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### Lesson Plan Title : Value of Coins
Age Range:
Kindergarten through Grade 2 (Primary / Elementary School)
Overview and Purpose:
This lesson will reinforce the value of pennies, nickels, and dimes. Students will need to decide which coins are needed to total different amounts written on index cards and then place those coins on the cards. The goal is to try to match all the coins with a card so there are no coins or empty cards left.
Objective:
The student will be able to choose the appropriate combination of pennies, nickels, and dimes needed to total the amounts listed on a series of five index cards.
Resources:
One snack size Ziploc bag for every three students
Two dimes, three nickels, and ten pennies for each bag
One set of five index cards for each set of three students.
Each set should have one card each with the following amounts written on it: 3¢ - 9¢ - 12¢ - 15¢ - 6¢
Optional: Dimes, nickels, and pennies copied onto transparency film and cut out
Activities:
Introduce the lesson by reviewing what a dime, nickel, and penny look like and the value of each one. Write an amount on the overhead and have students help you choose which coins equal the amount (i.e. 7¢ = 1 nickel and 2 pennies or 7 pennies). Point out that there are often at least two different combinations of coins that can be used.
Practice a few more times as a group and when you feel that the students understand the concept divide them into groups of three. Give each group a Ziploc bag with coins and a set of index cards. Have them work together to place the correct coins on each index card. (Each card has one combination that will allow for all the cards to be completed at the same time with no coins left over.) Walk around and help groups as needed.
When the groups are finished, come back together as a whole and discuss what solutions the students found.
Closure:
Beginning level students may need the correct coin shapes traced onto the index cards and more advanced students may need more cards or higher amounts listed on the cards.
Homework for this lesson can be for students to take home three index cards and write their own amounts on them with the solutions on the back. The cards can be collected and added to the math center in the classroom as a review game.<|endoftext|>
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A piano keyboard may look confusing at first, but the layout of piano keys is actually very simple. While it would be nearly impossible to identify individual keys if all we had were the white keys, the pattern of black keys makes it easy to identify the keyboard layout. The black keys are arranged in alternating groups of two and three:
(It’s not unimportant to point out that, in the piano’s early days, the “white” keys were black and the “black” keys were white! It was not until the 19th century that the now-common standard became universal.)
In fact, all you really need to recognize is this:
These are all the keys within an octave, and this pattern repeats itself seven and one third times on a modern piano for a total of 88 keys.
Now, most teachers will introduce you to the white keys at this point, starting from middle C and going up: D, E, F, G, A, B. But this isn’t the whole truth.
There’s one critical thing to note here. The keys are not the notes. That white key immediately to the left of each group of two black keys is not C! Well, it is and it isn’t. It could be C; in fact, this key spends most of its life playing C. But it could just as well be B-sharp, or D double flat. (See the Piano Key Chart for an overview.)
In order to understand this point it’s vital to understand something essential about the nature of the piano: The piano is inherently one giant compromise. You see, C, B-sharp and D double flat are not the same note, but they correspond to the same key on the piano. (English speakers might be confused by the fact that “key” is used for both the white and black buttons on a piano as well as in harmony, e.g., the “key” of C major or F-sharp minor. Other languages happen to use entirely different words for these concepts, which have no relation to one another.)
If you’re confused by sharps and flats, just remember that sharp means to play the next higher key and flat the next lower.
A set of notes that correspond to the same key on a piano is called enharmonic. (Don’t worry about this term for now if it sounds confusing; just know that you’ll encounter it later.) For example, F-sharp is enharmonic with G-flat. Enharmonic notes aren’t technically the same, but they’re “close enough” not to matter all that much to justify making them a discrete piano key. After all, you’d have to triple the number of keys just for the sharps and flats, and that would make the piano impossible to play, not to mention a piano tuner’s worst nightmare!
There is something special about the layout of piano keys of which even most experienced pianists are unaware. There are two points on the keyboard on which a mirror could be placed, and the pattern of white and black keys would be the same in either direction.
The first of these points is the white key right in the middle of the group of two black keys:
Place both hands on that key. Now play the next higher key with the right hand and the next lower one with the left. Keep doing this, and you’ll soon discover that you could go on forever like this, and both hands will always play either a black key or a white one. (Each hand just played a chromatic scale, by the way.)
Challenge: See if you can find the other mirror point!
Now let’s familiarize yourself with the piano keyboard. Take your index, middle and ring fingers in each hand and play every group of three black keys.
Next, using your index and middle fingers, find each group of two black keys.
Now try this exercise without looking at the keyboard! This is how blind pianists are able to play, incidentally: They know the layout of piano keys from the groups of two and three black keys. It is also the proper sight reading technique.
Now that you’ve familiarized yourself with the layout of the piano keyboard, you’re ready to move on to the Piano Notes Chart…. See you there!
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# Numerical Analysis Questions and Answers – Newton-Gregory Forward Interpolation Formula
This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Newton-Gregory Forward Interpolation Formula”.
1. Newton- Gregory Forward interpolation formula can be used _____________
a) only for equally spaced intervals
b) only for unequally spaced intervals
c) for both equally and unequally spaced intervals
d) for unequally intervals
Explanation: Newton – Gregory Forward Interpolation formula is given by
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3! + …..
This formula is obtained by the Newton’s Divided difference formula by substituting the intervals as h. This is done because we assume the intervals to be constant, that is, equally spaced.
2. Find n for the following data if f(0.2) is asked.
x 0 1 2 3 4 5 6 f(x) 176 185 194 203 212 220 229
a) 0.4
b) 0.2
c) 1
d) 0.1
Explanation: The formula is
x = x0 + nh.
Here x0 is 0 as 0 is the first element and h is 1.
Since in the question it is given that we have to find f(0.2), x= 0.2.
So, substituting the values in the formula we get,
0.2 = 0 + n(1) .
Hence, n= 0.2.
3. Find n for the following data if f(1.8) is asked.
x 0 0.5 1 1.5 2 f(x) 0.3989 0.3521 0.242 0.1295 0.054
a) 2.4
b) 3.4
c) 2.6
d) 3.6
Explanation: Here, x0 is 0, h is 0.5, x is 1.8.
Substituting the values in the formula
x = x0 + nh,
1.8 = 0 + n(0.5)
n = 3.6.
4. Find the polynomial for the following data.
x 4 6 8 10 f(x) 1 3 8 16
a) $$\frac{3x^2-22x+36}{8}$$
b) 3x2-22x+36
c) $$\frac{3x^2+22x+36}{2}$$
d) $$\frac{3x^2-19x+36}{8}$$
Explanation: Here,
x y Δy Δ2y Δ3y 4 1 2 3 0 6 3 5 3 8 8 8 10 16
y0 is 1 since it is forward interpolation formula.
Δy0 = 2
Δ2y0 = 3
Δ3y0 = 0
x = x0 + nh,
x = 4 + n(2)
Hence
n = (x-4)/2
Substituting these values in the formula,
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3!,
$$f(x)=1+\frac{(x-4)}{2} 2+\frac{(x-4)}{2} (\frac{x-4}{2}-1) \frac{3}{2}+0$$
$$=\frac{3x^2-22x+36}{8}.$$
5. Using Newton’s Forward formula, find sin(0.1604) from the following table.
x 0.16 0.161 0.162 f(x) 0.159318 0.160305 0.161292
a) 0.169713084
b) 0.159713084
c) 0.158713084
d) 0.168713084
Explanation: Here,
x0 = 0.160
x = 0.1604
h = 0.001
x = x0 + nh,
0.1604 = 0.160 + n(0.001)
n = 0.4
x y Δy Δ2y 0.160 0.1593182066 9.871475*10-4 -1.604*10-7 0.161 0.1603053541 9.869871*10-4 0.162 0.1612923412
y0 is 0.1593182066 since it is forward interpolation formula.
Δy0 = 9.871475*10-4
Δ2y0 = -1.604*10-7
Substituting in the formula,
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3! ,
f(0.1604) = 0.1593182066+(0.4)(9.871475*10-4)+(0.4)(-0.6) $$\frac{(-1.604*10^{-7})}{2}$$
= 0.159713084.
6. Find f(5) using Newton’s Forward interpolation formula from the following table.
x 0 2 4 6 8 f(x) 4 26 58 112 466
a) 71.109375
b) 61.103975
c) 70.103957
d) 71.103957
Explanation: Here,
x0 = 0
x = 5
h = 2
x = x0 + nh,
5 = 0 + n(2)
n = 2.5
x y Δy Δ2y Δ3y Δ4y 0 4 22 10 12 266 2 26 32 22 278 4 58 54 300 6 112 354 8 466
y0 is 4 since it is forward interpolation formula.
Δy0 = 22
Δ2y0 = 10
Δ3y0 = 12
Δ4y0 = 266
Substituting in the formula,
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3! +….
f(5)=4+(2.5)(22)+$$\frac{(2.5)(1.5)(10)}{2}+\frac{(2.5)(1.5)(0.5)(12)}{6}+\frac{(2.5)(1.25)(0.5)(-0.5)(266)}{24}$$
f(5) = 71.109375.
7. Find f(0.18) from the following table using Newton’s Forward interpolation formula.
x 0 0.1 0.2 0.3 0.4 f(x) 1 1.052 1.2214 1.3499 1.4918
a) 1.18878784
b) 1.8878784
c) 1.9878785
d) 0.8878784
Explanation: Here,
x0 = 0
x = 0.18
h = 0.1
x = x0 + nh,
0.18 = 0 + n(0.1)
n = 1.8
x y Δy Δ2y Δ3y Δ4y 0 1 0.052 0.1174 -0.1583 0.2126 0.1 1.052 0.1694 -0.0409 0.0543 0.2 1.2214 0.1285 0.0134 0.3 1.3499 0.1419 0.4 1.4918
y0 is 1 since it is forward interpolation formula.
Δy0 = 0.052
Δ2y0 = 0.1174
Δ3y0 = -0.1583
Δ4y0 = 0.2126
Substituting in the formula,
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3! +….
$$f(0.18)=1+(1.8)(0.052)+\frac{(1.8)(0.8)(0.1174)}{2}+\frac{(1.8)(0.8)(-0.2)(-0.1583)}{6}$$
$$+\frac{(1.8)(0.8)(-0.2)(-1.2)(0.2126)}{24}$$
f(0.18) = 1.18878784.
8. Find f(2.75) using Newton’s Forward interpolation formula from the following table.
x 1.5 2 2.5 3 3.5 4 y 3.375 7 13.625 24 38.875 59
a) 1.8296875
b) 18.296875
c) 22.296875
d) 24.296875
Explanation: Here,
x0 = 1.5
x = 2.75
h = 0.5
x = x0 + nh,
2.75 = 1.5 + n(0.5)
n = 2.5
x y Δy Δ2y Δ3y Δ4y Δ5y 1.5 3.375 3.625 3 0.75 0 0 2 7 6.625 3.75 0.75 0 2.5 13.625 10.375 4.5 0.75 3 24 14.875 5.25 3.5 38.875 20.125 4 59
y0 is 3.375 since it is forward interpolation formula.
Δy0 = 3.625
Δ2y0 = 3
Δ3y0 = 0.75
Δ4y0 = 0
Δ5y0 = 0
Substituting in the formula,
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3! +….
f(2.75)=3.375+(2.5)(3.625)+$$\frac{(2.5)(1.5)(3)}{2}+\frac{(2.5)(1.5)(0.5)(0.75)}{6}$$+0+0
f(2.75) = 18.296875.
9. Find n if x0 = 0.75825, x = 0.759 and h = 0.00005.
a) 1.5
b) 15
c) 2.5
d) 25
Explanation: Given
x0 = 0.75825
x = 0.759
h = 0.00005
Substituting in the formula,
x = x0 + nh,
0.759 = 0.75825 + n(0.00005)
Therefore, n = 15.
10. Find x if x0 = 0.6, n = 2.6 and h = 0.2.
a) 12
b) 1.2
c) 1.12
d) 1.22
Explanation: Given
x0 = 0.6
n = 2.6
h = 0.2
Substituting in the formula,
x = x0 + nh
x = 0.6+(0.2)(2.6)
x = 1.12.
Sanfoundry Global Education & Learning Series – Numerical Methods.
To practice all areas of Numerical Methods, here is complete set of 1000+ Multiple Choice Questions and Answers.<|endoftext|>
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# What happens to horizontal asymptote in reciprocal?
## What happens to horizontal asymptote in reciprocal?
This is called the horizontal asymptote of the graph. The two parts of the graph also get closer to the y-axis as x gets closer to 0. Again, the line never actually meets the y-axis because there is no value for y when x = 0.
Does a reciprocal function have a horizontal asymptote?
Given a function and the corresponding reciprocal function, the graph of the reciprocal function will have vertical asymptotes where the function has zeros (the x-intercept(s) of the graph of the function). f(x) = ( x – 3 )2 – 4. The graph of a function will never have more than one horizontal asymptote.
How do you find the horizontal asymptote of a reciprocal?
Let m=degree of p(x)n=degree of q(x) 1. If m”>n>m then the horizontal asymptote is y=0 2. If n=m then the horizontal asymptote is y=ab where a is the lead coefficient of p(x) and b is the lead coefficient of q(x) 3.
### What happens when you reciprocal a function?
The same concept applies when we find a function’s reciprocal function – we divide 1 by the function’s expression. Given a number, , its reciprocal is . Given a function, , its reciprocal function is 1 f ( x ) . The product of and its reciprocal is equal to · 1 k = 1 .
Why do reciprocal functions have asymptotes?
Ernest Z. Some functions have asymptotes because the denominator equals zero for a particular value of x or because the denominator increases faster than the numerator as x increases.
What is the end behavior of a reciprocal function?
What is the end behavior of a reciprocal function? The end behavior of a reciprocal function describes the value of ‘x’ in the graph approaching negative infinity on one side and positive infinity on the other side.
#### What are the horizontal asymptote rules?
The three rules that horizontal asymptotes follow are based on the degree of the numerator, n, and the degree of the denominator, m.
• If n < m, the horizontal asymptote is y = 0.
• If n = m, the horizontal asymptote is y = a/b.
• If n > m, there is no horizontal asymptote.
What creates a horizontal asymptote?
An asymptote is a line that a graph approaches without touching. Similarly, horizontal asymptotes occur because y can come close to a value, but can never equal that value. Thus, f (x) = has a horizontal asymptote at y = 0. The graph of a function may have several vertical asymptotes.
Why do horizontal asymptotes occur?
An asymptote is a line that a graph approaches without touching. Similarly, horizontal asymptotes occur because y can come close to a value, but can never equal that value. The graph of a function may have several vertical asymptotes. …
## Why do functions have horizontal asymptotes?
Often a function has a horizontal asymptote because, as x increases, the denominator increases faster than the numerator. The numerator has a constant value of 1 , but as x takes a very large positive or negative value, the value of y gets closer to zero.
What does the horizontal asymptote say about the end behavior?
While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term.
Which statement defines the horizontal asymptote?
Which statement defines the horizontal asymptote? B. m = n, so y = am / bn is the horizontal asymptote.
### How are horizontal asymptotes used in a graph?
Usually, functions tell you how y is related to x. Functions are often graphed to provide a visual. A horizontal asymptote is a horizontal line that tells you how the function will behave at the very edges of a graph. A horizontal asymptote is not sacred ground, however. The function can touch and even cross over the asymptote.
How to find the vertical asymptote of a reciprocal function?
Every reciprocal function has a vertical asymptote, and we can find it by finding the x value for which the denominator in the function is equal to 0. For example, the function y= 1 / (x+2) has a denominator of 0 when x=-2. Therefore, the vertical asymptote is x=-2. Likewise, the function y= 1 / (3x-5) has a denominator of 0 when x= 5 / 3.
How does a vertical shift change the asymptote of Y?
Any vertical shift for the basic function will shift the horizontal asymptote accordingly. For example, the horizontal asymptote of y= 1 / x +8 is y=8. The horizontal asymptote of y= 1 / x -6 is y=-6. The vertical asymptote is similar to the horizontal asymptote.
#### What is the symmetry of a reciprocal function?
If our reciprocal function has a vertical asymptote x=a and a horizontal asymptote y=b, then the two asymptote intersect at the point (a, b). Then, the two lines of symmetry are y=x-a+b and y=-x+a+b.<|endoftext|>
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VAK (Visual, Auditory and Kinesthetic) Methods:
Exploring the methods of learning is established that some children learn faster while receive information visually, others by listening, others by touch. There is individual preference of children for these different ways of learning yet, the latest research shows that if children at the same time receive and memorize the information coming from the visual (seeing), auditory (hearing) and kinetic (tactile / touch) neural circuits / senses in the brain, their learning and memorization is the largest and fastest, because it comes to the transfer of information from the left to the right hemisphere of the brain and overall brain activity.
Using those techniques in unique and innovative way, combining VAK methodology, the abacus tool and mental arithmetic, our Brainobrain programme carries out development and progress of the entire brain capacity of coordinating between right and left hemisphere in children, in order to realize the genius capacity .
NLP (Neuro – Linguistic Programming) techniques:
NLP (Neuro – Linguistic Programming) is the study of the process of thought. Increasing awareness of our words in creating thoughts, can greatly help us in realizing our personality, behavior, activities and our goals.
For children this awareness leads to the development of leadership skills, self-confidence, wise management of resources (time, space, conditions) for successfully achieving the highest set life goals.<|endoftext|>
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# Question Video: Using Fraction Models to Represent Hundredths Mathematics • 4th Grade
Each big square is one whole. Daniel has colored two wholes and seventeen-hundredths orange. Write this as a mixed number. Write this as a decimal. Tip: Use a place value table to help you.
02:24
### Video Transcript
Each big square is one whole. Daniel has colored two wholes and seventeen hundredths orange. Write this as a mixed number. Write this as a decimal. Tip: Use a place value table to help you.
We know that Daniel has colored two wholes and seventeen hundredths orange. We have to write this amount as a mixed number and as a decimal, and we’re told to use the place value table to help. How would we write two wholes and seventeen hundredths as a mixed number. We know that each big square is one whole, and we also know that Daniel has colored two whole squares or two big squares orange, so the whole part of his mixed number is two. We also know that the fractional part of his number is seventeen hundredths. The large square has been divided into 100 smaller squares or 100 equal parts, and each part is a hundredth. And we know that Daniel has colored seventeen hundredths, so two wholes and seventeen hundredths written as a mixed number is two and seventeen hundredths.
Now we need to write this as a decimal. We know that Daniel has colored two whole squares, which represent two ones. And the fractional part of the number is seventeen hundredths. So two wholes and seventeen hundredths as a decimal is 2.17, which we would say as two and seventeen hundredths. Daniel colored two wholes and seventeen hundredths orange. We would write this as a mixed number as two and seventeen hundredths. And to write this as a decimal, we would write 2.17, which is two and seventeen hundredths.<|endoftext|>
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Researchers writing in the journal Science describe a groundbreaking new way of genetically engineering plants, that could almost double yields in the world’s most important crops.
The technique, which was tested on tobacco plants as a proxy for crops, works by increasing the efficiency of photosynthesis. This enables plants to grow faster and bigger–and ultimately increases their yields by 40%, the study showed.
The scientists on the new study–who are part of an ambitious, long-term international research project called Realizing Increased Photosynthetic Efficiency—estimate that the boosted productivity could help to feed an additional 200 million people in the Midwestern United States alone, if applied to crops grown there. At a global scale, the genetic tweak could assist in feeding billions more people by 2050, they estimate.
Photosynthesis occurs with the help of an enzyme inside plants called RuBisCo, which latches onto molecules of carbon dioxide to convert them into energy-rich sugars for growth. But about a fifth of the time, RuBisCo grabs onto molecules of oxygen instead.
That presents a major lost opportunity for photosynthesis. That’s because oxygen molecules inside plants go on to create waste products, such as glycolate and ammonia. These have to be recycled in a process called photorespiration, otherwise they will harm the plant. But photorespiration involved a convoluted three-part pathway through a plants’ cells that is very energy-intensive. In turn, this “costs the plant precious energy and resources that it could have invested in photosynthesis to produce more growth and yield,” the researchers explain.
In some crops, in fact, photorespiration is estimated to cause a 50% loss in photosynthesis’ efficiency.
But in their investigations, the researchers were able to engineer tobacco plants (which are useful research subjects because they’re easy to genetically modify) to contain a much shorter photorespiration pathway with fewer steps. This saves on energy and crucially, increases the efficiency of photosynthesis, therefore boosting plant growth. In tests that ran for two years, the researchers showed that tobacco plants grown both in the laboratory and under real-world farming conditions consistently grew faster, taller, and had more biomass than their non-engineered counterparts.
In hot, dry regions of the world this discovery would be especially valuable: under high heat, RuBisCo struggles even more to differentiate between carbon dioxide and oxygen, leading to even more inefficient photosynthesis. These genetic tweaks would provide a useful way around that stumbling block, in countries that already have a high food security risk.
The researchers are careful to caution that their discovery isn’t a quick-fix, however. It will take several years to engineer crops with these energy-saving traits, and also to ensure that they are as safe to eat as their non-engineered counterparts. But with the rising twin challenges of increasing food production to feed billions more people, and doing so on less land to conserve biodiversity, the researchers believe this could be one powerful way of producing food more efficiently and sustainably.
To reach that goal, now they’re focusing on engineering globally-important staple crops such as soybeans, cowpeas, rice, and potatoes, to contain the beneficial traits. “Our goal is to build better plants that can take the heat today and in the future, to help equip farmers with the technology they need to feed the world,” the researchers say.
Source: South et. al. “Synthetic glycolate metabolism pathways stimulate crop growth and productivity in the field.” Science. 2018.<|endoftext|>
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1. ## Equation of line. Solving quadratic equations.
A straight line passes through the two points A and B with coordinates A(–3,–5) and B(5,1).
a) Determine an equation for the line
Problem 6: roots of quadratic equations
The following quadratic equations have been factorized for you. Read off the roots:
a. x2 – 40x + 300 = (x–10)(x–30) = 0
b. x2 + x – 1 = (x–½)(x+2) = 0
c. x2 – (+)x + = (x–)(x–) = 0
d. 6x2 + 6x – 72 = (2x–6)(3x+12) = 0
Factorise, and hence determine the roots of the quadratic equation:
e. x2 + 5x + 6 = 0
f. x2 + 6x – 7 = 0
Problem 7: throwing up…
A ball is thrown into the air. The height (over the ground, in metres) of the ball is called y, and a formula for y is y = . Here g denotes the acceleration of gravity, g = 9.8 m/s2, t is the time from when the ball is thrown (in seconds), and v is the speed with which the ball is thrown (in m/s).
a) With your graphing calculator, sketch y as a function of t for a ball thrown upwards with the speed v = 5 m/s. Copy the graph onto paper.
b) With your calculator, find the maximum height the ball reaches (2 sig. figs.)
c) After how many seconds is the ball back on the ground? (2 sig. figs.)
2. 1. The equation for a line can be represented as such:
$y = mx + b$
Where $m$ is the slope and $b$ is the y intercept.
The equation for slope is $\frac {rise}{run}$, or $\frac {y_1-y_2}{x_1-x_2}$. You have two points so you have two x coordinates and two y coordinates. It does not matter what you choose to be $x_1$ so long as you make the corresponding y value $y_1$.
Calculate the slope, then plug it in to the equation $y = mx + b$ using either of your original points to find $b$.
2. When you have an equation like $(x-10)(x-30) = 0$, the roots are whatever makes that equation true. So, for $(x-10)(x-30)$ to be equal to 0, either $(x-10) = 0$, or $(x-30) = 0$.
Now, just solve for $x$ in both cases.
Case 1: $x-10 = 0$, therefore $x = 10$
Case 2: $x-30 = 0$, therefore $x = 30$
So, the roots of the equation are 10 and 30.
3. To factorize a quadratic of the form $x^2 +bx +c = 0$, simply use the sum and product rule. What two numbers have a sum of $b$ and a product of $c$?
$x2 + 5x + 6 = 0
$
3 and 2 add up to 5 and multiply to 6, so your factored equation would be:
$(x+2)(x+3) = 0$
4. On your calculator there should be a y= button. Simply enter the equation into it and then press GRAPH. Press 2nd CALC then select MAXIMUM to find the maximum height. Find the two roots of the equations. The time the ball is the absolute value of $x_1 - x_2$
3. Thanks for the help dude<|endoftext|>
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Today, serpent enthusiasts throughout the world are sharing their appreciation for snakes. Roughly 3,000 species of these creatures inhabit the globe, from the icy climate of northern Canada to steamy rain forests and a majority of the world’s oceans. Ranging in size from a few inches to more than 30 feet (nine meters) long, about 375 snake species are venomous, but only a small portion are dangerous to human beings.
The following pictures are part of the National Geographic Photo Ark, in which Sartore sets out to capture every captive species on camera, with the goal of spreading awareness about animal conservation.
Follow Elaina Zachos on Twitter.
Corrections: July 18, 2016, 11 a.m.: Due to updates in animal taxonomy, the following captions have been edited to fit the animals' current scientific names:
Photo 1: Trimeresurus albolabris has been updated to Trimeresurus insularis.
Photo 3: The scientific name has been updated to Drymarchon melanurus erebennus.
Photo 6: The full species name is Azemiops feae.
Photo 12: The updated scientific name is Malayopython timoriensis.
Photo 19: The species has been split, and the animal depicted is now Lampropeltis californiae.<|endoftext|>
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We have come to a tipping point in math education when we must look to provide higher demanding math experiences for our students.
We have to change students’ relationship with math, and one way to do that is by changing the types of math experiences they have.
Perhaps the way to help students the most, both in terms of success and attitude, lies in the counter-intuitive notion of finding the right level of struggle or challenge—a level that is both constructive and instructive.
This course will help teachers with reflecting and creating math experiences that will help push students further and engage them with math that is more rigorous and meaningful for them. The resources and tools provided in this course will help to challenge students to engage in productive struggle, be more collaborative, participate in rigorous thinking, create a student-centered approach for learning math.
This course will provide instructional strategies which will focus on key areas in helping students to better understand math at a conceptual level. This interactive course is designed for job-embedded learning and practical classroom application of state assessment practices and instructional strategies.
This course is also fully aligned with Danielson’s Framework for Teaching Domains 1, 2, 3 and 4. Through a combination of lesson activities, online discussions, and instructional reflection processes participants will gain insight into a wide range of strategies and resources to incorporate into their own instructional practices.
This course has been divided into the following modules:
- State Assessments
- Using Effective Math Problems
- Implementing Skilled Math Instruction
- Leveraging Formative Assessments<|endoftext|>
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Seventh Capital, Houston
Houston became the seventh capital of Texas when President Houston ordered the seat of government to Houston on December 15, 1836. Houston was formed when the Allen brothers acquired a tract of land on Buffalo Bayou near the former town of Harrisburg. The brothers named the town Houston after Sam Houston in hopes of him choosing the town as the capital. As an incentive, a capital building was offered to be built. Congress moved to Houston in April 1837 even though the capital building was still unfinished. Early-day Houston had a reputation as a lawless town and then had a terrible yellow fever outbreak that killed over 10 percent of the population. All of this, along with Houston’s marshy location and the inadequate accommodations caused Houston to lose out in the battle to became the permanent capital of Texas.
The photo is a photograph of a reproduction of a map created in 1869 by Gail Borden, who was commissioned by the Allen brothers to produce a map of the city. The “Original Plan of Houston” shows a city hugging Buffalo Bayou with space reserved for a courthouse, churches, and schools. It can be found in special collections at the University of Houston Library.<|endoftext|>
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# Understanding Domain and Range Part 1
The domain of a function is the set of x-coordinates of the points in the function. The range of the function f is the set of y-coordinates of the points in the function. So if we have a function f with points (-3, -2), (-1, 3), (2, 3), and (5,4), then the domain of the function f is the set {-3, -1, 2, 5} and the range of f is the set {-2, 3, 4). Graphically, we can say that the domain is the “projection” of the points to the x-axis (see red points in the following figure).
The range of f is the projection of the points to the y-axis (see green points in the following figure).
Example 1
In the function g, we can see that the projection of the graph to the x-axis is the set of real numbers from -1 to 2 (see red line segment), however, small “empty circle” indicates that the domain does not include 2. Therefore, the domain of g is the set of real number
$\{x | x - 1 \leq x < 2 \}$.
In interval notation, that is $[-1, 2)$.
Recall: The interval notation where a is less than b, $[a, b]$ means from a to b including a and b and $(a, b)$ means from a to b excluding a and excluding b. The interval $(a, b]$ means from a to b excluding a and including b and $[a, b)$ means from a to b including a and excluding b.
Looking at the graph, the projection of the graph to the y-axis is from -2.7 up to 3.5 but not including 3.5 because of the empty circle (see green line segment). Therefore, the range is the set
$\{y | y \geq - 2.7 \leq y < 3.5 \}$
or $[2.7, 3.5)$ in interval notation.
Example 2
In the function h, the projection of the graph to the x-axis is from -3 to 2, except -2. Therefore, the domain of h is the set of real numbers from -3 to 2, excluding -2. Using the interval notation, we can them into two intervals which are from -3 to 2 and from -2 to 2 excluding -2 on both intervals.
In interval notation, we can say that the domain of h is $[-3, -2) \cup (-2, 2]$.
Example 3
The projection of the graph $p(x) = x^2$ on the x-axis is on the entire x-axis (can you visualize why?), so its domain is the set of real numbers. Its projection on the y-axis is from 0 extending upward indefinitely, so the range of f is the set of non-negative real numbers or $[0, \infty)$ in interval notation.
Example 4
The graph of the linear function $q(x) = x$ is a slanting line extending to the left and right indefinitely as well as up and down indefinitely. This means that the domain of q is the set of real numbers and the range is also the set of real numbers.
Example 5
The domain of the sine function $y = sin(x)$ whose graph is shown below is the set of real numbers (why?). Its range is from -1 to 1.
Now, you have a conceptual understanding of the graphical representation of domain and range. In the next post, we are going to learn some algebraic methods on how to find the domain and range of functions.<|endoftext|>
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Reading for Sure (primary school to adults)
When pupils have mastered the phonologic (sound by sound ) reading in stage 2 of STEPPING AHEAD, they can rightly announce “I can read!!”
Well, yes, if the printed text is in a phonetic language; that is, a language where each alphabet letter has one corresponding speech sound. That is not the case with English where three out of every four words cannot be pronounced by sounding out the individual letters. This seventy five percent of irregular words are often called ‘demons’ or ‘tricky’ and mustbe learned from repetition and memory. Of course this does not provide a satisfactory outcome for at least fifty percent of potential readers.
In other languages which were originally unphonetic, the alphabet has been changed or additional pronunciation marks have been added to eliminate ambiguity in written and spoken words.
The English language was subjected to such adaptation with the introduction of the International Pronunciation Alphabet (IPA) at the end of the nineteenth century and remains as a pronunciation code in English dictionaries.
However, the IPA is not adaptable for teaching English reading as the pronunciation of unfamiliar words requires dictionary use which is beyond the ability of most beginning readers.
How Reading for Sure began
Reading for Sure has solved the problem of the irregularities and ambiguities of English by producing a Universal Pronunciation Code which can be overprinted on any text and phased out when the reading becomes automatic through practice. This instruction, through coded phonics, is the unique feature of Reading for Sure and the reason for its twenty five year success.
Coded phonics – The System Teaching Kit
The pupil is introduced to the Universal Pronunciation Code in a step by step progression, beginning with the pronunciation signs for short vowels, followed by stories with words containing long and mixed vowels. Error free reading of continuous text begins immediately. The pupils read the supplementary readers- the Phonologic Fun Books –with confidence, knowing that the words will not ‘trick’ them. The reading practice, provided by the System Teaching Kit and the supplementary readers, leads to automatic reading without relying on the code.
Reading is the cornerstone required for all learning but other areas of literacy need to be addressed as well. Reading for Sure has this covered in its extension activities that are integrated at every step of the learning to read process. read more
Go to tutor directory here.<|endoftext|>
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# What do you call a graph with intersecting circles?
## What do you call a graph with intersecting circles?
Introduction. Venn diagrams are charts with overlapping circles that indicate how much different groups have in common. You specify the relative sizes of the circles and the amount of overlap between them.
### How do you use a Venn diagram for compare and contrast?
Simply draw two (or three) large circles and give each circle a title, reflecting each object, trait, or person you are comparing. Inside the intersection of the two circles (overlapping area), write all the traits that the objects have in common. You will refer to these traits when you compare similar characteristics.
What is intersection in Venn diagram?
A complete Venn diagram represents the union of two sets. ∩: Intersection of two sets. The intersection shows what items are shared between categories. Ac: Complement of a set. The complement is whatever is not represented in a set.
What does it mean to intersect a graph?
A point of intersection is a point where two lines or curves meet. We can find a point of intersection graphically by graphing the curves on the same graph and identifying their points of intersection.
## How do you represent an intersection on a graph?
All graphs are intersection graphs Any undirected graph G may be represented as an intersection graph: for each vertex vi of G, form a set Si consisting of the edges incident to vi; then two such sets have a nonempty intersection if and only if the corresponding vertices share an edge.
### What is the purpose of using a Venn diagram?
A Venn diagram is an illustration that uses circles to show the relationships among things or finite groups of things. Circles that overlap have a commonality while circles that do not overlap do not share those traits. Venn diagrams help to visually represent the similarities and differences between two concepts.
How can the Venn diagram help best the students in solving problem?
Venn diagrams enable students to organise information visually so they are able to see the relationships between two or three sets of items. They can then identify similarities and differences.
What is symbol used for intersection?
symbol ∩
The intersection operation is denoted by the symbol ∩.
## How do you find the intersection?
How Do I Find the Point of Intersection of Two Lines?
1. Get the two equations for the lines into slope-intercept form.
2. Set the two equations for y equal to each other.
3. Solve for x.
4. Use this x-coordinate and substitute it into either of the original equations for the lines and solve for y.
### Which is the intersection graph of a circle?
A circle graph is the intersection graph of a set of chords of a circle. The circle packing theorem states that planar graphs are exactly the intersection graphs of families of closed disks in the plane bounded by non-crossing circles.
Which is the intersection graph of a unit disk?
A unit disk graph is defined as the intersection graph of unit disks in the plane. A circle graph is the intersection graph of a set of chords of a circle. The circle packing theorem states that planar graphs are exactly the intersection graphs of families of closed disks in the plane bounded by non-crossing circles.
Is the intersection graph of a line segment nonplanar?
However, intersection graphs of line segments may be nonplanar as well, and recognizing intersection graphs of line segments is complete for the existential theory of the reals (Schaefer 2010). The line graph of a graph G is defined as the intersection graph of the edges of G, where we represent each edge as the set of its two endpoints.
## When do two circles touch in a graph?
To do this, you need to work out the radius and the centre of each circle. If the sum of the radii and the distance between the centres are equal, then the circles touch externally. If the difference between the radii and the distance between the centres are equal, then the circles touch internally.
Begin typing your search term above and press enter to search. Press ESC to cancel.<|endoftext|>
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## How do you calculate volumetric flow rate?
The measurement of volumetric flow is called the volumetric flow rate, and this is defined as the amount of volume that will be transferred in a unit period of time i.e. Q or V̇ = V / t.
## How do you calculate flow rate?
Q=Vt Q = V t , where V is the volume and t is the elapsed time. The SI unit for flow rate is m3/s, but a number of other units for Q are in common use. For example, the heart of a resting adult pumps blood at a rate of 5.00 liters per minute (L/min).
## Is volume flow rate constant?
The ” Volume flow rate would still remain constant. The sum of the flow rates in the two tubes is still equivalent to the flow rate in the original tube.
## What is the unit of flow rate?
Volumetric flow rate is sometimes measured in “standard cubic centimeters per minute” (abbreviation sccm), a unit acceptable for use with SI except that the additional information attached to the unit symbol. The SI standard would be m3/s (with any appropriate prefix, with temperature and pressure specified).
## What is normal water flow rate?
A toilet will normally use about 2-3 gallons per minute (gpm), a shower from 1.5 to 3.0 gpm, a bathroom or kitchen faucet from 2-3 gpm, a dishwasher from 2-4 gpm, and a washing machine from 3-5 gpm.
## How do you measure water flow rate?
For the best accuracy measure the flow 3 or 4 times and average the times together. The formula to find GPM is 60 divided by the seconds it takes to fill a one-gallon container (60 / seconds = GPM). Example: The one-gallon container fills in 5 seconds, breakdown: 60 divided by 5 equals 12 gallons per minute.
## How do you calculate flow rate with pressure?
Poiseuille’s Law states that flow rate F is given by F = π(P1-P2)r4 ÷ 8ηL, where r is the pipe radius, L is the pipe length, η is the fluid viscosity and P1-P2 is the pressure difference from one end of the pipe to the other.
## What is flow rate in nursing?
The formula for calculating the IV flow rate (drip rate) is… total volume (in mL) divided by time (in min), multiplied by the drop factor (in gtts/mL), which equals the IV flow rate in gtts/min. Let’s try an example. The provider has ordered 1,000 mL Lactated Ringers to infuse over 8 hours.
## What is the difference between volume flow rate and mass flow rate?
Volumetric flow rate is a measure of the 3-dimensional space that the gas occupies as it flows through the instrument under the measured pressure and temperature conditions. Mass flow rate is a measure of the number of molecules that flow through the instrument, regardless of how much space those molecules occupy.
## How do you convert mass flow rate to volume flow rate?
Divide the mass flow by the density. The result is the volumetric flow, expressed as cubic feet of material. An example is: 100 pounds (mass flow) / 10 pounds per cubic foot (density) = 10 cubic feet (volumetric flow).
## What unit is flow rate measured in?
Introduction. Flow is the volume of fluid that passes in a unit of time. In water resources, flow is often measured in units of cubic feet per second (cfs), cubic meters per second (cms), gallons per minute (gpm), or other various units.
## What is LPM in flow rate?
LPM is an abbreviation of litres per minute (l/min). When used in the context of a particle counter’s flow rate, it is a measurement of the velocity at which air flows into the sample probe. For example, a flow rate of 2.83 LPM means the particle counter will sample 2.83 litres of air per minute.
### Releated
#### Depreciation equation
What are the 3 depreciation methods? There are three methods for depreciation: straight line, declining balance, sum-of-the-years’ digits, and units of production. What do you mean by depreciation? Definition: The monetary value of an asset decreases over time due to use, wear and tear or obsolescence. This decrease is measured as depreciation. How do you […]
#### Polar to cartesian equation calculator wolfram
How do you convert polar to Cartesian? Summary: to convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) 😡 = r × cos( θ )y = r × sin( θ ) How do you find the polar Cartesian equation? Convert the polar equation r = 2sec θ to a rectangular equation, and draw its corresponding […]<|endoftext|>
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A recent study by researchers at the University of Hawaii — Manoa (UHM) School of Ocean and Earth Science and Technology (SOEST) and the University of Rhode Island (URI) changes the understanding of how the Hawaiian Islands formed. Scientists have determined that it is the eruptions of lava on the surface, extrusion, which grow Hawaiian volcanoes, rather than internal emplacement of magma, as was previously thought.
Before this work, most scientists thought that Hawaiian volcanoes grew primarily internally — by magma intruding into rock and solidifying before it reaches the surface. While this type of growth does occur, along Kilauea’s East Rift Zone (ERZ), for example, it does not appear to be representative of the overall history of how the Hawaiian Islands formed. Previous estimates of the internal-to-extrusive ratios (internally emplaced magma versus extrusive lava flow) were based on observations over a very short time frame, in the geologic sense.
Ashton Flinders (M.S. from UHM), lead author and graduate student at URI, and colleagues compiled historical land-based gravity surveys with more recent surveys on the Big Island of Hawaii (in partnership with Jim Kauahikaua of the U.S. Geological Survey — Hawaii Volcano Observatory) and Kauai, along with marine surveys from the National Geophysical Data Center and from the UH R/V Kilo Moana. These types of data sets allow scientists to infer processes that have taken place over longer time periods.
“The discrepancy we see between our estimate and these past estimates emphasizes that the short term processes we currently see in Hawaii (which tend to be more intrusive) do not represent the predominant character of their volcanic activity,” said Flinders.
“This could imply that over the long-term, Kilauea’s ERZ will see less seismic activity and more eruptive activity that previously thought. The 3-decade-old eruption along Kilauea’s ERZ could last for many, many more decades to come,” said Dr. Garrett Ito, Professor of Geology and Geophysics at UHM and co-author.
“I think one of the more interesting possible implications is how the intrusive-to-extrusive ratio impacts the stability of the volcano’s flank. Collapses occur over a range of scales from as large as the whole flank of a volcano, to bench collapses on the south coast of Big Island, to small rock falls. ” said Flinders. Intrusive magma is more dense and structurally stronger than lava flows. “If the bulk of the islands are made from these weak extrusive flows then this would account for some of the collapses that have been documented, but this is mainly just speculation as of now.”
The authors hope this new density model can be used as a starting point for further crustal studies in the Hawaiian Islands.
Note : The above story is based on materials provided by University of Hawaii ‑ SOEST, via EurekAlert!, a service of AAAS.<|endoftext|>
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When people talk about anodizing, they are actually talking about a very specific chemical process which is mainly used in the treatment of aluminum. The process of anodization is an electrochemical method to change a metal surface into a surface which is much more decorative and resistant to corrosion.
The metal which is the most conducive to anodizing is actually aluminum. This is because it is a nonferrous metal. There are other metals which can be put through an anodizing tank such as magnesium or titanium but aluminum is still the best. The oxide structure is made up completely of aluminum oxide. This structure is born in the substrate of the aluminum.
The way in which it is applied is not like paint or staining at all. In fact, it is fully sealed into the metal itself. The anodized surface is so porous in an orderly fashion that it is very suitable to be colored or sealed. The way anodization works are taking the metal and placing it into a bath of acidic electrolytes. Once the metal is fully submerged, an electrical current is passed through the liquid solution.
Inside the anodizing tank, a cathode is placed. The metal, which is usually aluminum, acts as an anode in the solution and so, therefore, the ions of the oxygen nature are combined with the atoms of the oxygenic nature and result in being anodized. As you can see, there are very many steps into properly transforming raw aluminum into anodized aluminum but the effort is well worth it.
The enhancement that the metal undergoes is far more strong than just the metal by itself. This means that it can be used for far more purposes than they would be able to without undergoing the treatment of the metal.<|endoftext|>
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The Iowa Watershed Approach
Download Full Text
What are buffers?
Buffers are established areas of permanent vegetation, within and around fields, and are designed to intercept and filter sediment and nutrients out of surface runoff and shallow groundwater before entering a water course. The vegetation provides habitat for wildlife and creates a recreational and aesthetically pleasing area. One of the primary functions of buffers is to slow surface runoff, trapping 41-100% of the sediment and significantly reducing the phosphorus load. By slowing surface runoff and promoting infiltration, buffers delay downstream flooding and reduce streamflow by 10%. Additionally, when shallow groundwater interacts with the buffer’s root zone, biological processes can remove 48-85% of its nitrate-nitrogen; however, the percent of shallow groundwater that interacts with the root zone could be small. There are many types of buffers which are distinguished by their design and vegetative species, including: riparian forest buffer, filter strips, field borders, field windbreaks, grassed waterways, among others.
Benning, Jamie; TeBockhorst, Kristina; and Johnson, Jason, "Buffers" (2018). Extension and Outreach Publications. 435.
Iowa State University Extension and Outreach publications in the Iowa State University Digital Repository are made available for historical purposes only. The information contained in these publications may be out of date. For current publications and information from Iowa State University Extension and Outreach, please visit<|endoftext|>
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# Box 5.3a: The Cubic Formula
## Solving a Cubic
Four centuries before European mathematicians solved the problems of the cubic and quartic equations, Indian mathematician and astronomer, Bhaskara, had solved these problems. You can see a biography of Bhaskara by clicking here.
There are also pdf converters available online if you need to have any pdf document in another format.)
Note that this material is not at all likely to be on the exam. The formula is not even all that useful. It's given here merely to satisfy your curiosity. So you may skip this box if you like.
If you are given a cubic equation in the form of
``` x3 + px2 + qx + r = 0 eq. 5.3a-1
```
and need to solve for x, then the first thing you do is substitute variables. Everywhere you see an x in the cubic, replace it with
``` p
x = u - eq. 5.3a-2
3
```
When you get done squaring and cubing this expression, then substituting stuff back in and gathering like terms, you will get
``` u3 + au + b = 0 eq. 5.3a-3a
```
where
``` p2
a = q - eq. 5.3a-3b
3
```
and
``` pq 2p3
b = r - + eq. 5.3a-3c
3 27
```
Now compute A and B by
``` A = ``` ``` ``` ``` eq. 5.3a-4a ```
and
``` B = ``` ``` ``` ``` eq. 5.3a-4b ```
Then you have the following solutions for u
``` u = A + B eq. 5.3a-5a
____
u = -(1/2)(A + B) + √-3/4 (A - B) eq. 5.3a-5b
____
u = -(1/2)(A + B) - √-3/4 (A - B) eq. 5.3a-5c
```
Of course, you know how convert from the u solutions to the x solutions (hint: look at 5.3a-2).
