across the country have dramatically decreased the white mice population; therefore helping tick infestations spread across the county, virtually unhindered and unchecked.
People can contract Lyme disease when a tick bites them and stays attached to their skin for at least 24 hours. Small ticks known as nymphs cause the most infections. Symptoms can show up any time from a few days to a month after infection and include fatigue, headache, fever and a slowly growing circular rash. Without treatment, the disease can impact joints, face muscles, your heartbeat and even your cognition. Ticks can also attach to household pets. Lyme disease can even be fatal.
Most Lyme disease is treated with two to four weeks of antibiotics. However, approximately 10 percent of Lyme patients can experience symptoms months or years later. Doctors have labeled the condition post-treatment Lyme disease syndrome, or PTLDS. Treatment of PTLDS is controversial. Some doctors and advocates promote long-term antibiotic treatment to manage the condition, while others claim the antibiotics have no real impact. Studies are ongoing to learn more about possible causes of and treatments for PTLDS.
The Department of Health Suggests Taking These Measures if You Find a Tick Attached to Your Skin:
- Using tweezers, grasp the tick as close to the skin as possible.
- Gently pull the tick in a steady, upward motion.
- Wash the area with a disinfectant.
When trying to remove the tick:
- DO NOT touch the tick with your bare hand
- DO NOT squeeze the body of the tick as this may increase your risk of infection.
- DO NOT put alcohol, nail polish remover or Vaseline on the tick
- DO NOT put a hot match or cigarette on the tick in an effort to make it “back out.
- DO NOT use your fingers to remove the tick.
Professional Tick Control: NJ Counties
Treating wooded areas with the most effective tick control NJ has to offer will help keep your property safe, especially during the incoming summer months when outdoor activity increases. The threat of Lyme disease is serious. Tick risks are high and Lyme disease can even be fatal if not treated properly. The most professional tick control NJ has to offer also provides same day service and free at-home inspections in most NJ counties including Bergen and Passaic. A thorough property evaluation can identify problem areas in your outdoor areas. Call today to talk to us about securing your property against the threat of harmful ticks (866) 971-2847 or contact us online.<|endoftext|>
| 3.890625 |
679 |
# 1985 AHSME Problems/Problem 29
## Problem
In their base $10$ representations, the integer $a$ consists of a sequence of $1985$ eights and the integer $b$ consists of a sequence of $1985$ fives. What is the sum of the digits of the base $10$ representation of $9ab$?
$\mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \ } 17865 \qquad \mathrm{(D) \ } 17874 \qquad \mathrm{(E) \ }19851$
## Solution
Notice that $a=8\cdot10^0+8\cdot10^1+\cdots+8\cdot10^{1984}=\frac{8\cdot10^{1985}-8}{9}$ by the formula for a geometric series.
Similarly, $b=\frac{5\cdot10^{1985}-5}{9}$.
Thus, $9ab=9\left(\frac{8\cdot10^{1985}-8}{9}\right)\left(\frac{5\cdot10^{1985}-5}{9}\right)=\frac{40(10^{1985}-1)^2}{9}$.
We can multiply out $(10^{1985}-1)^2$ to get $9ab=\frac{40(10^{3970}-2\cdot10^{1985}+1)}{9}=\frac{4(10^{3971}-2\cdot10^{1986}+10)}{9}$.
We now find this in decimal form. $10^{3971}=10000\cdots00$, where there is $1$ one and $3971$ zeroes.
$2\cdot10^{1986}=2000\cdots00$, where there is $1$ two and $1986$ zeroes.
We subtract to find that $10^{3971}-2\cdot10^{1986}=9999\cdots98000\cdots00$, where there are $1984$ nines, $1$ eight, and $1986$ zeroes.
We now add $10$ to get $999\cdots998000\cdots010$, where there are $1984$ nines, $1$ eight, $1984$ zeroes, $1$ one, and a final zero.
Next, we begin to divide by $9$. We get this to be $111\cdots110888\cdots890$, where there are $1984$ ones, $1$ zero, $1984$ eights, $1$ nine, and a final zero.
Finally, we have to multiply by $4$. Doing this, we find that the pattern continues, and the final outcome is $4444\cdots443555\cdots5560$, where there are $1984$ ones, $1$ three, $1984$ fives, $1$ six, and a final zero. Adding this up, the sum of the digits is $1984(4)+3+1984(5)+6+0=17865, \boxed{\text{C}}$.<|endoftext|>
| 4.6875 |
827 |
# How to write an equation from two points
Given Two Points When you are given two points, it is still possible to use the point-slope form of a line.
