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# How to solve sine
In this blog post, we will take a look at How to solve sine. Our website can solve math word problems.
## How can we solve sine
One of the most important skills that students need to learn is How to solve sine. In other words, all you need to do is find the number that when raised to a certain power equals the number under the radical. Let's say we want to solve for the cube root of 64. We would need to find a number that when multiplied by itself three times equals 64. That number is 4, because 4 x 4 x 4 = 64. So the cube root of 64 is 4. In general, solving radicals is a matter of finding numbers that when multiplied by themselves a certain number of times (the index) equals the number under the radical sign. With a little practice, you'll be able to solve radicals in your sleep!
There are a variety of online math graph calculators available, with different features and capabilities. However, all online math graph calculators have one thing in common: they allow users to perform calculations and visualize results using an online interface. This can be extremely helpful for students who are struggling to understand complex mathematics concepts. In addition, online math graph calculators can be used by educators to create custom teaching materials. As more and more people embrace digital learning, online math graph calculators are likely to become an essential tool for mathematics education.
The distance formula is generally represented as follows: d=√((x_2-x_1)^2+(y_2-y_1)^2) In this equation, d represents the distance between the points, x_1 and x_2 are the x-coordinates of the points, and y_1 and y_2 are the y-coordinates of the points. This equation can be used to solve for the distance between any two points in two dimensions. To solve for the distance between two points in three dimensions, a similar equation can be used with an additional term for the z-coordinate: d=√((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2) This equation can be used to solve for the distance between any two points in three dimensions.
The app, called Mathway, allows users to enter a problem and then see step-by-step instructions for solving it. In addition, the app includes a wide range of features that make it easy to use, including a built-in calculator and a library of solved problems. As a result, Mathway is an essential tool for any student who wants to improve their math skills.
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Calculator that solves any problem One step inequalities solver Quick math Solving 3x3 matrix Free math help online chat<|endoftext|>
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The crystal optics is concerned with the interaction of electromagnetic radiation, typically in the visible wavelength range, with crystalline or otherwise anisotropic solids, but also generalized with optically active liquids. It is a branch of optics, solid state physics and mineralogy.
The optical characteristics of crystals, which are responsible, inter alia, for reflection, refraction and absorption of the light are determined by its regular internal structure. Unlike the optically isotropic crystals glasses can be found in generally the phenomenon of the anisotropy: Important properties such as refractive index are dependent on the direction of light propagation in the crystal and its polarization.
More specifically, this applies to all the crystals, which do not have the cubic crystal system. To illustrate, carries a three dimensional diagram for any wave normal in the crystal direction of the light the value of the refractive indices in the two vibration directions perpendicular to this direction. This always results in an ellipsoid with three unequal generally mutually perpendicular principal axes, which are also called index ellipsoid, ellipsoid or indicatrix Fletcher.
- The crystal is cubic, the ellipsoid is reduced to the special case of a ball, since all three main axes have the same length. The propagation of light is isotropic in this case.
- In the case of the hexagonal, trigonal and tetragonal crystal system, only two of the principal axes are the same length, then one speaks of optically uniaxial or uniaxial crystals. Mentioned in the description, the axis is perpendicular to the two equally long major axes. When light is incident parallel to the axis of birefringence does not take place.
- Three different lengths of the principal axes can be found for the orthorhombic, monoclinic and triclinic crystal system, the crystal is now called optically biaxial or biaxially. These two axes do not coincide with the principal axes of the ellipsoid, rather they are clearly defined by the fact that they are perpendicular to the only two circles, which can be generated by intersection of a plane through the center of the ellipsoid with the indicatrix (all other sections can be derived ellipses and not circles ). The radius of this circle corresponds to the middle in length to the three main axes.
An important consequence of anisotropy of crystals is the birefringence, i.e. the splitting of incident light onto the crystal in an ordinary ray and an extraordinary ray having different polarization.
The optical activity of crystals can be attributed to their anisotropy: The plane of polarization of linearly polarized light is rotated by a proportional to the distance traveled in the crystal angle. It differs depending on whether the plane is rotated clockwise or counterclockwise if you look closely to the direction of light propagation, the right - and left-handed crystals, which are also referred to as visual modifications. Examples Links quartz and quartz rights are mentioned.
A third specific appropriate crystals to optical phenomenon is the so -called pleochroism. This means that light is absorbed to different extents, depending on the propagation and polarization direction. Additionally because the absorption depends on the wavelength, the Pleochronismus shows a direction-dependent color change of the irradiated light that is detectable in extreme cases, even with the naked eye.
The optical properties of the crystal can be influenced by external electric and magnetic fields, but also due to mechanical stress, in the former case one speaks of the electro-optic effect, in the second case of the magneto-optical effect. Conversely, they can be used for diagnosis of these external influences.
The basis of the mathematical formalism is the fact that the electric field strength and the electric displacement field are not the same direction. Thus, the dielectric function that combines the two formula sizes, not to be construed as a scalar, it must be treated as a second order tensor. The relationship between, and writes now:
Wherein the dielectric constant of vacuum is.
As an electromagnetic wave in an anisotropic medium spreads, can be calculated by solving the wave equation for anisotropic body:
Here represents a unit vector pointing in the propagation direction of the shaft, N is the refractive index.
Algebraically, the wave equation is a system of three coupled equations from which can be derived two refractive indices for the two different polarization directions. However, the equation system is not clearly shown in general in relation to the direction of polarization. Therefore, a method is used to reduce the three equations to two. First, we construct a system of three pairwise orthogonal vectors. Two of them are the direction of propagation and dislocation density, and the third is the magnetic field strength. Since no longer needs to stand as in isotropic solids at 90- degree angle, the wave equation is not suitable to determine the polarization character of the waves.
Is now utilized that is perpendicular to the propagation direction. It is
Said to inverse tensor. By selection of a new coordinate system with the coordinates A, B, C, which is selected such that the c- direction is parallel to, the system of equations can be reduced from three to two equations:
By solving this system of equations yields the two refractive indices and the polarization character for any direction.<|endoftext|>
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©G Dear 2010 – Not to be sold/Free to use
Presentation on theme: "©G Dear 2010 – Not to be sold/Free to use"— Presentation transcript:
©G Dear 2010 – Not to be sold/Free to use
Mathematic Extension 1 (Preliminary) Trigonometry Sum & Differences of angles Stage 6 - Year 11 Press Ctrl-A ©G Dear 2010 – Not to be sold/Free to use
Difference of Angles 1 -1 -1 A (cos y, sin y) (cos x, sin x) B 1 x-y x
Differences of Angles x - y We find the sum or difference of angles
to make the hard questions easier. 1 -1 A (cos y, sin y) (cos x, sin x) B 1 x-y x 1 y -1 End of Slide
cos(x – y) = cos x cos y + sin x sin y
Differences of Angles cos(x – y) = cos x cos y + sin x sin y Proof part 1 - distance formula d2 = (x2–x1)2 + (y2–y1)2 AB2 = (cos x – cos y)2 + (sin x – sin y)2 = cos2x – 2 cos x cos y + cos2y + sin2x – 2 sin x sin y + sin2y = (cos2x + sin2x) + (cos2y + sin2y) - 2cos x cos y - 2sin x sin y = 2 - 2(cos x cos y + sin x sin y) 1 Xo-yo End of Slide
cos(x – y) = cos x cos y + sin x sin y
Differences of Angles cos(x – y) = cos x cos y + sin x sin y Proof part 2 – cosine rule a2 = b2 + c2 – 2bc cos A AB2 = – 2(1)(1) cos(x - y) AB2 = 2 – 2 cos(x - y) 2 Xo-yo End of Slide
cos(x – y) = cos x cos y + sin x sin y
Differences of Angles cos(x – y) = cos x cos y + sin x sin y Proof part 3 – from 1 and 2 x x 2 – 2 cos(x-y) = 2 2 - 2(cos x cos y + sin x sin y) 1 x x –2 cos(x-y) = 2 -2(cos x cos y + sin x sin y) 1 cos(x-y) = cos x cos y + sin x sin y xo-yo End of Slide
Sum of Angles 1 -1 A (cos y, sin y) (cos x, sin x) B x+y x y 1 1 -1
Sum & Differences of Angles
xo + yo 1 -1 A (cos y, sin y) (cos x, sin x) B x+y x y 1 1 -1 End of Slide
Sum of Angles cos(x + y) = cos x cos y - sin x sin y Replace y with -y
xo+yo End of Slide
Sum of Angles sin(x + y) = sin x cos y + cos x sin y
Replace x with 90o-x cos(x–y) = cos x cos y + sin x sin y cos((90o-x)– y) = cos(90o-x) cos y + sin(90o-x) sin y cos(90o-(x+y)) = sin x cos y + cos x sin y sin(x+y) = sin x cos y + cos x sin y -1 xo+yo End of Slide
Differences of Angles sin(x - y) = sin x cos y - cos x sin y
Replace y with -y sin (x+y) = sin x cos y + cos x sin y sin (x+(-y)) = sin x cos (-y) + cos x sin (-y) sin (x-y) = sin x cos y + cos x (-sin y) sin(x-y) = sin x cos y - cos x sin y xo-yo End of Slide
Sum of Angles sin(x + y) = sin x cos y + cos x sin y
Replace x with 90o-x cos(x – y) = cos x cos y + sin x sin y cos((90o-x) – y) = cos (90o-x) cos y + sin (90o-x) sin y cos(90o- (x+y)) = sin x cos y + cos x sin y sin(x+y) = sin x cos y + cos x sin y xo+yo End of Slide
Sum of Angles 1 tan x + tan y tan (x+y) = 1 – tan x tan y sin(x+y)
cos(x+y) = sin x cos y + cos x sin y tan(x+y) = cos x cos y - sin x sin y = sin x cos y + cos x sin y cos x cos y = sin x cos y cos x cos y xxxx + cos x sin y cos x cos y xxxx cos x cos y + sin x sin y cos x cos y xxxx xxxx cos x cos y - sin x sin y cos x cos y 1 = tan x + tan y tan(x+y) = 1 - tan x tan y xo+yo End of Slide
Differences of Angles 1 tan x - tan y tan (x-y) = 1 – tan x tan y
sin(x-y) cos(x-y) = sin x cos y - cos x sin y tan(x-y) = cos x cos y + sin x sin y = sin x cos y - cos x sin y cos x cos y = sin x cos y cos x cos y xxxx - cos x sin y cos x cos y xxxx cos x cos y + sin x sin y cos x cos y xxxx xxxx cos x cos y + sin x sin y cos x cos y 1 = tan x - tan y tan(x-y) = 1 + tan x tan y xo-yo End of Slide
Sum & Differences of Angles
cos(x – y) = cos x cos y + sin x sin y cos(x + y) = cos x cos y - sin x sin y sin(x + y) = sin x cos y + cos x sin y sin(x - y) = sin x cos y - cos x sin y tan x + tan y 1 – tan x tan y tan (x+y) = tan x - tan y 1 – tan x tan y tan (x-y) = End of Slide
Similar presentations<|endoftext|>
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# Probability of Independent Events
## Two outcomes both occurring independently.
%
Progress
Practice Probability of Independent Events
Progress
%
Probability of Independent Events
Have you ever wondered if two things can happen at once?
Jana has two decks of cards. Each deck has ten cards in it. There are three face cards in the first deck and four in the second. What are the chances that Jana will draw a face card from both decks?
This is an independent event. In this Concept, you will learn how to figure out probabilities like this one.
### Guidance
Previously we worked on the difference between independent and dependent events. Think about independent events. If one event does not impact the result of a second event, then the two events are independent. For example, there are two different spinners $A$ and $B$ . The result of spinning spinner $A$ does not affect the result of spinning spinner $B$ .
But now we ask a new question. What is the probability of two completely independent events both occurring? For example, what is the probability of spinner $A$ landing on red and spinner $B$ landing on blue?
We could create a tree diagram to show all of the possible options and figure out the probability, but that is very complicated. There is a simpler way.
Notice that this probability equals the product of the two independent probabilities.
$P(\text{red-blue}) & = P(\text{red}) \cdot P(\text{blue})\\& = \frac{1}{4} \cdot \frac{1}{3}\\& = \frac{1}{12}$
Where did these fractions come from?
They came from the probability of the sample space of each spinner. The first spinner has four possible options, so the probability is $\frac{1}{4}$ . The second spinner has three possible options, so the probability is $\frac{1}{3}$ . The Probability Rule takes care of the rest.
In fact, this method works for any independent events as summarized in this rule.
Probability Rule: The probability that two independent events, $A$ and $B$ , will both occur is:
$P(A \ \text{and} \ B) = P (A) \cdot P (B)$
Write the Probability Rule in your notebook.
What is the probability that if you spin the spinner two times, it will land on yellow on the first spin and red on the second spin?
To find the solution, use the rule.
$P(\text{yellow and red}) & = P(\text{yellow}) \cdot P(\text{red})\\P (\text{yellow}) & = \frac{3}{5}\\P (\text{red}) & = \frac{2}{5}$
So:
$P(\text{yellow and red}) & = \frac{3}{5} \cdot \frac{2}{5}\\& = \frac{6}{25}$
The probability of both of these events occurring is $\frac{6}{25}$ .
Now it's time for you to try a few on your own.
#### Example A
What is the probability of spinner A landing on red and spinner B landing on red?
Solution: $\frac{1}{12}$
#### Example B
What is the probability of spinner A landing on blue or yellow and spinner B landing on blue?
Solution: $\frac{1}{6}$
#### Example C
What is the probability of spinner A landing on yellow and spinner B landing on red or green?
Solution: $\frac{1}{6}$
Here is the original problem once again.
Jana has two decks of cards. Each deck has ten cards in it. There are three face cards in the first deck and four in the second. What are the chances that Jana will draw a face card from both decks?
To figure this out, let's write the probability of picking a face card from the first deck.
$\frac{3}{10}$
Now let's write the probability of picking a face card from the second deck.
$\frac{4}{10} = \frac{2}{5}$
Now we can multiply and simplify.
$\frac{3}{10} \times \frac{2}{5}$
$\frac{3}{25}$
### Vocabulary
Independent Events
events where one event does not impact the result of another.
Probability Rule
$P(A) \cdot P(B) = \text{Probability of} \ A \ \text{and} \ B$
### Guided Practice
Here is one for you to try on your own.
The probability of rain tomorrow is 40 percent. The probability that Jeff’s car will break down tomorrow is 3 percent. What is the probability that Jeff’s car will break down in the rain tomorrow?
To find the solution, use the rule.
$P(\text{rain and break}) & = P (\text{rain}) \cdot P (\text{break})\\P (\text{rain}) & = 40\% = \frac{40}{100} = \frac{2}{5}\\P (\text{break}) & = 3\% = \frac{3}{100}$
So:
$P (\text{rain and break}) & = \frac{2}{5} \cdot \frac{3}{100}\\& = \frac{3}{250}$
You can see that the probability is very small.
### Practice
Directions : Solve each problem.
1. Mia spins the spinner two times. What is the probability that the arrow will land on 2 both times?
2. Mia spins the spinner two times. What is the probability that the arrow will land on 2 on the first spin and 3 on the second spin?
3. Mia spins the spinner two times. What is the probability that the arrow will land on an even number on the first spin and and an odd number on the second spin?
4. Mia spins the spinner two times. What is the probability that the arrow will land on an odd number on the first spin and a number less than 4 on the second spin?
5. A laundry bag has 8 black socks and 2 white socks. If you pull out a sock, then put it back in the bag and pull out a second sock, what is the probability that both socks will be black? Write your answer as a decimal.
6. A laundry bag has 8 black socks and 2 white socks. If you pull out a sock, then put it back in the bag and pull out a second sock, what is the probability that both socks will be white? Write your answer as a decimal.
7. A laundry bag has 8 black socks and 2 white socks. If you pull out a sock, then put it back in the bag and pull out a second sock, what is the probability that the first sock will be black and the second sock will be white? Write your answer as a decimal.
8. Dirk has two 52-card decks. Each deck has 4 Aces, 4 Kings, 4 Queens, and so on. What is the probability that Dirk will pick an Ace out of each deck?
9. Dirk has two 52-card decks. Each deck has 4 Aces, 4 Kings, 4 Queens, and so on. What is the probability that Dirk will pick a face card (Jack, Queen, King) out of each deck?
10. Dirk has two 52-card decks. Each deck has 4 Aces, 4 Kings, 4 Queens, and so on. What is the probability that Dirk will pick a card lower than a Jack out of each deck?
11. Karina flips a coin 3 times. What is the probability that she will flip heads 3 times in a row?
12. Karina flips a coin 4 times. What is the probability that she will flip heads 4 times in a row?
13. Karina flips a coin 4 times. What is the probability that she will NOT flip heads 4 times in a row?
14. Karina flips a coin 5 times. What is the probability that she will flip heads twice in a row?
15. Karina flips a coin 5 times. What is the probability that she will flip tails four times in a row?
### Vocabulary Language: English
Independent Events
Independent Events
Two events are independent if the occurrence of one event does not impact the probability of the other event.
Probability Rule
Probability Rule
The probability of two independent events A and B both occurring is P( A and B) = P(A) $\cdot$ P(B).<|endoftext|>
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Classrooms Curricula References Textbooks Site Map
# Definitions
## Exponents
Exponents are just a shorthand for long multiplication problems. Instead of writing out x · x · x · x · x · x we would write x6, i.e. the variable and the number of times we wanted it to be multiplied by itself. The superscript part of the expression is called the exponent; the lower part, the x in our example, is called the base. Of course, we can do this with numbers as well as variables so, for example, 5 · 5 · 5 · 5 would be 54.
Our definnition is based on multiplication so all of our exponents intially are natural numbers, 1, 2, 3, etc. The procedure has natural extensions to negative values and even fractions.
## Logarithms
Logarithms are a little trickier to define using English. The logarithm of a number is the number to which you'd have to raise ten to get the original number. The multiple uses of "a number" make the definition a little hard to follow. It's usually a little easier to look at some examples.
log(1000) = log(103) = 3
log(100) = log(102) = 2
log(10) = log(101) = 1
log(1) = log(100) = 0
log(.1) = log(10-1) = -1
So how about numbers that aren't a power of 10? For those you usually need to get out your calculator. The thinking goes like this:
log(12) = 1.079 because 101.079 = 12
log(.12) = -0.9208 because 10-.09208 = .12
Why does the logarithm always represent a power of 10? It doesn't, necessarily. If no other base is specified then the assumption is that it's 10 but you can specify other bases by using a subscript. For example log2x is "the number to which you have to raise 2 to get x". Revisiting our previous examples
log2(8) = log2(23) = 3
log2(4) = log2(22) = 2
log2(2) = log2(21) = 1
log2(1) = log2(20) = 0
log2(.5) = log2(2-1) = -1
So what about log25.4? Or some other number that isn't a power of 2? Here you can run into problems since calculators don't have keys for every possible base. To calculate these values you can use a simple formula:
lognx = log(x) log(n)
Applying this to my question at the beginning of the first paragraph, and using a calculator to get the base 10 logarithm values, gives us
log25.4 = log(5.4) = 0.7323 = 2.432 log(2) 0.3010
You can confirm with a scientific calculator that 22.423 = 5.4.<|endoftext|>
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```Chapter 2
Descriptive Statistics
§ 2.4
Measures of
Variation
Range
The range of a data set is the difference between the maximum and
minimum date entries in the set.
Range = (Maximum data entry) – (Minimum data entry)
Example:
The following data are the closing prices for a certain stock
on ten successive Fridays. Find the range.
Stock
56 56 57 58 61 63
63 67 67 67
The range is 67 – 56 = 11.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
3
Deviation
The deviation of an entry x in a population data set is the difference
between the entry and the mean μ of the data set.
Deviation of x = x – μ
Example:
The following data are the closing
prices for a certain stock on five
successive Fridays. Find the
deviation of each price.
The mean stock price is
μ = 305/5 = 61.
Stock
x
56
58
61
63
67
Σx = 305
Deviation
x–μ
56 – 61 = – 5
58 – 61 = – 3
61 – 61 = 0
63 – 61 = 2
67 – 61 = 6
Σ(x – μ) = 0
Larson & Farber, Elementary Statistics: Picturing the World, 3e
4
Variance and Standard Deviation
The population variance of a population data set of N entries is
(x μ )2
2
Population variance =
.
N
“sigma
squared”
The population standard deviation of a population data set of N
entries is the square root of the population variance.
2
Population standard deviation =
(x μ )2
“sigma”
Larson & Farber, Elementary Statistics: Picturing the World, 3e
N
.
5
Finding the Population Standard Deviation
Guidelines
In Words
In Symbols
1. Find the mean of the population
data set.
μ x
N
2. Find the deviation of each entry.
x μ
3. Square each deviation.
x μ2
4. Add to get the sum of squares.
SS x x μ
5. Divide by N to get the population
variance.
6. Find the square root of the
variance to get the population
standard deviation.
2
Larson & Farber, Elementary Statistics: Picturing the World, 3e
x μ
2
2
N
x μ
2
N
6
Finding the Sample Standard Deviation
Guidelines
In Words
In Symbols
1. Find the mean of the sample data
set.
x x
n
2. Find the deviation of each entry.
x x
3. Square each deviation.
x x 2
4. Add to get the sum of squares.
SS x x x
5. Divide by n – 1 to get the sample
variance.
6. Find the square root of the
variance to get the sample
standard deviation.
x x
s
n 1
2
s
Larson & Farber, Elementary Statistics: Picturing the World, 3e
2
2
x x
n 1
2
7
Finding the Population Standard Deviation
Example:
The following data are the closing prices for a certain stock on five
successive Fridays. The population mean is 61. Find the population
standard deviation.
Always positive!
Stock
x
56
58
61
63
67
Σx = 305
Deviation
x–μ
–5
–3
0
2
6
Σ(x – μ) = 0
Squared
(x – μ)2
25
9
0
4
36
Σ(x – μ)2 = 74
SS2 = Σ(x – μ)2 = 74
2
x μ
2
N
x μ
N
74
14.8
5
2
14.8 3.8
σ \$3.90
Larson & Farber, Elementary Statistics: Picturing the World, 3e
8
Interpreting Standard Deviation
When interpreting standard deviation, remember that is a measure
of the typical amount an entry deviates from the mean. The more
the entries are spread out, the greater the standard deviation.
14
=4
s = 1.18
12
10
8
6
4
Frequency
Frequency
14
10
8
6
4
2
2
0
0
2
4
Data value
6
=4
s=0
12
2
4
Data value
Larson & Farber, Elementary Statistics: Picturing the World, 3e
6
9
Empirical Rule (68-95-99.7%)
Empirical Rule
For data with a (symmetric) bell-shaped distribution, the
standard deviation has the following characteristics.
1. About 68% of the data lie within one standard
deviation of the mean.
2. About 95% of the data lie within two standard
deviations of the mean.
3. About 99.7% of the data lie within three standard
deviation of the mean.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
10
Empirical Rule (68-95-99.7%)
99.7% within 3
standard deviations
95% within 2
standard deviations
68% within
1 standard
deviation
34%
34%
2.35%
2.35%
13.5%
–4
–3
–2
–1
13.5%
0
1
2
3
Larson & Farber, Elementary Statistics: Picturing the World, 3e
4
11
Using the Empirical Rule
Example:
The mean value of homes on a street is \$125 thousand with a
standard deviation of \$5 thousand. The data set has a bell
shaped distribution. Estimate the percent of homes between
\$120 and \$130 thousand.
68%
105
110
115
120
125
130
μ–σ
μ
μ+σ
135
140
145
68% of the houses have a value between \$120 and \$130 thousand.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
12
Chebychev’s Theorem
The Empirical Rule is only used for symmetric
distributions.
Chebychev’s Theorem can be used for any distribution,
regardless of the shape.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
13
Chebychev’s Theorem
The portion of any data set lying within k standard
deviations (k > 1) of the mean is at least
1 12 .
k
For k = 2: In any data set, at least 1 12 1 1 3 , or 75%, of the
2
4
4
data lie within 2 standard deviations of the mean.
For k = 3: In any data set, at least 1 12 1 1 8 , or 88.9%, of the
3
9
9
data lie within 3 standard deviations of the mean.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
14
Using Chebychev’s Theorem
Example:
The mean time in a women’s 400-meter dash is 52.4
seconds with a standard deviation of 2.2 sec. At least 75%
of the women’s times will fall between what two values?
2 standard deviations
45.8
48
50.2
52.4
54.6
56.8
59
At least 75% of the women’s 400-meter dash times will fall
between 48 and 56.8 seconds.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
15
Standard Deviation for Grouped Data
(x x )2f
Sample standard deviation = s
n 1
where n = Σf is the number of entries in the data set, and x is the
data value or the midpoint of an interval.
Example:
The following frequency distribution represents the ages
of 30 students in a statistics class. The mean age of the
students is 30.3 years. Find the standard deviation of the
frequency distribution.
Continued.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
16
Standard Deviation for Grouped Data
The mean age of the students is 30.3 years.
Class
x
f
x–
(x –
)2
(x –
)2f
18 – 25 21.5
13
– 8.8
77.44
1006.72
26 – 33 29.5
8
– 0.8
0.64
5.12
34 – 41 37.5
4
7.2
51.84
207.36
42 – 49 45.5
3
15.2
231.04
693.12
50 – 57 53.5
2
23.2
538.24 1076.48
2988.80
n = 30
(x x )2f
2988.8
s
103.06 10.2
n 1
29
The standard deviation of the ages is 10.2 years.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
17
```
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We found 315 reviewed resources for conclusion
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How is the word assessment perceived in the eyes of students? Is it viewed as a tool such as a test, paper, or project, or a process to gather information? Furthermore, how is it perceived in the eyes of teachers? It is critical to get everyone on the same page with regard to the perception and purpose of assessment.
In some work this summer with a great group of educators in New Mexico, I had a participant come up to me with a look on her face that said nothing but excitement and enthusiasm. At that point, everything we had been working on had come together for her. She said, “What it really comes down to is that assessment is learning.” What a profound statement. I paused to reflect for a moment and my excitement grew just as hers had a few minutes ago. This simple statement encompasses so much in three short words. Assessment is learning.
When assessment is learning, it is an active process. It moves beyond the tools used to gather information about student learning to a place where students and teachers are able to take information and feedback to grow. Assessment becomes a conversation between students and teachers, as well as among classmates. It reveals where students are in order to progress with their learning. As a student myself, the word assessment was associated with the type of activity we were going to engage in instead of being a spark to light the fire of learning.
What can be done to realize the idea of assessment as learning in our classrooms? Here are five suggestions to get started:
- Involve students in the assessment process.
Kids need to feel part of the assessment process. Assessment becomes a much less fearful process when it is done with students, not to them. Nicole Dimich Vagle, in her book Design in 5 (2015), talks about two features to foster student investment in the assessment process. The first feature is that students “…clearly see and understand the connections among learning, homework, tests, instruction, grades, and improvement.” (p.11) Students may not make these important connections on their own; they can be made plain through classroom dialogue and discovery. Students engage in assessment and learning when relevant associations are made among all happenings in the classroom. They should easily be able to answer the question “Why am I doing this?” The second feature is this (Dimich Vagle, 2015): “Student investment is also built through seeking feedback from students.” (p.11) Feedback is not a one way street. When students are part of the feedback loop, the teacher and the student share a more robust picture of achievement and a more precise path forward.
- Infuse assessment into daily classroom happenings.
The learning process feels very disjointed when everything stops for assessment. Assessment practices should be infused in the process of teaching and learning so neither the student nor the teacher stops to give pause when it is happening. The pause comes when it is time to make decisions about next steps. Note that this is a pause, not a stop. It is a quick moment to make an informed decision and then move on. Assessment should be a familiar part of what happens when we learn. The less students feel like it is ‘time to be assessed’, the less high-stakes assessment becomes. The practice of finding out where students are with their learning, knowing where they need to go, and making choices based on those two pieces of information is a natural routine in the classroom, but the impact on learning is monumental.
- Show students how to interpret feedback and assessment results.
Students will not automatically know what to do with the feedback provided to them. For many, feedback (especially in written form) has been ignored over time because of a focus on grades. Moreover, the focus on grades can harm further learning. Providing specific, timely, and actionable feedback to students paired with time to act on the feedback shifts the focus from where the student was with their learning, to where they are going. Talking with students about what the feedback means and how to go about moving forward enhances the experience. According to Mark Barnes (2015), “The primary motivation behind discussions about feedback should be to clarify any misunderstanding of it and to help students make the feedback actionable” (p.70).
- Continuously learn more about assessment practices to support student learning.
Assessment practices are constantly evolving, and different ideas for classroom application abound. There are times when I feel in a rut with my assessment practices. The same tools keep coming up in my plans and it is time to shake things up a bit. Fortunate for all of us in education is the variety of places to find new ideas and strategies. Not only can you find ideas in books and blog posts, social media, and scholarly articles, but you can also get ideas for implementation and reviews from other educators. Now the challenge is not about finding the new ideas, but rather, determining the method that is the best fit.
- Work together with students to find success.
Learning is a collective and collaborative process. I cannot describe how much I learn from students each and every year I work with them. The focus of this collaborative process is success. This may seem like an easy correlation, but many times students don’t see things so clearly. It is our job as educators to show students that success is the goal – nothing more, nothing less. In his book Revolutionize Assessment, Rick Stiggins (2014) talks about the importance of getting students on winning streaks; the idea that success builds success. Stiggins states, “What, then, is the coach, teacher, or mentor to do to encourage success? Their job is to do whatever they can to convince the performer that success is within reach if he or she keeps trying” (p.44).
Involving students and teachers in an active assessment process where they can take information, interpret it, and move forward is the key to finding success in the classroom. When we think of assessment as the learning instead of as an endpoint, it creates a continuum rather than a sequence of starts and stops. Assessment is learning.
Barnes, M. (2015). Assessment 3.0: Throw Out Your Grade Book and Inspire Learning. United States of America: Corwin. 70.
Stiggins, R. J. (2014). Revolutionize Assessment: Empower Students, Inspire Learning. United States of America: Corwin. 44.
Vagle, N. D. (2015). Design in five: Essential Phases to Create Engaging Assessment Practice. Bloomington, IN: Solution Tree Press. 11.<|endoftext|>
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Скачать презентацию Area Perimeter Surface Area and Volume Review After
c587e3e81b2419dd840292ae3ec4159d.ppt
• Количество слайдов: 19
Area, Perimeter, Surface Area and Volume Review After this review you will be able to : ¡ Define area , perimeter, surface area and volume ¡ Explain how to calculate area, perimeter, surface area and volume ¡ Give real world situations when you would use area, perimeter, surface area and volume
Area is the measurement inside an object. It is the interior space.
How to Calculate the Area of Squares and Rectangles To calculate the area of a square or rectangle all you need to do is multiply the length times the width of the figure. 2 in Example: A=6 x 2=12 in ² 6 in The answer is ALWAYS labeled in units squared (u²). 5 cm Example: A=5 x 5=25 cm ²
Real World Uses of Area You would need to calculate the area for: ¡ Figuring how much carpeting for the floor area of a room ¡ How much wood for flooring in your house ¡ The number of tiles to cover the bathroom floor
Area of a Circle For finding the area for circles, you need to square the radius and multiply the answer times ∏ or 3. 14 *Remember the radius is half of the (diameter) distance across the center of a circle. * Don’t get punked! ____6 cm A= r²x 3. 14 A=6² x 3. 14=113. 04 cm² *Notice the answer is in units squared!
Perimeter The perimeter is the outside measurement of a shape like a square or a rectangle. The outside of a circle is called the CIRCUMFERENCE instead of the perimeter, but it still measures the outside of the shape.
How to Calculate the Perimeter of a Square or a Rectangle To figure out the perimeter of a square or rectangle all you have to do is add up all of the outside measurements. 5 cm Example: P= 5+5+5+5=20 cm or P=5 x 4=20 cm Example: P= 3+3+6+6=18 in or P= 2 x 3+ 2 x 6=18 in 3 in 6 in
Calculating Circumference of Circles ¡ To determine the circumference you multiply the diameter times ∏. ¡ *Don’t get punked!! ¡ Remember the diameter is the distance across the center of the circle. ___12 m___ Example: C=d x 3. 14 ¡ C=12 x 3. 14 C=37. 68 m
Real World Uses for Perimeter Knowing how much fence you need to put around your property ¡ Buying enough wallpaper border to make your bedroom look good ¡ How much fabric to put a border around a quilt you are making ¡ Figuring how much lumber you need to frame windows and doorways ¡
Surface Area Surface area is the total area of the outside faces of three dimensional objects like cubes prisms and cylinders.
Calculating Surface Area ¡ ¡ Since you already know how to determine the area of two-dimensional shapes it will be easy for you to figure out the surface area of 3 -d objects. All you have to do is calculate the area of each of the faces of cubes, prisms, or cylinders and then total up all of your measurements.
Calculating Surface Area ______10 cm So to calculate the surface area of this cylinder you would first have to find out the area of the two circles that make up the top and base. A = r²x ∏ A= 10² x 3. 14 A = 314 cm² * Don’t get punked! There are two circles so you have to double your area answer. A = 314 cm² x 2 = 628 cm² 50 cm
_____10 cm Calculating Surface Area ¡ Now that you have the area of the two circles, you have to figure out the area of: the rectangle that is formed by the height of the cylinder and the circumference of the circle if you were to unwrap the cylinder. _ 50 cm g l u e h e r e
_____10 cm Calculating Surface Area ¡ First find the circumference of the circle: ¡ *Don’t get punked by the radius! C = d x 3. 14 C = 20 x 3. 14 C = 62. 86 cm The circumference is equal to the length of the rectangle part of the cylinder net. _ 50 cm g l u Length e h of the cylinder
Calculating Surface Area The next step in finding the surface area of the cylinder is to calculate the area of the rectangle now that you have figured out the length by using the circumference of the circle. A=lxw A = 62. 8 x 50 A = 3140 cm² The final step is to add the area of the two circles from your first step. ____10 cm SA= 628 + 3140 = 3768 cm² 50 cm
Real World Surface Area Situations ¡ ¡ How much wrapping paper you need to wrap your best friend’s birthday present Figuring out how much paint you need to paint each of the rooms of your house It’s the number of square units needed to cover the outside of a figure How much siding or paint for the exterior of your house
Volume ¡ Volume is the measurement of the interior capacity of a 3 -dimensional figure like a cylinder, cube, or prism.
How to Calculate Volume • When you calculate the volume of a 3 -d object all you have to do is use your prior knowledge of finding area. • Volume is just the area of the base times the height of the shape. ____30 ft____ • The area of the base of this cylinder, as you know, is the area of the circle or ∏ times the radius² so here is the next step: *HINT: Don’t get punked by the diameter!!! A=3. 14 x r² A=3. 14 x 15² A= 706. 5 ft² V=706. 5 x 45 45 ft V=31792. 5 ft³ *NOTICE: VOLUME IS ALWAYS LABELED UNITS CUBED! (u³)
Real World Volume Situations Real world situations for needing volume include: ¡ Calculating the capacity (how much) fuel a tanker truck can carry ¡ How much water could be stored in a water buffalo…the tank, not the animal! ¡ The available space in a silo to store feed corn or grain for the winter ¡ How much water you need to fill your swimming pool for the summer<|endoftext|>
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You are currently browsing the tag archive for the ‘tetrahedral number’ tag.
Tetrahedral Numbers – Stacking Cannonballs
This is one of those deceptively simple topics which actually contains a lot of mathematics – and it involves how spheres can be stacked, and how they can be stacked most efficiently. Starting off with the basics we can explore the sequence:
1, 4, 10, 20, 35, 56….
These are the total number of cannons in a stack as the stack gets higher. From the diagram we can see that this sequence is in fact a sum of the triangular numbers:
S1 = 1
S2 1+3
S3 1+3+6
S4 1+3+6+10
So we can sum the first n triangular numbers to get the general term of the tetrahedral numbers. Now, the general term of the triangular numbers is 0.5n2 + 0.5n therefore we can think of tetrahedral numbers as the summation:
$\bf \sum_{k=1}^{n}0.5k+0.5k^2 = \sum_{k=1}^{n}0.5k+\sum_{k=1}^{n}0.5k^2$
But we have known results for the 2 summations on the right hand side:
$\bf \sum_{k=1}^{n}0.5k =\frac{n(n+1)}{4}$
and
$\bf \huge \sum_{k=1}^{n}0.5k^2 = \frac{n(n+1)(2n+1)}{12}$
and when we add these two together (with a bit of algebraic manipulation!) we get:
$\bf S_n= \frac{n(n+1)(n+2)}{6}$
This is the general formula for the total number of cannonballs in a stack n rows high. We can notice that this is also the same as the binomial coefficient:
$\bf S_n={n+2\choose3}$
Therefore we also can find the tetrahedral numbers in Pascals’ triangle (4th diagonal column above).
The classic maths puzzle (called the cannonball problem), which asks which tetrahedral number is also a square number was proved in 1878. It turns out there are only 3 possible answers. The first square number (1) is also a tetrahedral number, as is the second square number (4), as is the 140th square number (19,600).
We can also look at something called the generating function of the sequence. This is a polynomial whose coefficients give the sequence terms. In this case the generating function is:
$\bf \frac{x}{(x-1)^4} = x + 4x^2 + 10x^3 + 20x^4 ...$
Having looked at some of the basic ideas behind the maths of stacking spheres we can look at a much more complicated mathematical problem. This is called Kepler’s Conjecture – and was posed 400 years ago. Kepler was a 17th century mathematician who in 1611 conjectured that there was no way to pack spheres to make better use of the given space than the stack above. The spheres pictured above fill about 74% of the given space. This was thought to be intuitively true – but unproven. It was chosen by Hilbert in the 18th century as one of his famous 23 unsolved problems. Despite much mathematical efforts it was only finally proved in 1998.
If you like this post you might also like:
The Poincare Conjecture – the search for a solution to one of mathematics greatest problems.
IB Revision
If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:
The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!
The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.
I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.
### Website Stats
• 8,865,560 views
All content on this site has been written by Andrew Chambers (MSc. Mathematics, IB Mathematics Examiner).
### New website for International teachers
I’ve just launched a brand new maths site for international schools – over 2000 pdf pages of resources to support IB teachers. If you are an IB teacher this could save you 200+ hours of preparation time.
Explore here!
### Free HL Paper 3 Questions
P3 investigation questions and fully typed mark scheme. Packs for both Applications students and Analysis students.<|endoftext|>
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# If you flip a coin 9 times, you get a
If you flip a coin 9 times, you get a sequence of heads (H) and tails (T).
(a) How many different sequences of heads and tails are possible?
(c) How many different sequences have at most 2 heads?
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### Solve your problem for the price of one coffee
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Use Product rule: If one event can occur in m ways AND a second event can occur in n ways , then the number of ways that the two events can occur in sequence is then $$\displaystyle{m}\times{n}$$.
Definition of permutation:
$$\displaystyle{P}{\left({n},{r}\right)}={\frac{{{n}!}}{{{\left({n}−{r}\right)}!}}}$$
Definition of combination:
$$\displaystyle{C}{\left({n},{r}\right)}={\left({\frac{{{n}}}{{{r}}}}\right)}={\frac{{{n}!}}{{{r}!{\left({n}−{r}\right)}!}}}$$
With $$\displaystyle{n}\ne{n}\times{\left({n}-{1}\right)}\times\ldots\times{2}\times{1}$$
(a) If the coin is flipped 9 times and each flip has 2 possible outcomes, we can use the product rule:
$$\displaystyle{2}\times{2}\times{2}\times{2}\times{2}\times{2}\times{2}\times{2}\times{2}={29}={512}$$
(b)The order of the heads or tails is not important (we are interested in the number of heads, but not the order of the heads), hence we need to use the definition of combination.
$$\displaystyle{n}={9}$$
$$\displaystyle{r}={5}$$
Evaluate the definition of a combination:
$$\displaystyle{C}{\left({9},{5}\right)}={\frac{{{9}!}}{{{5}!{\left({9}−{5}\right)}!}}}={\frac{{{9}!}}{{{5}!{4}!}}}$$
$$\displaystyle{C}{\left({9},{5}\right)}={126}$$
(c)$$\displaystyle{n}={9}$$
$$\displaystyle{r}\leq{2}$$
Evaluate the definition of a combination:
$$\displaystyle{C}{\left({9},{0}\right)}={\frac{{{9}!}}{{{0}!{\left({9}−{0}\right)}!}}}={1}$$
$$\displaystyle{C}{\left({9},{1}\right)}={\frac{{{9}!}}{{{1}!{\left({9}−{1}\right)}!}}}={9}$$
$$\displaystyle{C}{\left({9},{2}\right)}={\frac{{{9}!}}{{{2}!{\left({9}−{2}\right)}!}}}={36}$$
Add the number of outcomes for each value of r :
$$\displaystyle{1}+{9}+{36}={46}$$<|endoftext|>
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The Pythagorean Theorem
Sure, we could cut triangles in two until the cows come home, but there's an even simpler way to calculate the sides of a right triangle.
Mr. Pythagoras has once again come back to haunt us, doing what he's always done: finding distances. Just picture him, in all his ancient wisdom, enlightening us about the wonders of right triangles and how to find their side lengths.
"I do say, my dear chap. Methinks I've come up with a theorem: if a triangle is a right triangle, the sum of the square of its two legs equals the square of its hypotenuse. Indubitably so."
What Mr. Pythagoras is trying to say is that for any right triangle, a2 + b2 = c2, where a and b are the lengths of the two sides and c is the hypotenuse. (Don't forget, the hypotenuse is always the side opposite the right angle.) This is called the Pythagorean Theorem.
Mr. Pythagoras is probably telling us the truth. After all, it's hard to doubt a gentleman in a top hat. Just to be sure, though, we should double-check his work and prove his theorem using this triangle.
We can make two geometric means if we compare the hypotenuse to the long and the short legs of the big triangle.
1. Comparing the hypotenuse to the short leg gives us:
2. Comparing the hypotenuse to the long leg gives us:
If we cross-multiply each of them, we get a2 = cx and b2 = cy. Adding the two equations gives us the first half of Mr. Pythagoras's claim.
a2+ b2 = cx + cy
Almost done. If we factor the right side, we get:
a2+ b2 = c(x + y)
Looking at our original triangle, x + y = c. Plug in the last piece of the puzzle, and we have our final equation.
a2 + b2 = c2
Way to go. Now we can sleep soundly at night, knowing Mr. Pythagoras wasn't bluffing. Or, instead of sleeping, we can use his theorem to figure out some side lengths.
Sample Problem
What is the length of the hypotenuse of this triangle?
We just proved the Pythagorean Theorem. That's a good place to start.
a2 + b2 = c2
The only values we have are the lengths of the legs. We can substitute those in.
(3)2 + (4)2 = c2
Now we'll simplify as much as we can.
9 + 16 = c2
25 = c2
Taking the square root of both sides gives us the answer. Even though we know that square roots have two answers (a positive and a negative answer), lengths can't be negative. That's why we only care about the positive version.
c = 5
We've found the length of the hypotenuse. It's 5, by the way.