You may be troubled by the expression, -3/4, which is not a real number, but is a complex number. Note that A and B will also nonreal complex numbers whenever
``` b2 a3
+ < 0 eq. 5.3a-6
4 27
```
So the cubic formula does require that you understand the arithmetic of complex numbers.
And if you are not up on complex numbers, don't worry. We will be reviewing them in a later section.
### How to Derive the Cubic Formula
If you have been trying to come up with a solution to the cubic on your own, my heart goes out to you. Generations of mathematicians searched for a cubic solution before Niccolo Fontana Tartaglia and Girolamo Cardano hit on it in the 16th century (Cardano published the solution in Ars Magna in 1545). So there is no shame in your not finding it (of course if you did find it on your own, my hat is off to you).
Since you solve the quadratic by completing the square, a lot of people who attack the cubic do so by trying to "complete the cube." Well that attack doesn't work. The trick is to convert the cubic, by a circuitous route, to a quadratic and then apply the quadratic formula.
We already saw in the previous paragraphs that it is sufficient to find a solution only to cubics in the form of
``` u3 + au + b = 0 eq. 5.3a-3a
```
This is because by suitable substitution of variables, any cubic can be brought into this form.
In order to find the solution to the cubic you would have had to have the insight of making the simplification of equation 5.3a-3a, and you would also have to have the insight to suppose that a solution to that equation, u, ought to be expressed as the difference of two new variables:
``` u = s - t eq. 5.3a-7
```
Now substitute that into equation 5.3a-3a:
``` (s - t)3 + a(s - t) + b = 0 eq. 5.3a-8a
```
Now multiply out the cubed term using the binomial formula.
``` s3 - 3s2t + 3st2 - t3 + as - at + b = 0 eq. 5.3a-8b
```
In order for this equation to hold we must be able to cancel all the terms on the left of the equal. If you substitute t3 - s3 = b, you can see that you cancel all the cubed terms:
``` s3 - 3s2t + 3st2 - t3 + as - at + t3 - s3 = 0 eq. 5.3a-8c
-3s2t + 3st2 + as - at = 0 eq. 5.3a-8d
```
Now just substitute 3st = a, and you cancel the remaining terms:
``` -3s2t + 3st2 + (3st)s - (3st)t = 0 eq. 5.3a-8e
```
The only problem that remains is to find a suitable s and t from a and b. We already made the substitutions:
``` 3st = a eq. 5.3a-9a
t3 - s3 = b eq. 5.3a-9b
```
So we solve these two equations simultaneously for s and t.
``` a
s = eq. 5.3a-10
3t
```
Substitute eq. 5.3a-10 into 5.3a-9b and you get
``` a3
t3 - = b eq. 5.3a-11
(3t)3
```
Multiply through by (3t)3 = 27t3, and it becomes:
``` 27t6 - a3 = 27t3b eq. 5.3a-12
```
Don't let the 6th power here scare you. This isn't as bad as it looks. All the powers of t are multiples of 3. So if substitute t3 = z, you find yourself in familiar territory.
``` 27z2 - a3 = 27bz eq. 5.3a-13
```
From here you use the quadratic formula to solve for z. For each solution you come up with for z, you will know that t = z1/3. From t you can find s = a/(3t), and from s you can find u = s - t, which is a solution to
``` u3 + au + b = 0
```
and from u you can find x (the solution to your original equation) using equation 5.3a-2.
If you thought that was interesting, click here to see how Cardano's assistant, Lodovico Ferrari, was able to solve the problem of the quartic (4th degree) polynomial.
Also see<|endoftext|>
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1. Moving or traveling from place to place.
2. Of or related to walking, moving, or traveling.
3. Of or related to Aristotle: his philosophy or his teaching method of conducting discussions while walking about.
ambulant, itinerant, itinerate, migrant, mobile, nomadic, perambulant, roaming, roving, vagabond, vagrant, wandering, wayfaring,pacing
[From Latin peripateticus, from Greek peripatetikos, from peripatein (to walk
about, to discourse while pacing as did Aristotle), from peri- (around) +
patein (to walk). Ultimately from the Indo-European root pent- (to tread)
that also gave us words such as English find, Dutch pad (path), Hindi path
(path), French pont (bridge), and Russian sputnik (traveling companion).
he Peripatetics were members of a school of philosophy in ancient Greece. Their teachings derived from their founder, the Greek philosopher, Aristotle, and Peripatetic (Greek: περιπατητικός) is a name given to his followers. The name refers to the act of walking, and as an adjective, "peripatetic" is often used to mean itinerant, wandering, meandering, or walking about. The school derives its name from the peripatoi (colonnades) of the Lyceum gymnasium in Athens where the members met, although a later legend claimed that the name came from Aristotle's alleged habit of walking while lecturing.
The school dates from around 335 BC when Aristotle began teaching in the Lyceum. It was an informal institution whose members conducted philosophical and scientific inquiries. Aristotle's successors Theophrastus and Strato continued the tradition of exploring philosophical and scientific theories, but after the middle of the 3rd century BC, the school fell into a decline, and it was not until the Roman era that there was a revival. Later members of the school concentrated on preserving and commentating on Aristotle's works rather than extending them, and the school eventually died out in the 3rd century AD, although the tradition of commentating on Aristotle's works was continued by the Neoplatonists. After the fall of the Roman empire, the works of the Peripatetic school were lost to the west, but in the east they were incorporated into early Islamic philosophy, which would play a large part in the revival of Aristotle's doctrines in Europe in the Middle Ages.
Private Reply to Tim Southernwood<|endoftext|>
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# The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger. What are the numbers?
Nov 11, 2016
$v \ge 3$ and $v + 1 \ge 4$
#### Explanation:
First let's get or define our variables. We can call the first variable $v$. Then because the problem states they are two consecutive integers which means the second integer is one more the first integer the second integer can be called $v + 1$.
"Nine times the smaller" can be written as $9 v$ and "5 times the larger" can be written as $5 \left(v + 1\right)$.
"is at most" means we have an inequality and specifically a $\le$ or less than or equal to inequality.
So, to write the inequality for the entire problem we have:
$v + \left(v + 1\right) \le 9 v - 5 \left(v + 1\right)$
Expanding the terms in parenthesis and then grouping like terms on each side of the inequality gives:
$v + v + 1 \le 9 v - 5 v - 5$
$2 v + 1 \le 4 v - 5$
Next we solve for $v$ while keeping the inequality balanced:
$2 v + 1 - 2 v + 5 \le 4 v - 5 - 2 v + 5$
$6 \le 2 v$
$\frac{6}{2} \le \frac{2 v}{2}$
$3 \le v$
To "flip" the inequality so the $v$ is on the left side we have to "flip " the inequality so $\le$ becomes $\ge$ and the inequality can be written as:
$v \ge 3$<|endoftext|>
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Concept: The question tests on the basic principles of counting/combinatorics
Solution:
In this question you have 2 numbers occurring the same number of times i.e twice and 1 number occurring once.
You can choose the one number occurring once in 9C1 = 9 ways and the 2 numbers occurring twice in 8C2 = 28 ways and then rearrange all these numbers in (5! / 2!2!) = 30 ways. Total number of ways= 9 * 28 * 30 = 7560
You can choose the two numbers occurring twice in 9C2 = 36 ways and the one number occurring once in 7C1 = 7 ways and then rearrange all these numbers in (5! / 2!2!) = 30 ways. Total number of ways= 36 * 7 * 30 = 7560
You can also choose on number which will occur twice in 9C1 = 9 ways, the next number occurring twice in 8C1 ways and the number occurring once in 7C1 ways.
What we need to remember here is that 2 of these numbers are occurring twice and hence we have to divide by 2! = 2 ways to avoid double counting.
Total number of ways = (9 * 8 * 7/2) * 30 = 7560 ways
Option E<|endoftext|>
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# What do the eigenvalues of a matrix tell us about the original matrix?
I have a problem of...
Let $A$ be a 2x2 matrix such that it is not invertible and 2 is an eigenvalue of $A$.
a) Find all eigenvalues of $A+I$.
b) Prove or disprove A+I is invertible.
Since it's not invertible, it has an eigenvalue of 0. So I can think of a matrix easily such as the one below with eigenvalues of 2 and 0...
$\begin{bmatrix} 0 & a \\ 0 & 2 \end{bmatrix}$
Where $a$ is just some unknown. However, I'm assuming there are many matrices that have eigenvalues of 2 and 0 for a 2x2. So I am having trouble even seeing what the eigenvalues will even tell me about the original matrix.
Do eigenvalues tell you anything about the original structure of the matrix?
Also I haven't learned about eigenvectors yet in class.
You don't have to find the original matrix to answer the question. As you said the eigenvalues must be $\lambda =0,2$. Claim, $\mu =1,3$ are the eigenvalues of $A+I$. Lets check, Let $x$ be an eigenvector of $A$ corresponding to the eigenvalue 2, then $$(A+I)x=Ax+x=2x+x=3x.$$Thus, $\mu=3$ is an eigenvalue for $A+I$. Similarly, we can conclude that $\mu=1$ is the other eigenvalue. Which means $A+I$ is invertible.
• You do need to mention for the final conclusion that a $2\times2$ matrix cannot have more than $2$ distinct eigenvalues, so with $1$ and $3$ already taken, $0$ cannot be an eigenvalue of $A+I$. – Marc van Leeuwen Oct 18 '16 at 6:18
The Given matrix has two distinct eigenvalues $0$ and $2$. Therefore it will have $2$ linearly independent eigenvectors and will be diagonalizable. That is, it will be similar to $$\begin{bmatrix} 2 & 0 \\ 0 & 0 \\ \end{bmatrix}$$ So we can right $A+I$ as \begin{bmatrix} 3 & 0 \\ 0 & 1 \\ \end{bmatrix} which is a invertible matrix with eigenvalues 3 and 1.<|endoftext|>
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least significant bit
most significant bit
and it is convergent for values of z that are larger than e
Similarly to what we saw for continuous-variable functions, the Fourier trans-
form for discrete-variable functions can be obtained as a specialization of the
Z transform where the values of the complex variable are restricted to the unit
Y () = Y
or, in detail,
Y () =
In this book, we use the symbol for the radian frequency in the case of discrete-
variable functions, leaving for the continuous-variable functions.
In order to fully understand the behavior of several hardware and software tools
for sound processing, it is important to know something about the internal
representation of numbers within computer systems. Numbers are represented
as strings of binary digits (0 and 1), but the specific meaning of the string
depends on the conventions used. The first convention is that of unsigned integer
numbers, whose value is computed, in the case of 16 bits, by the following
is the i-th binary digit starting from the right. The binary digits are
called bits, the rightmost digit is called least significant bit (LSB), and the
leftmost digit is called the most significant bit (MSB). For instance, we have
= 17190 ,
where the subscript 2 indicates the binary representation, being the usual deci-
mal representation indicated with no subscript.
The leftmost bit is often interpreted as a sign bit: if it is set to one it means
that the sign is minus and the absolute value is given by the bits that follow.
However, this is not the representation that is used for the signed integers. For
these numbers the two's complement representation is used, where the leftmost
bit is still a sign bit, but the absolute value of a negative number is recovered
by bitwise complementation of the following bits, interpretation of the result as
a positive integer, and addition of one. For instance, with four bits we have
+ 1) = -(5 + 1) = -6 .
The two's complement representation has the following advantages:
(x) is the real part of the complex number x<|endoftext|>
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The objective of each family who wishes to help develop their young members’ reading abilities must try, above everything, to generate curiosity for reading and interest and attraction for books and the written language. You can achieve this by using different activities.
- Read them a little each day. Choose a quiet time, like the moments before bedtime, and start reading your little ones stories appropriate for their age. This exercise will allow them to see reading as a pleasant activity and will start associating sounds with letters.
- Help them see the usefulness of reading. The written word does not only appear in books, but also on various objects and spaces we use every day. Parents can start teaching the children to make associations between the written and the spoken language, if they involve the little ones in these reading situations. You can read the street and shop signs or the labels on food together. Another option is to place labels written in large font on different objects in the house.
- Letter games: there are games that help children learn and recognize the letters of the alphabet, and associate them with complete words. Some interesting games that you can play with your child are “I spy…”, “Word chain”, or spelling names and words.
- Reading with pictures: stories with pictures include images that are easy to understand for children, and that can be easily replaced by an appropriate word. This type of reading is very motivating, as the children, though still unable to read properly, become more than mere spectators of the process of reading.
- Be an example yourself: growing up in an environment that encourages reading is one of the most influential background factors with direct implications on the child’s future reading skills. Parents must allow their children to join them when they read, talk to them about the text they are reading, help them create their own bookcase and introduce to them the public spaces where they can get books from – that is, libraries and bookstores.<|endoftext|>
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Why is reading so important?
Enjoying books and reading stories from a very early age is crucial in the development of children. It helps with their ability to understand words, use their imagination and develop their speech, as well as being something they really enjoy.
Teachers and parents play a huge part in the development of reading skills in young children. The more children experience books the more they will gain the interest and passion for them. Reading offers so much more than just quiet time in a cosy corner. It helps to develop spelling, listening, writing, literacy and social skills.
Young children need to be able to experience books; they need to be able to understand and enjoy stories, books, rhymes and songs and listen and respond to them with curiosity and enjoyment.
This will promote the value and pleasure of reading and encourage an interest in reading throughout school and in later life.
Even from a very young age child love books. It provides the opportunity to learn through touching and feeling different textures, along with experiences different actions.
There are some great early years books with pop up pages, lift-the-flap pages, noisy pages and different textured pictures to really get babies and toddlers engaged with books.
Children develop more rapidly during the first five years of their lives than at any other time. That's why they are called the foundation years – the building blocks for life! Throughout these important first years if they attend a nursery, pre-school or primary school, their progress will be monitored.
Below is a guide, set by early years’ foundation, to explain the various stages children are monitored and milestones they might reach in terms of literacy.
A child should have an interest in holding and looking at books by this age. It is important they can hold books so they can see what is going on whilst listening to you read the stories. They can learn what books are and play with pages.
By this age most children can identify their favourite books and stories they want to see and hear. They can recognise and mimic actions from their favourite songs and stories. Encourage this by always letting them join in with storytelling and songs, let them point to things they can recognise or make the sounds of things in the book.
By the age of three, children should be able to fill in missing gaps in stories and songs that they have heard repetitively. They have their favourite books and songs and they can help to tell the tale. By pausing during a story they know well you give them the opportunity to fill in the gaps and add more words to their vocabulary.
By four, children should be able to recognise the story being read to them, they can help with telling the story and can anticipate the end of the story. They should be able to join in with rhymes and be able to recognise words that start with the same sound such as ‘big boat’. They should also be able to recognise words that mean something to them, such as their own name or mummy, favourite shops and places.
They can sit and listen for longer and can hold the book correctly and turn pages by themselves.
By the age of five, a child is expected to be able to remember and speak words that rhyme like cat and hat and sat. Their imaginations and vocabulary mean they are able to tell their own made up stories and can make up their own songs. Story time doesn’t mean you just read to them – it’s far more interactive.
They can understand and read simple sentences and they use their phonics knowledge to sound out words to read them accurately. They can also demonstrate a real understanding of what has been read or said to them.
Reading At School
Children usually start school between 4-5 years old. By this time, it is likely they will be able to recognise letters, understand words and have fairly good listening skills and be able to deal with the changes in routine.
A typical school day in reception does feature some routine and structure, however, it still involves quite a lot of free play.
To encourage reading, teachers usually have story time which involves children sitting together on the classroom carpet. Group reading and reading out loud helps to boost confidence in children and encourages interest and interaction.
A fun group task is to read and act out the book together. Children will bounce off each other and shine. Imaginations run wild in young children and role play can also help make reading fun. Play is the best form of learning in early years as it develops a whole host of communication, reading and language skills.
Having a well-equipped school library or classroom reading area will also help to encourage reading. Most children, and even adults, like to sit down and get comfy so that they can free their mind and travel into their story book.
Creating The Perfect Classroom Reading Corner
A library or classroom reading corner should create excitement for reading before the child has even selected a book. It needs to be visually appealing and stimulating and have an interesting and age appropriate selection of books to read.
Setting up your reading area can be a challenge as you need to leave enough space for seating and movement with a wide variety of books as well as somewhere to store them. All this needs to look visually appealing to a child too!
We offer a wide range of book storage solutions that will help you to create an exciting, stimulating classroom reading corner.
It’s important the books are easily accessible and easy to view. Our mobile book trolleys are double sided which allows children to access books easily from each side. The Kinderboxes come in a range of shapes, styles, sizes and colours which make an interesting book display.
Our book spinners are ideal for a smaller area. They can hold many books and don’t take up too much room. Kids also love spinning them to find the right book.
Having somewhere for children to sit and look at books is important. You can provide floor cushions or bean bags with a low level book storage display such as a kinderbox, or book trolley. This will create the perfect reading area for young children.
If you have a large space for your reading area our animal-themed book browsers are great fun and provide plenty of space for book as well as areas for children to sit. Children can get snappy with the Crocodile Book Browser or blow the trumpet of the Elephant Book Browser!
Alternatively you can add the indoor outdoor wooden folding den to create a magical space for children. Add the rainbow accessory kit for colour and to spark imagination and fun.
Reading at home
With so many books to read and enjoy there just isn’t enough time in a school day to discover all there is to enjoy with reading. Support from parents and carers with reading at home is just as important to a child’s development.
Every child loves spending time with their parents and enjoys listening to them and interacting with them. Sharing a book together is a special time with the added benefit that it’s not just for entertainment but helps develop their concentration and understanding of language. Children who don’t get the reading support at home from parents and carers often come to reading at a disadvantage.
It is never too early to start sharing books with your child, even as young babies they enjoy, and will learn from, hearing stories and looking at books. Research shows that reading at home with their child is the single most important thing a parent can do to help their child’s education.
Try to set aside time each day to sit and share a book. Turn off distractions such as the TV and mobile phone and have some quiet time together getting lost in a story. Talk to your child about the pictures, ask them what they can see and what they think is going to happen in the story. Doing this little and often is best so that you don’t lose your child’s focus and attention, but by all means read for longer if that’s what your child wants.
Be a reading role model to your child – if your child sees you enjoying a book, they will show an interest too.
At ELF we are experts in early years’ educational resources and supply classroom furniture, library furniture and equipment to primary schools, pre-schools, nurseries and playgroups. For more information about our library furniture or any of our products please contact us or call our sales team on 01733 511121.<|endoftext|>
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In the plant kingdom, angiosperms, or plants that produce flowers, occupy the majority position, or 90 percent of all species. Named species of flowering plants number close to half a million, with many tropical varieties still unknown. Flowering plants cover all climate zones on earth, including rugged Arctic terrain. They range in size from the tiny Wolffia to the largest known flower, the Rafflesia, or corpse lily, which produces flowers three feet across. Plants that produce spores instead of flowers include ferns and mosses---they are classified as pteridophytes.
Plants flower in order to reproduce. Insects and other pollinators are attracted to flowers because of their appearance and their scent. The flower is simply a sexual organ, containing the male anther and the female stigma. Some plants are bisexual, having both organs, while others are unisexual, with only the male or female organ. Unisexual plants require another plant in order to form seeds, which are the end result of a successfully pollinated flower.
Largest Flowering Plant Families
Approximately 60,000 species of flowering plants, or 25 percent, are contained in three plant families. The sunflower family, or Asteraceae, includes 24,000 species, including sunflowers, asters, daisies, marigolds and chrysanthemums. The orchid family, Orchidaceae, has 20,000 species, and the legume or pea family (Fabaceae) has 18,000 species, which range from the common garden pea to trees.
Plants in the Asteraceae family of flowering plants include herbs, shrubs, trees and vines that grow in a wide variety of climate zones. Their common distinguishing characteristics are their alternate, opposite or whorled leaf structure, bisexual or unisexual flowers often called florets, often containing multiple florets on a receptacle that is surrounded by bracts. Thirteen tribes exist within this family.
Plants in the Orchidaceae family include up to 20,000 species of terrestrial flowering plants that live in many different climates. Orchids are epiphytes, meaning they need a host on which to live, but do not rely on that host for nutrition. Flowers are usually bisexual and rarely unisexual. They possess a single stamen and produce fruit, or seeds, that are capsular.
The Fabaceae, or legume, family contains approximately 18,000 species of flowering plants. Many, like green beans, peas and pinto beans serve as staple foods of the humans and animals that live where they grow. Many varieties of legumes are widely cultivated as food crops. In desert environments, legumes are important members of the environment because they provide food, shelter for birds and other animals, and nitrogen that enriches the soil.
- Features of Legumes
- Flowers Native to Ecuador
- Life Cycle of a Lily Flower
- What Part of a Plant Makes Pollen?
- Organisms That Aid in the Pollination of Flowers
- The Difference Between Orchids & Lilies
- What Phylum Are Ferns In?
- Plant Kingdom Classification
- Linnaeus Classification of Ferns
- Common Monocot Flowers
- Rare Plants in the Amazon
- Flowering Plant Identification<|endoftext|>
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Solve 3x^2+x+7=0
There are two best approaches you can take to obtain your solution. First, you could enter the function into a graphing calculator and have it solve for the roots---but most teachers are going to want you to do it by hand!
That being said, the quadratic equation will be your best bet. Remember the equation states for any quadratic ax^2 + bx +c,
x = (-b +/- sqrt(b^2-4ac) / 2a
(where sqrt(x) is a shorthand for square root, +/- is plus or minus, and the carat (^) denotes an exponent)
Looking at our equation, we obtain that a= 3, b =1, and c = 7. All we need to do is substitute in these values to our equation:
x = (-1 +/- sqrt[1^2-4(3)(7)]) / 2(3)
x= (-1 +/- sqrt[1-84]) / 6
x= (-1 +/- sqrt[-83]) / 6 --> x = (-1 +/- i*sqrt[83])/6.
So our solutions are x = (-1 + i*sqrt[83])/6 and (-1 - sqrt[83])/6.
Approved by eNotes Editorial Team
`3x^2 + x + 7= 0` .
The roots are given by the following formula:
`x= (-b+-sqrt(b^2-4ac))/(2a) `
`==gt x1= (-1+sqrt(1-4*7*3))/(2*3) `
`==gt x2= (-1+sqrt(-83))/6 = (-1+sqrt83*i)/6 `
`==gt x2= (-1-sqrt83*i)/6`
`` Then we have two complex roots: `(-1+sqrt83*i)/6 and (-1-sqrt83*i)/6.`
Approved by eNotes Editorial Team<|endoftext|>
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Pond/Aquatic PlantsPond/Aquatic plants play a very important role in the ecological system of your pond, water garden or koi pond. They also play an essential role in maintaining a healthy and balanced pond. Plants help filter your water by converting fish waste into plant food. Plants also provide oxygen for your koi and protect them from predators with surface coverage. All water plants perform a necessary function while beautifying your pond, and they add real value to their environment. During the day pond plants breathe in carbon dioxide and release oxygen. In the evening, pond plants breathe in oxygen and release carbon dioxide. Water garden pond plants convert carbon dioxide into oxygen. Plants use this carbon dioxide to produce energy or photosynthesis. All pond plants consume nitrogen chemicals, such as nitrates and phosphates that build up in the water. Let your water garden plants work for you to balance your pond environment. Plant foliage will absorb carbon dioxide and minerals from the water which will in turn, starve the algae. Pond plants are needed to prevent your pond water from looking green and murky, as algae will grow out of control without them. Removing nitrates can reduce weeds and algae so that they are less of a problem. Pond plants are necessary to achieve a quality water balance and provide surface coverage in your garden pond, and they will help create a beautiful pond appearance and give your water feature the finished look that you are striving for.
Pond plants fall into four basic groups: oxygenators, floaters, deep water aquatics, and bog/marginal plants. The health of your pond relies on the importance of each of these groups. Vite Greenhouses provide a wide range of pond plants for your water habitat.
Oxygenating Plants grow totally submerged and play a vital role in maintaining the pond's natural balance. These plants use waste nutrients and help purify the water, thus creating an environment that is unsuitable for algal growth. They also provide cover for protection and spawning grounds for pond fish. Submersed pond water plants are particularly good at oxygenating the pond water. Bubbles of oxygen can be observed coming from the leaves of these plants. It is best to include one bunch (these plants are sold by the 'bunch' or handful) for every two square feet of pond surface. Fewer bunches may be adequate once the natural balance is obtained. Grow a variety of species since each species grows at a different time of year and has different water depth requirements.
Floating Plants are some of the most beautiful water plants and play a vital role in any pond. Providing nourishment and shade for fish, promoting a balanced ecosystem, and increasing natural filtration, floaters are a must in any water garden! They just float on the surface of the pond and donít need to be rooted in a planter. Floating Plants such as Water Hyacinths and Water Lettuce provide shade and cover for your pond. Floating aquatic plant foliage should cover approximately 60% of the pond surface area in order to achieve a good balance. Floating plants will keep the water cool in the summer and help keep algae away.
Deep Water Plants include Lotus and Water Lilies. They grow on the base of the pond and send up leaves and blooms to the surface. Depending on the variety, they may grow a couple of inches to a few feet below the surface of the water. They provide valuable leaf cover to help shade the water, which reduces algae growth. Fish love to hide under their leaves, too. Lilies do not do well with strong water movement or splashing water. Most species need full sun 10 hours a day for best blooms. A pond should have approximately one lily for every 5-10 square feet of pond surface.These pond plants will bloom incredibly when fertilized with adequate amounts of fish safe aquatic plant fertilizer. Water Lilies are available in hardy and tropical varieties. The hardy varities provide the first colors in spring and help keep the water balanced. The tropical varieties are great for fast growing, vivid colors that last a long time during the summer season.
Bog Plants and Marginals are a broad category of plants that live at the edge of a pond or water feature. They include well known species such as cattails, cannas, grasses and reeds. Marginal plants grow well in the shallower areas of your pond or water garden from two to twelve inches of water depending on size and type. They provide color and interest to your pond but do so much more. They contribute to the healthy balance of your pond's eco-system by providing food for fish, and many species attract butterflies, bees, and hummingbirds. It is helpful if a shelf is incorporated in the pond design to support them. Bog Plants do well in the moist conditions and boggy areas around the rim of the pond. Choose plants for around the pond to blend in with the garden pond and surroundings.<|endoftext|>
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# Convert degree / (hour • second) to cycle / square minute
Learn how to convert 1 degree / (hour • second) to cycle / square minute step by step.
## Calculation Breakdown
Set up the equation
$$1.0\left(\dfrac{degree}{hour \times second}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{cycle}{square \text{ } minute}\right)$$
Define the base values of the selected units in relation to the SI unit $$\left(\dfrac{radian}{square \text{ } second}\right)$$
$$\text{Left side: 1.0 } \left(\dfrac{degree}{hour \times second}\right) = {\color{rgb(89,182,91)} \dfrac{π}{7.2 \times 10^{4}}\left(\dfrac{radian}{square \text{ } second}\right)} = {\color{rgb(89,182,91)} \dfrac{π}{7.2 \times 10^{4}}\left(\dfrac{rad}{s^{2}}\right)}$$
$$\text{Right side: 1.0 } \left(\dfrac{cycle}{square \text{ } minute}\right) = {\color{rgb(125,164,120)} \dfrac{π}{1.8 \times 10^{3}}\left(\dfrac{radian}{square \text{ } second}\right)} = {\color{rgb(125,164,120)} \dfrac{π}{1.8 \times 10^{3}}\left(\dfrac{rad}{s^{2}}\right)}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(\dfrac{degree}{hour \times second}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{cycle}{square \text{ } minute}\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{π}{7.2 \times 10^{4}}} \times {\color{rgb(89,182,91)} \left(\dfrac{radian}{square \text{ } second}\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{π}{1.8 \times 10^{3}}}} \times {\color{rgb(125,164,120)} \left(\dfrac{radian}{square \text{ } second}\right)}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} \dfrac{π}{7.2 \times 10^{4}}} \cdot {\color{rgb(89,182,91)} \left(\dfrac{rad}{s^{2}}\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{π}{1.8 \times 10^{3}}} \cdot {\color{rgb(125,164,120)} \left(\dfrac{rad}{s^{2}}\right)}$$
$$\text{Cancel SI units}$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{π}{7.2 \times 10^{4}}} \cdot {\color{rgb(89,182,91)} \cancel{\left(\dfrac{rad}{s^{2}}\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{π}{1.8 \times 10^{3}}} \times {\color{rgb(125,164,120)} \cancel{\left(\dfrac{rad}{s^{2}}\right)}}$$
$$\text{Conversion Equation}$$
$$\dfrac{π}{7.2 \times 10^{4}} = {\color{rgb(20,165,174)} x} \times \dfrac{π}{1.8 \times 10^{3}}$$
Cancel factors on both sides
$$\text{Cancel factors}$$
$$\dfrac{{\color{rgb(255,204,153)} \cancel{π}}}{7.2 \times {\color{rgb(99,194,222)} \cancelto{10}{10^{4}}}} = {\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{π}}}{1.8 \times {\color{rgb(99,194,222)} \cancel{10^{3}}}}$$
$$\text{Simplify}$$
$$\dfrac{1.0}{7.2 \times 10.0} = {\color{rgb(20,165,174)} x} \times \dfrac{1.0}{1.8}$$
Switch sides
$${\color{rgb(20,165,174)} x} \times \dfrac{1.0}{1.8} = \dfrac{1.0}{7.2 \times 10.0}$$
Isolate $${\color{rgb(20,165,174)} x}$$
Multiply both sides by $$\left(\dfrac{1.8}{1.0}\right)$$
$${\color{rgb(20,165,174)} x} \times \dfrac{1.0}{1.8} \times \dfrac{1.8}{1.0} = \dfrac{1.0}{7.2 \times 10.0} \times \dfrac{1.8}{1.0}$$
$$\text{Cancel}$$
$${\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{1.0}} \times {\color{rgb(99,194,222)} \cancel{1.8}}}{{\color{rgb(99,194,222)} \cancel{1.8}} \times {\color{rgb(255,204,153)} \cancel{1.0}}} = \dfrac{{\color{rgb(255,204,153)} \cancel{1.0}} \times 1.8}{7.2 \times 10.0 \times {\color{rgb(255,204,153)} \cancel{1.0}}}$$
$$\text{Simplify}$$
$${\color{rgb(20,165,174)} x} = \dfrac{1.8}{7.2 \times 10.0}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x} = 2.5 \times 10^{-2}$$
$$\text{Conversion Equation}$$
$$1.0\left(\dfrac{degree}{hour \times second}\right) = {\color{rgb(20,165,174)} 2.5 \times 10^{-2}}\left(\dfrac{cycle}{square \text{ } minute}\right)$$<|endoftext|>
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What scientists believed was a technical fault with their equipment turned out to be radiation left over from the Big Bang - one of the greatest astronomical discoveries of the 20th century.
- Radiation emitted when the Universe was first formed is called Cosmic Microwave Background Radiation.
- Cosmic Microwave Background Radiation was discovered by accident by Arno Penzias and Robert Wilson.
- Penzias and Wilson set up a horn antenna, and detected unexpected radiation originating from deep space.
- Penzias and Wilson were awarded the 1978 Nobel prize for physics.
In 1965, at the Horn Antenna in New Jersey, scientists Robert Wilson and
Arno Penzias were getting ready to map the radiation in our galaxy.
The Horn Antenna, New Jersey
But what they were about to discover was much bigger and would earn them a Nobel prize in the process.
The Horn ...
Please log in to view and download the complete transcript.<|endoftext|>
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The future of the ozone hole
There’s an urban legend floating around the isle of Tasmania: that the state sees more cases of sunburn than Queensland, because the ozone hole sits right above it. Though sun damage is a concern, the ozone hole has never reached Australia. And contrary to another popular belief – it is not a hole that exists all year round.
The ozone hole forms and disappears on an annual basis in springtime over Antarctica. The world has been acting since the 1980s to prevent it from spreading as far as Australia.
CSIRO scientists provide world-class monitoring and modelling of the ozone hole, examining its interactions with ozone-depleting substances, providing greater insights into its recovery and effects on climate, and strengthening global efforts to mitigate it.
But, what caused the hole in the first place, is it on the mend, what are its effects on climate, and what are the threats to its long-term recovery?
First, some history
It was the early 1970s when scientists first suggested that the human-produced chemicals, chlorofluorocarbons (CFCs), used in fridges, air conditioners and aerosol cans were destroying Earth’s ozone shield.
Then, in 1984 a significant sized ozone ‘hole’ seemed to appear quite suddenly. At the time of its discovery, the ozone hole wasn’t something that was specifically being looked for, and was even overlooked in NASA’s satellite data.
British physicist Dr Joe Farman had been visiting and measuring ozone levels and concentrations of trace gases at Halley Bay, Antarctica, each year. He and his team collected observations with instruments such as weather balloons and the Dobson spectrophotometer – a simple ozone-measuring device, which worked best when wrapped in a blanket.
It was these rudimentary instruments that alerted Farman to the ozone hole, highlighting the importance of comprehensive Earth observations.
At first, no one knew what to make of the discovery, until further examination of the hole revealed that it could bring about catastrophic consequences if it was not contained and ‘healed’.
By 1987 the Montreal Protocol was created to reduce the worldwide use of ozone-depleting chemicals. Often referred to as ‘the world’s most successful environmental agreement’, without the Protocol and its success, the Earth’s ozone layer would collapse by 2050.
Why is ozone important?
A layer of ozone, a colourless gas, sits within the stratosphere – the atmosphere typically 10-50 km above the Earth’s surface. At this level, ozone absorbs the Sun’s UV radiation, preventing most of it from penetrating the atmosphere and reaching the Earth.
UV radiation damages the DNA of most living organisms. It would be difficult to survive without an ozone layer, with the amount of UV radiation hitting the planet’s surface.
In regards to the urban legend, even though the ozone hole doesn’t reach as far as Tasmania, more than enough UV already arrives at the rest of the Earth’s surface to cause serious damage to unprotected skin.
The Montreal Protocol brought the world together to agree to phase out ozone damaging chemicals, and the hole never extended to Australia.
Climate change would also be far worse without the Protocol, as the ozone depleting CFCs are super-greenhouse gases as well, thousands of times more potent than carbon dioxide.
However, stratospheric ozone recovery is still linked to climate change, and climate change is also linked to stratospheric ozone recovery.
Dr Matthew Woodhouse, from the CSIRO Aerosol and Chemistry Modelling group says, as an example, stratospheric ozone loss is responsible for increasing the strength of the westerly winds that encircle Antarctica, pushing them further south.
“When these winds are pushed south, they contribute to delivering very dry conditions to Australia,” he says.
“As the ozone hole recovers, it is predicted that these westerly winds could move further north. However, increasing greenhouse gas concentrations are also pushing the westerly winds south, opposing the influence of the recovering ozone hole,” he explains.
Modelling the effects of ozone and climate change
Such opposing effects can only be fully evaluated with a chemistry-climate model (CCM), which Woodhouse is currently working on developing for the Australian science community.
The model will provide a world-leading forecasting and evaluation capability, complementing Australia’s position as a leader in ozone depleting substance monitoring.
“The CCM will include representation of the ocean, the atmosphere, and its chemistry, including stratospheric ozone. It will be able to forecast ozone hole recovery under future scenarios, and assess the impact of ozone hole recovery on climate,” Woodhouse says.
So, how is the recovery going so far?
CSIRO monitors and reports on the ozone loss that occurs over Antarctica in the southern hemisphere every spring, in the ‘ozone hole season’.
Dr Paul Krummel, from CSIRO’s Climate Science Centre, says satellite data are used to monitor the ozone hole development each year, and to estimate its annual size and ‘depth’.
“Thanks to the Montreal Protocol the hole in the ozone layer is slowly recovering,” Krummel says.
“Since around the year 2000, when the levels of ozone depleting chemicals in the stratosphere peaked, the annual ozone hole generally grew smaller as ozone-destroying gases were phased out,” he continues.
“However, since many of the ozone-destroying gases will remain in the atmosphere for tens to hundreds of years, recovery of the Antarctic ozone hole to pre-1980 levels will likely take another 40-60 years.”
There are few southern hemisphere observation points for stratospheric ozone and ozone-depleting substances, so Australian observations are essential inputs for global assessments.
CSIRO, along with the Bureau of Meteorology, have a long-term program to monitor the levels of ozone depleting substances in the atmosphere at the Cape Grim observatory, located on the northwest tip of Tasmania.
Woodhouse adds that the ability of the new chemistry-climate model to assess recovery timescales is a valuable application.
“Something that is rarely acknowledged is the impact the chemical substitutes for those banned under the Montreal protocol and subsequent amendments are having on ozone hole recovery,” he says.
“As industry adapts and creates new processes, new chemicals are released,” he explains.
“Only with continued monitoring can these emissions be identified, and only with modelling can the effect on ozone recovery of those new chemicals be forecast and understood.”
Threats and disruptions
A recent study identified dichloromethane (Ch2Cl2) as an unexpected and growing danger to ozone depletion.
Dichloromethane is not currently controlled by the Montreal Protocol, and, if left uncontrolled, could push ozone recovery into the next century.
The substance, widely used for paint stripping, in the manufacture of PVC, in agricultural fumigation and the production of pharmaceuticals, is increasing much faster than previously thought.
The main sources of the emissions come from rapidly developing areas such as China and India. China is currently the largest producer of PVC, used mainly in construction materials.
Volcanic eruptions can also have a huge effect on the recovery of the ozone hole. Woodhouse says it’s not just volcanologists who are closely watching Bali’s Mt Agung.
“Large volcanic eruptions release gases and small particles called aerosol. When these particles reach the stratosphere, they provide a surface area where chemical reactions can take place. These reactions release highly reactive chlorine, which is very effective at eating ozone,” he explains.
In the aftermath of the 1991 Pinatubo eruption, Woodhouse says large losses of stratospheric ozone occurred.
“Interestingly, and worryingly, these ozone losses don’t occur over Antarctica in the springtime like the now-familiar ozone hole, but they can occur year-round over large parts of the globe, and last for years after the eruption.”
Each amendment to the Montreal Protocol is intended to update and strengthen it, and is based on global observations of significant ozone depletion. There is currently no expected date for the next amendment.
Krummel says the 2016 Kigali amendment was the latest change, which managed to get HFCs (hydrofluorocarbons) reductions included.
“Even though HFCs don’t destroy ozone, they are replacements for chemicals that do destroy the ozone layer,” Krummel says.
“What is significant about this is that HFCs are potent greenhouse gases, so in effect, the Montreal Protocol is also the most successful international agreement for reducing the effects of climate change.
“CFCs are very potent greenhouse gases as well.”
The UN Environment Program and the World Meteorological Organization are preparing the next report required under the Montreal Protocol – the Scientific Assessment of Ozone Depletion: 2018.
Many Australian scientists are contributing to the report, which will examine the recent state of the ozone layer, the atmospheric concentration of ozone-depleting chemicals, future ozone layer projections, and links with climate.
As one of the first countries to ratify the Montreal Protocol, Australia is often ahead of targets and is seen as a leader in the efforts to phase out ozone depleting substances.
These efforts, along with CSIRO’s world-class observations and chemistry-climate model capabilities, strengthen and inform the Protocol to limit ozone depletion, ensuring that the ozone hole remains manageable.
For example from the Ozone Mapping and Profiler Suite (OMPS) of instruments on the Suomi National Polar-orbiting Partnership satellite.
Each Southern Hemisphere spring, the CSIRO provides weekly reports on the progress of the Antarctic Ozone Hole.
For more on the CSIRO Atmospheric Monitoring and Modelling Group.
A 2012 series of articles on the ozone layer, marking the 25th anniversary of the Montreal Protocol in The Conversation.<|endoftext|>
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Paul's Online Notes
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### Section 4-5 : Miscellaneous Functions
The point of this section is to introduce you to some other functions that don’t really require the work to graph that the ones that we’ve looked at to this point in this chapter. For most of these all that we’ll need to do is evaluate the function as some $$x$$’s and then plot the points.
#### Constant Function
This is probably the easiest function that we’ll ever graph and yet it is one of the functions that tend to cause problems for students.
The most general form for the constant function is,
$f\left( x \right) = c$
where $$c$$ is some number.
Let’s take a look at $$f\left( x \right) = 4$$ so we can see what the graph of constant functions look like. Probably the biggest problem students have with these functions is that there are no $$x$$’s on the right side to plug into for evaluation. However, all that means is that there is no substitution to do. In other words, no matter what $$x$$ we plug into the function we will always get a value of 4 (or $$c$$ in the general case) out of the function.
So, every point has a $$y$$ coordinate of 4. This is exactly what defines a horizontal line. In fact, if we recall that $$f\left( x \right)$$ is nothing more than a fancy way of writing $$y$$ we can rewrite the function as,
$y = 4$
And this is exactly the equation of a horizontal line.
Here is the graph of this function.
#### Square Root
Next, we want to take a look at $$f\left( x \right) = \sqrt x$$. First, note that since we don’t want to get complex numbers out of a function evaluation we have to restrict the values of $$x$$ that we can plug in. We can only plug in value of $$x$$ in the range $$x \ge 0$$. This means that our graph will only exist in this range as well.
To get the graph we’ll just plug in some values of $$x$$ and then plot the points.
$$x$$ $$f(x)$$
0 0
1 1
4 2
9 3
The graph is then,
#### Absolute Value
We’ve dealt with this function several times already. It’s now time to graph it. First, let’s remind ourselves of the definition of the absolute value function.
$f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}x&{{\mbox{if }}x \ge 0}\\{ - x}&{{\mbox{if }}x < 0}\end{array}} \right.$
This is a piecewise function and we’ve seen how to graph these already. All that we need to do is get some points in both ranges and plot them.
Here are some function evaluations.
$$x$$ $$f(x)$$
0 0
1 1
-1 1
2 2
-2 2
Here is the graph of this function.
So, this is a “V” shaped graph.
#### Cubic Function
We’re not actually going to look at a general cubic polynomial here. We’ll do some of those in the next chapter. Here we are only going to look at $$f\left( x \right) = {x^3}$$. There really isn’t much to do here other than just plugging in some points and plotting.
$$x$$ $$f(x)$$
0 0
1 1
-1 -1
2 8
-2 -8
Here is the graph of this function.
We will need some of these in the next section so make sure that you can identify these when you see them and can sketch their graphs fairly quickly.<|endoftext|>
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# What is Math?
What is math? It can be compared to a very useful tool, or maybe a collection of tools. Sometimes textbooks concentrate a lot on teaching about the small details of each and every type of tool. But it’s also really important to focus on how and when to use the different tools. This is my practical approach to teaching the subject. And it’s also important to note that math is much more than just numbers! If you’re really good with shapes and how they relate, you might enjoy geometry. And if you are good at solving puzzles, chances are that logic will be a great match for your skills.
NOTE: Be sure to pause the video when the timer reaches 6:30 to work on the Earn, Break Even, or Lose problem.
# How to Multiply by 12
Do you think you’ll need to know how to multiply by 12 or 11 more? Think of it this way: how often do you need to figure out how many dozen you need of something? It comes up a lot more than needing to know how many batches of 11, doesn’t it? That’s because of the way we’ve decided to group things mathematically as a society.
Here’s why: We picked 12 based on how we used to count on our fingers using the “finger segment” system. If you look at your hands, you’ll notice that your index finger has three segments to it. So does your middle finger, ring finger, and pinkie. Since you have four fingers, you actually have 12 sections for counting with (we’re not including your thumb, which is the pointer… your thumb rests on the section you’re currently on). When your thumb touches the tip of your index finger, that means “1”. When your thumb touches the middle segment, that’s “2”, and the base segment is “3”. The tip of your middle finger is “4”, and so on. That’s how we came to use the 12-in-a-batch system.
If you’re wondering why we didn’t use the 24-in-a-batch system (because you have two hands), that’s because one hand was for 1-12 and the second hand indicated the number of batches of 12. So if your left hand has your thumb on the ring finger’s base segment (9) and your right hand has the thumb touching the index finger’s middle segment (2 complete batches of 12, or 2 x 12), the number you counted to is: 24 + 9 = 33.
Fortunately we now have calculators and a base-ten system, so this whole thing worked out well. But still the number 12 persists! So I thought you’d like this video, which expands on the idea of quickly multiplying two-digit numbers and three-digit numbers by eleven. This is very similar to the shortcut used when multiplying by eleven, but it also involves some doubling. Are you ready?
# Divisibility
Can you look at a number and tell right away if it’s divisible by another number? Well, it’s pretty easy for 2 – if it’s an even number, it’s definitely divisible by two. Testing whether a number is divisible by five is easy as well. How can you tell?
In this video, I’ll show you some tricks to determine if a number is divisible by 3, 4, 6 and 7 before you start to divide. Some are simple and fast and some are a bit more complex. These can be very useful tricks for working with larger numbers (or just really fun to play with for a bit).
# What Day Were You Born On?
This is not only a neat trick but a very practical skill – you can figure out the day of the week of anyone’s birthday.