In this case we get an ellipse. We know that the new line must be parallel to the line given by the parametric equations in the problem statement.
Select Equations in the gallery list. Now simplify this expression into the form you need. To get b, all you do is take your point and slope and use this equation: If you need help calculating slope, click here for lessons on slope. The first step is to find the slope of the line that goes through those two points.
Read the assignment heading carefully and see whether it asks you to convert your answer to slope-intercept form. Right now, you have something that looks like this: A slope and an intercept.
Now substitute those values into the point-slope form of a line. Point-slope form is one approach for finding a line and it requires two things also: Both forms involve strategies used in solving linear equations.
Most students, since they have already labeled a and when finding the slope, choose to keep that labeling system. To change or edit an equation that was previously written, Select the equation to see Equation Tools in the ribbon.
We know a point on the line and just need a parallel vector.
Find the equation of the line that passes through 0, -3 and -2, 5. You can add or change the following elements to your equation. Other students will try to look ahead a few steps and see which point might be easiest to use. The vector that the function gives can be a vector in whatever dimension we need it to be.
Choose the down arrow and select Save as New Equation You need two things to use this form you guessed it! Find the equation of the line that goes through the point 4, 5 and has a slope of 2.
To see all the symbols, click the More button. Now you need to simplify this expression. We now have the following sketch with all these points and vectors on it. However, in this case it will. There is one more form of the line that we want to look at.
In this case we will need to acknowledge that a line can have a three dimensional slope. The strategy you use to solve the problem depends on the type of information you are given. Plug those values into the point-slope form of the line:STEP 2: Now, use the point-slope formula with one of our points, (1, 3), and Finding the Equation of a Line Given Two Points 2 | mi-centre.com welcome to coolmath.
Practice finding the equation of a line passing through two points If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *mi-centre.com and *mi-centre.com are unblocked. To write an equation in point-slope form, given a graph of that equation, first determine the slope by picking two points.
Then pick any point on the line and write it as an ordered pair (h, k). Algebra > Lines > Finding the Equation of a Line Given Two Points.
Page 1 of 2. Finding the Equation of a Line Given Two Points. No problem -- we'll just use the two points to pop the slope using this guy: Check it out: Let's find the equation of the line that passes through the points. Equation of a Line from 2 Points. First, let's see it in action.
Here are two points (you can drag them) and the equation of the line through them. Explanations follow. The Points. We use Cartesian Coordinates to mark a point on a. Just type the two points, and we'll take it form there Equation of line from 2 points Calculator. Enter 2 points and get slope intercept, point slope and standard forms.
How to write an equation from two points
Rated 3/5 based on 78 review<|endoftext|>
| 4.71875 |
645 |
## How can I measure 2/3 cup?
Related Posts
Use a 1/3 of a cup and fill it two times if you don’t own or can’t in finding your 2/3 measuring cup. You can also use 10 tablespoons plus 2 teaspoons in a pinch as a conversion for 2/3 of a cup.
## How can I measure 2/3 cup of water without a measuring cup?
Use an object as a reference level.
(*2*)
1. A teaspoon is concerning the size of the tip of your finger.
2. A tablespoon is about the size of an ice dice.
3. 1/4 cup is concerning the measurement of a large egg.
4. 1/2 cup is concerning the measurement of a tennis ball.
5. A full cup is about the dimension of a baseball, an apple or a fist.
How a lot is 2/3 a cup?
Original Amount Half the Amount One-Third the Amount
1 cup 1/2 cup 1/3 cup
3/4 cup 6 tbsp 1/4 cup
2/3 cup 1/3 cup 3 tbsp+ 1-1/2 tsp
1/2 cup 1/4 cup 2 tbsp + 2 tsp
What is a similar to 2 3?
Answer: 4/6, 6/9, 8/12, 10/15 are an identical to 2/3. All those fractions acquired by means of multiplying both the numerator and denominator of 2/3 by means of the similar number are identical to 2/3. All an identical fractions get reduced to the same fraction of their most simple form.
### What is a third cup?
6 tablespoons = 3/8 cup. 5 tablespoons + 1 teaspoon = 1/3 cup. Four tablespoons = 1/4 cup. 2 tablespoons = 1/8 cup. 2 tablespoons + 2 teaspoons = 1/6 cup.
### How do I measure 1/3 cup with no measuring cup?
Measurement Equivalents and Abbreviations
1. 3 teaspoons = 1 tablespoon.