The hypotenuse is only one side, though. The other two sides of a right triangle are the legs. That means that there's a two out of three chance that we'll have to calculate the length of one of the legs and not just the hypotenuse.
Don't worry. It's not that much different.
Sample Problem
What is the length of the missing leg of this triangle?
Starting with the Pythagorean theorem again, we have a2 + b2 = c2. If we substitute in the lengths of the sides we already know, we'll get a2 + (1.5)2 = (3)2. Let's simplify what we can.
a2 + 2.25 = 9
a ≈ 2.6
We've gotten the hang of it by now. All right triangles follow the Pythagorean Theorem. What happens if we don't know whether or not it's a right triangle? If the three sides satisfy the Pythagorean theorem, then the triangle must be a right triangle. That's called the Converse Pythagorean Theorem.
Easy enough.
A square root sign turns most numbers from nice, whole numbers into long decimals. After Mr. Pythagoras came up with his theorem, he noticed how rare it was for a right triangle to have sides that were only whole numbers.
He decided set up a special resort called the Pythagorean Triple Club, complete with squash courts and a hot tub. A triangle that's a Pythagorean triple must have sides lengths of three whole numbers and satisfy his a2 + b2 = c2 equation. That means any right triangle with three whole-numbered sides has access to this Pythagorean Triple Club.<|endoftext|>
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# 8: Solving Linear Equations
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Teetering high above the floor, this amazing mobile remains aloft thanks to its carefully balanced mass. Any shift in either direction could cause the mobile to become lopsided, or even crash downward. In this chapter, we will solve equations by keeping quantities on both sides of an equal sign in perfect balance.
• 8.1: Solve Equations Using the Subtraction and Addition Properties of Equality (Part 1)
The purpose in solving an equation is to find the value or values of the variable that make each side of the equation the same. Any value of the variable that makes the equation true is called a solution to the equation. We can use the Subtraction and Addition Properties of Equality to solve equations by isolating the variable on one side of the equation. Usually, we will need to simplify one or both sides of an equation before using the Subtraction or Addition Properties of Equality.
• 8.2: Solve Equations Using the Subtraction and Addition Properties of Equality (Part 2)
In most of the application problems we solved earlier, we were able to find the quantity we were looking for by simplifying an algebraic expression. Now we will be using equations to solve application problems. We’ll start by restating the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve. When assigning a variable, choose a letter that reminds you of what you are looking for.
• 8.3: Solve Equations Using the Division and Multiplication Properties of Equality
We can also use the Division and Multiplication Properties of Equality to solve equations by isolating the variable on one side of the equation. The goal of using the Division and Multiplication Properties of Equality is to "undo" the operation on the variable. Usually, we will need to simplify one or both sides of an equation before using the Division or Multiplication Properties of Equality.
• 8.4: Solve Equations with Variables and Constants on Both Sides (Part 1)
You may have noticed that in all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. This does not happen all the time—so now we’ll see how to solve equations where the variable terms and/or constant terms are on both sides of the equation.
• 8.5: Solve Equations with Variables and Constants on Both Sides (Part 2)
Each of the first few sections of this chapter has dealt with solving one specific form of a linear equation. It’s time now to lay out an overall strategy that can be used to solve any linear equation. We call this the general strategy. Some equations won’t require all the steps to solve, but many will. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.
• 8.6: Solve Equations with Fraction or Decimal Coefficients
The General Strategy for Solving Linear Equations can be used to solve for equations with fraction or decimal coefficients. Clearing the equation of fractions applies the Multiplication Property of Equality by multiplying both sides of the equation by the LCD of all fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions. When we have an equation with decimals, we can use the same process we used to clear fractions.
• 8.E: Solving Linear Equations (Exercises)
• 8.S: Solving Linear Equations (Summary)
Figure 8.1 - A Calder mobile is balanced and has several elements on each side. (credit: paurian, Flickr)<|endoftext|>
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Man has always looked to the skies for answers ... as if a celestial blueprint that tells the story of humanity's journey in time and space. Creation myths connect gods with the heavens, often the constellations, sometimes with a link to ancient aliens who came to Earth to seed humanity and will one day return. As the millennia progressed, so too did the use of telescopes to find answers to many of these age-old questions. That quest continues today with new and more advanced technologies as we venture into space.
A telescope is an instrument that aids in the observation of remote objects by collecting electromagnetic radiation (such as visible light). The first known practical telescopes were invented in the Netherlands at the beginning of the 17th century, using glass lenses. They found use in terrestrial applications and astronomy.
Within a few decades, the reflecting telescope was invented, which used mirrors. In the 20th century many new types of telescopes were invented, including radio telescopes in the 1930s and infrared telescopes in the 1960s. The word telescope now refers to a wide range of instruments detecting different regions of the electromagnetic spectrum, and in some cases other types of detectors.
The word "telescope" was coined in 1611 by the Greek mathematician Giovanni Demisiani for one of Galileo Galilei's instruments presented at a banquet at the Accademia dei Lincei. In the Starry Messenger Galileo had used the term "perspicillum". Read more
The earliest recorded working telescopes were the refracting telescopes that appeared in the Netherlands in 1608. Their development is credited to three individuals: Hans Lippershey and Zacharias Janssen, who were spectacle makers in Middelburg, and Jacob Metius of Alkmaar. Galileo heard about the Dutch telescope in June 1609, built his own within a month, and greatly improved upon the design in the following year.
Galileo's telescope reaches 400th anniversary - August 25, 2009
The idea that the objective, or light-gathering element, could be a mirror instead of a lens was being investigated soon after the invention of the refracting telescope. The potential advantages of using parabolic mirrors - reduction of spherical aberration and no chromatic aberration - led to many proposed designs and several attempts to build reflecting telescopes. In 1668, Isaac Newton built the first practical reflecting telescope, of a design which now bears his name, the Newtonian reflector.
The invention of the achromatic lens in 1733 partially corrected color aberrations present in the simple lens and enabled the construction of shorter, more functional refracting telescopes. Reflecting telescopes, though not limited by the color problems seen in refractors, were hampered by the use of fast tarnishing speculum metal mirrors employed during the 18th and early 19th century a problem alleviated by the introduction of silver coated glass mirrors in 1857, and aluminized mirrors in 1932. The maximum physical size limit for refracting telescopes is about 1 meter (40 inches), dictating that the vast majority of large optical researching telescopes built since the turn of the 20th century have been reflectors. The largest reflecting telescopes currently have objectives larger than 10 m (33 feet).
The 20th century also saw the development of telescopes that worked in a wide range of wavelengths from radio to gamma-rays. The first purpose built radio telescope went into operation in 1937. Since then, a tremendous variety of complex astronomical instruments have been developed. Read more ...
PHYSICAL SCIENCES INDEX
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# Given the equation x^2 + 6x + 4y - 7 = 0 determine: a) The Vertex V b) If the parabola opens up, down, left, right c) The Focus F
samhouston | Certified Educator
First, solve the equation for y.
Add -4y to both sides.
x^2 + 6x + 4y - 7 + (-4y) = 0 + (-4y)
x^2 + 6x - 7 = -4y
Divide both sides by -4.
-0.25x^2 + -1.5x + 1.75 = y
Second, identify the values of a, b, and c.
y = ax^2 + bx + c
a = -0.25
b = -1.5
c = 1.75
Next, calculate the vertex's x-value using the formula:
x = -b/2a
x = -(-1.5) / 2 * (-0.25)
x = 1.5 / (-0.5)
x = -3
Now substiute -3 in for x and solve for y. This will give you the vertex's y-value.
(-0.25) * -3^2 + (-1.5) * -3 + 1.75 = y
(-0.25) * 9 + 4.5 + 1.75 = y
(-2.25) + 4.5 + 1.75 = y
y = 4
The vertex is (-3, 4).
Since the value of a is negative, the parabola opens down.
The focus of a parabola is a fixed point on the interior of the parabola. Since the parabola opens down, the x-value of the focus is the same as the x-value of the vertex. To find the y-value of the focus, use this formula:
y = c - (b^2 - 1)/4a
y = 1.75 - (-1.5^2 - 1) / 4 * -0.25
y = 1.75 - (2.25 - 1) / 4 * -0.25
y = 1.75 - 1.25 / -1
y = 1.75 + 1.25
y = 3
The focus is (-3, 3).
Here is a graph of the parabola:
Summary:
a. Vertex = (-3, 4)
b. Parabola opens down
c. Focus = (-3, 3)<|endoftext|>
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# Chapter 10: Quadratic Equations and Functions
Difficulty Level: Basic Created by: CK-12
## Introduction
As you saw in Chapter 8, algebraic functions not only produce straight lines but curved ones too. A special type of curved function is called a parabola. Perhaps you have seen the shape of a parabola before:
• The shape of the water from a drinking fountain
• The path a football takes when thrown
• The shape of an exploding firework
• The shape of a satellite dish
• The path a diver takes into the water
• The shape of a mirror in a car’s headlamp
Many real life situations model a quadratic equation. This chapter will explore the graph of a quadratic equation and how to solve such equations using various methods.
Chapter Outline
## Summary
In this chapter, quadratic equations and functions are covered in detail. First, quadratic functions and their graphs are discussed, including vertical shifts of the graphs of quadratic functions and using graphs to solve quadratic equations. Next, other ways of solving quadratic equations are given, such as solving quadratic equations using square roots and completing the square to solve quadratic equations. Finding the vertex of a quadratic function by completing the square is also touched upon. The chapter then moves on to quadratic problems and the quadratic formula, and it talks about the discriminant. Finally, linear, exponential, and quadratic models are discussed, and advice is given on choosing a function model.
### Quadratic Equations and Functions Review
Define each term.
1. Vertex
2. Standard form for a quadratic equation
3. Model
4. Discriminant
Graph each function. List the vertex (round to the nearest tenth, if possible) and the range of the function.
1. \begin{align*}y=x^2-6x+11\end{align*}
2. \begin{align*}y=-4x^2+16x-19\end{align*}
3. \begin{align*}y=-x^2-2x+1\end{align*}
4. \begin{align*}y=\frac{1}{2} x^2+8x+6\end{align*}
5. \begin{align*}y=x^2+4x\end{align*}
6. \begin{align*}y=- \frac{1}{4} x^2+8x-4\end{align*}
7. \begin{align*}y=(x+4)^2+3\end{align*}
8. \begin{align*}y=-(x-3)^2-6\end{align*}
9. \begin{align*}y=(x-2)^2+2\end{align*}
10. \begin{align*}y=-(x+5)^2-1\end{align*}
Rewrite in standard form.
1. \begin{align*}x-24=-5x\end{align*}
2. \begin{align*}5+4a=a^2\end{align*}
3. \begin{align*}-6-18a^2=-528\end{align*}
4. \begin{align*}y=-(x+4)^2+2\end{align*}
Solve each equation by graphing.
1. \begin{align*}x^2-8x+87=9\end{align*}
2. \begin{align*}23x+x^2-104=4\end{align*}
3. \begin{align*}13+26x=-x^2+11x\end{align*}
4. \begin{align*}x^2-9x=119\end{align*}
5. \begin{align*}-32+6x^2-4x=0\end{align*}
Solve each equation by taking its square roots.
1. \begin{align*}x^2=225\end{align*}
2. \begin{align*}x^2-2=79\end{align*}
3. \begin{align*}x^2+100=200\end{align*}
4. \begin{align*}8x^2-2=262\end{align*}
5. \begin{align*}-6-4x^2=-65\end{align*}
6. \begin{align*}703=7x^2+3\end{align*}
7. \begin{align*}10+6x^2=184\end{align*}
8. \begin{align*}2+6x^2=152\end{align*}
Solve each equation by completing the square and then taking its square roots.
1. \begin{align*}n^2-4n-3=9\end{align*}
2. \begin{align*}h^2+10h+1=3\end{align*}
3. \begin{align*}x^2+14x-22=10\end{align*}
4. \begin{align*}t^2-10t=-9\end{align*}
Determine the maximum/minimum point by completing the square.
1. \begin{align*}x^2-20x+28=-8\end{align*}
2. \begin{align*}a^2+2-63=-5\end{align*}
3. \begin{align*}x^2+6x-33=4\end{align*}
Solve each equation by using the Quadratic Formula.
1. \begin{align*}4x^2-3x=45\end{align*}
2. \begin{align*}-5x+11x^2=15\end{align*}
3. \begin{align*}-3r=12r^2-3\end{align*}
4. \begin{align*}2m^2+10m=8\end{align*}
5. \begin{align*}7c^2+14c-28=-7\end{align*}
6. \begin{align*}3w^2-15=-3w\end{align*}
In 45-50, for each quadratic equation, determine:
(a) the discriminant
(b) the number of real solutions
(c) whether the real solutions are rational or irrational
1. \begin{align*}4x^2-4x+1=0\end{align*}
2. \begin{align*}2x^2-x-3=0\end{align*}
3. \begin{align*}-2x^2-x-1=-2\end{align*}
4. \begin{align*}4x^2-8x+4=0\end{align*}
5. \begin{align*}-5x^2+10x-5=0\end{align*}
6. \begin{align*}4x^2+3x+6=0\end{align*}
7. Explain the difference between \begin{align*}y=x^2+ 4\end{align*} and \begin{align*}y=-x^2+4\end{align*}.
8. Jorian wants to enclose his garden with fencing on all four sides. He has 225 feet of fencing. What dimensions would give him the largest area?
9. A ball is dropped off a cliff 70 meters high.
1. Using Newton’s equation, model this situation.
2. What is the leading coefficient? What does this value tell you about the shape of the parabola?
3. What is the maximum height of the ball?
4. Where is the ball after 0.65 seconds?
5. When will the ball reach the ground?
1. The following table shows the number of hours spent per person playing video games for various years in the United States. \begin{align*}& x && 1995 && 1996 && 1997 && 1998 && 1999 && 2000\\ & y && 24 && 25 && 37 && 43 && 61 && 70\end{align*}
1. What seems to be the best function for this data?
2. Find the best fit function.
3. Using your equation, predict the number of hours someone will spend playing video games in 2012.
4. Does this value seem possible? Explain your thoughts.
1. The table shows the amount of money spent (in billions of dollars) in the U.S. on books for various years. \begin{align*}& x && 1990 && 1991 && 1992 && 1993 && 1994 && 1995 && 1996 && 1997 && 1998\\ & y && 16.5 && 16.9 && 17.7 && 18.8 && 20.8 && 23.1 && 24.9 && 26.3 && 28.2\end{align*}
1. Find a linear model for this data. Use it to predict the dollar amount spent in 2008.
2. Find a quadratic model for this data. Use it to predict the dollar amount spent in 2008.
3. Which model seems more accurate? Use the best model to predict the dollar amount spent in 2012.
4. What could happen to change this value?
1. The data below shows the number of U.S. hospitals for various years. \begin{align*}& x && 1960 && 1965 && 1970 && 1980 && 1985 && 1990 && 1995 && 2000\\ & y && 6876 && 7123 && 7123 && 6965 && 6872 && 6649 && 6291 && 5810\end{align*}
1. Find a quadratic regression line to fit this data.
2. Use the model to determine the maximum number of hospitals.
4. In what years were there approximately 7,000 hospitals?
5. What seems to be the trend with this data?
1. A pendulum’s distance is measured and recorded in the following table. \begin{align*}& swing && 1 && 2 && 3 && 4 && 5 && 6\\ & length && 25 && 16.25 && 10.563 && 6.866 && 4.463 && 2.901\end{align*}
1. What seems to be the best model for this data?
2. Find a quadratic regression line to fit this data. Approximate the length of the seventh swing.
3. Find an exponential regression line to fit this data. Approximate the length of the seventh swing.
### Quadratic Equations and Functions Test
1. True or false? The vertex determines the domain of a quadratic function.
2. Suppose the leading coefficient of a quadratic equation is \begin{align*}a=-\frac{1}{3}\end{align*}. What can you conclude about the shape of the parabola?
3. Find the discriminant of the equation and determine the number of real solutions: \begin{align*}0=-2x^2+3x-2\end{align*}.
4. A ball is thrown upward from a height of four feet with an initial velocity of 45 feet/second.
1. Using Newton’s law, write the equation to model this situation.
2. What is the maximum height of the ball?
3. When will the ball reach 10 feet?
4. Will the ball ever reach 36.7 feet?
5. When will the ball hit the ground?
In 5–9, solve the equation using any method.
1. \begin{align*}2x^2=2x+40\end{align*}
2. \begin{align*}11j^2=j+24\end{align*}
3. \begin{align*}g^2=1\end{align*}
4. \begin{align*}11r^2-5=-178\end{align*}
5. \begin{align*}x^2+8x-65=-8\end{align*}
6. What is the vertex of \begin{align*}y=-(x-6)^2+5\end{align*}? Does the parabola open up or down? Is the vertex a maximum or a minimum?
7. Graph \begin{align*}y=(x+2)^2-3\end{align*}.
8. Evaluate the discriminant. How many real solutions does the quadratic equation have? \begin{align*}-5x^2-6x=1\end{align*}
9. Suppose \begin{align*}D=-14\end{align*}. What can you conclude about the solutions to the quadratic equation?
10. Rewrite in standard form: \begin{align*}y-7=-2(x+1)^2\end{align*}.
11. Graph and determine the function's range and vertex: \begin{align*}y=-x^2+2x-2\end{align*}.
12. Graph and determine the function's range and \begin{align*}y-\end{align*}intercept: \begin{align*}y=\frac{1}{2} x^2+4x+5\end{align*}.
13. The following information was taken from USA Today regarding the number of cancer deaths for various years.
14. Year Number of Deaths Per 100,000 men
1980 205.3
1985 212.6
1989 217.6
1993 212.1
1997 201.9
15. Cancer Deaths of Men (
16. Source: USA Today
17. )
1. Find a linear regression line to fit this data. Use it to predict the number of male deaths caused by cancer in 1999.
2. Find a linear regression line to fit this data. Use it to predict the number of male deaths caused by cancer in 1999.
3. Find an exponential regression line to fit this data. Use it to predict the number of male deaths caused by cancer in 1999.
4. Which seems to be the best fit for this data?
#### Texas Instruments Resources
In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9620.
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Language Processing Disorder Support Beyond Tutoring
Students with language processing disorders experience difficulties understanding spoken or written language, reading comprehension, finding the right words to say, or producing written work. This often causes stress and frustration with being unable to express thoughts, needs and opinions.
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Common Issues Associated with Learning Processing Disorders
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Instruction
1
Write the given function F(x), e.g. F(x) = (x3 + 15x +26). If the task explicitly specify the point through which the tangent is held, for example, its coordinate x0 = -2, we can dispense with plotting functions and additional straight lines on the Cartesian system OXY. Find the derivative of the first order from the given function F`(x). In this example, F`(x) = (3x2 + 15). Substitute the given value of the argument x0 in the derivative function and calculate its value: F`(-2) = (3(-2)2 + 15) = 27. So you found tg a = 27.
2
When considering tasks where it is required to determine the tangent of the angle of inclination of the tangent to the graph of the function at the point of intersection of this graph with the abscissa axis, you need first to find the numerical value of the coordinates of the point of intersection of the function with OH. For clarity, it is best to build a graph of a function on the two-dimensional plane OXY.
3
Specify the coordinate range for x, for example, from -5 to 5 with step 1. Substituting in the function values x, calculate the corresponding ordinates y and put on the coordinate plane plotting points (x, y). Connect the dots smooth line. You will see the completed chart, the intersection of function and the x-axis. The ordinate of the function at a given point is equal to zero. Find the numerical value of the corresponding argument. For this specified function, for example F(x) = (4x2 - 16), Paranaita to zero. Solve the resulting equation with one variable and calculate x: 4x2 - 16 = 0, x2 = 4, x = 2. Thus, according to the condition of the problem, the tangent of the angle of inclination of the tangent to the graph of the function is to find the point with coordinates x0 = 2.
4
Similarly to the previously described method determine the derivative function: F`(x) = 8*x. Then calculate its value at the point with x0 = 2, which corresponds to the intersection point of the original function with OH. Substitute the obtained value into the derivative function and calculate the tangent of an angle tangent: tg a = F`(2) = 16.
5
When finding the slope at the point of intersection of the function with the ordinate axis (Oh) do the same. Only the coordinate of the initial point x0 should be set to zero.<|endoftext|>
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American History Part 2 will help students understand and interpret the history of the United States. The course will begin with the Holocaust and World War II and move through time to the Cold War, the Civil Rights Movement, the Vietnam War, the Post-Vietnam Era, ending with the present. This course will deal with a vast scope of complex issues throughout American history.
It is recommended that students successfully complete American History Part 1 or equivalent course work before enrolling in this course.
Unit 1: Essential Content and Skills
- Analyze the events surrounding World War II, including the Holocaust and America’s entry into the war.
- Examine America’s role in World War II by summarizing the fighting in the European and Pacific theaters.
- Evaluate how America became a major force in world affairs by examining the outcome of World War II.
- Analyze the early stages of the Cold War by describing its origins and development.
- Examine the effects of the Cold War on the American population.
Unit 2: Essential Content and Skills
- Examine postwar America by analyzing the culture of the 1950s.
- Examine the problems postwar Americans faced.
- Analyze the New Frontier and the Great Society by examining the impact of their programs.
- Examine how African Americans fought discrimination during the Civil Rights era.
- Analyze the outcomes of the Civil Rights Movement by identifying the civil rights legislation passed in the 1960s.
Unit 3: Essential Content and Skills
- Examine the causes of the Vietnam War by explaining America’s desire to stop the spread of communism.
- Analyze the military and political events surrounding the Vietnam War.
- Evaluate the effectiveness and outcomes of the fight for civil rights equality by Latinos, Native Americans, and women.
- Examine the social protests of the 1960s.
- Examine the political, social, and economic events of the early 1970s by analyzing the Nixon administration.
Unit 4: Essential Content and Skills
- Examine the political, social, and economic events of the early 1970s by analyzing the Ford and Carter presidencies.
- Analyze the impact of environmentalism.
- Examine the key political and social events of the 1980s by analyzing American life during the Reagan and Bush presidencies.
- Evaluate the impact of the global economy on the American workforce.
- Analyze the important issues affecting the United States today by examining the changes that are occurring in the world.
- The Americans: Reconstruction to the 21st Century (textbook)
*Available as an iText course<|endoftext|>
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While solar radiation enables and sustains life on Earth, it also produces “space weather” that can profoundly impact different technologies, including telecommunications, satellite navigation, and the electric power grid. Solar flares can produce x-rays resulting in radio blackouts that block high-frequency radio waves. Solar Energetic Particles can penetrate satellite electronics and cause electrical failure. Coronal mass ejections (CMEs) can cause geomagnetic storms that induce ground currents and degrade power grid operations, sometimes catastrophically. The Sun, The Earth, and Near-Earth Space assembles concise explanations and descriptions—easily read and readily understood—of what we now know of the chain of events and processes that connect the Sun to the Earth, with especial emphasis on space weather and Sun-Climate. This 301-page text resource is made available courtesy of the National Aeronautics and Space Administration
and is not produced, owned or hosted by UCAR/COMET.<|endoftext|>
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# Condition of Constancy of the Function
Fact. Suppose function ${y}={f{{\left({x}\right)}}}$ is defined and continuous on interval ${X}$ and has finite derivative ${f{'}}{\left({x}\right)}$. Function ${y}={f{{\left({x}\right)}}}$ is constant if and only if ${f{'}}{\left({x}\right)}$ for all ${x}$ in ${X}$.
This fact means that if on some interval derivative of function equals 0 then function is constant their, its graph is just horizontal line.
Corollary. Suppose two function ${y}={f{{\left({x}\right)}}}$ and ${y}={g{{\left({x}\right)}}}$ are defined and continuous on interval ${X}$, and have finite derivatives ${f{'}}{\left({x}\right)}$ and ${g{{\left({x}\right)}}}$. Function ${y}={f{{\left({x}\right)}}}$ is constant if and only if ${f{'}}{\left({x}\right)}$ for all ${x}$ in ${X}$. If ${f{'}}{\left({x}\right)}={g{'}}{\left({x}\right)}$ on interval ${X}$ then ${f{{\left({x}\right)}}}={g{{\left({x}\right)}}}+{C}$, where ${C}$ is a constant for all ${x}$ in ${X}$.
This corollary means that if functions have same derivatives on interval ${X}$ then their difference is constant.
Example 1. Consider functions ${f{{\left({x}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}$ and ${g{{\left({x}\right)}}}={\operatorname{arcsin}{{\left(\frac{{x}}{{\sqrt{{{1}+{{x}}^{{2}}}}}}\right)}}}$.
They are defined and continuous on interval ${\left(-\infty,\infty\right)}$.
Since ${f{'}}{\left({x}\right)}=\frac{{1}}{{{1}+{{x}}^{{2}}}}$ and ${g{'}}{\left({x}\right)}=\frac{{1}}{{{1}-{{\left(\frac{{x}}{\sqrt{{{1}+{{x}}^{{2}}}}}\right)}}^{{2}}}}\cdot\frac{{\sqrt{{{1}+{{x}}^{{2}}}}-\frac{{{{x}}^{{2}}}}{{\sqrt{{{1}+{{x}}^{{2}}}}}}}}{{{1}+{{x}}^{{2}}}}=\frac{{1}}{{{1}+{{x}}^{{2}}}}$ then according to corollary ${\operatorname{arctan}{{\left({x}\right)}}}={\operatorname{arcsin}{{\left(\frac{{x}}{{\sqrt{{{1}+{{x}}^{{2}}}}}}\right)}}}+{C}$ on ${\left(-\infty,\infty\right)}$.
To find constant plug any value of ${x}$, for example, ${x}={0}$: ${\operatorname{arctan}{{\left({0}\right)}}}={\operatorname{arcsin}{{\left(\frac{{0}}{\sqrt{{{1}+{{0}}^{{2}}}}}\right)}}}+{C}$ or ${C}={0}$.
So, we have the following fact: ${\operatorname{arctan}{{\left({x}\right)}}}={\operatorname{arcsin}{{\left(\frac{{x}}{\sqrt{{{1}+{{x}}^{{2}}}}}\right)}}}$ for all ${x}$.
Example 2. Consider functions ${f{{\left({x}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}$ and ${g{{\left({x}\right)}}}=\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}$.
It can be easily proven that ${f{'}}{\left({x}\right)}={g{'}}{\left({x}\right)}$. However, function ${g{{\left({x}\right)}}}$ is not defined when ${x}=\pm{1}.$ So, $\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}+{C}$ on ${\left(-\infty,-{1}\right)},{\left(-{1},{1}\right)},{\left({1},\infty\right)}$.
It is interesting that constant will be different for different intervals.
For interval ${\left(-{1},{1}\right)}$ we plug ${x}={0}$: $\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}\cdot{0}}}{{{1}-{{0}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({0}\right)}}}+{C}$ or ${C}={0}$.
For interval ${\left(-\infty,-{1}\right)}$ we let ${x}\to-\infty$: ${0}=-\frac{\pi}{{2}}+{C}$ or ${C}=\frac{\pi}{{2}}$.
For interval ${\left({1},\infty\right)}$ we let ${x}\to\infty$: ${0}=\frac{\pi}{{2}}+{C}$ or ${C}=-\frac{\pi}{{2}}$.
So,
$\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}+\frac{\pi}{{2}}$ on ${\left(-\infty,-{1}\right)}$,
$\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}$ on ${\left(-{1},{1}\right)}$,
$\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}-\frac{\pi}{{2}}$ on ${\left({1},\infty\right)}$.
Graph confirms these facts.<|endoftext|>
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# How do you find three consecutive odd positive integers such that 3 times the sum of all three is 24 more than the product of the first and second integers?
Mar 16, 2016
$\left\{1 , 3 , 5\right\}$
#### Explanation:
Any odd number may be represented as $2 k + 1$ for some integer $k$. Then, let's represent our odd integers as $2 k + 1 , 2 k + 3 , 2 k + 5$ where $k$ is a non-negative integer. From our initial condition, we have that
$\left(2 k + 1\right) \left(2 k + 3\right) + 24 = 3 \left(\left(2 k + 1\right) + \left(2 k + 3\right) + \left(2 k + 5\right)\right)$
Simplifying both sides, we get
$4 {k}^{2} + 8 k + 27 = 18 k + 27$
$\implies 4 {k}^{2} - 10 k = 0$
$\implies 4 k \left(k - \frac{5}{2}\right) = 0$
$\implies k = 0$ or $k = \frac{5}{2}$
But, as we specified that $k$ must be an integer, we know that $k \ne \frac{5}{2}$. Thus we have $k = 0$ and so our odd integers are $\left\{1 , 3 , 5\right\}$
Checking this, we find that $1 \cdot 3 + 24 = 27 = 3 \left(1 + 3 + 5\right)$<|endoftext|>
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0
# What are the factor pairs for the number 10?
Updated: 9/26/2023
Wiki User
6y ago
1 and 10 are a factor pair for 10.
Wiki User
6y ago
Wiki User
6y ago
(1,10)(2,5)
Earn +20 pts
Q: What are the factor pairs for the number 10?
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Related questions
560 is one such.
### What are the factor pairs of 10?
The factor pairs of 10 are 1*10, 2*5
### How do you find factor pairs of a number?
Factor pairs are just the factors of a number listed in a different way. The factors of 20 are 1, 2, 4, 5, 10, 20 The factor pairs of 20 are (20,1)(10,2)(5,4)
### How does listing the factor pairs help?
It works by multiplying and diving for example: a factor of 10 would be 1,2,5 and the number itself 10
### What are factor pairs of 5000?
There are 10 factor pairs for 5,000:5000,12500,21250,41000,5625,8500,10250,20200,25125,40100,50
### What is the definition to factor pairs?
its a factor with two pairs of the same number
### What is the difference between factor pairs and distinct factor pairs?
The difference is between factor pairs and distinct factors. With square numbers, one of the factor pairs will be the same number twice. When listing the distinct factors, that number is only listed once.
### What are the factor pairs of 1080?
These are the 16 factor pairs for the number 1,080:1080,1540,2360,3270,4216,5180,6135,8120,9108,1090,1272,1560,1854,2045,2440,2736,30
### What is a number called that has an odd number of factor pairs?
A square number has an odd number of factors, but a number with an odd number of factor pairs is nothing special.
### What are the factor pairs of the number 100?
(1, 100) (2, 50) (4, 25) (5, 20) (10, 10)
### How is the list of factor pairs related to the rectangles that could be made to show 588?
Number of factor pairs = number of rectangles
### How do you find factor pairs?
You find a factor pair take the number that you want to find the factor pair of and divide it by a number. If the answer come out evenly then that's your factor pair EX. Factor pairs of 150 1 and 150 2 and 75 3 and 50 5 and 30 6 and 25 10 and 15<|endoftext|>
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What a p-Value Tells You about Statistical Data
When you perform a hypothesis test in statistics, a p-value helps you determine the significance of your results. Hypothesis tests are used to test the validity of a claim that is made about a population. This claim that’s on trial, in essence, is called the null hypothesis.
The alternative hypothesis is the one you would believe if the null hypothesis is concluded to be untrue. The evidence in the trial is your data and the statistics that go along with it. All hypothesis tests ultimately use a p-value to weigh the strength of the evidence (what the data are telling you about the population). The p-value is a number between 0 and 1 and interpreted in the following way:
A small p-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, so you reject the null hypothesis.
A large p-value (> 0.05) indicates weak evidence against the null hypothesis, so you fail to reject the null hypothesis.
p-values very close to the cutoff (0.05) are considered to be marginal (could go either way). Always report the p-value so your readers can draw their own conclusions.
For example, suppose a pizza place claims their delivery times are 30 minutes or less on average but you think it’s more than that. You conduct a hypothesis test because you believe the null hypothesis, Ho, that the mean delivery time is 30 minutes max, is incorrect. Your alternative hypothesis (Ha) is that the mean time is greater than 30 minutes. You randomly sample some delivery times and run the data through the hypothesis test, and your p-value turns out to be 0.001, which is much less than 0.05. In real terms, there is a probability of 0.05 that you will mistakenly reject the pizza place’s claim that their delivery time is less than or equal to 30 minutes. Since typically we are willing to reject the null hypothesis when this probability is less than 0.05, you conclude that the pizza place is wrong; their delivery times are in fact more than 30 minutes on average, and you want to know what they’re gonna do about it! (Of course, you could be wrong by having sampled an unusually high number of late pizza deliveries just by chance.)<|endoftext|>
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On Sunday, Death Valley temperatures reached 129 degrees F, a new June
The United States high temperature record was set in 1913, measured in Death Valley on July 13. Twenty-three of the 50 US state high temperature
The alarm about climate change is all about one degree. According to the National Oceanic and Atmospheric Administration (NOAA), global surface temperatures have increased about 1.3 degrees F (0.7oC) since 1880. Proponents of the theory of man-made warming claim that this is evidence that man-made greenhouse gases are raising global temperatures.
One degree over more than 130 years isn’t very much. In contrast, Chicago temperatures vary from about -5 degrees F to 95 degrees F, about 100 degrees, each year. When compared to this 100-degree annual swing, the rise in global temperatures since the 1800s is trivial.
Nevertheless, NOAA repeatedly raises concern about global temperatures. The NOAA website proclaims that “May 2013 global temperatures were the third highest on record.” This sounds alarming unless one understands that “on record” refers to the thermometer record, which only dates back to about 1880.
Climate changes over hundreds and thousands of years. Data from ice cores show several periods during the last 10,000 years that were warmer than today, including the Roman Climate Optimum at the height of the Roman Empire and the Medieval Warm Period, when the Vikings settled southwest Greenland. The warm and cool eras since the last ice age were due to natural climate cycles, not greenhouse gas emissions. The “on record” period that NOAA references is only a tiny part of the climatic picture.
Global average temperature is difficult to measure. The data sets of NOAA are an artificial estimate at best. They start with a patchwork collection of thousands of thermometer stations that inadequately cover the globe. Station coverage of the oceans and of the far northern and southern regions is inconsistent and poor. To cover areas without thermometers, averaging estimates are made from surrounding stations to try to fill in the holes.
In addition to coverage problems, gauge measurements often contain large errors. Man-made structures such as buildings and parking lots absorb sunlight, artificially increasing local temperatures. Cars, air conditioners, and other equipment generate heat when operating, creating what is called an Urban Heat Island effect.
The accuracy of the US temperature record is questionable. Meteorologist Anthony Watts, creator of the science website WattsUpWithThat, led a team of volunteers that audited more than 1,000 US temperature gauge stations from 2007 to 2011. Over 70 percent of the sites were found to be located near artificial heating surfaces such as buildings or parking lots, rated as poor or very poor by the site rating system of the National Climatic Data Center, a NOAA organization. These stations were subject to temperature errors as large as 3.6 degrees F (2oC).
Simple problems can throw off gauge readings. Temperature stations are louvered enclosures that are painted white to reflect sunlight and minimize solar heating. As the station weathers and the paint ages, gauge stations read artificially high temperatures. A study published last month found that after only five years of aging, temperature stations will record a temperature error of 2.9 degrees F (1.6oC) too high. This is greater than the one degree rise in the last 130 years that NOAA is alarmed about.
In addition to temperature measurement error, NOAA makes “adjustments” to the raw temperature data. According to a 2008 paper, after raw thermometer data is received, a computer algorithm “homogenizes” the data, adjusting for time-of-observation, station moves, thermometer types, and other factors to arrive at the official temperature data set.
This sounds good until one looks at the adjustment that NOAA has added. For temperature data from 1900 to 1960, very little adjustment is added. But after 1960, NOAA adds an upward adjustment to the thermometer data that rises to 0.5 degrees F (0.3oC) by the year 2000. This gives a whole new meaning to the phrase “man-made global warming.”
Heat waves are real just as climate change is real. But a heat record in Las Vegas or one degree of temperature rise since the Civil War is not evidence that humans are the cause.
See also these related NewsBlaze Climate Change Skeptic stories:<|endoftext|>
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Introduction to Objects
Databases in Access are made out of four objects: tables, queries, forms, and reports. Together, these articles enable you to enter, store, investigate, and incorporate your information any way you need.
In this exercise, you will find out about every one of the four objects and come to see how they communicate with each other to make a completely useful social database.
You may definitely realize that tables are sorted out into vertical columns and horizontal rows.
In Access, lines and segments are alluded to as records and fields. A field is something beyond a section; it’s a method for sorting out data by the type of information it is. Each snippet of data inside a field is of the same type. For instance, each section in a field called First Name would be a name, and each passage in the field called Street Address would be an address.
Likewise, a record is more than just a row; it’s a unit of information. Every cell in a given row is part of that row’s record.
Notice how each record spans several fields. Even though the information in each record is organized into fields, it belongs with the other information in that record. See the number at the left of each row? It’s the ID number that identifies each record. The ID number for a record refers to every piece of information contained on that row.
Tables are useful for storing closely related data. Suppose you claim a pastry kitchen and have a database that incorporates a table with your clients’ names and data, similar to their telephone numbers, personal residences, and email addresses. Since these snippets of data are on the whole points of interest on your clients, you’d incorporate them all in the same table.
Every client would be spoken to by a unique record, and each kind of data about these clients would be put away in its own particular field. On the off chance that you chose to include any more data—say, a client’s birthday—you would basically make another field inside a similar table.
To read more about Office Products visit WWW.OFFICE.COM/SETUP<|endoftext|>
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A research paper has drawn on data from NASA and ESA satellites to produce the most comprehensive assessment of ice sheet losses in Antarctica and Greenland to date.
The paper, published on 29 November in the journal Science, shows that melting Antarctic and Greenland ice sheets have added 11.1mm (0.43") to global sea levels since 1992. The paper, A reconciled estimate of ice sheet mass balance, was drawn together by an international team of 47 researchers, using data collected from ten satellite missions to produce the most consistent picture of polar ice sheets ever recorded. The resulting findings have been able to reconcile the differences that existed between previous ice sheet studies, by carefully matching observation periods and survey areas.
The findings of the international team—the Ice Sheet Mass Balance Inter-comparison Exercise (IMBIE)—give clarity to the Intergovernmental Panel on Climate Change's 2007 reports. The latter study's time scale was so broad that it was not possible to tell if Antarctica was growing or shrinking, but the use of data from NASA and ESA satellites have confirmed that both Antarctica and Greenland are losing ice.
The combined melting of both the Antarctic and Greenland ice sheets accounted for one-fifth of all sea level rises over the 20-year survey period—the remainder caused by the thermal expansion of the warming ocean, melting of mountain glaciers and small Arctic ice caps, and groundwater mining.
The rate at which the ice sheets are melting was also seen to rise over the study period, with both Antartica and Greenland shedding more than three times as much ice each year (the equivalent to a sea level rise of 0.95mm) as they were in the 1990s (0.27mm). Approximately two-thirds of the loss is coming from Greenland.
"The rate of ice loss from Greenland has increased almost five-fold since the mid-1990s" said NASA's Erik Ivins, co-author on the study. "In contrast, while the regional changes in Antarctic ice over time are sometimes quite striking, the overall balance has remained fairly constant—at least within the certainty of the satellite measurements we have to hand."
Andrew Shepherd of the University of Leeds led the study, which drew together teams from 26 laboratories from across the globe.
"It wasn't clear to us when we set out whether or not we would be able to reconcile the different satellite techniques that were involved," Shepherd told Wired.co.uk. The study was hoping to reconcile data sets from satellite altimetry, which measures changes in the ice sheet shape, interferometry, which measures changes in the speed of the ice flow, and gravimetry, in which satellites are able to sense the very small changes in attraction of the planet as a whole.
"In the past, the accuracy of climate models has been challenged by accuracy of the data that go into them. We would expect, by delivering a data model that's two or three times more reliable than the last, any model predictions will be two or three times more reliable than the last one as well.
"The caveat to that is that the great deal of uncertainty that exists over future sea level projections is not related to the observations of the ice sheets models -- it's the uncertainty associated to the emission scenario. There's a wide range of different trajectories that we might follow in terms of atmospheric emissions, and the variations in sea levels are really a consequence of those, and they're outside the control of the scientists."
The data from the study is now available on the IMBIE website for use in climate models.
This article originally appeared on Wired UK.<|endoftext|>
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Home » Learning About » Information and Resources » Sustainability » Sea Level Rise
Sea Level Rise - A Pembrokeshire Perspective
A change to sea level is expected as our climate changes. This will happen because:
- As the climate warms it warms the seas and oceans. The water droplets will expand (get bigger), and so will occupy more space.
- Glaciers and polar ice are melting. Much of this water has been locked away for thousands of years. If polar ice and glaciers melt there will be more water.
- Where the ice has melted it leaves bare earth or sea. This absorbs the heat from the sun, but before the ice reflected the heat back into space. If the sea and land are warmer the climate will get warmer.
Flooding at the car park in Lower Town, Fishguard.
Most people think that melting ice is the most important cause of sea level rise, but this is not true. It is the expansion of the sea water as it gets warmer that is causing the biggest change to sea levels.
According to the UK Government's UK Climate Change Risk Assessment 2017, sea levels in the UK are rising by approximately 3mm a year.
What might happen if sea level rises?
Most of us don't think that sea level rise will affect us much, but is that true?
- There will be a loss of coastal land. Beaches, roads, homes and businesses may disappear under the rising waters. Pembrokeshire depends on tourism for much of its prosperity. How will it manage if the beaches are gone?
Think of the land between Penally and Tenby. Much of it is low-lying, and just above sea level. If sea level rises the dunes will disappear, the sea will flood the coastal road and the railway line from Tenby to Pembroke.
- There will be a loss of estuary land. It will only take a small sea level rise to flood the estauries of the Cleddau Waterway. Many species rely on these habitats for food, and they will be lost.
The estuaries are also places where people keep their boats. Imagine the pill at Neyland or the harbour at Pembroke Dock. How will they change?
- Bigger tides. Wales has one of the biggest tidal ranges in the world. This will increase as sea levels rise. How many places are affected by high tides now?
High spring and autumn tides affect places such as Angle and Fishguard on the coast, but also reach far up the Daugleddau Estuary to flood Carew and even Haverfordwest.<|endoftext|>
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# What is the formula for distance over time?
## What is the formula for distance over time?
Calculate speed, distance or time using the formula d = st, distance equals speed times time.
### What is a change in distance called?
speed. Explanation: Speed is the rate of change of displacement with respect to its surrounding. Since the speed of a body is irrespective of its direction, therefore it is a scalar quantity.
How do you calculate change in distance?
Rate of change in position, or speed, is equal to distance traveled divided by time. To solve for time, divide the distance traveled by the rate. For example, if Cole drives his car 45 km per hour and travels a total of 225 km, then he traveled for 225/45 = 5 hours.
Is a measure of the change in distance over time?
Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (Δt), represented by the equation r = d/Δt.
## How do you solve time speed and distance questions?
Sol: Usual time taken = 50 minutes = 5/6 hours. The distance = 60 km. Usual Speed = Distance ÷ Usual Time → 60/(5/6) = 72 kmph. Speed on this occasion = Distance ÷ Time on this occasion = 60/1 = 60 kmph.
### What is the formula time?
FAQs on Time Formula The formula for time is given as [Time = Distance ÷ Speed].
What is the time rate change of distance called?
Speed is the rate of change of distance with time.