If you were born in the 20th Century, (1900-1999), we can use math to find out which day of the week you were born. If you’re a little too young for this, try it with a parent or grandparent’s birthday. Watch the video and I’ll teach you exactly how it works.
# \$1 Word Search
Have you ever heard of a dollar word search? It’s a special kind of puzzle where the letters in a word add up to a coin value. For example, an A is worth a penny, the letter B is worth two cents, C is worth three cents, and so on. Are you completely confused? That’s okay! Just watch the video and I’ll show you how it all works.
# Isn’t that SUM-thing?
This is a neat trick that you can use to really puzzle your friends and family. If someone gives you a three-digit number, you can actually figure out what the end result will be after you’ve received two additional numbers, but before you actually know what those numbers are. Does this sound confusing? Watch the video and I’ll show you how it works.
Want a peek under the ‘hood’ of my brain when I do a mental math calculation? This video is a slow-motion, step-by-step snapshot of what goes on when I add numbers in my head. The first thing you need to learn is how to add from LEFT to RIGHT, which is opposite from most math classes out there. I’ll show you how to do this – it’s easy, and essential to working bigger numbers in your head.
Here’s what you do:
# Multiplying 2-Digit Numbers by 11
Here’s our first MATH lesson. It is so easy that one night, I wound up showing it to everyone in the pizza restaurant. Well, everyone who would listen, anyway. We were scribbling down the answers right on the pizza boxes with such excitement that I couldn’t help it – I started laughing right out loud about how excited everyone was about math… especially on a Saturday night.
When you do this calculation in front of friends or family, it’s more impressive if you hand a calculator out first and let them know that you are ‘testing to see if the calculator is working right’. Ask for a two digit number and have them check the calculator’s answer against yours.
If you really want to go crazy, you can have math races against the calculator and its operator, just as the Arthur Benjamin video shows. (Only you don’t need to do the squaring of five-digit numbers in your head!) Have fun!
# Multiplying 3-Digit Numbers by 11
If you can multiply 11 by any 2-digit number, then you can easily do any three digit number. There’s just an extra step, and make sure you always start adding near the ones so you can see where to carry the extra if needed.
# Mental Mathemagic
We’re going to throw in a few math lessons here and there, so if math really isn’t your thing, free free to just watch the videos and see what you think. All of these lessons require only a brain, and once in awhile paper and pencil, so this area is ‘materials-free’ and jam-packed with great mathematical content. If you’re the parent, stick a calculator in your pocket and test out your kids as they go along.
Some of what we cover here is based on the book “Secrets of Mental Math” by Arthur Benjamin, an incredible professor at Harvey Mudd College. He’s also known as the “Lightning Human Calculator”. Here’s a video about him you may enjoy:
We’re going to break down the steps to really getting to know numbers and put it into a form that both you and your kids can use everyday, including shopping at grocery stores, baking in the kitchen, working on the car, and figuring out your taxes. It’s a useflu tool for flexing your mind as well as appreciating the simplicity of the numerical world.
# Squaring Two-Digit Numbers
This neat little trick shortcuts the multiplication process by breaking it into easy chunks that your brain can handle. The first thing you need to do is multiply the digits together, then double that result and add a zero, and then square each digit separately, and finally add up the results.
Slightly confused? Great – we made a video that outlines each step. There’s a definite pattern and flow to it. With practice, you will be able to do this one in your head within a very short time. Have fun!
Squaring three-digit numbers is one of the most impressive mental math calculations, and it doesn’t take a whole lot of effort after you’ve mastered two-digits. It’s like the difference between juggling three balls and five balls. Most folks (with a bit of practice) can juggle three balls. Five objects, however, is a whole other story (and WOW factor).
Once you get the hang of squaring two-digit numbers, three-digit numbers aren’t so hard, but you have to keep track as you go along. Don’t get discouraged if you feel a little lost. It’s just like anything you try for the first time… when you’re new at something, in the beginning you aren’t very good at it. But with practice, these steps will become second nature and you’ll be able to impress your friends, relatives, and math teachers.
The video below has two parts:<|endoftext|>
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# How do you solve (x+4)/(1-x)<=0 using a sign chart?
Oct 17, 2016
$x \in \left\{\left(- \infty , - 4\right] , \left(+ 1 , + \infty\right)\right\}$
(see below for method using sign chart)
#### Explanation:
First consider the "boundary values" of $x$ for $\frac{x + 4}{1 - x}$.
The "boundary values" are the values of $x$ for which the numerator or denominator become equal to $0$.
For this example
$\textcolor{w h i t e}{\text{XXX}} x + 4 = 0 \rightarrow x = - 4$
and
$\textcolor{w h i t e}{\text{XXX}} 1 - x = 0 \rightarrow x = 1$
So our boundary values are $\left\{- 4 , + 1\right\}$
and teh sign table looks like
{: (,"||",(-oo,-4),"|",-4,"|",(-4:+1),"|",+1,"|",(+1:+oo),"|"), (,,,,,,,,,,,), (x+4,"||",-ve,"|",0,"|",+ve,"|",+ve,"|",+ve,"|"), (1-x,"||",+ve,"|",+ve,"|",+ve,"|",0,"|",-ve,"|"), (,,,,,,,,,,,), ((x+4)/(1-x),"||",-ve,"|",0,"|",+ve,"|","undef'n","|",-ve,"|"), (,,,,,,,,,,,), ((x+4)/(1-x)<=0,"||","True","|","True","|","False","|","undef'n","|","True","|") :}
So $\frac{x + 4}{1 - x} \le 0 \textcolor{w h i t e}{\text{X}}$ for $x \le - 4 \textcolor{\setminus}{w h i t e} \left(\text{X}\right)$ or $\textcolor{w h i t e}{\text{X}} x > 1$<|endoftext|>
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GESTURES –voluntary or involuntary movements of a human body or its part that are usually performed simultaneously with the speech or serve as its substitute. In most cases, G. possess a certain meaning and show the speaker’s communicative intention. According to some linguists, only significant hand movements that directly contribute to communication and make the speech more expressive can be qualified as G.
In multimodal communication (which means that people simultaneously use several channels to transmit and receive information) the role of G. is extremely important. Speech and G. act as parts of an integral multimodal system, and their interaction contributes to a fuller understanding of the process of communication.
Numerous works of Russian and foreign linguists are dedicated to G. R. Jakobson states that oral communication is primal, and gestures can only add a certain meaning in appropriate circumstances [Jakobson 1972]. According to A. Kendon, speech and gestures represent an indivisible unit, and G. complement the verbal part of the utterance, making it more expressive and thus helping the speaker to fulfill a particular communicative intention [Kendon 2004]. C. Müller says that G., being a certain combination of hand movements and representing visual information, help create a combination of knowledge and experience on a conceptual level [Müller 2013].
As far as various G. classifications are concerned, scholars consider different G. characteristics to be the most important ones, which prevents the appearance of a typology that would be generally accepted. G. can be classified according to their function in the process of communication, their correlation with the speech, their orientation in space, etc. One of the most widespread G. classifications was introduced by P. Ekman and W. Friesen [Ekman, Friesen 1969]; it is partially based on D. Efron’s typology [Efron 1941/1972] and highlights four G. types:
1) “Emblems” possess a certain lexical meaning and can be easily understood by all representatives of a given group (class, society). They often replace a certain word or phrase that would otherwise be expressed with the help of verbal means. A particular hand movement used when greeting or saying goodbye to a person, shrugging one’s shoulders or showing a “thumb up” (a sign of approval in most countries) belong to this gesture type;
2) “Regulators” support and guide the communication between interactants. These G. serve as a certain hint at the continuation of the utterance; they can also express a demand to repeat what has been said or to give the floor to another speaker. Among the most popular G. of this type are a handshake, a head nod (a sign of agreement in most countries), eyebrow raising (shows surprise) or a slight body movement towards the interlocutor during a conversation;
3) “Illustrators” help describe what is being said, in other words, they serve as visual “illustration” of the utterance. They can repeat or substitute particular sentences or phrases, contradict the given information or intensify the meaning of the utterance, if the speaker wants to emphasize it. “Illustrators” appear with the speech, and do not have any common standard form.
4) G. belonging to “adaptors” do not contribute to the communication directly, and can serve as a certain psychological characteristic of a person. Fixing one’s hair, touching the interlocutor, holding and touching small objects in one’ hands can be qualified as “adaptors”. This G. type is closely related to the speaker’s emotional state, helping him adjust to a particular situation and feel more comfortable. Such G. are spontaneous and are not perceived by the speaker on a conscious level.
There also exist other G. typologies. For instance, D. McNeill distinguishes between iconic G. (having a direct connection with the referent, resembling the object described) and metaphoric G. (used to express abstract notions) [McNeill 1992/1995]. E. Grishina examines and pays special attention to single and recurrent G., as well as to brief (equal to one word) and prolonged G. (lasting more than one accentuated word) [Гришина 2014].
As far as G. functions are concerned, one of the most significant ones is the communicative function. According to G. Kreydlin, G. can also perform the regulative function, controlling the interactants’ verbal behavior, as well as the representational function (showing the psychological state of the speaker and his attitude towards the interlocutor). The deictic function of G., serving to describe certain objects is also of great significance [Kreydlin 2002].
Гришина Е. Жесты и прагматические характеристики высказывания // Мультимодальная коммуникация: теоретические и эмпирические исследования. – М.: Буки-Веди, 2014.
Крейдлин Г. Невербальная семиотика: Язык тела и естественный язык. – М.: «Новое литературное обозрение», 2002.
Якобсон Р. К вопросу о зрительных и слуховых знаках //Семиотика и искусствометрия. – М: Мир, 1972.
Efron D. Gesture and environment. // N.Y., 1941. 2nd edition: Gesture, race and culture. – The Hague: Mouton, 1972.
Ekman P., Friesen W.V. The Repertoire of nonverbal behavior: categories, origins, usage, and coding. – Semiotica 1 (1), 1969.
Kendon A. Gesture: Visible action as utterance. – Cambridge: Cambridge University Press, 2004.
McNeill D. Hand and Mind: What gestures reveal about thought. – Chicago: Chicago University Press, 1992/1995.
Müller C. Gestural modes of representation as techniques of depiction // Body – Language – Communication: An International Handbook on Multimodality in Human Interaction. Volume 2. / Ed. Müller C., Cienki A., Frickle E., Ladewig S., McNeill D., Bressem J. – Berlin: De Gruyter Mouton, 2013. – P. 1687–1702.<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 12.11: Multiplying Binomials
Difficulty Level: At Grade Created by: CK-12
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Practice Multiplying Binomials Mentally
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Have you ever been to a community garden? Take a look at this dilemma.
The students loved walking through the community garden which was adjacent to the town hall. All of the flowers were in full bloom. One of the community workers was there and he presented the students with the following problem. The students suspected that this was the work of their teacher Mr. Travis, but they began to work on solving the problem anyway. Here it is.
A farmer has two rectangular fields. One measures 3x+7\begin{align*}3x + 7\end{align*} by 2x4\begin{align*}2x - 4\end{align*}. The other measures x2+1\begin{align*}x^2 + 1\end{align*} by 6x+5\begin{align*}6x + 5\end{align*}. Find the combined area of the two fields.
To figure this out, you will need to understand multiplying binomials. Because we find the area of a rectangle through multiplication and both of these fields have measurements written in binomials, understanding binomials will be important in solving this problem.
### Guidance
We defined binomials as two-term polynomials.
When we added and subtracted polynomials, we were careful to combine like terms. When we multiply polynomials we will carefully apply the rules of exponents, as well.
When we multiply binomials, we can use a table to help us to organize and keep track of the information. This table will help you to organize and keep track of your work. Let’s take a look.
Multiply the binomials (x+5)(x+3)\begin{align*}(x+5)(x+3)\end{align*}.
We can use a table like a rectangle, as if each of the binomials were a dimension of the rectangle.
We will insert the two binomials along the sides of the table like a rectangle.
Now, we will find the area of the four separate rectangles.
The dimensions of the first rectangle is xx\begin{align*}x \cdot x\end{align*}, the second is 5x\begin{align*}5 \cdot x\end{align*}, the third is 3x\begin{align*}3 \cdot x\end{align*}, and the fourth is 35\begin{align*}3 \cdot 5\end{align*}.
In order to find the total, we will add the four areas: x2+5x+3x+15\begin{align*}x^2 + 5x +3x +15\end{align*}
Now, combine like terms carefully: x2+8x+15\begin{align*}x^2 + 8x + 15\end{align*}.
Here is one that is a little different.
Multiply (5x8)2\begin{align*}(5x - 8)^2\end{align*}.
Remember that the exponent applies to the entire binomial such that (5x8)2=(5x8)(5x8)\begin{align*}(5x -8)^2 = (5x - 8)(5x - 8)\end{align*}.
5x\begin{align*}5x\end{align*} 8\begin{align*}-8\end{align*}
5x\begin{align*}5x\end{align*} 25x2\begin{align*}25x^2\end{align*} 40x\begin{align*}-40x\end{align*}
8\begin{align*}-8\end{align*} 40x\begin{align*}-40x\end{align*} 64\begin{align*}64\end{align*}
25x240x40x+64=25x280x+64\begin{align*}& 25x^2-40x-40x+64 \\ &=25x^2-80x +64\end{align*}
A second method for multiplying binomials is similar to the algorithm that we commonly use for multiplying two-digit numbers.
thousandshundredstensones73 70 +3 ×81 81 +173 70 +3 +4580 4000+ 500+ 80 +04653 4000+ 500+150 +3\begin{align*}& \qquad \qquad \qquad \quad \text{thousands} \qquad \quad \text{hundreds} \qquad \quad \text{tens} \qquad \quad \text{ones} \\ & \quad \quad 73 \ \quad \rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 70 \ \quad + \qquad 3 \\ & \ \underline{\quad \times 81} \ \quad \rightarrow \ \ \underline{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \quad 81 \ \quad + \qquad 1} \\ & \quad \quad 73\ \quad \rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 70 \ \quad + \qquad 3 \\ & \ \underline{+4580} \ \quad \rightarrow \ \ \underline{\ \ \qquad 4000 \qquad + \qquad \ 500 \quad + \ \quad 80 \ \quad + \qquad 0} \\ & \quad 4653 \ \quad \leftarrow \ \quad \qquad 4000 \qquad + \ \qquad 500 \quad + \quad 150 \ \quad + \qquad 3 \end{align*}
When you expand the multiplication like is done on the right, you can see that in our multiplication algorithm for two-digit numbers, we line up numbers by similar places.
We will use this same idea to multiply binomials but instead of using decimal places, we will line up the products by like terms.
Take a look at this one.
Multiply (3x+2)(5x+4)\begin{align*}(3x + 2)(5x + 4)\end{align*}
2ndpower 1stpower0 power 3x+ 25x+ 4 12x+ 815x2+ 10x +15x2+ 22x+ 815x2+22x+8\begin{align*}& 2^{nd} \text{power} \qquad \ \quad 1^{st} \text{power} \qquad \quad 0 \ \text{power} \\ & \qquad \qquad \qquad \qquad \ \ \ \quad 3x \quad + \qquad \ \quad 2 \\ & \underline{\qquad \qquad \qquad \qquad \qquad 5x \quad + \qquad \ \quad 4 \;\;} \\ & \qquad \qquad \qquad \qquad \ \ \quad 12x \quad + \qquad \ \quad 8 \\ & \underline{\qquad 15x^2 \quad + \qquad \ \quad 10x \ \quad + \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \qquad 15x^2 \quad + \qquad \ \quad 22x \quad + \qquad \ \quad 8 \quad \rightarrow \quad 15x^2 + 22x + 8\end{align*}
Just as in multiplication with numerals, we first multiply 4 by 2 and get 8. We then multiply 4 by 3x\begin{align*}3x\end{align*} and get 12x\begin{align*}12x\end{align*}. When we multiply 5x\begin{align*}5x\end{align*} by 2, we get 10x\begin{align*}10x\end{align*}. This is not a like term of 8 but of 12x\begin{align*}12x\end{align*} so the 10x\begin{align*}10x\end{align*} was aligned beneath the 12x\begin{align*}12x\end{align*}. Finally, the product of 5x\begin{align*}5x\end{align*} and 3x\begin{align*}3x\end{align*} is 15x2\begin{align*}15x^2\end{align*}. This is to the 2nd\begin{align*}2^{nd}\end{align*} power so how no like terms. Our sum on the bottom-right shows the final product.
A third way is to use “FOIL”. “FOIL” is an acronym which tells us which terms to multiply in order to get our product—
F—First terms in the binomials
O—Outside terms in the binomials
I—Inside terms in the binomials
L—Last terms in the binomials
Let’s multiply (2x+8)(5x13)\begin{align*}(2x+8)(5x-13)\end{align*} using the FOIL method.
F First terms are 2x\begin{align*}2x\end{align*} and 5x\begin{align*}5x\end{align*} \begin{align*}(\mathbf{\underline{2x}}+8)(\mathbf{\underline{5x}}-13)\end{align*} \begin{align*}10x^2\end{align*}
O Outside terms are \begin{align*}2x\end{align*} and -13 \begin{align*}(\mathbf{\underline{2x}}+8)(5x\underline{\mathbf{-13}})\end{align*} \begin{align*}-26x\end{align*}
I Inside terms are 8 and \begin{align*}5x\end{align*} \begin{align*}(2x\underline{\mathbf{+8}})(\mathbf{\underline{5x}}-13)\end{align*} \begin{align*}40x\end{align*}
L Last terms are 8 and -13 \begin{align*}(2x\underline{\mathbf{+8}})(5x\underline{\mathbf{-13}})\end{align*} -104
The table above helps to illustrate which terms must be multiplied. However, we don’t need to make a table like that for each multiplication. We can show it like this:
\begin{align*}(2x+8)(5x-13)=10x^2 - 26x + 40x - 104 = 10x^2 + 14x - 104\end{align*}
Of the three methods we’ve seen for multiplication, you might agree that this is the quickest method. Of course, all three methods should give us the same product.
Write the acronym FOIL and what each letter stands for in your notebook.
Take a look at one more.
Multiply \begin{align*}(5x^3+2x)(7x^2+8)\end{align*}
\begin{align*}&(5x^3+2x)(7x^2+8) \\ &= 5x^3 \cdot 7x^2 + 5x^3 \cdot 8 + 2x \cdot 7x^2 + 2x \cdot 8 \\ &= 35x^5 +40x^3 +14x^3 +16x \\ &= 35x^5 +54x^3 +16x \end{align*}
Multiply the following binomials.
#### Example A
\begin{align*}(x+2)(x+4)\end{align*}
Solution: \begin{align*}x^2+6x+8\end{align*}
#### Example B
\begin{align*}(x-6)(x+5)\end{align*}
Solution: \begin{align*}x^2-x-30\end{align*}
#### Example C
\begin{align*}(x+3)(x-3)\end{align*}
Solution: \begin{align*}x^2-9\end{align*}
Now let's go back to the dilemma from the beginning of the Concept.
Remember, we need to work on figuring out the area of both rectangles and then the sum of those two areas will give us the total area. Here is how we can solve this problem.
\begin{align*}&(3x+7)(2x-4)+(x^2+1)(6x+5) \\ &= (3x \cdot 2x + 3x \cdot -4 + 7 \cdot 2x + 7 \cdot -4) + (x^2 \cdot 6x +x^2 \cdot 5 + 1 \cdot 6x + 1 \cdot 5) \\ &= (6x^2 -12x +14x -28) + (6x^3 +5x^2 + 6x +5) \\ &= 6x^2 + 2x - 28 +6x^3 + 5x^2 +6x +5 \\ &= 6x^3 + 11x^2 + 8x -23 \end{align*}
### Vocabulary
Binomials
polynomials with two terms in them.
Perfect Square Trinomial
a trinomial where the first term and the last term are perfect squares because the binomial was to the second power or squared.
FOIL
firsts, outers, inners, lasts – binomial multiplication
### Guided Practice
Here is one for you to try on your own.
Multiply by using a table.
Multiply \begin{align*}(x-4)(x+6)\end{align*}
Solution
\begin{align*}x\end{align*} \begin{align*}-4\end{align*}
\begin{align*}x\end{align*} \begin{align*}x^2\end{align*} \begin{align*}-4x\end{align*}
\begin{align*}+6\end{align*} \begin{align*}6x\end{align*} \begin{align*}-24\end{align*}
\begin{align*}& x^2-4x+6x-24 \\ &= x^2+2x-24\end{align*}
Notice the careful work with the negative and positive signs.
### Practice
Directions: Use a table to multiply the following binomials.
1. \begin{align*}(x+3)(x+5)\end{align*}
2. \begin{align*}(x-3)(x-5)\end{align*}
3. \begin{align*}(x+3)(x-3)\end{align*}
4. \begin{align*}(x+2)(x-8)\end{align*}
5. \begin{align*}(3x^2+3x)(6x-2)\end{align*}
6. \begin{align*}(2x-7y)(5x+4y)\end{align*}
7. \begin{align*}(2x-9)^2\end{align*}
Directions: Multiply the following binomials vertically.
1. \begin{align*}(d+2)(4d-1)\end{align*}
2. \begin{align*}(5x+7)(5x-7)\end{align*}
3. \begin{align*}(4b^2+3c)(2b-5c^2)\end{align*}
Directions: Multiply the following binomials using the FOIL method:
1. \begin{align*}(p+6)(5p+2)\end{align*}
2. \begin{align*}(-7y^2-4y)(6y+2)\end{align*}
3. \begin{align*}(x^3+3x)^2\end{align*}
4. \begin{align*}(2x+1)(x-4)\end{align*}
5. \begin{align*}(3x-3)(5x+9\end{align*})
6. \begin{align*}(x+5)^2\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Binomial
A binomial is an expression with two terms. The prefix 'bi' means 'two'.
FOIL
FOIL is an acronym used to remember a technique for multiplying two binomials. You multiply the FIRST terms, OUTSIDE terms, INSIDE terms, and LAST terms and then combine any like terms.
Perfect Square Trinomial
A perfect square trinomial is a quadratic expression of the form $a^2+2ab+b^2$ (which can be rewritten as $(a+b)^2$) or $a^2-2ab+b^2$ (which can be rewritten as $(a-b)^2$).
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Authors:
Tags:
Subjects:<|endoftext|>
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Blue water supplies – those from rain water that flows over land and becomes surface water and groundwater – have been regularly studied and are fairly well documented. Green water supplies, however, aren’t as well studied.
Our friend and the father of the water footprint concept Arjen Hoekstra and his colleagues have gone a long way toward closing that knowledge gap by assessing green water scarcity throughout the world. They mapped out their results, which were recently published in the journal PNAS, and offered a regional and country-by-country analysis of green water use.
The study looked at green water uses by humans and nature and examined the potential limitations between the two. People typically use rainwater for food and fiber crops, timber, bioenergy resources and raising livestock. Human uses of water can often happen at the expense of the ecosystems that also rely on that water for biodiversity and survival.
According to the authors, ” The limit to our direct rainwater use has been reached or even exceeded in many places.” [Phys.org]<|endoftext|>
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# Application - Solving System of Equations Using Matrices
## Application - Solving System of Equations Using Matrices - How it Works - Video
### First Steps
First Steps:
Our first steps is with the system as a coefficient matrix. Then our next step is add our constant terms as another column to form an augmented matrix. Now we have to do some row operations to make our coefficient part of the augmented matrix into an identity matrix.
### Example 1 - Step 1
Example 1 - Step 1:
Here we had 2 * R1 - R2 to replace R2 and the reason is that we want all the numbers beneath a1,1 or 1 in this case to be 0.
The main number that we are focusing on here is the 2 underneath the 1. We multiplied the first row by 2 so that we can get 2, -4, 6, 8. So we get 0, -5, 10, 5 for our new R2. Our main focus is 2 - 2 = 0 for the element a2,1.
If we get any other 1s or 0s, then that is great. If not, more arithmetic.
### Example 1 - Step 2
Example 1 - Step 2:
Here we had 3 * R1 + R3 to replace R3 and the reason is that we want all the numbers beneath a1,1 or 1 in this case to be 0.
The main number that we are focusing on here is the -3 underneath the 1. We multiplied the first row by 3 so that we can get 3, -6, 9, 12. So we get 0, -2, 10, 14 for our new R3. Our main focus is 3 + (-3) = 0 for the element a3,1.
If we get any other 1s or 0s, then that is great. If not, more arithmetic.
### Example 1 - Step 3
Example 1 - Step 3:
Here we had (-1/5) * R2 to replace R2 and the reason is that we want the first number to be 1 or a2,2.
The main number that we are focusing on here is the -5 underneath the -2. We multiplied the second row by (-1/5) so that we can get 0, 1, -2, -1, which is our new R2. Our main focus is (-1/5) * -5 = 1 for the element a2,2.
Since all of the elements in this row are multiples of 5, we don't get any fractions. Otherwise I would wait until all the numbers not in the principal diagonal are 0. Then multiply by a scalar to get 1.
If we get any other 1s or 0s, then that is great. If not, more arithmetic.
### Example 1 - Step 4
Example 1 - Step 4:
Here we had (-1/2) * R3 to replace R3 and the reason is that we want the first number to be 1 or a3,2.
The main number that we are focusing on here is the -2 underneath the 1. We multiplied the third row by (-1/2) so that we can get 0, 1, -5, -7, which is our new R3. Our main focus is (-1/2) * -2 = 1 for the element a3,2.
Since all of the elements in this row are multiples of 2, we don't get any fractions. Otherwise I would wait until all the numbers not in the principal diagonal are 0. Then multiply by a scalar to get 1.
If we get any other 1s or 0s, then that is great. If not, more arithmetic.
### Example 1 - Step 5
Example 1 - Step 5:
Here we had R2 - R3 to replace R3 and the reason is that we want all the numbers beneath a2,2 or 1 in this case to be 0.
The main number that we are focusing on here is the 1 underneath the 1. We subtract the third row from the second row so that we can get 0, 0, 3, 6 which is our new R3. Our main focus is 1 - 1 = 0 for the element a3,2.
If we get any other 1s or 0s, then that is great. If not, more arithmetic.
### Example 1 - Step 6
Example 1 - Step 6:
Here we had (1/3) * R3 to replace R3 and the reason is that we want the first number to be 1 or a3,3.
The main number that we are focusing on here is the 3 underneath the -2. We multiplied the third row by (1/3) so that we can get 0, 0, 1, 2, which is our new R3. Our main focus is (1/3) * 3 = 1 for the element a3,3.
Since all of the elements in this row are multiples of 2, we don't get any fractions. Otherwise I would wait until all the numbers not in the principal diagonal are 0. Then multiply by a scalar to get 1.
If we get any other 1s or 0s, then that is great. If not, more arithmetic.
### Example 1 - Answer with Substitution
Example 1 - Answer with Substitution:
Since we row echelon here, we can use substitution to solve for the variables. The last row is z = 2. Once we substitute that back into for the second row, we get y = 3. Finally, we substitute y and z to find x and we get x = 4.
Weirdly the answers are the same as the constant that rarely happens, but on occasion math is cool like that.
### Example 1 - Step 7
Example 1 - Step 7:
Here we had R1 + 2 * R2 to replace R1 and the reason is that we want all the numbers above a2,2 or 1 in this case to be 0.
The main number that we are focusing on here is the -2 above the 1. We multiply the second row to get 1, -2, 3, 4. Now we add the first row and 2 times the second row. So we get 1, 0, -1, 2 for our new R1. Our main focus is (-2) + 2 = 0 for the element a1,2.
If we get any other 1s or 0s, then that is great. If not, more arithmetic.
### Example 1 - Step 8
Example 1 - Step 8:
Here we had R1 + R2 to replace R1 and the reason is that we want all the numbers above a3,3 or 1 in this case to be 0.
The main number that we are focusing on here is the -1 above the -2. We add the first row and the third row so that we can get 1, 0, 0, 4 which is our new R1. Our main focus is (-1) + 1 = 0 for the element a1,3.
If we get any other 1s or 0s, then that is great. If not, more arithmetic.
### Example 1 - Step 9
Example 1 - Step 9:
Here we had R2 + 2 * R3 to replace R2 and the reason is that we want all the numbers above a3,3 or 1 in this case to be 0.
The main number that we are focusing on here is the -2 above the 1. We multiply the third row by 2 to get 0, 0, 2, 4. Now we add the second row and two times the third row so that we can get 1, 0, 0, 4 which is our new R1. Our main focus is (-2) + 2 = 0 for the element a2,3.
If we get any other 1s or 0s, then that is great. If not, more arithmetic.
### Example 1 - Answer with Gauss-Jordan
Example 1 - Answer with Gauss-Jordan:
After all of that, we found the answers for each variable; x = 2; y = 3; and z = 2.
Weirdly the answers are the same as the constant that rarely happens, but on occasion math is cool like that.<|endoftext|>
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Broad spectrum revolution
The broad spectrum revolution (BSR) hypothesis, proposed by Kent Flannery in a 1968 paper presented to a London University symposium, suggested that the emergence of the Neolithic in southwest Asia was prefaced by increases in dietary breadth among foraging societies. The broad spectrum revolution followed the most recent ice age around 15,000 BP in the Middle East and 12,000 BP in Europe. During this time, there was a transition from focusing on a few main food sources to gathering/hunting a "broad spectrum" of plants and animals.
Flannery's hypothesis was meant to help explain the adoption of agriculture in the Neolithic Revolution. Unpersuaded by "the facile explanation of prehistoric environmental change" Flannery suggested (following Lewis Binford's equilibrium model) that population growth in optimal habitats led to demographic pressure within nearby marginal habitats as daughter groups migrated. The search for more food within these marginal habitats forced foragers to diversify the types of food sources harvested, broadening the subsistence base outward to include more fish, small game, waterfowl, invertebrates (such as snails and shellfish), as well as previously ignored or marginal plant sources. Most importantly, Flannery argues that the need for more food in these marginal environments led to the deliberate cultivation of certain plant species, especially cereals. In optimal habitats, these plants naturally grew in relatively dense stands, but required human intervention in order to be efficiently harvested in marginal zones. Thus, the broad spectrum revolution set the stage for domestication and rise of permanent agricultural settlement.
A BSR is likely to manifest as both an increased spectrum of food resources and an evenness in the exploitation of high- and low-value prey. Under a broad spectrum economy a greater amount of low-value prey (i.e. high cost-to-benefit ratio) would be included because there are insufficient high-value prey to reliably satisfy a population's needs. In terms of plants, it would be expected that foodstuffs that had once been ignored because of difficulty of extraction were now included in a diet. In terms of fauna, animal prey which was previously considered an inefficient use of resources (particularly small, fast mammals or fish) could now also be worthwhile. In other words, increasing scarcity made the extra effort necessary for survival.
In the Middle East, the broad spectrum revolution led to an increase in the production of food. The growth and reproduction of certain plants and animals became vastly popular. Because large animals became quite scarce, people had to find new resources of food and tools elsewhere. Interests focused on smaller game like fish, rabbits, and shellfish because the reproduction rate of small animals is much greater than that of large animals.
The most commonly accepted stimulation for the BSR is demographic pressures on the landscape, under which over-exploitation of resources meant narrow diets restricted to high-value prey could no longer feed the expanding population.
The broad-spectrum revolution has also been linked to climatic changes, including sea level rises during which:
- Conditions became more inviting to marine life offshore in shallow, warm waters.
- Quantity and variety of marine life increased drastically as did the number of edible species.
- Because the rivers' power weakened with rising waters, the currents flowing into the ocean were slow enough to allow salmon and other fish ascend upstream to spawn.
- Birds found refuge next to riverbeds in marsh grasses and then proceeded to migrate across Europe in the wintertime.
The Japanese site Nittano (inlet near Tokyo) was occupied several times between 6000 and 5000 BP. The Jōmon culture occupied Nittano at over 30,000 sites known in Japan. People hunted deer, pigs, bears, antelope, fish, shellfish, and gathered plants. Sites have yielded over 300 sample remains of shellfish and 180 sample remains of plants.
The broad spectrum revolution has been a subject of intense debate since it was first proposed, but its basic arguments are well-supported.
- Flannery 1969
- Flannery 1969: 75
- Stiner 2004: 6993
- Steiner 2001 and Weiss 2004
- Kent Flannery, "Origins and Ecological Effects of Early Domestication in Iran and the Near East," The Domestication and Exploitation of Plants and Animals, eds. Peter J. Ucko and G.W. Dimbleby (Chicago: Aldine Publishing Co., 1969), 73-100
- Mary Stiner, "Thirty Years on the 'Broad Spectrum Revolution' and Paleolithic Demography," PNAS, 98, no. 13 (2001): 6993-6996;
- Ehud Weiss et al., "The Broad Spectrum Revisited: Evidence from Plant Remains," PNAS, 101, no. 26 (2004): 9551-9555<|endoftext|>
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1. ## Simplify
Simplify the expression
My final answer is $\displaystyle \frac{2\sqrt{16-x^2}}{2x^2-16}$but its telling me I am wrong
$\displaystyle \frac{2\tan(\alpha)}{1-\tan^2(\alpha)}$ this then works down eventually to my answer or is my frist step wrong?
I found my mistake the answer is $\displaystyle \frac{2x\sqrt{16-x^2}}{2x^2-16}$
2. ## Re: Simplify
Hello, M670!
You are missing an $\displaystyle x.$
$\displaystyle \text{Simplify: }\;\tan\left[2\cos^{-1}\left(\frac{x}{4}\right)\right]$
$\displaystyle \text{Let }\theta = \cos^{-1}\left(\frac{x}{4}\right) \quad\Rightarrow\quad \cos\theta \:=\:\frac{x}{4} \:=\:\frac{adj}{hyp}$
$\displaystyle \theta\text{ is in a right triangle with: }\,adj = x,\;hyp = 4$
$\displaystyle \text{Pythagorus says: }\,opp = \sqrt{16-x^2}$
$\displaystyle \text{Hence: }\:\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{\sqrt{16-x^2}}{x}$
$\displaystyle \text{The problem becomes: }\:\tan2\theta \;=\;\dfrac{2\tan\theta}{1 - \tan^2\!\theta} \;=\;\dfrac{2\frac{\sqrt{16-x^2}}{x}}{1 - \frac{16-x^2}{x^2}}$
$\displaystyle \text{Multiply by }\frac{x^2}{x^2}\!:\;\;\dfrac{2x\sqrt{16-x^2}}{2x^2-16} \;=\;\frac{x\sqrt{16-x^2}}{x^2-8}$
You found your error . . . Good!
3. ## Re: Simplify
Originally Posted by M670
Simplify the expression
My final answer is $\displaystyle \frac{2\sqrt{16-x^2}}{2x^2-16}$but its telling me I am wrong
$\displaystyle \frac{2\tan(\alpha)}{1-\tan^2(\alpha)}$ this then works down eventually to my answer or is my frist step wrong?
I found my mistake the answer is $\displaystyle \frac{2x\sqrt{16-x^2}}{2x^2-16}$
First, I would write \displaystyle \displaystyle \begin{align*} \tan{(\theta)} = \frac{\sin{(\theta)}}{\cos{(\theta)}} = \frac{\sqrt{1 - \cos^2{(\theta)}}}{\cos{(\theta)}} \end{align*} (ignoring the plus/minus signs for the moment). Then that means
\displaystyle \displaystyle \begin{align*} \tan{\left[ 2\arccos{\left( \frac{x}{4} \right)} \right]} &= \frac{\sqrt{1 - \cos^2{\left[ 2\arccos{\left( \frac{x}{4} \right)} \right]}}}{\cos{\left[ 2\arccos{\left(\frac{x}{4}\right)}\right]}} \end{align*}
Then make use of \displaystyle \displaystyle \begin{align*} \cos{(2\theta)} = 2\cos^2{(\theta)} - 1 \end{align*} so that
\displaystyle \displaystyle \begin{align*} \frac{\sqrt{1-\cos^2{\left[2\arccos{\left(\frac{x}{4}\right)}\right]}}}{\cos{\left[2\arccos{\left(\frac{x}{4}\right)}\right]}} &= \frac{\sqrt{1 - \left\{ 2\cos^2{\left[\arccos{\left(\frac{x}{4}\right)}\right]} - 1 \right\}^2}}{2\cos^2{\left[\arccos{\left(\frac{x}{4}\right)}\right]}-1} \\ &= \frac{\sqrt{1 - \left[ 2\left(\frac{x}{4}\right)^2 - 1 \right]^2}}{2\left(\frac{x}{4}\right)^2-1} \\ &= \frac{\sqrt{1 - \left[ 2\left( \frac{x^2}{16} \right) - 1 \right]^2}}{2\left( \frac{x^2}{16} \right) - 1} \\ &= \frac{\sqrt{1 - \left( \frac{x^2 - 8}{8} \right)^2}}{\frac{x^2 - 8}{8}} \\ &= \frac{8\sqrt{\frac{8^2 - \left( x^2 - 8 \right)^2}{8^2}}}{x^2 - 8} \\ &= \frac{\sqrt{8^2 - \left( x^2 - 8 \right)^2}}{x^2 - 8} \\ &= \frac{\sqrt{\left[ 8 - \left(x^2 - 8 \right) \right] \left[ 8 + \left( x^2 - 8 \right) \right] }}{x^2 - 8} \\ &= \frac{\sqrt{x^2 \left( 16 - x^2 \right) }}{x^2 - 8} \\ &= \frac{ x \sqrt{ 16 - x^2 } }{ x^2 - 8 } \end{align*}<|endoftext|>
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# PROBABILITY THEORY
```SINGLE RANDOM VARIABLES
AND PROBABILITY DISTRIBUTION
Probability
Trial and Event: Consider an experiment, which though repeated under essential and
identical conditions, does not give a unique result but may result in any one of the
several possible outcomes. The experiment is known as Trial and the outcome is
called Event.
E.g. (1) Throwing a dice experiment getting the n’s 1,2,3,4,5,6 (event)
(2) Tossing a coin experiment and getting head or tail (event)
Probability
Introduction
Probability is the likelihood or chance of an event occurring.
Probability = the number of ways of achieving success
the total number of possible outcomes
For example, the probability of flipping a coin and it being heads is ½, because there
is 1 way of getting a head and the total number of possible outcomes is 2 (a head or
tail). We write P(heads) = ½ .
The probability of something which is certain to happen is 1.
The probability of something which is impossible to happen is 0.
The probability of something not happening is 1 minus the probability that it
will happen.
Single Events:
Example
There are 6 beads in a bag, 3 are red, 2 are yellow and 1 is blue. What is the
probability of picking a yellow?
The probability is the number of yellows in the bag divided by the total number of
balls, i.e. 2/6 = 1/3.
Example
There is a bag full of colored balls, red, blue, green and orange. Balls are picked out
and replaced. John did this 1000 times and obtained the following results:
Number of blue balls picked out: 300
Number of red balls: 200
Number of green balls: 450
Number of orange balls: 50
a) What is the probability of picking a green ball?
For every 1000 balls picked out, 450 are green. Therefore P(green) = 450/1000 =
0.45
b) If there are 100 balls in the bag, how many of them are likely to be green?
The experiment suggests that 450 out of 1000 balls are green. Therefore, out of 100
balls, 45 are green (using ratios).
Multiple Events
Independent and Dependent Events
Suppose now we consider the probability of 2 events happening. For example, we
might throw 2 dice and consider the probability that both are 6's.
We call two events independent if the outcome of one of the events doesn't affect
the outcome of another. For example, if we throw two dice, the probability of getting
a 6 on the second die is the same, no matter what we get with the first one- it's still
1/6.
On the other hand, suppose we have a bag containing 2 red and 2 blue balls. If we
pick 2 balls out of the bag, the probability that the second is blue depends upon what
the color of the first ball picked was. If the first ball was blue, there will be 1 blue
and 2 red balls in the bag when we pick the second ball. So the probability of getting
a blue is 1/3. However, if the first ball was red, there will be 1 red and 2 blue balls
left so the probability the second ball is blue is 2/3. When the probability of one
event depends on another, the events are dependent.
Exhaustive Events:
The total no. of possible outcomes in any trial is called exhaustive event.
E.g.: (1) In tossing of a coin experiment there are two exhaustive events.
(2) In throwing an n-dice experiment, there are 6n exhaustive events.
Favorable event:
The no of cases favorable to an event in a trial is the no of outcomes which entities
the happening of the event.
E.g. (1) In tossing a coin, there is one and only one favorable case to get either
Mutually exclusive Event: If two or more of them cannot happen simultaneously
in the same trial then the event are called mutually exclusive event.
E.g. In throwing a dice experiment, the events 1,2,3, ------6 are M.E. events
Equally likely Events: Outcomes of events are said to be equally likely if there is
no reason for one to be preferred over other.
E.g. tossing a coin. Chance of getting 1,2,3,4,5,6 is equally likely.
Independent Event:
Several events are said to be independent if the happening or the non-happening of
the event is not affected by the concerning of the occurrence of any one of the
remaining events.
An event that always happen is called Certain event, it is denoted by “S‟. An event
that never happens is called Impossible event, it is denoted by “ “.
Eg: In tossing a coin and throwing a die, getting head or tail is independent of
getting no’s 1 or 2 or 3 or 4 or 5 or 6.
```<|endoftext|>
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The familiar odor of a just-struck match is caused by sulfur dioxide (SO2), a heavy, colorless, poisonous gas. Its chief uses are in the preparation of sulfuric acid and other sulfur-containing compounds (see sulfur; sulfuric acid). It is also used as a disinfectant, a refrigerant, a bleach, and as a food preservative, especially in dried fruits.
Large quantities of sulfur dioxide are formed and released into the air during the combustion of sulfur-containing fuels. Consequently, sulfur dioxide has become the most widespread man-made pollutant. As such, measurements of the gas are often used as an index of air pollution. In the atmosphere, sulfur dioxide is converted to sulfur trioxide (SO3), which is absorbed by water vapor. When it rains or snows, this acidic moisture falls to the Earth as acid rain. Acid rain poses a serious threat to the environment. It increases the acidity of soils, lakes, and rivers, endangering the organisms that live there. Acid rain also attacks limestone and marble and has severely damaged many buildings and outdoor sculptures in cities that have high levels of air pollution (see acid rain). In the second half of the 20th century, measures to control sulfur-dioxide pollution were being widely adopted. (See also pollution, environmental).
Sulfur dioxide occurs naturally in volcanic gases and in solution in the waters of some warm springs. It is prepared industrially by burning sulfur or certain sulfur compounds, such as iron pyrite or copper pyrite, in the presence of air or oxygen. In the laboratory sulfur dioxide may be prepared by reducing sulfuric acid (H2SO4) to sulfurous acid (H2SO3), which decomposes into water and sulfur dioxide. It is also sometimes prepared by treating sulfites (salts of sulfurous acid) with strong acids, such as hydrochloric acid, to form sulfurous acid. As in the previous process, the sulfurous acid is then decomposed into water and sulfur dioxide.
At room temperature sulfur-dioxide gas can be turned into a liquid if the gas is subjected to moderate pressures. Under atmospheric pressure liquid sulfur dioxide freezes at –99.4° F (–73° C) and boils at +14° F (–10° C).<|endoftext|>
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# Thread: Show that y(x) satisfies equation
1. ## Show that y(x) satisfies equation
Assume that the implicit function $e^{xy} = y$ determines y as a differentiable function of x in some interval. Without attempting to solve the equation for y(x), show that y(x) satisfies the differential equation $(1-xy)y' - y^2 = 0$.
I started by finding the derivative of y.
$y' = y'e^x \Rightarrow$
$e^{xy} = 1$
I am unsure of what to do next to answer the question.
EDIT: Looking at this problem more, I realized that I forgot to use the multiplication rule when finding the derivative of $xy$
2. ## Re: Show that y(x) satisfies equation
Originally Posted by Mike9182
Assume that the implicit function $e^{xy} = y$ determines y as a differentiable function of x in some interval. Without attempting to solve the equation for y(x), show that y(x) satisfies the differential equation $(1-xy)y' - y^2 = 0$.
I started by finding the derivative of y.
$y' = y'e^x \Rightarrow$
$e^{xy} = 1$
I am unsure of what to do next to answer the question.
I think your derivative is incorrect... Let $\displaystyle u = xy \implies y = e^u$, then
\displaystyle \begin{align*} \frac{du}{dx} &= x\frac{dy}{dx} + y \\ \\ \frac{dy}{du} &= e^u \\ &= e^{xy} \\ \\ \frac{dy}{dx} &= e^{xy}\left(x\frac{dy}{dx} + y\right) \\ \frac{dy}{dx} &= x\,e^{xy}\frac{dy}{dx} + y\,e^{xy} \\ \frac{dy}{dx} - x\,e^{xy}\frac{dy}{dx} &= y\,e^{xy} \\ \frac{dy}{dx}\left(1 - x\,e^{xy}\right) &= y\,e^{xy} \\ \frac{dy}{dx} &= \frac{y\,e^{xy}}{1 - x\,e^{xy}} \end{align*}
Now substitute this into the original DE.
3. ## Re: Show that y(x) satisfies equation
I substituted the result into the DE and then played around with re-arranging things, however I am still having trouble proving the equality after I substitute.