2. Four tablespoons = 1/4 cup.
3. 5 tablespoons + 1 teaspoon = 1/3 cup.
4. Eight tablespoons = 1/2 cup.
5. 1 cup = 1/2 pint.
6. 2 cups = 1 pint.
7. 4 cups (2 pints) = 1 quart.
8. Four quarts = 1 gallon.
How can I measure 2 cups of water with no measuring cup?
For liquids, because there is a difference between measuring dry and liquid ingredients, use a mason jar. Just calculate the quantity of cups based on how many fluid oz. the jar holds. (For instance: a 16-ounce mason jar equals about 2 cups.)
How do you in finding two thirds of a number?
Divide the number via 3 to search out ‘one-third,, then multiply through 2 to find ‘two-thirds’.
(*3*)<|endoftext|>
| 4.40625 |
532 |
St Paul's C of E Primary School
Heathside Grove
LEARNING to make a difference
# Week 5
Maths Activity 1
Division This week we are going to divide numbers by grouping objects. EG: Crayons come in packs of 16. We need to put 4 in each pot. How many pots will we need? There are 16 4 in each group I have made 4 groups Have a go at this one: Remember to put them in groups of 5 and then count how many groups you have made. Use objects to help you for example pencils, paper clips, counters, pegs, pieces of lego. We can write this as a number sentence using the division sign. Remember the division sign looks like this: ÷ 20 ÷ 5 = 4 Or 20 ÷ 4 = 5 Now try these: Farmer Ted has 24 eggs. He needs to put 6 eggs in each box. How many boxes will he need? There are _____ eggs altogether. There are _____ eggs in each box. There are _____ boxes. Can you fill in the missing numbers to write the number sentence to match: 24 ÷ ___ = 4 24 ÷ 4 = ___
Maths Activity 3
Revisit What time is it? Division reasoning and problem solving For example 10 groups of 3
Maths Activity 2
Revisit
Name the 3D shapes:
cylinder cone cuboid cube sphere
Division
We are going to draw number lines to help work out the number of equal groups.
How long should your number line be?
What will you count up in?
His number line should go up to 12
He needs to count up in 2’s
6 groups of 2 makes 12
Now have a go at these using a number line:
1. How many equal groups of 5 can you make from 30?
1. How many equal groups of 2 can you make from 28?
1. How many equal groups of 3 can you make from 12?
1. How many equal groups of 10 can you make from 30?
1. How many equal groups of 4 can you make from 16?
1. How many equal groups of 5 can you make from 20?
Top<|endoftext|>
| 4.625 |
647 |
Tips for Real-World, Age-Appropriate Environmental Education
There are so many real-world issues we're coping with these days, and it's important to get kids involved in finding, designing and building solutions. However, whether your an educator, parent or caregiver, it can be hard to know where to start, or which projects are age appropriate so that kids are inspired rather than overwhelmed. This diagram, based largely off of research by David Sobel and others, was developed by Aubrey Nelson, former NHEE president, to help communicate which real-world projects might be cognitively appropriate for kids. Below that are some tips that we have found helpful in designing real-world projects in schools. We hope that you find it helpful in tailoring your own projects to a scope that your audience can grasp!
Tips for Keeping Education “Real-World”: Incorporating Meaningful Projects and Citizen Science
#1. Make your students do the work. Don't bend over backwards, calling all sorts of folks and building materials to arrange your project if your students are old enough to make calls and build things themselves. And ask the students what matters to them. They can even be in charge of finding and directing a project just based on their curiosity or interests. They will learn valuable life-skills, and you might regain a little time for your own life.
#2. Start small and local. Your project doesn't have to save the world. Just going outside with your students is a start to building sense of place. Children’s understanding of the world expands as they grow, and curriculum shouldn’t outpace this natural, cognitive expansion of sense of place and community.
#3. Connect. Contact local government, organizations, officials and citizens, and ask what you can do them AND what they can do for you. But also contact the folks in your building: teachers of very different subjects, staff, volunteers, administrators. Oftentimes teachers don't have time to look around for what the most urgent problems are. Our local governments, conservation groups and even businesses have problems they're trying to solve, and might lack the (wo)man-power to solve them. Students can be a resource to these organizations, but they won't get fired up if the actions they're taking aren't 'real.' Encourage folks to contact you, or your school, when there's a problem that needs solving, data that needs collecting, or even marketing opportunities.