What is the formula of time?
## What is rate of change of distance?
Answer: speed is the rate of change of distance.
### Is speed distance over time?
Speed tells us how fast something or someone is travelling. You can find the average speed of an object if you know the distance travelled and the time it took. The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time.
What is change in velocity called?
Acceleration is defined as the rate of change of velocity. Velocity is a vector, which means it contains a magnitude (a numerical value) and a direction.
What is the rate of change in distance per hour?
List all given rates and the unknown rate. As Car A travels north, the distance y is growing at 50 miles per hour. That’s a rate, a change in distance per change in time. So, As Car B travels west, the distance x is shrinking at 40 miles per hour.
## How do you calculate distance from speed and time?
To solve for distance use the formula for distance d = st, or distance equals speed times time. distance = speed x time Rate and speed are similar since they both represent some distance per unit time like miles per hour or kilometers per hour.
### What is the equivalent formula for distance divided by time?
You can use the equivalent formula d = rt which means distance equals rate times time. distance = rate x time To solve for speed or rate use the formula for speed, s = d/t which means speed equals distance divided by time.
How do you find the speed of a moving object?
To solve for speed or rate use the formula for speed, s = d/t which means speed equals distance divided by time. To solve for time use the formula for time, t = d/s which means time equals distance divided by speed. Related Calculators.<|endoftext|>
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# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३६८
एतत् पृष्ठम् परिष्कृतम् अस्ति
172
GAŅITASĀRASAŃGRAHA.
difference between the first term and the common difference (in the series). (Then) the square of the sum (of the series) is multiplied by the common difference. If the first term is smaller than the common difference, then (the first of the products obtained above is) subtracted (from the second product). If, however, (the first term is) greater (than the common difference), then (the first product above-mentioned is added (to the second product). (Thus) the (required) sum of the cubes is obtained.
Examples in illustration thereof.
304. What may be the sum of the cubes when the first term is 3, the common difference 2, and the number of terms 5; or, when the first term is 5, the common difference 7, and the number of terms 6 ?
The rule for arriving at the sum of (a number of terms in a series wherein the terms themselves are successively) the sums of the natural numbers (from 1 up to a specified limit, these limiting numbers being the terms in the given series in arithmetical progression):-
305–305${\displaystyle {\tfrac {1}{2}}}$.[1] Twice the number of terms (in the given series in arithmetical progression) is diminished by one and (then) multiplied by the square of the common difference. This product is divided by six and increased by half of the common difference and (also) by the product of the first term and the common difference. The sum (so obtained) is multiplied by the number of terms as diminished by one and then increased by the product obtained by multiplying the first term as increased by one by the first term itself. The quantity (so resulting) when multiplied by half the number of terms (in the given series) gives rise to the required sum of the series wherein the terms themselves are sums (of specified series)
305-305${\displaystyle {\tfrac {1}{2}}}$.^ Algebraically, ${\displaystyle \left[\left\{{\tfrac {(2n-1)^{2}}{6}}+{\tfrac {b}{2}}+ab\right\}(n-1)+a(a+1)\right]{\tfrac {n}{2}}}$ is the sum of the series in arithmetical progression, wherein each term represents the sum of a series of natural numbers up to a limiting number, which is itself a member in a series in arithmetical progression.<|endoftext|>
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# Some important points to be remembered for calendar and clock questions
Feb 7, 2018 17:17 IST
SSC Clock and calendar tips
The experts of Jagranjosh.com have come forward to help the aspirants in attempting all the questions speedily with basic preparatory strategy. By providing basic concepts, we are trying to make the calculation faster than doing from long and traditional ones. Here we, come up with Some Important Point to be remembered about Calendar and Clock Questions in Reasoning Section.
• A year divisible by 4 is a leap year
• In case of century, leap year is that which is divisible by 400
• There are 365 days in an ordinary year, so there are 52 weeks+1 day. Hence, an ordinary year contain 1 odd day.
• There are 366 days in a leap year. Hence a leap year contain 2 odd days.
• There are 28 days in February in an ordinary year while in leap year there are 29 days in February.
• There are 28 days in February in an ordinary year while in a leap year there are 29 days in February.
• The day of week on 1st Jan 1 AD is Monday
• After 11 years the calendar is repeated
Take Online Quiz
• The minute hand is also called, the long hand while hour hand is known as short hand.
• In every one hour, the minute hand gains 55 minutes on the hour hand
• The hands are in the same straight-line when they are opposite to each other or coincident
• In every hours hands coincide once
• The hands coincides 11 times in every 12 hours
• The minute hand moves 360in 1 hour while hour moves 30 in 1 hour
Example 1: Sita remembers that her brother Anuj’s birthday was after 15th but before 20thSeptember, while her mother remembers that Anuj’s birthday was after 17th but before 19thSeptember. On which date of September was Anuj’s birthday.
Solution: According to Sita:
According to her mother:
According to both common days is 18th.
Hence, Anuj's birth was on 18th September.
Example 2: If 22nd August is Sunday, What day was on 22 days before?
Solution: Difference of No. of days = 22
No. of odd days = 1
[Because 22/7 = 3 + 1]
1 day before Sunday is Saturday.
Hence, it will be Saturday before 22 days.
DISCLAIMER: JPL and its affiliates shall have no liability for any views, thoughts and comments expressed on this article.
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Factors the 39 are the perform of positive and an adverse whole numbers that have the right to be separated evenly right into 39. There space a complete of 4 determinants of 39. Did friend know, 39 civilization signed the United says Constitution and also North Dakota to be the 39th state to it is in admitted come the Union? In this lesson, us will find out to calculation the factors of 39, its prime factors, and also its determinants in pairs together with solved instances for a better understanding.
You are watching: What are the factors of 39
Factors that 39: 1, 3, 13, and also 39Factors of -39: -1, -3, -13, and also -39Prime administrate of 39: 39 = 13 × 3
1 What are determinants of 39? 2 Important Notes 3 How come Calculate factors of 39? 4 Factors the 39 by element Factorization 5 Factors that 39 in Pairs 6 FAQs on determinants of 39 7 Tips and Tricks
## What are factors of 39?
The number 39 is an odd composite number. As it is odd, it will certainly not have 2 or any type of multiples that 2 as its factor. To recognize why that is composite, let"s recall the an interpretation of a composite number. A number is claimed to it is in composite if that has much more than two factors.
Consider the number 11. It has only two components 1 and 11. Now, let"s take an instance of the number 60. The factors of 60 room 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 over there are an ext than two factors of 60. Thus, 60 is a composite number vice versa, 11 is not.
Coming back to 39, the factors of 39 room all the integers that 39 deserve to be separated into.
## How to calculate the components of 39?
Let"s start calculating the factors of 39, beginning with the smallest entirety number, i.e., 1
Divide 39 v this number. Is the remainder 0? correct! So, we will certainly get, 39/1 = 39.The next totality number is 2Now division 39 with this number. Is the remainder 0? No, the is not.
As currently discussed in the previously section, even number cannot divide 39. Hence, we need to examine odd numbers only.
39/3 = 13
3 × 13 = 39
Hence, the factors of 39 space 1, 3, 9, and 13. Explore components using illustrations and also interactive examples:
## Factors that 39 by prime Factorization
Prime factorization method to to express a composite number together the product the its prime factors. To acquire the prime factorization of 39, we divide it by its smallest prime element which is 3.
39/3 = 13
Since 13 is a element number, it is divisible through 13 only. This procedure goes top top till we obtain the quotient together 1 The prime factorization of 39 is presented below:
## Factors that 39 in Pairs
The pair of number which provides 39 when multiplied is known as factor pairs that 39. The complying with are the determinants of 39 in pairs.
If us consider an adverse integers, both the number in the pair components will it is in negative. 39 is a confident number and we recognize that the product of two an unfavorable numbers is a positive number.
Therefore, we deserve to have factor pairs that 39 together (-1,-39) ; (-3,-13).
Important Notes:
As 39 is an odd number, every its factors will be odd.39 is a non-perfect square number. Thus, that will have actually an even number of factors. This building holds true for every non-perfect square number.Factors are never fractions or decimals.
Ms. Evans has actually 39 package of crayons. There space 3 teams of students in she class. She plan to provide the package to the three teams for an task so that they room evenly distributed.How numerous packs will each team get?Determine the following:Factors of 35Factors that 18Factors the 49
Example 1: James has 39 systems of a cutlery set. He wants to fill it in cartons such that these units are evenly distributed. There are two size of cartons available. The very first size has a capacity of 9 units and also the 2nd size has actually a capacity of 13 units.
Which type of carton will certainly he pick so the there is no unit left and maximum units room filled in the cartons?How countless cartons will be used?
Solution:
The problem that over there is no unit left means that 39 need to be split by one of those two numbers, the is, 9 or 13, and also the remainder need to be 0
That means the number must be a element of 39, the end of the two provided numbers, 13 is a factor of 39Thus, that will pick cartons of the second size that the volume of 13 units.To discover the number of cartons that the 2nd size used, we have to divide 39 through 13 i.e. 39/13 = 3.
See more: Watch Boku No Hero Season 2 Episode 14 English Subbed, Watch My Hero Academia 2 Episode 14 Online<|endoftext|>
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# If the product of two consecutive odd integers is decreased by 3, the result is 12. How do you find the integers?
the integers are $3$ and $5$
Also the integers are $- 3$ and $- 5$
#### Explanation:
The solution
Let $2 n + 1$ be the odd integer
Let $2 n + 3$ be the next odd integer
$\left(2 n + 1\right) \left(2 n + 3\right) - 3 = 12$
$4 {n}^{2} + 8 n + 3 - 3 = 12$
$4 {n}^{2} + 8 n = 12$
${n}^{2} + 2 n = 3$
${n}^{2} + 2 n - 3 = 0$
Solution by factoring
$\left(n - 1\right) \left(n + 3\right) = 0$
$n - 1 = 0$
$n = 1$
Let $2 n + 1$ be the odd integer which is equal $= 3$
Let $2 n + 3$ be the next odd integer which is equal $= 5$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
At $n + 3 = 0$
$n = - 3$
Let $2 n + 1$ be the odd integer which is equal $= - 5$
Let $2 n + 3$ be the next odd integer which is equal $= - 3$
Checking: using odd numbers 3, and 5
$\left(2 n + 1\right) \left(2 n + 3\right) - 3 = 12$
$\left(3\right) \left(5\right) - 3 = 12$
$12 = 12 \text{ }$correct
Checking: using odd numbers -5, and -3
$\left(2 n + 1\right) \left(2 n + 3\right) - 3 = 12$
$\left(- 5\right) \left(- 3\right) - 3 = 12$
$12 = 12 \text{ }$correct
God bless....I hope the explanation is useful.<|endoftext|>
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Ancient Iranian religion
Ancient Iranian religion, diverse beliefs and practices of the culturally and linguistically related group of ancient peoples who inhabited the Iranian plateau and its borderlands, as well as areas of Central Asia from the Black Sea to Khotan (modern Hotan, China).
The northern Iranians (referred to generally as Scythians [Saka] in Classical sources), who occupied the steppes, differed significantly from the southern Iranians. In religion and culture, both the northern and southern Iranians had much in common with the ancient Indo-Aryan-speaking peoples of the Indian subcontinent, although there was much borrowing from Mesopotamia as well, especially in western Iran. From at least the time of the rise of the Median empire, Iranian religion and culture has had a profound influence upon the Middle East, as also the Middle East upon Iran.
This account will take the conquest of the Achaemenian dynasty by Alexander the Great as a somewhat arbitrary date for the close of the period of ancient Iranian religion, even though these influences have continued through later history and some forms of Iranian religion have persisted to the present day. It will also treat ancient Iranian religion, insofar as possible, apart from Zoroastrianism. Unless otherwise indicated, all spellings of Iranian names and terms are given in reconstructed forms that often differ from the Avestan spellings of the Zoroastrian canon.
Sources of knowledge
Modern understanding of ancient Iranian religion is impeded by the limitations of the available sources, which are inevitably of two sorts: textual and material.
Textual sources are both indigenous and foreign, the latter being primarily Greek, although for purposes of historical reconstruction the ancient Indian Vedic literature is indispensable. The main problem with the Greek sources, the most important of which is Herodotus, is that the information they contain is not always very reliable, either because it is outright erroneous or because it is based on misunderstandings. The main indigenous sources are the Achaemenian royal inscriptions in the Old Persian language (with Akkadian, Elamite, and Aramaic translations) and the Avesta, the Zoroastrian sacred scriptures, in a language called Avestan. The royal inscriptions, especially those of Darius (522–486 bce) and his son Xerxes I (486–465 bce), for the most part eloquent pieces of propaganda, are rich in references to religion. In addition to the information they contain, they have the great advantage of being fixed in time and place. Matters are quite otherwise in the case of the Avesta, which is the principal source of knowledge of ancient Iranian religions.
Like the Bible, the Avesta is a collection of a variety of texts composed over what appears to be a considerable span of time by different authors, which has endured editing and redaction at several points during the history of its development. The text that is now extant represents only a fragment of what remained in the 9th century of the late Sāsānian Avesta compiled under the direction of Khosrow I (531–579 ce). Summaries of the contents of the Sāsānian Avesta show that it was an enormous collection containing texts in Avestan as well as in—and predominantly so—Pahlavi, the language of Sāsānian Zoroastrianism. In spite of the relatively recent date of the existing Avesta, it contains matter of great antiquity, of which the Gāthās (“Songs”) of the Prophet Zarathustra (also known by his Greek name, Zoroaster) and much of the Yashts are among the oldest. The Gāthās contain expressions of Zarathustra’s religious vision which, in many ways, is a complicated reinterpretation of inherited Iranian religious ideas. The Yashts are collections of verses dedicated to the various deities. Most of the Yashts, though touched up with Zoroastrian terminology and ideas, have little to do with anything specifically Zoroastrian. The gods invoked are essentially the gods of pre-Zoroastrian Iran. Unfortunately, there is little agreement as to when Zarathustra lived, though most scholars agree that he lived sometime between about 1200 and 600 bce. It does not seem possible to date the Yashts much more precisely, except to believe that their redaction (not necessarily composition) may have first taken place in the 5th century bce.
The earliest religious texts of the closely related Indo-Aryan speakers (principally the Rigveda) are indispensable for making historical reconstructions of the development of Iranian religion. The Rigveda, a collection of more than 1,000 hymns to various deities, can be dated to a period from approximately 1300 to 900 bce. Apart from the Achaemenian inscriptions, there is no secure evidence that religious compositions were reduced to writing until the late Arsacid or early Sāsānian periods. Thus, unlike the other religions of the Middle East, the Iranian religions had no written texts in the ancient period. All religious “literature” was oral, in both composition and transmission.
Material sources are much more limited and are, for the most part, restricted to western Iran. The remains of Achaemenian architecture and art, by far the most important of the material sources, provide abundant evidence of imperial articulation of religious symbols and show a thorough dependence on Middle Eastern precedents.
Origin and historical development
During the second half of the 2nd millennium bce, two groups of culturally and linguistically related peoples who called themselves arya (“nobles”) migrated from the steppes down into the Middle East, the Iranian plateau, and the northwestern part of the Indian subcontinent. One group settled in Anatolia and India. The other settled in greater Iran. These people were originally seminomadic pastoralists whose chief economic base was cattle, primarily bovines but also sheep and goats. They bred horses, which they used for riding and pulling chariots in warfare and sport. It is not at all clear how rigidly their society was originally segmented. There were specialists in religious matters, and men who could afford horses and chariots were reckoned as warriors and leaders. By the Achaemenian period there developed a more rigid division of society into four basic classes: priests, nobles, farmers/herdsmen, and artisans. Society generally was patriarchal, and male dominance was strongly reflected in the religion. Like the ancient Israelites, as the Iranians occupied the land, they became increasingly dependent upon agriculture and settled in villages and towns. During this process they were certainly influenced by the indigenous populations, of whose religion almost nothing is known, except by inference from elements of Iranian religion that have no reflex in the Veda or among other Indo-European-speaking peoples.
Owing to their common origin, Iranian and Indo-Aryan religions are very similar. From a comparative study of both groups, it is possible to reconstruct, in general features, the early forms of Iranian religion for which there is no direct documentation. The pantheon, similar to those of other Indo-European-speaking peoples, embraced a large number of deities, both female and, predominantly, male. Some of these were personifications of natural phenomena, others of social norms or institutions. There appear to have been two major groups of deities, the daivas and the ahuras. Daiva (literally “heavenly one”; Vedic deva, Latin deus) is derived from the common Indo-European word for “god,” and this is the meaning it has in the Vedas. Among many Iranians and in Zoroastrianism the daivas were regarded as demons, but this belief was not pan-Iranian. The ahuras (“lords”; Vedic asura) were certain lofty sovereign deities, in contradistinction to the other deities called bagha (Vedic bhaga, “the one who distributes”) and yazata (“the one to be worshipped”). At the head of the pantheon stood Ahura Mazdā, the “Wise Lord,” who was particularly connected with the principle of cosmic and social order and truth called arta in Vedic (asha in Avestan). Closely associated with him was another ahura named Mithra (Vedic Mitra), the god who presided over covenants. In Iran there were two gods with martial traits quite similar to those of Vedic Indra, Mitra, and Vrthraghna. Among female deities the Earth, Spantā Aramati, and the sacred river, Ardvī Sūrā, were most prominent. A sacrificial ritual yazna (Avestan yasna, Vedic yajna) was performed in which fire and the sacred drink hauma (Avestan haoma, Vedic soma) played an important part. The principal officiant at the sacrifice was the zautar (Vedic hotar).
As with other ancient religions, the cosmological dichotomy of chaos and cosmos figured in both myth and worldview. The most prominent and unique feature of ancient Iranian religion was the development of dualism, primarily expressed in the opposition of truth (arta) and falsehood (drug, drauga). Originally confined to ideas of social and natural order opposed by disorder and chaos, a dualistic ideology came to permeate all aspects of life. The pantheon was divided between the gods and demons. Especially under the influence of the magi, members of a priestly tribe of Median origin, the animal kingdom was divided into two classes: beneficent animals and noxious creatures. Even in vocabulary there developed a system of “ahuric” and “daivic” words for such things as body parts: for example, the word zasta was used for the hand of a righteous person and gava for the hand of an evil person. It is important to note that this was not a gnostic system, like those that flourished in the Middle East during the early centuries of the Common Era, as there was no myth of evil matter coming into being through the corruption and fall of a spiritual being.
Except for a mostly legendary line of eastern Iranian kings, the kavis, the last of whom was Zarathustra’s patron Vishtāspa (Greek Hystapes), the only historical information on the relation of religion to political authority comes from the Achaemenian period in western Iran. The ideology of kingship was closely tied to the supreme deity, Ahura Mazdā, through whose will the kings ruled. The Achaemenian kings had to contend with the power of the Median priesthood, the magi. Their origin is unclear, but, according to Classical sources, they presided at all religious ceremonies, where they chanted “theogonies.” That they were deeply involved in politics is seen from the attempt of the magus Gaumāta to seize the throne upon the death of Cambyses II. Although Darius persecuted the magi, they remained powerful and eventually became the official priesthood of the empire. They were probably responsible for articulating a thoroughly dualist ideology and contributing to Zoroastrianism its zealous preoccupation with ritual purity. In addition, they were famous throughout the ancient world as wonder-workers.
Because all the sources for Iranian myths, whether those of Classical authors or indigenous texts, are post-Zoroastrian, it is often difficult to discern what elements of the myths are Zoroastrian innovations and what elements are inherited. It is particularly hard to resolve this problem owing to the nature of Zoroastrianism itself as a religion that has always drawn heavily on already existing ideas and that has accommodated itself to various forms of Iranian religions. As is the case with ancient religions generally, Iranian religions did not possess one unified collection of myth. What one finds are fragments of a wide variety of myths exhibiting many variations on common themes.
Creation of the cosmos
Both the Avesta and the Achaemenian inscriptions have little to say about creation in the sense that they contain nothing comparable to the Babylonian Enuma elish or to the traditions of the first three chapters of the book of Genesis. What is primarily emphasized is the power and majesty of Ahura Mazdā as creator of heaven and earth. However, beside Ahura Mazdā is an ancient Indo-Iranian god called Thvarshtar (Vedic Tvashtar; “Artisan”), who also appears in Zarathustra’s system of the Beneficent Immortals under the name Spenta Mainyu (“the Beneficent Spirit”). Thvarshtar functions in many ways as Ahura Mazdā’s creative aspect. While in the Gāthās and the Younger Avesta Spenta Mainyu is paired with his evil antagonist Angra Mainyu (“the Evil Spirit”; Ahriman in Middle Persian), in other, later sources it is Ohrmazd (Middle Persian for Ahura Mazdā) himself who is paired with Ahriman. Although there are cryptic allusions in the Avesta to creations of the two antagonistic spirits, the first discursive exposition of the creation of the world by the two spirits occurs in Plutarch (De Iside et Osiride 47), where he says that the Persians “tell many mythical tales about the gods, such as the following. Oromazes (i.e., Ahura Mazdā) born from the purest light, and Areimanios (i.e., Ahriman) born from gloom, strive in war with one another.” The dualistic idea of two primordial spirits, called twins by Zarathustra, goes back to an Indo-European prototype. As far as this myth can be reconstructed, it seems that there were primordial twins before the creation of the world. They came into conflict. One, whose name was “Man” (Iranian *Manuʾ, meaning “man”), killed the other, “Twin” (Iranian Yama; Avestan Yima), and from the dismembered body he fashioned the world, using the skull for the sky, the flesh for the earth, the bones for mountains, and so forth. In another Iranian variant of this myth, Yama appears as the first mortal and the first ruler. The period of his rule, described as a golden age when there was neither death nor old age, neither hot nor cold, and so on, comes to an end when falsehood enters Yama’s speech. The royal Glory (Khvarnah) flees from Yama and takes refuge in the cosmic sea. Yama is then overthrown by a serpentine tyrant named Azhi Dahāka (“Dahāka the Snake”), whose rule ushers in a period of drought, ruin, and chaos. In turn, Azhi Dahāka is defeated by the hero Thraitauna, who establishes the legendary line of kings called kavis.
Zarathustra seems to have been the first religious thinker to conceive an eschatological myth concerning a future saviour who will rescue the world from evil, an idea that has been greatly elaborated in Zoroastrianism. It may have been influential in the development of the concept of the messiah in postexilic Judaism. Iranian religion also had a variant of the Noah’s Ark myth. In this myth Yama appears as the first herdsman and leader of humankind. After a long rule during which he has to enlarge the earth three times because of overcrowding, Ahura Mazdā tells him that a great winter is coming and advises him to prepare for it by building a gigantic three-story barnlike structure (vara) to hold pairs of animals and seeds of plants. From the fragmentary version of the myth that has survived, it appears that the vara is actually a sort of paradise or island of the blessed, though this story originated as a pastoralist myth about the building of the first winter cattle station by the culture hero.
The Iranians conceived of the cosmos as a three-tiered structure consisting of the earth below, the atmosphere, and the stone vault of heaven above. Beyond the vault of heaven was the realm of the Endless Lights, and below the earth was the realm of darkness and chaos. The earth itself rested on the cosmic sea called Varu-Karta. In the centre of the earth was the cosmic mountain Harā, down which flowed the river Ardvī. The earth was divided into six continents surrounding the central continent, Khvaniratha, the locus of Aryāna Vaijah, the Aryan land (i.e., Iran).
Cultic practices, worship, and festivals
In sharp contrast to the peoples of the Middle East, the Iranians did not make images of their deities, nor did they build temples to house them, preferring to worship in the open. Worship of the gods was performed primarily in the context of a central ritual called yazna, which corresponds in a great many details to the Vedic yajña. It is interesting to note that both rituals, though they have undergone some changes over the millennia, are still performed by Zoroastrians and Hindus in what must be the oldest continuously enacted ritual known. The plan of the yazna, as far as it can be reconstructed, was essentially that of a highly stylized festive meal offered to an honoured guest, the sacrificer being the host and the deity the guest. Although it is not known precisely when or how frequently the yazna was held (in Zoroastrianism it became a daily ritual), the reason for holding a yazna was to enter into communion with a divine being either for a specific purpose (to obtain offspring or a victory, and for example) or for general welfare or as an expression of piety. As a ritual meal, the yazna followed the established rules of hospitality: the guest was sent an invitation; on his arrival from afar he was greeted, shown to a comfortable seat, given meat and a refreshing and invigorating drink, and entertained with song extolling his great deeds and virtues. Finally, the guest was expected to return the hospitality in the form of a gift.
Of utmost importance was fire. In ancient Iran, fire was at once a highly sacred element and a deity. Thus, the word ātar denoted simultaneously “fire” and “Fire,” every instance of fire being a manifestation of the deity. Since burned offerings were not made, the role of Ātar, like that of his Vedic counterpart Agni, was principally that of intermediary between heaven and earth, between humans and gods. Beyond the sphere of the yazna, fire was always treated with utmost care as a sacred element. Whether in the household hearth or, at a later period, in fire temples, the sacred fire had to be maintained with proper fuel, kept free from polluting agents, and above all never permitted to go out or be extinguished.
More important than the meat offering of an animal victim was the preparation of the divine drink hauma. As with fire, hauma was regarded as both a sacred drink and as a powerful deity. Probably the greatest part of the yazna was devoted to the pressing of the hauma. Although there have been a number of proposals for the identification of the plant whose juices were extracted for the ritual drink, all have fallen short of absolute proof, and some, such as the toxic mushroom Amanita muscaria, are not worth consideration. In any case, an understanding of the religious meaning of hauma does not depend on a botanical identification. The word hauma itself derives from a verb “to press, extract” and thus literally means the juice that has been pressed out of the stems or stalks (ãsu) of whatever plant was being used. In this process the stalks were first soaked in water, then pounded. In Zoroastrianism this has been done with a metal mortar and pestle, but originally the stalks were pounded between two pressing stones, a lower and an upper. The juice, described as yellow, was filtered and mixed with milk, to cut the bitter taste, and perhaps with water too. Since the resulting drink was consumed immediately, it is clear that it was not alcoholic but rather was a mind-altering drug. The Yasht to Hauma says, “All other intoxicants are accompanied by Wrath with the horrible club, but that intoxication which is Hauma’s is accompanied by gladdening Truth (arta).” This brief statement can be amplified by the far more extensive descriptions of the Rigveda, where soma is not only offered to the gods but also taken by poets to enhance their insight and creative powers in their search for truth. Also, hauma, invoked for victory, was taken as a stimulant by warriors going into battle, and various heroes of Iranian myth and legend are remembered as primary practitioners of its cult.
As mentioned above, a comfortable seat was provided for the god or gods invited to yazna. Originally this consisted of special grasses strewn on the ground in front of the altar. In Vedic terminology this seat was called the barhish (Avestan barzish, “cushion”), while in Zoroastrianism a cognate word, Avestan barəsman (Iranian barzman), is used for a bundle of sticks—later thin metal rods—that is manipulated by the priest.
It is likely that from a very early period a priest, the zautar (Vedic hotar), was required to properly carry out the yasna. The zautar might be assisted by a number of other ritual specialists. With the priest or priests acting on behalf of the sacrificer, the god or gods were invoked through the intermediary of Fire. The sacred drink was prepared and the victim led up. When the god arrived, he was seated on the barzman and given food (parts of the slaughtered victim) and drink, after which he was entertained with song. Finally, the host-sacrificer put in a request for a return gift: heroic sons, long life, health, or victory, for example. In a certain sense, then, the entire ritual followed the old Latin dictum do ut des (“I give so that you may give”), in providing a means of inducing the powerful deities to act with favour toward human beings. Yet it also made possible communion between the divine and human realms. Deities could also be addressed directly in prayer with the supplicant standing erect with upraised arms; prostration was unknown.
Of further importance is the song of praise directed to the divine guest. Much of the poetic portions of the Avesta and almost all of the Rigveda must be understood in this ritual context. That is to say, ancient Indo-Iranian poetry was religious in nature and specifically composed for those ritual occasions when the gods required songs of praise to make them well disposed to their worshippers. The obscurity of Zarathustra’s Gāthās and of many Vedic hymns can best be understood when it is realized that the intended audience was not humans but rather the gods.
During the year there were various festivals, mostly relating to the agricultural and herding cycles. By far the most important was that of the New Year, which is still celebrated by Iranians with great festivity.<|endoftext|>
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# Population of state X increased by x% and the population of state Y increased by y% from 2001 to 2011. Assume that x is greater than y. Let P be the ratio of the population of state X to state Y in a given year. The percentage increase in P from 2001 to 2011 is _______.
Free Practice With Testbook Mock Tests
## Options:
1. $$\frac{x}{y}$$
2. x – y
3. $$\frac{{100\left( {x - y} \right)}}{{100 + x}}$$
4. $$\frac{{100\left( {x - y} \right)}}{{100 + y}}$$
### Correct Answer: Option 4 (Solution Below)
This question was previously asked in
GATE CE 2019 Official Paper: Shift 2
## Solution:
Let,
Population of state X in 2001 = a
∴ Population of state X in 2011 = a + 0.01 × x × a
Population of state Y in 2001 = b
∴ Population of state Y in 2011 = b + 0.01 × y × b
∵ P is the ratio of the population of state X to state Y in a given year
$${P_{2001}} = \frac{a}{b}$$
$${P_{2011}} = \frac{{\left( {1{\rm{\;}} + {\rm{\;}}0.01{\rm{\;}} \times {\rm{\;x\;}}} \right) \times {\rm{\;a}}}}{{\left( {1 + {\rm{\;}}0.01{\rm{\;}} \times {\rm{\;y}}} \right) \times {\rm{\;b}}}}$$
$${\rm{\Delta }}P\left( \% \right) = \frac{{{P_2} - {P_1}}}{{{P_1}}} \times 100 = \frac{{\frac{{\left( {1{\rm{\;}} + {\rm{\;}}0.01{\rm{\;}} \times {\rm{\;x\;}}} \right) \times {\rm{\;a}}}}{{\left( {1 + {\rm{\;}}0.01{\rm{\;}} \times {\rm{\;y}}} \right) \times {\rm{\;b}}}} - \frac{a}{b}}}{{\frac{a}{b}}} \times 100$$
$${\rm{\Delta }}P\left( \% \right) = \left( {\frac{{\left( {1{\rm{\;}} + {\rm{\;}}0.01{\rm{\;}} \times {\rm{\;x\;}}} \right)}}{{\left( {1 + {\rm{\;}}0.01{\rm{\;}} \times {\rm{\;y}}} \right)}} - 1} \right) \times 100 = \frac{{100\left( {{\rm{x}} - {\rm{y}}} \right)}}{{100 + {\rm{y}}}}$$<|endoftext|>
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# Thread: How do I begin to solve this problem?
1. ## How do I begin to solve this problem?
The question is as follows:
"A Farmer encloses three adjacent rectangular corrals with 1500 feet of fencing. Determine the length and width that will yield a maximum area."
2. ## Re: How do I begin to solve this problem?
The perimeter of the fence is a rectangle, of length x and width y, so we have 2x + 2y = 1500. The area is A = xy.
How can you put the area of the corral in terms of one variable? How do you go about finding a maximum area?
-Dan
3. ## Re: How do I begin to solve this problem?
it's 3 adjacent corrals though (imgur: the simple image sharer). Don't I have to somehow account for the fence that goes between the corrals?
4. ## Re: How do I begin to solve this problem?
There are two long fences. Calling each length "x", their lengths total 2x. There are four fences connecting them, two at the ends, two separating the three pens. If we call the length of each "y", their length total 4y. The total fencing is 2x+ 4y= 1500 feet and the area is xy square feet. You want to maximize xy subject to the constraint 2x+ 4y= 1500 feet.
One way to do that is to solve 2x+ 4y= 1500 for x= 750- 2y and write the area as $xy= (750- 2y)y= 1500y- 2y^2$. You can find the maximum by completing the square.
5. ## Re: How do I begin to solve this problem?
How are the corrals set up?
$\begin{tabular}{|c|c|c|}\hline & & \\ \hline\end{tabular}$
Or like this:
$\begin{tabular}{|c|c|}\hline & \\ \hline \multicolumn{2}{|c|}{ } \\ \hline \end{tabular}$
Should the corrals all be the same size?
Edit: I just saw the picture posted by the OP. I guess this post is moot.
6. ## Re: How do I begin to solve this problem?
Originally Posted by HallsofIvy
There are two long fences. Calling each length "x", their lengths total 2x. There are four fences connecting them, two at the ends, two separating the three pens. If we call the length of each "y", their length total 4y. The total fencing is 2x+ 4y= 1500 feet and the area is xy square feet. You want to maximize xy subject to the constraint 2x+ 4y= 1500 feet.
One way to do that is to solve 2x+ 4y= 1500 for x= 750- 2y and write the area as $xy= (750- 2y)y= 1500y- 2y^2$. You can find the maximum by completing the square.
That should be $xy= (750- 2y)y= 750y- 2y^2$, of course.
7. ## Re: How do I begin to solve this problem?
So I solved it and got the length as 375 feet and the width as 187.5 feet. Am I correct?
8. ## Re: How do I begin to solve this problem?
Originally Posted by promack
So I solved it and got the length as 375 feet and the width as 187.5 feet. Am I correct?
Yes, looks good to me.<|endoftext|>
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Apraxia is a motor disorder caused by damage to the brain (specifically the posterior parietal cortex) in which the individual has difficulty with the motor planning to perform tasks or movements when asked, provided that the request or command is understood and the individual is willing to perform the task. The nature of the brain damage determines the severity, and the absence of sensory loss or paralysis helps to explain the level of difficulty.
|Apraxia is characterized by loss of the ability to execute or carry out learned purposeful movements|
There are several types of apraxia including:
- Ideomotor apraxia: These patients have deficits in their ability to plan or complete motor actions that rely on semantic memory. They are able to explain how to perform an action, but unable to "imagine" or act out a movement such as "pretend to brush your teeth" or "pucker as though you bit into a sour lemon." However, when the ability to perform an action automatically when cued remains intact, this is known as automatic-voluntary dissociation. For example, they may not be able to pick up a phone when asked to do so, but can perform the action without thinking when the phone rings.
- Ideational/conceptual apraxia: Patients have an inability to conceptualize a task and impaired ability to complete multistep actions. Consists of an inability to select and carry out an appropriate motor program. For example, the patient may complete actions in incorrect orders, such as buttering bread before putting it in the toaster, or putting on shoes before putting on socks. There is also a loss of ability to voluntarily perform a learned task when given the necessary objects or tools. For instance, if given a screwdriver, the patient may try to write with it as if it were a pen, or try to comb their hair with a toothbrush.
- Buccofacial or orofacial apraxia: Non-verbal oral or buccofacial ideomotor apraxia describes difficulty carrying out movements of the face on demand. For example, an inability to lick one's lips or whistle when requested suggests an inability to carry out volitional movements of the tongue, cheeks, lips, pharynx, or larynx on command.
- Constructional apraxia: The inability to draw or construct simple configurations, such as intersecting shapes.
- Gait apraxia: The loss of ability to have normal function of the lower limbs such as walking. This is not due to loss of motor or sensory functions.
- Limb-kinetic apraxia: voluntary movements of extremities are impaired. For example, a person affected by limb apraxia may have difficulty waving hello.
- Oculomotor apraxia: Difficulty moving the eye, especially with saccade movements that direct the gaze to targets. This is one of the 3 major components of Balint's syndrome.
- Apraxia of speech (AOS): Difficulty planning and coordinating the movements necessary for speech (e.g. Potato=Totapo, Topato.) AOS can independently occur without issues in areas such as verbal comprehension, reading comprehension, writing, articulation or prosody.
Apraxia is most often due to a lesion located in the dominant (usually left) hemisphere of the brain, typically in the frontal and parietal lobes. Lesions may be due to stroke, acquired brain injuries, or neurodegenerative diseases such as Alzheimer's disease or other dementias, Parkinson's disease, or Huntington's disease. It is also possible for apraxia to be caused by lesions in other areas of the brain including the non-dominant (usually right) hemisphere.
Ideomotor apraxia is typically due to a decrease in blood flow to the dominant hemisphere of the brain and particularly the parietal and premotor areas. It is frequently seen in patients with corticobasal degeneration.
Ideational apraxia has been observed in patients with lesions in the dominant hemisphere near areas associated with aphasia; however, more research is needed on ideational apraxia due to brain lesions. The localization of lesions in areas of the frontal and temporal lobes would provide explanation for the difficulty in motor planning seen in ideational apraxia as well as its difficulty to distinguish it from certain aphasias.
Constructional apraxia is often caused by lesions of the inferior non-dominant parietal lobe, and can be caused by brain injury, illness, tumor or other condition that can result in a brain lesion.
Although qualitative and quantitative studies exist, there is little consensus on the proper method to assess for apraxia. The criticisms of past methods include failure to meet standard psychometric properties as well as research-specific designs that translate poorly to non-research use.
The Test to Measure Upper Limb Apraxia (TULIA) is one method of determining upper limb apraxia through the qualitative and quantitative assessment of gesture production. In contrast to previous publications on apraxic assessment, the reliability and validity of TULIA was thoroughly investigated. The TULIA consists of subtests for the imitation and pantomime of non-symbolic (“put your index finger on top of your nose”), intransitive (“wave goodbye”) and transitive (“show me how to use a hammer”) gestures. Discrimination (differentiating between well- and poorly performed tasks) and recognition (indicating which object corresponds to a pantomimed gesture) tasks are also often tested for a full apraxia evaluation.
However, there may not be a strong correlation between formal test results and actual performance in everyday functioning or activities of daily living (ADLs). A comprehensive assessment of apraxia should include formal testing, standardized measurements of ADLs, observation of daily routines, self-report questionnaires and targeted interviews with the patients and their relatives.
As stated above, apraxia should not be confused with aphasia; however, they frequently occur together. It has been stated that apraxia is so often accompanied by aphasia that many believe that if a person displays AOS; it should be assumed that the patient also has some level of aphasia.
Treatment for individuals with apraxia includes speech therapy, occupational therapy, and physical therapy. Generally, treatments for apraxia have received little attention for several reasons, including the tendency for the condition to resolve spontaneously in acute cases. Additionally, the very nature of the automatic-voluntary dissociation of motor abilities that defines apraxia means that patients may still be able to automatically perform activities if cued to do so in daily life. Nevertheless, research shows that patients experiencing apraxia have less functional independence in their daily lives, and that evidence for the treatment of apraxia is scarce. However, a literature review of apraxia treatment to date reveals that although the field is in its early stages of treatment design, certain aspects can be included to treat apraxia. One method is through rehabilitative treatment, which has been found to positively impact apraxia, as well as activities of daily living. In this review, rehabilitative treatment consisted of 12 different contextual cues, which were used in order to teach patients how to produce the same gesture under different contextual situations. Additional studies have also recommended varying forms of gesture therapy, whereby the patient is instructed to make gestures (either using objects or symbolically meaningful and non-meaningful gestures) with progressively less cuing from the therapist. It may be necessary for patients with apraxia to use a form of alternative and augmentative communication depending on the severity of the disorder. In addition to using gestures as mentioned, patients can also use communication boards or more sophisticated electronic devices if needed. No single type of therapy or approach has been proven as the best way to treat a patient with apraxia, since each patient's case varies. However, one-on-one sessions usually work the best, with the support of family members and friends. Since everyone responds to therapy differently, some patients will make significant improvements, while others will make less progress. The overall goal for treatment of apraxia is to treat the motor plans for speech, not treating at the phoneme (sound) level. Research suggests that individuals with apraxia of speech should receive treatment that focuses on the repetition of target words and rate of speech. Research rerouted that the overall goal for treatment of apraxia should be to improve speech intelligibility, rate of speech and articulation of targeted words.
The prognosis for individuals with apraxia varies. With therapy, some patients improve significantly, while others may show very little improvement. Some individuals with apraxia may benefit from the use of a communication aid. However, many people with apraxia are no longer able to be independent. Those with limb-kinetic and/or gait apraxia should avoid activities in which they might injure themselves or others.
Occupational therapy, physical therapy, and play therapy may be considered as other references to support patients with apraxia. These team members could work along with the SLP to provide the best therapy for people with apraxia. However, because people with limb apraxia may have trouble directing their motor movements, occupational therapy for stroke or other brain injury can be difficult.
No medication has been shown useful for treating apraxia.
- "Definition of APRAXIA". www.merriam-webster.com. Retrieved 2017-05-02.
- Sathian, K; et al. (Jun 2011). "Neurological and rehabilitation of action disorders: common clinical deficits". Neurorehabilitation and Neural Repair. 25 (5): 21S–32S. doi:10.1177/1545968311410941. PMC 4139495. PMID 21613535.
- Gross, RG; Grossman, M. (Nov 2008). "Update on apraxia". Current Neurology and Neuroscience Reports. 8 (6): 490–496. doi:10.1007/s11910-008-0078-y. PMC 2696397. PMID 18957186.
- Nadeau SE (2007). "Gait apraxia: further clues to localization". Eur. Neurol. 58 (3): 142–5. doi:10.1159/000104714. PMID 17622719.
- Treatment Resource Manual for Speech Pathology 5th edition
- Heilman KM, Watson RT, Gonzalez-Rothi LJ. Praxis. In: Goetz CG. Goetz: Textbook of Clinical Neurology. 3rd ed. Philadelphia, PA: Saunders Elsevier; 2007:chap 4.
- Duffy, Joseph R. (2013). Motor Speech Disorders: Substrates, Differential Diagnosis, and Management. St. Louis, MI: Elsevier. p. 269. ISBN 978-0-323-07200-7.
- Tonkonogy, Joseph & Puente, Antonio (2009). Localization of clinical syndromes in neuropsychology and neuroscience. Springer Publishing Company. pp. 291–323. ISBN 0826119670.
- Vanbellingen, T.; Bohlhalter, S. (2011). "Apraxia in neurorehabilitation: Classification, assessment and treatment". NeuroRehabilitation. 28 (2): 91–98. doi:10.3233/NRE-2011-0637. PMID 21447909.
- Vanbellingen, T.; Kersten, B.; Van Hemelrijk, B.; Van de Winckel, A.L.J.; Bertschi, M.; Muri, R.; De Weerdt, W.; Bohlhalter, S. (2010). "Comprehensive assessment of gesture production: a new test to measure upper limb apraxia". European Journal of Neurology. 17 (1): 59–66. doi:10.1111/j.1468-1331.2009.02741.x. PMID 19614961.
- (Manasco, 2014)
- "NINDS Apraxia Information Page". Retrieved 8 March 2012.
- Hanna-Pladdy, B; Heilman, K.M.; Foundas, A.L. (Feb 2003). "Ecological implications of ideomotor apraxia: evidence from physical activities of daily living". Neurology. 60 (3): 487–490. doi:10.1212/wnl.60.3.487. PMID 12578932.
- West, C; Bowen, A.; Hesketh, A.; Vail, A. (Jan 2008). "Interventions for motor apraxia following stroke". Cochrane Database of Systematic Reviews. 23 (1): CD004132. doi:10.1002/14651858.CD004132.pub2. PMID 18254038.