4. ## Re: Show that y(x) satisfies equation
Show us your work, and we might be able to see where the problem is.
5. ## Re: Show that y(x) satisfies equation
$(1 - xy)(y' - y^2) = 0 (Original DE) \Rightarrow$
$(1 - xe^{xy})(\frac{ye^{xy}}{1-xe^{xy}} - y^2) = 0 \Rightarrow$
$ye^{xy} - (e^{xy})^2(1-xe^{xy}) = 0 \Rightarrow$
$ye^{xy}-(e^{xy})^2 + x(e^{xy})^3 = 0 \Rightarrow$
$e^{xy}(y - e^{xy} + x(e^{xy})^2) = 0 \Rightarrow$
$e^{xy} - e^{xy} + x(e^{xy})^2 = 0 \Rightarrow$
$e^{xy}(1 - 1 + x(e^{xy}) = 0 \Rightarrow$
$e^{xy}(xe^{xy}) = 0$
6. ## Re: Show that y(x) satisfies equation
Originally Posted by Mike9182
$(1 - xy)(y' - y^2) = 0 (Original DE) \Rightarrow$
This is not the original DE. The original DE is
$(1 - xy)y' - y^{2} = 0.$
Another suggestion: instead of writing all those exponentials, I would use your implicit equation for y to get rid of them in the derivative. Less writing that way, though it should work out in either case.
$(1 - xe^{xy})(\frac{ye^{xy}}{1-xe^{xy}} - y^2) = 0 \Rightarrow$
$ye^{xy} - (e^{xy})^2(1-xe^{xy}) = 0 \Rightarrow$
$ye^{xy}-(e^{xy})^2 + x(e^{xy})^3 = 0 \Rightarrow$
$e^{xy}(y - e^{xy} + x(e^{xy})^2) = 0 \Rightarrow$
$e^{xy} - e^{xy} + x(e^{xy})^2 = 0 \Rightarrow$
$e^{xy}(1 - 1 + x(e^{xy}) = 0 \Rightarrow$
$e^{xy}(xe^{xy}) = 0$
,
,
,
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### allinurl: show Y????????? ?? Y??????? ?????: ?????????????? ?????
Click on a term to search for related topics.<|endoftext|>
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Jump to navigation Jump to search
In linguistics and semiotics, a notation is a system of graphics or symbols, characters and abbreviated expressions, used in artistic and scientific disciplines to represent technical facts and quantities by convention.
- Quotes are arranged alphabetically by author
A - F
- There was yet another disadvantage attaching to the whole of Newton’s physical inquiries, … the want of an appropriate notation for expressing the conditions of a dynamical problem, and the general principles by which its solution must be obtained. By the labours of LaGrange, the motions of a disturbed planet are reduced with all their complication and variety to a purely mathematical question. It then ceases to be a physical problem; the disturbed and disturbing planet are alike vanished: the ideas of time and force are at an end; the very elements of the orbit have disappeared, or only exist as arbitrary characters in a mathematical formula.
- George Boole in: Artificial Intelligence and Symbolic Computation:International Conference AISC 2000 Madrid, Spain, July 17-19, 2000, Springer Science & Business Media, 25 April 2001, p. 3
- Aryabhata's work, called Aryabhatiya, is composed of three parts, in only the first of which use is made of a special notation of numbers. It is an alphabetical system in which the twenty-five consonants represent 1-25, respectively; other letters stand for 30, 40, …., 100 etc. The other mathematical parts of Aryabhata consists of rules without examples. Another alphabetic system prevailed in Southern India, the numbers 1-19 being designated by consonants, etc.
- The process of writing has something infinite about it. Even though it is interrupted each night, it is one single notation.
- A term used in Linguistics and Phonetics to refer to any system of Graphic representation of speech (as in Phonemic notation, where the term transcription is widely used). Specifically it refers to the set of symbols which represent a mode of linguistic analysis, as in the ‘Phrase-structure notation’ in general grammar. An analytic convention, in this sense, which is introduced into an analysis to facilitate the formulation of a statement such as a Rule, is often termed a notational device, e.g. the use of ( ) to indicate optionality in generative syntax.
- An English sentence like 'every whale is a mammal' transcribes directly into algebraic notation as '-W+M'. Similarly, its equivalent, 'no non-mammals are whales', transcribes directly as '-(+(-M)+W)'. Sentences that come ready made for direct transcription are called “Canonical”.
- George Englebretsen, in: Something to Reckon with: The Logic of Terms, University of Ottawa Press, 1 January 1996, p. xiv
- The fact is that, while Leibniz did insist on a subject-predicate analysis of statements, this attitude was well thought out, and the failures of his logic are due only to his inability to devise an appropriate system of notation for his logical algorithm.
- George Englebretsen in: "Something to Reckon with: The Logic of Terms”, p. 34
- Instead, he [De Morgan] introduced his "spicular" notation. In this notational scheme, parentheses are used to indicate the quantity (distribution) of a term, and negation is indicated by a dot or the use of a lowercase term letter. Distribution is indicated by a parenthesis facing the term, otherwise parenthesis faces away from the term.
- George Englebretsen in: "Something to Reckon with: The Logic of Terms”, p. 45
- More specifically, we must not say this in a language of "adequate notation" (6.122; also 3.325, 5.533, 5.534). In other words, in a logically adequate, or correct, notation—-a Begriflrchrift (4. l273)...a notation in which sentences consist only of simple names naming simple objects.
- George Englebretsen, in: "Something to Reckon with: The Logic of Terms”, p. 112
- Peano's scheme [system of symbolization for logic] became the dominant system of notation for the majority of mathematical logicians; Russell's popularity and the influence of Principia Mathematica, which made use of Peano's notation, were mainly responsible for this.
- George Englebretsen, in: "Something to Reckon with: The Logic of Terms”, P. 233
- We could, of course, use any notation we want; do not laugh at notations; invent them, they are powerful. In fact,mathematics is, to a large extent, invention of better notations.
G - L
- Any phase of the growth of mathematical notation is an interesting study, but the chief educational lesson to be derived is that notation always grows too slowly. Older and inferior forms possess remarkable longevity, and the newer and superior forms appear feeble and backward. We have noted the state of transition in the sixteenth century from the Roman to the Hindu system of characters, the introduction of symbols of operation +, -, and the slow growth towards decimal notation.
- Lambert Lincoln Jackson in: Florian Cajori :A History of Mathematical Notations", p. 228
- There are now two systems of notations, giving the same formal results, one of which gives them with self-evident force and meaning, the other by dark and symbolic processes. The burden of proof is shifted, and it must be for the author or the supporter of the dark system to show that it is in some way superior to the evident system.
- Jevons in: “Something to Reckon with: The Logic of Terms”, p. 99
- All my paintings are usually done in drawing form, very small. I make notations in drawings first, and then I make a collage for color. But drawing is always my notation. And I think artists all work that way really. I'm not special. But I like plants, and I don't think anyone else draws like this, today. I'm special in that way.
- Such is the advantage of a well-constructed language that its simplified notation often becomes the source of profound theories.
- It is India that gave us the ingenious method of expressing all numbers by means of ten symbols, each symbol receiving a value of position as well as an bsolute value; a profound and important idea which appears so simple to us now that we ignore its true merit. But its very simplicity and the great ease with which it has lent to all computations put our arithmetic in the first rank of useful inventions and we shall appreciate the grandeur of this achievement the more when we remember that it escaped the genius of Archimedes and Apollonius, two of the greatest men produced by antiquity.
- Piere Simon Laplace, in Joseph Mazur Enlightening Symbols: A Short History of Mathematical Notation and Its Hidden Powers, Princeton University Press, 23 March 2014, p. 38
- It is worth noting that the notation facilitates discovery. This, in a most wonderful way, reduces the mind's labour.
- Gottfried Wilhelm Leibniz in: Eberhard Zeidler Quantum Field Theory III: Gauge Theory: A Bridge between Mathematicians and Physicists, Springer Science & Business Media, 17 August 2011, p. 439
- As the mother teaches her children how to express themselves in their language, so one Gypsy musician teaches the other. They have never shown any need for notation.
- Franz Liszt in: Arthur Friedheim Life and Liszt: The Recollections of a Concert Pianist, Courier Dover Publications, 13 June 2013, p. 270
- There is also general agreement that notations operating on a level higher than the word fall outside writing proper, although there are exceptions. In his typology Hill (1967, 93) includes a category 'discourse systems' which he considers as one of the main three divisions of writing systems.
- Hartmut Guenther, Otto Ludwig in: Schrift und Schriftlichkeit (Writing and Its Use): Ein Interdisziplinaeres Handbuch Internationaler Forschung - An Interdisciplinary Handbook of International Research, Volume 2, Walter de Gruyter, 1 January 1996, p. 1381
- A writing system is wider than a spelling system. English uses many w:Logogramslogograms which have as unambiguous correspondence with a single morpheme but give no indications of pronunciations:1,2,3 etc, L, @, &, %, +. Many more are in use in mathematics and formal logic. Other specialized uses include proofreading symbols (e.g. for delete, insert). Many of these symbols are not specific to English, and more detailed discussion would shade off into special purpose writing systems such as shorthands and scientific notations. However, “normal” uses of English also include a wide range of such forms.
M - R
- We are driven by the usual insatiable curiosity of the scientist, and our work is a delightful game. I am frequently astonished that it so often results in correct predictions of experimental results. How can it be that writing down a few simple and elegant formulae, like short poems governed by strict rules such as those of the sonnet or the waka, can predict universal regularities of Nature? Perhaps we see equations as simple because they are easily expressed in terms of mathematical notation already invented at an earlier stage of development of the science, and thus what appears to us as w:Eleganceelegance of description really reflects the interconnectedness of Nature's laws at different levels.
- Roman-based alphabetic notations make use of Roman letters, sometimes with diacritical marks, graphic transformations, and other letters added. The alphabet adopted by the International Phonetic Association (IPA) in 1888 is Roman based. This system has become widely recognized as the international phonetic alphabet. Following Daniel Jones (1914) phoneticians have distinguished between the more detailed narrow and the less specific broad transcription. Among the latter systems are phonemic notations, which transcribe only phonemes and no phonetic variants.
- Winfried Nöth in: Handbook of Semiotics, Indiana University Press, 1 January 1995, P.258
- Alphabetic notations represent the articulatory features of speech sounds by a composite formula of symbols indicating the active speech organs. One of the Phoenicians who has developed such symbols is Pike.
- Winfried Nöth in: "Handbook of Semiotics", p. 258
- An important step in solving a problem is to choose the notation. It should be done carefully. The time we spend now on choosing the notation carefully may be repaid by the time we save later by avoiding hesitation and confusion.
- G.Polya in : “Something to Reckon with: The Logic of Terms”, p. 99
- Imagine someone so infatuated by a band that they have every different pressing of every album the band made. Most of the time, the only difference in the album is the matrix number or a different 'made in' notation on the back cover or label. This is enough to make some people extremely excited. Actually, much more than excited.
- Henry Rollins in: Henry Rollins: The Column! Are You Collector Scum?, LA Weekly, Jan 19, 2012
- Language is music. Written words are musical notation. The music of a piece of fiction establishes the way in which it is to be read, and, in the largest sense, what it means. It is essential to remember that characters have a music as well, a pitch and tempo, just as real people do. To make them believable, you must always be aware of what they would or would not say, where stresses would or would not fall.
- A good notation [is]... like a live teacher.
- Bertrand Russel in : : “Something to Reckon with: The Logic of Terms”, p. 99
S - Z
- After all, quantity, like quality, is oppositional, which suggests a +/- notation. To determine which quantity is + and which - Sommers relies on the equivalences established by valid conversion.
- F. Sommers in: "Something to Reckon with: The Logic of Terms”, p. 112
- Traditional logic's failure to do so [incorporating relations into term logic] was due not to its syntax, but to (i) the absence of an adequate system of formal notation, and (ii) the failure to make explicit the logical forms of pronouns.
- F.Sommers in: "Something to Reckon with: The Logic of Terms”, p. 128
- My being a teacher had a decisive influence on making language and systems as simple as possible so that in my teaching, I could concentrate on the essential issues of programming rather than on details of language and notation.<|endoftext|>
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# Sum of Two Odd Powers/Examples/Sum of Two Cubes
## Theorem
$x^3 + y^3 = \paren {x + y} \paren {x^2 - x y + y^2}$
## Proof 1
$\ds a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
Let $x = a$ and $y = -b$.
Then:
$\ds x^3 + y^3$ $=$ $\ds x^3 - \paren {-y^3}$ $\ds$ $=$ $\ds x^3 - \paren {-y}^3$ $\ds$ $=$ $\ds \paren {x - \paren {-y} } \sum_{j \mathop = 0}^2 x^{n - j - 1} \paren {-y}^j$ $\ds$ $=$ $\ds \paren {x + y} \paren {x^2 + x \paren {-y} + \paren {-y}^2}$ $\ds$ $=$ $\ds \paren {x + y} \paren {x^2 - x y + y^2}$
$\blacksquare$
## Proof 2
$a^{2 n + 1} + b^{2 n + 1} = \paren {a + b} \paren {a^{2 n} - a^{2 n - 1} b + a^{2 n - 2} b^2 - \dotsb + a b^{2 n - 1} + b^{2 n} }$
We have that $3 = 2 \times 1 + 1$.
Hence setting $n = 1$ gives the result.
$\blacksquare$<|endoftext|>
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The log band saw, which Schrenkeisen was one of the first firms to adopt, made it possible for large boards to be cut straight more quickly and efficiently than the older frame and sash saws, although its major benefit was in its economy of wood. The smaller band saw, which could easily maneuver small pieces of wood that required specific curves, was also a relatively recent development.
Despite the proliferation of machinery at large-scale factories such as that operated by M. & H. Schrenkeisen, some work still required significant manpower. The process of sanding was eventually simplified with the introduction of belt-powered sanders, as seen here, but the sanding process continued to be finished by hand until the 1870s. Likewise, hand-carving, the best paid of the specialized furniture tasks, was often necessary. In the case of Schrenkeisen, the firm prided itself on providing affordable hand-carved details as late as the 1890s.
The finishing and upholstery of furniture were typically carried out on the first floor of large-scale furniture factories. Upholstery and carving were the more highly paid professions, and mid-century workers could earn up to thirty dollars a week. In this image, we see two finishers checking for sound construction in a fashionable settee and a side table. A figure in the background completes the upholstery, probably with nails and hammer, on the back and seat of another settee, which is set on a simple worker’s bench. Women workers took on particular upholstery work in some factories, such as Herter Brothers. This side chair, with its original upholstery, was most likely the product of a New York factory.
The two six-story buildings of the German furniture makers M. & H. Schrenkeisen produced an enormous array of parlor furniture. This image depicts many factory activities in a sequence from the top left, where logs are first cut, to the finishing and upholstery of furniture at the bottom right. This idealized image, however, elides the distinction between factory owners and laborers, whose relationships were often vexed. By 1880, the date of this Scientific American illustration and article, large factories like that of M. & H. Schrenkeisen utilized the latest technologies for the production of furniture.
The greatest threat to furniture workers came not from replacement by machinery but from the availability of poorly trained immigrants who were willing to work for as little as five dollars a week. The New York Times covered the strikes of Schrenkeisen workers hoping for higher wages and fewer hours from the 1860s through the 1880s. Just months before that article appeared, Schrenkeisen workers protesting for a reduction of the requisite sixty-hour work week were fired. The image of the idyllic factory setting and its committed workers may have served as a booster for Schrenkeisen at a time when they were losing employees.
M. & H. Schrenkeisen
M. & H. Schrenkeisen, which specialized in the production of parlor furniture for the middle class, was one of many immigrant furniture makers located on the Lower East Side, known in the nineteenth century as Klein Deutschland (Little Germany). Although many smaller German workshops dotting the streets of Little Germany faded in and out over the course of the second half of the century, factories like that of Schrenkeisen on Elizabeth Street successfully produced on a large scale, employing the latest technologies and maintaining a predominantly male workforce.<|endoftext|>
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The cucumber is a plant whose eponymous fruit is mainly eaten raw in Europe. The cucumber belongs to the Cucumis sativus species, which has been cultivated for over 3000 years. Dozens of cucumbers can grow on one stalk. The present, cultivated cucumber only has female flowers from which the cucumber grows without pollination or fertilisation.
The cucumber plant is a creeper with self-adhesive vines, but can also be cultivated horizontally in the ground. In the day before there were greenhouses, the cucumbers were cultivated in trenches with hay and cow manure under flat glass. Later on the flat glass was erected and became a greenhouse. In glasshouses the plants are led up along binding wire. Since around 1960, the cucumber has been cultivated in heated greenhouses in the Netherlands which means that cucumbers cultivated in the Netherlands are available all year round. In these greenhouses the cucumbers will, after they’ve bloomed, grow into full-fledged cucumbers in 10 to 14 days in these greenhouses; so from a cucumber of less than 50 grams into a cucumber of approximately 350-400 grams. Two to three cultivations a year are planted and harvested each year to ensure yearlong delivery.
Cucumbers From Drenthe
Drenthe Growers BV cultivates cucumbers on 12,5 hectares. Various fungi, virus diseases and pests form big risks for cultivation. Insect pests such as aphids, spider mites or trips are controlled biologically. The cucumber mosaic virus is also a big problem. That’s why Drenthe Growers BV has strict hygiene rules to prevent spreading of diseases and pests as much as possible.
Continuous innovation is important for Drenthe Growers BV as well: to stand still is to move backwards. Various advisers and keeping up to date with the latest developments enables Drenthe Growers BV to remain a modern company.<|endoftext|>
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# Difference between revisions of "Graph Theory"
Jump to: navigation, search
Many problems are naturally formulated in terms of points and connections between them. For example, a computer network has PCs connected by cables, an airline map has cities connected by routes, and a school has rooms connected by hallways. A graph is a mathematical object which models such situations.
## Overview
A graph is a collection of vertices and edges. An edge is a connection between two vertices (sometimes referred to as nodes). One can draw a graph by marking points for the vertices and drawing lines connecting them for the edges, but the graph is defined independently of the visualrepresentation. For example, the following two drawings represent the same graph:
## Terminology
The precise way to represent this graph is to identify its set of vertices {A, B, C, D, E, F, G, H, I, J, K, L, M}, and its set of edges between these vertices {AG, AB, AC, LM, JM, JL, JK, ED, FD, HI, FE, AF, GE}.
A path from vertex x to y in a graph is a list of vertices, in which successive vertices are connected by edges in the graph. For example, BAFEG is path from B to G in the graph above. A simple path is a path with no vertex repeated. For example, BAFEGAC is not a simple path.
A graph is connected if there is a path from every vertex to every other vertex in the graph. Intuitively, if the vertices were physical objects and the edges were strings connecting them, a connected graph would stay in one piece if picked up by any vertex. A graph which is not connected is made up of connected components. For example, the graph above has three connected components: {I, H}, {J, K, L, M} and {A, B, C, D, E, F, G}.
A cycle is a path, which is simple except that the first and last vertex are the same (a path from a point back to itself). For example, the path AFEGA is a cycle in our example. Vertices must be listed in the order that they are traveled to make the path; any of the vertices may be listed first. Thus, FEGAF and GAFEG are different ways to identify the same cycle. For clarity, we list the start / end vertex twice: once at the start of the cycle and once at the end.
We’ll denote the number of vertices in a given graph by V, the number of edges by E. Note that E can range anywhere from V to V2 (or 1/2 V(V-1) in an undirected graph). Graphs will all edges present are called complete graphs; graphs with relatively few edges present (say less than V log(V)) are called sparse; graphs with relatively few edges missing are called dense.
## Trees
A graph with no cycles is called a tree. There is only one path between any two nodes in a tree. A tree on N vertices contains exactly N-1 edges. A spanning tree of a graph is a subgraph that contains all the vertices and forms a tree. A group of disconnected trees is called a forest.
Directed graphs are graphs which have a direction associated with each edge. An edge xy in a directed graph can be used in a path that goes from x to y but not necessarily from y to x. For example, a directed graph similar to our example graph is drawn below:
There is only one directed path from D to F. Note that there are two edges between H and I, one each direction, which essentially makes an undirected edge. An undirected graph can be thought of as a directed graph with all edges occurring in pairs in this way. A dag (directed acyclic graph) is a directed graph with no cycles.
## Adjacency Matrices
It is frequently convenient to represent a graph by a matrix, as shown in the second sample problem below. If we consider vertex A as 1, B as 2, etc., then a “one” in M at row i and column j indicates that there is a path from vertex i to j. If we raise M to the pth power, the resulting matrix indicates which paths of length p exist in the graph. The value in \$M^p(i,j)\$ is the number of paths from vertex i to vertex j.
## Sample Problems
### Sample Problem 1
Find the number of different cycles contained in the directed graph with vertices {A, B, C, D, E} and edges {AB, BA, BC, CD, DC, DB, DE}.
Solution: The graph is as follows:
By inspection, the cycles are: {A,B}, {B,C,D} and {C,D}. Thus, there are 3 cycles in the graph.
### Sample Problem 2
In the following directed graph, find the total number of different paths from vertex A to vertex C of length 2 or 4.
Solution:
Let matrix M represent the graph. Recall that the number of paths from vertex i to vertex j of length p equals M p[i,j]. The values of M, M 2 and M 4 are:
There is 1 path of length 2 (M 2[1,3]) and 3 paths of length 4 (M 4[1,3]).
### Sample Problem 3
Given the adjacency matrix, draw the directed graph.
Solution:
The vertices do not have to be in those exact locations, but the edges must be the same. In particular, there must be exactly 7 edges: AB, AD, BA, BD, CA, DB, and DC. Here are two valid drawings of the graph:
## Video Resources
### ACSL Videos
The following YouTube videos show ACSL students and advisors working out some ACSL problems that have appeared in previous contests. Some of the videos contain ads; ACSL is not responsible for the ads and does not receive compensation in any form for those ads.
ACSL - Graph Theory Worksheet 1 (misterminich) Shows the solution to the ACSL problem: 'Draw the directed graph with vertices A, B, C, D, E and the directed edges AB, BC, AD, BC, DC, ED, and EA. ACSL Graph Theory Worksheet 3 (misterminich) Shows the solution to an ACSL problem asking to find how many paths of a specific length there are in a specific directed graph. Graph Theory ACSL Example Problem (Tangerine Code) Shows the solution to an ACSL problem asking to find how many different cycles there are from a specific vertex in a specific directed graph.
### Other Videos
There are many dozens of YouTube videos that cover introductions to graph theory. The following are a couple that we found particularly nicely done. Be aware thqt the videos contain ads; ACSL is not responsible for the ads and does not receive compensation in any form for those ads.
CS 106B Lecture: Graphs: basic concepts This lecture from Stanford University is s wonderful introduction to graph theory. The lecture starts out with many examples of real world problems that are modeled by graphs, and then moves on to review basic terminology relating to graph theory. Data structures: Introduction to Graphs (mycodeschool) A nice introduction to the fundamental concepts of graphs.<|endoftext|>
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A review of integration techniques
We review common techniques to compute indefinite and definite integrals.
Area between curves
We introduce the procedure of “Slice, Approximate, Integrate” and use it study the area of a region between two curves using the definite integral.
We can also use the procedure of “Slice, Approximate, Integrate” to set up integrals to compute volumes.
What is a solid of revolution?
We define a solid of revolution and discuss how to find the volume of one in two different ways.
The washer method
We use the procedure of “Slice, Approximate, Integrate” to develop the washer method to compute volumes of solids of revolution.
The shell method
We use the procedure of “Slice, Approximate, Integrate” to develop the shell method to compute volumes of solids of revolution.
Length of curves
We can use the procedure of “Slice, Approximate, Integrate” to find the length of curves.
Surface areas of revolution
We compute surface area of a frustrum then use the method of “Slice, Approximate, Integrate” to find areas of surface areas of revolution.
We apply the procedure of “Slice, Approximate, Integrate” to model physical situations.
Integration by parts
We learn a new technique, called integration by parts, to help find antiderivatives of certain types of products by reexamining the product rule for differentiation.
We can use substitution and trigonometric identities to find antiderivatives of certain types of trigonometric functions.
We can use limits to integrate functions on unbounded domains or functions with unbounded range.
Representing sequences visually
We can graph the terms of a sequence and find functions of a real variable that coincide with sequences on their common domains.
The alternating series test
Alternating series are series whose terms alternate in sign between positive and negative. There is a powerful convergence test for alternating series.
Absolute and Conditional Convergence
The basic question we wish to answer about a series is whether or not the series converges. If a series has both positive and negative terms, we can refine this question and ask whether or not the series converges when all terms are replaced by their absolute values. This is the distinction between absolute and conditional convergence, which we explore in this section.
Slope fields and Euler’s method
We describe numerical and graphical methods for understanding differential equations.
Separable differential equations
Separable differential equations are those in which the dependent and independent variables can be separated on opposite sides of the equation.
The Dot Product
The dot product is an important operation between vectors that captures geometric information.
Projections and orthogonal decomposition
Projections tell us how much of one vector lies in the direction of another and are important in physical applications.<|endoftext|>
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When George Eastman’s Kodak box camera was introduced in 1888, its popularity spawned an identity crisis of sorts within the photographic community. The widespread availability and relatively low cost of the Kodak camera and its ease of use resulted in an explosive surge in the number of new photographers. This led to the creation of millions of photographs, characterized by a small number of strong images obscured in a sea of mediocrity. While the huge “snapshot” market provided the financial strength to sustain the growing photography industry, it challenged serious amateur and professional photographers to differentiate their work from that of casual shooters. It seemed the solution could be found in two parts: through either technical excellence, or through artistic merit.
Ironically, this divided answer to photographic distinction paralleled the differences between the two most popular processes in early photography. The daguerreotype exhibited exquisitely high image sharpness, resolution, and detail, arguably capturing a more accurate rendering of the subject. In contrast, the calotype presented a softer, grainier, more ethereal image quality, perhaps encouraging a more artistically interpretive approach to the subject. The debate over these two approaches, to either document a technically accurate record of the subject, or to exercise aesthetic interpretation, has fueled contentious photographic art movements from the 1880’s to the modern day.
In the earliest days of photography, many saw the discipline as a technical one requiring expertise in chemistry and optics, and the mastery of an arcane workflow. They viewed its output as a scientific, or clinical record of the physical world. To a great extent, photography’s early adherents supported this view, by photographing landscapes, architecture, personalities, and events with the intent to factually inform the viewing public. Whereas paintings had previously served this function, they had also been conduits for artistic expression. With the advent of photography, many saw it as a superior recording medium, but without artistic value. As the field of photography grew and matured, a growing faction came to think otherwise, seeing photography as a new means of artistic expression.
The call for photography to be viewed as art came early in the medium’s history. From 1840 to 1860, the main emphasis on photography was on improving the tools and processes to yield sharp images, directly from nature and not manipulated, as faithful as possible to the original scene. This could be considered the original approach to “straight” photography. The first call for naturalism, or straight photography, came from England’s Peter Henry Emerson, in his 1890 book, Naturalistic Photography for Students of the Art, in which he advocated that a “proper” photograph demonstrate accurate and unaltered images of nature; the book’s title page quotes Keats, “Beauty is truth, truth beauty, –that is all Ye know on earth, and all ye need to know.” To a great extent, the members of Britain’s Royal Photographic Society and other early national photographic bodies shared this view. This naturalistic approach called for realistic versus painterly images, and promoted taking pictures of real, or natural scenes, as opposed to staged or contrived poses to re-enact classic paintings or literary scenes. On this point, Emerson essentially dismissed some earlier evocative work, such as Julia Margaret Cameron’s biblical and literary scenes, and Oscar Gustav Rejlander’s elaborate photomontages, as clumsy clichés.
By the late 1800’s, the technology and knowledge of photography had advanced and broadened to enable a variety of techniques to capture and manipulate images on negatives, plates, and prints. This led to another view, first articulated in Henry Peach Robinson’s 1869 book, Pictorial Effect in Photography: Being Hints On Composition And Chiaroscuro For Photographers. Robinson contended that pictorial photography was primarily symbolic or allegorical; an image could convey a message beyond its technical record of a subject. For example, evocative lighting could imply divinity, or a beautiful woman could symbolize Love. In this vein, manipulation of the image to serve this metaphoric message could be a key characteristic. This combination of artistic intent and the photographer’s hand-on manipulation of the image generally defined the idea underlying what came to be known as pictorialism. In pictorialism, the aesthetics, emotional impact, and artistic quality of the final image take primacy over the subject matter in front of the lens. The concept of photographic pictorialism developed in parallel with that of the Tonalism style of painting, exemplified by artists such as James McNeill Whistler. Both Tonalism and pictorialism encouraged the artist to generate an emotional response in the viewer through the use of atmospheric elements and subdued tonalities to convey a mood in the image.
The hand of man on the final image was another important concept underlying pictorialism. If the final image was purely the result of the mechanical properties of the camera and lens, and the chemical processes employed in development, i.e., a scientific outcome, then how could it be art? On the other hand, if the photographer employed creative techniques at the time of capture, combined with manipulation of the image in the darkroom, a unique artistic vision could be realized – only through the artist’s intervention could the final image be achieved. These techniques might include: multiple exposures; the use of soft focus, dramatic lighting, negative manipulation (scratching or painting over the negative; and such nuanced alternative printing techniques as, platinum printing (improved tonality); direct carbon printing; photogravure; combination printing (from multiple negatives); gum bichromate printing (colored prints); brushed watercolor pigments mixed with light-sensitized gum Arabic; and bromoil and oil pigment printing. These various “ennobling” techniques could yield images that exhibited colored tints, textures, charcoal or graphite drawing effects, or impressionist painterly effects. Regardless of the specific technique employed, the key was creating a handcrafted unique art object equivalent to a painting.
Pictorialism was the first photographic ideology recognized as an artistic movement, but it was not accepted without debate. The emphasis of established bodies like the Royal Photography Society on scientific and commercial aspects of photography, their view of photography as a primarily documentary medium, and their rejection of photography as an art form, led to various groups “seceding” from the established photographic orthodoxy to follow the pictorialist ideology. The first such group, England’s Linked Ring, was founded by Robinson and others in 1892, and included invited-only photographers such as Frank M. Sutcliffe, Frederick H. Evans, and Alvin Langdon Colburn. Americans Alfred Stieglitz and Clarence H. White were later admitted, and in 1900 Gertrude Käsebier was among the first women invited.
The success of The Linked Ring in advancing pictorialism, along with similar secessionist movements in France, Germany, and Austria led to the development in 1902 of the American Photo-Secession movement in New York under the leadership of Stieglitz, who had chafed under the conservative intransigence of the Camera Club of New York. The Photo-Secession’s membership was also by invitation only, and Stieglitz tightly controlled the membership and the tone of the group, which included Edward Steichen, Clarence H. White, Gertrude Käsebier, Frank Eugene, and Alice Boughton, among others. The Photo-Secession brought visibility to the pictorialist genre though exhibitions at a gallery space donated by Steichen in New York City called the Little Galleries of the Photo-Secession, or “291” (for its address on Fifth Street). To give the pictorialism movement a longer reach, Stieglitz published and edited a quarterly magazine, Camera Work, from 1903 to 1917 that featured exquisite photogravures hand-tipped by Stieglitz into each issue that showcased members’ work. The publication also served as a soundboard for Stieglitz’s critical views on photography and fine art, showing an occasional openness to non-pictorialist photographic fine art approaches, and increased coverage of international versus American-only aspects of fine art photography.
By 1909, new ideas had permeated the art world, and enthusiasm for pictorialism began to wane. The Linked Ring disbanded due to disinterest, and Stieglitz recognized this shift in artistic development. In 1910 he began featuring non-photographic modern art in Camera Work, and showcased avant-garde modern artists such as Rodin, Cézanne, Matisse, and Picasso in its pages. While Stieglitz’s efforts to promote pictorialism had succeeded in garnering acknowledgement of photography as an art form by art museums, his expansion into non-photographic arenas in the magazine alienated his photography-centered subscriber base, and the publication gradually declined. Ironically, in 1917 the final two issues of Camera Work introduced a young new photographer, Paul Strand, whose stunning work eschewed the soft focus and symbolic approach of pictorialism for a perspective that emphasized the beauty of clear lines and forms of ordinary objects. Stieglitz, one of pictorialism’s strongest sponsors, was now hinting at a new direction for the photographic art world.
This new direction manifested itself in an exhibit in 1932 at the M.H. de Young Memorial Museum in San Francisco, where a group of 11 photographers announced themselves as F.64. Formed in opposition to pictorialism, F.64 embodied a Modernism aesthetic for straight photography, based on precisely exposed images of natural forms and found objects. F.64 centered around Edward Westin, an already accomplished photographer, who saw a need to break free from the pervasive pictorialism construct in favor of a new vision; its core members included Ansel Adams, Imogen Cunningham, John Paul Edwards, Sonya Noskowiak, Henry Swift, and Willard Van Dyke.
The name of the group referred to the smallest aperture opening of contemporary large format view cameras, suggesting that photographs should embrace the medium’s unrivaled capacity to see the world versus apologizing for it. The group espoused the use of large depth of field, sharp focus, and contact printing of large negatives on glossy paper; all to more precisely render the image of the subject. The goal was to transform the photographer from a print maker to an image selector; the quality of the image was a result of the photographer’s choice of subject form and framing, a concept that came to be known as “previsualization.”
Whether the subject was a landscape, an architectural or industrial feature, or a nude study, the guiding tenets of this new movement were the shared emphasis on capturing the exact features of the subject in front of the lens and the emotional experience of form. As the group’s manifesto declared,
“The Group will show no work at any time that does not conform to its standards of pure photography. Pure photography is defined as possessing no qualities of technique, composition or idea, derivative of any other art form. The production of the “Pictorialist,” on the other hand, indicates a devotion to principles of art which are directly related to painting and the graphic arts.”
So where does the struggle between Pictorialism (anything is permitted to create a beautiful image) and Modernism, or straight photography (the image must represent reality as closely as possible), find us today? Which perspective is superior or correct? Which movement has prevailed? How much of a pictorialist and how much of a straight photographer are you – and does it matter?
(Next Time: The Story of Leica – Short Version)
This is the twelfth installment of an ongoing series on the history and development of the art of photography. It is inspired by the History of Photography class previously taught by Professor Jeff Curto in the College of DuPage Photography Program. While not a slavish copy of his work, I freely admit to following Curto’s general course outline and sharing many of the perspectives he has developed. I would encourage anyone with a greater interest in this subject to follow his course online via video podcasts, at http://photohistory.jeffcurto.com.
A World History of Photography, 4th Ed, 2007 by Naomi Rosenblum
History of Photography Podcasts, class lectures with Jeff Curto from College of DuPage http://photohistory.jeffcurto.com
The Art of Photography, “Episode 130:: Pictorialism”, text and video, http://theartofphotography.tv/episodes/pictorialism/
Barris, Roann, Dr., Radford University, “The End of Fact? Pictorialism”, http://www.radford.edu/rbarris/art451%20Hist%20of%20Photog/pictorialism.html
Barris, Roann, Dr., Radford University, “Overview of Styles and Movements, 19th – 20th centuries: Photography and the 1920’s”, http://www.radford.edu/rbarris/art451%20Hist%20of%20Photog/styles%20overview.html
JM Colberg, Conscientious Photography Magazine, “The New Pictorialism”, http://cphmag.com/the-new-pictorialism/
Emerson, Peter Henry, Naturalistic Photography for Students of the Art, 1890, https://archive.org/stream/naturalisticphot00emerrich#page/n11/mode/2up
Johnson, Steve, The Minimalist Photographer, “Pictorialism vs Modernism”, http://theminimalistphotographer.com/pictorialism-vs-modernism/
Heyman, Therese Thau, University of California Press, “On the Edge of America: Modernist Photography and the Group f.64”, http://publishing.cdlib.org/ucpressebooks/view?docId=ft5p30070c&chunk.id=d0e6989&toc.id=d0e6989&brand=ucpress
King, Sandy, Sandy King Photography, “Notes on Pictorialism”, http://www.sandykingphotography.com/resources/technical-writing/notes-on-pictorialism
Maloney, Meghan, In The In-Between: Journal of Digital Imaging Artists, “Henry Peach Robinson and the Combination Print- Before Digital”, http://www.inthein-between.com/henry-peach-robinson-and-the-combination-print-before-digital-2/
Metropolitan Museum of Art, “Heilbrunn Timeline of Art History: Group f/64”, http://www.metmuseum.org/toah/hd/f64/hd_f64.htm
Metropolitan Museum of Art, “Heilbrunn Timeline of Art History: Pictorialism in America”, http://www.metmuseum.org/toah/hd/pict/hd_pict.htm
Metropolitan Museum of Art, “Heilbrunn Timeline of Art History: International Pictorialism”, http://www.metmuseum.org/toah/hd/ipic/hd_ipic.htm
Museum of Modern Art, The Collection, Art Terms, “Pictorial photography”, http://www.moma.org/collection/details.php?theme_id=10161
The Phillips Collection, “Truth Beauty: Pictorialism and the Photograph as Art, 1845-1945”, http://www.phillipscollection.org/events/2010-10-9-exhibition-truthbeauty
Restrepo, Victoria, Intersections of Photography and Painting, “Pictorialism”, http://www.vrestrepo.com/two/page7.html
Taylor, Kim, 180 Degree Imaging, “Workshop Resources: A Brief History of Photographic Art Movements”, http://180degreeimaging.com/workshops/article-movements.html
Wikipedia, “Camera Work”, http://en.wikipedia.org/wiki/Camera_Work
Wikipedia, “Group f/64”, http://en.wikipedia.org/?title=Group_f/64
Wikipedia, “The Linked Ring”, http://en.m.wikipedia.org/wiki/The_Linked_Ring
Wikipedia, “Photo-Secession”, http://en.m.wikipedia.org/wiki/Photo-Secession
Wikipedia, “Pictorialism”, http://en.m.wikipedia.org/wiki/Pictorialism
Women Photographers, UCR/California Museum of Photography, “Photography as Art: Pictorialism to Photo Secession”, http://22.214.171.124/collections/permanent/object_genres/photographers/women/pictorialism.html<|endoftext|>
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# Top 10 SSAT Lower Level Math Practice Questions
The best way to prepare your students for the SSAT Lower Level Math test is to work through as many SSAT Lower Level Math practice questions as possible. Here are the top 10 SSAT Lower Level Math practice questions to help your students review the most important SSAT Lower Level Math concepts. These SSAT Lower Level Math practice questions are designed to cover mathematics concepts and topics that are found on the actual test. The questions have been fully updated to reflect the latest 2021 SSAT Lower Level guidelines. Answers and full explanations are provided at the end of the post.
Help your students start their SSAT Lower Level Math test prep journey right now with these sample SSAT Lower Level Math questions.
## SSAT Lower Level Math Practice Questions
1- Which of the following statement is False?
☐A. $$2×2=4$$
☐B. $$(4+1)×5=25$$
☐C. $$6÷(3−1)=1$$
☐D. $$6×(4−2)=12$$
☐E.$$(10+23)×10=330$$
2- If all the sides in the following figure are of equal length and length of one side is 4, what is the perimeter of the figure?
☐A. 15
☐B. 18
☐C. 20
☐D. 24
☐E. 28
3- $$\frac{4}{5}\ −\ \frac{3}{5}$$= ?
☐A. 0.3
☐B. 0.35
☐C. 0.2
☐D. 0.25
☐E. 0.1
4-If $$?=2$$ and $$\frac{64}{?}+4= ☐$$, then $$☐= …..$$
☐A. 30
☐B. 32
☐C. 34
☐D. 36
☐E. 38
5- If (?) is an even number, which of the following is always an odd number?
☐A. $$\frac{N}{2}$$
☐B. $$N+4$$
☐C. $$2N$$
☐D. $$(2×?)+2$$
☐E. $$N+1$$
6- In the following figure, the shaded squares are what fractional part of the whole set of squares?
☐A. $$\frac{1}{2}$$
☐B. $$\frac{5}{8}$$
☐C. $$\frac{2}{3}$$
☐D. $$\frac{3}{4}$$
☐E. $$\frac{5}{6}$$
7- Which of the following is greater than $$\frac{3}{2}$$ ?
☐A. $$\frac{1}{2}$$
☐B. $$\frac{5}{2}$$
☐C. $$\frac{3}{4}$$
☐D. 1
☐E. 1.4
8- If (\frac{1}{3}) of a number is greater than 6, the number must be …..
☐A. Less than 3
☐B. Equal to 3
☐C. Equal to 18
☐D. Greater than 18
☐E.Equal to 6
9- If $$4×(?+?)=20$$ and M is greater than 0, then N could Not be …..
☐A. 1
☐B. 2
☐C. 3
☐D. 4
☐E. 5
10- Which of the following is closest to $$5.03$$?
☐A. 6
☐B. 5.5
☐C. 5
☐D. 5.4
☐E. 6.5
## Best SSAT Lower Level Math Exercise Resource for 2021
1- C
A)$$2×2=4$$ This is true!
B)$$(4+1)×5=25$$ This is true!
C)$$6÷(3-1)=1→6÷2=3$$ This is NOT true!
D)$$6×(4-2)=12→6×2=12$$ This is true!
E)$$(10+23)×10=330→33×10=330$$ This is true!
2- D
The shape has 6 equal sides. And is side is 4. Then, the perimeter of the shape is: $$4×6=24$$
3- C
$$\frac{4}{5}-\frac{3}{5}=\frac{1}{5}=0.2$$
4- D
$$N=2$$ , then: $$\frac{64}{2}+4=32+4=36$$
5- E
N is even. Let’s choose 2 and 4 for N. Now, let’s review the options provided.
A. $$\frac{N}{2}=\frac{2}{2}=1, \frac{N}{2}=\frac{4}{2}=2$$, One result is odd and the other one is even.
B. $$N+4=2+4=6,4+4=8$$
Both results are even.
C. $$2N=2×2=4,4×2=8$$
Both results are even.
D. $$(2×N)+2=(2×2)+2=6,(4×2)+2=10$$
Both results are even.
E. $$N+1=2+1=3,4+1=5$$
Both results are odd.
6- A
There are 10 squares and 5 of them are shaded. Therefore, 5 out of 10 or $$\frac{5}{10}=\frac{1}{2}$$ are shaded.
7- B
$$\frac{3}{2}=1.5$$, The only option that is greater than 1.5 is $$\frac{5}{2}$$
$$\frac{5}{2}=2.5 \ , \ 2.5>1.5$$
8- D
If $$\frac{1}{3}$$ of a number is greater than 6, the number must be greater than 18.
$$\frac{1}{3} x>6→x>18$$
9- E
$$4×(M+N)=20$$, then $$M+N=5. M>0→N$$ could not be 5
10- C
The closest to 5.03 is 5 in the options provided.
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Help save the environment<|endoftext|>
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# Thread: how to prove the inequality (1-x^2)^n>=1-nx^2?
1. ## how to prove the inequality (1-x^2)^n>=1-nx^2?
Let $\displaystyle x\in [0,1]$ and $\displaystyle n$ be any positive integer, prove $\displaystyle (1-x^2)^n\geq1-nx^2$.
It can be easily proved by induction on $\displaystyle n$, but I would like to use binomial theorem to prove it. By binomial theorem, $\displaystyle (1-x^2)^n=1-nx^2+...$. The proof is completed if the remaining "..." part in the above expansion can be proved to be $\displaystyle \geq0$, but I have no idea how to prove it. Could anyone help me with this problem? Thanks!
2. Hello,
As you said, we have :
$\displaystyle (1-x^2)^n=\sum_{k=0}^n {n\choose k} (-x^2)^{k}=1-nx^2+\sum_{k=2}^n {n\choose k} (-x^2)^{k}$
Now let's consider $\displaystyle S=\sum_{k=2}^n {n\choose k} (-x^2)^{k}$
Try to make the sum start by 0, but not by adding terms, otherwise we'd be back to the beginning.
Instead, let $\displaystyle j=k-2$
Thus $\displaystyle S=\sum_{j=0}^{n-2} {n\choose j+2} (-x^2)^{j+2}$
$\displaystyle S=\sum_{j=0}^{n-2} \frac{n!}{(j+2)!(n-j-2)!} \cdot (-x^2)^{j+2}$
Now factor out some things in order to get a binomial expansion :
$\displaystyle S=n(n-1)(-x^2)^2\sum_{j=0}^{n-2} \frac{(n-2)!}{j!(n-2-j)!}\cdot (-x^2)^j$
$\displaystyle S=n(n-1)x^4\sum_{j=0}^{n-2} {n-2 \choose j} (-x^2)^j$
Which is exactly $\displaystyle n(n-1)x^4(1-x^2)^{n-2}$
And this is positive, since $\displaystyle 1-x^2\geq 0$
Looks clear to you ?
3. Thank you but, during the factoring, how to turn the $\displaystyle (j+2)!$ in the denominator into $\displaystyle j!$ ?<|endoftext|>
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A reconstructed roundhouse in the former museum's courtyard
Our ancestors lived in roundhouses, a style of building that dates back more than 3000 years to the Bronze Age. Each farm would have had a number of houses, probably for different members of the same family, parents, grandparents and cousins, or for the animals.