#4 Build a team. Work as a team. Support the team. Who are the other people in your school, organization or community who may be excited about this? Who can help you? Parents, volunteers, another teacher from another school? Who has expertise? How can your project help them, too?
#5. Share what you’re doing. Spread the good word. Be a model of what’s possible.
#6. Why does this matter? Make sure this is made explicit for you and the students. Take time to reflect. If the kids aren’t digging it, or it isn’t aligned with your goals, ditch it.<|endoftext|>
| 3.859375 |
677 |
# Math Homework Activities: Using The Number Line
By Finn Orfano
Parents can help their elementary school students learn to add and subtract on a math/number line by planning home activities. Rather than working with a traditional number line written on paper, parents can get creative with a number line race car game, a vertical number line, and a walking line.
In early elementary school, children begin learning math/number line activities as a way of familiarizing themselves with basic addition and subtraction. During homework time, parents can assist their children in learning math facts with the following activities (which can be done with a 1-10 or 1-20 number line):
## Race Up And Down The Number Line
This homework activity combines practicing addition and subtraction with creative game play. Parents who do not have toy cars on hand can substitute board game pieces or other small items.
(1) Place a number line on the table (the number line can be hand-drawn or a printed image) and give your child a toy car. Explain that the car will be moved up and down the number line as math facts are read.
(2) Gather a set of addition/subtraction flash cards. If the first card reads "5+2", for example, ask your child to place the toy car on the number 5 to start. Then say "5+2" out loud, and instruct your child to move the car two places up on the number line, resting at 7. Repeat this process with a few flash cards until your child understands the concept.
(3) As the game progresses, allow your child to independently figure out where the car needs to go. As your child becomes more familiar with basic math facts, increase the speed of the game.
## Create A Number Line On The Wall
Vertical number lines that are attached to a wall or door can be used as visual aids during math homework time. When making this type of number line, be sure that your child is able to reach the highest number with his or her hand.
(1) Once the number line is displayed in the room where your child normally does homework, ask your child to stand by the line and listen for the first addition or subtraction fact.
(2) Begin reading from the math flash cards, and guide your child in moving his or her hand up the line for addition problems and down the line for subtraction problems. Encourage you child to count numbers out loud if they need to--for example, "Seven minus three. I need to move down three spaces...one, two, three. The answer is four."
## Life-Size Number Line Activity
Another way to make math homework fun is to make a life-size number line that your child can walk on. In warm weather, create an outdoor line with sidewalk chalks. When studying indoors, tape a roll of paper to the floor and color in the numbers with crayons or markers.
This activity is similar to the one above, except that your child will stand on the number line and walk up and down according to which math fact is given.
By organizing math/number line activities that are enjoyable for young children, parents can encourage learning without having to worry about a child becoming bored or restless. Math homework activities that involve both parent and child are also beneficial in giving families an active role in education and development.<|endoftext|>
| 4.53125 |
860 |
From BBC Blue Planet
This documentary explores the unknown depths of the ocean.
Over 60% of the sea is more than a mile deep and it forms the planet's most mysterious habitat.
A sperm whale descends 1000 metres to look for food and is followed.
On the way down, a number of unusual creatures are witnessed, such as transparent squid and jellies, whose photophores give pulsating displays of colour.
In such dark places, both being able to see (or sense movement) and the means of quick concealment are equaly desrirable.
To that end, some use bioluminescence as a means of detecting food or evading predators.
A descend to the very bottom of the ocean - some 4,000 metres - reveals life even at such cold temperatures, much of it new to science.
It is dominated by echinoderms that sweep the sea bed; however, there are occasional large hunters, such as chimaera.
Life in the Deep Oceans (part I)
The term deep sea refers to organisms that live below the photic zone of the ocean.
These creatures must survive in extremely harsh conditions, such as hundreds of atmospheres of pressure, small amounts of oxygen, very little food, no sunlight and constant extreme cold.
Most creatures have to depend on food floating down from above.
These creatures live in very harsh environments such as abyssal or hadal zones, which, being thousands of meters below the surface, are almost completely devoid of light.
The water is very cold (between 3 and 10C) and has low oxygen levels.
Due to the depth, the pressure is between 20 to 1000 atmospheres.
These animals have evolved to survive the extreme pressure of the sub-photic zones.
The pressure increases by about one atmosphere every ten meters.
To cope with the pressure, many fish are rather small, usually not exceeding 25 cm in length.
Also, scientists have discovered that the deeper these creatures live, the more gelatinous their flesh and more insignificant their skeletal structure is.