- Buxbaum LJ, Haaland KY, Hallett M, et al. (February 2008). "Treatment of limb apraxia: moving forward to improved action" (PDF). Am J Phys Med Rehabil. 87 (2): 149–61. doi:10.1097/PHM.0b013e31815e6727. PMID 18209511.
- Smania, N; et al. (Dec 2006). "Rehabilitation of limb apraxia improves daily life activities in patients with stroke". Neurology. 67 (11): 2050–2052. doi:10.1212/01.wnl.0000247279.63483.1f. PMID 17159119.
- "ASHA, Apraxia of Speech in Adults".
- Dovern, A.; Fink, GR.; Weiss, PH. (Jul 2012). "Diagnosis and treatment of upper limb apraxia". J Neurol. 259 (7): 1269–83. doi:10.1007/s00415-011-6336-y. PMC 3390701. PMID 22215235.
- Wambaugh, JL; Nessler, C; Cameron, R; Mauszycki, SC (2012). "Acquired apraxia of speech: the effects of repeated practice and rate/rhythm control treatments on sound production accuracy". American Journal of Speech-Language Pathology. 2: S5–S27. doi:10.1044/1058-0360(2011/11-0102.
- Fish, Margaret; "Here's How to Treat Childhood Apraxia of Speech, Second Edition". San Diego: Plural Publishing, Inc., 2015.
- Kasper, D.L.; Braunwald, E.; Fauci, A.S.; Hauser, S.L.; Longo, D.L.; Jameson, J.L.. Harrison's Principles of Internal Medicine. New York: McGraw-Hill, 2005. ISBN 0-07-139140-1.
- Manasco, H. (2014). Introduction to Neurogenic Communication Disorders. Jones & Bartlett Publishers.<|endoftext|>
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# Convert full circle to circle
Learn how to convert 1 full circle to circle step by step.
## Calculation Breakdown
Set up the equation
$$1.0\left(full \text{ } circle\right)={\color{rgb(20,165,174)} x}\left(circle\right)$$
Define the base values of the selected units in relation to the SI unit $$\left(radian\right)$$
$$\text{Left side: 1.0 } \left(full \text{ } circle\right) = {\color{rgb(89,182,91)} 2.0 \times π\left(radian\right)} = {\color{rgb(89,182,91)} 2.0 \times π\left(rad\right)}$$
$$\text{Right side: 1.0 } \left(circle\right) = {\color{rgb(125,164,120)} 2.0 \times π\left(radian\right)} = {\color{rgb(125,164,120)} 2.0 \times π\left(rad\right)}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(full \text{ } circle\right)={\color{rgb(20,165,174)} x}\left(circle\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} 2.0 \times π} \times {\color{rgb(89,182,91)} \left(radian\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} 2.0 \times π}} \times {\color{rgb(125,164,120)} \left(radian\right)}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} 2.0 \times π} \cdot {\color{rgb(89,182,91)} \left(rad\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} 2.0 \times π} \cdot {\color{rgb(125,164,120)} \left(rad\right)}$$
$$\text{Cancel SI units}$$
$$1.0 \times {\color{rgb(89,182,91)} 2.0 \times π} \cdot {\color{rgb(89,182,91)} \cancel{\left(rad\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} 2.0 \times π} \times {\color{rgb(125,164,120)} \cancel{\left(rad\right)}}$$
$$\text{Conversion Equation}$$
$$2.0 \times π = {\color{rgb(20,165,174)} x} \times 2.0 \times π$$
Cancel factors on both sides
$$\text{Cancel factors}$$
$${\color{rgb(255,204,153)} \cancel{π}} \times {\color{rgb(99,194,222)} \cancel{2.0}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(255,204,153)} \cancel{π}} \times {\color{rgb(99,194,222)} \cancel{2.0}}$$
$$\text{Simplify}$$
$$1.0 = {\color{rgb(20,165,174)} x}$$
Switch sides
$${\color{rgb(20,165,174)} x} = 1.0$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x} = 1$$
$$\text{Conversion Equation}$$
$$1.0\left(full \text{ } circle\right) = {\color{rgb(20,165,174)} 1}\left(circle\right)$$<|endoftext|>
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The concept of recovery was conceived by, and for, people with mental health issues to describe their own experiences and journeys and to affirm personal identity beyond the constraints of diagnosis.
The recovery movement began in the 1970s primarily as a civil rights movement aimed at restoring the human rights and full community inclusion of people with mental health issues.
Recovery approaches are viewed by the consumer movement as an alternative to the medical model with its emphasis on pathology, deficits and dependency. There is no single description or definition of recovery because recovery is different for everyone. However, central to all recovery paradigms are hope, self-determination, self-management, empowerment and advocacy. Also key is a person's right to full inclusion and to a meaningful life of their own choosing, free of stigma and discrimination.
Some characteristics of recovery commonly cited are that it is:
- a unique and personal journey
- a normal human process
- an ongoing experience and not the same as an end point or cure
- a journey rarely taken alone
- nonlinear—frequently interspersed with both achievement and setbacks.
Personal recovery is defined within this framework as 'being able to create and live a meaningful and contributing life in a community of choice with or without the presence of mental health issues'.
Recovery approaches will be different depending upon where a person is on their recovery journey. During an acute phase of illness, the person's capacity may be impaired to the extent that alleviation of distress and the burden of symptoms, as well as safety, is the primary focus of treatment and care. Regaining capacity for self-determination or deeper engagement should be a focus in the next stage of treatment and support. At later stages, when capacity is improved, there are opportunities for the person to consider broader recovery strategies.
The personal view of recovery is viewed as a journey that is a unique and personal experience for each individual. It has often been said to be about: gaining and retaining hope, understanding of ones abilities and limitations, engagement in an active life, personal autonomy, social identity, meaning and purpose in life, and a positive sense @of self. Essentially, the personal view of recovery is about a life journey of living a meaningful and satisfying life.
NSW Consumer Advisory Group (2012)
The concept of recovery is represented in Figure 2.Top of page
Figure 2: The concept of recovery
Text version of figure 2A series of concentric circles represents the concept of recovery. The innermost circle contains personal recovery characteristics of resilience, strength, optimism and hope and represents individuals with a lived experience. Surrounding the inner circle are recovery strategies such as advocacy, treatment, support, connection, acceptance and inclusion. The outermost circle contains support networks such as services, practitioners, peer specialists, community, friends and family.Top of page
Conceptual models of recovery processesIn recent years, mental health services and programs throughout Australia have adopted different models for helping staff to understand personal recovery processes and how they might enable and support personal recovery. While this new national framework is not seeking to standardise the use of particular models, the following models are highlighted as useful examples.
Andresen, Oades and Caputi (2003, 2006 & 2011)By studying personal accounts of recovery, this Australian team of researchers developed a conceptual model of recovery processes to guide research and training and to inform clinical practices. The team identified four processes involved with personal recovery.
- Finding and maintaining hope—believing in oneself; having a sense of personal agency; optimistic about the future
- Re-establishment of positive identity—incorporates mental health issues or mental illness, but retains a positive sense of self
- Building a meaningful life—making sense of illness or emotional distress; finding a meaning in life beyond illness; engaged in life
- Taking responsibility and control—feeling in control of illness and distress and in control of life.
Glover (2012)Glover's model reflects the efforts that people undertake in their personal recovery journeys through a set of five processes.
- From passive to active sense of self—moving from the passive position of being a recipient of services to reclaiming one's strengths, attributes and abilities to restore recovery
- From hopelessness and despair to hope—moving from hopelessness and despair to one of hope
- From others' control to personal control and responsibility—moving from others takingresponsibility for recovery to the person taking, holding and retaining responsibility
- From alienation to discovery—'finding meaning and purpose in the journey; doing more of what works and less of what does not work; learning from past experiences and incorporating that lesson into the present; acknowledging that journeys always have something to teach us and contribute to our sense of discovery'
- From disconnectedness to connectedness—moving from an identity of illness or disability to an appreciation of personal roles and responsibilities and to 'participating in life as a full citizen and not through the powerlessness of illness'.
This is a challenging concept for workers in helping and caring professions. Their impulse is to 'do for another' who is experiencing distress, pain, illness or disability. However, constantly 'doing for another' can contribute to a state of impotence and inability. A recovery approach encourages people to take an active role and reclaim responsibility for the direction of their life (Glover 2012).Top of page
Le Boutillier, Leamy, Bird, Davidson, Williams and Slade (2011)This study analysed 30 international documents to identify the key characteristics of!2 recovery-oriented practice guidance. The researchers developed an overarching conceptual framework to aid the translation of recovery guidance into practice.
In terms of people's recovery processes, this research team identified similar, but differently worded, processes to those proposed by Andresen, Oades and Caputi and by Glover.
Viewing recovery as a normal human process 'demystifies' the process of recovery from mental health problems and puts people in a better position to support someone in their recovery journey.
NSW Mental Health Coordinating Council (2008).
[Recovery is] a deeply personal, unique process of changing one's attitudes, values, feelings, goals, skills and/or roles. It is a way of living a satisfying, hopeful, and contributing life even within the limitations caused by illness. Recovery involves the development of new meaning and purpose in one's life as one grows beyond the catastrophic effects of mental illness.
William Anthony (1993).<|endoftext|>
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# 1.23 Lesson 23
```# Lesson 23. Power Series
# ------------------------
> restart:
# A power series about the point x=c is a series of the form
> Sum(a[n]*(x-c)^n, n=0..infinity);
infinity
-----
\ n
) a[n] (x - c)
/
-----
n = 0
# A power series has a radius of convergence R (possibly 0 or
# infinity), such that the series converges absolutely when
# |x-c|<R and diverges when |x-c|>R.
# The Taylor series of f(x) about the point x=c is the power
# series
> Sum((D@@n)(f)(c)/n!*(x-c)^n, n=0..infinity);
infinity
----- (n) n
\ D (f)(c) (x - c)
) -------------------
/ n!
-----
n = 0
# If f is analytic at x=c, this series converges to f(x) for x
# in some interval around x=c.
# Maple has the "taylor" command to find a given number of terms
# of the Taylor series of an expression.
> S:=taylor(exp(x), x=2, 5);
2 3
S := exp(2) + exp(2) (x - 2) + 1/2 exp(2) (x - 2) + 1/6 exp(2) (x - 2)
4 5
+ 1/24 exp(2) (x - 2) + O((x - 2) )
# The "5" tells Maple to what order to compute the series, i.e.
# up to but not including the term in (x-2)^5.
# The result of "taylor" looks like an ordinary polynomial with
# O((x-2)^5), but it's really something a bit different: a
# special "series" data structure.
> whattype(S);
series
# For just about any manipulations that you want to do with a
# series, you'll need the Taylor polynomial rather than this
# "series" structure. You can't just subtract the O((x-2)^5),
# you must use "convert":
> Sp:= convert(S, polynom);
2 3
Sp := exp(2) + exp(2) (x - 2) + 1/2 exp(2) (x - 2) + 1/6 exp(2) (x - 2)
4
+ 1/24 exp(2) (x - 2)
# Let's look at the Taylor series for exp about x=0
# and its convergence to exp. It's convenient to define a
# function that will calculate the n'th degree Taylor polynomial
# at a given point.
> P:= (n,t) -> subs(x=t, convert(taylor(exp(x), x=0, n+1), polynom));
P := (n,t) -> subs(x = t, convert(taylor(exp(x), x = 0, n + 1), polynom))
> P(4,t);
2 3 4
1 + t + 1/2 t + 1/6 t + 1/24 t
> plot({ exp(x), seq(P(nn, x), nn=1 .. 10) },
> x=-6 .. 2, y=-2 .. 8, colour = black);
** Maple V Graphics **
# For x > 0, it's more convenient to look at the remainders.
> plot({ seq(exp(x)-P(nn, x), nn=1 .. 12) },
> x = -6 .. 6, y = -3 .. 3, colour=black);
** Maple V Graphics **
# An animation is another possibility. There is an
# "animate" command, but here it's simplest to use
# "display" from the "plots" package, with the option
# "insequence=true".
> with(plots,display):
> display([ seq(
> plot(exp(x)-P(nn,x), x=-7 .. 7, y=-3 .. 3),
> nn=1 .. 16) ], insequence=true);
# It's almost (but not quite) true that the curves for x > 0
# march off to the right at a constant rate of 1/E per step.
> plot({ seq((exp(x+nn/E) - P(nn, x+nn/E)), nn = 1 .. 20) },\
x=-1 .. 1.5, y = -1 .. 2, color=black);
** Maple V Graphics **
# The y value here turns out to be approximately proportional to
# 1/sqrt(n). Here is an animation where we multiply by
# that factor sqrt(n).
> display([ seq(\
plot((exp(x+nn/E) - P(nn, x+nn/E))*sqrt(nn), x=-1 .. 1.5, y = -1 .. 2), nn=1..25) ], insequence=true);
# This can be analyzed as follows. We are looking at
> R:= (n,x) -> Sum((x+n/E)^k/k!, k=n+1..infinity);
infinity
----- k
\ (x + n/E)
R := (n,x) -> ) ----------
/ k!
-----
k = n + 1
# At least for x > -n/E, the terms are all positive, so R(n,x)
# is greater than the first term, which is
> FirstTerm:= (x+n/E)^(n+1)/(n+1)!;
(n + 1)
(x + n/E)
FirstTerm := ----------------
(n + 1)!
# On the other hand, for x < n/E, each successive term is
# multiplied by (x+n/E)/k < 2/E, so
> 'R(n,x)' < (x+n/E)^(n+1)/(n+1)!* sum((2/E)^k, k=0 .. infinity);
(n + 1)
(x + n/E) E
R(n, x) < ------------------
(n + 1)! (- 2 + E)
# Thus, up to a factor that's between 1 and E/(E-2), R(n,x)
# is the first term. As n -> infinity, Stirling's formula
# says n! is asymptotically
> asympt(n!, n,2);
1/2 1/2
2 Pi 1/2 1/2 1/2 3/2
---------- + 1/12 2 Pi (1/n) + O((1/n) )
1/2
(1/n)
---------------------------------------------------
n
exp(n) (1/n)
> asympt(FirstTerm,n,3);
1/2 1/2
exp(x E - 1) 2 exp(-2) exp(2) (1/n)
1/2 ----------------------------------------- + O(1/n)
1/2
Pi
# So let's look at R(n,x) divided by this. The result should
# be approximately constant (or at least between 1 and E/(E-2)).
> display([ seq(\
plot((exp(x+nn/E) - P(nn, x+nn/E))*sqrt(2*nn*Pi)*exp(1-x*E), x=-1 .. 1.5, y = -1 .. 2), nn=1 .. 25) ], insequence=true);
#
# We'd get a very different picture for, e.g., arctan(x). The
# Taylor series in this case:
> P:= subs(exp=arctan, eval(P));
P := (n,t) -> subs(x = t, convert(taylor(arctan(x), x = 0, n + 1), polynom))
> P(10,t);
3 5 7 9
t - 1/3 t + 1/5 t - 1/7 t + 1/9 t
> plot({ seq(arctan(x)-P(2*nn-1, x), nn=1 .. 10) }, x=-2..2, \
y=-1..1, colour=black);
** Maple V Graphics **
# Here the Taylor series only converges on the interval
# [-1 .. 1], and outside that interval P(n,x) is useless as an
# approximation to arctan(x).
#
# Various operations can be done to obtain new series from old
# series:
# the basic operations of arithmetic, as well as substitution,
# differentiation and integration.
#
# Example: Starting with "well-known" series, obtain the degree
# 10 Taylor polynomial for ln(1+x^2) sin(cos(x))) in powers of x.
#
#
> taylor(1/(1+t), t=0);
2 3 4 5 6
1 - t + t - t + t - t + O(t )
> int(", t);
2 3 4 5 6 7
t - 1/2 t + 1/3 t - 1/4 t + 1/5 t - 1/6 t + O(t )
# This is the Taylor series for ln(1+t) in powers of t.
> s1:= subs(t=x^2, convert(", polynom));
2 4 6 8 10 12
s1 := x - 1/2 x + 1/3 x - 1/4 x + 1/5 x - 1/6 x
> s2:= taylor(cos(x), x=0, 11);
2 4 6 8 10 11
s2 := 1 - 1/2 x + 1/24 x - 1/720 x + 1/40320 x - 1/3628800 x + O(x )
# Now I want sin(cos(x)), which is approximately sin(s2).
# But it would be wrong to use the power series of sin x in
# powers of x: we need it in powers of x-1.
> s3:= taylor(sin(t), t=1, 11);
2 3
s3 := sin(1) + cos(1) (t - 1) - 1/2 sin(1) (t - 1) - 1/6 cos(1) (t - 1)
4 5 6
+ 1/24 sin(1) (t - 1) + 1/120 cos(1) (t - 1) - 1/720 sin(1) (t - 1)
7 8
- 1/5040 cos(1) (t - 1) + 1/40320 sin(1) (t - 1)
9 10 11
+ 1/362880 cos(1) (t - 1) - 1/3628800 sin(1) (t - 1) + O((t - 1) )
> subs(t=convert(s2, polynom), convert(s3,polynom));
2 3 4
sin(1) + cos(1) %1 - 1/2 sin(1) %1 - 1/6 cos(1) %1 + 1/24 sin(1) %1
5 6 7
+ 1/120 cos(1) %1 - 1/720 sin(1) %1 - 1/5040 cos(1) %1
8 9 10
+ 1/40320 sin(1) %1 + 1/362880 cos(1) %1 - 1/3628800 sin(1) %1
2 4 6 8 10
%1 := - 1/2 x + 1/24 x - 1/720 x + 1/40320 x - 1/3628800 x
> taylor(", x=0, 11);
2 4
sin(1) - 1/2 cos(1) x + (1/24 cos(1) - 1/8 sin(1)) x
6 / 209 \ 8
+ (7/360 cos(1) + 1/48 sin(1)) x + |- ----- cos(1) + 1/960 sin(1)| x
\ 40320 /
/ 1259 193 \ 10 12
+ |------- cos(1) - ------ sin(1)| x + O(x )
\3628800 241920 /
> taylor(s1*convert(", polynom), x=0, 11);
2 4 6
sin(1) x + (- 1/2 cos(1) - 1/2 sin(1)) x + (7/24 cos(1) + 5/24 sin(1)) x
/ 121 \ 8 /143 4999 \ 10 12
+ |- --- cos(1) - 1/6 sin(1)| x + |--- sin(1) + ----- cos(1)| x + O(x )
\ 720 / \960 40320 /
# Find the first six terms of the Taylor series for f(x) in
# powers of x-1, if y=f(x) satisfies
# the equation x^4 + y^4 = 2 x y with y(1) = 1.
#
# We expect the answer to be 1 plus a sum of coefficients times
# powers of x-1.
> ys:= 1 + sum(a[n] * (x-1)^n, n=1..5);
ys :=
2 3 4 5
1 + a[1] (x - 1) + a[2] (x - 1) + a[3] (x - 1) + a[4] (x - 1) + a[5] (x - 1)
# Plug that in to the equation.
> x^4 + ys^4 = 2 * x * ys;
4
x + (
2 3 4 5
1 + a[1] (x - 1) + a[2] (x - 1) + a[3] (x - 1) + a[4] (x - 1) + a[5] (x - 1)
)^4 = 2 x (
2 3 4 5
1 + a[1] (x - 1) + a[2] (x - 1) + a[3] (x - 1) + a[4] (x - 1) + a[5] (x - 1)
)
# Take the difference between the two sides, and take the Taylor
# series (discarding terms in (x-1)^n for n > 5).
> g:= taylor(op(1,") - op(2,"), x=1, 6);
2 2
g := (2 a[1] + 2) (x - 1) + (6 a[1] + 2 a[2] + 6 - 2 a[1]) (x - 1)
2 3
+ (4 + 4 (a[1] + 2 a[2]) a[1] + 4 a[2] a[1] + 2 a[3] - 2 a[2]) (x - 1) +
2
(1 + 4 a[3] a[1] + 2 a[2] + 2 a[4] + 4 (2 a[2] a[1] + 2 a[3]) a[1]
2 2 4
+ (a[1] + 2 a[2]) - 2 a[3]) (x - 1) + (2 a[5] + 4 a[1] a[4]
2
+ 4 a[2] a[3] + 4 a[1] (2 a[3] a[1] + a[2] + 2 a[4])
2 5
+ 2 (a[1] + 2 a[2]) (2 a[2] a[1] + 2 a[3]) - 2 a[4]) (x - 1)
6
+ O((x - 1) )
# The "coeff" command can be used to extract any coefficient.
> coeff(g, x-1, 2);
2
6 a[1] + 2 a[2] + 6 - 2 a[1]
# The coefficient of (x-1)^n for each n from 1 to 5 should give
# us an equation.
> eqs:= { seq( coeff(g, x-1, nn) = 0, nn=1..5) };
2
eqs := {6 a[1] + 2 a[2] + 6 - 2 a[1] = 0,
2
1 + 4 a[3] a[1] + 2 a[2] + 2 a[4] + 4 (2 a[2] a[1] + 2 a[3]) a[1]
2 2
+ (a[1] + 2 a[2]) - 2 a[3] = 0,
2
2 a[5] + 4 a[1] a[4] + 4 a[2] a[3] + 4 a[1] (2 a[3] a[1] + a[2] + 2 a[4])
2
+ 2 (a[1] + 2 a[2]) (2 a[2] a[1] + 2 a[3]) - 2 a[4] = 0,
2
4 + 4 (a[1] + 2 a[2]) a[1] + 4 a[2] a[1] + 2 a[3] - 2 a[2] = 0,
2 a[1] + 2 = 0}
# Now solve this set of equations for the variables a[1] to a[5].
> solve(eqs, {a[1], a[2], a[3], a[4], a[5]});
{a[1] = -1, a[2] = -7, a[3] = -49, a[5] = -4627, a[4] = -449}
# So here is our solution:
> subs(", ys);
2 3 4 5
2 - x - 7 (x - 1) - 49 (x - 1) - 449 (x - 1) - 4627 (x - 1)
>
```<|endoftext|>
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# Converting 11/12 to a Percentage – Easy Guide
Did you know that fractions can be converted into percentages with simple calculations? Understanding how to convert fractions to percentages is a valuable skill that can be applied in a variety of mathematical and real-world scenarios. Today, we’ll explore the process of converting the fraction 11/12 into a percentage. Whether you’re a student working on a math problem or someone trying to calculate a practical percentage, this guide will provide you with the necessary steps to convert 11/12 to a percentage.
### Key Takeaways:
• To convert a fraction to a percentage, you can adjust the fraction or convert it to a decimal.
• Method 1: Adjusting the fraction involves finding a multiple that adjusts the denominator to 100 and multiplying both the numerator and denominator accordingly.
• Method 2: Converting the fraction to a decimal involves dividing the numerator by the denominator and multiplying the result by 100.
• Converting fractions to percentages is a valuable skill in mathematics and everyday life, allowing you to calculate equivalent percentages in various scenarios.
• Whether you’re solving math problems or making practical calculations, understanding fraction to percentage conversion will help you accurately determine percentages.
## Understanding Fractions and Percentages
Before diving into converting fractions to percentages, it’s important to understand the basics. A fraction consists of a numerator (the number above the fraction line) and a denominator (the number below the fraction line). Percentages, on the other hand, represent a fraction of 100.
To convert a fraction to a percentage, we need to determine how many parts of the whole (100) the fraction represents. We can do this by finding a multiple that adjusts the fraction’s denominator to 100 and then converting the adjusted fraction to a decimal or multiplying the fraction by 100.
## Method 1: Adjusting the Fraction
One method to convert a fraction to a percentage is by adjusting the fraction. In the case of 11/12, we can adjust the denominator (12) to 100 by dividing 100 by 12, resulting in a multiple of 8.3333333333333. Then, we multiply both the numerator (11) and denominator (12) by this multiple to get 91.666666666667/100. Simplifying this fraction gives us 91.6667%, which is the equivalent percentage of 11/12.
In the process of adjusting the fraction, we can use a table to illustrate the steps:
NumeratorDenominatorMultiply by MultipleSimplified FractionPercentage
11128.333333333333391.666666666667/10091.6667%
## Method 2: Converting to Decimal
Another method to convert a fraction to a percentage is by converting the fraction to a decimal. This method can be more straightforward and requires fewer steps than the previous method.
To convert 11/12 to a decimal, we divide the numerator (11) by the denominator (12), resulting in 0.91666666666667. We then multiply this decimal by 100 to obtain the percentage. In this case, 0.91666666666667 multiplied by 100 equals 91.6667%.
Using the fraction percent formula, we can understand the conversion process:
Decimal = Fraction ÷ Denominator
FractionDecimalPercentage
11/120.9166666666666791.6667%
## Real-World Examples
Converting fractions to percentages is not limited to mathematical exercises. It can be used in real-world situations as well. Let’s explore some practical examples that demonstrate the application of converting fractions to percentages.
### Tyler’s Final Exam Score
Imagine Tyler earned a score of 35 out of 40 on his final exam. To calculate the equivalent percentage, we can convert the fraction 35/40 to a decimal, which is 0.875. Multiplying this decimal by 100 gives us 87.5%. Therefore, Tyler earned an impressive 87.5% of the total score.
### Catherine’s Toilet Paper Purchase
Now, let’s consider Catherine, who recently went shopping for toilet paper. Out of a total of 17 packs available, she bought 3 packs. To determine the percentage of toilet paper she purchased, we can convert the fraction 3/17 to a decimal, which is approximately 0.176. Multiplying this decimal by 100 gives us 17.6%. Hence, Catherine purchased 17.6% of the available toilet paper.
These examples showcase how converting fractions to percentages is not only a mathematical exercise but also a practical skill used in everyday scenarios. By understanding fraction to percentage conversion, individuals can apply this knowledge to various real-world situations.
ScenarioOriginal FractionDecimalPercentage
Tyler’s Final Exam Score35/400.87587.5%
Catherine’s Toilet Paper Purchase3/170.17617.6%
## Percentage Change
Percentage change is a useful concept when comparing two quantities. It allows us to quantify the difference between the original value and the new value in terms of a percentage. This calculation can be applied to various scenarios, such as calculating sales growth, investment returns, or changes in stock prices.
To calculate the percentage change, we follow a simple formula. First, subtract the original value from the new value. Then, divide the difference by the original value. Finally, multiply the result by 100 to express it as a percentage.
“The percentage change formula: ((New Value – Original Value) / Original Value) * 100”
Let’s consider an example: the decrease in a stock price from \$67 to \$52. By subtracting \$52 from \$67, we get a difference of \$15. Dividing \$15 by \$67 and multiplying by 100 gives us a decrease of approximately 22.39%. This means that the stock price has decreased by 22.39%.
The percentage change formula is versatile and can be used in various contexts. Whether you’re analyzing financial data, tracking performance metrics, or monitoring market trends, understanding percentage change allows you to measure and compare differences effectively.
Here’s a visual representation to illustrate the calculation:
Original ValueNew ValueDifferencePercentage Change
\$67\$52\$1522.39%
As you can see, the percentage change provides a clear representation of the difference between two quantities. It’s a valuable tool for analyzing data, making informed decisions, and understanding the impact of changes. Whether you’re an investor, a business owner, or simply curious about analyzing trends, mastering the calculation of percentage changes is an essential skill.
## Method 3: Finding the Missing Whole
Sometimes, we encounter scenarios where we have a given percentage and a known “part,” but we need to determine the missing “whole.” In such cases, we can utilize a simple formula known as the fraction percent formula to find the answer we seek.
P/100 = Part/Whole
By rearranging this formula, we can solve for the missing value of the “whole.” Let’s illustrate this with an example:
If we are told that 39% of a number is equal to 89, we can set up the following equation:
39/100 = 89/Whole
By applying algebraic principles and solving for the “whole,” we can determine its actual value. This method proves tremendously helpful in cases where we only know a percentage and a specific part but seek to uncover the complete picture.
Percentage (P)Part (Part)Whole (Whole)
39%89228.21
## Method 4: Percentage Increase
When we want to compare a new value that is larger than the original value, we use the concept of percentage increase. This calculation allows us to determine the difference between the two values as a percentage of the original value. Here’s how you can calculate the percentage increase:
1. Subtract the original value from the new value. For example, if the price of an item increased from \$13.99 to \$15.75, the difference would be \$1.76.
2. Divide the difference by the original value. In this case, dividing \$1.76 by \$13.99 gives us approximately 0.12618870550468.
3. Multiply the result by 100 to get the percentage. Multiplied by 100, the result is approximately 12.62%.
This means that the price of the item increased by approximately 12.62%. This calculation can be useful in various scenarios, such as analyzing price changes, sales growth, or investment returns.
By understanding the percentage increase, you can effectively analyze and interpret changes in values, whether they are related to prices, quantities, or other measurements.
## Method 5: Percentage Decrease
When comparing a new, smaller value to an original value, we use percentage decrease. This method allows us to quantify the decrease between the two values.
To calculate the percentage decrease, we follow a similar process to calculating the percentage increase. First, subtract the new value from the original value. Then, divide the difference by the original value and multiply the result by 100. This will give us the percentage decrease.
For example, let’s say the price of a stock dropped from \$67 to \$52. By subtracting \$52 from \$67, we find that the difference is \$15. Dividing \$15 by \$67 and multiplying by 100 gives us a percentage decrease of approximately 22.39%.
In this example, the percentage decrease helps us understand the drop in stock price and evaluate the impact on investments. By utilizing this method, we can accurately assess and analyze changes in values.
## Conclusion
Converting fractions to percentages is an essential skill that finds applications in both mathematics and everyday life. Whether you choose to adjust the fraction or convert it to a decimal, the process is straightforward and allows for accurate calculations of equivalent percentages.
By mastering this skill, individuals gain confidence in their ability to convert any fraction to a percentage, enabling them to solve a wide range of problems across various real-world scenarios. Whether it’s determining the percentage of a score on an exam or calculating the portion of a total quantity, the conversion from fractions to percentages provides a clear and concise representation of the data.
Whether you’re a student, professional, or simply someone looking to enhance their mathematical knowledge, learning how to convert fractions to percentages is a valuable asset. It not only boosts your mathematical prowess but also equips you with a practical and usable skill that can be applied in numerous situations. With this skill in your toolbox, you can confidently convert fractions to percentages with ease, facilitating accurate analysis and efficient decision-making.
## FAQ
### What is 11/12 as a percentage?
To convert the fraction 11/12 to a percentage, multiply the fraction by 100. 11/12 as a percentage is 91.6667%.
### How do you calculate the percentage of 11/12?
There are two methods to calculate the percentage of 11/12. The first method is adjusting the fraction by multiplying the numerator and denominator by a multiple that converts the denominator to 100. The second method is converting the fraction to a decimal by dividing the numerator by the denominator and then multiplying by 100. Both methods yield the same result – 11/12 as a percentage is 91.6667%.
### How do you convert a fraction to a percentage?
To convert a fraction to a percentage, multiply the fraction by 100. This will give you the equivalent percentage value of the fraction.
### What is the formula to convert a fraction to a percentage?
The formula to convert a fraction to a percentage is (Numerator / Denominator) * 100, where Numerator is the number above the fraction line and Denominator is the number below the fraction line.
### How can I use the percentage of 11/12 in real life?
The percentage of 11/12 can be used in various real-life scenarios. For example, if you have 11 out of 12 correct answers on a test, you can say that you scored 91.6667%. This can also be applied to sales percentages, discounts, or any situation where you want to represent a fraction as a percentage.
### How do you calculate percentage change?
To calculate the percentage change, subtract the original value from the new value, divide the difference by the original value, and multiply by 100.
### How do you find the missing whole using a percentage?
To find the missing whole using a percentage and a part, you can use the formula P/100 = Part/Whole. By rearranging the formula, you can solve for the missing value.
### How do you calculate a percentage increase?
To calculate a percentage increase, subtract the original value from the new value, divide the difference by the original value, and multiply by 100.
### How do you calculate a percentage decrease?
To calculate a percentage decrease, subtract the new value from the original value, divide the difference by the original value, and multiply by 100.
### How can I convert fractions to percentages?
To convert fractions to percentages, you can multiply the fraction by 100. This will give you the equivalent percentage value of the fraction.<|endoftext|>
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# 2010 AMC 10B Problems/Problem 6
## Problem
A circle is centered at $O$, $\overbar{AB}$ (Error compiling LaTeX. Unknown error_msg) is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ$. What is the degree measure of $\angle CAB$?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65$
## Solution 1
Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since $O$ is the center, $OC$ and $OA$ are radii and they are congruent. Thus, $\triangle COA$ is an isosceles triangle. Also, note that $\angle COB$ and $\angle COA$ are supplementary, then $\angle COA = 180 - 50 = 130^{\circ}$. Since $\triangle COA$ is isosceles, then $\angle OCA \cong \angle OAC$. They also sum to $50^{\circ}$, so each angle is $\boxed{\textbf{(B)}\ 25}$.
## Solution 2
An inscribed angle is always half its central angle, so therefore, half of 50 is 25, or B. We can see this when we graph the problem.
(Solution by Flamedragon)<|endoftext|>
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# Solving equations by elimination
When Solving equations by elimination, there are often multiple ways to approach it. We can solve math word problems.
## Solve equations by elimination
Are you struggling with Solving equations by elimination? In this post, we will show you how to do it step-by-step. math word problem solver online can be a great resource for students who are struggling with word problems. This type of resource can provide step-by-step instructions on how to solve a problem, as well as offer tips and tricks for solving similar problems. Additionally, many online math word problem solvers will allow users to input their own problems to get customized solutions. This can be a great way for students to practice solving problems and see how different types of problems can be approached.
Algebra 1 math problems can be difficult to solve, but there are a few tips and tricks that can help. First, make sure that you understand the problem and all of the given information. Next, identify any key words or phrases that will help you solve the problem. Lastly, work through the problem step by step, using the information you have gathered to solve the equation. With a little practice, you will be solving algebra 1 math problems like a pro!
In mathematics, ln is the natural logarithm of a number, typically denoted by ln(x), loge(x), or log_e(x). If the natural logarithm is applied to a number greater than one, the result is a positive real number. For example, ln(2)= 0.693 and ln(3)= 1.097 . If the natural logarithm is applied to a number
To solve equations with both x and y variables, you'll need to use algebraic methods. First, you'll need to identify which equation is the "x equation" and which is the "y equation." Then, you'll use algebraic methods to solve for each variable in turn. Once you have both the x and y values, you can plug them into either equation to check your work.
There are a number of online word problem solvers that can be very helpful for students who are struggling with math. These solvers can provide step-by-step solutions to problems, as well as explanations of the concepts involved. This can be a great resource for students who are struggling to understand word problems or who need extra practice.
## Math solver you can trust
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Fraction solver with steps Math worker Geometry app solver Multiplication solver Math answers with work How to do pre algebra word problems<|endoftext|>
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# RD Sharma Class 12 Ex 19.16 Solutions Chapter 19 Indefinite Integrals
Here we provide RD Sharma Class 12 Ex 19.16 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.16 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.
## RD Sharma Class 12 Ex 19.16 Solutions Chapter 19 Indefinite Integrals
### Question 1. Evaluate ∫ sec2x/ 1 – tan2x dx
Solution:
Let us assume I = ∫ sec2x/ 1 – tan2x dx …..(i)
Now, put tan x = t
sec2x dx = dt
So, put all these values in eq(i)
= ∫ dt/ 1– t2
On integrating the above equation then, we get
= 1/ 2(1) log|1 + t/1 – t| + c
Since, ∫ 1/ a– x2 dx = 1/ 2a log|a + x/a – x| + c]
Hence, I = 1/2 log|1 + tanx/1 – tanx| + c
### Question 2. Evaluate ∫ ex/ 1 + e2x dx
Solution:
Let us assume I = ∫ ex/ 1 + e2x dx …..(i)
Now, put ex = t
ex dx = dt
So, put all these values in eq(i)
= dt/ 1 + t2
On integrating the above equation then, we get
= tan-1t + c
Since, ∫ 1/ 1 + x2dx = tan-1x + c
Hence, I = tan-1ex + c
### Question 3. Evaluate ∫ cosx/ sin2x + 4sinx + 5 dx
Solution:
Let us assume I = ∫ cosx/ sin2x + 4sinx + 5 dx …..(i)
Now, put sinx = t
cosx dx = dt
So, put all these values in eq(i)
= ∫ dt/ t+ 4t + 5
= ∫ dt/ t+ 2t(2) + (2)– (2)+ 5
= ∫ dt/ (t + 2)+ 1 …..(ii)
Again, Put t + 2 = u
dt = du
Now, put all these values in eq(ii)
= ∫ du/ u+ 1
On integrating the above equation then, we get
= tan-1u + c
Since, ∫1/ x+ 1 dx = tan-1x + c
= tan-1(t + 2) + c
Hence, I = tan-1(sinx + 2) + c
### Question 4. Evaluate ∫ ex/e2x + 5ex + 6 dx
Solution:
Let us assume I = ∫ ex/e2x + 5e+ 6 dx …..(i)
Now, put ex = t
ex dx = dt
So, put all these values in eq(i)
= ∫ dt/ t+ 5t + 6
= ∫ dt/ t + 2t(5/2) + (5/2)– (5/2)+ 6
= ∫ dt/ (t + 5/2)– 1/4 …..(ii)
Put t + 5/2 = u
dt = du
Now, put the above value in eq(ii)
= ∫ du/ u– (1/2)2
On integrating the above equation then, we get
= 2/2 log|u – (1/2)/u + (1/2)| + c
Since, ∫ 1/ x2 – a2 dx = 1/ 2alog|x – a/x + a| + c
= log|2u – 1/2u + 1| + c
= log|2(t + 5/2) – 1/2(t + 5/2) + 1| + c
Hence, I = log|e+ 2/e+ 3| + c
### Question 5. Evaluate ∫ e3x/ 4e6x – 9 dx
Solution:
Let us assume I = ∫ e3x/ 4e6x– 9 dx …..(i)
Now, put e3x = t
3e3x dx = dt
e3x dx = dt/3
Now, put the above value in eq(i)
= 1/3 ∫ dt/ 4t– 9
= 1/12 ∫ dt/ t– (3/2)2
On integrating the above equation then, we get
= 1/12 x 1/ 2(3/2) log|t – 3/2/t + 3/2| + c
Since, ∫1/ x– a2 dx = 1/2a log|x – a/x + a| + c]
= 1/36 log|2t – 3/2t + 3| + c
Hence, I = 1/36 log|2e3x – 3/2e3x + 3| + c
### Question 6. Evaluate ∫ dx/ex + e-x
Solution:
Let us assume I = ∫ dx/e+ e-x
= dx/e+ 1/ex
= ∫ exdx/ (ex)2 + 1 …..(i)
Now, put ex = t
exdx = dt
Now, put the above value in eq(i)
= ∫ dt/ t+ 1
On integrating the above equation then, we get
= tan-1t + c
Since ∫ 1/ 1 + x2 dx = tan-1x + c
Hence, I = tan-1(ex) + c
### Question 7. Evaluate ∫ x/ x4 + 2x2 + 3 dx
Solution:
Let us assume I = ∫ x/ x+ 2x+ 3 dx …..(i)
Now, put x2 = t
2x dx = dt
x dx = dt/2
Now, put the above value in eq(i)
= 1/2 ∫ dt/ t+ 2t + 3
= 1/2 ∫ dt/ t+ 2t + 1 – 1 + 3
= 1/2 ∫ dt/ (t + 1)2 + 2 …..(ii)
Now put t + 1 = u
dt = du
So, put the above value in eq(ii)
= 1/2 ∫ du/ u+ (√2)2
On integrating the above equation then, we get
= 1/2 x 1/√2 tan-1(u/√2) + c
Since ∫1/ x+ a2dx = 1/a tan-1(x/a) + c
= 1/2√2 tan-1(t + 1/ √2) + c
Hence, I = 1/2√2 tan-1(x+ 1/ √2) + c
### Question 8. Evaluate ∫ 3x5/ 1 + x12 dx
Solution:
Let us assume I = ∫ 3x5/ 1 + x12 dx
= ∫ 3x5/ 1 + (x6)2dx …..(i)
Now, put x= t
6x5dx = dt
x5dx = dt/6
Now, put the above value in eq(i)
= 3/6 ∫ dt/ 1 + t2
On integrating the above equation then, we get
= 1/2 tan-1(t) + c
Since ∫ 1/ x+ 1 dx = tan-1x + c
Hence, I = 1/2 tan-1(x6) + c
### Question 9. Evaluate ∫ x2/ x6 – a6 dx
Solution:
Let us assume I = ∫ x2/ x– a6 dx
= ∫ x2/ (x3)– (a3)2 dx …..(i)
Now, put x3 = t
3x2 dx = dt
x2 dx = dt/3
Now, put the above value in eq(i)
= 1/3 ∫ dt/ t– (a3)2
On integrating the above equation then, we get
= 1/3 x 1/2a3 log|t – a3/t + a3| + c
Since ∫1/ x– a2 dx = 1/2a log|x – a/x + a| + c
= 1/6a3 log|x– a3/x+ a3| + c
Hence, I = 1/6a3 log|x– a3/x+ a3| + c
### Question 10. Evaluate ∫ x2/ x6 + a6 dx
Solution:
Let us assume I = ∫ x2/ x+ a6 dx
= ∫ x2/ (x3)+ (a3)2 dx …..(i)
Now, put x3 = t
3x2 dx = dt
x2 dx = dt/3
Now, put the above value in eq(i)
= 1/3 ∫ dt/ t+ (a3)2
On integrating the above equation then, we get
= 1/3 x (1/a3) tan-1(t/a3) + c
Since, ∫1/ x+ a2 dx = 1/a tan-1(x/a) + c
Hence, I = 1/3a3 tan-1(x3/a3) + c
### Question 11. Evaluate ∫ 1/ x(x6 + 1) dx
Solution:
Let us assume I = ∫ 1/ x(x+ 1) dx
= ∫ x5/ x6(x+ 1) dx …..(i)
Now, put x6 = t
6x5 dx = dt
x5 dx = dt/6
Now, put the above value in eq(i)
= 1/6 ∫dt/ t(t + 1)
= 1/6 ∫dt/ t+ t
= 1/6 ∫dt/ t+ 2t(1/2) + (1/2)– (1/2)2
= 1/6 ∫dt/ (t + 1/2)– (1/2)2 …..(ii)
Let t + 1/2 = u
dt = du
So, put the above value in eq(ii)
= 1/6 ∫du/ (u)– (1/2)2
On integrating the above equation then, we get
= 1/6 x 1/ 2(1/2) log|u – (1/2)/u + (1/2)| + c
Since ∫ 1/ x– a2dx = 1/2a log|x – a/x + a| + c
= 1/6 log|{(t + 1/2) – 1/2}/(t + 1/2) + 1/2| + c
Hence, I = 1/6 log|x6/ x+ 1| + c
### Question 12. Evaluate ∫ x/ (x4 – x2 + 1) dx
Solution:
Let us assume I = ∫ x/ (x– x+ 1) dx …..(i)
Let x2 = t
2x dx = dt
x dx = dt/2
Now, put the above value in eq(i)
= 1/2 ∫dt/ t– t + 1
= 1/2 ∫dt/ t– 2t(1/2) + (1/2)– (1/2)+ 1
= 1/2 ∫dt/ (t – 1/2)+ (3/4) …..(ii)
Let t – 1/2 = u
dt = du
So, put the above value in eq(ii)
= 1/2 ∫du/ (u)+ (√3/2)2
On integrating the above equation then, we get
= 1/2 x 1/(√3/2) tan-1(u/(√3/2)) + c
Since, ∫ 1/ x+ a2dx = 1/a tan-1(x/a) + c
= 1/√3 tan-1(t – 1/2/ (√3/2)) + c
Hence, I = 1/√3 tan-1(2x– 1/ √3) + c
### Question 13. Evaluate ∫ x/ (3x4 – 18x2 + 11) dx
Solution:
Let us assume I = ∫ x/ (3x– 18x+ 11) dx
= 1/3 ∫ x/ (x– 6x+ 11/3) dx …..(i)
Let x2 = t
2x dx = dt
x dx = dt/2
So, put the above value in eq(i)
= 1/3 x 1/2 ∫dt/ t– 6t + 11/3
= 1/6 ∫dt/ t– 2t(3) + (3)– (3)+ 11/3
= 1/6 ∫dt/ (t – 3)– (16/3) …..(ii)
Let t – 3 = u
dt = du
Now, put the above value in eq(ii)
= 1/6 ∫du/ (u)– (4/√3)2
On integrating the above equation then, we get
= 1/6 x 1/ 2(4/√3) log|u – (4/√3)/u + (4/√3)| + c
Since, ∫ 1/ x– a2dx = 1/2a log|x – a/x + a| + c
= √3/48 log|(t – 3 – 4/√3)/(t – 3 + 4/√3)| + c
Hence, I = √3/48 log|(x– 3 – 4/√3)/(x– 3 + 4/√3)| + c
### Question 14. Evaluate ∫ ex/ (1 + ex)(2 + ex) dx
Solution:
Let us assume I = ∫ ex/ (1 + ex)(2 + ex) dx …..(i)
Let ex = t
ex dx = dt
So, put the above value in eq(i)
= ∫ dt/ (1 + t)(2 + t)
= ∫dt/ (1 + t) – ∫dt/(2 + t)
On integrating the above equation then, we get
= log|1 + t| – log|2 + t| + c
= log|1 + t/2 + t| + c
Hence, I = log|1 + ex/2 + ex| + c
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.