Excavation gives us very few clues about what happened inside the houses. Objects that survive tell us what kind of items people owned and used, and suggest the kind of activities that went on in the farms. Some houses had a quern for grinding cereals and a hearth in the centre for heat and cooking, and a clay oven for baking bread.
A typical roundhouse was reconstructed in the museum courtyard for this exhibition. Built using traditional methods, the roundhouse showed how houses in this area may have looked in the Roman period. You can see step-by-step photos and a time-lapse video of the construction on the building the roundhouse page.
David Freeman building the roundhouse, dressed in Romano-British costume
How did we know how to build the roundhouse?
No original roundhouse survives so we need to use various clues to reconstruct what their houses may have looked like.
We know from excavations on sites such as Irby, Wirral, the size and shape of the foundations. These survive as postholes in the ground, into which upright posts were set to form the walls. We know from other sites that stakes of wood were set between the posts and hazel twigs were woven through the posts to make a wattle wall. The walls were then plastered with clay and straw. The roof was almost certainly covered with straw or heather, built in a conical shape. There is no smoke-hole as smoke gradually filters through the thatch. This roundhouse has been constructed using original methods and materials.
Alongside the roundhouse in the museum courtyard a garden was created with plant types known from Roman Britain. It featured a variety of species during the course of the exhibition, from native plants such as mint and the turnip to ‘new’ foods like onion and rosemary. The Romans introduced many new plants to Britain, including garlic, white mustard and dill.
Some of the species grown in Roman Britain were food plants, others were used to flavour food or as medicines. People also collected wild foods such as berries, nuts and fruit.
Follow the links below to see the construction of the roundhouse and find out more about who would have lived there<|endoftext|>
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## Book: RS Aggarwal & V Aggarwal - Mathematics
### Chapter: 7. Areas
#### Subject: Mathematics - Class 9th
##### Q. No. 23 of Exercise 7A
Listen NCERT Audio Books - Kitabein Ab Bolengi
23
##### Find the area of the parallelogram ABCD in which AB = 14 cm, BC = 10 cm and AC = 16cm. [ Given = 1.73]
According to the question,
In order to find the area of quadrilateral ABCD,
At first,
Let us consider triangle ABC,
Say,
a = 10 cm, b = 16 cm and c = 14 cm
Now,
Semi perimeter of ΔABC, s =
=
=
= 20 cm
Now,
Area of ΔABC =
=
=
= 40√3 cm2
We know that, the diagonal of a parallelogram divides it into two triangles of equal areas.
Hence,
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD
= Area of ΔABC × 2
= 40√3 × 2
= 80√3 cm2
= 138.4 cm2
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# Lesson 3, Part C
## lesson 3, part c
Expand +
Michelle challenges her students to individually select one grouping of representations from their chart and move them over to a poster without the justifications. Each student then writes responses to several questions, each of which has a sentence stem prompt:
• Why did you select this grouping? / I selected this grouping because...
• How do you know that all the cards represent the same quantity? / I know that all the cards represent the same quantity because...
• When sharing your work with other pairs, what did you learn from the discussion? / When I shared my chart with another pair I learned... from the discussion.
## lesson 3, part c
4th grade math - Understanding Fractions
Michelle Makinson, Bagby Elementary School, Cambrian School District, San José, California
Next Up: Summative Reflection
Previous: Lesson 3, Part B
MICHELLE MAKINSON: As we spent time putting together the chart of all the factual...the five ways to represent the fractions, right? And so now you're going to pick with your partner. So Michael and Jakob, uh, Nathan and Giovanni, come to the front again. You can leave your materials there, just bring a pencil. I have paper up here already. Okay. So you're going to, based upon part of the presentation we saw, based upon your work so far, um, everything you've heard about fractions, you're going to have an opportunity to share out, change or modify one grouping of your choice from your chart. So each partner is going to pick a different grouping and you're...literally, because they're on the tape loops, you can move them onto your poster. Does that make sense? So you're going to move the whole set over to your poster, but you're not going to move the gold justification cards. You're going to leave those on your chart. Does that make sense? So how many cards will you move?
STUDENT: Three.
STUDENTS: Five.
MICHELLE MAKINSON: Five. Five cards without the justification cards. Okay? And you'll put them all on your construction paper, and then you're going to answer some questions related to those. Each one of the partners is going to pick one row that they want to change, modify, explain. And you're going to answer these questions: Why did you select this grouping? And so on the actual, you're basically making a poster about that row. So you're not going to write the question down but you're going to use the sentence stem “I selected this grouping because,” and then finish the thought. Does that make sense? Complete sentences.
Then the second question is: How do you know that all the cards represent the same quantity? “I know that all the cards represent the same quantity because,” and then finish that sentence. Nod your head if that is making sense. No, it's not making sense?
STUDENTS: Yeah.
MICHELLE MAKINSON: So nod your head. Okay. When sharing your work with other pairs, what did you learn from the discussion? Remember when you shared your charts with another group? What did you learn from that? So when I shared my chart with another partnership, I learned "blank:" multiple things, possibly, from the discussion. Does that make sense? So you're picking a row from your own chart that you think you want to change, or modify, explain. Right? To show your thinking. Does that make sense?
STUDENT: Yeah.
MICHELLE MAKINSON: Okay. So we're going to get your charts to you. All right, so you need to be opening up your chart and looking at the different rows. And each partner will take a different row to put onto their yellow construction paper, and then answer those questions in complete sentences. Explaining your thinking. Good job, JT. Okay, so you can pick any. You can move it over so he can start working. Make sure you get your name and number on this yellow paper, please.
STUDENT: I selected this grouping because it's the first one I saw. I know this card...these cards represent the same quantity because they all have to do with one out of two equal parts. When I shared my...I'm still working on the next one.
STUDENT: I selected this grouping because it has equivalent fractions. I know that all the cards represent the same quantity because they all represent two fourths, which equals one half. When I shared my chart with other partner, partners...I'm not finished with that one yet.
STUDENT: And I know because you need to really think about what you're doing when you...when you, like, do all these answers because you don't want to get them wrong.
STUDENT: I selected this grouping because this is easy to explain. I know the cards represent the same quantity because all of them are equal to two...equal to two sixths, except the number line, and it is equivalent to two sixths. When I shared my chart I learned that the number line doesn't have to be two sixths. It can be, uh, it can be something else as long as it's equivalent to two sixths.
STUDENT: And I know that all the cards represent the same quantity because they are all eight twenty-fourths. And when I shared my chart with another partnership I learned that not all the answers were the same.
STUDENT: I selected this grouping because it is the most basic -- well, in my point of view. And I know all these match because there is five, five here that are four-sided, and it's a...and there was kind of here in total. So there's five out of ten, and right here is ten out of five, and right here is...this is zero, this is one. And right there is the middle and five tenths -- that's pretty much obvious. So, would...I'm still... So I know all of these cards match because, because they're...if you just look at them all, they mean the exact same thing, or in our case, five tenths.
STUDENT: I selected this grouping because I'm familiar with these fractions. I know that all the cards represent the same quantity because each card has the same numerator of two and the denominator of four.
STUDENT: I selected this grouping because there is great quantity fractions, and I know that all the cards represent the same quantity because (inaudible) and then I'm going to finish.
MICHELLE MAKINSON: I often encourage my students to get up out of their seats and move, to keep them from becoming frustrated. This clip is after we’d begun to do our partner sharing outside, but ultimately they were too distracted by other things that were going on, so we came back in. My thinking was, they were feeling cooped up. I like to move them around to different environments because it kept their interest. It kept them from becoming fatigued and annoyed, and when they become fatigued and annoyed they do especially disruptive things to each other, verbally. I could see the meltdown happening. It was not a very good day. I thought, "Okay, I have a little, tiny window before the younger kids come out for recess.” They’re actually more focused in this sharing than they were outside, maybe in part because they got to move.
Because of all the emotional issues and the turmoil in that class, I had to do a lot of work. Every day was a struggle. There was so much stuff. If one person would calm down, another one would pop off. There were at least 10 of them that were extremely high needs, and few of those had any official diagnosis of special needs.
I see them using academic vocabulary well here. There was some anchoring in that. Just constantly using the same academic vocabulary, insisting on it, not moving forward until they use the right vocabulary and questioning them towards it.
I'm working as a teacher trying to figure out what is the right balance. I also want to create reference books that they have organized, so then it becomes a writing thing, an organization thing, so they can look up content for math. We have a pretty well developed math journal. We're used to documenting our own thinking and then going back and seeing what you have to do in order to make the thing we created viable and useful.<|endoftext|>
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Reading and writing often go hand in hand; students need to engage with written material, digest it, critique it and deploy what they have read, in shaping their own ideas, usually in writing. But being able to read effectively – and to use reading effectively - is not necessarily something students are equipped to do. The texts students are asked to read in their disciplinary courses are often highly challenging - in the way they are written, the language they use, their structure and content - and in their sheer volume.
We’ve worked with a number of colleagues to find ways of offering more overt support to/developing students’ reading in their subject courses. As reading is a foundation element of everything we do you will find these resources elsewhere on our site but we feel that they are usefully gathered together here too:
· The first year Geographical Ideas in Practice module included worksheet guidance and tasks to get students thinking about what they were reading and practice developing responses in writing
· First year Politics students were encouraged to use a reading grid to help them extract and analyse salient aspects from a limited number of texts and organise their material to highlight similarities and differences across texts. This was part of a series of tasks integrated into a first year Politics module that help students read effectively and use what they read in their writing.
· Other politics students were helped in engaging with primary sources from Parliament and the government as part of a module developed to encourage research-based learning
· Some ideas from a module on Catalan Culture to that were found to very helpful to students for engaging with the reading they did in their subsequent writing involve designing the activities before ascribing texts to them, writing a series of progressively shorter summaries of a key text, and shortening and lengthening a text using negative summary, adding detail. More details and ideas around this can be found on our Countering Plagiarism page.
· Offering Reading Spaces and retreats. Another simple approach that involves scheduling time for students to read in the company of others with some light guidance including writing questions before reading and summing up key points once the text has been put to one side.
· Other ideas for developing reading through critical and creative rewriting, and examples of students responses to these ideas, can be found on our page Provoking critical engagement.<|endoftext|>
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Precalculus (6th Edition) Blitzer
The simplest form of the expression is ${{\left( {{x}^{2}}-1 \right)}^{4}}={{x}^{8}}-4{{x}^{6}}+6{{x}^{4}}-4{{x}^{2}}+1$
Thus, we get \begin{align} & {{\left( {{x}^{2}}-1 \right)}^{4}}=\left( \begin{matrix} 4 \\ 0 \\ \end{matrix} \right){{\left( {{x}^{2}} \right)}^{4}}+\left( \begin{matrix} 4 \\ 1 \\ \end{matrix} \right){{\left( {{x}^{2}} \right)}^{3}}\left( -1 \right)+\left( \begin{matrix} 4 \\ 2 \\ \end{matrix} \right){{\left( {{x}^{2}} \right)}^{2}}{{\left( -1 \right)}^{2}}+\left( \begin{matrix} 4 \\ 3 \\ \end{matrix} \right)\left( {{x}^{2}} \right){{\left( -1 \right)}^{3}}+\left( \begin{matrix} 4 \\ 4 \\ \end{matrix} \right){{\left( -1 \right)}^{4}} \\ & =\frac{4!}{0!\left( 4-0 \right)!}\left( {{x}^{8}} \right)+\frac{4!}{1!\left( 4-1 \right)!}\left( -{{x}^{6}} \right)+\frac{4!}{2!\left( 4-2 \right)!}\left( {{x}^{4}} \right)+\frac{4!}{3!\left( 4-3 \right)!}\left( -{{x}^{2}} \right)+\frac{4!}{4!\left( 4-4 \right)!}\left( 1 \right) \\ & ={{x}^{8}}-4{{x}^{6}}+6{{x}^{4}}-4{{x}^{2}}+1 \end{align}<|endoftext|>
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# Term
A term is an algebraic expression which can form a separable part of another expression such as an algebraic equation or a sequence. Terms are a specific part of the symbolic language of algebra. The symbols of this language were primarily developed during the sixteenth and seventeenth centuries and are used to represent otherwise lengthy expressions. They can be as simple as using the single character, +, to mean addition, or as complicated as y = 4x2 + 2x - 3 to represent an algebraic polynomial equation.
In general, there are three types of algebraic expressions which can be classified as terms. These include expressions made up of a single variable or constant, ones that are the product or quotient of two or more variables and/or constants, and those that are the product or quotient of other expressions. For example, the number 4 and the variable x are both terms because they consist of a single symbol. The expression 2z is also a term because it represents the product of two symbols. It should be noted that terms like 2z, in which a number and a variable are written together, are indicated products because multiplication is implied. Therefore, the symbol 2z means 2 × z. Finally, an expression like 2pq(a + 5)n is a term because it represents a quotient (the result of division) of two expressions.
The symbols that make up a term are known as coefficients. In the term 4x, the number 4 is known as a numerical coefficient and the letter x is known as the literal coefficient. For this expression, we could say that 4 is the coefficient of x or x is the coefficient of 4.
Terms should be thought of as a single unit that represents the value of a particular number. This is particularly useful when discussing the terms of a larger expression such as an equation. In the expression 5x3 + 2x2 + 4x - 7, there are four terms. Numbering them from left to right, the first term is 5x3, the second is 2x2, the third is 4x, and the fourth is -7. Notice that the sign in front of a term is actually part of it.
Some expressions contain terms which can be combined to form a single term. These "like terms" contain the same variable raised to the same power. For example, the like terms in the expression 3x + 2x can be added and the equation simplifies to 5x. Similarly, the expression 7y2 - 3y2 can be simplified to 4y2. Expressions containing unlike terms can not be simplified. Therefore, 4x2 2x is in its simplest form because the differences in the power of x prevents these terms from being combined.<|endoftext|>
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What were the environmental impacts of the huge coal ash spill in Tennessee this past December? — Dave S, Lynnfield, MA
Environmentalists’ call for an end to the age of coal—one of the dirtiest and most common of all the fossil fuels we now use—took on new urgency this past December when some 525 million gallons of wet coal ash, enough toxic slurry to flood more than 3,000 acres of nearby land, spilled into the nearby Tennessee River and surrounding areas when a retaining wall at a power plant in the town of Harriman gave way.
The sludge destroyed 12 homes, though no one was directly injured. However, an unprecedented fish kill occurred in the Tennessee River and area tributaries in the aftermath of the spill. According to John Moulton, a spokesman for the Tennessee Valley Authority which owns the plant, a test of river water near the spill site found elevated levels of lead and thallium, both of which have been linked to birth defects and nervous and reproductive system disorders. He reassured locals that, although these substances exceeded safety limits for drinking water, they would be filtered out by normal water treatment processes.
But some area residents aren’t so sure that they are safe from the effects of the spill, which is estimated to have been over 40 times bigger by volume than the infamous Exxon Valdez oil spill of 1989. Calling it an “environmental disaster of epic proportions,” Carol Kimmons, a local resident who works at the non-profit Sequatchie Valley Institute, told reporters that the nasty black ash flowed into “the water supply for Chattanooga and millions of people living downstream in Alabama, Tennessee and Kentucky.” She added that the spill was 70 percent bigger than a similar one in Kentucky in October 2000 (306 million gallons) that the U.S. Environmental Protection Agency (EPA) referred to at the time as “one of the worst environmental disasters in the Southeastern United States.”
More than a year after that Kentucky spill, researchers found levels of lead downstream from where the spill took place that were 400 times higher than the EPA’s safe limit. And levels of Beryllium were 160 times higher than acceptable EPA levels. “Coal contains huge amounts of heavy metals, and when coal is burned, the organic matter burns off, but many of the nasty chemicals stick around, in higher concentrations,” said Kimmons. “Also, coal is ‘washed’ using some really nasty chemicals, which are also left over in coal slurry.” The bottom line, she concluded, is that “coal slurry is really, really toxic stuff.”
Ironically, on the very same day as the huge Tennessee spill, a coalition of 39 non-profit groups delivered a letter to then President-elect Barack Obama asking him to overturn a pending Bush administration rule change that would ease regulations on coal waste disposal. The groups contend that coal ash has already polluted 23 states and that the proposed new rule would only allow more pollution and more risks to human health and the environment. Now-President Obama has pledged to undertake a comprehensive inventory of liquid coal ash waste and propose new regulations to ensure its safe disposal.
“This disaster proves that regulations around coal slurry impoundments need to be tightened, and not loosened,” says Kimmons. Only time will tell if verbal commitments from Washington materialize into help on the ground.
CONTACTS: Sequatchie Valley Institute, svionline.org; Tennessee Valley Authority, tva.gov.
SEND YOUR ENVIRONMENTAL QUESTIONS TO: EarthTalk, P.O. Box 5098, Westport, CT 06881; [email protected]. Read past columns at: www.emagazine.com/earthtalk/archives.php. EarthTalk is now a book! Details and order information at: www.emagazine.com/earthtalkbook.<|endoftext|>
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##### Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX - plane.
The vector equation of a line passing through two points with position vectors & is
The position vector of point A(5,1,6) is given as -
.
The position vector of point B(3,4,1) is given as -
...(1)
Let the coordinates of the point where the line crosses the ZX plane be (0,y)
So, …(2)
Since point lies in line, it will satisfy its equation,
Putting (2) in (1)
Two vectors are equal if their corresponding components are equal
So,
x = 5 - 2λ …(3)
0 = 1 + 3λ …(4)
and, z = 6 - 5λ …(5)
From equation (4), we get -
λ = -1/3
Substitute the value of λ in equation (3) and (5), we get -
x = 5 - 2λ = 5 - 2 × (-1/3) = 5 + (2/3) = 17/3
and
z = 6 - 5λ = 6 - 5 × (-1/3) = 6 + (5/3) = 23/3
Therefore, the coordinates of the required point is
(17/3, 0, 23/3).
11<|endoftext|>
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# Count Good Nodes in Binary Tree Leetcode Solution
Difficulty Level Medium
algorithms Binary Tree coding Depth First Search Interview interviewprep LeetCode LeetCodeSolutionsViews 247
## Problem Statement
In this problem a binary tree is given with its root. A node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X.
We have to return the number of good nodes in the given binary tree.
### Example
``` 3
/ \
1 4
/ / \
3 1 5```
`4`
Explanation:
Nodes in blue are good.
Root Node (3) is always a good node.
Node 4 -> (3,4) is the maximum value in the path starting from the root.
Node 5 -> (3,4,5) is the maximum value in the path
And Node 3 -> (3,1,3) is the maximum value in the path.
``` 3
/
3
/ \
4 2```
`3`
Explanation:
Node 2 -> (3, 3, 2) is not good, because “3” is higher than it.
## Approach
To find whether a node is good or not we must traverse path from root to the that node and check if its value is not smaller than maximum in this path.
To find the number of good nodes we have to check like this for each node of the given binary tree. But here observe one thing,
if we find the answer for a particular node by traversing its path from root, we can go to its child node also from there itself because we have already traversed almost path of child node also and we also have maximum value traversed till yet. We just have to update the maximum with current node value to transfer to its both children node.
So this looks like a DFS or recursion where to go to a particular node we traverse the path from root to that node. Thus in this problem recursion will be very helpful. The steps are:
• Create a recursive function with two arguments as its parameter. One is the address of the node and second is the maximum value we found till here.
• Now whenever we will be at a particular node we will check if current node value is smaller than current max. If it is not smaller we will add this node to our ans and call the same function for its child nodes after updating the maximum value.
## Implementation
### C++ Program for Count Good Nodes in Binary Tree Leetcode Solution
```#include <bits/stdc++.h>
using namespace std;
struct TreeNode {
int val;
TreeNode *left,*right;
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};
int rec(TreeNode* root, int mx)
{
if(!root) return 0;
int cur=0;
if(mx <= root->val) cur++;
mx=max(mx,root->val);
return rec(root->left,mx) + rec(root->right,mx) + cur;
}
int goodNodes(TreeNode* root) {
int mx= INT_MIN;
return rec(root,mx);
}
int main()
{
TreeNode* root= new TreeNode(3);
root->left= new TreeNode(1);
root->right= new TreeNode(4);
root->left->left= new TreeNode(3);
root->right->left= new TreeNode(1);
root->right->right= new TreeNode(5);
cout<< goodNodes(root) ;
return 0;
}
```
`4`
### Java Program for Count Good Nodes in Binary Tree Leetcode Solution
```class Rextester{
static class TreeNode {
int val;
TreeNode left,right;
TreeNode(int x) {
val=x;
left=null;
right=null;
}
}
static int rec(TreeNode root, int mx)
{
if(root==null) return 0;
int cur=0;
if(mx <= root.val) cur++;
mx = Math.max(mx,root.val);
return rec(root.left,mx) + rec(root.right,mx) + cur;
}
public static int goodNodes(TreeNode root)
{
int mx= Integer.MIN_VALUE;
return rec(root,mx);
}
public static void main(String args[])
{
TreeNode root= new TreeNode(3);
root.left= new TreeNode(1);
root.right= new TreeNode(4);
root.left.left= new TreeNode(3);
root.right.left= new TreeNode(1);
root.right.right= new TreeNode(5);
System.out.println( goodNodes(root) );
}
}
```
`4`
## Complexity Analysis for Count Good Nodes in Binary Tree Leetcode Solution
#### Time Complexity
O(n) : where n is the total number of nodes in the given binary tree. We are visiting each node once.
#### Space Complexity
O(n) : Space used will be the maximum size of recursion stack. At worst it can go to O(n) size in case of skewed binary tree.
Translate »<|endoftext|>
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# Points A and B are at (3 ,7 ) and (4 ,2 ), respectively. Point A is rotated counterclockwise about the origin by pi and dilated about point C by a factor of 5 . If point A is now at point B, what are the coordinates of point C?
Jan 13, 2018
After Point A is rotated counterclockwise about the origin by $\pi$, its new coordinates are $\left(- 3 , - 7\right)$.
The difference between the $x$ coordinates of Point A and B now is $4 - \left(- 3\right) = 7$ and $y$ coordinates: $2 - \left(- 7\right) = 9$
Since Point A was dilated about Point C by a factor of 5, we can find out by how much the coordinates change with each integer increase in factor.
For the $x$ coordinate:
$\frac{7}{4} = 1.75$
and $y$ coordinate:
$\frac{9}{4} = 2.25$
So for every integer increase in factor, the point moves 1.75 to the right and 2.25 upwards.
Point C is therefore
$\left(- 3 - 1.75 , - 7 - 2.25\right)$
$= \left(- 4.75 , - 9.25\right)$
Jan 13, 2018
$C = \left(- \frac{19}{4} , - \frac{37}{4}\right)$
#### Explanation:
$\text{under a counterclockwise rotation about the origin of } \pi$
• " a point "(x,y)to(-x,-y)
$\Rightarrow A \left(3 , 7\right) \to A ' \left(- 3 , - 7\right) \text{ where A' is the image of A}$
$\Rightarrow \vec{C B} = \textcolor{red}{5} \vec{C A '}$
$\Rightarrow \underline{b} - \underline{c} = 5 \left(\underline{a} ' - \underline{c}\right)$
$\Rightarrow \underline{b} - \underline{c} = 5 \underline{a} ' - 5 \underline{c}$
$\Rightarrow 4 \underline{c} = 5 \underline{a} ' - \underline{b}$
$\textcolor{w h i t e}{4 \underline{c} \times} = 5 \left(\begin{matrix}- 3 \\ - 7\end{matrix}\right) - \left(\begin{matrix}4 \\ 2\end{matrix}\right)$
$\textcolor{w h i t e}{\times \times} = \left(\begin{matrix}- 15 \\ - 35\end{matrix}\right) - \left(\begin{matrix}4 \\ 2\end{matrix}\right) = \left(\begin{matrix}- 19 \\ - 37\end{matrix}\right)$
$\Rightarrow \underline{c} = \frac{1}{4} \left(\begin{matrix}- 19 \\ - 37\end{matrix}\right) = \left(\begin{matrix}- \frac{19}{4} \\ - \frac{37}{4}\end{matrix}\right)$
$\Rightarrow C = \left(- \frac{19}{4} , - \frac{37}{4}\right)$<|endoftext|>
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Tomato (Lycopersicon esculentum) plants produce small yellow flowers in early summer that grow into delectable, nutritious fruit. Hardy to U.S. Department of Agriculture zones 10 through 11, tomatoes can be grown as annuals in all zones, although many varieties require heat to ripen. Blossom drop prevents tomatoes from setting fruit but can be circumvented most of the time.
Tomatoes are self-fertile, but bud drop can occur when insects that are instrumental in pollen distribution aren't active in the garden. Recent declines honeybee and bumblebee populations have caused many home gardeners to seek alternate pollination solutions. Gardeners can circumvent pollination problems by gently shaking tomato plants to loosen and distribute pollen. Tomatoes set fruit in a temperature window of 60 and 70 degrees Fahrenheit. Shaking plants should occur on warm, sunny days between the hours of 10 a.m. and 4 p.m.
Proper Watering Practices
Fluctuations in soil moisture can cause blossom drop, so water tomato plants deeply at regular intervals to keep the soil consistently moist. Soak the soil around the plants to a depth of between 8 to 10 inches. Adding several inches of mulch will help to retain soil moisture. Don't let plants wilt from lack of water, and keep in mind that although many varieties of tomatoes can be grown in containers, they will need more water than their garden-grown counterparts.
Tomato fertilizer should have a low amount of nitrogen, a high amount of phosphorus and a medium-to-high amount of potassium. Both 8-32-16 and 6-24-24 formulas are recommended for healthy tomatoes. Work fertilizer 6 inches into soil at time of planting at the rate of 2 tablespoons per plant. Fertilize again when fruit first sets and every 10 days thereafter by side dressing each plant with 1 to 2 tablespoons of 8-32-16 fertilizer.
Plants under stress caused by temperature extremes will abort fruit to save their resources for survival. Blossom drop frequently occurs when nighttime temperatures consistently drop below 55 degrees Fahrenheit. Heat spells involving nighttime temperatures of above 75 degrees F and daytime temperatures of over 90 degrees F can also adversely affect pollination and cause blossom drop. Pollen also becomes nonviable at temperatures of above 85 degrees F or below 55 degrees F, because it turns tacky and won't stick to the stigma at these temperatures.
- Tomato Dirt: Blossom Drop: Why Fruit Doesn’t Set and What To Do About It
- The Pollination Home Page: Tomato Pollination Graphics
- University of Idaho College of Agriculture and Life Sciences: Short-Season, High-Altitude Gardening: Growing Tomatoes in Cool, Short-Season Locations
- Colorado State University Extension: Growing Tomatoes in the Home Garden
- The University of Missouri Extension: Growing Home Garden Tomatoes
- Proceedings of the National Academy of Sciences: Patterns of Widespread Decline in North American Bumblebees
- Rutgers New Jersey Agricultural Experiment Station: Jersey Fresh Information Exchange: What Causes Blossom Drop in Tomatoes?
- Texas A&M University Aggie Horticulture: Tomato, Part I
- Michael Blann/Digital Vision/Getty Images<|endoftext|>
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Dairy allergy, or milk allergy, refers to any allergic reaction caused by a component of cow's milk. The three components of cow's milk that cause dietary reactions are casein protein, whey protein, and lactose sugar. Casein and whey are considered more likely to cause true allergies, while lactose causes a well-known intolerance in many adults (and some children) due to the body's lack of an enzyme known as lactase.
Similar components to cow's milk are found in the milk of other ruminants, including goats and sheep, so any patient with a dairy allergy who is considering using other animal milk as a substitute for cow's milk should talk to their allergist before proceeding.
Dairy allergies may appear with a wide variety of symptoms, including hives (urticaria), eczema, chronic congestion, and diarrhea. Lactose intolerance, like many other dietary intolerances, causes gastrointestinal symptoms, such as bloating, cramping, and diarrhea. As always, if you suspect you or your child has a food allergy, contact your physician.
Preventing Dairy Allergies:
Because dairy allergies are especially prevalent among babies, parents with atopic families - that is, families with a history of severe allergies - should discuss feeding options with their pediatricians before delivery, if at all possible. There is some evidence that nursing exclusively until six months and delaying the introduction of solid foods until that time can help prevent the development of allergies.
Foods Containing Dairy Products:
Cheese, butter, yogurt, cream, kefir, sour cream, and ice cream, unless specifically formulated to be dairy-free, always contain milk. Milk is also present in many types of processed food. Processed foods that are likely to contain dairy products include chocolate, salad dressings, pastries, snack foods with butter or cheese flavorings (even if they're artificial), soups, and even canned tuna and deli meats. As with any food allergy, never eat any processed food unless you have read the label, and always be aware of cross-contamination risks from utensils or surfaces where dairy products may have been prepared.
Dairy Products and Labeling Laws:
Dairy is one of the eight most common allergens in the United States, and as such, current food labeling laws require that the presence of milk be clearly marked on ingredient labels. However, it's best to learn the myriad names dairy products appear on in labels. While FDA laws require that the presence of milk be marked in plain English, it's safest to rely on that in conjunction with your own knowledge of dairy-containing ingredients. This printable list includes aliases for milk on food labels plus some foods that are especially likely to contain milk.
Living with Dairy Allergies and Lactose Intolerance:
You'll find substitutes for milk products in many supermarkets and health-food stores. Always check these for the presence of dairy, however; some may include traces of milk and thus be unsuitable for someone with allergies. With that caveat, try the many milk substitutes on the market for baking, drinking, and cooking. Soy milk, rice milk, and nut milks are but a few of the varieties available, and each has different properties. Rice milk is low in protein (so it acts quite differently than cow's milk in baking) but has a mild taste; in its vanilla flavor it is delicious on cereal and good for drinking plain. Soy milk and nut milks have a stronger flavor and can work well in baked goods.
Milk has a somewhat outsized reputation as a nutritional powerhouse. However, with planning, you can easily replace the nutrients in milk. Be especially aware of calcium, protein, and vitamins A, D, E, and K, which are found in abundance in dairy products.
You'll find an even wider variety of dairy-free products at natural foods stores, co-ops, and specialty chains like Whole Foods Market and Fresh Market. There are also online dairy-free retailers like The Wheat- and Dairy-Free Supermarket.
Whether you've got a dairy allergy or lactose intolerance in your family, or you're simply giving up milk for a while as part of an elimination diet, you may find yourself checking out the rows of milk alternatives in the grocery store. There are quite a few options available, and most have slightly different properties for drinking and cooking.
Non-Dairy Substitutes / Dairy Alternatives
The following dairy alternatives / non-dairy substitutes are increasingly available from health shops and some supermarkets :
- Soya milk, Rice Dream, and oat milk drink. Soya milk drinks are also available flavoured with strawberry or chocolate and rice milk drinks are available in some flavours. Rice milk, although sweet, contains no added sugar.
- Soya cheese, both in hard and soft forms is available and also tofu cheese. Soya soft cheese and Veeze cheese spread can be used like Dairylea. Toffutti is a good cream cheese alternative, and cheese slices are now available.
- Soya yoghurts make an excellent snack. They come in many flavours and are sweetened with fruit juice. Many soya yoghurts and desserts have a long shelf life and do not need refrigerating, making them ideal to keep in the drawer at work. Live soya yoghurts containing beneficial cultures are now available from some supermarkets. These are fresh, and so kept in the chilled cabinets with other fresh yoghurts.
- For desserts, use soya cream, soyasun desserts, soya ice cream and soya vanilla dessert (great substitute for custard) and other Provomel deserts.
- Yeast flakes can be delicious sprinkled onto dishes or on toast, giving quite a cheesy taste and lots of B vitamins. Not suitable for a yeast free diet though.
- For sandwich fillers and toppings, use avocado, humus, tahini (mix 1 tablespoon of tahini in water and lemon juice until creamy), salads, guacamole, unhydrogenated vegetable margerines such as Pure, Vitquel, Vitaseig, and Suma.
- You can make your own nut milk by blending almonds in a food processor with water until smooth. Adjust the consistency as required.
If you are avoiding dairy foods, it is wise to avoid goats and sheeps milk products, unless you have been otherwise advised during your consultation with a qualified Nutritionist.
Here's a cheat sheet to your options. Just be prepared to sample a variety to figure out which taste suits you best; most are quite distinctive.
Lactose-free milk is only suitable for people with lactose intolerance; it contains the same proteins as milk and is just as allergenic for people with dairy allergies.
That said, for people with an intolerance, lactose-free milk is almost indistinguishable from "regular" milk. The lactase enzyme added to regular milk to break down lactose into simpler sugars makes it taste slightly sweeter to most people. Lactose-free milk is available in both conventional and organic varieties.
Goat Milk and Other Ruminant Milk
Goat, sheep, and other ruminant milks contain similar proteins to cow's milk and are considered to have a high degree of cross-reactivity. That means that people with an allergy to cow's milk are likely to react to other ruminant milks, too.
If you or a loved one have a dairy allergy and you're considering trying goat milk (say, drinking it yourself, or giving it to a toddler), consult an allergist first. These milks do contain lactose and are not suitable for those who are lactose-intolerant without prior use of an over-the-counter lactase supplement.
The most widely available dairy-free milk alternative is soy milk, which can be found both in cartons on supermarket shelves as well as alongside milk in dairy cases. Competition from national brands, like 8th Continent and Silk, has lowered prices across the board, making soy milk one of the more cost-effective milk alternatives.
Soy milk is high in protein, making it an attractive alternative to milk for cooking and baking. Soy itself has a strong, distinctive taste, so make sure you like it before adding it to a sauce or to your favorite cereal.
Almond milk is among the most common nut milks. Like soy milk, nut milks are high in protein and are useful for baking. You may find their taste blends in with baked goods, coffee, or nutty cereals better than soy milk, although personal tastes vary. Nuts are also high in "good fats" and Vitamin E. One drawback to both soy and nut milk: both of these are common allergens in and of themselves.
Unlike soy and nut milks, rice milk is not especially allergenic, making it an attractive choice for families concerned about avoiding allergens in young children.
Rice milk, especially in its vanilla flavor, is quite sweet. But its texture is the most watery of all milk alternatives, and it is not particularly useful for cooking. Being low in protein, it does not make a good nutritional replacement for milk unless heavily fortified. It is best used as a beverage and for pouring on cereal.
A newer milk alternative, hemp milk may be difficult to find in some places. Its protein level and texture fall in between that of rice and soy milk. It is more watery than regular milk when poured, but has enough protein for use in some cooking applications -- sauces that don't rely on large amounts of protein, for example.
Like hemp milk, oat milk has a moderate amount of protein, making it more useful than rice milk for cooking. However, it's still not a true drop-in replacement for cow's milk in baking.
Oat milk may not be suitable for those with celiac disease, who may be sensitive to avenin protein found in oats. Oat milk is fairly mild and nutty tasting, and is a natural match for hot cereals and many breakfast foods.<|endoftext|>
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# Quick Answer: Why Is 91 Not A Prime Number?
## Which is the smallest prime number?
The smallest prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23.
The number 2 is the only even prime number.
The number 7 has only two factors: 1 and itself.
The number 11 has only two factors: 1 and itself..
## What is the smallest 4 digit prime number?
elevenThe smallest four-digit prime number in an integer base is eleven.
## Is 91 divisible by any number?
Solution: The number 91 is divisible by 1, 7, 13 and 91. Therefore 91 is composite since it has more than two factors. Summary: Divisibility tests can be used to find factors of large whole numbers quickly, and thus determine if they are prime or composite.
## Why is 1001 not a prime number?
A prime number is a natural number, greater than one, that can only be divided by 1 and itself. The number 1001 can be evenly divided by 1, 7, 11, 13, 77, 91, 143 and 1,001, with no remainder. Since 1,001 cannot be divided by just 1 and 1,001, it is not a prime number.
## Is 13 a prime number Yes or no?
When a number has more than two factors it is called a composite number. Here are the first few prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, etc.
## What can 91 be divided by?
The list of all positive divisors (i.e., the list of all integers that divide 91) is as follows: 1, 7, 13, 91. For 91 to be a prime number, it would have been required that 91 has only two divisors, i.e., itself and 1.
## Is 51 a prime numbers?
Yes, 17 is a prime number because it only has two factors, 1 and 17. Is 51 a prime number? No, 51 is NOT a prime number because it has more than two factors. 51 is a composite number and can be factored by any of the following numbers: 1, 3, 17, 51.
## Is 29 divisible by any number?
The numbers that 29 is divisible by are 1 and 29 and 29. You may also be interested to know that all the numbers that 29 is divisible by are also known as the factors of 29. Not only that, but all the numbers that are divisible by 29 are the divisors of 29.
## How can you tell a prime number?
Prime numbers are numbers that have only 2 factors: 1 and themselves. For example, the first 5 prime numbers are 2, 3, 5, 7, and 11. By contrast, numbers with more than 2 factors are call composite numbers.
## What is the easiest way to find a prime number?
To prove whether a number is a prime number, first try dividing it by 2, and see if you get a whole number. If you do, it can’t be a prime number. If you don’t get a whole number, next try dividing it by prime numbers: 3, 5, 7, 11 (9 is divisible by 3) and so on, always dividing by a prime number (see table below).
## Is 91 and 97 a prime number?
The previous prime number is 89. All numbers between 91 and 97 are composite. Among them, 91 is divisible by 7 and 13, 93 by 3, and 95 by 5.
## Why is 91?
The +91 is called the country code. As the phone system was expanded outside of the United States, we realized that we needed to either expand the number of digits for everyone, or to add a prefix to identify the country to which the call should be routed.
## Why is 97 a prime number?
For 97, the answer is: yes, 97 is a prime number because it has only two distinct divisors: 1 and itself (97).
## Is 13 divisible by any number?
650: 65+0*4=65 and number 65 is divisible by 13 and gives divisor as 5. Therefore, 650 is also divisible by 13. … 728: 7*4-28=28-28=0 and number 0 is divisible by 13 giving the result as 0. Divisibility Rule 4: Multiply the last digit by 9 of a number N and subtract it from the rest of the number.
## How do you work out 91 divided by 7?
The answer to the question: What is 91 divided by 7 is as follows:91 / 7 = 13. Instead of saying 91 divided by 7 equals 13, you could just use the division symbol, which is a slash, as we did above. … 91 ÷ 7 = 13.91 over 7 = 13.91⁄7 = 13.
## Is 2.5 a natural number?
Since it is infinite, N can never be exhausted by removing its members one at a time. The set of natural numbers is closed with respect to addition and multiplication, which means that if you add (or multiply) two natural numbers together, you get another natural number. … −2 and 2.5 are not natural numbers.
## Is 0 a real number?
Real numbers consist of zero (0), the positive and negative integers (-3, -1, 2, 4), and all the fractional and decimal values in between (0.4, 3.1415927, 1/2). Real numbers are divided into rational and irrational numbers.
## What is the formula to find prime numbers?
Every prime number can be written in the form of 6n + 1 or 6n – 1 (except the multiples of prime numbers, i.e. 2, 3, 5, 7, 11), where n is a natural number. Method 2: To know the prime numbers greater than 40, the below formula can be used.
## What are not prime numbers?
Nonprime numbers are integers that are not prime numbers, i.e. zero (0), units (e.g. one (1), minus one (–1)), composite numbers and the associates of prime numbers (negated primes). Some nonprime numbers are –2563, 1, 48 and 1729.
## What type of number is 91?
91 (number)← 90 91 92 →Cardinalninety-oneOrdinal91st (ninety-first)Factorization7 × 13Divisors1, 7, 13, 918 more rows
## Is 91 a prime factor?
91 is not prime as it has more than 2 factors. … Factors of 91 are: 1, 7, 13, and 91.<|endoftext|>
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# Subtraction of Numbers without using Number Line
Learn the rules used for subtraction of numbers without using number line. When the numbers are big, the use of number line is not convenient for their subtraction. Also, it is very time consuming to draw a number line every time, and perform the operation of subtraction.
Rules for subtraction of numbers without using number line in different situation:
To subtract a number from another number, the sign of the number (which is to be subtracted) should be changed and then this number with the changed sign, should be added to the first number.
For Example:
(i) Evaluate (+6) – (+2)
= (+6) + (-2) (charging the sign of the number to be subtracted and then adding)
On subtracting smaller number 2 from bigger number 6; we get 6 – 2 = 4
Since, the sign of bigger number is + (positive)
= +4 or 4
Therefore, (+6) – (+2) = 4
(ii) Evaluate (+5) – (-3)
= (+5) + (+3) (charging the sign of the number to be subtracted and then adding)
We know, to add a positive (+ ve) number to a positive (+ ve) number, the numbers should be added and positive sign should be attached to the sum obtained.
= +8
Therefore, (+5) – (-3) = 8
(iii) Evaluate (-7) – (+2)
= (-7) + (-2) (charging the sign of the number to be subtracted and then adding)
We know, to add a negative number to a negative number, the numbers should be added and negative sign should be attached to the sum obtained.
= -9
Therefore, (-7) – (+2) = -9
(iv) Evaluate (-9) – (-6)
= (-9) + (+6) (charging the sign of the number to be subtracted and then adding)
On subtracting smaller number 6 from bigger number 9; we get 9 – 6 = 3
As the sign of bigger number is negative (-)
= -3
Therefore, (-9) – (-6) = -3
6th Grade Math Practice
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Write-up #6
Exploring Triangles and Medians Triangles
by
Holly Anthony
Fall 2001
Problem: Construct a triangle and its medians. Construct a second triangle with the three sides having the lengths of the three medians from your first triangle. Find some relationship between the two triangles. (E.g., are they congruent? similar? have same area? same perimeter? ratio of areas? ratio of perimeters?) Prove whatever you find.
Let's begin our exploration with the construction of the triangle ABC and its medians AD, BE, and CF.
From this triangle, let's construct a second triangle with the three sides having the lengths of the three medians from our first triangle ABC. This gives us one construction for its Medians triangle. Using segment CF as the base of the new triangle, a line parallel to AD through point F and a line parallel to BE through point C can be constructed. We see that a triangle is created by the intersection of the parallel lines. The new triangle is the Medians triangle.
Observations that I made from exploring this construction include:
• The point C is a vertex for both triangles.
• The centroid of the medians triangle is the midpoint of BC in the original triangle.
• The midpoint of AB is a vertex of the medians triangle.
After some investigations of several triangles and their medians triangles, I also found that when triangle ABC is isosceles, its Medians triangle DBE is also an isosceles triangle.
However, I did not find any similar relationships when ABC was a right triangle, an obtuse triangle, or an acute triangle.
Now, let's examine the following questions.
Are the two triangles congruent?
Are they similar?
Do they have the same perimeter?
Do they have the same area?
What about a ratio of perimeters?
What about a ratio of areas?
After some investigations, I found that there are basically no relationships between the two triangles regarding perimeter. I also found that the two triangles are not congruent or similar.
However, after explorations, and prompting from Dr. Wilson, I found that the area of the median triangle is 3/4 the area of its original triangle ABC.
A Proof of this follows and was adopted from a write-up done by Robyn Bryant and Kaycie Maddox.
The proof will be based on the following picture:
Proof: We want to show that the area of triangle EIC is 3/4 the area of triangle ABC. BICD is a parallelogram, therefore triangle BIC and triangle BCD have congruent areas.
The area of triangle IJC is 1/2 * JC *h
The area of triangle BIC is 1/2 *BC*h
Because EJ is parallel to AF and EJ bisects AB then EJ also bisects BF. As a result BJ = JF and therefore JC =3/4 (BC). We will now substitute 3/4 (BC) for JC. We now have the following equations.
The area of triangle IJC = 1/2 [3/4(BC)]h
The area of triangle BIC = 1/2(BC)h
The area of triangle ABC is twice the area of triangle BDC because of the definition of medians. (Remember that the area of triangle BDC is = the area of triangle BIC.)
The area of triangle ABC = 2[1/2(BC)]h
The area of triangle EIC is twice the area of triangle IJC.
The area of triangle EIC = 2[1/2 * 3/4(BC)]h
After cancelling out the fractions we are left with the fact that:
The area of triangle EIC = 3/4 [BC * h]
The area of triangle ABC = BC * h
If those 2 formulas are placed in a ratio the [BC * h] cancels out leaving us with the ratio of areas of triangle EIC and ABC to be three-fourths.
End of proof<|endoftext|>
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Abnormally Slow Heartbeat (Bradycardia)
What is bradycardia?
Bradycardia means that the heart beats too slowly, usually less than 60 beats per minute. As a result, your body doesn’t get enough oxygen and nutrients.
It may be caused by a problem with specialized tissue called the sinus or S-A node. This tissue is your heart's “natural pacemaker”. Or the problem could relate to your heart's electrical pathways.
Some people don’t have any symptoms, and if they do, they may be subtle. Others may feel tired, light-headed or dizzy — and may faint.
Your doctor will probably do an electrocardiogram (ECG) test to look for heart rhythm problems.
How is a slow heartbeat treated?
If you have symptoms, you will likely need treatment. Doctors may recommend medications, a pacemaker or both. A pacemaker is an electronic device that sends electrical impulses to the heart to restore a normal heartbeat.<|endoftext|>
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As transistors get smaller, they also grow less reliable. Increasing their operating voltage can help, but that means a corresponding increase in power consumption.