Lack of light
Because there's no light most animals have very large eyes with retinas constructed only of cones, which increases sensitivity.
Many animals have also developed large feelers to replace peripheral vision.
To be able to reproduce, many of these fish have evolved to be hermaphroditic, eliminating the need to find a mate.
Many creatures have developed very strong senses of smell to detect the chemicals released by mates.
Since at such deep levels, there is little to no sunlight, photosynthesis is impossible as a means of energy production, leaving some creatures with the quanday of how to produce food for themselves.
For the giant tube worm, the answer comes in the form of bacteria that live inside of it.
These bacteria are capable of chemosynthesis and live inside of the giant tube worm, which lives on hydrothermal vents.
These vents spew high amounts of chemicals that these bacteria can transform into energy.
These bacteria can also grow freely of a host and create mats of bacteria on the sea floor around hydrothermal vents, where they serve as food to other creatures.
Bacteria are a key energy source in the food chain.
This source of energy creates large populations in areas around hydrothermal vents, which provides scientists an easy stop for research.
Whales can dive to about 1.200m deep in search of their prey.
The giant squid is one of the very few deep ocean creatures that can visit the ocean surface.
The viperfish have long sharp clear teeth that they use to catch their prey.
The hatchet fish has a light that attracts their prey, gulper eels have huge heads and mouths so they can swallow their prey easily.
They also have elastic stomachs, which allows them to eat fish larger than themselves. Anglerfishes use a light on top of their head to catch their prey, the rattail fish detects its prey with a whip like tail, a sea pen is like a worm like creature that lives and crawls on the ocean flood.
Many fish larger than the sea pen make it their lunch.<|endoftext|>
| 4.125 |
877 |
For all their extraordinary power, black holes are not immortal.
They have a life cycle just like we do. Forty years ago Stephen Hawking, the world’s foremost expert on black holes, announced that they evaporate and shrink because they emit radiation.
But if a black hole evaporates and shrinks, what happens to everything it devoured during its lifetime?
Most mathematical calculations have suggested that the information and everything else inside the black hole simply vanishes, a conclusion that raised more questions than it has answered.
“The issue was never laid to rest because Hawking’s calculation was not able to capture the effect that the radiation, called Hawking radiation, has on the black hole itself,” says Chris Adami, professor of physics and astronomy. “Physicists assumed that the black hole would shrink in time as the Hawking radiation carries away the black hole’s mass, but no one could verify this through mathematical calculations.”
A calculation of the black hole’s evaporation seemed impossible, unless a full theory of quantum gravity that unites Einstein’s general relativity with the framework of quantum field theory could be found.
Adami’s new paper, published in Physical Review Letters, changes that premise.
Hawking’s theory, but with a twist
Adami and colleague Kamil Bradler of the University of Ottawa have developed a new theory that allows them to follow a black hole’s life over time. What they find is striking: Whatever quantum mysteries were hiding behind the black hole event horizon—the invisible boundary of a black hole—slowly leak back out during the later stages of the black hole’s evaporation.
With this finding, a major black hole physics problem is avoided. Physicists have argued strenuously that it was not possible that all quantum information could remain hidden within the black hole when it shrunk to minute sizes.
It turns out that to show that black holes do not destroy information forever as they evaporate, Adami and Bradler did not have to create the elusive theory of quantum gravity. Instead, they used Hawking’s own theory, but with a twist.
To understand how a black hole would interact directly with the Hawking radiation it generates, Adami and Bradler used a set of sophisticated mathematical tools and high-performance computers to evolve the black holes over sufficiently long times until they were able to find quantum information outside of the black holes.
Signs of Page curves
“To perform this calculation, we had to guess how a black hole interacts with the Hawking radiation field that surrounds it,” Adami says. “This is because there currently is no theory of quantum gravity that could suggest such an interaction. However, it appears we made a well-educated guess because our model is equivalent to Hawking’s theory in the limit of fixed, unchanging black holes.”
“While our model is just that—a model—we were able to show that any quantum interaction between black holes and Hawking radiation is very likely to have the same properties as our model,” Bradler says.
The theory was able to reproduce a feature long searched for in black hole physics, the so-called “Page curves,” named after University of Alberta physicist Don Page. His model predicted the curves that show the quantum information first entering, then exiting the black hole. Adams and Bradler’s calculation is the first that yielded curves just like those Page had anticipated.