If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.<|endoftext|>
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New Defender's Study Bible Notes
Introduction to Nehemiah
Since Ezra and Nehemiah were once considered to be one book, and since Ezra and Nehemiah were contemporaries in post-exilic Jerusalem, there is much in common between the two books (see “Introduction” to Ezra). In fact, many of the ancient scribes believe Ezra actually wrote the first few chapters of Nehemiah, but the internal evidence strongly favors Nehemiah as the author.
Nehemiah was a high official in the court of Artaxerxes, king of Persia. As a Jew, however, he was greatly concerned about the reestablishment of Jerusalem and the temple back in Israel. Approximately fourteen years after Ezra received his decree from the king Artaxerxes, Nehemiah obtained another decree from the same king, giving him authority to rebuild the wall and the city as a whole. This was almost certainly the decree prophesied by Daniel as the beginning of the “seventy weeks” in Daniel’s famous prediction of the coming of the Messiah (Daniel 9:24-27).
Under Nehemiah’s dynamic leadership, the walls were quickly rebuilt, despite much opposition from the previous inhabitants of the land. Under Ezra’s spiritual leadership, and Nehemiah’s governmental leadership, the remnant nation experienced a significant religious revival, though it never again gained complete independence.
1:1 twentieth year. That is, the twentieth year of Artaxerxes’ reign (Nehemiah 2:1). This would be thirteen years after this same king sent Ezra to Jerusalem (Ezra 7:7).
1:1 Shushan. Shushan is the same as Susa, the winter capital of the great Persian empire, about 250 miles east of Babylon, well identified and confirmed archaeologically.
1:2 Hanani. Hanani was evidently Nehemiah’s brother (Nehemiah 7:2) and so would naturally report back to Nehemiah after making the thousand-mile, four-month (Ezra 7:9) journey from Jerusalem to Shushan. He arrived in the month Chisleu (Nehemiah 1:1), corresponding to our November-December.
1:3 broken down. The wall of Jerusalem had apparently been at least partially built by Zerubbabel (Ezra 4:12; 5:3,8). Many years had passed, however, and the returning exiles had encountered continuing serious opposition from the people of the land, so their wall had fallen into decay by Nehemiah’s time. Therefore, Nehemiah sought and obtained approval to rebuild and finish the wall (Nehemiah 2:8).
1:6 we have sinned. As had Ezra (Ezra 9:5-15), so Nehemiah assumes that he personally is associated with the sins of Israel as a nation.
1:8 scatter you abroad. God’s ancient prophetic warnings through Moses (Leviticus 26:33, etc.), as well as His more recent warnings through His prophets (e.g., Jeremiah 25:11), had been fulfilled. Nevertheless, as Nehemiah “reminded” God, He had also promised to restore them to their land if they would return to the Lord (e.g., Deuteronomy 30:1-5).
l:11 the king’s cup bearer. This position was one of high trust and responsibility, as well as close fellowship with the king. The cupbearer risked his life daily for the king, tasting all his beverages first to thwart any attempted assassination of the king by poison.<|endoftext|>
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While silicon (Si) is not considered a macronutrient for plants, it is an important nutrient and one that helps plants adapt to hostile environments. Various published research papers have concluded that Si is an effective antidote to pests and diseases caused by fungi and bacteria. Si has also been found to positively impact plants under various abiotic stresses including salt stress, metal toxicity, drought stress, radiation damage, nutrient imbalance, high temperature, freezing etc.
Silicon is an important nutrient and one that can help your plants thrive.
What Silicon Does
Until recently, Si was not recognized as an essential element for plant growth. However, recently, Epstein and Bloom (2003) have ventured to offer a new basis for including Si in the list of essential elements as Si deficiency causes various abnormalities in the plant. Beneficial effects of Si include increased photosynthetic activity, increased insect and disease resistance, reduced mineral toxicity, improvement of nutrient imbalance, and enhanced drought and frost tolerance.
Si has been reported to prevent the incidence of powdery mildew disease. In strawberry, when the Si content of leaves increased, the incidence of powdery mildew decreased (Kanto 2002). Silicon deficiency in barley and wheat leads to a poor growth habit and increased powdery mildew susceptibility (Zeyen 2002).
Foliar application of Si has been reported to be effective in inhibiting powdery mildew development on cucumber, muskmelon, and grape leaves (Bowen et al. 1992. In turfgrass, several diseases were also suppressed by Si application (Datnoff et al. 2002).
Water deficiency (drought stress) leads to the closure of stomata and subsequent decrease in the photosynthetic rate. Si can alleviate water stress by decreasing transpiration. Transpiration from the leaves occurs mainly through the stomata and partly through the cuticle. As Si is deposited beneath the cuticle of the leaves forming a Si-cuticle double layer, the transpiration through the cuticle may decrease by Si deposition. Silicon can reduce the transpiration rate by 30% in rice, which has a thin cuticle (Ma et al. 200la).
How Silicon Works in Plants
How does Silicon help plants adapt to environments and do these amazing things? Silicon like carbon can form four chemical bonds. Carbon is found in all living things used to make proteins, sugars, and DNA. Some scientists believe that life on another planet might be silicon-based instead of carbon-based (all life on earth is carbon based).
To protect plants from pathogen and pests plants incorporate silicon into their cell wall making it stronger. This stronger cell wall is harder for insects to pierce and bite.
Silicon also helps the plant adapt to abiotic stresses like salt and drought. Both of these stresses have to do with water. Because of its electrons silicon binds with water, holding on to it. When the plant is drying out more water is held in the plant by the silicon than if the plant didn’t have it. Plants with silicon have more water in them, allowing them to keep their shape, continue conducting photosynthesis, and increase the chance of survival.
There has been a considerable amount of work on the effects of Si under chemical stresses including nutrient imbalance, metal toxicity, salinity and other environmental stresses. Silicon can help plants adapt to these stresses.
13Essentials is one of the first nano scale (particle size of 1-30 nm) foliar fertilizers to become available in the US market. It contains a balanced mix of primary nutrients (P and K), secondary nutrients (Ca, Mg and S) and micronutrients (Fe, Mn, Zn, Cu, B, Mo and Co) adsorbed on a nano-silica base.
In 13Essentials, Si is both a nutrient as well as a carrier for other nutrients and micronutrients, preventing them from complexing and making them bioavailable to the plant.
13Essentials Foliar Fertilizer Spray
13Essentials is a balanced formulation of nano-scale silica plus 12 other nutrients to maximize results of plants. Compatible with all fertilizer programs, 13Essentials is a highly concentrated formula that gives plants everything needed to remain healthy throughout the growth cycle.<|endoftext|>
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Did you know that there are currently an estimated 442 million people living with diabetes in the world? According to the World Health Organization (WHO), diabetes caused 1.5 million deaths in 2012 and higher-than-optimal blood glucose caused an additional 2.2 million deaths by increasing the risks of cardiovascular and other diseases.
Those numbers will continue to increase unless we make some drastic changes.
That is why on World Health Day on April 7, WHO is issuing a call to action campaign, “Stay Super, Beat Diabetes” and launching its first Global Report on Diabetes. The new report highlights the main factors driving the increase of diabetes; challenges in countries to address these epidemics; and measures required to prevent, detect, and treat diabetes.
WHO is also calling on governments to ensure that people are able to make healthy choices and that health systems are able to diagnose, treat, and care for people with diabetes. It encourages individuals to eat healthy, be physically active, and avoid excessive weight gain.
As part of the Sustainable Development Goals, WHO Member States have set an ambitious target to reduce premature mortality from non-communicable diseases – including diabetes – by one third by 2030.
Here’s what you need to know and how you can help the world achieve that target.
Diabetes is a chronic disease with three main types: type 1 – when the pancreas does not produce enough insulin (a hormone that regulates blood glucose); type 2 – when the body cannot effectively use the insulin it produces; and gestational diabetes – when women develop hyperglycemia (raised blood sugars) during pregnancy. The increased glucose of all three types attacks the body’s organs and, over time, can lead to heart attacks, strokes, kidney failure, leg amputations, vision loss, and nerve damage.
The biggest risk factor for type 2 diabetes is excess body fat. Other risk factors include family history of diabetes, unhealthy diet, physical inactivity, increasing age, high blood pressure, and ethnicity.
Most type 2 diabetes cases are preventable, and all diabetes cases are treatable. Simple lifestyle measures have been shown to be effective in preventing or delaying the onset of type 2 diabetes. Maintaining normal body weight, engaging in regular physical activity, eating a healthy diet, and avoiding tobacco use can reduce the risk of diabetes.
Most diabetes deaths occur in low- and middle-income countries, which is an equity issue because many doctors in low-income countries do not even have access to the basic tests and services to diagnose and treat people with diabetes. Increasing access to diagnosis, self-management education, and affordable treatment are vital components of the response to beat diabetes.
Want to help raise awareness so we can beat diabetes? Check out WHO’s Stay Super, Beat Diabetes campaign and WHO’s World Health Day information to learn more. Also watch and share WHO’s Halt to Rise video to learn more about how diabetes works. You can also join and share the conversation online with #Diabetes.<|endoftext|>
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This is a comprehensive 12 week course that teaches the big ideas behind life science. Beginning with plant science and ending with DNA study, students will discover the science behind the barely visible parts of life and the environment, such as plant structure, plant processes, tiny insects, cell division, and genetics. Students will construct and use apparatuses such as a water cycle column, a terraqua column and a carnivorous greenhouse to investigate and identify components of different ecosystems. They will also learn proper use and care of sophisticated equipment such as the compound microscope to further observe and record data in the living world around us.
You will find 24 lessons outlined to take you from an introduction of life science on through several advanced projects complex enough to win a prize at the science fair. Includes students worksheets, instructional videos, daily lessons, and tests and quizzes… all in one complete package.<|endoftext|>
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# Lesson 4
Coordinate Moves
Let’s transform some figures and see what happens to the coordinates of points.
### 4.1: Translating Coordinates
Select all of the translations that take Triangle T to Triangle U. There may be more than one correct answer.
1. Translate $$(\text-3,0)$$ to $$(1,2)$$.
2. Translate $$(2,1)$$ to $$(\text-2,\text-1)$$.
3. Translate $$(\text-4,\text-3)$$ to $$(0,\text-1)$$.
4. Translate $$(1,2)$$ to $$(2,1)$$.
### 4.2: Reflecting Points on the Coordinate Plane
1. Five points are plotted on the coordinate plane.
1. Using the Pen tool or the Text tool, label each with its coordinates.
2. Using the $$x$$-axis as the line of reflection, plot the image of each point.
3. Label the image of each point with its coordinates.
4. Include a label using a letter. For example, the image of point $$A$$ should be labeled $$A’$$.
2. If the point $$(13,10)$$ were reflected using the $$x$$-axis as the line of reflection, what would be coordinates of the image? What about $$(13, \text-20)$$? $$(13, 570)$$? Explain how you know.
3. The point $$R$$ has coordinates $$(3,2)$$.
1. Without graphing, predict the coordinates of the image of point $$R$$ if point $$R$$ were reflected using the $$y$$-axis as the line of reflection.
2. Check your answer by finding the image of $$R$$ on the graph.
3. Label the image of point $$R$$ as $$R’$$.
4. What are the coordinates of $$R’$$?
4. Suppose you reflect a point using the $$y$$-axis as line of reflection. How would you describe its image?
### 4.3: Transformations of a Segment
The applet has instructions for the first 3 questions built into it. Move the slider marked “question” when you are ready to answer the next one. Pause before using the applet to show the transformation described in each question to predict where the new coordinates will be.
Apply each of the following transformations to segment $$AB$$. Use the Pen tool to record the coordinates.
1. Rotate segment $$AB$$ 90 degrees counterclockwise around center $$B$$ by moving the slider marked 0 degrees. The image of $$A$$ is named $$C$$. What are the coordinates of $$C$$?
2. Rotate segment $$AB$$ 90 degrees counterclockwise around center $$A$$ by moving the slider marked 0 degrees. The image of $$B$$ is named $$D$$. What are the coordinates of $$D$$?
3. Rotate segment $$AB$$ 90 degrees clockwise around $$(0,0)$$ by moving the slider marked 0 degrees. The image of $$A$$ is named $$E$$ and the image of $$B$$ is named $$F$$. What are the coordinates of $$E$$ and $$F$$?
4. Compare the two 90-degree counterclockwise rotations of segment $$AB$$. What is the same about the images of these rotations? What is different?
Suppose $$EF$$ and $$GH$$ are line segments of the same length. Describe a sequence of transformations that moves $$EF$$ to $$GH$$.
### Summary
We can use coordinates to describe points and find patterns in the coordinates of transformed points.
We can describe a translation by expressing it as a sequence of horizontal and vertical translations. For example, segment $$AB$$ is translated right 3 and down 2.
Reflecting a point across an axis changes the sign of one coordinate. For example, reflecting the point $$A$$ whose coordinates are $$(2,\text-1)$$ across the $$x$$-axis changes the sign of the $$y$$-coordinate, making its image the point $$A’$$ whose coordinates are $$(2,1)$$. Reflecting the point $$A$$ across the $$y$$-axis changes the sign of the $$x$$-coordinate, making the image the point $$A’’$$ whose coordinates are $$(\text-2,\text-1)$$.
Reflections across other lines are more complex to describe.
We don’t have the tools yet to describe rotations in terms of coordinates in general. Here is an example of a $$90^\circ$$ rotation with center $$(0,0)$$ in a counterclockwise direction.
Point $$A$$ has coordinates $$(0,0)$$. Segment $$AB$$ was rotated $$90^\circ$$ counterclockwise around $$A$$. Point $$B$$ with coordinates $$(2,3)$$ rotates to point $$B’$$ whose coordinates are $$(\text-3,2)$$.
### Glossary Entries
• clockwise
Clockwise means to turn in the same direction as the hands of a clock. The top turns to the right. This diagram shows Figure A turned clockwise to make Figure B.
• coordinate plane
The coordinate plane is a system for telling where points are. For example. point $$R$$ is located at $$(3, 2)$$ on the coordinate plane, because it is three units to the right and two units up.
• counterclockwise
Counterclockwise means to turn opposite of the way the hands of a clock turn. The top turns to the left.
This diagram shows Figure A turned counterclockwise to make Figure B.
• image
An image is the result of translations, rotations, and reflections on an object. Every part of the original object moves in the same way to match up with a part of the image.
In this diagram, triangle $$ABC$$ has been translated up and to the right to make triangle $$DEF$$. Triangle $$DEF$$ is the image of the original triangle $$ABC$$.
• reflection
A reflection across a line moves every point on a figure to a point directly on the opposite side of the line. The new point is the same distance from the line as it was in the original figure.
This diagram shows a reflection of A over line $$\ell$$ that makes the mirror image B.
• rotation
A rotation moves every point on a figure around a center by a given angle in a specific direction.
This diagram shows Triangle A rotated around center $$O$$ by 55 degrees clockwise to get Triangle B.
• sequence of transformations
A sequence of transformations is a set of translations, rotations, reflections, and dilations on a figure. The transformations are performed in a given order.
This diagram shows a sequence of transformations to move Figure A to Figure C.
First, A is translated to the right to make B. Next, B is reflected across line $$\ell$$ to make C.
• transformation
A transformation is a translation, rotation, reflection, or dilation, or a combination of these.
• translation
A translation moves every point in a figure a given distance in a given direction.
This diagram shows a translation of Figure A to Figure B using the direction and distance given by the arrow.
• vertex
A vertex is a point where two or more edges meet. When we have more than one vertex, we call them vertices.
The vertices in this polygon are labeled $$A$$, $$B$$, $$C$$, $$D$$, and $$E$$.<|endoftext|>
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# How do you solve -2x - 7 + 3x =10?
May 29, 2018
$x = 17$
#### Explanation:
$\text{collect like terms on left side of equation}$
$x - 7 = 10$
$\text{add 7 to both sides}$
$x \cancel{- 7} \cancel{+ 7} = 10 + 7$
$x = 17$
$\textcolor{b l u e}{\text{As a check}}$
Substitute thus value into the left side of the equation and if equal to the right side then it is the solution.
$\left(- 2 \times 17\right) - 7 + \left(3 \times 17\right) = - 34 - 7 + 51 = 10$
$x = 17 \text{ is the solution}$<|endoftext|>
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Groos argued that whilst the exact form of the play is impacted by the culture around the child the categories of play cross cultures. So for example whilst play fighting occurs in all cultures playful sword fighting only occurs in cultures familiar with swords.
These categories of play are:
- Physical Play
- Language Play
- Exploratory Play
- Constructive Play
- Fantasy Play
- Social Play
These types of play do not exist in isolation and as Peter Gray points out in his thought provoking book Free to Learn " A lively outdoor group game may be physical play, language play, exploratory play, constructive play, fantasy play, and social play all at once." Gray goes on to make a definition of what play is and defines it in five ways:
- Play is self chosen and self directed
- Play is an activity in which means are more valued than ends
- Play has structure or rules that are not dictated by physical necessity but emanate from the minds of players
- Play is imaginative
- Play involves an active alert, but non-stressed frame of mind
What leaps out to me about this list is that if we replaced the word play with learning virtually all educators I know would agree whole heatedly that this was a wonderful list of what we seek to achieve in our classrooms.
- Entirely teacher directed
- Valued only as a means to achieve a grade or other external reward
- Has a structure and rules emanating from the minds of the teachers rather than the students
- Lacks imagination
- Involves passive students in a stressed frame of mind
I would not go as far as Gray does in his book in rejecting notions of traditional schooling, and would question whether the success he attributes to the alternative schooling system at Sudbury Valley, is not, in part, due to the Cultural Capital that the parents and students who go there possess.
I do however feel that all educators need to look reflectively at how we can promote a playful state of mind in our students. Why is it that so many students and adults alike view learning not as play but as work.
I would argue that often the reason we are unable to achieve this is that, whilst intellectually we may agree, on a core level many people do not accept Vygotsky's argument that , “A child’s greatest achievements are possible in play, achievements that tomorrow will become her basic level of real action.”
As adults we have to resist our impulses to intervene too early and allow children time to explore and play with ideas. Whilst observing closely and being supportive we need to trust that given this freedom children will push themselves to the limit of their ability.
Too often this lack of trust is obvious in the place where free play should be the most valued, the playground. In particular teachers concern about rough and tumble play where,"the roughness of play is perceived by teachers and playground supervisors as potential problems. However, the potential benefits, such as conflict resolution training and motor skills training, are overlooked by many teachers." (On the Child’s Right to Play Fight), means that even in the playground play is heavily regulated.
Once in the classroom the time and space to play with objects and ideas is often limited or closely prescribed leaving students to feel that learning is something that is done to them not something they do.
Don't get me wrong I am not arguing that teachers should not have a clear idea of where they will take student learning (see my previous post Four implications of structured inquiry). What I am arguing is that learning can be done most effectively if teachers have the confidence to give the students increased ownership over their learning and time to play with ideas without fear of the final product being 'wrong'. In this way students will be empowered to go above and beyond the intended learning of the teacher and really reach their maximum level of achievement.
Whilst important things like high stakes external exams do place limitations on how far this can be achieved, hopefully the playful mindedness questions below will help you reflect upon this in your classroom, whatever age of student you teach.
I'd love to hear back from you whether they have enabled you to move learning in a more playful direction or whether you have other questions to provoke playful learning.
- How do I let my students know that I have confidence in them as self motivated learners?
- How much choice do your students have about how and what they learn?
- Is the learning space your classroom or the students classroom?
- How much input do students have in setting up the rules and norms of the classroom?
- Is there time and space for students to 'play' with objects or ideas?
- How is it clear that imagination and creativity is valued?
- Which word do I use most in class 'play' or 'work'?<|endoftext|>
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ENG 1A T/TH (3:30-4:50)
17 April 2018
The American Dream has many different definitions throughout literacy and is different for many Americans and non-Americans throughout the United States. The literary definition is, the ideal that every US citizen should have an equal opportunity to achieve success and prosperity through hard work, determination, and initiative. While this is the literary definition, it does not include non-US citizens which have to work just as hard if not harder than US born Americans because of the culture change and different opportunities. Many immigrants come to the United States to paint their own picture of the American Dream and be successful as they see it, however, there are struggles that come with them and working to be successful for them sometimes proves even harder. “The Book of Unknown Americans” is a story that shows examples of families struggling to create their own version of the American Dream and the different paths they take to accomplish their goals and be successful in their own eyes. There are also current examples of immigrants today that have their own struggles coming to America and also want their stories shared because everyone’s American Dream is different.
Within the novel “The Book of Unknown Americans” a scene early on in the story is about a family, called the Riveras, coming from Mexico and struggling to get adjusted to the new culture and also the language barrier. The Riveras arrive by truck not knowing where to go for food which they stop at a gas station, they struggle paying for the food because they didn’t know
how much money it costs. This immediately shows the change coming to America because the family didn’t know where to go for food which many American citizens don’t have that problem because of being already familiar with the surroundings. The housing conditions in which they live in are horrible as they only have one mattress, the bathroom is in bad condition as the toilet is rusted, and the shower doesn’t have a curtain or door. The mother, Alma, describes the feeling of having nothing in America “We had bundled up our old life and left it behind, and then hurtled into a new one with only a few of our things, each other, and hope”(Henriquez, 6). Many Americans have well established homes and live well already as they try to provide for their own families that don’t have to worry about conditions, this part of the book shows that immigrants have to start with almost nothing and have to be grateful that they even get the opportunity to live in America,while still working on their dream to be successful in America.
Everyone has their own story of why they come to the United States, which is why everyone’s American Dream is different in a way. While being successful in the workplace and owning a house is part of the American Dream, some immigrants come in hopes to get help and seek guidance through situations that only make sense for them to come to the United States. In the story, the Riveras, come to America because their daughter has brain problems that occured from an accident and struggles with learning. “Her body smacked against the mud, sending it splattering into the air, all over me, all over Luis. Her neck snapped back. Her eyes closed.”(Henriquez, 100), Alma describes the accident in which her daughter falls off a ladder. This forces the family to make the decision to come to America in hopes that the schools in America can help their daughter and hopefully make her better. Many immigrants are sometimes forced to try and come to America because they see no other option, an example would be families leaving their home countries because of war and they seek safety which they believe the United States can provide.
Another family within the story, the Toros, are another example of leaving their home because of war and looking for safety in the United States. The father, Rafael Toro, explains that their country was invaded and the fighting was too much to bare, so him and his wife made the decision to leave the country they love to go to America in hopes of safety, he was lucky enough to get a job and provide for his family. He states “I feel gratified when I see Enrique and Mayor doing well here. Maybe they wouldn’t have done so well in Panama. Maybe they wouldn’t have had the same opportunities. So that makes coming here worth it”(Henriquez, 23) This shows that some immigrants aren’t looking for the same type of success as others when they come to America, as long as they can provide for their family and give their children a chance to be even more successful, than they are living their own dream.
The article “The Transformation of the American Dream” focuses on what the current President of the United States sees the American Dream and compares it to what it’s definition was back in history. The article writes that President Trump has suggested that the American Dream is about owning a nice house and running a successful business, which wasn’t the case in past years depending on the situation at the time. In past years, the American Dream focused on freedom, mutual respect and equality of opportunity. This idea that owning a home and trying to be successful by running their own business had consequences in the past which affected many families believing the definition by the Presidents who encourage huge house spending. The article gives definitions of the American Dream from the past which changes over time, in the 1930s it focused on ideals values, in the 1950s it focused on freedom and equality after World War Two, and it keeps changing depending on current events around the world.
All of these examples of the American Dream show that it is different between every family and generation throughout the United States and will be constantly changing. “The Book of Unknown Americans” is a great book showcasing the ideas of the American Dream being different for everyone in the story because they all come to the United States for different reasons. The American Dream has changed over periods of time given how the world is shaped and what is the major concern at the time which impacts how Americans think they should live their lives. Both US citizens and immigrants have their own ideas of how success looks like in the United States and want equal opportunity to live out their own dream even if it doesn’t mean they will be wealthy. The American Dream is what makes the United States special because there is a mix of culture like no other country that everyone should embrace and have equal opportunity to get what they want in the United States.
HenriÌquez, Cristina. The Book of Unknown Americans. Vintage Books, a Division of Random House LLC, 2015.
Shiller, Robert J. “The Transformation of the ‘American Dream’.” The New York Times, The New York Times, 4 Aug. 2017, http://www.nytimes.com/2017/08/04/upshot/the-transformation-of-the-american-dream.html.<|endoftext|>
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What is generally referred to as an ear infection, is medically known as otitis media or AOM. These ear infections can have an effect on adults and children alike, particularly after a sinus infection or a cold. Even a bad tooth can trigger an ear infection.
When you have an infection in the middle ear you will most likely have at least some hearing loss, but will it go away? To find a precise answer can be rather complicated. There are many things happening with ear infections. You should learn how the damage caused by ear infections can have an impact on your hearing.
Otitis Media, Exactly What is it?
To put it simply, otitis media is an infection of the middle ear. Bacteria is the most common cause, but it may be caused by any micro-organism.
Ear infections are defined by where they appear in the ear. The outer ear, which is medically known as the pinna, is the part of the ear where swimmer’s ear develops, which is called otitis externa. If the bacterial growth occurs in the cochlea, the term is labyrinthitis or inner ear infection.
The area in front of the cochlea but behind the eardrum is known as the middle ear. This area houses the three ossicles, or very small bones, that vibrate the membranes of the inner ear. An infection in this area tends to be very painful because it puts a lot of pressure on the eardrum, in most cases until it actually breaks. That pressure is also why you can’t hear very well. The infectious material accumulates and blocks the ear canal enough to obstruct the movement of sound waves.
A middle ear infection has the following symptoms:
- Leakage from the ear
- Pain in the ear
- Reduced hearing
Usually, hearing will come back eventually. The pressure goes away and the ear canal opens up. The issue will only be resolved when the infection is resolved. Sometimes there are complications, however.
Chronic Ear Infections
At least once in their life, most people get an ear infection. For others, the issues become chronic, so they have infections over and over. Because of complications, these people’s hearing loss is more serious and can possibly become permanent.
Conductive Hearing Loss From Ear Infections
Ear infections can lead to conductive hearing loss. In other words, sound waves can’t make it to the inner ear at the proper intensity. The ear has mechanisms along the canal that amplify the sound wave so by the time it gets to the tiny hair cells of the inner ear, it is strong enough to create a vibration. Sometimes things change along this route and the sound is not correctly amplified. This is known as conductive hearing loss.
Bacteria don’t simply sit and do nothing in the ear when you have an ear infection. They must eat to survive, so they break down those mechanisms that amplify sound waves. Usually, this kind of damage includes the eardrum and those tiny little bones. It doesn’t take very much to break down these fragile bones. Once they are gone, they stay gone. That’s permanent damage and your hearing won’t return on its own. In some cases, surgeons can put in prosthetic bones to repair hearing. The eardrum may have some scar tissue once it repairs itself, which will affect its ability to vibrate. Surgery can deal with that, also.
Can This Permanent Damage be Avoided?
It’s essential to see a doctor if you think you might have an ear infection. The sooner you get treatment, the better. Always get chronic ear infection examined by a doctor. The more serious the infections you have, the more harm they will cause. Finally, take the appropriate steps to lessen colds, allergies, and sinus infections because that is where ear infections normally start. If you are a smoker, now is the right time to quit, too, because smoking multiplies your risk of getting chronic respiratory troubles.
If you are still having difficulty hearing after getting an ear infection, consult a doctor. Other things can cause conductive hearing loss, but it may be possible that you may have some damage. Hearing aids are very helpful if you have permanent loss of hearing. You can schedule an appointment with a hearing specialist to get more info about hearing aids.<|endoftext|>
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- stapler, ball, paper, aluminum foil, blocks, toys, books, and other everyday objects to make identifiable sounds that are different from each other
This activity center helps children identify an object by the sound it makes, even when they can’t see the object. Line the objects up and put them on the floor against a wall or on a shelf. One child makes a sound with one of the objects, while the rest of the children, with their backs turned, try to guess what the object is and how the sound was made. Ask questions such as: Was the paper waved back and forth, or was it crunched into a ball? Model the activity, then have children take turns making a sound for the others to guess.
Extending Sound Detectives: Every other day, change some of the objects.<|endoftext|>
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# ESTIMATING QUOTIENTS
## About "Estimating quotients"
Estimating quotients :
When we a divide a number by another number, the result what we get is called quotient.
For example,
If \$250 is divided equally to four people, then the share of each person is \$50. This \$50 is called quotient.
Because, when we divide 250 by 50, we get the result 4.
We can estimate quotient using the formula given below.
In case, we use long division to divide a number by another a number, we can estimate quotient using the formula given below.
## Estimating quotients - Practice problems
Problem 1 :
Estimate quotient when 200 is divided by 50.
Solution :
Hence, the estimation of quotient when 200 divided by 50 is 4.
Problem 2 :
Estimate quotient when 78 is divided by 13.
Solution :
78 ÷ 13 = 78/13 = 6
Because 13 goes into 78 by 6 times.
Hence, the estimation of quotient when 78 divided by 13 is 6.
Problem 3 :
A local zoo had a total of 98,464 visitors last year. The zoo was open every day except three holidays. On average, how many visitors did the zoo have each day ?
Solution :
Total number of days in an ordinary year = 365.
The zoo was open every day except three holidays.
So, total number of days that zoo opened = 362
Total number of visitors in the year = 98646
Therefore, number of visitors visited the zoo each day is nothing but the quotient when we divide 98464 by 362.
Let us use long division to estimate the quotient when we divide 98464 by 362.
Step 1 :
362 is greater than 9 and 98, so divide 984 by 362. Place the first digit in the quotient in the hundreds place. Multiply 2 by 362 and place the product under 984. Subtract.
Step 2 :
Bring down the tens digit. Divide 2,606 by 362. Multiply 7 by 362 and place the product under 2,606. Subtract.
Step 3 :
Bring down the ones digit. Divide the ones.
When we divide 98464 by 362, we get the quotient 272.
Hence, the average number of visitors per day last year was 272.
Problem 4 :
Marco paid \$39 for 8 tubes of oil paint. About how much did each tube cost ?
Solution :
Cost of each tube is nothing but the quotient when we divide 39 by 8.
Let us use long division to estimate the quotient when we divide 39 by 8.
Since 39 is not evenly divisible by 8, place a decimal point in the quotient. Add zeros to the right of the decimal point as needed.
Round to the nearest hundredth to find the dollar amount.
Therefore, the cost of each tube is about \$4.88.
Problem 5 :
A school track is 9.76 meters wide. It is divided into 8 lanes of equal width. How wide is each lane ?
Solution :
The width of each lane is nothing but the quotient when we divide 9.76 by 8.
Let us use long division to estimate the quotient when we divide 9.76 by 8.
When we use the digit which comes after the decimal point, place a decimal point in the quotient directly above the decimal point in the dividend.
Therefore, each lane is 1.22 meters wide
Problem 6 :
Aerobics classes cost \$153.86 for 14 sessions. What is the fee for one session ?
Solution :
The fee for one session is nothing but the quotient when we divide 153.86 by 14.
Let us use long division to estimate the quotient when we divide 153.86 by 14.
When we use the digit which comes after the decimal point, place a decimal point in the quotient directly above the decimal point in the dividend.
Therefore, the fee for one aerobics class is \$10.99.
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Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6<|endoftext|>
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# Adding Unlike Rational Numbers Worksheet
A Reasonable Numbers Worksheet will help your child become more knowledgeable about the principles right behind this rate of integers. In this worksheet, college students can remedy 12 various troubles related to rational expressions. They are going to discover ways to grow a couple of numbers, group them in pairs, and figure out their products. They will also practice simplifying realistic expression. Once they have perfected these methods, this worksheet will be a important instrument for advancing their scientific studies. Adding Unlike Rational Numbers Worksheet.
## Realistic Numbers really are a ratio of integers
There are two forms of phone numbers: rational and irrational. Realistic phone numbers are understood to be entire phone numbers, while irrational phone numbers do not recurring, and also have an endless number of digits. Irrational figures are no-zero, low-terminating decimals, and square origins which are not excellent squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To establish a rational number, you must know just what a realistic quantity is. An integer can be a complete quantity, and a rational quantity is actually a proportion of two integers. The percentage of two integers may be the variety ahead divided by the amount at the base. If two integers are two and five, this would be an integer, for example. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be made in to a portion
A reasonable variety includes a numerator and denominator which are not no. Because of this they can be conveyed being a small fraction. Together with their integer numerators and denominators, realistic numbers can furthermore have a unfavorable importance. The negative importance must be placed to the left of and its total importance is its length from no. To make simpler this instance, we shall state that .0333333 is a portion that can be published like a 1/3.
Together with adverse integers, a realistic number can even be produced in to a small percentage. For instance, /18,572 can be a realistic variety, whilst -1/ is not. Any fraction comprised of integers is realistic, given that the denominator fails to consist of a and will be created for an integer. Also, a decimal that ends in a point can be another realistic variety.
## They are sensation
Despite their label, logical numbers don’t make much sense. In mathematics, they are solitary entities by using a special size in the amount line. Which means that once we count up anything, we can easily purchase the dimensions by its rate to its original volume. This holds accurate even though there are endless rational phone numbers involving two certain phone numbers. In other words, numbers should make sense only if they are ordered. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To get the duration of a pearl, for example, we could count up its breadth. Just one pearl weighs twenty kgs, which is a realistic quantity. Moreover, a pound’s bodyweight is equal to 15 kilos. Hence, we should certainly split a lb by 15, without the need of be worried about the length of an individual pearl.
## They may be indicated as being a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal amount could be composed as being a numerous of two integers, so 4 times 5 is equal to 8-10. An identical issue involves the repetitive fraction 2/1, and either side ought to be separated by 99 to find the right respond to. But how can you have the conversion? Here are several examples.
A logical variety may also be designed in many forms, which include fractions and a decimal. A great way to stand for a logical variety within a decimal is to break down it into its fractional comparable. There are three ways to split a rational variety, and all these ways brings its decimal equal. One of these brilliant approaches is usually to break down it into its fractional equal, and that’s what’s referred to as a terminating decimal.<|endoftext|>
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What is nausea?
Nausea is the sensation of having to vomit or having the urge to vomit. Vomiting may or may not occur as a result of nausea. Some describe the sensation of nausea as unsettled feeling in the stomach or queasiness. Nausea is not a disease but a symptom of a variety of conditions, ranging from viral and bacterial infections to motion sickness, to food poisoning, to abscesses of the brain. Certain medications can cause nausea.
If a child has nausea, it is essential to find the underlying cause of the symptom. Physical exam and blood tests are usually the first steps. Additional tests and procedures may be needed depending on the findings from the physical exam and the blood analysis.
Because nausea is a symptom rather than a condition, treatment depends on the underlying cause.
When to Call for Help
Call your pediatrician if your child’s nausea is accompanied by vomiting that doesn’t go away after 24 hours, if the nausea persists or if the child complains of stomach pains or a headache and stiff neck. Nausea and/or vomiting with pain in the lower right corner of the abdomen may mean that your child has appendicitis, an inflammation of the appendix, which usually requires prompt surgical removal. Severe headache with stiff neck and nausea or vomiting can be symptoms of meningitis.<|endoftext|>
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# Closest-Pair Problem: Divide and Conquer Brute force approach requires comparing every point with every other point Brute force approach requires comparing.
## Presentation on theme: "Closest-Pair Problem: Divide and Conquer Brute force approach requires comparing every point with every other point Brute force approach requires comparing."— Presentation transcript:
Closest-Pair Problem: Divide and Conquer Brute force approach requires comparing every point with every other point Brute force approach requires comparing every point with every other point Given n points, we must perform 1 + 2 + 3 + … + n-2 + n-1 comparisons. Given n points, we must perform 1 + 2 + 3 + … + n-2 + n-1 comparisons. Brute force O(n 2 ) Brute force O(n 2 ) The Divide and Conquer algorithm yields O(n log n) The Divide and Conquer algorithm yields O(n log n) Reminder: if n = 1,000,000 then Reminder: if n = 1,000,000 then n 2 = 1,000,000,000,000 whereas n 2 = 1,000,000,000,000 whereas n log n = 20,000,000 n log n = 20,000,000
Closest-Pair Algorithm Given: A set of points in 2-D
Closest-Pair Algorithm Step 1: Sort the points in one D
Lets sort based on the X-axis O(n log n) using quicksort or mergesort 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm
Step 2: Split the points, i.e., Draw a line at the mid-point between 7 and 8 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm Sub-Problem 1Sub-Problem 2
Advantage: Normally, we’d have to compare each of the 14 points with every other point. (n-1)n/2 = 13*14/2 = 91 comparisons 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm Sub-Problem 1Sub-Problem 2
Advantage: Now, we have two sub-problems of half the size. Thus, we have to do 6*7/2 comparisons twice, which is 42 comparisons 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm d1 d2 Sub-Problem 1Sub-Problem 2 solution d = min(d1, d2)
Advantage: With just one split we cut the number of comparisons in half. Obviously, we gain an even greater advantage if we split the sub-problems. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm d1 d2 Sub-Problem 1Sub-Problem 2 d = min(d1, d2)
Problem: However, what if the closest two points are each from different sub-problems? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm d1 d2 Sub-Problem 1Sub-Problem 2
Here is an example where we have to compare points from sub-problem 1 to the points in sub- problem 2. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm d1 d2 Sub-Problem 1Sub-Problem 2
However, we only have to compare points inside the following “strip.” 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm d1 d2 Sub-Problem 1Sub-Problem 2 dd d = min(d1, d2)
Step 3: But, we can continue the advantage by splitting the sub-problems. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm
Step 3: In fact we can continue to split until each sub-problem is trivial, i.e., takes one comparison. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm
Finally: The solution to each sub-problem is combined until the final solution is obtained 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm
Finally: On the last step the ‘strip’ will likely be very small. Thus, combining the two largest sub- problems won’t require much work. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm
1 2 3 4 5 6 7 8 9 10 11 12 13 14 Closest-Pair Algorithm In this example, it takes 22 comparisons to find the closets-pair. The brute force algorithm would have taken 91 comparisons. But, the real advantage occurs when there are millions of points.
Closest-Pair Problem: Divide and Conquer Here is another animation: Here is another animation: http://www.cs.mcgill.ca/~cs251/ClosestPair/Close stPairApplet/ClosestPairApplet.html http://www.cs.mcgill.ca/~cs251/ClosestPair/Close stPairApplet/ClosestPairApplet.html http://www.cs.mcgill.ca/~cs251/ClosestPair/Close stPairApplet/ClosestPairApplet.html http://www.cs.mcgill.ca/~cs251/ClosestPair/Close stPairApplet/ClosestPairApplet.html
Similar presentations<|endoftext|>
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Order Fractions with Equal Numerators
Start Practice
## How to Order Fractions With the Same Numerators
So far, you've learned that a fraction tells us about parts of a whole.
Fractions have a denominator and a numerator.
The denominator tells us about the total number of equal parts.
This rectangle is divide into 3 parts.
1/3 of the rectangle is blue.
### Comparing Fractions
Which fraction is larger?
The numerators are the same, but their denominators are not.
Here's a drawing for each fraction:
By looking at the colored parts, we see that 1/4 is greater.
1/4 > 1/8
Another way to compare fractions is by looking at their numerators and denominators.
1️⃣ First, we look at their numerators.
They have the same numerators.
2️⃣ Next, we look at their denominators and compare them.
A smaller denominator means the whole is divided into less parts.
If a whole is divided into less parts, then each part is bigger.
So a smaller denominator means a bigger fraction.
That's why:
1/4 > 1/8.
Now let's try to order, or sort, some fractions.
### Ordering Fractions
Sort these fractions from smallest to greatest.
They have the same numerators.
So, we compare their denominators.
How do we order them from smallest to greatest?
1/8 is smallest and 1/4 is greatest.
Great job! 👏
Think you got it? Try the practice to see.
Start Practice
Complete the practice to earn 1 Create Credit
Teachers:
Assign to students
Duplicate
Edit Questions
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Assign
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Edit
Duplicate<|endoftext|>
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A noun is a word that represents a thing, either concrete (e.g., a chair, a dog) or abstract (an idea, happiness).
In Spanish, all nouns have a gender – they are either masculine or feminine. It is very important to learn a noun’s gender along with the noun itself because definite articles, indefinite articles, adjectives, and pronouns have to agree with nouns; that is, they change depending on the gender of the noun they modify or replace. The gender of some nouns makes sense (hombre f[man] is masculine, mujer [woman] is feminine) but others don’t (persona [person] is always feminine, even if the person is a man!) The best way to learn the gender of nouns is to make your vocabulary lists with the definite or indefinite article. That is,
Rather than lists like this …
- libro = book
- flor = flower BAD LIST 🙁
Make lists like this …
- un libro = book
- una flor = flower GOOD LIST 🙂
so that you learn the gender with the noun. The gender is part of the noun and you will be much better off learning it now, as a beginner, than trying to go back after years of study and memorizing the genders of all the words you’ve already learned (I speak from experience).