With information technology consuming a steadily growing fraction of the world's energy supplies, some researchers and hardware manufacturers are exploring the possibility of simply letting chips botch the occasional computation. In many popular applications -- video rendering, for instance -- users probably wouldn't notice the difference, and it could significantly improve energy efficiency.
At this year's Object-Oriented Programming, Systems, Languages and Applications (OOPSLA) conference, researchers from MIT's Computer Science and Artificial Intelligence Laboratory presented a new system that lets programmers identify sections of their code that can tolerate a little error. The system then determines which program instructions to assign to unreliable hardware components, to maximize energy savings while still meeting the programmers' accuracy requirements.
The system, dubbed Chisel, also features a tool that helps programmers evaluate precisely how much error their programs can tolerate. If 1 percent of the pixels in an image are improperly rendered, will the user notice? How about 2 percent, or 5 percent? Chisel will simulate the execution of the image-rendering algorithm on unreliable hardware as many times as the programmer requests, with as many different error rates. That takes the guesswork out determining accuracy requirements.
The researchers tested their system on a handful of common image-processing and financial-analysis algorithms, using a range of unreliable-hardware models culled from the research literature. In simulations, the resulting power savings ranged from 9 to 19 percent.
The new work builds on a paper presented at last year's OOSPLA, which described a programming language called Rely. Each paper won one of the conference's best-paper awards.
Rely provides the mechanism for specifying the accuracy requirements, and it features an operator -- a period, or dot -- that indicates that a particular instruction may be executed on unreliable hardware. In the work presented last year, programmers had to insert the dots by hand. Chisel does the insertion automatically -- and guarantees that its assignment will maximize energy savings.
"One of the observations from all of our previous research was that usually, the computations we analyzed spent most of their time on one or several functions that were really computationally intensive," says Sasa Misailovic, a graduate student in electrical engineering and computer science and lead author on the new paper. "We call those computations 'kernels,' and we focused on them."
Misailovic is joined on the paper by his advisor, Martin Rinard, a professor in the Department of Electrical Engineering and Computer Science (EECS); by Sara Achour and Zichao Qi, who are also students in Rinard's group; and by Michael Carbin, who did his PhD with Rinard and will join the EECS faculty next year.
In practice, Misailovic says, programs generally have only a few kernels. In principle, Chisel could have been designed to find them automatically. But most developers who work on high-performance code will probably want to maintain a degree of control over what their programs are doing, Rinard says. And generally, they already use tools that make kernel identification easy.
A single kernel, however, may still consist of 100 or more instructions, any combination of which could be assigned to unreliable hardware. Manually canvassing all possible combinations and evaluating their effects on both computational accuracy and energy savings would still be a prohibitively time-consuming task.
But the researchers developed three separate mathematical expressions that describe accuracy of computation, reliability of instruction execution, and energy savings as functions of the individual instructions. These expressions constrain the search that the system has to perform to determine which instructions to assign to unreliable hardware. That simpler -- though still complex -- problem is one that off-the-shelf software can handle.
Written by Larry Hardesty, MIT News Office
How to program unreliable chips
New mathematical framework formalizes oddball programming techniques<|endoftext|>
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At the Ross Ice Shelf in Antarctica, scientists used a hot-water drill hose to create a hole through the thick ice until they reached the perpetually dark water. What they found surprised them. This is the Ross Ice Shelf – the biggest floating ice shelf in Antarctica. Such shelves are important because they hold back a vast amount of ice.
image/text credit: National Geographic
If all such West Antarctic shelves were to collapse and spill the ice into the ocean then global sea level would rise by 10 feet. Beneath the Ross Ice Shelf is one of the least explored bits of ocean on Earth.
New Zealand scientists used a hot-water drill hose to create a hole through the thick ice until they reached the perpetually dark water. They hoped to study the health and history of the shelf.
Their findings surprised them. They found that the ice in the hole itself and along the base of the shelf was crystalizing and freezing rather than melting. Measurements will be taken for the next few years, to see how the Ross Ice Shelf is changing over time.<|endoftext|>
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Encryption key management is the administration of tasks involved with protecting, storing, backing up and organizing encryption keys.
High-profile data losses and regulatory compliance requirements have caused a dramatic increase in the use of encryption in the enterprise. A single enterprise might use several dozen different and possibly incompatible encryption tools, resulting in thousands of encryption keys. Each key must be securely stored, protected and retrievable.
There are several encryption key management standards efforts underway. The best known is the Key Management Interoperability Protocol (KMIP) developed by vendors and submitted to the Organization for the Advancement of Structured Information Standards (OASIS). The goal of KMIP is to allow users to attach any encryption device to a key management system.
Cryptographic key management types
Encryption keys apply complicated algorithms to data and then convert that data into streams of apparently random alphanumeric characters. There are two basic types of cryptographic algorithms: symmetric (private key) and asymmetric (public key).
Symmetric key algorithms use a single key to secure communications and achieve confidentiality, integrity and authentication. A private key is a variable that is used with an algorithm to encrypt and decrypt code. While the algorithm doesn't need to be kept secret, the key does.
Asymmetric algorithms provide each user with a public key and a private key. Users can freely distribute the public key, while keeping the private key secret. These key pairs achieve all four goals of cryptography: confidentiality, integrity, authentication and nonrepudiation. A public key infrastructure supports the distribution and identification of public encryption keys, which allow users and computers to securely exchange data over networks such as the internet and verify the other party's identity.
Encryption key storage and backup
Key management means protecting encryption keys from loss, corruption and unauthorized access. Many processes can be used to control key management, including changing the keys regularly, and managing how keys are assigned and who gets them. In addition, organizations must evaluate whether one key should be used for all backup types or if each type should have its own key.
The importance of encryption key management cannot be overstated. Unless the creation, secure storage, handling and deletion of encryption keys are carefully monitored, unauthorized parties can gain access to them. When keys are lost or corrupted, it can lead to loss of access to systems and data, as well as make a system completely useless unless it is reformatted and reinstalled.
Encryption key management tips
- Key management is not simple: If encrypting data, somebody must manage the keys, and you must have key recovery procedures in place.
- Within large businesses, key management processes must be capable of being distributed across multiple business functions with the same standards, rules and quality levels.
- Have one point of contact for cryptography; don't spread it among operational users.
- Ensure the central key repository is well-protected.
- Decide whether your outsourcer will have any role in key management, such as key pair generation, recovery of keys and escrow access.
- Decide whether you will have a "two person" rule for aspects of key management.
- Decide whether information security should manage keys, as well as encryption policy. In practice, during development, key management is the responsibility of the project, but handed over to security at implementation.
Source: The Corporate IT Forum's Information Security Service
Having more than one person in charge of storing, backing up, referencing and rotating encryption keys is essential. The roles of key players should be defined, and the encryption key management policy should be accessible to everyone on an intranet site.
A major challenge is the lack of unified tools to reduce management overhead. A key management system purchased from one vendor can't manage the keys from another vendor because each one implements encryption in a slightly different way.
Amazon Web Services Key Management Service enables users to create and manage cryptographic keys that protect data in AWS. Encryption keys are generally used in a single region. For example, data encrypted in one region would be encrypted with a different key than a replicated version of that data stored in a different region. The challenge is keeping track of the keys used to encrypt data across regions.
Key aliases alleviate some of the cloud encryption key management burden by allowing administrators to associate a name string with each key. These aliases can be used in multiple regions. Code running in multiple regions doesn't need to manage multiple keys, and can instead refer to a key alias.
Key aliases prove helpful when rotating keys, as it lessens the risk of data leaks due to compromised keys. Key rotation serves the same purpose as changing passwords. If others learn an administrative password, they can access that account for as long as that password is in place. Likewise, a former employee with a copy of an encryption key or someone who obtained a copy of an encryption key can access all the data encrypted with that key.
The AWS Key Management Service allows admins to update a key alias without deleting the previous key first. While key aliases are not keys, their names should be protected. Aliases should be descriptive and easily distinguishable from one another so keys are not applied to the wrong data due to an unclear key alias.<|endoftext|>
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# High School Math : How to find the length of a radius
## Example Questions
← Previous 1
### Example Question #1 : How To Find The Length Of A Radius
In a large field, a circle with an area of 144π square meters is drawn out. Starting at the center of the circle, a groundskeeper mows in a straight line to the circle's edge. He then turns and mows ¼ of the way around the circle before turning again and mowing another straight line back to the center. What is the length, in meters, of the path the groundskeeper mowed?
24 + 36π
24 + 6π
12 + 36π
24π
12 + 6π
24 + 6π
Explanation:
Circles have an area of πr2, where r is the radius. If this circle has an area of 144π, then you can solve for the radius:
πr2 = 144π
r 2 = 144
r =12
When the groundskeeper goes from the center of the circle to the edge, he's creating a radius, which is 12 meters.
When he travels ¼ of the way around the circle, he's traveling ¼ of the circle's circumference. A circumference is 2πr. For this circle, that's 24π meters. One-fourth of that is 6π meters.
Finally, when he goes back to the center, he's creating another radius, which is 12 meters.
In all, that's 12 meters + 6π meters + 12 meters, for a total of 24 + 6π meters.
### Example Question #2 : How To Find The Length Of A Radius
Two concentric circles have circumferences of 4π and 10π. What is the difference of the radii of the two circles?
5
6
7
4
3
3
Explanation:
The circumference of any circle is 2πr, where r is the radius.
Therefore:
The radius of the smaller circle with a circumference of 4π is 2 (from 2πr = 4π).
The radius of the larger circle with a circumference of 10π is 5 (from 2πr = 10π).
The difference of the two radii is 5-2 = 3.
### Example Question #352 : Geometry
In the figure above, rectangle ABCD has a perimeter of 40. If the shaded region is a semicircle with an area of 18π, then what is the area of the unshaded region?
96 – 36π
204 – 18π
96 – 18π
336 – 36π
336 – 18π
96 – 18π
Explanation:
In order to find the area of the unshaded region, we will need to find the area of the rectangle and then subtract the area of the semicircle. However, to find the area of the rectangle, we will need to find both its length and its width. We can use the circle to find the length of the rectangle, because the length of the rectangle is equal to the diameter of the circle.
First, we can use the formula for the area of a circle in order to find the circle's radius. When we double the radius, we will have the diameter of the circle and, thus, the length of the rectangle. Then, once we have the rectangle's length, we can find its width because we know the rectangle's perimeter.
Area of a circle = πr2
Area of a semicircle = (1/2)πr2 = 18π
Divide both sides by π, then multiply both sides by 2.
r2 = 36
Take the square root.
r = 6.
The radius of the circle is 6, and therefore the diameter is 12. Keep in mind that the diameter of the circle is also equal to the length of the rectangle.
If we call the length of the rectangle l, and we call the width w, we can write the formula for the perimeter as 2l + 2w.
perimeter of rectangle = 2l + 2w
40 = 2(12) + 2w
Subtract 24 from both sides.
16 = 2w
w = 8.
Since the length of the rectangle is 12 and the width is 8, we can now find the area of the rectangle.
Area = l x w = 12(8) = 96.
Finally, to find the area of just the unshaded region, we must subtract the area of the circle, which is 18π, from the area of the rectangle.
area of unshaded region = 96 – 18π.
The answer is 96 – 18π.
### Example Question #1 : How To Find The Length Of A Radius
Consider a circle centered at the origin with a circumference of . What is the x value when y = 3? Round your answer to the hundreths place.
10.00
5.77
5.8
None of the available answers
5.778
5.77
Explanation:
The formula for circumference of a circle is , so we can solve for r:
We now know that the hypotenuse of the right triangle's length is 13.5. We can form a right triangle from the unit circle that fits the Pythagorean theorem as such:
Or, in this case:
### Example Question #111 : Plane Geometry
What is the radius of a circle with a circumference of ?
Explanation:
To find the radius of a circle given the circumference we must first know the equation for the circumference of a circle which is
Then we plug in the circumference into the equation yielding
We then divide each side by giving us
The answer is .
### Example Question #1 : How To Find The Length Of A Radius
A circle has an area of 36π inches. What is the radius of the circle, in inches?
9
36
18
6
6
Explanation:
We know that the formula for the area of a circle is πr2. Therefore, we must set 36π equal to this formula to solve for the radius of the circle.
36π = πr2
36 = r2
6 = r
### Example Question #162 : Plane Geometry
Circle X is divided into 3 sections: A, B, and C. The 3 sections are equal in area. If the area of section C is 12π, what is the radius of the circle?
Circle X
7
√12
6
4
6
Explanation:
Find the total area of the circle, then use the area formula to find the radius.
Area of section A = section B = section C
Area of circle X = A + B + C = 12π+ 12π + 12π = 36π
Area of circle = where r is the radius of the circle
36π = πr2
36 = r2
√36 = r
6 = r
### Example Question #2 : How To Find The Length Of A Radius
The specifications of an official NBA basketball are that it must be 29.5 inches in circumference and weigh 22 ounces. What is the approximate radius of the basketball?
3.06 inches
5.43 inches
9.39 inches
14.75 inches
4.70 inches
4.70 inches
Explanation:
To Find your answer, we would use the formula: C=2πr. We are given that C = 29.5. Thus we can plug in to get [29.5]=2πr and then multiply 2π to get 29.5=(6.28)r. Lastly, we divide both sides by 6.28 to get 4.70=r. (The information given of 22 ounces is useless)
### Example Question #112 : Plane Geometry
If the circumference of a circle is , what is the radius?
Explanation:
The formula for circumference is .
Plug in our given information.
Divide both sides by .
### Example Question #92 : Circles
Find the radius of a circle with area<|endoftext|>
| 4.625 |
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The Insular Celts are the speakers of the Insular Celtic languages, which comprise all the living Celtic languages as well as their precursors, but the term is mostly used in reference to the peoples of the British Iron Age prior to the Roman conquest, and their contemporaries in Ireland.
According to older theories, the Insular Celtic languages spread throughout the islands in the course of the insular Iron Age. But this is now doubted by most scholars, who see the languages as already present, and possibly dominant, in the Bronze Age, and perhaps earlier. At some point the languages split into the two major groups, Goidelic in Ireland and Brittonic in Great Britain, corresponding to the population groups of the Goidels (Gaels) on one hand and the Britons and the Picts on the other. The extent to which these peoples ever formed a distinct ethnic group remains unclear. While there are early records of the Continental Celtic languages, allowing a comparatively confident reconstruction of Proto-Celtic, Insular Celtic languages become attested in connected texts only at the end of the Dark Ages, from around the 7th century AD, by which time they had become mutually incomprehensible.
Celtic settlement of Britain and Ireland
In older theories, the arrival of Celts, defined as speakers of Celtic languages, which derive from a Proto-Celtic language, roughly coincided with the beginning of the European Iron Age. In 1946 the Celtic scholar T. F. O'Rahilly published his influential model of the early history of Ireland, which postulated four separate waves of Celtic invaders, spanning most of the Iron Age (700 to 100 BCE). However the archaeological evidence for these waves of invaders proved elusive. Later research indicated that the culture may have developed gradually and continuously between the Celts and the indigenous populations. Similarly in Ireland little archaeological evidence was found for large intrusive groups of Celtic immigrants, suggesting to archaeologists such as Colin Renfrew that the native late Bronze Age inhabitants gradually absorbed European Celtic influences and language.
In the 1970s a "continuity model" was popularized by Colin Burgess in his book The Age of Stonehenge which theorised that Celtic culture in Great Britain "emerged" rather than resulted from invasion and that the Celts were not invading aliens, but the descendants of, or culturally influenced by, figures such as the Amesbury Archer, whose burial included clear Continental connections.
The archaeological evidence is of substantial cultural continuity through the 1st millennium BCE, although with a significant overlay of selectively adopted elements of the "Celtic" La Tène culture from the 4th century BCE onwards. There are claims of continental-style states appearing in southern England close to the end of the period, possibly reflecting in part immigration by élites from various Gallic states such as those of the Belgae. Evidence of chariot burials in England begins about 300 BC and is mostly confined to the Arras culture associated with the Parisii.
Remnants of pre-Celtic languages may remain in the names of some geographical features, such as the rivers Clyde, Tamar and Thames, whose etymology is unclear but possibly derive from a pre-Celtic substrate (Gelling).
It is thought that by about the 6th century BCE most of the inhabitants of the isles of Ireland and Britain were speaking Celtic languages. A controversial phylogenetic linguistic analysis of 2003 puts the age of Insular Celtic a few centuries earlier, at 2,900 years before present, or slightly earlier than the European Iron Age.
It is not entirely clear if there was ever a "Common Insular Celtic" language, the alternative being that the Celtic settlement of Ireland and Great Britain was undertaken by separate populations speaking separate Celtic dialects from the beginning. However, the "Insular Celtic hypothesis" has been favoured as the most probable scenario in Celtic historical linguistics since the later 20th century (supported by e.g. Cowgill 1975; McCone 1991, 1992; and Schrijver 1995). This would point to a single wave of immigration of early Celts (Hallstatt D) to both Great Britain and Ireland, which however divided into two isolated groups (one in Ireland and one in Great Britain) soon after their arrival, placing the split of Insular Celtic into Goidelic and Brythonic close to 500 BCE. However, this is not the only possible interpretation. In an alternative scenario, the migration could have brought early Celts first to Britain (where a largely undifferentiated Insular Celtic was spoken initially), from whence Ireland was colonised only later. Schrijver has pointed out that according to the absolute chronology of sound changes found in Kenneth Jackson's "Language and History in Early Britain", British and Goidelic were still essentially identical as late as the mid-1st century CE apart from the P/Q isogloss, and that there is no archaeological evidence pointing to Celtic presence in Ireland prior to about 100 BCE.
The Goidelic branch would develop into Primitive Irish, Old Irish and Middle Irish, and only with the historical (medieval) expansion of the Gaels would it split into the modern Gaelic languages (Modern Irish, Scottish Gaelic, Manx). Common Brythonic, on the other hand, split into two branches, British and Pritenic as a consequence of the Roman invasion of Britain in the 1st century. By the 8th century, Pritenic had developed into Pictish (which would be extinct during the 9th century or so), and British had split into Old Welsh and Old Cornish.
Genetic studies have supported the prevalence of native populations. A study in 2003 by Christian Capelli, David Goldstein and others at University College London showed that genetic markers associated with Gaelic names in Ireland and Scotland are also common in certain parts of Wales and England (in most cases, the Southeast of England with the lowest counts of these markers), and are similar to the genetic markers of the Basque people and most different from Danish and North German people. This similarity supported earlier findings in suggesting a large pre-Celtic genetic ancestry, likely going back to the original settlement of the Upper Paleolithic. The authors suggest, therefore, that Celtic culture and the Celtic language may have been imported to Britain at the beginning of the Iron Age by cultural contact, not "mass invasions". In 2006, two popular books, The Blood of the Isles by Bryan Sykes and The Origins of the British: a Genetic Detective Story by Stephen Oppenheimer discuss genetic evidence for the prehistoric settlement of the British Isles, concluding that while there is evidence for a series of migrations from the Iberian Peninsula during the Mesolithic and, to a lesser extent, the Neolithic eras, there is comparatively little trace of any Iron Age migration.
Later genetic studies regarding Y-DNA Haplogroup I-M284 did find evidence for some Late Iron Age migration of Celtic (La Tène) people to Britain and onto north-east Ireland.
Iron Age Britain
The British Iron Age is a conventional name in the archaeology of Great Britain, typically excluding prehistoric Ireland, which had an independent Iron Age culture of its own. The parallel phase of Irish archaeology is termed the Irish Iron Age.
The British Iron Age lasted in theory from the first significant use of iron for tools and weapons in Britain to the Romanisation of the southern half of the island. The Romanised culture is termed Roman Britain and is considered to supplant the British Iron Age.
The only surviving description of the Iron Age populations of the British Isles is that of Pytheas, who travelled to the region in about 325 BC. The earliest tribal names on record date to the 1st century AD (Ptolemy, Caesar; to some extent coinage), representing the situation at the moment of Roman conquest.
Roman era and Dark Ages
Roman Britain existed for about four centuries, from the mid 1st to the mid 5th century. This led to the formation of a syncretized Romano-British culture in the southern part of Great Britain, comparable in some aspects to Gallo-Roman culture on the continent. However, while in Gaul Roman influence was sufficient to almost wholly replace the Gaulish language with Vulgar Latin, this was nowhere near the case in Roman Britain. Although a British Latin dialect was presumably spoken in the population centres of Roman Britain, it did not become influential enough to displace British dialects spoken throughout the country. There did presumably remain pockets of Romance-speaking populations in Britain as late as the 8th century.
Northern Britain (north of the Antonine Wall) and Ireland would essentially remain in the prehistoric period until after the end of the Roman period. The "protohistoric" period of Ireland can be argued to begin around 400 AD, due to cultural diffusion from Roman Britain, importing writing (ogham, reflecting the earliest records of Primitive Irish) and Christianity. The populations north of Roman Britain are summarized under the term Caledonians (the ancestors of the Picts of later centuries). Very little is known about them other than they posed a constant military threat to the Roman border.
With the Anglo-Saxon invasion and settlement of Great Britain in the 5th and 6th centuries, the British languages were gradually marginalised to the western parts of the island, to what is now Wales and Cornwall. The transition may not necessarily present itself as a mass immigration with a substantial replacement of population, but rather could involve the arrival of a new elite installing their culture and language as a superstrate. A similar process happened as the Gaels installed themselves over the formerly Pictish-speaking populations in Northern Britain. There seems to have been a period of British-Saxon syncretism during the 6th century, with British rulers bearing Saxon names (as in Tewdrig) and Saxon rulers bearing British names (as in Cerdic).
- Bede in the early 8th century is explicit on the Britons, the Irish, and the Pict speaking distinct languages; he also notes that Columba, a Gael, required an interpreter to communicate with the Picts.
- Cunliffe, Barry (2008). A Race Apart: Insularity and Connectivity in Proceedings of the Prehistoric Society 75, 2009, pp. 55–64. The Prehistoric Society. p. 61.
- Koch, John (2005). Celtic Culture : A Historical Encyclopedia. ABL-CIO. pp. 197–198. ISBN 978-1-85109-440-0. Retrieved March 12, 2011.
- Russell D. Gray & Quentin D. Atkinson, "Language-tree divergence times support the Anatolian theory of Indo-European origin", Nature, 2003.
- Capelli, Cristian; et al. (May 27, 2003). "A Y Chromosome Census of the British Isles" (PDF). Current Biology. 13: 979–984. doi:10.1016/S0960-9822(03)00373-7. PMID 12781138.
- McEvoy and Bradley, Brian P and Daniel G (2010). Celtic from the West Chapter 5: Irish Genetics and Celts. Oxbow Books, Oxford, UK. p. 117. ISBN 978-1-84217-410-4.
- Cunliffe (2005) page 27.
- Raftery, Barry (2005). "Iron-age Ireland". In O Croinin, Daibhi. Prehistoric and Early Ireland: Volume I. Oxford University Press. pp. 134–181. ISBN 0-19-821737-4. ISBN 978-0-19-821737-4.
- Loyn, Anglo-Saxon England and the Norman Conquest, 2nd ed. 1991:11
- Pattison, John E. (2008), Is it Necessary to Assume an Apartheid-like Social Structure in Early Anglo-Saxon England?, Proceedings of the Royal Society B 275(1650) 2423-2429; doi:10.1098/rspb.2008.0352
- "Records of the West Saxon dynasties survive in versions which have been subject to later manipulation, which may make it all the more significant that some of the founding 'Saxon' fathers have British names: Cerdic, Ceawlin, Cenwalh." in: Hills, C., Origins of the English, Duckworth (2003), p. 105. Also "The names Cerdic, Ceawlin and Caedwalla, all in the genealogy of the West Saxon kings, are apparently British." in: Ward-Perkins, B., Why did the Anglo-Saxons not become more British? The English Historical Review 115.462 (June 2000) p513. P. Sims-Williams, Religion and Literature [in Western England, 600–800], Cambridge 1990, p. 26.<|endoftext|>
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26. 2 pipes can fill a tank in 20 and 24 mins respectively. And a waste pipe can empty 3 gallons per minute.All the 3 pipes working together can fill the tank in 15 minutes.The capacity of the tank is ?
60 gallons
100 gallons
120 gallons
180 gallons
```Answer
Answer: Option C Explanation:Work done by the waste pipe in 1 minute = 1/15 - (1/20 + 1/24) = ( 1/15 - 11/120) = (- 1/40 (negative sign means emptying)
Therefore volume of 1/40 part = 3 gallons.
Volume of whole = (3*40) gallons = 120 gallons. ```
Report
27. 3 tapes A,B & C can fill a tank in 12,15 and 20 hrs respectively.If A is open all the time and B and C are open for 1 hr each alternately ,the tank shall be full in?
6 hrs
6 2/3 hrs
5 hrs
7 hrs
```Answer
Answer: Option D Explanation:(A+B)'s 1 hour work = (1/12 + 1/15) = 9/60 = 3/20
(A+C)'s 1 hour work = (1/12+1/20) = 8/60 = 2/50
Part filled in 2 hrs = (3/20+2/15) = 17/60 : part filled in 6 hrs = (3*17/60) = 17/20
Remaining part =( 1-17/20) = 3/20
Now it is the turn off A and B and 3/20 part is filled by A & B in 1 hr
therefore total time taken to fill the tank = (6+1) hrs = 7 hrs. ```
Report
28. Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full?
4
4 1/2
5
5 1/2
```Answer
Answer: Option C Explanation:A's work in 1 hour = 1/6, B's work in 1 hour = 1/4.
(A+B)'s 2 hour's work when opened alternately = (1/6+1/4) = 5/12.
(A+B)'s 4 hour's work when opened alternately = 10/12 = 5/6.
Remaining part = (1- 5/6) = 1/6.
Now, it is A's turn and 1/6 part is filled by A in 1 hour.
Therefore total time taken to fill the tank = (4+1) hrs = 5 hrs. ```
Report
29. 2 pipes A & B can fill a cistern in 12 mins and 15 mins respectively while a 3'rd pipe C can empty the full tank in 6 mins.A & B are kept open for 5 mins in the beginning then C is also opened. In what time is the cistern emptied?
30 mins
33 mins
37 1/2 mins
45 mins
```Answer
Answer: Option D Explanation:Part filled in 5 minute = 5 (1/12+1/15) = (5*9/60) = 3/4
Part emptied in 1 minute when all the pipes are opened
=1/6 - (1/12 + 1/15 ) = ( 1/6-3/20) = 1/60
now 1/60 part is emptied in 1 minute
Therefore 3/4 part will be emptied in ( 60*3/4) = 45 mins ```
Report
30. A large tanker can be filled by 2 pipes A & B in 60 minutes and 40 minutes respectively.How many minutes will it take to fill the tanker from empty state.If B is used for half the time and A & B fill it together for the half ?
15 minutes
20 minutes
27.5 minutes
30 minutes
```Answer
Answer: Option D Explanation:Part filled by (A+B) in 1 minute = ( 1/60+1/40) = 1/24
suppose the tank is filled in x minutes
Then x/2 (1/24 + 1/40) = 1 = x/2 * 1/15 = 1 = x= 30 minutes ```
Report<|endoftext|>
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Melanoma is a type of skin cancer. It is the less common form of skin cancer but it can be more serious because it is more likely to spread to other parts of the body. Melanoma arises from the type of cells that give moles their dark colors. These cells can be found in the skin, eyes, digestive system, nail beds, or lymph nodes. Although melanoma is most common in the skin it may also arise in these other areas.
Cancer occurs when cells in the body divide without control or order. Eventually these uncontrolled cells form a growth or tumor. The term cancer refers to malignant growths. These growths invade nearby tissues and spread to other parts of the body. It is not clear exactly what causes these problems in the cells but is probably a combination of genetics and environment.
The most common risk factor for melanoma is exposure to ultraviolet radiation. The most common source of this radiation comes form the sun but it is also found in sun lamps and tanning booths.
Melanoma is found most often in older adults but it can happen in young adults. Other factors that increase the risk of melanoma include:
- Certain types of moles called dysplastic nevi, or atypical moles
- Large nevi present at birth
- Fair skin, freckling
- Red or blonde hair
- Light-colored eyes
- Caucasian race
- Family members with melanoma
- Excessive skin exposure to the sun without protective clothing or sunscreen
- Suppressed immune system
Melanomas are not usually painful. They often have no symptoms at first.
The first sign is often a change in the size, shape, color, or feel of an existing mole. Melanoma may also appear as a new, dark, discolored, or abnormal mole. Remember that most people have moles. Almost all moles are benign.
The following are signs that a mole may be a melanoma:
- Uneven shape
- Ragged edges
- Uneven color
- Large size
(See image on the right side below which shows an irregular border compared to the image on the left which is round with an even border.)
Your doctor will look at your skin and moles. A sample of area will be removed and sent to a lab for closer examination. Your doctor may also examine lymph nodes. Enlarged lymph nodes may suggest the spread of melanoma. A sample of lymph node tissue may also be removed for testing.
Treatment options may include one or more of the following:
- Radiation Therapy
To reduce your chance of getting melanoma:
- Avoid spending too much time in the sun
- Protect your skin from the sun:
- Wear a shirt, wide-brim hat, and sunglasses
- Use sunscreens with a sun protection factor (SPF) of at least 15
- Avoid exposure from:
- 10 a.m. and 2 p.m. (standard time)
- 11 a.m. and 3 p.m. (daylight savings time)
- Avoid sun lamps and tanning booths
Early diagnosis and treatment is important. Take the following steps to find melanoma in its early stages:
- If you have many moles or a family history of melanoma, have your skin checked regularly for changes in moles.
- Ask your doctor to show you how to do a skin self-exam. Do self-exams to look for any new or changing moles.
Content was created using EBSCO’s Health Library. Edits to original content made by Rector and Visitors of the University of Virginia. This information is not a substitute for professional medical advice.<|endoftext|>
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An electrolyte is a substance that produces an electrically conducting solution when dissolved in a polar solvent, such as water. The dissolved electrolyte separates into cations and anions, which disperse uniformly through the solvent. Electrically, such a solution is neutral. If an electric potential is applied to such a solution, the cations of the solution are drawn to the electrode that has an abundance of electrons, while the anions are drawn to the electrode that has a deficit of electrons. The movement of anions and cations in opposite directions within the solution amounts to a current. This includes most soluble salts, acids, and bases. Some gases, such as hydrogen chloride, under conditions of high temperature or low pressure can also function as electrolytes. Electrolyte solutions can also result from the dissolution of some biological (e.g., DNA, polypeptides) and synthetic polymers (e.g., polystyrene sulfonate), termed "polyelectrolytes", which contain charged functional groups. A substance that dissociates into ions in solution acquires the capacity to conduct electricity. Sodium, potassium, chloride, calcium, magnesium, and phosphate are examples of electrolytes.
In medicine, electrolyte replacement is needed when a person has prolonged vomiting or diarrhea, and as a response to strenuous athletic activity. Commercial electrolyte solutions are available, particularly for sick children (such as oral rehydration solution, Suero Oral, or Pedialyte) and athletes (sports drinks). Electrolyte monitoring is important in the treatment of anorexia and bulimia.
The word electrolyte derives from the Greek lytós, meaning "able to be untied or loosened".
Svante Arrhenius put forth, in his 1884 dissertation, his explanation of the fact that solid crystalline salts disassociate into paired charged particles when dissolved, for which he won the 1903 Nobel Prize in Chemistry.
Arrhenius's explanation was that in forming a solution, the salt dissociates into charged particles, to which Michael Faraday had given the name "ions" many years earlier. Faraday's belief had been that ions were produced in the process of electrolysis. Arrhenius proposed that, even in the absence of an electric current, solutions of salts contained ions. He thus proposed that chemical reactions in solution were reactions between ions.
Electrolyte solutions are normally formed when a salt is placed into a solvent such as water and the individual components dissociate due to the thermodynamic interactions between solvent and solute molecules, in a process called "solvation". For example, when table salt (sodium chloride), NaCl, is placed in water, the salt (a solid) dissolves into its component ions, according to the dissociation reaction
- NaCl(s) → Na+(aq) + Cl−(aq)
It is also possible for substances to react with water, producing ions. For example, carbon dioxide gas dissolves in water to produce a solution that contains hydronium, carbonate, and hydrogen carbonate ions.
Molten salts can also be electrolytes as, for example, when sodium chloride is molten, the liquid conducts electricity. In particular, ionic liquids, which are molten salts with melting points below 100 °C, are a type of highly conductive non-aqueous electrolytes and thus have found more and more applications in fuel cells and batteries.
An electrolyte in a solution may be described as "concentrated" if it has a high concentration of ions, or "diluted" if it has a low concentration. If a high proportion of the solute dissociates to form free ions, the electrolyte is strong; if most of the solute does not dissociate, the electrolyte is weak. The properties of electrolytes may be exploited using electrolysis to extract constituent elements and compounds contained within the solution.
Alkaline earth metals form hydroxides that are strong electrolytes with limited solubility in water, due to the strong attraction between their constituent ions. This limits their application to situations where high solubility is not required.
In physiology, the primary ions of electrolytes are sodium (Na+), potassium (K+), calcium (Ca2+), magnesium (Mg2+), chloride (Cl−), hydrogen phosphate (HPO42−), and hydrogen carbonate (HCO3−). The electric charge symbols of plus (+) and minus (−) indicate that the substance is ionic in nature and has an imbalanced distribution of electrons, the result of chemical dissociation. Sodium is the main electrolyte found in extracellular fluid and potassium is the main intracellular electrolyte; both are involved in fluid balance and blood pressure control.
All known higher lifeforms require a subtle and complex electrolyte balance between the intracellular and extracellular environments. In particular, the maintenance of precise osmotic gradients of electrolytes is important. Such gradients affect and regulate the hydration of the body as well as blood pH, and are critical for nerve and muscle function. Various mechanisms exist in living species that keep the concentrations of different electrolytes under tight control.
Both muscle tissue and neurons are considered electric tissues of the body. Muscles and neurons are activated by electrolyte activity between the extracellular fluid or interstitial fluid, and intracellular fluid. Electrolytes may enter or leave the cell membrane through specialized protein structures embedded in the plasma membrane called "ion channels". For example, muscle contraction is dependent upon the presence of calcium (Ca2+), sodium (Na+), and potassium (K+). Without sufficient levels of these key electrolytes, muscle weakness or severe muscle contractions may occur.
Electrolyte balance is maintained by oral, or in emergencies, intravenous (IV) intake of electrolyte-containing substances, and is regulated by hormones, in general with the kidneys flushing out excess levels. In humans, electrolyte homeostasis is regulated by hormones such as antidiuretic hormones, aldosterone and parathyroid hormones. Serious electrolyte disturbances, such as dehydration and overhydration, may lead to cardiac and neurological complications and, unless they are rapidly resolved, will result in a medical emergency.
Measurement of electrolytes is a commonly performed diagnostic procedure, performed via blood testing with ion-selective electrodes or urinalysis by medical technologists. The interpretation of these values is somewhat meaningless without analysis of the clinical history and is often impossible without parallel measurements of renal function. The electrolytes measured most often are sodium and potassium. Chloride levels are rarely measured except for arterial blood gas interpretations, since they are inherently linked to sodium levels. One important test conducted on urine is the specific gravity test to determine the occurrence of an electrolyte imbalance.
In oral rehydration therapy, electrolyte drinks containing sodium and potassium salts replenish the body's water and electrolyte concentrations after dehydration caused by exercise, excessive alcohol consumption, diaphoresis (heavy sweating), diarrhea, vomiting, intoxication or starvation. Athletes exercising in extreme conditions (for three or more hours continuously, e.g. a marathon or triathlon) who do not consume electrolytes risk dehydration (or hyponatremia).
Electrolytes are commonly found in fruit juices, sports drinks, milk, nuts, and many fruits and vegetables (whole or in juice form) (e.g., potatoes, avocados).
When electrodes are placed in an electrolyte and a voltage is applied, the electrolyte will conduct electricity. Lone electrons normally cannot pass through the electrolyte; instead, a chemical reaction occurs at the cathode, providing electrons to the electrolyte. Another reaction occurs at the anode, consuming electrons from the electrolyte. As a result, a negative charge cloud develops in the electrolyte around the cathode, and a positive charge develops around the anode. The ions in the electrolyte neutralize these charges, enabling the electrons to keep flowing and the reactions to continue.
For example, in a solution of ordinary table salt (sodium chloride, NaCl) in water, the cathode reaction will be
- 2H2O + 2e− → 2OH− + H2
and hydrogen gas will bubble up; the anode reaction is
- 2NaCl → 2 Na+ + Cl2 + 2e−
and chlorine gas will be liberated. The positively charged sodium ions Na+ will react toward the cathode, neutralizing the negative charge of OH− there, and the negatively charged hydroxide ions OH− will react toward the anode, neutralizing the positive charge of Na+ there. Without the ions from the electrolyte, the charges around the electrode would slow down continued electron flow; diffusion of H+ and OH− through water to the other electrode takes longer than movement of the much more prevalent salt ions. Electrolytes dissociate in water because water molecules are dipoles and the dipoles orient in an energetically favorable manner to solvate the ions.
In other systems, the electrode reactions can involve the metals of the electrodes as well as the ions of the electrolyte.
Electrolytic conductors are used in electronic devices where the chemical reaction at a metal-electrolyte interface yields useful effects.
- In batteries, two materials with different electron affinities are used as electrodes; electrons flow from one electrode to the other outside of the battery, while inside the battery the circuit is closed by the electrolyte's ions. Here, the electrode reactions convert chemical energy to electrical energy.
- In some fuel cells, a solid electrolyte or proton conductor connects the plates electrically while keeping the hydrogen and oxygen fuel gases separated.
- In electroplating tanks, the electrolyte simultaneously deposits metal onto the object to be plated, and electrically connects that object in the circuit.
- In operation-hours gauges, two thin columns of mercury are separated by a small electrolyte-filled gap, and, as charge is passed through the device, the metal dissolves on one side and plates out on the other, causing the visible gap to slowly move along.
- In electrolytic capacitors the chemical effect is used to produce an extremely thin dielectric or insulating coating, while the electrolyte layer behaves as one capacitor plate.
- In some hygrometers the humidity of air is sensed by measuring the conductivity of a nearly dry electrolyte.
- Hot, softened glass is an electrolytic conductor, and some glass manufacturers keep the glass molten by passing a large current through it.
Solid electrolytes can be mostly divided into four groups:
- Gel electrolytes - closely resemble liquid electrolytes. In essence, they are liquids in a flexible lattice framework. Various additives are often applied to increase the conductivity of such systems.
- Dry polymer electrolytes - differ from liquid and gel electrolytes in the sense that salt is dissolved directly into the solid medium. Usually it is a relatively high dielectric constant polymer (PEO, PMMA, PAN, polyphosphazenes, siloxanes, etc.) and a salt with low lattice energy. In order to increase the mechanical strength and conductivity of such electrolytes, very often composites are used, and inert ceramic phase is introduced. There are two major classes of such electrolytes: polymer-in-ceramic, and ceramic-in-polymer.
- Solid ceramic electrolytes - ions migrate through the ceramic phase by means of vacancies or interstitials within the lattice. There are also glassy-ceramic electrolytes.
- Organic ionic plastic crystals - are a type organic salts exhibiting mesophases (i.e. a state of matter intermediate between liquid and solid), in which mobile ions are orientationally or rotationally disordered while their centers are located at the ordered sites in the crystal structure. They have various forms of disorder due to one or more solid–solid phase transitions below the melting point and have therefore plastic properties and good mechanical flexibility as well as improved electrode|electrolyte interfacial contact. In particular, protic organic ionic plastic crystals (POIPCs), which are solid protic organic salts formed by proton transfer from a Brønsted acid to a Brønsted base and in essence are protic ionic liquids in the molten state, have found to be promising solid-state proton conductors for fuel cells. Examples include 1,2,4-triazolium perfluorobutanesulfonate and imidazolium methanesulfonate.
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- Jiangshui Luo; Annemette H. Jensen; Neil R. Brooks; Jeroen Sniekers; Martin Knipper; David Aili; Qingfeng Li; Bram Vanroy; Michael Wübbenhorst; Feng Yan; Luc Van Meervelt; Zhigang Shao; Jianhua Fang; Zheng-Hong Luo; Dirk E. De Vos; Koen Binnemans; Jan Fransaer (2015). "1,2,4-Triazolium perfluorobutanesulfonate as an archetypal pure protic organic ionic plastic crystal electrolyte for all-solid-state fuel cells". Energy & Environmental Science. 8 (4): 1276–1291. doi:10.1039/C4EE02280G.
- "The Roll-to-Roll Battery Revolution". Ev World. Retrieved 20 August 2010.
- Syzdek J, Borkowska R, Perzyna K, Tarascon JM, Wieczorek W (2007). "Novel composite polymeric electrolytes with surface-modified inorganic fillers". Journal of Power Sources. 173 (2): 712–720. Bibcode:2007JPS...173..712S. doi:10.1016/j.jpowsour.2007.05.061. ISSN 0378-7753.
- Syzdek J, Armand M, Marcinek M, Zalewska A, Żukowska G, Wieczorek W (2010). "Detailed studies on the fillers modification and their influence on composite, poly(oxyethylene)-based polymeric electrolytes". Electrochimica Acta. 55 (4): 1314–1322. doi:10.1016/j.electacta.2009.04.025. ISSN 0013-4686.
- Syzdek J, Armand M, Gizowska M, Marcinek M, Sasim E, Szafran M, Wieczorek W (2009). "Ceramic-in-polymer versus polymer-in-ceramic polymeric electrolytes—A novel approach". Journal of Power Sources. 194 (1): 66–72. Bibcode:2009JPS...194...66S. doi:10.1016/j.jpowsour.2009.01.070. ISSN 0378-7753.
- Jiangshui Luo; Olaf Conrad; Ivo F. J. Vankelecom (2013). "Imidazolium methanesulfonate as a high temperature proton conductor". Journal of Materials Chemistry A. 1 (6): 2238–2247. doi:10.1039/C2TA00713D.<|endoftext|>
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- Explain that waves transfer energy, not matter.
- Distinguish between mechanical and electromagnetic waves.
- Summarize the major properties and behavior of waves, including (but not limited to) wavelength, frequency, amplitude, speed, refraction, reflection and diffraction.
Course Digital Resources:
Event #1 (THIS WEDNESDAY)
3-13-2019: Students must be at school by 8:30 AM for the Cabrillo Trip! Bus leaves at 9 sharp and we will be back at 2pm at the latest.
Event #2 (Next Wednesday)
3-20-2019: Engineering Design Challenge at 9AM.
For those of you who did Physical Science A with me in Fall 2018, remember the lunar drop we did with the egg? Well, this design challenge will be very similar. Both Biology and Physical Science Students will be working together in the large classroom here at the Wilson Center starting at 9AM on this challenge.
amplitude: How far the medium (crests and troughs, or compressions and rarefactions) moves from rest position (the place the medium is when not moving).
compression: When the particles of a longitudinal wave are close together.
compressional (longitudinal) wave: A wave in which the medium moves back and forth in the same direction as the wave.
crest: The highest point on a transverse wave.
diffraction: The bending of waves around an object.
electromagnetic wave: A wave that does not require a medium to travel, for example, it can travel through a vacuum. Also called an EM wave.
energy: The capacity to do work.
frequency: How many waves go past a point in one second. Measured in hertz (Hz).
mechanical wave: A wave that requires a medium to travel.
rarefaction : When the particles of a longitudinal wave are far apart.
reflection: When a wave bounces off a surface.
refraction: When a wave bends.
transverse wave: A wave in which the medium moves at right angles to the direction of the wave.
trough: The lowest point on a transverse wave.
wave: A disturbance that carries energy from one place to another.
wavelength: Distance between one point on a wave and the exact same place on the next wave.
additive color system: Involves light emitted directly from a source, before an object reflects the light. Mixes various amounts of red, green and blue light to produce other colors. Examples include computer monitors and TVs.
concave or negative lens: A lens that diverges or spreads out light rays.
convex or positive lens: A lens that converges or focuses light, and can form images.
cyan: A highly saturated green-blue that is the complementary color of red and forms, with magenta and yellow, a set of primary colors.
laser: Acronym for light amplification by stimulated emission of radiation. Lasers only produce one wavelength of light, resulting in a beam of light that is very distinct and does not spread out.
opaque: A characteristic of an object that does not allow light to pass through; it absorbs or reflects all light.
prism : A transparent optical object that refracts light.
subtractive color system: Creates color by subtracting or absorbing certain wavelengths of color while reflecting other wavelengths back to the viewer. Examples include photographs and printed magazines.
translucent: A characteristic of an object that can be seen through, but not clearly; it absorbs, reflects and transmits light, such as wax paper or frosted glass.
transparent: A characteristic of an object that allows almost all the light to pass through, so it can be seen through clearly, such as glass or clear plastic.
electromagnetic radiation: A phenomenon that takes the form of self-propagating waves in a vacuum or matter. It is comprised of electric and magnetic field components that oscillate in phase perpendicular to each other and the direction of energy propagation. All travel the speed of light.
electromagnetic spectrum: The range of all possible frequencies of electromagnetic radiation.