But much work remains to be done. In principle, the team’s guess should follow from the yet-to-be-discovered fundamental unified theory of quantum gravity. But in the absence of that theory, the success of Adami and Bradler’s theory may give hints as to just how such a theory—one that goes beyond Hawking’s—could be constructed.
In the new era of gravitational wave observatories that the LIGO discovery just ushered in, such a theory may even one day be tested.
This text is published here under a Creative Commons License.
Author: Kim Ward-Michigan State University
Check here the article’s original source with the exact terms of the license to reproduce it in your own website<|endoftext|>
| 3.84375 |
876 |
# Analytic Geometry
Share on
Analytic geometry creates a connection between graphs and equations. For example, the linear function f(x) = x2 – 2 (an equation) can also be represented by a graph:
Euclidean Geometry is based solely on geometric axioms without formulas or co-ordinates; Analytic geometry is the “marriage” of algebra and geometry with axes and co-ordinates [1].
## Calculus and Analytic Geometry
Calculus and analytic geometry have become so intertwined, it’s rare nowadays to find a course in pure “Analytic Geometry”. It’s more common to take a course in Calculus and Analytic Geometry, which blends the principles of basic analytic geometry with concepts like functions, limits, continuity, derivatives, antiderivatives, and definite integrals.
## Topics in Analytic Geometry A to Z
Related articles:
## X Y Plane
In 3D space (also called xyz space), the xy plane contains the x-axis and y-axis:
The xy plane can be described as the set of all points (x, y, z) where z = 0. In other words, any point (x, y, 0). For example, all of the following points are on the xy plane:
• (1, 5, 0)
• (-2, 19, 0)
• (π, -1, 0)
• (.5, .2, 0)
This fact gives us the equation for the xy plane: z = 0.
This is just an extension of the same idea of the x-axis (in the Cartesian plane) being the place where y = 0:
(Left) The x-axis in 2D is where points are located at y = 0. In 3D, the xy plane has points where z = 0.
The xy plane, together with the yz plane and xz plane, divide space into eight octants. The O in the center of the diagram is the origin, which is a starting point for the 3D-coordinate system. The points are described by an ordered triple of real numbers (x, y, z). For example, the point (2, 3, 0) can be found at:
• x = 2,
• y = 3,
• z = 0.
As z is zero, we know this point must be somewhere on the xy plane.
## Distance Formula for Points in the XY Plane
The distance between any two points in xyz-space can be found with a generalization of the distance formula:
Example question: What is the distance between the points (4, 3, 0) and (2, 9, 0)?
Step 1: Identify the coordinate components that we need to put into the formula. We know our coordinates are always ordered (x, y, z), so:
1. (4, 3, 0):
• x1 = 4
• y1 = 3
• z1 = 0.
2. (2, 9, 0):
• x2 = 2
• y2 = 9
• zz = 0.
Don’t worry about which coordinate is which (e.g. does x = 4 go into x1 or x2?). The distance formula squares these values, so you’ll get the same answer no matter which way you choose.
Step 2: Plug your values from Step 1 into the distance formula:
If you aren’t good with algebra, head over to Symbolab and just replace the x, y, z values with your inputs.
## References
[1] Analytic Geometry and Calculus. Retrieved May 3, 2021 from: math.uci.edu/~ndonalds/math184/analytic.pdf
CITE THIS AS:
Stephanie Glen. "Analytic Geometry" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/calculus-problem-solving/analytic-geometry/
---------------------------------------------------------------------------
Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free!<|endoftext|>
| 4.46875 |
282 |
Volunteering is generally considered an altruistic activity where an individual or group provides services for no financial gain “to benefit another person, group or organization”. Volunteering is also renowned for skill development and is often intended to promote goodness or to improve human quality of life. Volunteering may have positive benefits for the volunteer as well as for the person or community served. It is also intended to make contacts for possible employment. Many volunteers are specifically trained in the areas they work, such as medicine, education, or emergency rescue. Others serve on an as-needed basis, such as in response to a natural disaster.
The verb was first recorded in 1755. It was derived from the noun volunteer, in C.1600, “one who offers himself for military service,” from the Middle French voluntaire. In the non-military sense, the word was first recorded during the 1630s. The word volunteering has more recent usage—still predominantly military—coinciding with the phrase community service. In a military context, a volunteer army is a military body whose soldiers chose to enter service, as opposed to having been conscripted. Such volunteers do not work “for free” and are given regular pay.
This article uses material from the Wikipedia article Volunteering, which is released under the Creative Commons Attribution-Share-Alike License 3.0.<|endoftext|>
| 3.84375 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.