There are some tendencies in the gender of nouns, but there are always exceptions. I will list the patterns that I have noticed, but please don’t use these as a way to avoid learning the genders of nouns – just learn each word as gender + noun and then you’ll know them forever.
|This ending…||is usually|
- Gender of Nouns (Worksheet, 6th grade)
- The Gender Nouns (Coloring page, 8th-9th grade)
- Making Nouns Plural (Worksheet, 7th-8th grade)
- Sustantivos (PPT, 6th-12th grade)<|endoftext|>
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Women and Girls in Science is key to breaking gender stereotyping
As the world marks the International Day of Women and Girls in Science on 11 February it is crucial to demand inclusion of women in the science to achieve gender equality and the Sustainable Development Goals.
Humana People to People is for inclusion of girls and young women in education.
The International Day of Women and Girls in Science was adopted by the United Nations General Assembly to promote full and equal access to and participation in science for women and girls. This Day is a reminder that women and girls play a critical role in science and technology that their participation should be strengthened. The celebration is led by UNESCO and UN-Women, in collaboration with institutions and civil society partners that promote women and girls' access to and participation in science.
Humana People to People is for the active participation of women and girls inorder to bring in other science research perspectives which can advance humanity. Women and girls continue being excluded from participating fully in science: a UNESCO background paper on women and girls in science shows that less than 30% of researchers worldwide are women.
Tackling some of the greatest challenges of the Agenda for Sustainable Development, from improving health to combating climate change will rely on harnessing all talent. That means getting more women working in these fields. Diversity in research expands the pool of talented researchers, bringing in fresh perspectives, talent and creativity.
The members of Humana People to People in India and Mozambique are empowering girls and women to acquire confidence and venture in male dominated science fields. The starting point is to inspire girls to stay in school in marginalized and rural communities. Measures being taken build an enabling environment for equal participation of the women and girls and thus actions on addressing gender imbalance. The Humana People to People members improve the learning environment at the local schools by strengthening their water and sanitation conditions and by sensitizing teachers, students, and school council and community members about gender issues, including gender roles and stereotypes, violence against women and girls, sexual abuse and bullying, thereby making schools safer, healthier and more child-friendly.
ADPP Mozambique is running a programme benefiting 3 250 girls to remain in school, finish their primary education and transit to secondary education. United States Agency for International Development (USAID) is funding the programme. The programme’s scholarship support initiative is making it possible for girls to pursue their dream of education.
Humana People to People India is running a similar programme called Girls’ Bridge Education Course in Rajasthan and Haryana states with an aim to educate out of school girls of the 9 to 14 years age group and enroll them in the main stream education. Besides mitigating gender disparity and preventing girls from dropping out of school, the Girls Bridge Course is providing skills training. Young girls are given a chance to carry out science education among other principal subjects. A second programme called “Illam Mottukal” is raising the quality of girls education in two districts of Tamil Nadu state of India providing remedial training to more than 8 000 girls as well as building capacity of 243 teachers involved with the programme. Each year science fairs are conducted with full participation of the “Illam Mottukal” girls who have come up with science inspired basic solutions to some everyday life matters such as models on how to clean dirty water, low-cost models to sun drying vegetables among others.<|endoftext|>
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# Sin And Cos Algorithms
Quick synopsis of formulas:
In [0,pi/2], you can estimate:
sin(x) ~= 0.01 * x * ( 100 - 15(x^2) )
cos(x) ~= 0.01 * [100 + {y-25}{y}] .........where y = 2*(x^2)
In [-pi/4,pi/4], you can estimate:
tan(x) ~= x + 0.43 * (x^3)
also remember, tan(x)=sin(x)/cos(x)
Error range: sin(x) within 0.011, cos(x) within 0.010, tan(x) within 0.009
In the range [0, pi/2], there are two simple Curta formulas that will give you a good estimate of sin(x) and cos(x) where x is in radians, usually within 0.01.
****
Recall how to do serial multiplication (A*B*C*D...)
****
First, let's recall how to do serial multiplication without clearing:
5*7*12*36
First, enter 5 in SR, and crank 7 times to get 35 in RR
Now enter 12 in SR. Since we want to get 35 *twelve* times, and we already have 35 *one* time in the RR, we need (12-1) *more* 35s. So decrease 12 by one in the SR to 11.
Now so you don't forget how many times to turn, move the registers so the leftmost digit of current result (35) lines up with rightmost digit of SR (11). See a 3, crank three times. Move one to the right, see a 5, crank 5 times. Now you have 420.
Enter (36-1)=35 in SR. Line up the 5 with the 4 in 420. Crank 4 times, twice in the next position, 0 times in the last position. Now you have 15120.
*****
Sin(x)
*****
Recall that the Taylor series expansion of sin(x) is
x - (x^3)/3! + (x^5)/5! - (x^7)/7! ...
A very good estimate for this in the range [0, pi/2] is
x - (x^3)/6.66666667
which can be rewritten as
(1/100) * x * (100-15(x^2))
So all you need to do is
A. square x
B. multiply by -15 and add 100
C. multiply the result by x again
D. move the decimal 2 spaces to the left.
Accuracy is within + or - 0.011
Example with x=1.2 radians
A. x^2 = 1.44
B. Use (-15-1)=-16 to do a serial multiplication: 1.44*-15. Then add 100 to get 78.4.
C. Do another serial multiplication with 12 - 1 = 11 (for the 1.2.) Result is 94.08.
D. Result is 0.9408. Actual sin(1.2)=0.9320. Error is 0.0088.
*****
Cos(x)
*****
Recall that the Taylor series expansion of cos(x) is
1 - (x^2)/2! + (x^4)/4! - (x^6)/6! ...
A very good estimate for this in the range [0, pi/2] is
1 - (x^2)/4 + (x^4)/25
which can be rewritten as
(1/100) * [100 + {2(x^2)-25}{2(x^2)}]
and if you define y=2(x^2), you have
(1/100) * [100 + {y-25}{y}]
So all you need to do is
A. square x and multiply by 2. This is y, put this in SR.
B. multiply by (y-25)
C. add 100
D. move the decimal 2 spaces to the left.
Accuracy is within + or - 0.010
Example with x=1.2 radians
A. 2(x^2) = 2.88 = y
B. Multiply by y-25.
C. Add 100. You should get 36.2944
D. Result is 0.362944. Actual cos(1.2)=0.362358. Error is 0.000586.
*****
Tan(x)
*****
Recall that the Taylor series expansion of cos(x) is
x + (x^3)*2/3! + (x^5)*16/5! ...
A very good estimate for this in the range [0, pi/4] is
x + 0.43 * (x^3)
Edit Text of this page (last edited July 18, 2006 by adsl-69-219-41-94.dsl.chcgil.ameritech.net)
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KS1 Maths Quiz - Year 2 Calculation - Addition (Questions)
This quiz addresses the requirements of the National Curriculum KS1 Maths and Numeracy for children aged 6 and 7 in year 2. Specifically this quiz is aimed at the section dealing with addition of numbers including: a two-digit number and ones, a two-digit number and tens, two two-digit numbers and three one-digit numbers.
Calculating totals means looking at the numbers in an addition calculation and thinking about a sensible way to add them together. When three one-digit numbers are to be added, for example in the calculation 5 + 7 + 5, it might make sense to add the 5 and 5 to make 10, before adding on the 7. If two two-digit numbers are to be added, such as 12 + 13, adding the 2 and 3 to make 5 and then adding the two tens to make 20 might be wise. The same sort of logic applies when adding a two-digit number and a one-digit number. Recognising that additions can be completed in any order could help with this.
1. What is 10 + 9 + 10?
[ ] 11 [ ] 39 [ ] 19 [ ] 29
2. What is the total of 2, 15 and 3?
[ ] 20 [ ] 19 [ ] 25 [ ] 15
3. What is 10 + 27 + 10?
[ ] 45 [ ] 37 [ ] 47 [ ] 54
4. What is 23 + 45?
[ ] 75 [ ] 68 [ ] 65 [ ] 57
5. What is 3 + 9 + 4?
[ ] 16 [ ] 15 [ ] 21 [ ] 19
6. What is the total of 7 and 26?
[ ] 34 [ ] 43 [ ] 23 [ ] 33
7. What is 30 and 48 added together?
[ ] 87 [ ] 78 [ ] 76 [ ] 68
8. What is the sum of 27 and 43?
[ ] 70 [ ] 80 [ ] 60 [ ] 56
9. What is 8 + 9 + 7?
[ ] 26 [ ] 25 [ ] 23 [ ] 24
10. When adding numbers together...
[ ] you should always put the largest number first [ ] you should always put the smallest number first [ ] you should guess the answer [ ] you can add them in any order
KS1 Maths Quiz - Year 2 Calculation - Addition (Answers)
1. What is 10 + 9 + 10?
[ ] 11 [ ] 39 [ ] 19 [x] 29
Adding the 10 and 10 up first, then adding the 9 might have helped
2. What is the total of 2, 15 and 3?
[x] 20 [ ] 19 [ ] 25 [ ] 15
Additions can be done in any order - putting the 15 first could be useful
3. What is 10 + 27 + 10?
[ ] 45 [ ] 37 [x] 47 [ ] 54
Starting with 27 and counting up in tens is a good option to try
4. What is 23 + 45?
[ ] 75 [x] 68 [ ] 65 [ ] 57
Partitioning means splitting up the tens and units, making this addition easier
5. What is 3 + 9 + 4?
[x] 16 [ ] 15 [ ] 21 [ ] 19
Adding up in any order will give the same answer
6. What is the total of 7 and 26?
[ ] 34 [ ] 43 [ ] 23 [x] 33
Starting with the larger number and counting on might be a good plan
7. What is 30 and 48 added together?
[ ] 87 [x] 78 [ ] 76 [ ] 68
Adding the 30 and 40 first and then adding on the units is a good strategy here
8. What is the sum of 27 and 43?
[x] 70 [ ] 80 [ ] 60 [ ] 56
Using the number bond of 7 and 3 might be useful
9. What is 8 + 9 + 7?
[ ] 26 [ ] 25 [ ] 23 [x] 24
Trying to make 10 and then counting on, or using a near double of 7 + 7 could have helped
10. When adding numbers together...
[ ] you should always put the largest number first [ ] you should always put the smallest number first [ ] you should guess the answer [x] you can add them in any order
Addition calculations can be done in any order, but subtractions cannot<|endoftext|>
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Homogeneous coordinates
The introduction of ideal points provides many advantages,
but it means we can no longer represent a point by its cartesian coordinates.
To allow us to use algebraic methods to study the euclidean plane with ideal points,
we need a new model. We begin with a purely formal definition of a new geometry,
which we call RP2.
Definition 1
The set RP2 consists of all objects of the form [u] = {ku, k non-zero},
where u = (u,v,w) is a non-zero vector. The elements are RP2-points .
For any choice of k, the triple (ku,kv,kw) are homogeneous coordinates for U.
In other words, an element of RP2 is a line through the origin O in R3 (but with O deleted).
Note that [u] = [v] if and only if they give the same line, i.e. u=kv, for some non-zero k.
Definition 2
If U = [u] and V = [v] are RP2-points, then the RP2-line UV is defined to be
the set {ku + lv, k,l not both zero}
Then, in R3, UV is the plane through O generated by u and v (but with O deleted).
Thus an RP2-line has equation of the form ax + by + cz = 0.
Now suppose that L and M are distinct RP2-lines.
Then the defining planes in R3 intersect (since each is through O),
and so meet in a line through O, i.e. in an RP2-point.
Thus, the incidence rules for the geometry RP2 are,
two RP2-points define an RP2-line,
two RP2-lines define an RP2-point.
These are exactly the same as those of the euclidean plane with ideal points!
In fact the two geometries are very closely related indeed.
Let Ê denote the euclidean plane E, together with its ideal points.
As usual, we introduce x,y coordinates on E.
Now the clever bit - we think of E as lying on the plane z=1 in R3,
so the point (a,b) on E lies on (a,b,1).
Let P be the point (a,b) on E. Then P lies on the RP2-point [(a,b,1)]. Thus,
each point of E defines an RP2-point with non-zero third coordinate.
Now suppose that the lines L and M lie on E and are parallel.
As parallel lines on E, they must have equations ax + by + c =0 and ax + by + c' = 0.
It is easy to see that L and M are, respectively, the intersections of
ax + by + cz =0 and of ax + by + c'z =0 with z=1.
Now, where ax + by + cz =0 and ax + by + c'z =0 intersect, we have
ax + by = 0 and z=0. These define the RP2-point [(b,a,0)], so
each family of parallel lines on E defines an RP2-point with third coordinate zero.
But a family of parallel lines on E correspond to an ideal point of E, so
each ideal point for E defines an RP2-point with third coordinate zero.
Thus, the points of RP2 correspond to those of Ê.
Lines in RP2.
Let L be an RP2-line with equation ax + by + cz = 0.
• if a and b are not both zero, then L meets z=1 in the line ax + by + c = 0, z=1,
so L meets E in the euclidean line ax + by + c = 0.
Also, L meets z=0 in the line ax + by = 0, z=0. This is the RP2-point [b,-a,0].
In this case, L corresponds to a line of E, together with its ideal point.
• if a=b=0, then L is z=0, so corresponds to the set of ideal points of E,
i.e. to the ideal line for E.
Thus, the lines of RP2 correspond to the lines of Ê.
Conics revisited.
Let C be the solution set of the equation ax2 + bxy + cy2 +fxz +gyz +hz2 = 0.
If (x,y,z) is a solution, then so is (kx,ky,kz) for any non-zero k.
Thus, C is an object in RP2, i.e. a set of RP2-points.
We are interested cases where C does not contain an RP2-line.
If a=b=c=0, the equation is 0 = fxz + gyz + hz2 = (fx + gy + hz)z, so C contains z=0.
Thus, we may assume that a, b, c are not all zero.
As in the case of an RP2-line, we shall look at how C meets Ê by considering
separately its intersections with z=1 and with z=0.
• C meets z=1 in the curve ax2 + bxy + cy2 +fx +gy +h = 0, z=1,
i.e. C corresponds to the curve ax2 + bxy + cy2 +fx +gy +h = 0 on E.
This is the equation of a conic on E.
• C meets z=0 in the curve F(x,y)=ax2 + bxy + cy2= 0, z=0,
There are three possible cases
• F(x,y) does not factorize over R. Then C does not meet z=0.
In other words, C gives rise to no ideal points for E.
• F(x,y) = p(qx + ry)^2. Then C meets z=0 in [r,-q,0],
a single RP2-point (which corresponds to one ideal point for E).
• F(x,y)=(px + qy)(rx + sy). Then C meets z=0 in [q,-p,0] and in [s,-r,0].
These correspond to two ideal points for E.
Thus, C gives rise to a conic on E plus 0, 1 or 2 ideal points.
It is not difficult to see that these correspond to an ellipse, a parabola and a hyperbola.
A proof can be obtained from these pictures.<|endoftext|>
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Practice is an important part of becoming skilled at anything, which may be why there are so many axiom's like "practice makes perfect" floating around common parlance. But what's happening in the brain when we practice?
Researchers believe that practice helps build up the protective layer of myelin, the fatty substance that protects axons in the brain. Axons move electrical signals from the brain to our muscles and when they are better protected by thick myelin they move more efficiently, creating an "information superhighway" between the brain and muscles.
To get the most out of practicing it must be consistent, intensely focused, and target the edge of one's ability. That's why educators target the zone of proximal development in every learner. There are several things individuals can do to make practicing more effective:
1. Focus when engaged in practice
2. Minimize distractions
3. Start slow and increase speed later
4. Practice repeatedly with frequent breaks
5. Visualize the skill to help reinforce practice
Check out this TED-Ed video for more info on the fascinating processes that neuroscientists are beginning to unlock.<|endoftext|>
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Several time in your discovering of mathematics, you have actually been introduced to brand-new kinds the numbers. Every time, these numbers made feasible something that appeared impossible! before you learned about negative numbers, you couldn’t subtract a higher number indigenous a lesser one, but an adverse numbers provide us a means to carry out it. When you were discovering to divide, you initially weren"t may be to carry out a trouble like 13 separated by 5 because 13 isn"t a many of 5. You then learned how to perform this trouble writing the answer as 2 remainder 3. Eventually, you were able to express this answer as
. Using fractions enabled you to make sense of this division.
You are watching: Write in terms of i
Up to now, you’ve well-known it was difficult to take it a square root of a an adverse number. This is true, using just the genuine numbers. Yet here you will certainly learn about a brand-new kind the number that allows you work with square root of negative numbers! choose fractions and an unfavorable numbers, this new kind of number will let you perform what was formerly impossible.
You really need only one new number to begin working with the square roots of an unfavorable numbers. The number is the square root of −1, . The genuine numbers space those that can be displayed on a number line—they seem pretty genuine to us! as soon as something’s not real, you regularly say the is imaginary. So let’s contact this new number i and use that to represent the square root of −1.
Because
, us can likewise see the
or
, so we deserve to conclude the .
The number i permits us to occupational with root of all an adverse numbers, not just . There room two crucial rules come remember: , and . You will usage these rules to rewrite the square root of a an adverse number together the square root of a optimistic number times . Next you will simplify the square root and rewrite as i. Let’s shot an example.
Example Problem Simplify. Use the dominance to rewrite this together a product utilizing . Since 4 is a perfect square (4 = 22), you deserve to simplify the square root of 4. Use the meaning of i to rewrite as i. Answer
Example Problem Simplify. Use the preeminence to rewrite this together a product using . Since 18 is no a perfect square, use the same rule to rewrite it using determinants that space perfect squares. In this case, 9 is the just perfect square factor, and the square root of 9 is 3. Use the meaning of i to rewrite as i. Remember to write i in front of the radical. Answer
Example Problem Simplify. Use the preeminence to rewrite this together a product making use of . Since 72 is not a perfect square, usage the same preeminence to rewrite the using determinants that room perfect squares. Notice that 72 has actually three perfect squares as factors: 4, 9, and also 36. It’s simplest to usage the largest variable that is a perfect square. Use the meaning of i to rewrite as i. Remember to write i in front of the radical. Answer
You may have actually wanted to leveling
using various factors. Part may have thought of rewriting this radical together
, or
, or
for instance. Each of this radicals would certainly have eventually yielded the exact same answer of
.
Rewriting the Square root of a negative Number · find perfect squares within the radical. · Rewrite the radical making use of the rule . · Rewrite as i. Example:
Simplify.
A) 5
B)
C) 5i
D)
A) 5
Incorrect. You may have actually noticed the perfect square 25 together a factor of 50, yet forgot the remainder of the number under the radical. The exactly answer is:
B)
Incorrect. While
C) 5i
Incorrect. You may have correctly i found it the perfect square 25 together a variable of 50, and correctly offered , but you forgot the remaining element of −50, which is 2. The correct answer is:
D)
Correct.
You can produce other numbers by multiplying i by a genuine number. An A number in the kind bi, where b is a actual number and i is the square source of −1.
")">imaginary number
is any variety of the type bi, wherein b is genuine (but no 0) and i is the square root of −1. Look in ~ the adhering to examples, and notice that b have the right to be any kind of kind of real number (positive, negative, entirety number, rational, or irrational), yet not 0. (If b is 0, 0i would simply be 0, a real number.)
Imaginary Numbers 3i (b = 3) −672i (b = −672) (b = ) (b = )
You can use the normal operations (addition, subtraction, multiplication, and also so on) through imaginary numbers. You’ll see much more of that, later. As soon as you add a actual number to an imaginary number, however, you acquire a A number in the form a + bi, whereby a and also b are actual numbers and also i is the square source of −1.
")">complex number
. A complicated number is any kind of number in the type a + bi, where a is a real number and bi is an imagine number. The number a is sometimes referred to as the The actual term, a, in a complex number a + bi.
")">real part
that the facility number, and also bi is sometimes referred to as the The imagine term, bi, in a facility number a + bi.
Complex Number Real part Imaginary part 3 + 7i 3 7i 18 – 32i 18 −32i
In a number v a radical as component of b, such together above, the imaginary i must be composed in prior of the radical. Though composing this number as
is technically correct, it renders it lot more difficult to tell whether ns is within or external of the radical. Placing it before the radical, as in , gets rid of up any kind of confusion. Watch at these last 2 examples.
Number Number in complicated form: a + bi Real part Imaginary part 17 17 + 0i 17 0i −3i 0 – 3i 0 −3i
By do b = 0, any type of real number can be expressed as a facility number. The real number a is created a + 0i in complex form. Similarly, any type of imaginary number have the right to be expressed together a complex number. By making a = 0, any type of imaginary number bi is created
0 + bi in complex form.
Example Problem Write 83.6 together a facility number. a + bi 83.6 + bi Remember the a complex number has the type a + bi. You require to figure out what a and b have to be. Since 83.6 is a real number, the is the real component (a) that the facility number a + bi. A genuine number does no contain any kind of imaginary parts, so the value of b is 0. Answer 83.6 + 0i
Example Problem Write −3i together a facility number. a + bi a – 3i Remember that a facility number has actually the form a + bi. You need to figure out what a and also b have to be. Since −3i is an imagine number, it is the imaginary part (bi) of the complicated number a + bi. This imaginary number has actually no real parts, so the value of a is 0. Answer 0 – 3i
A) 9
Incorrect. The number 9 is in the imaginary part (9i) that this facility number. In a facility number a + bi, the real component is a. In this case, a = −35, for this reason the real part is −35.
B) −35
Correct. In a facility number a + bi, the real component is a. In this case, a = −35, so the real part is −35.
C) 35
Incorrect. In a complex number a + bi, the real component is a. In this case, a = −35, so the real part is −35. The real part can be any kind of real number, including an adverse numbers.
See more: 1691 Carney Lane, Wimberley, Tx, Hays County Transfer Station Wimberley Tx, 78676
D) 9 and −35
Incorrect. The number 9 is in the imaginary part (9i) that this complex number. In a complicated number a + bi, the real part is a. In this case, a = −35, therefore the real part is just −35.
Complex numbers have actually the type a + bi, wherein a and b are actual numbers and i is the square root of −1. All actual numbers can be composed as facility numbers by setup b = 0. Imaginary numbers have actually the type bi and can likewise be created as complex numbers by setup a = 0. Square roots of negative numbers can be streamlined using and<|endoftext|>
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1. ## three distinct
If a,b,c are three distinct nonzero real numbers and
$\displaystyle a + (1 / b) =b + (1 / c) =c + (1 / a) =t$
for some real number t
prove that $\displaystyle abc + t=0$ .
2. Originally Posted by perash
If a,b,c are three distinct nonzero real numbers and
$\displaystyle a + (1 / b) =b + (1 / c) =c + (1 / a) =t$
for some real number t
prove that $\displaystyle abc + t=0$ .
so we have: $\displaystyle ab+1=bt, \ \ bc+ 1 = ct, \ \ ac + 1 = at,$ which we will call 1), 2), and 3) respectively.
now first subtract 2) from 1), then 3) from 2), and finally 1) from 3) to get:
$\displaystyle (a-c)b=(b-c)t, \ \ (b-a)c=(c-a)t, \ \ (c-b)a=(a-b)t,$ and multiply the equations together:
$\displaystyle (b-a)(c-b)(a-c)abc=(a-b)(b-c)(c-a)t^3,$ which, since $\displaystyle a \neq b \neq c,$ gives us: $\displaystyle abc+t^3=0. \ \ \ \ (*)$
this time multilpy 1) by $\displaystyle c$, 2) by $\displaystyle a$, and 3) by $\displaystyle b$ to get: $\displaystyle abc+c=bct, \ \ abc+a=act, \ \ abc+b=abt,$
which we'll call 4), 5), and 6) respectively. now subtract 5) from 4), then 6) from 5) and finally 4) from 6)
to get: $\displaystyle c-a=(b-a)ct, \ \ a-b=(c-b)at, \ \ b-c=(a-c)bt,$ and multiply the equations together:
$\displaystyle (a-b)(b-c)(c-a)=(b-a)(c-b)(a-c)abct^3,$ which, since $\displaystyle a \neq b \neq c,$ gives us: $\displaystyle abct^3 = -1. \ \ \ \ \ (**)$
now by $\displaystyle (*)$ and $\displaystyle (**): \ t^6 = 1.$ thus: $\displaystyle t = \pm 1,$ since $\displaystyle t \in \mathbb{R}.$ so: $\displaystyle t^3=t,$ and hence by $\displaystyle (*): \ \ abc+t=0. \ \ \ \square$
3. Originally Posted by perash
If a,b,c are three distinct nonzero real numbers and
$\displaystyle a + (1 / b) =b + (1 / c) =c + (1 / a) =t$
for some real number t
prove that $\displaystyle abc + t=0$ .
I found an alternative, simpler way:
$\displaystyle ab+1 = bt, bc+1 = ct, ac+1 = at$
$\displaystyle abc+c = bct, abc+a = act, abc+b = abt$
$\displaystyle abc + t= bct+t - c = act+t - a = abt+t - b$
$\displaystyle abc + t= ct^2 - c = at^2 - a = bt^2 - b$
$\displaystyle ct^2 - c = at^2 - a \Rightarrow (t^2 - 1)(c-a) = 0 \Rightarrow t^2 = 1$
This implies $\displaystyle abc + t= ct^2 - c = c(t^2 - 1) = 0$<|endoftext|>
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Throughout the school, the use of Edward de Bono's Thinking Hats are central to the children's learning. In every aspect of learning, the Thinking Hats are used to stimulate discussion and to compartmentalise thinking. The children take a very active role in the use of the Thinking Hats and will use the visual prompts around the school to support their learning.
The Thinking Hats
Each of the six hats represent a different element of thinking.
Red (Feelings): How do I feel about this? What do I like about this? What don't I like about this? What are my likes, dislikes, worries and concerns?
Yellow (Strengths): What are the good points? Why can this be done? Why is this a good thing? What are the strengths/benefits?
Black (Judgement): What is wrong with this? Will this work? Is it safe? Can it be done?
Blue (Planning): What thinking is needed? Where are we now? What is the next step? How far have we come?
White (Information): What are the key facts? What information do I have? What information do I need? What do I want to know?
Green (Creativity): What new ideas are possible? What is my suggestion? Can I create something new? What are the weaknesses?
A summary card can be downloaded by clicking the image below.<|endoftext|>
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Fractions-Solutions Ex-7.1
CBSE Class –VI Mathematics
NCERT Solutions
Chapter 7 Fraction (Ex. 7.1)
Question 1. Write the fraction representing the shaded portion:
Answer: (i)$\frac{2}{4}$ (ii) $\frac{8}{9}$ (iii) $\frac{4}{8}$ (iv)$\frac{1}{4}$ ( v)$\frac{3}{7}$
(vi)$\frac{9}{12}$ (vii)$\frac{10}{10}$ (viii)$\frac{4}{9}$ (ix)$\frac{4}{8}$ (x)$\frac{1}{2}$
Question 2. Colour the part according to the given fraction:
Question 3. Identify the error, if any?
Answer: All the figures are not equally divided. For making fractions, it is necessary that figure is divided into equal parts.
Question 4. What fraction of a day is 8 hours?
Answer: Since, 1 day = 24 hours.
Therefore, the fraction of 8 hours = $\frac{8}{24}=\frac{1}{3}$
Question 5. What fraction of an hour is 40 minutes?
Answer: Since, 1 hour = 60 minutes.
Therefore, the fraction of 40 minutes = $\frac{40}{60}=\frac{2}{3}$
Question 6. Arya, Abhimanyu and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich.
(a)How can Arya divide his sandwiches so that each person has an equal share?
(b)What part of a sandwich will each boy receive?
Answer:(a) Arya will divide each sandwich into three equal parts and give one part of each sandwich to each one of them.
(b) $1×\frac{1}{3}=\frac{1}{3}$
Question 7. Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?
Answer: Total number of dresses to dye= 30
Work completed = 20
Fraction of completed work = $\frac{20}{30}=\frac{2}{3}$
Question 8. Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Answer: Natural numbers from 2 to 12: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Prime numbers from 2 to 12: 2, 3, 5, 7, 11
Hence, fraction of prime numbers = $\frac{5}{11}$
Question 9. Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Answer: Natural numbers from 102 to 113: 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113
Prime numbers from 102 to 113: 103, 107, 109, 113
Hence fraction of prime numbers = $\frac{4}{12}=\frac{1}{3}$
Question 10. What fraction of these circles has ‘X’s in them?
Answer: Total number of circles = 8 and number of circles having ‘X’ = 4
Hence, the fraction = $\frac{4}{8}$
Question 11. Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?
Answer: Total number of CDs = 3 + 5 = 8
Number of CDs purchased = 3
Fraction of CDs purchased = $\frac{3}{8}$<|endoftext|>
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# 1954 AHSME Problems/Problem 28
## Problem 28
If $\frac{m}{n}=\frac{4}{3}$ and $\frac{r}{t}=\frac{9}{14}$, the value of $\frac{3mr-nt}{4nt-7mr}$ is:
$\textbf{(A)}\ -5\frac{1}{2}\qquad\textbf{(B)}\ -\frac{11}{14}\qquad\textbf{(C)}\ -1\frac{1}{4}\qquad\textbf{(D)}\ \frac{11}{14}\qquad\textbf{(E)}\ -\frac{2}{3}$
## Solution 1
From $\frac{m}{n}=\frac{4}{3}$, we have $3m=4n$. From $\frac{r}{t}=\frac{9}{14}$, we have $14r=9t\implies 7r=4.5t$
This simplifies the fraction to $\frac{4nr-nt}{4nt-7r\cdot m}\implies \frac{4nr-nt}{4nt-4.5mt}\implies \frac{4nr-nt}{4nt-1.5t\cdot3m}\implies \frac{4nr-nt}{4nr-1.5\cdot t\cdot 4n}\implies \frac{\frac{4\cdot7r}{7}t-nt}{4nt-6nt}\implies \frac{\frac{4\cdot9t}{7\cdot2}n-nt}{-2nt}\implies \frac{nt(\frac{36}{14}-1)}{-2(nt)}\implies\frac{\frac{22}{14}}{-2}\implies \boxed{\frac{-11}{14} (\textbf{B})}$
## Solution 2
Because the ratio works for any set of integers satisfying $\frac{m}{n}=\frac{4}{3}$ and $\frac{r}{t}=\frac{9}{14}$, it has to satisfy $m=4$, $n=3$, $r=9$, and $t=14$. From here it is just simple arithmetic.
$\frac{3mr-nt}{4nt-7mr}\implies\frac{3\cdot4\cdot9-3\cdot14}{4\cdot3\cdot14-7\cdot4\cdot9}\implies \frac{3(36-14)}{4(42-63)}\implies \frac{3(22)}{4(-21)}\implies \boxed{\frac{-11}{14} (\textbf{B})}$<|endoftext|>
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Skimming and Scanning
English Teaching Resources: Skimming and Scanning skills.
(PowerPoint teaching resource).
English Resources: Skimming and Scanning skills is a 6 slide PowerPoint presentation designed to teach pupils how to develop the key reading skills of skimming and scanning. The resource includes:
An introduction explaining the difference between skimming and scanning
An interactive activity* which allows pupils to apply their skimming and scanning skills and enables teachers to assess their understanding.
The use of bold clear fonts, simple language, bright colours and 'lower school friendly' imagery.
To preview the PowerPoint presentation English Teaching Resources: Skimming and Scanning please click on the images.
*please note that you will need to have Flash Player installed and a device capable of playing Flash files to be able to use the interactive elements of this resource. You can download Flash Player free of charge here - http://get.adobe.com/flashplayer/Our Price : £2.99 / 3 Credits<|endoftext|>
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Severe epilepsy is sometimes treated surgically by cutting through the skull and performing a lobectomy in the area causing seizures, which understandably comes with a bit of risk. However, a team of engineers led by Eric Barth of Vanderbilt University have developed a new technique in which a robot performs the surgery by entering through the cheek. This could minimize how invasive the procedure is and reduce recovery time. Some epilepsy surgeries target the hippocampus near the base of the skull. By going through the cheek to get to the brain, surgeons could essentially take a shortcut in getting to the affected area. Leaving the skull intact would reduce healing time and potential complications. While getting to the hippocampus from the top of the skull might be invasive, it can be done going straight down. Going through the cheek would require the needle to steer around obstacles. The needle itself is composed of 1.4 millimeter tubes that get slowly pushed out pneumatically. Some sections are curved, allowing the needle to be steered directly to the hippocampus and avoid damaging unaffected structures; a major advance. After each addition of a segment, the area is scanned via MRI to ensure it is on the right path.
“I’ve done a lot of work in my career on the control of pneumatic systems,” Barth said in a press release. “We knew we had this ability to have a robot in the MRI scanner, doing something in a way that other robots could not. Then we thought, ‘What can we do that would have the highest impact?’”
Because the needles are non-magnetic, they do not interfere with the MRI. Other parts of the device are made out of 3D-printed plastic, which is also able to be used. Accurately tracking the needle’s progress millimeter by millimeter through MRI allows surgeons to make much smaller holes, rather than having to go down through the brain and visually guide the tools. According to Joseph Neimat, a neurosurgeon on Barth’s team, accessing the brain through the cheek is not an entirely novel idea. Before a lobectomy can be performed, doctors need to identify the origin of an epileptic’s seizures. Electrodes that monitor brain activity are placed near the back end of the brain, and many doctors will also go through the cheek in order to place those probes. The ability to perform a surgery in this manner dramatically changes the game.
“The systems we have now that let us introduce probes into the brain—they deal with straight lines and are only manually guided,” Neimat explained. “To have a system with a curved needle and unlimited access would make surgeries minimally invasive. We could do a dramatic surgery with nothing more than a needle stick to the cheek.” Test procedures using a prototype device showed that the surgery was successful and accurate to 1.18 millimeters. However, it probably won’t be hitting an operating room near you for quite some time, as there is much more testing that needs to be done. The researchers note that after their successful preliminary testing, they will move on and try the procedure on human cadavers. They hope it will be approved within the next decade.<|endoftext|>
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# What are the special types of conic sections?
## What are the special types of conic sections?
Defining Conic Sections The three types of conic sections are the hyperbola, the parabola, and the ellipse. The circle is type of ellipse, and is sometimes considered to be a fourth type of conic section.
### What kinds of lines are intersecting lines?
Two non-parallel lines may meet at a point and those lines are called intersecting lines. Intersecting lines are two lines that share exactly one point. This shared point is called the point of intersection.
What are the characteristics of intersecting lines?
Properties of intersecting lines
• The intersecting lines (two or more) meet only at one point always.
• The intersecting lines can cross each other at any angle. This angle formed is always greater than 0o and less than 180o .
• Two intersecting lines form a pair of vertical angles.
What is the best definition of a conic section?
conic section, also called conic, in geometry, any curve produced by the intersection of a plane and a right circular cone. Depending on the angle of the plane relative to the cone, the intersection is a circle, an ellipse, a hyperbola, or a parabola.
## What are the 4 conic section?
A conic is the intersection of a plane and a right circular cone. The four basic types of conics are parabolas, ellipses, circles, and hyperbolas. Study the figures below to see how a conic is geometrically defined. In a non-degenerate conic the plane does not pass through the vertex of the cone.
### What are the two examples of intersecting lines?
Two examples of intersecting lines are listed below: Crossroads: When two straight roads meet at a common point they form intersecting lines. Scissors: A pair of scissors has two arms and both the arms form intersecting lines.
Why are conic sections called conic sections?
They are called conic sections because they can be formed by intersecting a right circular cone with a plane. When the plane is perpendicular to the axis of the cone, the resulting intersection is a circle. When the plane is slightly tilted, the result is an ellipse.
Why is a conic section called a tangent line?
In the complex projective plane. If the intersection point is double, the line is said to be tangent and it is called the tangent line . Because every straight line intersects a conic section twice, each conic section has two points at infinity (the intersection points with the line at infinity ).
## When does a conic section have a double point?
If the intersection point is double, the line is a tangent line . Intersecting with the line at infinity, each conic section has two points at infinity. If these points are real, the curve is a hyperbola; if they are imaginary conjugates, it is an ellipse; if there is only one double point, it is a parabola.
### Which is the line joining the vertices of a conic?
The line joining the foci is called the principal axis and the points of intersection of the conic with the principal axis are called the vertices of the conic. The line segment joining the vertices of a conic is called the major axis, also called transverse axis in the hyperbola.
What are the three types of conic sections?
The three types of conic section are the hyperbola, the parabola, and the ellipse; the circle is a special case of the ellipse, though historically it was sometimes called a fourth type. The ancient Greek mathematicians studied conic sections, culminating around 200 BC with Apollonius of Perga ‘s systematic work on their properties.<|endoftext|>
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Early Math Skills
You can help your child build early math skills by talking about numbers and how we use them in everyday life. This will help your child develop basic problem solving skills, understand patterns and sequences, and support early literacy. Counting every day objects and finding patterns and shapes around them are important ways that children understand how to use early math.
Below are tips to support your child’s early math skills development:
- Ask children to count items in picture books: How many cats do you see? How many windows are there?
- Suggest children count real, everyday, objects using one number for each item.
Sorting and matching
- Ask children to match pictures of things that go together in storybooks: “Show me the picture of
the bed for papa bear. Show me baby bear’s bed.”
- Ask children to help with simple household tasks such as sorting and matching socks.
Put events in order
- Talk about the fact that stories have a beginning, middle and end.
- Use words that help children understand the order in which things happen, such as “first” and
“second.” “First we need to wash our hands. Second we will eat apples.”
Identify simple shapes
- Look for different shapes in your house and neighborhood: What shape are the windows? What
shape are the plates?
- Help children point out circles, squares, and triangles in books.
Understand the Number Line
- Play board games like Candy Land or Hopscotch in which children count the number of moves.<|endoftext|>
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Discover the cosmos! Each day a different image or photograph of our fascinating universe is featured, along with a brief explanation written by a professional astronomer.
2012 May 13
Explanation: Many spiral galaxies have bars across their centers. Even our own Milky Way Galaxy is thought to have a modest central bar. Prominently barred spiral galaxy NGC 1672, pictured above, was captured in spectacular detail in image taken by the orbiting Hubble Space Telescope. Visible are dark filamentary dust lanes, young clusters of bright blue stars, red emission nebulas of glowing hydrogen gas, a long bright bar of stars across the center, and a bright active nucleus that likely houses a supermassive black hole. Light takes about 60 million years to reach us from NGC 1672, which spans about 75,000 light years across. NGC 1672, which appears toward the constellation of the Dolphinfish (Dorado), is being studied to find out how a spiral bar contributes to star formation in a galaxy's central regions.
Authors & editors:
Jerry Bonnell (UMCP)
NASA Official: Phillip Newman Specific rights apply.
A service of: ASD at NASA / GSFC
& Michigan Tech. U.<|endoftext|>
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Lesson/Unit Planning Resources
California State Standards and Frameworks:
Educational standards describe what students should know and be able to do in each subject in each grade. In California,the State Board of Education decides on the standards for all students, from kindergarten through high school.
Curriculum frameworks provide guidance for implementing the standards adopted by the State Board of Education
Additional Planning Resources:<|endoftext|>
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Andrea Palmer PROSPECT HILL ACADEMY CHARTER SCHOOL, SOMERVILLE, MA
6th Grade Math : Unit #5 - Proportional Reasoning: Ratios and Rates : Lesson #11
Calculating Speed Day 1
Objective: SWBAT: • Make conversions to determine my speed using different units • Use multiplication and division to calculate rates
Standards: 6.RP.A.3 6.RP.A.3b 6.RP.A.3d MP1 MP2 MP3 MP5 MP6
Subject(s): Math
60 minutes
1 Do Now - 7 minutes
See my Do Now in my Strategy folder that explains my beginning of class routines.
Often, I create do nows that have problems that connect to the task that students will be working on that day. Today I want students to start thinking about what speed really means.
I ask students, “What does it mean for Usain Bolt to run at a speed of 23 miles per hour?” I want students to set up the rate, recognizing that “per hour” means their denominator needs to be 1 hour. Just because he was traveling at a speed of 23 mph does not mean that he could maintain that speed for an hour.
2 Setting Expectations - 5 minutes
I tell students that today they will be working to figure out their sprinting and walking speeds. I tell students to copy their data from the previous day’s packet. I tell them that by the end they should know their speed in miles per hour. I ask students what conversion facts they think they will need. I copy them on the board and students put them in their packet to refer to later.
I tell students that this task is complex. They have skills and strategies for working with rates and measurements and all of these skills will help them to work through the task.
3 Part 1: What was your sprinting speed? - 28 minutes
Notes:
• Before this class, I use the data from the previous lesson’s TTG to Create Homogeneous Groups of 3-4 students.
• Each group receives a Group Work RubricI use this data to give each student a citizenship grade for the day.
• My goal is for students to at least complete the questions in part 1 today. We will be working on extending this task during the next lesson, so students will have some time to work on part 2.
I tell students about the groups they are working in. Students move into groups and begin working. Each student has unique data, but they will be able to ask each other questions about how to change their rates and measurements. I walk around to monitor student progress and behavior. Students are engaging in MP1: Make sense of problems and persevere in solving them, MP2: Reason abstractly and quantitatively, MP5: Use appropriate tools strategically, and MP6: Attend to precision.
If students are struggling, I may ask the following questions:
• How long did it take you to sprint 40 feet? How can we figure out how many feet you traveled in 1 second?
• How far did you travel in one second? At this rate, how far would you travel in 2 seconds? In 3 seconds?
• How can you use what you have to create a rate showing how far you would travel in one minute? What units do you have? What units do you need to have in your answer?
• If you know how far you travel in a minute, how far will you travel in one hour?
• How far would you travel in one hour? What units do you have? What units do you need to have?
If students are struggling with the actual calculations, I have calculators for students to use. If a student uses a calculator, I still require them to set up the rates and show the calculations he/she is making to get the answer.
If students need extension, I may ask them the following questions:
• What if your walking speed is 1 meter per 1.5 seconds, what is your speed in feet per second?
• How long do you think you could keep up this speed?
If students successfully complete part 1, they move onto part 2. I took away the scaffolding in part 2 so students need to complete multiple steps to answer the questions. If they struggle, I have them look at their process in the part 1 problems as a reference.
If students successfully complete part 2, they move on to the challenge questions. If students ask me how many kilometers are in a mile, I will give them the conversion. I want students to figure out on their own what information they need to know to complete the problem.
Student Work: Converting measures of speed
Real World Applications
Students really enjoyed using their data to calculate their speed. By breaking down the more complex question of “How fast were you running in miles per hour?” into smaller questions, more students were able to access and solve the problems. I ended up passing out calculators to each group during this part of the lesson. Some students used them and some students did not. I collected students’ work at the end of the lesson and here are some of my observations.
Unit 5.10 Accurate Calculation.jpg
Most students who used a calculator to were able to accurately calculate their speed in miles per hour. This student had a speed of 27,720 feet per hour. She was able to use the conversion fact of 1 mile = 5280 feet to set up a proportion. Using the calculator she was able to find that 27,720 feet is equivalent to 5.25 miles.