Task #1 (Last Unit 8 Project):
The Philadelphia Experiment is an alleged military experiment supposed to have been carried out by the U.S. Navy at the Philadelphia Naval Shipyard in Philadelphia, Pennsylvania, sometime around October 28, 1943. The U.S. Navy destroyer escortUSS Eldridge (DE-173) was claimed to have been rendered invisible (or “cloaked“) to enemy devices.
The story first appeared in 1955, in letters of unknown origin sent to a writer and astronomer, Morris K. Jessup. It is widely understood to be a hoax; the U.S. Navy maintains that no such experiment was ever conducted, that the details of the story contradict well-established facts about USS Eldridge, and that the alleged claims do not conform to known physical laws.
The experiment was allegedly based on an aspect of some unified field theory, a term coined by Albert Einstein to describe a class of potential theories; such theories would aim to describe — mathematically and physically — the interrelated nature of the forces of electromagnetism and gravity, in other words, uniting their respective fields into a single field.
According to some accounts, unspecified “researchers” thought that some version of this field would enable using large electrical generators to bend light around an object via refraction, so that the object became completely invisible. The Navy regarded this of military value and it sponsored the experiment.
Philadelphia Experiment Paper
Go to Google and create a Google Doc entitled – “Unit 8 Waves Paper – Philadelphia Experiment”. Share it to me at [email protected]
If you don’t remember how to create a Google Doc click HERE to be taken to a How To Guide.
If you need help sharing your Google Doc click HERE to be taken to a How To guide.
After you have created and shared your paper. You will begin writing a 5 paragraph paper addressing the question “Can waves be manipulated to the point they could purposefully hide a massive object?” Below is my recommended outline for your paper. You can can set up your paper however you like, the outline below is just a recommendation.
Paragraph 1 –
Introduce your reader to what the Philadelphia Experiment is in general.
Paragraph 2 –
Introduce your reader to transverse waves. What are they? Where can they be seen? Do they use a medium?
Paragraph 3 –
Introduce your reader to electromagnetic waves.What are they? Where can they be seen? Do they use a medium?
Paragraph 4 –
Given what you know, is it possible for waves to hide a massive object purposefully? Why or why not? Explain your answer.
Paragraph 5 –
Why should the reader care about this topic? Why should the average person know what waves are, how they work, and their potential?
YouTube Video -> Background on the Philadelphia Experiment
YouTube Video -> Philadelphia Experiment was Real!
YouTube Video -> Experiment a Hoax
Upon completion of the Philadelphia Experiment project, make sure to go over this check list so you can close your Unit 8.
Unit 8 Items:
- Philadelphia Experiment Analysis Paper – Printed and handed to Mr. Tyler (New as of today)
- Powerpoint Presentation on Waves – Printed and handed to Mr. Tyler(Previously done)
- Vocabulary Test – Passed with above a 70% – Mr. Tyler will confirm and print (Previously done)
- Have Mr. Tyler confirm your handout work is in and complete – Remember Mr. Tyler collects most of your work before you leave. (Previously done)
- Unit 8 Test – Passed with above a 70% (Last thing before closing unit)<|endoftext|>
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# Inverse Tan
View Notes
## Tan Inverse
The inverse in Trigonometry for tangent is known as Inverse Tan. It is the basic function that we use in real-world problems and solve them. However, the primary use of Tan Inverse is to apply the tangent ratio for a specified angle. Thus, with the help of this function, you can quickly find any value to tangent. Examples are tan 1, tan 10, arctan 1 and more. Here arc is the basic way to name any inverse formula in trigonometry. Thus we can say that the inverse of the tangent is also known as arctan. In the article below, you will understand why and how to use this formula to solve different problems. Also, you will learn about its practical applications.
### Tan Inverse Formula
The inverse of tan or anti-tan is the arcus of tan. Here, we will define the formula of the tan inverse.
Suppose we are given as x =tan y
then, y =tan-1 x
Here y can be any real number.
This is also known as tan inverse x.
When you add two different tan inverse functions, the formula will be represented as below.
tan a $\pm$ b = $\frac{tan a \pm tan b}{1 \mp tan(a) tan(b)}$
This is also known as the additional formula for inverse tan. It is derived for the addition of two inverses of tan.
If in the above equation, we put a = arctan x and b = arctan y, we get the equation as presented below.
arctan(x) $\pm$ arctan(y) = arctan($\frac{x\pm y}{1 \mp xy}$) (mod π), xy ≠ 1
This is achieved after substituting the values of a and b.
### The Derivative of Tan Inverse
To find out the derivative for the inverse of tan, we will find the derivative for tan inverse x.
The formula of derivative of the tan inverse is given by:
d/dx(arctan(x)). Hence, we define derivatives as 1/ (1 + x2). Here x does not belong to i or -i. This is also known as differentiation of tan inverse.
Let us take an example for a graph of the tan inverse.
We will define it with the help of the graph plot between π/2 and –π/2.
### Graph of Tan Inverse x
In the above graph of tan inverse x, the points are plotted between π/2 and –π/2. The plotting is along the real axis.
### Integration of Tan Inverse x
If we want to give the values of definite integral for the inverse of tan, we use the concept of integration. The derivative of tan inverse will be integrated back to get the normal value of tan inverse. However, for a definite point, the value will be fixed. Thus the below expression defines the integral of inverse tan.
arctan(x) = $\int_{0}^{x}$ $\frac{1}{y^{2}+1}$ dy
The above pictures describe the integral of tan inverse x.
The basic graph for tan inverse is given by
The above graph is defined for tangent inverse give by equation 1/ (1 + x2)
### How Can You Relate Tan Inverse Infinity With Other Trigonometric Equations?
We take an example of a triangle with one side as x and another as 1. The hypotenuse becomes
√1+ x2
Suppose a triangle has an arctan angle as θ then we can define the below relation for finding other trigonometric functions:
Sin (arctan(x)) = x/ (√1+ x2)
Cos (arctan(x)) = 1 / (√1+ x2)
tan (arctan(x)) = x
### What are The Different Properties of Inverse Tan?
Below are some mentioned properties of the tan inverse:
Suppose = arctan (x)
Also, x = arctan (y)
Here we can use the real number as a domain.
Here -π/2 < y< π/2
This is the defined property of tan inverse used widely.
### How Will You Give Value For tan-1 Infinity?
We know that a tan of 90 degrees is defined as infinity. Thus for tan-1 the value is 90 degrees.
tan 90° = ∞ or tan π/2 = ∞
Hence, tan-1 (∞) = π/2 or tan-1 (∞) = 90°
Q1. What are the Six Different Inverse Functions of Trigonometry?
In trigonometry, there are mainly six functions that are widely used. These include sine, cosine, cos, sectant, tangent, cosecant and cotangent. The use of these functions is to get the value of the sides of a triangle when the angle is given to us. However, in the case of the inverse of these six functions, the concept is reversed. The actual working of the inverse of all trigonometric functions is to find the angle of the triangle when sides are given to us. The concept of inverse trigonometric functions is widely operated in real-time examples, which include physics, navigation, engineering and geometry. The three primary notations for trigonometry that are widely in use are:
1. tan-1 (x)
2. sin-1 (x)
3. cosine-1 (x)
Q2. How Do You Define Tan x?
Usually, sin and cosine are the main trigonometric functions used in solving real-time problems. However, there are more functions that are generated by the ratio of both sin and cosine. There are four functions: cosecant, tangent, cotangent and secant. If you know the values of sin and cosine of an angle, then it is easy to find the values of other trigonometric equations. Therefore, the tangent of x is given by:
tan x = sin (x)/ cosine (x)
Similarly, you can find solutions to other functions as tan(x) = inverse of cotangent (x).
Secant (x) in inverse of cosine(x) and cosecant (x) is inverse of sin (x). With the given formulas, you can easily relate them to the concept of functions and thus find their domain, codomain, and range. Thus, tan x is valid when cosine (x) is not equal to zero. Hence the value of x must not be equal to π/2.<|endoftext|>
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Information Report by: Tupou Tupi
Treaty of Waitangi
The Treaty of Waitangi is an agreement signed in 1840 between the Maori and the British settlers. New Zealand celebrates the signing of The Treaty of Waitangi every year on February 6th. The Maori people were the first to discover New Zealand also known as Aotearoa (meaning the Land of the Long White Cloud).
Signing Of The Treaty Of Waitangi
The Treaty of Waitangi was first signed on the 6th of February, 1840. The Treaty of Waitangi was an agreement between the British people and the Maori people. On February 5th, it was the day the Pakeha people presented the Treaty of Waitangi to the Maori people. In New Zealand, there is a town called Waitangi and that is where the Treaty of Waitangi was signed.
The Two Different Versions Of The Treaty Of Waitangi
In the English text, Māori leaders gave Queen Victoria all the rights and the supremacy over their land. It was also written in the Māori text that the Māori leaders gave the Queen complete government over their land.
In the Māori text, the Māori people were guaranteed that they still had leadership over their lands, villages, and all their property and treasures. Māori also agreed to give the Crown the right to buy their land if they wished to sell it.
The Treaty of Waitangi is an important agreement that was signed by representatives of the British Crown and Māori in 1840. The Treaty was aimed to protect the rights of Māori to keep their land, forests, fisheries, and treasures while handing over authority to the Pakeha.<|endoftext|>
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Categories
## What is 75% of the area of a circle with a circumference of 10 units? Round the solution to the nearest square unit.
75% of the area of the circle is 6 square units.
Step-by-step explanation:
Given:
Circumference of a circle = 10 units
Now we have to find the radius of the circle, then we have to find the area of the circle.
Circumference of a circle = 2*π*r [π = 3.14]
2*3.14*r = 10
6.28r = 10
Dividing both side by 6.28, we get
r = 10/6.28
r = 1.59
Now let’s find the area of the circle.
The area of the circle = π*r^2
= 3.14*1.59*1.59
The area of the circle = 7.94, which is 100 %
Now we have to find the 75% of the area of the circle.
75% = 0.75
75% of the area of the circle = 0.75*7.94
= 5.95
To round off to the nearest whole number, we get 6 square units.
Categories
## Find the general solution of the given system. dx dt = x + y ? z dy dt = 2y dy dt = y ? z
Find the general solution of the given system. dx dt = x + y ? z dy dt = 2y dy dt = y ? z
Categories
## An 8.65-g sample of an unknown group 2a metal hydroxide is dissolved in 85.0 ml of water. an acid-base indicator is added and the resulting solution is titrated with 2.50 m hcl(aq) solution. the indicator changes color signaling that the equivalence point has been reached after 56.9 ml of the hydrochloric acid solution has been added.
The molar mass of metal hydroxide = 121.66 g/ mole
The formula is Sr(OH)2
Explanation;
M(OH)2 + 2 HCl = MCl2 + 2 H2O
Moles of the acid = 2.5 x 56.9 / 1000 = 0.14225
Therefore; moles of hydroxide at the eq point = 0.0711 moles
These have a mass of 8.65g
Thus; mass of 1 mole = 8.65 / 0.0711
= 121.66 g / mole
That means the compound is Sr (OH)2 ..
Categories
## Which of the following would be considered a solution ? A salt in the sugar bowl B sugar added to water C oil added to water D any fat-soluble substance mixed with water soluble substance
Which of the following would be considered a solution ? A salt in the sugar bowl B sugar added to water C oil added to water D any fat-soluble substance mixed with water soluble substance
Categories
## A 150ml sample of hydrochloric acid completely reacted with 60.0mL of a 0.100 M NaOH solution. What was the original concentration of the HCl solution
A 150ml sample of hydrochloric acid completely reacted with 60.0mL of a 0.100 M NaOH solution. What was the original concentration of the HCl solution
Categories
## In a situation where handicapped person can only input data into the computer using a stylus or light pen, which keyboard configuration might be the solution?
In a situation where handicapped person can only input data into the computer using a stylus or light pen, which keyboard configuration might be the solution?
Categories
## Given the system of equations, what is the solution? x + 2y = 11 x – 2y = -1 {(-5, -3)} {(1, 1)} {(5, 3)}
Given the system of equations, what is the solution? x + 2y = 11 x – 2y = -1 {(-5, -3)} {(1, 1)} {(5, 3)}
Categories
## To determine the pH of a solution using a pH indicator, you need a? A. Color key B. neutral Solution C. Range of acids and bases
To determine the pH of a solution using a pH indicator, you need a? A. Color key B. neutral Solution C. Range of acids and bases
Categories
## Which of the following angles is in the solution set of sec2θ – 3secθ – 2 = 0 for 0° ≤ θ < 360°
Which of the following angles is in the solution set of sec2θ – 3secθ – 2 = 0 for 0° ≤ θ < 360°
Categories
## What kind of change occurs when salt dissolves in water a.) solution change b.) physical change c.) chemical change d.) substance change
What kind of change occurs when salt dissolves in water a.) solution change b.) physical change c.) chemical change d.) substance change
Categories
## Why does changing the PH of a solution affect the ability of enzymes?
Why does changing the PH of a solution affect the ability of enzymes?
Categories
## When a solution of LiC2H3O2 reacts with a solution of MgSO4, what is the net ionic equation? No reaction Mg+2 (aq) + C2H3O2- (aq) → Mg(C2H3O2)2 (s) Mg+2 (aq) + 2C2H3O2- (aq) → Mg(C2H3O2)2 (s) 2Li+ (aq) + SO4-2 (aq) → Li2SO4 (s)
I think the correct answer would be the first option. When a solution of LiC2H3O2 is mixed with a solution of MgSO4, no reaction would be observed from the mixture. This is because magnesium cannot replace lithium from the acetate compound. Lithium is the least reactive substance based from the activity series. So, it can only react with a few substances and it does involve the reaction with magnesium sulfate. It is the very least reactive of all the alkali metals.
Categories
## What value of x is in the solution set of 9(2x + 1)
The answer is: ” x < -3 ” .
_____________________
Explanation:
_____________________
Given:
______________________
” 9(2x + 1)
First , factor out a “9” in the expression on the right-hand side of the inequality:
9x – 18 = 9(x – 2) ;
and rewrite the inequality:
_____________________
9(2x + 1) < 9(x – 2) ;
Now, divide EACH SIDE of the inequality by “9” ;
[9(2x + 1)] / 9 < [9(x – 2)] / 9 ;
to get:
2x + 1 < x – 2 ;
Now, subtract “x” and add “2” to each side of the inequality:
2x + 1 – x + 2 < x – 2 – x + 2 ;
to get:
x + 3 < 0 ;
Subtract “3” from EACH SIDE ;
x + 3 – 3 < 0 – 3 ;
to get:
” x < -3 ” .
____________________________________
Categories
## A set of equations is given below: Equation H: y = –x + 2 Equation J: y = 3x – 4 Which of the following steps can be used to find the solution to the set of equations? (4 points) A.–x = 3x – 4 B.–x +2 = 3x C.–x + 2 = 3x – 4 D.–x + 1 = 3x + 2
A set of equations is given below: Equation H: y = –x + 2 Equation J: y = 3x – 4 Which of the following steps can be used to find the solution to the set of equations? (4 points) A.–x = 3x – 4 B.–x +2 = 3x C.–x + 2 = 3x – 4 D.–x + 1 = 3x + 2
Categories
## The minimum and maximum temperature on a cold day in Lollypop Town can be modeled by the equation below:2|x − 6| + 14 = 38What are the minimum and maximum temperatures for this day? A:x= -9,x=21 B:X= -6,X=18 C:X=6,X= -18 D:no solution
The minimum and maximum temperature on a cold day in Lollypop Town can be modeled by the equation below:2|x − 6| + 14 = 38What are the minimum and maximum temperatures for this day? A:x= -9,x=21 B:X= -6,X=18 C:X=6,X= -18 D:no solution
Categories
## Consider the following set of equations: Equation M: y = x + 4 Equation P: y = 3x + 6 Which of the following is a step that can be used to find the solution to the set of equations? (4 points) A. x = 3x + 6 B. x + 4 = 3x + 6 C. x + 6 = 3x + 4 D.x = 3x
Consider the following set of equations: Equation M: y = x + 4 Equation P: y = 3x + 6 Which of the following is a step that can be used to find the solution to the set of equations? (4 points) A. x = 3x + 6 B. x + 4 = 3x + 6 C. x + 6 = 3x + 4 D.x = 3x
Categories
## f(x) = 4x f(x) = −x + 5 what is the solution
f(x) = 4x f(x) = −x + 5 what is the solution
Categories
## What is the solution to this equation 4x+x-15+3-8x=13
What is the solution to this equation 4x+x-15+3-8x=13
Categories
## What is the concentration of Sr2+ in a saturated solution of SrSO4? SrSO4 has a Ksp of 3.2 x 10–7
What is the concentration of Sr2+ in a saturated solution of SrSO4? SrSO4 has a Ksp of 3.2 x 10–7
Categories
## Consider the following pair of equations: y = –2x + 8 y = x – 1 Explain how you will solve the pair of equations by substitution. Show all the steps and write the solution in (x, y) form. (5 points)
There is a special kind of substitution which some books call it by comparison.
When the two equations are both
y=some function of x
y=some other function of x, then
we can substitute the second equation into the first giving
some function of x = some other function of x and start solving.
For example,
y = –2x + 8
y = x – 1
substitute second into first
x-1=-2x+8
isolate x on left,
3x=9
x=3
second step is to substitute x=3 into second equation to get
y=x-1=3-1=2
Therefore the solution is (3,2)
Categories
## Do you think it is always safe to pour waste chemicals and solution down the sink with lots of water? Explain.
No, it is not always safe to pour waste chemicals and solution down the sink with lots of water because it could be harmful to the environment. When you pour materials down the drain they either enter a septic system or a municipal sewer system. If the material enter a septic system, it goes into a tank buried underground where the solids settle and decompose. The rest of the wastewater is than released into a drain field where it goes into the soil. Sometimes the wastewater kills the good bacteria and can contaminate ground water or surface water.
A municipal sewer system the wastewater goes to a central sewage plant, where after treatment, it is then released into lakes, rivers and streams. Some hazardous household waste can pass through the system unchanged and pollute the waters. Also the wastewater may affect the plumbing or collect in the trap and release fumes through the drains. Most sewage facilities aren’t capable of removing such toxic substances.
So, No, it is not always safe to pour chemicals and solutions down the sink even with a lot of water.
Categories
## I NEEED HELP PLEASEEEEEE I WILL RATE BRAINLIEST!!!!!!!!!!!! Select the properly written and balanced equation for the following chemical reaction: Aqueous solutions of sodium hydroxide and iron (III) nitrate react to produce a precipitate of iron (III) hydroxide in an aqueous solution of sodium nitrate. ——-3NaOH (aq) + Fe(NO3)3 (aq) Fe(OH)3 (s) + 3NaNO3 (aq) NaOH (aq) + FeNO3 (aq) FeOH (s) + NaNO3 (aq) NaOH (aq) + Fe(NO3)3 (aq) FeOH3 (s) + NaNO3 (aq) 3NaOH (aq) + FeNO3 (aq) FeOH (s) + 3NaNO3 (aq) I believe that it is the first choice but I am not sure….
I NEEED HELP PLEASEEEEEE I WILL RATE BRAINLIEST!!!!!!!!!!!! Select the properly written and balanced equation for the following chemical reaction: Aqueous solutions of sodium hydroxide and iron (III) nitrate react to produce a precipitate of iron (III) hydroxide in an aqueous solution of sodium nitrate. ——-3NaOH (aq) + Fe(NO3)3 (aq) Fe(OH)3 (s) + 3NaNO3 (aq) NaOH (aq) + FeNO3 (aq) FeOH (s) + NaNO3 (aq) NaOH (aq) + Fe(NO3)3 (aq) FeOH3 (s) + NaNO3 (aq) 3NaOH (aq) + FeNO3 (aq) FeOH (s) + 3NaNO3 (aq) I believe that it is the first choice but I am not sure….
Categories
## What is the colligative property that means a solution typically has a measurably higher boiling point than a pure solvent alone? A) osmotic pressure B) boiling point elevation C) boiling point alleviation D) vapor-pressure lowering
What is the colligative property that means a solution typically has a measurably higher boiling point than a pure solvent alone? A) osmotic pressure B) boiling point elevation C) boiling point alleviation D) vapor-pressure lowering
Categories
## What is the solution of log4x − 10^32 = 5? x = −2 x = 2 x = 3 x = 4
What is the solution of log4x − 10^32 = 5? x = −2 x = 2 x = 3 x = 4
Categories<|endoftext|>
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# A box with an initial speed of 8 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 3/5 and an incline of pi /4 . How far along the ramp will the box go?
Feb 17, 2018
Consider this scenario,
Moreover, assess the coordinate system as to be parallel with the incline. We are given this data,
${\mu}_{k} = 0.60$
theta = pi/4 = 45°
${v}_{0} = \frac{8 m}{s}$
We need three vectors, and resolve gravity into its components due to the preceding coordinate system. A vertical vector pointing down, (i) gravity, a vector pointing up perpendicular to the incline, (ii) normal force, and a vector pointing to the left, (iii) ${F}_{k}$ (friction).
Now, using Newton's second law, we resolve the gravity vector into its x and y components. (something like this, but friction is pointing left is well)
$\Sigma {F}_{y} = {F}_{N} - m g \cos \theta = 0$
$\therefore {F}_{N} = m g \cos \theta$
Recall,
${F}_{k} = {\mu}_{k} \cdot {F}_{N}$
$\Sigma {F}_{x} = - {\mu}_{k} \cdot m g \cos \theta - m g \sin \theta = m a$
$\implies - {\mu}_{k} \cdot g \cos \theta - g \sin \theta = a$
Hence,
a = -0.60 * (9.8m)/s^2 * cos(45°) - (9.8m)/s^2sin(45°) approx (-11.09m)/s^2
Furthermore, recall,
${\nu}^{2} = {\nu}_{0}^{2} + 2 a \Delta x$,
and assume the $\nu = 0$, since when the box reaches the top it will stop and begin to slide back down as a result of the acceleration via gravity.
Hence,
$0 = \frac{8.0 m}{s} - 2 \cdot \frac{11.09 m}{s} ^ 2 \cdot \Delta x$
$\therefore \Delta x \approx 0.36 \text{m}$
Not very far, the kinetic friction is very high!<|endoftext|>
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# Thread: Money Math-Trick Question
1. ## Money Math-Trick Question
Kane has $50.00. He goes to the supermarket and buys football boots and then he buys a soccer ball, which is half the price of the football boots.He then spends half of what she has left on a belt, leaving him with$15.00.
How much did the soccer ball cost?
How much did the football boots cost?
I started it like this:
Let x= price of football boots
Let 1/2(x) = price of soccer ball
the combined cost of soccer ball and football boots = x + 1/2(x)
Amount spent on the football boots and soccer ball :
$50-( I am stuck The text gives a clue that the price of the belt =$15
y-y/2=$15 can you solve it from here? 3. ## Re: Money Math-Trick Question Originally Posted by KayPee Kane has$50.00. He goes to the supermarket and buys football boots and then he buys a soccer ball, which is half the price of the football boots.He then spends half of what she has left on a belt, leaving him with $15.00. How much did the soccer ball cost? How much did the football boots cost? I started it like this: Let x= price of football boots Let 1/2(x) = price of soccer ball the combined cost of soccer ball and football boots = x + 1/2(x) Amount spent on the football boots and soccer ball :$50-(
I am stuck
The text gives a clue that the price of the belt = $15 (How did they arrive at the conclusion that the price of the belt equals$15?)
Anyone has any ideas as to how I can move on?
Thanks
Let x be the price of the boots
After buying the football and the football boots he is left with: $y = \left[50 - \left(x+ \dfrac{x}{2}\right)\right]$
He then spends half of this on the belt and is left with $15: $y - \dfrac{y}{2} = 15$ Of course we have an expression for y in x so sub that in: $\left[50 - \left(x+ \dfrac{x}{2}\right)\right] - \dfrac{1}{2}\left[50 - \left(x+ \dfrac{x}{2}\right)\right] = 15$ And after simplification: $\left(50 - \dfrac{3x}{2}\right) - \left(25 - \dfrac{3x}{4}\right) = 15$ Solve for x, which is the price of the boots 4. ## Re: Money Math-Trick Question Originally Posted by KayPee Kane has$50.00. He goes to the supermarket and buys football boots and then he buys a soccer ball, which is half the price of the football boots.He then spends half of what she has left on a belt, leaving him with $15.00. Well, you said this was a trick question! Who is "she" and how much money does she have? How much did the soccer ball cost? How much did the football boots cost? I started it like this: Let x= price of football boots Let 1/2(x) = price of soccer ball the combined cost of soccer ball and football boots = x + 1/2(x) Amount spent on the football boots and soccer ball :$50-(
I am stuck
The text gives a clue that the price of the belt = $15 (How did they arrive at the conclusion that the price of the belt equals$15?)
Anyone has any ideas as to how I can move on?
Thanks
You were told that he spent "half of what [s]he has left on a belt" and then had $15 left. If you spend "half", you have "half" left. That means that before buying the belt, he had$30. That means that the boots and ball cost 50- 30= \$20. Taking x to be the cost of the boots, the ball cost x/2 so x+ x/2= 3x/2= 20.<|endoftext|>
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Albert Teen
YOU ARE LEARNING:
Expanding Brackets
# Expanding Brackets
### Expanding Brackets
Expanding brackets is an important step for manipulating expressions, and involves multiplying the terms inside a bracket by those outside of a bracket.
1
What does this equal?
2
What if we add brackets? What does this equal now?
3
So brackets matter for the result!
By the way, notice that when you multiply brackets, you don't need to write the multiplication sign.
$3 \times (4+2)$ is the same as just $3(4+2)$
1
Now, you could also work out this problem in a different way
You could also change the expression so that the brackets disappear. That is called expanding the brackets.
2
Basically, the $3$ wants to be multiplied by both terms inside the brackets. What is $3 \times 4$?
3
And what is $3\times2$?
4
Now, you have $12+6$
So $12+6$ is the same as $3(4+2)$, but you have expanded the brackets - you have made them disappear.
1
But why would you want to expand brackets?
It can help you simplify expressions to make them easier to read. This example shows you how.
2
To expand the brackets here, you first have to say $2 \times x$. What is that?
3
Then you say $2\times 4$. What is that?
4
Now the brackets are expanded
Now you can simplify the expression.
5
You have $3x$ and another $2x$. How many $x$ is that in total?
6
So you expanded the brackets to simplify this expression
$5x+8$ is a lot easier to read than $3x+2(x+4)$
Expand $5(3+x)$
1
Recap! You can expand brackets to simplify expressions
For example, $6x+20$ is a lot easier to read than $2x+4(x+5)$
2
To expand brackets, you need to multiply all the terms inside the brackets by the number outside the brackets
First you say $4 \times x=4x$ Then you say $4\times5=20$ Then you put the two together $4x+20$
3
Finally, you can see if you can simplify the whole expression
In this case, you have $2x$ and $4x$ which you can combine to $6x$
Expand $6(x+2)$
Expand $(x+6)8$
Now, what is $2(x-4)$ when you expand the brackets?
Let's try this one $x(3x+6)$.
1
What happens when the number outside the bracket is also a variable?
We apply the same principle.
2
First we multiply $x\times 3x$. What does this give us?
3
Next we multiply the second term in the bracket. What is $x\times 6$?
4
We have $x\times 3x=3x^2$ and $x\times 6=6x$.
Putting these together we have $3x^2+6x$.
Expand $(x-2)4x$
Expand and simplify $(2+3x)5-4x$
Expand and simplify $x(8+3x)-2x$
1
Summary! You can expand brackets to simplify expressions
For example, $6x+3x^2$ is easier to read than $x(8+3x)-2x$
2
So how do you expand brackets?
You multiply both terms inside the bracket by the number outside the bracket.
3
The number might follow the brackets rather than stand before them
Don't be confused by that.
4
If you multiply a positive and a negative, then remember to keep the minus!
For example, $2 \times -4$ is $-8$, so this becomes $2x-8$, not $2x+8$
5
If you multiply $x\times x$ it becomes $x^2$
So if you multiply $x \times 3x$ as in this example, it becomes $3x^2$<|endoftext|>
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The Boötes Dwarf Galaxy, also known as Boötes I, is a satellite galaxy to the Milky Way. It lies approximately 197,000 light years from Earth. It is located in the constellation Boötes, the Herdsman.
Boötes I is a rather faint galaxy. It has an apparent visual magnitude of 13.1 and an absolute magnitude of -5.8. The galaxy’s luminosity is only 100,000 times that of the Sun.
It is one of the least luminous galaxies ever discovered. In fact, when it was discovered in 2006, it was the single least luminous galaxy ever seen, beating the previous record holder, an extremely dim galaxy in the constellation Ursa Minor.
The Boötes Dwarf is also one of the Milky Way’s most remote companion galaxies.
Boötes I is a dwarf spheroidal galaxy that seems to be tidally disrupted by our own galaxy, the Milky Way, which it orbits.
It is a metal and gas-poor galaxy. Its stars are spread out from one another and they look like stars in a very old globular cluster, suggesting that the galaxy itself is ancient.<|endoftext|>
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- Generalize logical clocks to offer non-causality information along with causality information.
- Implement along with values drawn from the partially ordered set rather than an entirely ordered set.
- Assign the value V( e ) in order to every computation event e in an execution so that the a ® b if and only if V( a ) < V( b ).
- Each pi keeps an n-vector Vi, initially all 0's
- Entry j in Vi is pi 's estimate associated with how many steps pj has taken
- Every msg pi sends is timestamped along with current value associated with Vi
- At each and every step, increment Vi[ i ] through 1
- When finding a message along with vector timestamp T, update Vi 's components j ≠ i so that Vi[ j ] = max( T[ j ],Vi[ j ])
- If a is an event at pi, after that allocate V( a ) to be value associated with Vi at end of a.
- Equality: V = W iff V[ i ] = W[ i ] for all i.
- Example: ( 3, 2, 4 ) = ( 3, 2, 4 )
- Less than or equal: V ≤ W iff V[ i ] ≤ W[ i ] for all i.
- Example: ( 2, 2, 3 ) ≤ ( 3, 2, 4 ) and ( 3, 2, 4 ) ≤ ( 3, 2, 4 )
- Less than: V < W iff V ≤ W but V ≠ W.
- Example: ( 2, 2, 3 ) < ( 3, 2, 4 )
- Incomparable: V || W iff ! ( V ≤ W ) and ! ( W ≤ V ).
- Example: ( 3, 2, 4 ) || ( 4, 1, 4 )
- The partial order on n-vectors simply described is actually different then lexicographic ordering.
- Lexicographic ordering is really a complete order on vectors.
- Consider ( 3, 2, 4 ) versus ( 4, 1, 4 ) within the 2 approaches.
Vector Timestamps Algorithm:
Manipulating Vector Timestamps:
Let V and W be two n-vectors of integers.
Vector Timestamps Example:
V(g) = ( 0, 0, 1 ) and V(b) = ( 2, 0, 0 ), which are incomparable. Compare with logical clocks L(g) = 1 and L(b) = 2.<|endoftext|>
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Encouraging Classroom Language Use - Part 2
Nov 14, 2008 English Language Teaching (ELT) 3270 Views
This article will show how an activity can be modified to encourage the four kinds of classroom language (requests, choices, leadership, and manners and values) described in part 1.
The Basic Activity: Peephole Cards for Vocabulary Review
Stack of large picture cards of vocabulary for review, several A-3-sized, opaque sheets of paper with a hole cut in the middle about half a centimeter to a centimeter square in size (larger hole for younger students). I usually just ran A-3 paper through the copy machine with the cover up (although you will be scolded, like I always was, for doing this).
"What is it?", "It's a ____," "Is it a ____?", "No, it isn't," "Yes, it is."
Students have already been taught the classroom language and are familiar with the non-verbal prompts (gestures, etc.).
Teacher holds up a card with the peephole screen in front of it. "What is it?" she asks mysteriously. Students are perplexed. She moves the card behind the screen so that, through the hole, students can see different parts of the picture. Students yell guesses, teacher replies, until someone gets it right. Teacher demonstrates two more times, using different picture cards, then divides students into pairs and they take turns quizzing each other.
When dividing classes into pairs with different roles, designate one student A and one student B. Explain that all the A's are the quizzers and all the B's the guessers. Call the A's to the front to pick up the cards. As pairs finish, tell the B's to take the peephole screen and choose a new card (as necessary).
Students will not get X unless they request it appropriately (in English, of course). Students must desire X, or they will not be motivated to make the effort of requesting it, and they must have the ability and aids to make that request.
Motivating young children is simple and fun: show them something, make a big deal over it, show them there is enough for everyone, and then blatantly fail to give them any. With some classes, lording it over them, and then crying and feigning agony when you are "forced" to distribute it because they have asked appropriately, is also a great motivator.
A's come to the teacher to get the peephole screens, perhaps carelessly requesting the screen in L1. After fiercely ordering, "Line up," teacher studies the ceiling casually or admires her nails while casually prompting "____, please," with her outstretched hand.
Baffled by what the screens are called, the first A will point to it and ask, "What is it?" (If this question doesn't come, the teacher can remind the student to ask by shrugging her shoulders.) The teacher says, "Black peephole paper" (or whatever). The student says, "Black peephole paper, please," and, after getting the screen, goes to choose a picture card.
This requires students to make choices and requests based on them so shy or reticent students may be unfairly discriminated against. (This can be avoided by letting students take turns choosing first, for example.) It is helpful to demonstrate how to request the choices first, such as by holding up each item and saying what it is before failing to give them away.
The peephole screens can be diversified by having the holes in different shapes, or, rather than being all black, can be of construction paper of different colors. Students then request, "Heart peephole paper, please," or "Blue peephole paper, please." The request can also be simplified if necessary to "Heart paper, please," or just "Heart, please." More peephole screens than pairs can be available, and the quizzers are free to change their screen a limited number of times. Of course, they must first request it: "Change paper, please," for example.
My students were usually in their first or second year of learning English and very young, so I kept the language structure for requests very simple. With students of more experience or older age, I would require longer, more correct requests ("May I have...?" and so on).
Perhaps the more accurate phrase should be "Being in Charge," or "Bossing Your Classmates Around." It's been my experience that nothing excites a child more than power (except, perhaps, causing pain). Minor adjustments to almost any activity can open it up to letting students have more control over certain aspects.
After the second or third time the class has done the activity (and so is comfortable, perhaps even a little bored with it), review the directions up, down, left, right, and stop. Then, as quizzer, communicate that you will not move the hole unless they tell you. (Be sure to clarify whose perspective will be used for left and right.) This makes the activity more fun for both A and B as one gets to choose the card while the other gets to give orders.
Manners and Values
If nothing else, students leave the class knowing "Thank you" and "Please." As much as the subject of the class, teachers embody certain values, and it's always been important to me–especially with the current problems of bullying and classroom collapse–to emphasize respect for each other as well as the teacher and fair play. Students also do better knowing they are in a safe and just environment. (For a discussion of rules and classroom policy, see Effective Classroom Rules).
The first A successfully requests a peephole screen from the teacher and has it in hand, but the teacher does not let go. A tugs and tugs, and is rewarded with a dark look or raised eyebrows from the teacher. A remembers–or is prompted by those behind him–to say, "Thank you!" The teacher says, "You're welcome," and lets go of the peephole screen.
During excited quizzing, A's partner B is unable to guess the picture. A forgets himself and cries out, "Baka! (Stupid!)" The classroom goes silent as the teacher immediately stops all other activity, walks up to A, and says sternly, "We don't say baka in class. No baka. Tell B you're sorry." A tells B, "I'm sorry." B replies, "That's okay." They shake hands. (This may be prompted as necessary).<|endoftext|>
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Understanding why governments issue debt
The fundamental principles that arise in a fiat monetary system are as follows.
- The central bank sets the short-term interest rate based on its policy aspirations.
- Government spending is independent of borrowing which the latter best thought of as coming after spending.
- Government spending provides the net financial assets (bank reserves) which ultimately represent the funds used by the non-government agents to purchase the debt.
- Budget deficits put downward pressure on interest rates contrary to the myths that appear in macroeconomic textbooks about "crowding out".
- The "penalty for not borrowing" is that the interest rate will fall to the bottom of the "corridor" prevailing in the country which may be zero if the central bank does not offer a return on reserves, For example Japan easily maintains a zero interest rate policy with record budget deficits simply by spending more than it borrows.
- Government debt-issuance is a "monetary policy" consideration rather than being intrinsic to "fiscal policy", although in a modern monetary paradigm the distinctions between monetary and fiscal policy as traditionally defined are moot.
Accordingly, debt is issued as an interest-maintenance strategy by the central bank. It has no correspondence with any need to fund government spending. Debt might also be issued if the government wants the private sector to have less purchasing power.
Further, the idea that governments would simply get the central bank to "monetise" treasury debt (which is seen by orthodox economists as the alternative "financing" method for government spending) is highly misleading. So debt monetisation is usually referred to as a process whereby the central bank buys government bonds directly from the treasury. In other words, the federal government borrows money from the central bank rather than the public. Debt monetisation is the process usually implied when a government is said to be printing money. Debt monetisation, all else equal, is said to increase the money supply and can lead to severe inflation.
However, fear of debt monetisation is unfounded, not only because the government doesn?t need money in order to spend but also because the central bank does not have the option to monetise any of the outstanding federal debt or newly issued federal debt.
As long as the central bank has a mandate to maintain a target short-term interest rate, the size of its purchases and sales of government debt are not discretionary. Once the central bank sets a short-term interest rate target, its portfolio of government securities changes only because of the transactions that are required to support the target interest rate. The central bank's lack of control over the quantity of reserves underscores the impossibility of debt monetisation.
The central bank is unable to monetise the federal debt by purchasing government securities at will because to do so would cause the short-term target rate to fall to zero or to the support rate. If the central bank purchased securities directly from the treasury and the treasury then spent the money, its expenditures would be excess reserves in the banking system. The central bank would be forced to sell an equal amount of securities to support the target interest rate. The central bank would act only as an intermediary. The central bank would be buying securities from the treasury and selling them to the public. No monetisation would occur.
Further, the concept of debt monetisation is inapplicable. However, the central bank may agree to pay the short-term interest rate to banks who hold excess overnight reserves. This would eliminate the need by the commercial banks to access the interbank market to get rid of any excess reserves and would allow the central bank to maintain its target interest rate without issuing debt.
Accordingly, the concept of fiscal sustainability should never make any financing link between debt issuance and net government spending. There is no inevitability for debt to rise as deficits rise. Voluntary decisions by the government to make such a link have no basis in the fundamentals of the fiat monetary system.
Setting budget targets
Any financial target for budget deficits or the public debt to GDP ratio can never be sensible for all the reasons outlined above. It is highly unlikely that a government could actually hit some previously determined target if it wasn't consistent with the public purpose aims to create full capacity utilisation. As long as there are deficiencies in aggregate demand (a positive spending gap) output and income adjustments will be downwards and budget balances and GDP will be in flux.
The aim of fiscal policy should always be to fulfill public purpose and the resulting public debt/GDP ratio will just reflect the accounting flows that are required to achieve this basic aspiration.
Accordingly, the concept of fiscal sustainability cannot be sensibly tied to any accounting entity such as a debt/GDP ratio.
First, we have learned that exports are a cost and imports provide benefits. This is not the way that mainstream economists think but reflect the fact that if you give something away that you could use yourself (export) that is a cost and if you are get something that you do not previously have (import) that is a benefit. The reason why a country can run a trade deficit "more imports than exports" is because the foreigners (who sell us imports) want to accumulate financial assets in $dollars relative to our desire to accumulate their currencies as financial assets.
This necessitates that they send more real goods and services to us than they expect us to send to them. For as long as that lasts this real imbalance provides us with net benefits. If the foreigners change their desires to hold financial assets in $s then the trade flows will reflect that and our terms of trade (real) will change accordingly. It is possible that foreigners will desire to accumulate no financial assets in $s which would mean we would have to export as much as we import.
When foreigners demand less $s, its value declines. Prices rise to some extent in the domestic economy but our exports become more competitive. This process has historically had limits in which the fluctuations vary. At worst, it will mean small price rises for imported goods. If we think that depreciation will be one consequence of achieving full employment via net government spending then we are actually saying that we value having access to cheaper foreign travel or luxury cars more than we value having all people in work. It means that we want the unemployed to '[pay]' for our cheaper holidays and imported cars.
We might want to have those values embedded in public policy but I don't think the concept of fiscal sustainability should reflect these perverse ethical standards.
Further, foreigners do not fund the spending of a sovereign government. If the Chinese do not want to buy US Government bonds then they will not. The US government will still go on spending and the Chinese will have less $USD assets. No loss to the US.
Accordingly, the concept of fiscal sustainability does not include any notion of foreign 'financing' limits or foreign worries about a sovereign government's solvency.
Understanding what a cost is
The deficit-debt debate continually reflects a misunderstanding as to what constitutes an economic cost. The numbers that appear in budget statements are not costs! The government spends by putting numbers into accounts in the banking system.
The real cost of any program is the extra real resources that the program requires for implementation. So the real cost of a Job Guarantee is the extra consumption that the formerly unemployed workers can entertain and the extra capital etc that is required to provide equipment for the workers to use in their productive pursuits. In general, when there is persistent and high unemployment there is an abundance of real resources available which are currently unutilised or under-utilised. So in some sense, the opportunity cost of many government programs when the economy is weak is zero.
But in general, government programs have to be appraised by how they use real resources rather than in terms of the nominal $-values involved.
Accordingly, the concept of fiscal sustainability should be related to the utilisation rates of real resources, which takes us back to the initial point about the pursuit of public purpose.
Fiscal sustainability will never be associated with underutilised labour resources.
This blog aimed to bring together the last two Parts of the discussion. There is a lot of repetition across the mini-series and across all my blogs in general. You cannot say these things enough. Once government policy reflects an understanding of the things that I write about I will turn my blog into a daily surf report! I don't plan on doing that anytime soon.
But in defining a working conceptualisation of fiscal sustainability I have avoided, as you can see, very much analysis of debt, intergenerational tax burdens and other debt-hysteria concepts used by the deficit nazis. They are largely irrelevant concepts and divert our attention from the essential nature of fiscal policy practice which is to pursue public purpose and the first place to start is to achieve and sustain full employment.<|endoftext|>
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# Using the State Curriculum: Mathematics, Grade 8
Clarifications: Each clarification provides an explanation of the indicator/objective to help teachers better understand the concept. Classroom examples are often included to further illustrate the concept. While classroom examples could be shared with the students, the intended audience for the explanation/clarification is the classroom teacher-not the student. In addition, classroom examples may or may not reflect the assessment limits.
Standard 1.0 Knowledge of Algebra, Patterns, and Functions Topic A. Patterns and Functions Indicator 1. Identify, describe, extend, and create patterns, functions and sequences Objective b. Determine the recursive relationship of geometric sequences represented in words, in a table, or in a graph
### Clarification
A sequence is an ordered set of related numbers. This relationship is expressed recursively when each number in the sequence is determined using the previous number, or term. Members of the set of numbers in the sequence are called terms. A geometric sequence is an ordered set of numbers in which each member in the set is determined by multiplying the preceding term by a constant value. This constant value is the ratio of the second term to the first term or the ratio of third term to the second term, etc.
### Classroom Example 1
Determine the 8th term of the sequence: 3, 6, 12, 24, ...
Answer: 384
To determine the common ratio find .
Multiply each successive term by 2 until you get to the 8th term, 384.
### Classroom Example 2
What is the value of x in the table?
1 2 3 4 5 7500 1500 300 60 x
Answer: 12
Use what you know about recursive relationships of geometric sequences to justify why your answer is correct. Use words, numbers, and/or symbols in your justification.
Sample correct response: The constant ratio is because . Each term is of the previous term, so the answer should be of 60, or 12.
/toolkit/vsc/clarification/mathematics/grade8/1A1b.xml
Resources for Objective 1.A.1.b: CLARIFICATIONS | Lesson Seeds | Sample Assessments |<|endoftext|>
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# Percent: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
(Redirected to Percentage article)
In mathematics, a percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred"). It is often denoted using the percent sign, "%", or the abbreviation "pct". For example, 45% (read as "forty-five percent") is equal to 45 / 100, or 0.45.
Percentages are used to express how large/small one quantity is, relative to another quantity. The first quantity usually represents a part of, or a change in, the second quantity, which should be greater than zero. For example, an increase of \$ 0.15 on a price of \$ 2.50 is an increase by a fraction of 0.15 / 2.50 = 0.06. Expressed as a percentage, this is therefore a 6% increase.
Although percentages are usually used to express numbers between zero and one, any dimensionless proportionality can be expressed as a percentage. For instance, 111% is 1.11 and -0.35% is -0.0035.
## Proportion
Percentages are correctly used to express fractions of the total. For example, 25% means 25 / 100, or one quarter, of some total.
Percentages larger than 100 can be meant literally (such as "a family must earn at least 125% over the poverty line to sponsor a spouse visa").