Unit 5.10 Labeling Mistake.jpg
Some students were able to find the correct speed, but made mistakes in labeling their answers. This student used the calculator to divide 28,800 feet by 5280 feet. She found her sprinting speed was about 5.45 miles per hour. She mistakenly wrote 28,800 miles/1 hour as part of her answer.
Unit 5.10 Conversion Trouble.jpg
Other students made mistakes converting their speed from one unit to another. This student accurately calculated his sprinting speed as 41,040 feet/ hour. In trying to change his speed into miles per hour he starts adding groups of 5280 feet. He ends up adding 3 groups of 10,560 feet and gets 42,240 feet. Since his amount is more than his total of 41,040 feet, he estimates to about 3 miles per hour. He counted each group of 10,560 feet as one mile, instead of two miles. He did not stop to make an estimate or use the calculator and ended up with a speed that was much lower than his actual speed of 7.77 miles per hour.
At the beginning of the next lesson I gave students their work back. I asked how students could estimate to make a reasonable guess of their sprinting speed. I also asked students how they could use a calculator to find a more precise measure of their speed.
4 Closure and Ticket to Go - 10 minutes
I ask students to share out how they found their speed in feet per second. I have a student come up to the front to show and explain their thinking under the document camera. I ask students to come up to show and explain their work for problem 3 and 4. Students are engaging with MP3: Construct viable arguments and critique the reasoning of others. I ask students to show me with their hands if they have completed the problems in part 1.
I ask students to share out what obstacles they encountered and whether/how they overcame them. I remind students that they will be working on a task using this data in the next lesson.
With a few minutes left, I pass out the Ticket to Go and the HW Calculating Speed Day 1.<|endoftext|>
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# BINOMIAL THEOREM EXAMPLES
Binomial theorem examples :
Here we are going to see some example problems on binomial theorem.
Binomial expansion for (x + a)n is,
nc0xna+ nc1xn-1a+ nc2xn-2a+ .........+ ncnxn-na0
## Binomial theorem examples
Example 1 :
Find the expansion of (2x + 3y)5
Solution :
Comparing the given question with (x + a)n
we get x = 2x, a = 3y and n = 5
Since we have power 5, we are going to have 6 terms in the expansion
(2x+3y)5 = 5c0(2x)5(3y)+ 5c1(2x)4(3y)+ 5c2(2x)3(3y)+ 5c3(2x)2(3y)+ 5c4(2x)1(3y)4+ 5c5(2x)0(3y)5
While calculating the values of 5c0, 5c1 ,......... we can follow the given tricks
The value of nc0 and ncn is 1.
The value of nc1 is n.
The value of 5c2 means, in the numerator we should write 5 as 5 4 and in the denominator we should write 2 1
5c0 = 1, 5c1 = 5, 5c2 = (54)/(21) ==> 20/2 = 10
5c3 = (543)/(321) ==> 60/6 = 10
5c4 = (5432)/(4321) ==> 5, 5c5 = 1
= 32x+ 5(16x4)(3y)+ 10(8x3)(9y) + 10(4x2)(27y3) + 5(2x)(81y4) + 1(243 y5)
= 32x+240x4y + 720x3y2 + 1080x2y3 + 810xy4 + 243 y5
Let us see the next example on "Binomial theorem examples"
Example 2 :
Find the expansion of (3a + 5b)5
Solution :
Comparing the given question with (x + a)n
we get x = 3a, a = 5b and n = 5
(3a + 5b)5
Since we have power 5, we are going to have 6 terms in the expansion
= 5c0(3a)5(5b)+ 5c1(3a)4(5b)+ 5c2(3a)3(5b)+ 5c3(3a)2(5b)+ 5c4(3a)1(5b)4+ 5c5(3a)0(5b)5
5c0 = 1, 5c1 = 5, 5c2 = (54)/(21) ==> 20/2 = 10
5c3 = (543)/(321) ==> 60/6 = 10
5c4 = (5432)/(4321) ==> 5, 5c5 = 1
= 5c0(3a)5(5b)+ 5c1(3a)4(5b)+ 5c2(3a)3(5b)+ 5c3(3a)2(5b)+ 5c4(3a)1(5b)4+ 5c5(3a)0(5b)5
= 243a5 + 5(81a4)(5b) + 10 (27a3)(25b2) + 10(9a2)(125b3) + 5(3a) (625b4) + 1 (3125b5)
= 243a+ 2025a4b + 6750a3b+ 11250a2b+ 9375ab4+ 3125b5
Let us see the next example on "Binomial theorem examples"
Example 3 :
Find the expansion of (a - 2b)5
Solution :
Comparing the given question with (x + a)n
we get x = a, a = -2b and n = 5
(a - 2b)5
Since we have power 5, we are going to have 6 terms in the expansion
= 5c0(a)5(-2b)+ 5c1(a)4(-2b)+ 5c2(a)3(-2b)+ 5c3(a)2(-2b)+ 5c4(a)1(-2b)4+ 5c5(a)0(-2b)5
= a- 10 ab + 40 ab2 - 80 ab+ 80 b4-32 b5
Example 4 :
Find the expansion of (2x – 5y)7
Solution :
Since we have power 7, we are going to have 8 terms in the expansion
(2x – 5y)7
= 7C0 (2x)7(–5y)0 + 7C1 (2x)6(–5y)1 + 7C2 (2x)5(–5y)2
+ 7C3 (2x)4(–5y)3 + 7C4 (2x)3(–5y)4 + 7C5 (2x)2(–5y)5
+ 7C6 (2x)1(–5y)6 + 7C7 (2x)0(–5y)7
7c0 = 1, 7c1 = 7, 7c2 = (7⋅6)/(2⋅1) ==> 56/2 = 28
7c3 = (7⋅6⋅5)/(3⋅2⋅1) ==> 210/6 = 35
7c4 = (7⋅6⋅5⋅4)/(4⋅3⋅2⋅1) = 35
7c5 = (7⋅6⋅5⋅4⋅3)/(5⋅4⋅3⋅2⋅1) = 21
7c6 = (7⋅6⋅5⋅4⋅3⋅2)/(6⋅5⋅4⋅3⋅2⋅1) = 7
7c7 = 1
= (1)(128x7)(1) + (7) (64x6) (–5y) + (21) (32x5) (25y2) + (35) (16x4) (–125y3) + (35) (8x3) (625y4) + (21) (4x2)(–3125y5) + (7) (2x) (15625y6) + (1)(1)(–78125y7)
= 128x7 – 2240x6y + 16800x5y2 – 70000x4y3 + 175000x3y4 – 262500x2y+ 218750xy6 – 78125y7
After having gone through the stuff given above, we hope that the students would have understood, "Binomial theorem examples".
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# Invert vs inverse relationship
### Inverse Relationship of Addition and Subtraction
In statistical terminology, an inverse correlation is denoted by the correlation coefficient "r" having a value between -1 and 0, with r = There is an inverse relationship between the price of a good and the quantity demanded of a good. When the price goes up, demand falls. When the price goes. inverse meaning, definition, what is inverse: if there is an inverse relationship betw: Learn This circle self-inverts; that is, its inverse is the same circle. Yes , this is the inverse of what is known as the mutation rate, and it can be measured.
We want to give you a y and get an x. So all we have to do is solve for x in terms of y. So let's do that. If we subtract 4 from both sides of this equation-- let me switch colors-- if we subtract 4 from both sides of this equation, we get y minus 4 is equal to 2x, and then if we divide both sides of this equation by 2, we get y over 2 minus 4 divided by 2 is is equal to x.
So what we have here is a function of y that gives us an x, which is exactly what we wanted. We want a function of these values that map back to an x.
So we can call this-- we could say that this is equal to-- I'll do it in the same color-- this is equal to f inverse as a function of y. Or let me just write it a little bit cleaner.
We could say f inverse as a function of y-- so we can have 10 or so now the range is now the domain for f inverse. So all we did is we started with our original function, y is equal to 2x plus 4, we solved for-- over here, we've solved for y in terms of x-- then we just do a little bit of algebra, solve for x in terms of y, and we say that that is our inverse as a function of y.
Which is right over here.
And then, if we, you know, you can say this is-- you could replace the y with an a, a b, an x, whatever you want to do, so then we can just rename the y as x. So all you do, you solve for x, and then you swap the y and the x, if you want to do it that way.
That's the easiest way to think about it.
And one thing I want to point out is what happens when you graph the function and the inverse. So let me just do a little quick and dirty graph right here. And then I'll do a bunch of examples of actually solving for inverses, but I really just wanted to give you the general idea. Function takes you from the domain to the range, the inverse will take you from that point back to the original value, if it exists.
### What is Inverse Relationship? definition and meaning
So if I were to graph these-- just let me draw a little coordinate axis right here, draw a little bit of a coordinate axis right there. This first function, 2x plus 4, its y intercept is going to be 1, 2, 3, 4, just like that, and then its slope will look like this. It has a slope of 2, so it will look something like-- its graph will look-- let me make it a little bit neater than that-- it'll look something like that. That's what that function looks like. What does this function look like?
What does the inverse function look like, as a function of x? Remember we solved for x, and then we swapped the x and the y, essentially. We could say now that y is equal to f inverse of x. The slope looks like this. Let me see if I can draw it. The slope looks-- or the line looks something like that. And what's the relationship here? I mean, you know, these look kind of related, it looks like they're reflected about something. It'll be a little bit more clear what they're reflected about if we draw the line y is equal to x.
So the line y equals x looks like that. I'll do it as a dotted line. And you could see, you have the function and its inverse, they're reflected about the line y is equal to x.
And hopefully, that makes sense here. Because over here, on this line, let's take an easy example. Our function, when you take so f of 0 is equal to 4.
• Intro to inverse functions
• Inverse curve
Our function is mapping 0 to 4. The inverse function, if you take f inverse of 4, f inverse of 4 is equal to 0. Or the inverse function is mapping us from 4 to 0. Which is exactly what we expected.
### InverseFunction—Wolfram Language Documentation
The function takes us from the x to the y world, and then we swap it, we were swapping the x and the y. We would take the inverse. And that's why it's reflected around y equals x. So this example that I just showed you right here, function takes you from 0 to maybe I should do that in the function color-- so the function takes you from 0 to 4, that's the function f of 0 is 4, you see that right there, so it goes from 0 to 4, and then the inverse takes us back from 4 to 0.
This moves price from the Y vertical axis to the X horizontal axis and demand from the X axis to the Y axis. Here is an example inverse calculation: Graph Structure The primary change between a demand curve and an inverse demand curve is the shape of the graphs.
Direct and inverse variation - Rational expressions - Algebra II - Khan Academy
The two functions are complete opposites of each other. The independent and dependent variables are reversed. Demand moves to the Y axis and price to the X axis on an inverse function.
## Differences Between Demand Curve and Inverse
The slopes are opposite as well. A steep demand curve has a flat inverse demand curve and vice versa. Use The demand curve was originally designed when economies were based primarily on agriculture.
Farmers grew as much crop as possible, and the market price was determined by how much crop was produced.
## InverseFunction
This is why quantity is the independent variable on the demand curve. Today, production is driven more by price. Businesses get an idea of the price of their good and this sets their production goals. The economics field uses the original demand curve out of respect for tradition.<|endoftext|>
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#### Need Help?
Get in touch with us
# Multiplication of Decimal Numbers with a Power of 10
Aug 4, 2023
#### Introduction:
Have you ever thought about what exactly is power of 10 is? Here are some examples where we use the powers of 10.
1. In St. Paul School, there are 10,000 students. Mr. David informed John tomorrow there would be a fest. John went to the class and told this news secretly to 10 of his friends, his 10 friends told the same to their 10 friends and the chain continued.
2. Ellie performed in a singer’s contest and got her video. She sent the video to 10 of her friends and those 10 friends forwarded the video to 10 of their friends.
3. Rita used lens to enlarge the specimen 100 times.
In all the above examples, we are dealing with powers of 10.
Can you now give any other example of using the power of 10?
## What is a product of decimal numbers with a power of 10?
Example:
Jason and Sarah each get 4 apples.
Now, if Jason and Sarah each get 2 apples.
Suppose we are going to half the number of apples to Jason and Sarah again.
What if, again, I want to give them half of the previous amount?
### Product of Decimal Number and a Power of 10:
Cut the apple into 2 equal parts
Distribute 3 apples evenly equal to two persons:
Now, let’s distribute one and a half apples to Jason and Sarah.
Now distribute 1.5 apples to 10 people. How many apples do you need?
What if we have 100 kids, and we want to distribute 1.5 apples to each kid?
How many apples do you need for the same?
Distribute 1.5 apples to 1000 people.
Find the pattern in all the 3 cases.
When 1.5 apples were distributed to 1,000 people:
#### Pattern for the Power of 10:
1.5 x 1 = 1.5
1.5 x 101 =15
1.5 x 102 =150
1.5 x 103 =1,500
In the pattern, we meant that
If 1.5 apple is distributed to 1 person, we need 1.5 apples.
If the same 1.5 apple is to be given to each person among 101 we will need 15 apples.
If 1.5 apple is to be given to each person among 102 we will need 150 apples.
If 1.5 apple is to be given to each person among 103 we will need 1500 apples.
It is observed that:
A digit in one place is worth 10 times more when moved to the place on its left. Every time a number is multiplied by 10, the digits of the number shift to the left.
Let’s understand this better by solving some examples.
Example 1:
What pattern follows when we multiply any decimal number say 4.98 and a power of 10?
Solution:
Hence,
Every time it seems as if the decimal is shifting towards the right.
Example 2:
Saniya went to the ice-cream parlor and found that each ice-cream costs 12.5 dollars. If she wants to give the same to 100 of her friends as a treat. How much money will she require?
Solution:
Cost of an ice-cream = 12.5 dollars
Cost of 100 ice-creams = 12.5 × 100 = 12.5 × 102 = \$1250
So, Saniya requires \$1,250.
Example 3: Jose uses a microscope to observe specimens in science class. The microscope enlarges objects to 100 times their actual size. Find the size of each specimen as seen in the microscope.
Solution:
As the microscope enlarges the image 100 times, i.e., 102 times. Hence, the decimal point will shift to its right 2 times. Therefore, the table will be as follows.
Example 4: James wants to distribute dictionaries to 1000 of his students. If the cost of 1000 books is \$14400, what is the price of each dictionary?
Solution:
Let the cost of 1 dictionary be ‘d’.
Cost of 1000 (= dictionaries is = 14400 dollars
d x 103 = 14400
Since the power of 10 is 3. To find the value of ‘d’ keep the decimal point before 3 digits of a number.
d = 14.4
So, the price of each dictionary is \$14.4.
Example 5: During a fest, gifts were distributed to some people. If the cost of a single gift was \$11.34 and Michel spent \$113400. How many people were there?
Solution:
Cost of single gift = \$11.34
Let the total number of persons be ‘m’.
Amount spent = \$113400
\$11.34 \$113400
The decimal point has moved to four digits right from the original place.
11.34 x 104 = 113400
m= 104 =10,000
#### Exercise:
1. 1.275 × __ = 1275. Find the missing number.
2. Which of the following is correct?
• 681 × = 6817
• 42.3 × = 4230
• 0.678 × = 6.78
• 0.086 × = 0.00086
3. 26.014 × __= 26014. Find the missing number.
4. If 0.056 × = 56. Find .
5. Which one has the greatest value?
• 0.1
• 0.001
• 0.01
• 0.0001
6. Complete the chart.
7. When multiplying by a power of 10, like 4.58 x 103 how do you know you are moving the decimal in the correct direction?
(8-10) Find the missing numbers.
8. 0.629 x 10? = 62.5
9. 10? × 0.056 = 560
10. 94.6 x 103 = ?
#### What We Have Learned:
• Solve problems related to product of a decimal number and a power of 10.
• Perform multiplication on power of 10.
• Know the pattern followed in multiplication for decimal numbers with power of 10.
• Use problems of product of decimals with power of 10 in day-to-day life.
#### 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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# How do you solve the following system: 6x + y = 2, 3x - 4y = -10?
Mar 15, 2016
$x = - \frac{2}{27}$ and $y = \frac{22}{9}$
#### Explanation:
$6 x + y = 2$ ---------Eqn $1$
$3 x - 4 y = - 10$ ---------Eqn $2$
Multiplying Eqn $2$ by $2$
$6 x - 8 y = - 20$ ---------Eqn $3$
Subtracting Eqn $1$ from Eqn $3$
Note: Subtracting changes the bottom equation's sign
$\cancel{6 x} + y = 2$
$\cancel{- 6 x} + 8 y = 20$
$9 y = 22 \implies y = \frac{22}{9}$
put $y = \frac{22}{9}$ in Eqn $2$
$3 x - 4 \left(\frac{22}{9}\right) = - 10$
$3 x - \frac{88}{9} = - 10$
$27 x - 88 = - 90$
$27 x = - 90 + 88$
$27 x = - 2 \implies x = - \frac{2}{27}$<|endoftext|>
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# How do you solve x^2 + 6x + 10 = 0 by completing the square?
##### 2 Answers
Apr 3, 2016
${\left(x + 3\right)}^{2} + 1 \to x = \pm i - 3$
#### Explanation:
Completing the square is a method of getting as close as you can with one set of brackets and adding or subtracting the rest.
It ends up in a form like
${x}^{2} + a x + b = {\left(x + c\right)}^{2} + d$
$c$ is always half of $a$, so that when you expand out the brackets you have the right coefficients for ${x}^{2}$ and $x$.
$a = 6 \to c = 3$
${\left(x + 3\right)}^{2} = {x}^{2} + 6 x + 9$
As you can see this is pretty close to the original quadratic. All you need to do is add $1$ and it equates perfectly.
${x}^{2} + 6 x + 10 = {\left(x + 3\right)}^{2} + 1$
To then solve, rearrange like so
${\left(x + 3\right)}^{2} + 1 = 0$
${\left(x + 3\right)}^{2} = - 1$
$x + 3 = \sqrt{- 1} = \pm i$
$x = \pm i - 3$
Apr 3, 2016
There is no solution for the given equation for any $x$ in the set of 'Real Numbers'.
$\textcolor{b l u e}{x = - 3 \pm i}$
#### Explanation:
Consider the standard form of $y = a {x}^{2} + b x + c$
The by completing the square we have:
$y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c - \left[{\left(\frac{b}{2}\right)}^{2}\right]$
In your case $a = 1$ so we have:
$y = {\left(x + \frac{b}{2}\right)}^{2} + c - \left[{\left(\frac{b}{2}\right)}^{2}\right]$
$\implies y = {\left(x + 3\right)}^{2} + 10 - \left({3}^{2}\right)$
$\implies y = {\left(x + 3\right)}^{2} + 1$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
${x}_{\text{vertex}} = \left(- 1\right) \times \left(+ 3\right) = - 3$
${y}_{\text{vertex}} = + 1$
The coefficient of ${x}^{2}$ = +1 so the graph is of shape type $\cup$
Thus the vertex is a minimum and above the x-axis
Thus there is no solution for $x$ at $y = 0$ where $x \in \mathbb{R}$
However their will be a solution for $x \in \mathbb{C}$ (Complex numbers)
Given that $0 = {\left(x + 3\right)}^{2} + 1$
$\implies \sqrt{{\left(x + 3\right)}^{2}} = \sqrt{- 1}$
$\implies \pm \left(x + 3\right) = i$
Updated: The $\pm$ is for the $x \text{ and } 3$. Not for the $i$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(brown)("Suppose we had "-(x+3)= i
This becomes: $- x - 3 = i$
$\implies - x = + 3 + i$
Multiply by (-1)
$\textcolor{b l u e}{\implies + x = - 3 - i}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Suppose we had } + \left(x + 3\right) = i}$
This becomes: $x = - 3 + i$
$\textcolor{b l u e}{\implies + x = - 3 + i}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Solution for $x$ at $y = 0$ is
$\textcolor{m a \ge n t a}{x = - 3 \pm i}$<|endoftext|>
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# Basic Basics of Statistics(Descriptive) & Probability for Data Scientists-A Quick Refresher
Probability deals with predicting the likelihood of future events, while statistics involves the analysis of the frequency of past events.
# PROBABILITY
In probability theory, an event is a set of outcomes of an experiment to which a probability is assigned. If `E` represents an event, then `P(E)` represents the probability that `E`will occur. A situation where `E` might happen (success) or might not happen (failure) is called a trial.
This event can be anything like tossing a coin, rolling a die or pulling a colored ball out of a bag. In these examples the outcome of the event is random, so the variable that represents the outcome of these events is called a random variable.
The empirical probability of an event is given by number of times the event occurs divided by the total number of incidents observed. If for`n`trials and we observe `s`successes, the probability of success is s/n.
Theoretical probability on the other hand is given by the number of ways the particular event can occur divided by the total number of possible outcomes.
## JOINT AND CONDITIONAL PROBABILITY
Joint Probability: Probability of events A and B denoted by`P(A and B) or P(A ∩ B)`is the probability that events A and B both occur.
`P(A ∩ B) = P(A). P(B)` . This only applies if `A`and `B`are independent, which means that if `A`occurred, that doesn’t change the probability of `B`, and vice versa.
Conditional Probability: When A and B are not independent, it is useful to compute the conditional probability, P (A|B), which is the probability of A given that B occurred: `P(A|B) = P(A ∩ B)/ P(B)`.
The probability of an event A conditioned on an event B is denoted and defined `P(A|B) = P(A∩B)/P(B)`
Similarly, `P(B|A) = P(A ∩ B)/ P(A)` . We can write the joint probability of as A and B as `P(A ∩ B)= p(A).P(B|A)`, which means : “The chance of both things happening is the chance that the first one happens, and then the second one given the first happened.”<|endoftext|>
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That’s the claim, but hydrogen still needs an energy source for its manufacture, so any sustainability depends on what that source is. And looking a bit closer, the report says: “If the extracted hydrogen gas is ultimately used as fuel, for example in a fuel cell of a car, the hydrogen reacts back to water with oxygen gas from the atmosphere.” So what happens when a hydrogen powered fuel cell vehicle trails water from its ‘exhaust’ on to a road in sub-zero temperatures? An icy, or more icy, surface seems likely to be the undesirable result.
The research group led by Leiden chemist Marc Koper has discovered a catalyst that minimizes the production of chlorine gas during salt water electrolysis, reports Phys.org.
The invention can enable the direct production of hydrogen from seawater.
The article has been published in the Journal of the American Chemical Society.
“In the electrolysis of salt water, such as seawater, the ultimate goal is to produce hydrogen at the cathode,” explains Ph.D. student Jan Vos from the Leiden Institute of Chemistry. “The product formed at the anode is ideally oxygen, because that is harmless to the environment.” However, during salt water electrolysis toxic chlorine gas can also form at the anode.
The researchers have now produced a catalyst that minimizes the formation of chlorine gas in favour of oxygen formation. Vos explains: “The catalyst consists of two metal oxides: iridium oxide with a layer of manganese oxide only a dozen nanometers thick. Iridium is a material that exhibits high catalytic activity for the formation of both oxygen gas and chlorine gas; the manganese oxide acts as a kind of membrane that prevents the supply of chloride ions and suppresses the formation of chlorine gas.”
The electrolysis of water is an important step for the production and use of hydrogen as an alternative energy carrier. An anode that counteracts the formation of chlorine gas enables water electrolysis where it is not necessary to first rid the water of dissolved salt, the process of which still costs significant amounts of energy and capital. It would allow the direct production of hydrogen from seawater, thereby relieving the rare freshwater reserves on earth.
According to Vos, a useful side effect of salt water electrolysis is the production of very pure fresh water. “If the extracted hydrogen gas is ultimately used as fuel, for example in a fuel cell of a car, the hydrogen reacts back to water with oxygen gas from the atmosphere. That way, the large-scale application of water electrolysis and hydrogen in fuel cells will lead to large quantities of this ‘waste product’: pure water. In a future where water shortages become an ever more acute problem, this would certainly not be undesirable.”
via Tallbloke’s Talkshop
August 11, 2018 at 11:46AM<|endoftext|>
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# Algebric Formulas
## 1. Binomial Theorem
(a + b)^0 = 1
(a + b)^1 = a + b
(a + b)^ 2 = a^2 + 2ab + b^2
(a-b)^ 2 = a^2-2ab + b^2
(a + b)^ 3 = a^3 + 3a^2 b + 3ab^2 + b^3
(a-b)^ 3 = a^3 -3a^2 b + 3ab^2-b^3
(a + b)^4 = a^4 + 4a^3 b + 6a^2 b^2 + 4ab^3 + b^4
And so on. These formulas are easy to remember because of their symmetry and these are used very frequently in Algebra. We will cover a number of examples that involve these formulas.
## 2. Difference of Two Squares Formula
x^2-y^2 = (x + y)(x-y)
This formula indicates that we can determine the difference of two squares simply my taking a product of the sum and the difference of the variables involved.
## 3. Sum / Difference of Two Cubes
x^3 + y^3 = ( x + y ) ( x^2-xy + y^2 )
x^3-y^3 = ( x-y ) ( x^2 + xy + y^2 )
## Evaluate (x + 5)^2
#### Explanation:
Using the formula for a perfect square involving the summation sign:
(a + b) ^2 = a^2 + 2ab + b^2
That is we need to square the first term, and then square the second term, and then take twice the product of the two terms and then combine the three results.
=> (x + 5)^2 = x^2 + 2(x)(5) + (5)^2
=> (x + 5)^2 = x^2 + 10x + 25
## Evaluate (2x-6)^2
#### Explanation:
Using the formula for binomial difference squared:
(a-b)^ 2 = a^2 -2ab + b^2
=> (2x-6)^2 = (2x)^2 -2(2x)(6) + (6)^2
=> (2x-6)^2 = 4x^2 -24x + 36
## Evaluate (2x-2)^3
#### Explanation:
Using the binomial theorem for cube
(a-b) ^3 = a^3 -3a^2 b + 3ab^2-b^3
=> (2x-2)^3 = (2x)^3-3(2x)^2 (2) + 3(2x)(2)^2-(2)^3
=> (2x-2)^3 = 8x3-3(4x2) (2) + 3(2x)(4)-8
=> ( 2x-2 )^3 = 8x^3-24x^2 + 24x-8
## Evaluate (2x + 5) (2x-5)
#### Explanation:
From the Difference of the Squares Formula, we know that
x^2-y^2 = (x + y) (x-y)
=> (2x + 5) (2x-5) = (2x)^2 -(5)^2
=> (2x + 5) (2x-5) = 4x^2 - 25
## Factorize x^3 + 27
#### Explanation:
To find total score for the two innings we have to add runs made in both innings.
We can express the given expression as the sum of cubes form:
x^3 + 27 = x3 + (3)3
=> x^3 + 27 = (x + 3) (x^2-(x)(3) + (3)2)
=> x^3 + 27 = (x + 3) (x^2-3x + 9)
## WorkSheets
#### Become a member today!
Register (it’s Free)<|endoftext|>
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Auxiliary verbs, also called helping verbs, are used with a main verb to create compound verb forms.
The simple present of the verb BE can be contracted in two ways: with a subject pronoun or with the negative adverb not
The simple present and simple past of the auxiliary verb DO are frequently contracted with the negative adverb not.
This lesson explains how to make contractions with the verb HAVE and a subject pronoun or the negative adverb not.
The phrase could have refers to something that was possible but did not occur in the past. In informal speech, it is contracted to could've, not could of.
This lesson introduces the various forms English uses to indicate the future.
Verbs are the words in a sentence that indicate an action, a state of being, or possession.
A list of the most common irregular English verbs, including their simple past conjugations and present and past participles.
This lesson explains the grammar of modal verbs and where they are placed in a sentence or question.
This lesson explains the meaning of each modal verb and provides example sentences.
The verb BE can be negated with negative adverbs, adjectives, and prefixes.
This lesson explains one of the very important roles of the verb DO: the formation of negative phrases.
The past participle is a verb form that indicates a completed action. It is used in perfect aspects, adverb clauses, and the passive voice.
The past perfect, or pluperfect, is a compound verb form which requires two verbs: had and a past participle.
The past perfect progressive, also called the pluperfect progressive is a compound verb form. It requires three verbs: had + been + verb+ing.<|endoftext|>
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These free printable First Grade Math Sheets will help your child to achieve their Elementary Math benchmark set out by Achieve, Inc. You may freely use any of the first grade math worksheets below in your classroom or at home. First grade math Here is a list of all of the math skills students learn in first grade! These skills are organized into categories you can move your mouse over any skill name to preview the skill. This is a suitable resource page for 1st graders teachers parents. Free 1st grade math coloring pages. First grade math worksheets include subtraction mixed operation addition subtraction worksheets to help build a strong math foundation. 1st grade math worksheets – Printable PDF activities for math practice. This page contains math worksheets for First grade children spatial sense, Time, Math Signs, Fractions, covers all topics of Kindergarten such as Graphs, Addition, Data, Shapes, patterns, Subtractions, Comparisons Mixed operations & more.
Exercises begin with simple subtraction facts using pictures progress to subtraction of 2- digit numbers in columns , first number lines followed by 1- digit subtraction facts , subtraction word problems. Kindergarten 4th Grade, 1st Grade, 3rd Grade, 5th Grade , 2nd Grade more! Grade 1 - first Math Worksheets ( Horizontal Subtraction) The worksheets are printable and the questions on the math worksheets change each time you visit. Cover all corners of first grade math from coins to measurement to two- digit numbers with our first grade math worksheets. With our math sheet generator you can easily create Grade 1 Subtraction worksheets that are never the same , always different providing you with an unlimited supply of math sheets to use in the. Math Subtraction Number Line Subtraction Subtraction Activities Math Games Math Activities Math Worksheets Number Line Activities 1st Grade Math Kindergarten Math Forward Subtraction on a Number Line - differentiated freebie sample first - Inspired in. First Grade Addition and Subtraction worksheets from Kids Academy offers your child valuable practice first learning new skills! MATH WORKSHEETS FOR FIRST GRADE first - PDF. These math sheets can be printed as extra teaching material for teachers extra math practice for kids as homework material parents can use.
Grade 1 subtraction worksheets Our grade 1 subtraction worksheets provide practice in solving basic subtraction problems. Just click on the math worksheet title and. math sheets work take away worksheets math problems online Single Digit first Subtraction Worksheets Free. Kids solve addition two- digit numbers to crack the code , subtraction problems with one- find the mystery word on this first first grade math worksheet. The best source for free subtraction worksheets. First grade math subtraction sheets. First grade subtraction worksheets for use at home or in the classroom. All the free First Grade first Math Worksheets in this section support the Elementary Math Benchmarks for First Grade. Here you will find a selection of First Grade Subtraction sheets designed to help your child improve their mental subtraction skills. First grade math subtraction sheets. The first following worksheets involve using the First Grade Math skills of subtracting. We offer 40 puzzles practicing addition subtraction, place value, shapes more. Printable First Grade Math Worksheets Math Fractions Worksheets Subtraction Activities Math Worksheet Answers Addition And Subtraction Games. 1st Grade Math Subtraction Showing top 8 worksheets in the category - 1st Grade Math Subtraction. Addition Subtraction in a Variety of Ways Endless math drills problems are boring for kids! First grade is a big year for math moving past counting into simple addition.
First Grade Math Worksheets. In the UK, 1st Grade is equivalent to Year 2. Using these sheets will help your child to: learn to subtract 2 digit numbers. Some of the worksheets displayed are Homework practice Sample work from, problem solving practice workbook, 1st grade, Time subtractionspeed drill 0 9, Single digit subtraction, subtraction work, Vertical addition , Single digit subtraction Touchmath second grade. Easier to grade more in- depth best of all.
Our subtraction worksheets are free to download, easy to use, and very flexible. These subtraction worksheets are a great resource for children in Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade. Click here for a Detailed Description of all the Subtraction Worksheets. grade 1 worksheets grade 1 printables grade one math Addition and Subtraction Worksheets for Grade 5.
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Maths Homework Sheets Activity Sheets For Class 1 Printable Math Sheets For 1st Grade Addition Subtraction Mixed Math Worksheets. First grade math worksheets. First grade math worksheets for children to supplement their math activities at home or in school.<|endoftext|>
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in
# How To Find Slope Of A Line
Finding the equation of a line given a point and a slope. Slope, sometimes referred to as gradient in mathematics, is a number that measures the steepness and direction of a line, or a section of a line connecting two points, and is usually denoted by m.generally, a line's steepness is measured by the absolute value of its slope, m.the larger the value is, the steeper the line.
Slope Intercept Form Vertical Line 8 Fantastic Vacation
### The slope of this line, which is often denoted by the letter m, is your rate of change of y with respect to x.
How to find slope of a line. Doing the manipulative mathematics activity “slope of lines between two points” will help you develop a better understanding of how to find the slope of a line from its graph. So, the equation of the line is x = a x = a. The slope of a line is the direction in which the line goes, and its steepness.
But where do we start? And the y value over here is y sub 1. Find the slope of the followingline.
If the equation of a line is given general form, we can find the slope of the line using the formula given below. Finding the equation of a line given two points. It gives you the slope of a curve at a single point.
To find the slope of a line, pick any two coordinates on the line, and choose one of the points to be the dominant point. An equation for a zero slope line will be y = b, where the line’s slope is 0 (m = 0). Let’s take a look at a few examples!
The slope is…• orange line 0• green line undefined 22. Y = mx + b. Input the values into the formula.
We can write the equation of a line in slope intercept form and find the slope 'm' which is being as the coefficient of x. Find the slope of these lines. To find the slope, we must count out the rise and the run.
Slope intercept form equation of a line : Finding the slope of a line from the equation. Find the slope of the line below.
A line with a positive slope increases if you look at it from left to right. Zero slope, m = 0 m = 0, if a line y = mx + b y = m x + b is horizonal. Given any two points on a line, (x 1, y 1) and (x 2, y 2), we can calculate the slope of the line by using this formula:.
Formula to find the slope of the line passing through two points is. The direction can be either positive or negative. The slope formula calculator uses the simple and smart formula for \((m)\) or gradient to do calculations.
Here a is the slope of the line. We could plot the points on grid paper, then count out the rise and the run, but there is a way to find the slope without graphing. Y = mx + b.
This worksheet includes some points located at the origin of the coordinate graph. And the best way to view it, slope is equal to change in y over change in x. Find the slope of the line in the graph.
And just as a bit of a review, slope is just telling us how steep a line is. We locate any two points on the line. You can see that the slope of the parabola at (7, 9) equals 3, the slope of the […]
And for a line, this will always be constant. The line makes angle of 30° from positive y axis hence it makes angle of (90° + 30°) from positive x axis i.e. Find the slope of a line given two points on the line sometimes we need to find the slope of a line between two points and we might not have a graph to count out the rise and the run.
General form equation of a line : A slope can be positive or negative. How to find slope with this slope calculator:
So just as a review, the slope of this line, and a line by definition, has a constant slope between any two points that you pick. Simplify the fraction if possible. So this is the point x sub 1, y sub 1.
You can find the slope of any line by following these three easy steps: Determine if the slope if positive (increasing) or negative (decreasing) step two: A line can be represented with a linear function y = ax + b.
Using two points on the line, calculate the rise and the run and express it as a fraction (rise over run). First, look at this figure. You can perform calculations with the following:
This is because division by zero leads to infinities. Remember, if the numerator and denominator are both negative, then the negative signs cancel out, and the fraction (and slope) is positive. Find the slope of these lines 21.
Undefined slope, if a line y = mx + b y = m x + b is vertical. How do you find a perpendicular slope? A line with a negative slope is decreasing.
Find the slope of the line. In this case, the equation of the line is y = b y = b; How to find the slope of line when its equation is given,ax+by=c calculator, ax+by+c=0, meaning,ax+by+c=0 solve for y, ax+by=c given two points, ax+by=c what is c, slope formula, ax+by=c meaning,slope of a line formula, slope of a line calculator, how to find the slope of a graph,how to find slope from an equation, slope of a line definition,slope formula example,slope definition,slope of a.
The Finding Slope and yintercept from a Linear Equation
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Slope Card Sort Them, Equation and An<|endoftext|>
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# I Proof of an inequality
1. May 29, 2017
### stevendaryl
Staff Emeritus
I'm pretty sure that the following is true, but I don't see an immediate compelling proof, so I'm going to throw it out as a challenge:
Let $A,A', B, B'$ be four real numbers, each in the range $[0,1]$. Show that:
$AB + AB' + A'B \leq A' B' + A + B$
(or show a counter-example, if it's not true)
This inequality was inspired by Bell's Theorem, but that's not relevant to proving or disproving it.
2. May 29, 2017
### Staff: Mentor
Consider the region $A+A' \leq 1$ first.
For B=0, the inequality simplifies to $AB' \leq A'B'+A$ which is true.
The derivatives with respect to B are $A+A'$ and $1$, respectively, which means the derivative for the left hand side is smaller or equal. For the chosen region this means the inequality stays true for all B.
Now consider the region $A+A' > 1$.
For B=1, the inequality simplifies to $AB'+A' \leq A'B'+1$. As $B'+A' \leq A'B'+1$, this inequality is satisfied.
The derivatives with respect to B are $A+A'$ and $1$, respectively, for the chosen region the derivative on the left hand side is larger, but now we are going backwards with B. For the chosen region this means the inequality stays true for all B.
For $A=1-A'$, the inequality can be written as $AB+AB'+(1-A)B \leq (1-A)B'+A+B$ which simplifies to $2AB' \leq B'+A$, that inequality does not depend on B and it is true with enough margin to make the derivatives work on both sides.
3. May 29, 2017
### stevendaryl
Staff Emeritus
Thanks! That's a little more elegant than the only way I found to prove it, which was to do an exhaustive case split.<|endoftext|>
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Important Formula and Equations
Minute Spaces: The face or dial of watch is a circle whose circumference is divided into 60 equal parts, called minute spaces.
Hour Hand and Minute Hand: A clock has two hands, the smaller one is called the hour hand or short hand while the larger one is called minute hand or long hand .
In 60 minutes, the minute hand gains 55 minutes on the hour on the hour hand.
In every hour, both the hands coincide once.
The hands are in the same straight line when they are coincident or opposite to each other.
When the two hands are at right angles, they are 15 minute spaces apart.
When the hands are in opposite directions, they are 30 minute spaces apart.
Angle traced by hour hand in 12 hrs = 360°
Angle traced by minute hand in 60 min. = 360°.
If a watch or a clock indicates 8.15, when the correct time is 8, it is said to be 15 minutes too fast .
On the other hand, if it indicates 7.45, when the correct time is 8, it is said to be 15 minutes too slow .
Odd Days: We are supposed to find the day of the week on a given date. For this, we use the concept of ‘odd days’. In a given period, the number of days more than the complete weeks are called odd days .
Leap Year: (i). Every year divisible by 4 is a leap year, if it is not a century. (ii). Every 4th century is a leap year and no other century is a leap year.
Ordinary Year: The year which is not a leap year is called an ordinary years . An ordinary year has 365 days.
Counting of Odd Days: 1 ordinary year = 365 days = (52 weeks + 1 day.) . 1 ordinary year has 1 odd day.
1 leap year = 366 days = (52 weeks + 2 days) 1 leap year has 2 odd days.
100 years = 76 ordinary years + 24 leap years
= (76 x 1 + 24 x 2) odd days = 124 odd days.
= (17 weeks + days) = 5 odd days.
Number of odd days in 100 years = 5
Number of odd days in 200 years = (5 x 2)= 3 odd days.
Number of odd days in 300 years = (5 x 3) = 1 odd day.
Number of odd days in 400 years = (5 x 4 + 1) = 0 odd day.
Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days.
Day of the Week Related to Odd Days (Assuming that 1AD January 1st is a Sunday):
No. of days: 0 1 2 3 4 5 6
Day: Sun. Mon. Tues. Wed. Thurs. Fri. Sat.
Key Notes:
Clocks Concepts :
The dial of the clock is circular in shape and was divided into 60 equal minute spaces
60 minute spaces traces an angle of 3600. Therefore, 1minute space traverses an angle of 60
In 1 hour, Minute hand traverses 60 minute space or 3600 ,Hour hand traverses 5 minute space or 300
The hands of the clock are perpendicular in 15 minute spaces apart
The hands of the clock are in straight line and opposite to each other in 30 minute spaces apart.
The hands of the clock are in straight line when they coincide or opposite to each other.
The hands of the clock are perpendicular to each other for 22 times in 12 hours and for 44 times in a day.
The hands of the clock are opposite to each other for 11 times in 12 hours and 22 times in a day.
The hands of the clock coincides with each other for 11 times in 12 hours and 22 times per day
The hands of the clock are 44 times in a straight line per day
The minute hand gain 55 minutes over hour hand per hour.
Hence x minute space to be gained by minute hand over hour hand can be calculated as x.(60/55) or x.(12/11)
Ex : At what time between 2’O clock and 3’O clock the hands of the clock
are opposite to each other.
1. 34( 6/11 ) past 2’Oclock 2. 43( 7/11 ) past 2’Oclock
3. 56( 8/11 ) past 2’Oclock 4. 64(9/11past 2’Oclock
Sol At 2’O clock the minute hand will be at 12 as shown below
The minutes hand to coincide with the hour hand it should trace at first 10 minute spaces
And then the hands of the clocks to be opposite to each other minute hand should trace 30 minute spaces i.e. totally it should gain 10+30=40 minute spaces to be opposite to that of hour hand
We know that,
Minute hand gains 55 minute spaces over hour hand in 1 hour
Therefore, Minute hand gain 40 minute spaces over hour hand in 40 × (60/55) = 43(7/11)
Hence the hand of the clock will minutes be opposite to each at 43( 7/11 ) past 2’Oclock
Therefore, Correct option is 2′
When clock is too fast, too slow
If a clock or watch indicates 6 hr 10 min when the correct time is 6, it is said that the clock is 10 min too fast
If it indicates 6. 40 when the correct time is 7, it is said to be
20 min too slow.
Now let us have an example based on this concept
Ex. My watch, which gains uniformly, is 2 min, & show at noon on Sunday, and is 4 min 48 seconds fast at 2 p.m on the following Sunday when was it correct ?
Sol: From Sunday noon to the following Sunday at 2 p.m there are 7 days 2 hours or 170 hours.
The watch gains 2+4 4/5 min in 170 hrs.
Therefore, the watch gains 2 min in 2 *170 hrs i.e., 50 hours
6 4/5
Now 50 hours = 2 days 2 hrs.
Therefore, 2 days 2 hours from Sunday noon = 2 p.m on Tuesday.
Calendars Concept :
The time in which the earth travels round the sun is a solar year and is equal to 365 days 5 hrs. 48 minutes and 47 1/2 seconds
Year is 365.2422 days approximately.
The common year consists of 365 days.
The difference between a common year and a solar year is therefore 0.2422 of a day and we consider it by adding a whole day to every fourth year.
Consequently in every 4th year there are 366 days.
The years which have the extra day are called leap years. The day is inserted at the end of February, The difference between 4 common years and 4 solar years is 0.969 of a day.