## Calculations
The fundamental concept to remember when performing calculations with percentages is that the percent symbol can be treated as being equivalent to the pure number constant 1 / 100 = 0.01. , for an example 35% of 300 can be written as (35 / 100) × 300 = 105.
To find the percentage that a single unit represents out of a whole of N units, divide 100% by N. For instance, if you have 1250 apples, and you want to find out what percentage of these 1250 apples a single apple represents, 100% / 1250 = (100 / 1250)% provides the answer of 0.08%. So, if you give away one apple, you have given away 0.08% of the apples you had. Then, if instead you give away 100 apples, you have given away 100 × 0.08% = 8% of your 1250 apples.
To calculate a percentage of a percentage, convert both percentages to fractions of 100, or to decimals, and multiply them. For example, 50% of 40% is:
(50 / 100) × (40 / 100) = 0.50 × 0.40 = 0.20 = 20 / 100 = 20%.
It is not correct to divide by 100 and use the percent sign at the same time. (E.g. 25% = 25 / 100 = 0.25, not 25% / 100, which actually is (25 / 100) / 100 = 0.0025.)
### Example problems
Whenever we talk about a percentage, it is important to specify what it is relative to, i.e. what is the total that corresponds to 100%. The following problem illustrates this point.
In a certain college 60% of all students are female, and 10% of all students are computer science majors. If 5% of female students are computer science majors, what percentage of computer science majors are female?
We are asked to compute the ratio of female computer science majors to all computer science majors. We know that 60% of all students are female, and among these 5% are computer science majors, so we conclude that (60 / 100) × (5/100) = 3/100 or 3% of all students are female computer science majors. Dividing this by the 10% of all students that are computer science majors, we arrive at the answer: 3% / 10% = 30 / 100 or 30% of all computer science majors are female.
This example is closely related to the concept of conditional probability.
Here are other examples:
1. What is 200% of 30?
Answer: 200% × 30 = (200 / 100) × 30 = 60.
2. What is 13% of 98?
Answer: 13% × 98 = (13 / 100) × 98 = 12.74.
3. 60% of all university students are male. There are 2400 male students. How many students are in the university?
Answer: 2400 = 60% × X, therefore X = (2400 / (60 / 100)) = 4000.
1. There are 300 cats in the village, and 75 of them are black. What is the percentage of black cats in that village?
Answer: 75 = X% × 300 = (X / 100) × 300, so X = (75 / 300) × 100 = 25, and therefore X% = 25%.
2. The number of students at the university increased to 4620, compared to last year's 4125, an absolute increase of 495 students. What is the percentual increase?
Answer: 495 = X% × 4125 = (X / 100) × 4125, so X = (495 / 4125) × 100 = 12, and therefore X% = 12%.
## Percent increase and decrease
Sometimes due to inconsistent usage, it is not always clear from the context what a percentage is relative to. When speaking of a "10% rise" or a "10% fall" in a quantity, the usual interpretation is that this is relative to the initial value of that quantity. For example, if an item is initially priced at \$200 and the price rises 10% (an increase of \$20), the new price will be \$220. Note that this final price is 110% of the initial price (100% + 10% = 110%).
Some other examples of percent changes:
• An increase of 100% in a quantity means that the final amount is 200% of the initial amount (100% of initial + 100% of initial = 200% of initial); in other words, the quantity has doubled.
• An increase of 800% means the final amount is 9 times the original (100% + 800% = 900% = 9 times as large).
• A decrease of 60% means the final amount is 40% of the original (100% − 60% = 40%).
• A decrease of 100% means the final amount is zero (100% − 100% = 0%).
In general, a change of x percent in a quantity results in a final amount that is 100 + x percent of the original amount (equivalently, 1 + 0.01x times the original amount).
It is important to understand that percent changes, as they have been discussed here, do not add in the usual way, if applied sequentially. For example, if the 10% increase in price considered earlier (on the \$200 item, raising its price to \$220) is followed by a 10% decrease in the price (a decrease of \$22), the final price will be \$198, not the original price of \$200. The reason for the apparent discrepancy is that the two percent changes (+10% and −10%) are measured relative to different quantities (\$200 and \$220, respectively), and thus do not "cancel out".
In general, if an increase of x percent is followed by a decrease of x percent, and the initial amount was p, the final amount is p((1 + 0.01x)(1 − 0.01x)) = p(1 − (0.01x)2); thus the net change is an overall decrease by x percent of x percent (the square of the original percent change when expressed as a decimal number). Thus, in the above example, after an increase and decrease of x = 10 percent, the final amount, \$198, was 10% of 10%, or 1%, less than the initial amount of \$200.
This can be expanded for a case where you do not have the same percent change. If the initial percent change is x and the second percent change is y, and the initial amount was p, then the final amount is p((1 + 0.01x)(1 + 0.01y)). To change the above example, after an increase of x = 10 and decrease of y = − 5 percent, the final amount, \$209, is 4.5% more than the initial amount of \$200.
In the case of interest rates, it is a common practice to state the percent change differently. If an interest rate rises from 10% to 15%, for example, it is typical to say, "The interest rate increased by 5%" — rather than by 50%, which would be correct when measured as a percentage of the initial rate (i.e., from 0.10 to 0.15 is an increase of 50%). Such ambiguity can be avoided by using the term "percentage points". In the previous example, the interest rate "increased by 5 percentage points" from 10% to 15%. If the rate then drops by 5 percentage points, it will return to the initial rate of 10%, as expected.
### Change in sign
When the first number is negative and second number is positive, the percentage change from first number to second number is negative. This often occurs in financial statements that changes from a period of loss to period in profit.
```Acme Company EBIT
First quarter (100)
Second quarter 100
Change in profitability (100 - (-100))/(-100) = -200%
```
In expressing a number as a percentage, the base of the comparison cannot be negative. The First number in the above example is the base of the comparison when it is expressed as a positive amount becomes Loss of 100. The change from First quarter loss to Second quarter profit becomes Percentage change in loss by -200% to turn a profit of 100.
## Word and symbol
In British English, percent is sometimes written as two words (per cent, although percentage and percentile are written as one word). In American English, percent is the most common variant (but cf. per mille written as two words). In EU context the word is always spelled out in one word percent. In the early part of the twentieth century, there was a dotted abbreviation form "per cent.", as opposed to "per cent". The form "per cent." is still in use as a part of the highly formal language found in certain documents like commercial loan agreements (particularly those subject to, or inspired by, common law), as well as in the Hansard transcripts of British Parliamentary proceedings. While the term has been attributed to Latin per centum, this is a pseudo-Latin construction and the term was likely originally adopted from the French pour cent. The concept of considering values as parts of a hundred is originally Greek. The symbol for percent (%) evolved from a symbol abbreviating the Italian per cento. In some other languages, the form prosent is used instead. Some languages use both a word derived from percent and an expression in that language meaning the same thing, e.g. Romanian procent and la sută (thus, 10 % can be read or sometimes written ten for [each] hundred, similarly with the English one out of ten).
Grammar and style guides often differ as to how percentages are to be written. For instance, it is commonly suggested that the word percent (or per cent) be spelled out in all texts, as in "1 percent" and not "1%." Other guides prefer the word to be written out in humanistic texts, but the symbol to be used in scientific texts. Most guides agree that they always be written with a numeral, as in "5 percent" and not "five percent," the only exception being at the beginning of a sentence: "Ninety percent of all writers hate style guides." Decimals are also to be used instead of fractions, as in "3.5 percent of the gain" and not "3 ½ percent of the gain." It is also widely accepted to use the percent symbol (%) in tabular and graphic material.
There is no consensus as to whether a space should be included between the number and percent sign in English. Style guides – such as the Chicago Manual of Style – commonly prescribe to write the number and percent sign without any space in between.[1] The International System of Units and the ISO 31-0 standard, on the other hand, require a space.[2][3]
## Other uses
It is a well known cliché that many in the media and sports worlds use terms such as 'The players are giving 110% out on the pitch tonight' despite the fact that such phrases do not make any mathematical sense.
The word "percentage" is often misused in the context of sports statistics, when the referenced number is expressed as a decimal proportion, not a percentage: "The Phoenix Suns' Shaquille O'Neal led the NBA with a .609 field goal percentage (FG%) during the 2008-09 season." (O'Neal made 60.9% of his shots, not 0.609%.) The practice is probably related to the similar way that batting averages are quoted.
As "percent" it is used to describe the steepness of the slope of a road or railway.
## References
1. ^ "The Chicago Manual of Style". University of Chicago Press. 2003. Retrieved 2007-01-05.
2. ^
3. ^
# Simple English
Percent means out of one hundred. It is often shown with the symbol "%". It is used even if there are not a hundred items. The number is then scaled so it can be compared to one hundred. For example, I have a bowl of fruit with three apples and one orange. The percentage of apples is 3 out of 4 = 3/4 = 75/100 = 75%.
## Uses
Percentages are useful because people can compare things that are not out of the same number. For example, exam marks are often percentages, so people can compare them even if there are more questions on one exam paper than the other. In coding it can mean modulo.<|endoftext|>
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After North Carolina joined the other future states during the American Revolution, it gave up its territory west of the Appalachian mountains to the new American government. Those lands became a territory belonging to the United States, with the clunky name “Territory of the United States, South of the River Ohio.” The people who had settled those western lands asked the Continental Congress to admit their territory as a State.
While North Carolina was working toward becoming a state, which they accomplished in 1789, they temporarily regained control over the territory. When they became a State, though, the new government of North Carolina confirmed its transfer of the western territory to the federal government. The Congress of the United States, using the powers given by the Territorial Clause in Art. IV, Sec. 3, Cl. 2 of the Constitution, organized the territory in the same way the Northwest Territory of greater Ohio had been organized. The Northwest Ordinance became law under the newly ratified U.S. Constitution, and it was used as the pattern for moving territories toward statehood.
So the land which is now Tennessee became the “Southwest Territory” and was governed under the Northwest Ordinance. This meant that Congress expected the Southwest Territory to become a state eventually, and was prepared to support the people living there along that path.
But the people of Tennessee were not satisfied. They had fought in the Revolutionary War, they wanted to become a state, and the process Congress had in mind was just too slow for them. They held a vote and 73% of the people voted for immediate statehood. The Governor and the local legislature held a convention to establish a constitutional government — not as a territorial government but for government as a State of the Union. The convention approved a state constitution, declared the end of territorial government on March 28, 1796, and said Tennessee would become a State on that same day.
The legislature also established two Congressional districts, authorized four presidential electors, sponsored elections for two members of the U.S. House, and elected two Senators. The U.S. Senate opposed admission; Tennessee was being too bold for them. The Tennessee Senators went to the Senate and demanded their seats, but the Senate refused.
The House supported admission, though. On June 1, 1796, Congress yielded and passed an admission act allowing Tennessee one seat in the House until the next census. They also insisted on new elections, since the citizens of a territory did not have the power to elect members of Congress. Only citizens of a state can do that. Tennessee had declared itself a state, but only Congress can do that.
Tennessee accepted the compromise and became a state just a few months after they said they would.
Until that time, territories had asked for statehood and then waited for Congress to declare them states. Tennessee’s bold move essentially meant that they decided they were a state, declared themselves a state, and persuaded Congress to agree. This approach has been called “the Tennessee Plan.”
Six other territories organized under a version of the Tennessee Plan and forced their admission into statehood. These were Michigan (1837), Iowa (1846), California (1850), Oregon (1859), Kansas (1861), and Alaska (1959). New Mexico attempted the Tennessee Plan in 1850, but it failed. New Mexico was finally admitted in 1912 after a waiting through the Congressional process for territorial status resolution for 62 more years.
Alaska studied the New Mexico experience before committing itself to the Tennessee Plan. The people of Alaska noticed that the Tennessee Plan worked best when it had local law behind it and when it was clearly the will of the people.
Proving the will of the people can take different forms. In the case of Tennessee, it was a locally sponsored referendum on how to work toward statehood. In other cases, it was the vote to ratify a state constitution or to elect Congressional representatives.
Tennessee’s plan worked for them and for states after them, and it might work for Puerto Rico, too. Read more about the Tennessee Plan as an option for Puerto Rico.
More of our series on paths to statehood:
This post was originally written in English and may be being auto-translated by Google.<|endoftext|>
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Conditioning for the young athlete
The majority of kids will have a good time playing sports, especially when the prepare for the activity. Kids should acquire skills, train their bodies, and learn to protect themselves so they are less at risk for injury. This is important for every sport.
Young athletes need to learn the skills for their sport and also how to warm up, stretch, strength train, and develop power, agility, toughness and explosiveness.
Conditioning, however, should also be a big part of their athletic career. Warming up, stretching, and developing power, agility, toughness, and explosiveness. Unlike strength training and weight lifting, conditioning is all about body movements that increase athletic skill and physical fitness and decrease the chance of injury. The types of exercises may be different depending on the fitness goals.
To improve athletic performance, the athlete uses targeted, specific movement to mimic the moves used on the court of field. A basketball player can practice shooting baskets repeatedly to help with sports conditioning. However, aerobic exercises – an important part of any general athletic program – increases cardiovascular endurance and lung capacity.
Usually the aerobic part of the training includes low-intensity, long-duration exercises such as running or cycling. The heart rate should be increased and sustained for a time period that challenges the heart and lungs so they get stronger.
There is also anaerobic exercises for sports that require intense, sudden burst of speed and strength. Weigh training and sprinting are anaerobic exercises. One of the main goals of a conditioning program is to increase the amount of stress that the body can handle before it gets injured. Therefore, conditioning programs play an important role in preventing injury.
A propper warm-up for sports should increase the blood flow to the working muscles, which will in turn lead to a decrease in muscle stiffness and less risk of injury. The warm-up will also improve performance and add to the athletes’ preparation mentally before practice or a game.
Th muscle temperature will rise, and, therefore, speed and strength are improved. Muscle elasticity is improved with the increase in body temperature. The blood vessels dilate, and resistance to blood flow decreases. A goo way to warm up cold muscles and get the circulation going is about ten minutes of light running or cycling before practice.<|endoftext|>
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Communication means interaction and therefore both production and comprehension. Oral comprehension (listening and viewing) is very important for effective communication and often regarded as a very challenging skill. Listening (and viewing) is a key step in communication and this activity aims to help you reflect on your listening and viewing comprehension skills in order to find some strategies that can enhance your ability to communicate effectively. The text used for this activity refers to learners of English as a foreign language, but it can be applied to foreign language learning in general.
Upon completion of this task you will be able to:
- have a better understanding of how comprehension works when listening/viewing in the target language
- reflect on your listening and viewing skills
- use strategies to improve your listening and viewing skills
Computer, tablet, internet
Material and links:
Before the activity:
- What is your level of comprehension in your target language?
- Is oral comprehension (listening and/or viewing) in the target language difficult for you? If yes, why? How? What areas would you highlight as problematic?
- Do you use any strategies when listening to someone talking in the target language? If yes, make a list.
- Compare how you listen in your native language and in the target language. Are there any differences?
- The following sections are included in the article Five essential listening skills for English learners
- Predicting content
- Listening for the gist
- Detecting signposts
- Listening for details
- Inferring meaning
What do you think the titles refer to?
- Open the link and read the text, comparing your answers to the contents of the text.
- What did you find in the text that you were already familiar with?
- What new information did you find?
- How do you think the new information could be helpful for your language learning experience?
Once you have done this activity you may wish to try a MOOC (Massive Open Online Course)
Or the following activities to apply listening and viewing strategies:
Source/attribution: Digilanguages. Author: Valentina Rizzo
Mein Weg nach Deutschland and other playlists on the Goethe Institute youtube channel might be helpful.
Das Deutschlandlabor is an interesting playlist on A2 level by Deutsche Welle.
Léigh an chomhairle ag an nasc seo a leanas: Forbairt Scileanna Léitheoireachta agus Éisteachta / Developing your reading and listening skills.<|endoftext|>
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Metacognition (the thinking about thinking) refers to how children are able to monitor and direct their own learning processes.
To develop metacognitive awareness and strategies, and know when and how to use the strategies, students should have opportunities to solve non-routine and open-ended problems, to discuss their solutions, to think aloud and reflect on what they are doing, and to keep track of how things are going and make changes when necessary.
To develop metacognitive awareness, children typically go through the following steps (Pressley, Borkowski, & Schneider, 1987):
- They establish a motivation to learn a metacognitive process. This occurs when either they themselves or someone else points gaives them reason to believe that there would be some benefit to knowing how to apply the process. (Their readiness to learn)
- They focus their attention on what it is that they or someone else does that is useful. This proper focusing of attention puts the necessary information into working memory. Sometimes this focusing of attention can occur through modelling, and sometimes it occurs during personal experience.
- They talk to themselves about the metacognitive process. This talk can arise during their interactions with others, but it is their talk to themselves that is essential. This self talk serves several purposes:
- It enables them to understand and encode the process.
- It enables them to practice the process.
- It enables them to obtain feedback and to make adjustments regarding their effective use of the process.
- It enables them to transfer the process to new situations beyond those in which it has already been used.
Eventually, children begin to use the process without even being aware that they are doing so.
Some example of questions to ask oneself during phrases of learning:
- During the ‘readiness’ phase
- What am I supposed to learn?
- What prior knowledge will help me with this task?
- What should I do first?
- What should I look for in this reading?
- How much time do I have to complete this?
- In what direction do I want my thinking to take me?
- During the ‘engagement’ phase
- How am I doing?
- Am I on the right track?
- How should I proceed?
- What information is important to remember?
- Should I move in a different direction?
- Should I adjust the pace because of the difficulty?
- What can I do if I do not understand?
- During the ‘mastery’ phase
- How well did I do?
- What did I learn?
- Did I get the results I expected?
- What could I have done differently?
- Can I apply this way of thinking to other problems or situations?
- Is there anything I don’t understand—any gaps in my knowledge?
- Do I need to go back through the task to fill in any gaps in understanding?
- How might I apply this line of thinking to other problems?<|endoftext|>
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## EvaluateiExponential Functions
### Learning Outcomes
• Identify the base of an exponential function and restrictions for its value.
• Evaluate exponential growth functions.
• Use the compound interest formula.
• Evaluate exponential functions with base e.
India is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year.[1] If this rate continues, the population of India will exceed China’s population by the year 2031. When populations grow rapidly, we often say that the growth is “exponential.” To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions, which model this kind of rapid growth.
The base of an exponential function must be a positive real number other than 1. Why do we limit the base b to positive values? This is done to ensure that the outputs will be real numbers. Observe what happens if the base is not positive:
• Consider a base of –9 and exponent of $\frac{1}{2}$. Then $f\left(x\right)=f\left(\frac{1}{2}\right)={\left(-9\right)}^{\frac{1}{2}}=\sqrt{-9}$, which is not a real number.
Why do we limit the base to positive values other than 1? This is because a base of 1 results in the constant function. Observe what happens if the base is 1:
• Consider a base of 1. Then $f\left(x\right)={1}^{x}=1$ for any value of x.
To evaluate an exponential function with the form $f\left(x\right)={b}^{x}$, we simply substitute x with the given value, and calculate the resulting power. For example:
Let $f\left(x\right)={2}^{x}$. What is $f\left(3\right)$?
$\begin{array}{llllllll}f\left(x\right)\hfill & ={2}^{x}\hfill & \hfill \\ f\left(3\right)\hfill & ={2}^{3}\text{}\hfill & \text{Substitute }x=3. \hfill \\ \hfill & =8\text{}\hfill & \text{Evaluate the power}\text{.}\hfill \end{array}$
To evaluate an exponential function, it is important to follow the order of operations. For example:
Let $f\left(x\right)=30{\left(2\right)}^{x}$. What is $f\left(3\right)$?
$\begin{array}{c}f\left(x\right)\hfill & =30{\left(2\right)}^{x}\hfill & \hfill \\ f\left(3\right)\hfill & =30{\left(2\right)}^{3}\hfill & \text{Substitute }x=3.\hfill \\ \hfill & =30\left(8\right)\text{ }\hfill & \text{Simplify the power first}\text{.}\hfill \\ \hfill & =240\hfill & \text{Multiply}\text{.}\hfill \end{array}$
Note that if the order of operations were not followed, the result would be incorrect:
$f\left(3\right)=30{\left(2\right)}^{3}\ne {60}^{3}=216,000$
### Example: Evaluating Exponential Functions
Let $f\left(x\right)=5{\left(3\right)}^{x+1}$. Evaluate $f\left(2\right)$ without using a calculator.
### Try It
Let $f\left(x\right)=8{\left(1.2\right)}^{x - 5}$. Evaluate $f\left(3\right)$ using a calculator. Round to four decimal places.
Because the output of exponential functions increases very rapidly, the term “exponential growth” is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.
### A General Note: Exponential Growth
A function that models exponential growth grows by a rate proportional to the amount present. For any real number x and any positive real numbers and b such that $b\ne 1$, an exponential growth function has the form
$\text{ }f\left(x\right)=a{b}^{x}$
where
• a is the initial or starting value of the function.
• b is the growth factor or growth multiplier per unit x.
In more general terms, an exponential function consists of a constant base raised to a variable exponent. To differentiate between linear and exponential functions, let’s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function $A\left(x\right)=100+50x$. Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function $B\left(x\right)=100{\left(1+0.5\right)}^{x}$.
A few years of growth for these companies are illustrated below.
Year, x Stores, Company A Stores, Company B
0 100 + 50(0) = 100 100(1 + 0.5)0 = 100
1 100 + 50(1) = 150 100(1 + 0.5)1 = 150
2 100 + 50(2) = 200 100(1 + 0.5)2 = 225
3 100 + 50(3) = 250 100(1 + 0.5)3 = 337.5
x A(x) = 100 + 50x B(x) = 100(1 + 0.5)x
The graphs comparing the number of stores for each company over a five-year period are shown below. We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.
The graph shows the numbers of stores Companies A and B opened over a five-year period.
Notice that the domain for both functions is $\left[0,\infty \right)$, and the range for both functions is $\left[100,\infty \right)$. After year 1, Company B always has more stores than Company A.
Now we will turn our attention to the function representing the number of stores for Company B, $B\left(x\right)=100{\left(1+0.5\right)}^{x}$. In this exponential function, 100 represents the initial number of stores, 0.5 represents the growth rate, and $1+0.5=1.5$ represents the growth factor. Generalizing further, we can write this function as $B\left(x\right)=100{\left(1.5\right)}^{x}$ where 100 is the initial value, 1.5 is called the base, and x is called the exponent.
### Example: Evaluating a Real-World Exponential Model
At the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013 with an annual growth rate of about 1.2%. This situation is represented by the growth function $P\left(t\right)=1.25{\left(1.012\right)}^{t}$ where t is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031?
### Try It
The population of China was about 1.39 billion in the year 2013 with an annual growth rate of about 0.6%. This situation is represented by the growth function $P\left(t\right)=1.39{\left(1.006\right)}^{t}$ where t is the number of years since 2013. To the nearest thousandth, what will the population of China be in the year 2031? How does this compare to the population prediction we made for India in the previous example?
## Using the Compound Interest Formula
Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest. The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account.
The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly interest rate earned by an investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal rate! This is a powerful tool for investing.
We can calculate compound interest using the compound interest formula which is an exponential function of the variables time t, principal P, APR r, and number of times compounded in a year n:
$A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt}$
For example, observe the table below, which shows the result of investing $1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases. Frequency Value after 1 year Annually$1100
Semiannually $1102.50 Quarterly$1103.81
Monthly $1104.71 Daily$1105.16
### A General Note: The Compound Interest Formula
Compound interest can be calculated using the formula
$A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt}$
where
• A(t) is the accumulated value of the account
• t is measured in years
• P is the starting amount of the account, often called the principal, or more generally present value
• r is the annual percentage rate (APR) expressed as a decimal
• n is the number of times compounded in a year
A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child’s future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to $40,000 over 18 years. She believes the account will earn 6% compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now? ### Try It Refer to the previous example. To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly? ## Evaluating Exponential Functions with Base e As we saw earlier, the amount earned on an account increases as the compounding frequency increases. The table below shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue. Examine the value of$1 invested at 100% interest for 1 year compounded at various frequencies.
Frequency $A\left(t\right)={\left(1+\frac{1}{n}\right)}^{n}$ Value
Annually ${\left(1+\frac{1}{1}\right)}^{1}$ $2 Semiannually ${\left(1+\frac{1}{2}\right)}^{2}$$2.25
Quarterly ${\left(1+\frac{1}{4}\right)}^{4}$ $2.441406 Monthly ${\left(1+\frac{1}{12}\right)}^{12}$$2.613035
Daily ${\left(1+\frac{1}{365}\right)}^{365}$ $2.714567 Hourly ${\left(1+\frac{1}{\text{8766}}\right)}^{\text{8766}}$$2.718127
Once per minute ${\left(1+\frac{1}{\text{525960}}\right)}^{\text{525960}}$ $2.718279 Once per second ${\left(1+\frac{1}{31557600}\right)}^{31557600}$$2.718282
These values appear to be approaching a limit as n increases without bound. In fact, as n gets larger and larger, the expression ${\left(1+\frac{1}{n}\right)}^{n}$ approaches a number used so frequently in mathematics that it has its own name: the letter $e$. This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below.
### A General Note: The Number $e$
The letter e represents the irrational number
${\left(1+\frac{1}{n}\right)}^{n},\text{as }n\text{ increases without bound}$
The letter e is used as a base for many real-world exponential models. To work with base e, we use the approximation, $e\approx 2.718282$. The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first investigated and discovered many of its properties.
### Example: Using a Calculator to Find Powers of $e$
Calculate ${e}^{3.14}$. Round to five decimal places.
### Try It
Use a calculator to find ${e}^{-0.5}$. Round to five decimal places.
## Contribute!
Did you have an idea for improving this content? We’d love your input.<|endoftext|>
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Simplifying Negative Square Roots
Lesson Features »
Lesson Priority: High
Algebra Two $\longrightarrow$
Complex Numbers $\longrightarrow$
Objectives
• Learn how to simplify square roots of negative numbers
• Apply what we already know about reducing square roots of positive integers
Lesson Description
Now that we understand what the imaginary number is and how it works, we can simplify the square root of negative numbers in a very similar way to how we simplify square roots of positive ones.
Practice Problems
Practice problems and worksheet coming soon!
I Thought We Couldn't Root Negatives?
By far, the most common and useful applications of the imaginary number involve simplifying square root expressions that contain negative numbers. For better or worse, this process is virtually a carbon copy of the skill of simplifying positive root expressions - so if you aren't caught up on how to simplify integer square roots » then get with the program, and then read on.It's worth mentioning that we probably have had at least one teacher who has specifically told us that we cannot take the square root of a negative number. Whether or not you want to consider that a lie depends on your perspective - it's true that you cannot square root a negative number if you are only going to work with real numbers, but if you allow for the existence of imaginary numbers » , then you can actually square root anything (including square rooting imaginary numbers » , but that's not something we'll look at until after we know some trigonometry!). For all students taking Algebra 2 and beyond, it is necessary to work with imaginary numbers, at least at the basic level, and this lesson is the skill you absolutely need to know.Fortunately, as we said this is a straight-forward adaptation of what we already know about simplifying square roots. Specifically, here's how it works.
Theorem: Negative Square RootsSimplifying an expression involving the square root of a negative quantity is done by simplifying the quantity as if it was positive, and placing an $i$ in front as a coefficient. In other words,$$\sqrt{-x} = i\sqrt{x}$$For all $x$ such that $x$ is a positive number, so that $-x$ is a negative number.
Let's take a look at an example.
Example 1Simplify.$$\sqrt{-40}$$$\blacktriangleright$ Digging back to our knowledge of simplifying integer square roots, we realize that $\sqrt{40}$ is simplified as $2\sqrt{10}$. Therefore, applying Theorem 1, we have$$\sqrt{-40} = i\sqrt{40}$$$$= 2i\sqrt{10}$$
It's a simple adaptation to what we already know. We know $\sqrt{40} = 2\sqrt{10}$, so $\sqrt{-40} = 2i\sqrt{10}$.
Pro Tip
Sharpen up on your regular root simplification skills. Once you get in the groove on practicing this skill, you'll see that you're really just using that regular root skill that you already have, except you're simply sticking an $i$ in front of your answer. All the "hard work" revolves completely around the root simplification process you already know!
Here's another example.
Example 2Simplify.$$\sqrt{-72}$$$\blacktriangleright$ Via the square root simplification process we already know:$$\sqrt{72} = \sqrt{(36)(2)}$$$$=6 \sqrt{2}$$Once again, Theorem 1 above tells us that$$\sqrt{-72} = i \sqrt{72}$$It follows that$$\sqrt{-72} = i \cdot 6\sqrt{2}$$$$6i \sqrt{2}$$
Warning!
The Root Product Property fails for square roots of products of negative numbers due to the properties of the imaginary number!!! Be careful not to apply it verbatim! For example, let's try and apply the Root Product Property to the following example:$$\sqrt{(-10)(-7)}$$According to the Root Product Property:$$\sqrt{(-10)(-7)} = \sqrt{(-10)} \cdot \sqrt{(-7)}$$$$\longrightarrow = i\sqrt{10} \cdot i \sqrt{7} = i^2 \sqrt{10}\sqrt{7}$$$$= -\sqrt{10} \sqrt{7} = -\sqrt{70}$$But if we instead multiplied under the root first, then simplified, we would have got a completely different answer:$$\sqrt{(-10)(-7)} = \sqrt{70}$$As you can see, with the first perspective, the imaginary numbers created a negative sign outside of the radical. However, under the second perspective, the two negatives multiplied to a positive result before we even started trying to take square roots. It is the latter case that is correct. It looks dumbfounding, but the good news is, teacher and professors rarely try to use this apparent slight of hand against you. It is incredibly rare that this mishap is actually presented to you in a situation where you're getting graded.
Overall, the takeaway of this lesson is that usually you will simply be given a square root expression with a negative sign to simplify. Accomplish this by applying what we already know about simplifying square roots to get the number part correct, and use the idea that we end up with the same result as we would in the case of a positive square root, but with an $i$ factored out.
Put It To The Test
Try a few problems out on your own. Remember that there's nothing new here from what you already should know about simplifying numeric roots, in terms of the numbers. If you're looking for practice problems about other things we do with $i$, check out the other lessons on Imaginary Numbers ».
Example 3Simplify.$$\sqrt{-100}$$
Show solution
$$\blacktriangleright \,\, \sqrt{-100} = i \sqrt{100}$$$$= 10i$$
Example 4Simplify.$$\sqrt{-75}$$
Show solution
$$\blacktriangleright \,\, \sqrt{-75} = i \sqrt{75}$$$$=i \sqrt{25 \cdot 3} = 5i \sqrt{3}$$
Example 5Simplify.$$\sqrt{-56}$$
Show solution
$$\blacktriangleright \,\, \sqrt{-56} = i \sqrt{56}$$$$=i \sqrt{4 \cdot 14} = 2i \sqrt{14}$$
Example 6Simplify.$$\sqrt{-91}$$
Show solution
$$\blacktriangleright \,\, \sqrt{-91} = i \sqrt{91}$$$91$ is prime, so there is nothing further to simplify.
Lesson Takeaways
• Be able to simplify square roots when the radical contains a negative number
• Although it is unlikely to appear on a test, understand why the Root Product Property fails for square roots of the product of negative numbers.
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At the top of the lesson, you can see the lesson title, as well as the DNA of Math path to see how it fits into the big picture. You can also see the description of what will be covered in the lesson, the specific objectives that the lesson will cover, and links to the section's practice problems (if available).
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Headlines - Every lesson is subdivided into mini sections that help you go from "no clue" to "pro, dude". If you keep getting lost skimming the lesson, start from scratch, read through, and you'll be set straight super fast.
Examples - there is no better way to learn than by doing. Some examples are instructional, while others are elective (such examples have their solutions hidden).
Perils and Pitfalls - common mistakes to avoid.
Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!
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Remember! - Remember notes need to be in your head at the peril of losing points on tests.
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Warning! - Something you should be careful about.<|endoftext|>
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# ps9b23 - Ph 111.01 PS 9b(24 Problem 1 A 2 kg mass and a 5...
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Ph 111.01 October 29, 2008 PS 9b (24) Problem 1. A 2 kg mass and a 5 kg mass are attached to either end of a 3 m long massless rod. a.) Find the center of mass of the system. m, from the 2 kg mass. Find the rotational inertia (I) of the system when rotated about: b.) the end with the 2 kg mass. kg m 2 c.) the end with the 5 kg mass. kg m 2 d.) the center of the rod. kg m 2 e.) the center of mass of the system. kg m 2 (Compare this to parts b-d. Is this what you expect?) f.) If the system is rotated about the center of mass by a force of 8 N acting on the 5 kg mass, (perpendicular to the rod), what will the size of the angular acceleration of the system be? rad/s 2 m 1 m 2 x 0L Solution. (a) We will calculate the x -coordinate of the CM, with x = 0 at the position of the smaller mass, as shown. () ( ) ( ) 11 2 2 12 2 kg 0 5 kg 3 m 2.143 m 7 kg CM Xm x m x mm =+ = + = + (b) To find the rotational inertia (the moment of inertia I ) about various points, we will use the simple expression for point masses, 2 I MR = . ( )( ) 2 22 2 5 kg 3 m 45 kg-m Im L == = (c) Now about the big mass: ( )( ) 2 1 2 kg 3 m 18 kg-m L = (d) Now about the point x = L /2: ( ) ( ) 2 2 kg 1.5 m 5 kg 1.5 m 15.75 kg-m LL m ⎛⎞ =+= + = ⎜⎟ ⎝⎠ (e) About x = 2.143 m: ( ) ( )( ) ( )( ) 2222 2 2.143 m 2.143 m 2 kg 2.143 m 5 kg 0.857 m 12.86 kg-m mL = + = It is interesting that the minimum moment of inertia is about the CM.
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A major study for the UK government has cast doubt over claims that rising temperatures are causing soil to pump greater amounts of carbon dioxide into the atmosphere, further fuelling global warming.
In 2005 it was reported in the science journal Nature that over the past 25 years 100m tonnes of carbon dioxide had been released by the soil of England and Wales. The figure cancelled out all emissions cuts in the UK since 1990.
However, a national survey of the soils of Great Britain, funded by the department for environment food and rural affairs, claims to have found no net loss of carbon over approximately the same period.
Scientists have now proposed that a special study group, with an independent statistical expert, should examine why the reports differ and which result is more likely to be correct.
The latest questions follow weeks of claims that predictions about the impacts of climate change have been overstated or miscalculated, including the melting of Himalayan glaciers, and separate allegations of bias based on leaked emails from scientists at the climate research unit at the University of East Anglia.
The author of the latest report, Professor Bridget Emmett of the UK Centre for Ecology and Hydrology (CEH), warned that finding there had been no loss of carbon so far should not be taken to mean the absence of a threat. In the long term, scientists predict a "tipping point" when the faster activity of microbes in warmer soils starts to generate more CO2 than can be absorbed by plants.
"That's when you start losing carbon as a whole," said Emmett. "Most of the models say that will be later this century."
The 2005 report in Nature was based on the National Soil Inventory, carried out initially between 1978 and 1983, and again from 1994 to 2003, by the National Soil Resources Institute at Cranfield University. That study said that from 1978 to 2003 there had been an estimated loss of 4m tonnes of carbon a year from the soils of England and Wales, and the researchers estimated that, because of the higher carbon content of Scotland's peaty soils, the annual loss from the UK as a whole was 13m tonnes a year. The fact that the losses occurred across all types of land use suggested a link to climate change, said the team.
At that time, one of the research team, Professor Guy Kirk of Cranfield University, told a conference: "It had been reckoned that the CO2 fertilisation effect was offsetting about 25% of the direct human-induced carbon dioxide emissions. It was reckoned that the soil temperature emission effect would catch up in maybe 10 to 50 years' time. We are showing that it seems to be happening rather faster than that."
The latest report by the CEH, just released as part of the ongoing analysis of the 2007 Countryside Survey of Great Britain, compared studies between 1978 and 2007. It found carbon concentration in the top 15cm of soil increased over the first two decades, and decreased between 1998 and 2007. The only exception was arable land, where there was a net loss of carbon, probably because of disruption by ploughing.
"Overall there was no change in carbon concentration ... and [we] cannot confirm the loss reported by the National Soil Inventory," states the report.
Kirk told the Guardian that the Cranfield team were still "confident in our results [that] there was a net loss of carbon". But he said subsequent studies had suggested that "at best" 10% of the loss of carbon was due to climate change, and the rest was due to changes in land use and management, such as conversion of grassland to crops.
Reasons being examined for the difference in results include where and how samples were chosen and analysed and how the data was compiled.
"The amount of carbon in topsoils across England and Wales is about 2bn tonnes, so detecting a change of even 4m tonnes per year is very challenging," said Emmett. "Small differences in methods between the two surveys can therefore have a large effect."<|endoftext|>
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Clark's Grebes are large and slender with long necks and long, thin bills. Their plumage is dark gray above and white below, with a clear color division. The top of the face is black and the bottom white. In Clark's Grebe, the black does not extend below the eye. In the closely related and similar-appearing Western Grebe, the black ends below the eye. The bill of Clark's Grebe is bright yellow to orange-yellow.
In winter Clark's Grebes are found mostly on saltwater bays. During the breeding season they prefer freshwater wetlands with a mix of open water and emergent vegetation. Breeding areas are located in the central arid steppe and Big Sage/Fescue zones that stretch from California north and east to south-central Canada. Clark's Grebes tend to forage farther from shore and in deeper water than Western Grebes. They are commonly found in mixed flocks with Western Grebes, but even in these flocks they tend to associate preferentially with other Clark's Grebes.
Clark's Grebes are highly gregarious in all seasons, wintering in large flocks and nesting in colonies. The neck structure of the Clark's Grebe allows it to thrust its beak forward, like a spear, which it does to catch prey. As a family, grebes are known for their elaborate courtship displays. Clark's Grebes and the closely related Western Grebes perform the most spectacular displays of the family, and arguably the most complex known for any birds. The rituals of Western and Clark's Grebes are almost identical; the only difference is that one of many calls differs in the number of notes.
In all seasons and habitats, the primary food of the Clark's Grebe is fish.
Both the male and female build a floating nest made from a heap of plant material anchored to emergent vegetation in a shallow area of a marsh. The female lays three to four eggs, and both parents incubate. Once hatched, the young leave the nest almost immediately and ride on the backs of the parents. Both parents feed the young. Young are plain gray and white, not striped like the young of other grebes.
Birds from the northern part of the population migrate at night to the Pacific Ocean. Others, mainly in the central California valleys, are year-round residents.
Tens of thousands of Western and Clark's Grebes were killed at the turn of the 20th Century for their feathers. With protection, they have recovered and can now be found breeding in new areas not occupied historically. However, fluctuating water levels, oil spills, gill nets, and poisons such as rotenone (used to kill carp) may have negative effects on the population. When approached by humans, Clark's Grebes will leave their nest, leaving eggs vulnerable to predation and the elements. For this reason, areas frequently disturbed by humans may have low productivity. Even though they are extremely similar and closely related, Western and Clark's Grebes are separate species. Although they breed side by side in the same area, hybrids are surprisingly rare. Clark's Grebes are far less common in Washington than Westerns.
When and Where to Find in Washington
Clark's Grebes are found on both small and large bodies of water in and around Grant County, including Potholes Reservoir, Moses Lake, Banks Lake, and on ponds within the Columbia and Saddle Mountain National Wildlife Refuges. Very rarely, Clark's Grebes are also found on lakes in western Washington in winter.
Washington Range Map
North American Range Map
|Federal Endangered Species List||Audubon/American Bird Conservancy Watch List||State Endangered Species List||Audubon Washington Vulnerable Birds List|
|Yellow List||Candidate||Early Warning|<|endoftext|>
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# Solve System of Equations Using Elimination Method
Aug 23, 2023
## Solving System of Equations by Elimination:
The addition (or elimination) method for solving a system of equations can be used to find the solution to a system. The goal is to obtain one equation containing only one variable.
#### Steps to Solve a System of Equations by the Addition Method:
1. If necessary, rewrite each equation in standard form, ax + by = c.
2. If necessary, multiply one or both equations by a constant(s) so that when the equations are added, the sum will contain only one variable.
3. Add the respective sides of the equations. This will result in a single equation containing only one variable.
4. Solve the equation obtained in step 3.
5. Substitute the value found in Step 4 into either of the original equations. Solve that equation to find the value of the remaining variable.
6. Write the solution as an ordered pair.
7. Check your solution in all equations in the system.
### Solve a System of Equations by Adding
Example 1:
What is the solution to the system of equations?
2x + = 11
x + 3y = 18
Solution:
Since adding the equations at this point would not eliminate a variable, the first equation is multiplied by –2.
Now solve for x by substituting 5 for y in either of the original equations.
The solution is (3, 5).
Example 2:
What is the solution to the system of equations?
2x + 5y = 3
3x – 5y = 17
Solution:
2x + 5y = 3
3x – 5y = 17
The x-variable can now be obtained using the same steps used for the substitution method.
2x + 5y = 3
3x – 5y = 17
Substitute x = 2 into either equation:
2x + 5y = 3
2(4) + 5y= 3
8 + 5y = 3
5y = -5
y = -1
The solution of the system is (4, -1).
### Understand Equivalent Systems of Equations
Example 3:
What is the solution to the system of equations?
3x – 2y = 8
4x + 10y = -2
Solution:
Before adding equations, multiply each side of one of the equations by a constant that makes either the x or y terms opposites.
3x – 2y = 8
4x + 10y = -2
Multiply the top equation by 5:
5(3x – 2y) = 5(8)
Now substitute 2 for y in either of the two equations in the system.
2 (2) + 5y = -1
5y = -5
y = -1
The solution is (2, -1).
### Apply Elimination
Example 4:
Ashley wants to use milk and orange juice to increase the amount of calcium and vitamin A in her daily diet. An ounce of milk contains 37 milligrams of calcium and 57 micrograms of vitamin A. An ounce of orange juice contains 5 milligrams of calcium and 65 micrograms of vitamin A. How many ounces of milk and orange juice should Ashley drink each day to provide exactly 500 milligrams of calcium and 1,200 micrograms of vitamin A?
Solution:
Formulate:
Let x = number of ounces of milk
Let y = number of ounces of orange juice
Calcium: 37x + 5y = 500
Vitamin A: 57x + 65y = 1,200
Compute:
Multiply each equation by constants to eliminate one variable.
37x + 5y = 500……………….Multiply by –13
-481x – 65y = -6,500
-481x – 65y = -6,500
57x + 65y = 1,200
___________________________
-424x = -5,300
x = 12.5
Solve for y:
37(12.5) + 5y = 500
5y = 37.5
y = 7.5
The solution is (12.5, 7.5)
Interpret:
Drinking 12.5 ounces of milk and 7.5 ounces of orange juice each day will provide Ashley with the required amounts of calcium and vitamin A.
### Choose a Method of Solving
Example 5:
What is the solution to the system of equations?
A.
3x – 2y = 38
x = 6 – y
Solution:
Since the equation is already solved for one of the variables, you can easily substitute 6 – y for x.
3(6 – y)- 2y = 38
18 – 3y – 2y = 38
18 – 5y = 38
–5y = 38 – 18
–5y = 20
y = 20/–5
y = –4
Solve for x,
x = 6 – (–4)
x = 10
The solution is (10, –4)
B.
6x + 12y = –6
3x – 2y = –27
Solution:
The coefficient of y in the second equation is an integer multiple of the coefficient of y in the first equation. This makes it easy to eliminate the y variable.
6x + 12y = –6
3x – 2y = –27………..Multiply by 6 18x – 12y = –162
Add the equation and solve for x,
6x + 12y = –6
18x – 12y = –162
_____________________
24x = –168
x = –7
Now solve for y
3(–7)– 2y = –27
–21 –2y = –27
–2y = –27 + 21
y = 3
#### Exercise:
1. What is the solution to the system of equations?
2x – 4y = 2
x + 4y = 3
2. What is the solution to the system of equations?
2x + 3y = 1
–2x + 2y = –6
3. What is the solution to the system of equations?
x + 2y = 4
2x – 5y = –1
4. What is the solution to the system of equations?
2x + y = 2
x – 2y = –5
5. A florist is making regular bouquets and mini bouquets. The florist has 85 roses and 163 peonies to use in the bouquets. How many of each type of bouquets can the florist make?
6. What is the solution to the system of equations? Explain your choice of the solution method.
6x + 12y = –6
3x – 2y = –27
7. What is the solution to the system of equations? Explain your choice of the solution method.
3x – 2y = 38
x = 6 – y
8. The sum of 5 times the width of a rectangle and twice its length is 26 units. The difference of 15 times the width and three times the length is 6 units. Write and solve a system of equations to find the length and width of the rectangle.
9. Ella is a landscape photographer. One weekend at her gallery she sells a total of 52 prints for a total of \$2,975. How many of each size print did Ella sell?
10. Is the given pair of systems of equations equivalent? Explain.
3x – 9y = 5
6x + 2y = 18
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]<|endoftext|>
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