If therefore, we add a whole day to every 4th year, we add too much by 0.0312 of a day. To take account of this, we omit the extra day three times every 400 years,
The thing is to ensure that each season may fall at the same time of the year in all years. In course of time, without these corrections, we should have winter in July and summer in January also .
With the very small variation, the present divisions of the year are those given in B.C 46 by Julius Caesar . The omission of the extra day three times in 400 years is called the Gregorian Correction. This correction was adopted at once in 1582 in Roman Catholic Countries. but not in England until, 1752.
The Gregorian mode of reckoning is called the New Style, the former, the Old Style.
The New Style has not yet been adopted in Russia , so that they are now 13 days behind us as an Example What we call Oct. 26th they call 13th Oct . They have Christmas day on 7th of January and we have on 25th December every year.
In an ordinary year there are 365 days i.e., 52 weeks + 1 day
Therefore an ordinary years contains 1 odd day.
A leap year contains 2 odd days.
100 year = 76 ordinary years + 24 leap years.
= 76 odd days + 48 odd days
= 124 odd days = 17 weeks + 5 days. (in the consideration of weeks)
Therefore, 100 years contain 5 odd days.
200 years contains 3 odd days.
300 years contain 1 odd days
Since there are 5 odd days in 100 years, there will be 20 days in 400 years. But every 4th century is a leap year.
Therefore, 400 years contain 21 days. Here 400 years contain no odd days.
As First January 1 AD was Monday. we must count days from Sunday
i.e. Sunday for 0 odd days , Monday for 1 odd day , Tuesday for 2 odd days and so on.
Last day of a century cannot be either Tuesday. Thursday or Saturday.
The first day of a century must either be Monday. Tuesday, Thursday or Saturday.
Now let us observe the Examples
Ex How many times does the 29th days of the month occur in 400 consecutive years
1) 97 times 2) 4400 times 3) 4497 times 4) none
Sol: In 400 consecutive years there are 97 leap years. Hence in 400 consecutive years, February has the 29th day 97 times, and the remaining 11 months have the 29th day 400 x 11 or 4400 times.
Therefore, 29th day of the month occurs (4400 + 97) or 4497 times
Ex Given that on 10th November 1981 is Tuesday, what was the day on 10th November 1581
1) Monday 2) Thursday 3) Sunday 4) Tuesday
Sol: After every 400 years, the same day comes.
Thus if 10th November1981 was Tuesday, before 400 years i.e on 10th November 1581, it has to be Tuesday.
10 questions Exercise questions
1. What is the angle between the two hands of a clock when the time shown by the clock is 6.30 p.m. ?
a) 00
b) 50
c)30
d)150
Ans: option d
Explanation: q = 11/2 m – 30h
= 11/2 *30 – 30 *6
= /165-180/ = 150
2. At what time between 3 and 4 o’clock will the minute hand and the hour hand are on the same straight line but facing opposite directions.
a) 3:49
b)3:15
c)3:39 1/11
d)3:49 1/11
Ans: option d
Explanation : On straight line means 180 degree angle.
180 = 11/2m- 30h
180 = 11/2 m –30*3
180 = 11/2m-90
(180+90)2 = 11m
m = 540/11 = 49 1/11
3. By how many degrees does the minute hand move in the same time, in which the hour hand move by 280 ?
a)168
b)336
c) 196
d) 376
Ans: option b
Explanation: 28*2 *6 = 3360
4. At what time, between 3 o’clock and 4 o’clock, both the hour hand and minute hand coincide each other?
a) 3:30
b) 3:16 4/11
c) 3:1611/4
d) 3:16 7/11
Ans: Option b
Explanation : Coincide means 00 angle.
0 =11/2m –30*3
11m = 90*2 = 180
m= 180/11 = 16 4/11
So time = 3 : 16 4/11
5. How many degrees will the minute hand move, in the same time in which the second hand move 4800 ?
a ) 60
b) 90
c) 40
d) 80
Ans: Option d
Explanation : Minute hand covers 480/60= 80
6. How many years have 29 days in February from 2001 to 2100.
a)26
b)25
c)23
d)24
Ans: option d
Explanation: 100th year is not a leap year. So 24 February’s has 29 days
7. 2012 January 1st is Sunday, then which day is the Indian Independence day of the same year.
a) Saturday
b) Wednesday
c) Thursday
d) Friday
Ans: Option b
Explanation: 30+ 29+ 31 + 30 + 31 + 30 + 31+15 = 227/7 = reminder = 3
So Independence day is Wednesday
8. Which year has the same calendar as 1700 ?
a) 1705
b)1706
c)1707
d)1708
Ans: Option b
Explanation:
Year : 1700 1701 1702 1703 1704 1705
Odd days : 1 1 1 1 2 1
9. If Arun’s birthday is on May 25 which is Monday and his sister’s birthday is on July 13. Which day of the week is his sister’s birthday?
a) Monday
b) Wednesday
c) Thursday
d) Friday
Ans: option a
Explanation: Reference day : May 25th Monday
Days from May 25th to July 13 = 6 + 30 +13 = 49
No of odd days : 49/7 = 0
10. March 1st is Wednesday. Which month of the same year starts with the same day?
a) October
b) November
c) December
d) None of these
Ans: Option b
Explanation:
Month : Mar April May June July Aug Sept. Oct.
Odd dys : 3 2 3 2 3 3 2 3
Total 21 odd days. 21/7 = 0. So November has start with the same day<|endoftext|>
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# Matrices
• Jan 25th 2013, 08:17 AM
MMCS
Matrices
A transformation in the plane is represented by a 2 x 2 matrix.
Under this transformation, the point (7,7) is transformed to (-5,-6) and the point (-2,8) is transformed to (-4,2)
Can someone help me to find the transformation matrix for this?
I dont have any working to show as i dont know where to start
Thanks
• Jan 25th 2013, 09:01 AM
Soroban
Re: Matrices
Hello, MMCS!
Quote:
A transformation in the plane is represented by a 2 x 2 matrix.
Under this transformation, the point (7,7) is transformed to (-5,-6) and the point (-2,8) is transformed to (-4,2).
Find the transformation matrix.
You can start by letting the transformation matrix be: .$\displaystyle A \;=\;\begin{bmatrix} a&b \\ c&d\end{bmatrix}$
Then: .$\displaystyle (7,7)\begin{bmatrix}a&b\\c&d\end{bmatrix} \;=\;(\text{-}5,\text{-}6) \quad\Rightarrow\quad \begin{Bmatrix}7a + 7c &=& \text{-}5 \\ 7b + 7d &=& \text{-}6 \end{Bmatrix}$
Also: .$\displaystyle (\text{-}2,8)\begin{bmatrix}a&b\\c&d\end{bmatrix}\;=\;( \text{-}4,2) \quad\Rightarrow\quad \begin{Bmatrix}\text{-}2a + 8c &=& \text{-}4 \\ \text{-}2b + 8d &=& 2 \end{Bmatrix}$
We have two systems of equations:
. . . . . . . . . . . $\displaystyle \begin{Bmatrix}7a + 7c &=& \text{-}5 \\ \text{-}2a+8c &=& \text{-}4\end{Bmatrix} \qquad\begin{Bmatrix}7b+7d &=& \text{-}6 \\ \text{-}2b + 8d &=& 2 \end{Bmatrix}$
Their solutions are: .$\displaystyle \begin{Bmatrix} a&=& \text{-}\frac{6}{35} \\ \\[-4mm] c &=& \text{-}\frac{19}{35} \end{Bmatrix} \qquad \begin{Bmatrix}b &=& \text{-}\frac{31}{35} \\ \\[-4mm] d &=& \frac{11}{35} \end{Bmatrix}$
Therefore: .$\displaystyle A \;=\;\begin{bmatrix}\text{-}\frac{6}{35} & \text{-}\frac{31}{35} \\ \\[-3mm] \text{-}\frac{19}{35} & \frac{1}{35} \end{bmatrix}$
• Jan 25th 2013, 09:30 AM
ILikeSerena
Re: Matrices
Here's an alternative method:
We have:
$\displaystyle A \begin{pmatrix}7 \\ 7\end{pmatrix} = \begin{pmatrix}-5 \\ -6\end{pmatrix} \qquad A \begin{pmatrix}-2 \\ 8\end{pmatrix} = \begin{pmatrix}-4 \\ 2\end{pmatrix}$
Therefore:
$\displaystyle A \begin{pmatrix}7 & -2 \\ 7 & 8\end{pmatrix} = \begin{pmatrix}-5 & -4 \\ -6 & 2\end{pmatrix}$
$\displaystyle A = \begin{pmatrix}-5 & -4 \\ -6 & 2\end{pmatrix} \begin{pmatrix}7 & -2 \\ 7 & 8\end{pmatrix}^{-1}$
• Jan 25th 2013, 09:45 AM
HallsofIvy
Re: Matrices
Yet a third way: (1, 0)= a(7, 7)+ b(-2, 8) for some a and b. That gives the two equations 7a- 2b= 1 and 7a+ 8b= 0. If we subtract the first equation from the second we get 10b= -1 so that b= -1/10. Putting that back into the first equation, 7a+ 2/10= 1 so 7a= 8/10 and a= 8/70. That is (1, 0)= (8/70)(7, 7)+ (-1/10)(-2, 8) and then A(1, 0)= (8/70)A(7, 7)+ (-1/10)A(-2, 8)= (8/70)(-5, -6)+ (-1/10)(-4, 2)= (8/70, -31/35). The point of reducing to (1, 0) is that $\displaystyle \begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ c \end{bmatrix}$. That is the first column of the matrix is whatever A maps (1, 0) to. This tells us that the first column of the matrix is $\displaystyle \begin{bmatrix}\frac{8}{70} \\ -\frac{31}{35}\end{bmatrix}$.
Now, find a, b such that a(7, 7)+ b(-2, 8)= (0, 1) to find the second column.<|endoftext|>
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Our Language Arts instruction is divided into the following areas: phonics, fluency, vocabulary, writing, and comprehension. Comprehension skills and strategies are introduced during a shared reading experience. All students have equal access to the text and strategies such as predicting, questioning and making connections are modeled. Students are learning to think as “good readers” do. The modeled skills are then practiced in small guided reading groups in which the text is at the student’s reading level. The ultimate goal is for every reader to apply these thinking and organizing skills to their independent reading.
Phonics instruction is sequential. Each new skill builds on previous skills taught. We are currently reviewing short vowel sounds. It is essential that they can distinguish between these sounds. We will also be checking to be sure that the phonemic and phonological skills that were taught in kindergarten are now being independently applied in class. This includes rhyming, counting syllables, and blending and segmenting sounds to make words. Phonics instruction is made fun with word sorts, games, and hands on activities. In first grade, phonics in explicitly taught for twenty minutes a day.
Students are practicing reading with fluency and expression. Fluency instruction is done primarily through reading and rereading of poetry and reader’s theater in partners and small groups. Our vocabulary instruction is focusing on the meaning of new words. We are learning to use context clues and prior knowledge to figure out the meaning of new words.
The final component of our language arts instruction is writing. We will have a strong focus on the six traits of writing (ideas, organization, conventions, word choice, sentence fluency, and voice). Students will work through the writing process (prewriting, first draft, editing, revising, and publishing) to publish their writing. They will be writing to persuade, inform, and entertain this year.
We are reviewing number concepts and creating a mathematical community in which students can talk about their thinking and solve problems in ways that make sense to them. The primary guide for the teaching and learning in mathematics this year is Investigations. This curriculum is designed to engage students in making sense of mathematical ideas. There is a focus on mathematical fluency and making connections between math concepts. Investigations is based on experience from research and practice. Based on extensive classroom testing, it takes seriously the time students need to develop a strong foundation in math concepts and skills.
Social Studies / Science
We are spending some time getting to know each other and talking about how character counts in school. We will be beginning our study of safety. We will explore ways to stay safe in school, on the bus, and at home.
- We would love any donations of zip lock baggies, tissues, or stickers.
- Be sure to return your child’s homework folder and reading book each day.
- We appreciate our classroom volunteer helpers. A schedule of volunteers will be coming home soon. Let us know if you did not sign up but would like to come in and help.
We look forward to an exciting year. We know that working together we can ensure a fun and successful experience in first grade for your child.
The First Grade Team:<|endoftext|>
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Question
Q&A Session
1. WHAT IS THE FACTORIAL OF 100 IN VOICE? What is the Factorial of Hundred (100)?
The factorial of a number is the product of the factorials of its individual digits. For example, the factorial of 4 is 120.
What is a Factorial?
In mathematics, the factorial of a non-negative integer n is the product of all integers between 1 and n, inclusive. For example, 3! = 3*2*1 = 6. The factorial of 100 is 10 since 100=10*9*8*7…
What is the Factorial of Hundred (100)?
The factorial of a number is the product of all the factorials of that number.
For example, the factorial of 100 is 1000 because 100 x 101 = 1000 and 100 x 102 = 10002. The factorials are always written in parentheses, like this: (100).
How to Calculate the Factorial of Hundred (100)?
The factorial of a hundred is 1,000. To calculate this, take the product of all the integers from one (1) to 100.
2. WHAT IS THE FACTORIAL OF 100 IN VOICE
Have you ever wondered what the factorial of 100 is? The answer is a mind-blowing number! Calculating the factorial of any number can be quite challenging, but luckily there are tools out there that make this process easier. In this blog post, we will explore what a factorial is and how to calculate it in voice. We will also discuss different techniques such as recursion and iteration that can be used to get the desired result. Finally, we will look at an example given in voice to explain the concept more clearly. By the end of this post, you should have a better understanding of what a factorial is and how to calculate it in voice.
What is the factorial of 100?
The factorial of 100 is the product of all positive integers less than or equal to 100. The factorial of a number is often denoted by the symbol n!.
For example,
5! = 5 x 4 x 3 x 2 x 1 = 120.
The factorial of 100 is 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000.
The different types of factorials
There are three different types of factorials: complete, partial, and rising.
A complete factorial is when all the factors of a number are multiplied together. For example, the complete factorial of 5 would be 5 x 4 x 3 x 2 x 1 = 120.
A partial factorial is when only some of the factors of a number are multiplied together. For example, the partial factorial of 5 with two terms would be 5 x 4 = 20.
A rising factorial is when a number is multiplied by successive numbers starting at that number. For example, the rising factorial of 5 would be 5 x 6 x 7 x 8 = 1680.
How to calculate the factorial of 100
To calculate the factorial of 100, you would need to multiply all integers from 1 up to and including 100. This can be done using a calculator, or by writing out the calculation in long hand.
If you are doing the calculation in long hand, you would start by multiplying 1 x 2 to get 2. Then, you would take that answer (2) and multiply it by 3 to get 6. You would continue this pattern until you reach 100 x 101, which would give you the final answer of 10,206.
The benefits of knowing the factorial of 100
While many people may not know what the factorial of 100 is, there are actually quite a few benefits to knowing this number. For one, it can help you understand math concepts better. Additionally, it can help you solve problems more quickly and easily. Finally, it can serve as a useful tool for memory recall.<|endoftext|>
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Members of Antioch Baptist Church in 2010
A drive through Atlanta’s Vine City/English Avenue neighborhood is remarkable for both the abundance of black churches present and the impression of their vitality in an otherwise depressed cityscape. It seems as if there is at least one well-kept church per block in the area, whereas over 40% of houses are abandoned and many businesses have shut their doors (2010 Census). High membership and daily programming also conveys church vitality. Dr. Cameron Alexander, the pastor of Antioch Baptist Church, even had a street renamed after him in 2010, when his church had 12,000 members (Atlanta City Council). Since the population of English Avenue and Vine City was 6,148 according to that year’s census, some churches are drawing congregants from outside the neighborhood (2010 Census). Why are the black churches in Vine City/English Avenue flourishing in spite of the neighborhood’s deterioration?
The black Christian church was established due to the proselytism of slave owners. Social, political, and economic institutions were purposefully destroyed through enslavement. As it was the only institution, the church fulfilled the role of political, economic, and social institution in African American society after emancipation. Free blacks in the North formed churches separate from white churches in the late 18th century, and in the South, whites relinquished control over black congregations after emancipation (DuBois). After slavery ended, whites prevented blacks from participating in political and social institutions in the larger American public realm. This exclusion from public life, coupled with the intentional destruction of African American institutions, reinforced the church’s role as a political and social institution as well as a religious one. Virtually all public services and future institutions were created through the church, including schools, vocational training, libraries, athletic clubs, and insurance companies (Higgenbotham, 7-9). In the late 1800s, during the worst period of American race relations, the church became a place where African Americans could exercise political action, create political and social structures, and safely express dissent with the U.S. government and repressive dominant institutions.
In the early to mid-20th century, many African Americans moved from the rural South to urban areas across America. Urbanization changed the class dynamics of African Americans. Many joined the middle class or strove to do so, which led to a rise in competing religious groups and educational gulfs between denominations. At the same time, African Americans were making strides in their civil rights, leading to greater integration and participation in the public sphere. As they gained the right to vote and as schools were desegregated, the black church lost its place as the primary political and social institution of the African American population. Giving up political power to secular organizations was a strategic move that allowed churches to “develop secular vehicles in order to cope with more complex and pluralistic urban environments” (ibid, 9). Membership lists and goals often overlapped. These organizations allowed church leaders to continue influencing American politics after public attention turned toward the separation of church and state.
The fact that the church continues partially to function as an economic, social, and political institution, and that it has successfully worked toward racial equality in changing social circumstances, is pivotal to understanding its importance in Vine City/English Avenue. The neighborhood began to decline in the wake of suburban flight in the 1970s; empty houses gave way to crime, addiction, and a high rate of poverty. Today, Vine City/English Avenue residents are trying to overcome this challenge. Many of the social services in the neighborhood are created or administered by church leaders and laity. The New Life Covenant Church established the Mattie Freeland Community Garden to serve as a safe recreation space. Antioch Baptist Church created housing for recovering addicts and those who are HIV positive. Many smaller churches in the neighborhood offer daycare and credit unions. Black churches in the neighborhood continue to function as spaces for resistance to oppressive social orders, primarily through institution building and the implementation of social service programs.
Why are the churches in Vine City/English Avenue thriving when secular organizations could build institutions and implement social services equally well? Black theology’s emphasis on freedom for all African Americans, rather than the freedom of individual choice, connects otherworldly salvation directly to this-worldly action (ibid, 5). Salvation comes to those attempting to promote God’s vision of freedom and equality. Freedom is seen as a condition for spiritual willingness, so to be saved one must work toward social justice. Everyday activities that are tied to the church, such as tithing, can function on multiple levels (Frederick). Tithing can be seen as a way of honoring God, working toward salvation, and contributing to the success of African Americans through the programming which results from this funding. Prioritizing tithing over other financial obligations and tithing to purposefully support the Vine City/English Avenue residents can also be a way of expressing agency in the face of a repressive social order. Similar arguments can be made for other church involvement in Vine City/English Avenue, from cooking meals in the church kitchen to joining a men’s club. In this neighborhood, the church provides a religious imperative to advance social justice, a means to express agency through manageable everyday actions, and the social cohesion that reinforces the behavior.
The history of the black church has prepared these neighborhood churches to thrive in the absence of other infrastructure and meet the demands of being a primary social, political, and economic institution in the neighborhood. Black churches in this neighborhood also flourish because they meet their laity’s religious needs through legitimating core values of African Americans (such as freedom and equality) and providing the means to reshape the social structures that constrain them.
City of Atlanta 2010 Census Summary Report Neighborhood Planning Unit L.” City of Atlanta. Accessed Oct 21, 2013.
Du Bois, W. E. B, and Phil Zuckerman. Du Bois on Religion. Walnut Creek, CA: AltaMira Press, 2000. and Frazier, Edward Franklin. The Negro Church in America. New York: Schocken Books, 1969.
Frederick, Marla F. Between Sundays: Black Women and Everyday Struggles of Faith. Berkeley: University of California Press, 2003.
Higginbotham, Evelyn Brooks. Righteous Discontent: The Women’s Movement in the Black Baptist Church, 1880-1920. Cambridge: Harvard University Press, 1993.
Lincoln, C. Eric, Mamiya, Lawrence H. The Black Church in the African-American Experience. Durham: Duke University Press, 1990.
Ordinance to Rename Kennedy Street NW.” Atlanta City Council. Accessed Oct 21, 2013.<|endoftext|>
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The Real Number System
The Real Number System. Rational Numbers. Irrational Numbers. Real Numbers (all numbers are real). …any number that is not rational. …-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5…. Natural Numbers. Example: = 3.14159…… e= 2.71828….. . Whole Numbers. Integers.
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The Real Number System
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Presentation Transcript
1. The Real Number System Rational Numbers Irrational Numbers Real Numbers (all numbers are real) …any number that is not rational …-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5… Natural Numbers • Example: • = 3.14159…… e= 2.71828….. Whole Numbers Integers rational number is a number that can be written as one integer over another: …-2/3, 5/1, 17/4….
2. Inequality notation Read left to right … a<ba is lessthanba<b a is less than or equal to ba >ba is greater thanba >b a is greater than or equal to b a = b a is not equal to b
3. The Real Number Line Any real number corresponds to a point on the real number line. Order Property for Real Numbers Given any two real numbers a and b, - if a is to the left of b on the number line, then a < b. - if a is to the right of b on the number line, then a > b.
4. Basic Properties of Real Numbers Did you ever notice when you perform SOME arithmetic operations the order which you perform them does not affect the answer? Example • Add: 4 + 3 = • Add: 3 + 4 = Commutative Property of Addition & Multiplication • Multiply: 4 x 3 = • Multiply: 3 x 4 = a + b = b + a a x b = b x a Think: Does this property apply to subtraction and division. Give an Example!
5. Basic Properties of Real Numbers 2 x ( 4 + 3 ) = 2 x ( 7 ) = 14 Consider the following example solved in two ways: 2 x ( 4 + 3 ) = 2 x 4+2 x 3= 8 + 6 = 14 Same result Distributive Property of Real Numbers a (b + c) = ab+ac a (b – c ) = ab–ac
6. Absolute Value of a Number • The absolute valueof x, notated | x |, measures theDISTANCEthat x is away from the origin(0) on the real number line. Example: absolute value of 3
7. is the reciprocal of 3 Clearly 3 is the reciprocal of The Reciprocal of a Number One number is the reciprocal of another if their product is 1. In general: The reciprocal of a fraction is obtained by interchanging the numerator and the denominator, i.e. by inverting the fraction. Example: The reciprocal of is
8. Absolute Value of a Number Note:Distance is always going to be positive(unless it is0) whether the number you are taking the absolute value of is positive or negative. Example: absolute value of - 3
9. Wrap up - Objectives • Identify what numbers belong to the set of • natural numbers, whole numbers, integers, rational numbers, • irrational numbers, and real numbers. • Use the Order Property for Real Numbers. • Find the absolute value of a number. • Write a mathematical statement with an equal sign or an • inequality. • Know and understand scientific notation.
10. Fractions • A numeric FRACTION is a quotient of two numbers. Numerator , denominator = 0 Fraction = Denominator Fraction Improper Fraction Proper Fraction Numerator is smaller than denominator Numerator is larger than denominator
11. Fractions • Mixed numbers expression consisting of a whole number and a proper fraction: Convert mixed number to improper fraction: • Equivalent fractions fractions that represent the same number are called equivalent fractions.
12. Fractions Reducing fractions Steps: • List the prime factors of the numerator and denominator. • Find the factors common to both the numerator and denominator and divide the numerator and denominator by all common factors (called canceling). • Reduce to the lowest terms. Example: Reduce to the lowest terms. Step 1 Step 2 Step 3
13. Fractions • LCD – Least common denominator of two fractions: is the smallest number that can be divided by both denominators. Note: we need to find the LCD before we add or subtract two fractions. LCD of 3 and 5 = 15 (smallest number that can be divided by both 3 and 5)
14. the factors here are 2 and 3 LCD = Fractions Find LCD and add the fractions 1. We write each denominator as a product of its prime factors. 1. 2. LCD = product of all factors taken at the biggest power
15. Fractions • Write a fraction as a decimal: use a calculator to divide numerator by the denominator. If the division comes to an end , the decimal is a terminating decimal, if the division never ends the decimal is a repeating decimal. 1.Terminating decimal : 2. Repeating decimal: • Write a fraction as a percent: • A percent is a ratio of a number to 100. Percent means "per hundred." Thus, 20 • percent, or 20%, means . Fraction Decimal Percent x 100 = =
16. Wrap up - Objectives • Know what the numerator and denominator of a fraction are. • Simplify a fraction. • Find the least common denominator of given fractions. • Multiply, divide, add and subtract fractions. • Write a fraction in decimal form and as a percent. • Know how to do percent increase and percent decrease word problems
More Related<|endoftext|>
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Over the course of the last few weeks, I’ve dove into the wide world of algorithms in computer programming. Although they can be intimidating at first, constructing algorithms is just the coding version of solving a puzzle, which activates the creative problem-solving capacity of your brain and can actually become quite fun.
At a high level, an algorithm is just a procedure for taking an input of data (which can consist of words, numbers, or other data types), and transforming that data into some form of desired end result.
Approaching an algorithm question for the first time is a daunting task. Your first reaction might be to start writing code, but more often than not that will not prove useful. Instead, I like to follow a process that delays the actual code writing until the problem is fully understood.
Let’s step through solving an algorithm with a practice problem.
“Given an input n, where n is less than or equal to 50, calculate that input’s value in the Fibonacci sequence."
The Fibonacci sequence is a famous mathematical construct of a group of integers where each number is the sum of the previous two. Here's an example of the sequence revealing itself in nature.
1. Read the Problem
Maybe this sounds stupid, but it’s important to read the problem meticulously and truly understand the task at hand. Ask yourself what the desired end result is and think about the different paths with which you might be able to get there. In this instance, we want to calculate a number based on the two numbers that came before it in a certain sequence.
For more challenging problems, this process can actually take some time, which is OK. It will probably end up saving you time in the long run!
2. Solve an example manually
An example for our problem here could be n = 7. Or, in English, what is the value of the 7th integer in the Fibonacci sequence?
The important part of going through this process is not just to achieve the desired end result, but also to identify any repeatable rules or tasks that we can imply. These rules will become important aspects of our code once we start writing.
- First, we’ll have to start build the Fibonacci sequence, starting with 1 and 1 as our first 2 integers.
- Second, we’ll have to add the previous two numbers to calculate the next value.
- 1 + 1 = 2 (This is our 3rd value)
- 1 + 2 = 3 (4th)
- 2 + 3 = 5 (5th)
- 3 + 5 = 8 (6th)
- 5 + 8 = 13 (7th)
- Third, we’ll want to take that final value, 13 and print it or return it to the program to use it somewhere else.
At this point, we’ll also want to consider edge cases: situations where our normal rules might need to be broken. In this relatively simple example, I can't think of any right now, but it's a good thing to keep in the back of your head as you go through the problem-solving process.
Based on our manual example, let’s write out some comments that describe exactly what we want our program to do.
Initiating the process with some pseudo-code can help alleviate coder's block and get the ball rolling on a tricky problem. It's also a great way to ensure you don't skip any steps while translating the manual example to a working program.
4. Code it!
Our Ruby code below simply follows the comments laid out by the pseudo code. If you’re relatively new to programming, you might be confused about how we’re able to call the method fibonacci from inside itself. This utilizes a recursive strategy, which you can learn more about here.
The simple program above outputs the following:
At this point, we've achieved our initial goal! We're able to correctly produce the desired output of our test case, and by changing the input method from n = 7 to n = rand(1..50), we can test our program for all of the problem's cases.
If you’re a programmer dealing with algorithms, a helpful mantra to maintain is “Can we do it better?” Any initial code written to fix a bug or build a feature is unlikely to be the best way of approaching solving a problem.
In this example, we’ve actually written some downright troubling code. Our fibonacci method has an exponential runtime, meaning that as the input n increases, it will take more and more time to run the method. Let's add in Ruby's benchmark module to output not only the result but also how long it took to run.
As n increases, we can now evaluate our program's runtime and determine whether or not it's a workable solution we'd feel comfortable using in a production app. Unfortunately, somewhere between n = 35 and n = 45, the runtime takes a dramatic turn for the worse. Evaluating fibonacci(45) actually takes over 3 minutes!
This graphical demonstration of runtimes (also known as Big-O notation) demonstrates that as long as n is a relatively small number, we won’t really run into any problems. But once we pass a certain tipping point, the lag in our application will actually become quite drastic.
At this point, let’s return to our code and evaluate how we might be able to reduce the runtime to a more realistic timeframe. Our current problem is that each operation contains two sub-operations, unnecessarily complicating the code and making each evaluation longer than the once before it.
One way to approach this is to cache our data as we move through the method, meaning we'll sacrifice some disk space as we hold on to data longer, but it'll be worth it because we're able to evaluate inputs significantly faster. I'll also refactor the first part of the simple if statement into a more concise cache statement. Here's the potential solution:
The important part here is the output. We can now evaluate the higher numbers in the range almost instantaneously!
At this point, I'll leave the algorithm alone for now. Although the code isn't perfect, the performance is solid and it accurately completes the original desired output. When some potential improvement comes to me later, I'll definitely refactor it again.<|endoftext|>
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Communicable diseases kill more people and destroy more livelihoods than natural disasters and conflicts together. There are two ways in which communicable disease and disasters are related: 1) the outbreak of communicable disease may overwhelm community coping capacities and spark a disaster and 2) disaster events themselves are often feared to create situations where the spread of communicable disease is more likely.
Communicable diseases are caused by contamination hazards. Most of these diseases can be passed from one person to another. Methods of transmission include mucus, blood, bites (from insects or animals capable of transmitting the disease), breath, saliva and sexual contact.
How do I prepare?
- Protect sources of clean water.
- Vaccinate children against killer diseases.
- Do not re-use needles.
- Use clean and protected water sources.
- Keep water clean and protected from contamination.
- Use clean vessels and closed containers for transportation and storage.
- Keep water clean during collection, transportation and storage.
- Remove standing water that may attract insects and become contaminated.
- Learn and practice reliable methods for purifying water.
- Wash hands very well with soap.
- Dispose of waste safely for humans, animals and the environment.
- If latrines are not available, defecate well away from houses, water sources and places where children play.
- Bury faeces immediately or cover with earth, sand or ash.
- Locate trench and pit latrines away from water sources and keep them clean.
- Make sure pit latrines are emptied or replaced regularly by trained people with mechanical and protective equipment, rather than manually.
- Practice good sanitation
- Dispose of any waste that attracts flies and insects.
- Dispose of waste properly without contaminating water or soil.
- Keep food clean and food safe for eating.
For airborne diseases:
- Wash your hands very well using clean water and soap.
- Always cover your coughs and sneezes.
- When there are contagious diseases going around, keep a safe distance and avoid crowds if possible.
- Use good ventilation.
- Separate and care for your sick – Use only one caregiver – preferably a family member who is not at high risk and uses safety measures.
- Other family members should monitor themselves daily for fever and cough.
- Make sure that young children, pregnant women and people who have another disease receive medical care if they get sick.
For body-fluid borne diseases:
- Wash your hands very well.
- Practice safe sex.
For vector (animal) borne diseases:
- Be careful handling and slaughtering animals.
- Sleep under mosquito netting.
Local and National Government
In each emergency situation, the lead agency for health is responsible for preparation for and response to a sharp increase in the numbers of cases of disease. To prepare for such an eventuality, it is essential that:
- a surveillance system is put in place to ensure early warning of an increase in the incidence or numbers of cases of diseases;
- an outbreak response plan is written for the disease – covering the resources, skills and activities required;
- standard treatment protocols for the disease are available to all health facilities and agencies and that clinical workers are trained;
- stockpiles of essential treatment supplies (medication and material) and laboratory sampling kits are available for the priority diseases, such as oral rehydration salts, intravenous fluids, vaccination material, tents, transport media and water purification supplies;
- a competent laboratory is identified for confirmation of cases;
- sources of relevant vaccines are identified in the event that a mass vaccination campaign is required, and that supplies of needles and syringes are adequate;
- sources of additional treatment supplies are identified for non vaccine-preventable diseases in case of expansion and outbreak.
- the availability and security of a cold chain are established.
There are a limited number of diseases with epidemic potential that pose a major threat to the health of a population facing an emergency situation. These diseases should be identified during the rapid assessment. In addition, the lead health agency should draw up a list of the main risk factors for outbreaks in the emergency-affected population.
[Source: WHO “Communicable disease control in emergencies”]
There is great vulnerability in poorer and more remote communities to a range of communicable diseases including diarrhea, dengue, HIV/AIDS, and Avian Influenza. All of this amounts to a need for good prevention, preparedness and response systems.
Local leaders need to find ways to ensure that their communities can report outbreaks as soon as possible. Additionally, civil national societies should be as well a driving force in advocating change in disaster legislation in the country to improve disaster and communicable disease preparedness in their communities. Legal preparedness is ensuring that national laws and policies:
- Facilitate fast mobilization and response to disaster response and communicable disease emergencies
- Ensure good coordination and information exchange between different partners – local, national and international
- Encourage good quality and accountability standards
- Integrate key international and regional agreements and standards
Consider creating a or revising your current workplace's Business Continuity Program.
What do I need to know?
Infectious diseases kill 13 million people every year. The outbreak of disease is certainly one among a range of hazards or threats that has the potential to cause a catastrophic disaster. Poverty is the primary cause of the spread of disease, while in turn, poor health exacerbates poverty. Therefore, communicable diseases have a profound effect on the lives of people, in particular vulnerable groups, and their overall development. [Source:IFRC]
With the significant movement of people in and out of countries, cities, and communities on a daily basis as the result of globalization and modernization trends, disease outbreaks have the potential to spread fast and cause significant negative impacts health, livelihoods, and well-being.
The outbreak of communicable diseases is also often presumed to be a high risk following the impacts of significant disaster events. Humanitarian emergencies often involve the displacement of large numbers of people. Those affected are frequently settled in temporary locations with high population densities, inadequate food and shelter, unsafe water, poor sanitation and lack of infrastructure. These circumstances can increase the risk of transmission of communicable diseases and other conditions, and can thus lead to increased mortality (death). In particular, diseases that have a tendency to become epidemic (referred to as epidemic-prone diseases) can be a major cause of morbidity (disease) and mortality during emergencies. Rapid detection and prompt response to epidemics among the affected population is a key priority during humanitarian crises. [Source: WHO]
However experience has often shown that increases in communicable diseases are dependent upon many factors and that disaster events alone are not a sufficient predictor of significant changes in communicable disease prevalence. Instead, the risks of outbreaks must be evaluated in the full context of exposure and vulnerability in disaster-affected communities through a comprehensive risk assessment. This allows the prioritization of interventions to reduce the most significant factors that may contribute to the spread of communicable diseases after a disaster.
Communicable diseases can be prevented through a variety of measures, such as:
- public practice of good hygiene and sanitation
- access to clean water
- hand washing
- proactive surveillance
- vector control
Communicable diseases are caused by contamination hazards. Examples are:
- airborne (such as flu, typhus, tuberculosis, smallpox, measles, Severe Acute Respiratory Syndrome (SARS))
- conveyed by body fluids (such as polio or HIV)
- water borne (cholera, e. coli, dysentery)
- food borne (such as salmonella, e. coli, listeria, hepatitis)
- soil borne (such as anthrax)
- vector borne (transmitted from animals to humans (such as the H5N1 avian flu virus, malaria, dengue or letospira).
Communicable disease: It is an illness caused by an infectious agent, such as bacteria, virus, fungi or parasites and/or toxin. Most of these diseases can be passed from one person to another. Methods of transmission include mucus, blood, bites (from insects or animals capable of transmitting the disease), breath, saliva and sexual contact.
Glossary of communicable diseases preventable by vaccination:
Diphtheria: A bacterial disease marked by the formation of a false membrane, especially in the throat, which can cause death.
Haemophilus influenzae type b (Hib): A bacterial infection that may result in severe respiratory infections, including pneumonia, and other diseases such as meningitis.
Measles (also called Rubeolla): A contagious viral disease marked by the eruption of red circular spots on the skin.
Mumps: Acute contagious viral illness marked by swelling, especially of the parotid glands.
Pertussis (also called whooping cough): Bacterial infectious disease marked by a convulsive spasmodic cough, sometimes followed by a crowing intake of breath.
Poliomyelitis (also called polio): An acute infectious viral disease characterized by fever, paralysis, and atrophy of skeletal muscles.
Glossary of viral hepatitis:
Hepatitis A: A minor viral disease, that usually does not persist in the blood; transmitted through ingestion of contaminated food or water.
Hepatitis B: A viral disease transmitted by infected blood or blood products, or through unprotected sex with someone who is infected.
Hepatitis C: is a liver disease caused by the Hepatitis C virus (HCV), which is found in the blood of persons who have the disease. HCV is spread by contact with the blood of an infected person.
Glossary of other communicable diseases:
Avian influenza: Influenza viruses circulating in animals pose threats to human health. Humans can become ill when infected with viruses from animal sources, such as avian influenza virus subtypes H5N1 and H9N2 and swine influenza virus subtypes H1N1 and H3N2. The primary risk factor for human infection appears to be direct or indirect exposure to infected live or dead animals or contaminated environments.
Cholera: Cholera is an acute intestinal infection caused by ingestion of food or water contaminated with the bacterium Vibrio cholerae. It has a short incubation period, from less than one day to five days, and produces an enterotoxin that causes a copious, painless, watery diarrhea that can quickly lead to severe dehydration and death if treatment is not promptly given. Vomiting also occurs in most patients.
Malaria: Malaria is a life-threatening parasitic disease transmitted by mosquitoes. The parasite is transmitted from person to person through the bite of a female Anopheles mosquito, which requires blood to nurture her eggs.
Rabies: Rabies is a serious infection of the nervous system caused by a virus, known as Rabies virus. Rabies almost always results in death if a bite or scratch from a rabid animal (an animal infected with rabies) is not treated at the time of exposure and symptoms of an infection develop.
The Human Immunodeficiency Virus (HIV): HIV, is the virus that causes AIDS, the Acquired Immune Deficiency Syndrome in the late stages of infection. HIV is in the semen, vaginal fluid and blood of infected persons. Unprotected sex and shared needles or syringes with HIV or AIDS carriers are the main methods of disease transmission.
Tuberculosis (TB): Tuberculosis is a communicable disease that is caused by bacteria (germs) that attack the lungs or other parts of the body such as the kidney, spine or brain. If not treated properly, TB can be fatal. Other: Epidemic: The occurrence of disease within a specific geographical area or population that is in excess of what is normally expected. Pandemic: An epidemic occurring over a very large geographic area.
Epidemic: The occurrence of disease within a specific geographical area or population that is in excess of what is normally expected.
Pandemic: An epidemic occurring over a very large geographic area.
Get the latest videos and photos, case studies, and training materials contributed by practitioners from around the globe. Visit our Resource Library for more.<|endoftext|>
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- Gene – Cells contain genes, which are pieces of DNA that contain information for making proteins. Genes contain information on hereditary characteristics such as hair color, eye color, and height. as well as whether one is at higher risk for developing certain diseases.
- Grade – Grade is the measurement of a cancer, reflecting how abnormal the cells look under a microscope. There are several grading systems for cancer, but all divide cancers into those with:
- Esophagus: esophagoscopy
- abnormality (grade 1 or well differentiated)
- Intermediate features (grade 2 or moderately differentiated)
- Reatest abnormality (grade 3 or poorly differentiated)
A specialist called a pathologist performs the grading by examining the biopsy specimen. Knowing the grade is important because higher-grade cancers tend to grow and spread more quickly and have a worse prognosis. A cancer’s nuclear grade is based on features of the central part of its cells. the nucleus. The histologic grade is based on features of individual cells as well as how the cells are arranged together.
- Growth factors – A substance that is normally produced in the body that is involved in cell division, maturation, or survival. Growth factors may also be produced in a laboratory to mimic the growth factors naturally produced by the body. These synthetic growth factors may be used as biologic therapy to stimulate the immune system to fight cancer or lessen side effects of treatment.<|endoftext|>
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UTMB RESEARCHERS HAVE DEVELOPED a less expensive way to produce vaccines that cuts the cost of vaccine production and storage by 80 percent without decreasing safety or effectiveness.
Vaccines are the most effective way to prevent and eradicate infectious diseases. Currently, many vaccines have to be manufactured in cell culture or eggs, which is expensive and carries the risk of contaminations. In addition, most vaccines must be kept refrigerated during transportation from manufacturers to health care clinics.
In tropical and subtropical regions, such cold storage requirements could contribute to more than 80 percent of the vaccine cost.
“The ability to eliminate cell culture or eggs and cold storage will change the process of vaccine development,” said UTMB’s Pei-Yong Shi, professor in the Department of Biochemistry and Molecular Biology. “Importantly, this vaccine technology could potentially serve as a universal platform for development of vaccines made from live virus for many viral pathogens.”
To achieve these goals, the UTMB team engineered a live-attenuated Zika vaccine (in which a weakened form of the virus is used to create immunity without causing disease) in the DNA form, rather than by traditional methods. Once the DNA is delivered into the body, it launches the vaccine into cells, leading to antibody production and other protective immunity. With this production method, there is no need to manufacture the vaccine in a cell culture or eggs at factories. Because DNA molecules are shelf-stable, the vaccine will not expire at warm temperatures and could be stockpiled at room temperature for years.
Using UTMB’s Zika vaccine as a model, the research group showed that the DNA platform worked very efficiently in mice. After a single low dose, the DNA vaccine protected mice from Zika virus infection, mother-to-fetus transmission during pregnancy, and male reproductive tract infection and damage.
“This is the first study to demonstrate that, after a single low dose, a DNA vaccine could induce saturated protective immunity,” Shi said. “We will continue testing this promising Zika vaccine platform and then apply the platform to other viruses.”
Other authors include UTMB Drs. Jing Zou, Xuping Xie, Huanle Luo, Chao Shan, Antonio Muruato, Scott Weaver and Tian Wang.<|endoftext|>
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Lesson 6.8: Music: Influences Into and Out of the Mediterranean
• Students will be able to identify the roots and influence of raï music from North Africa and hip-hop from the United States
• They will trace the evolution of these styles as popular music in Europe and in North Africa and beyond.
• They will explain the ways in which these music styles crossed cultural lines and hypothesize reasons for their transmission
• They will gain appreciation for the music and its role in cross-cultural exchanges
• Projection device and speakers for viewing/listening to music videos
• Student Handout 6.8.1
• Student Handout 6.8.2
Lesson Plan Text
1. Activity A: Raï music This lesson traces the musical evolution and influence of Andalusian and North African music to becoming a popular phenomenon in France, creating a dialogue with other forms of music such as jazz, rock, and punk. Beginning with definitions and background sample the music that traces a path from Andalusian, Berber, and Arab roots to the urban music of the twentieth century and its crossover to the immigrant and pop music scenes in France, concluding with crossovers to rock & punk music, in the example of “Rock the Casbah” by the Clash and the cover “with a twist” by Rachid Taha “Rock El-Casbah.”
2. Distribute Student Handout 6.8.1 in electronic form, or project it in the classroom with a speaker setup for listening to the music and podcasts and viewing the videos. Beginning with the definition of Raï music from Wikipedia and the article provided, have students list the influences mentioned and locate their origins on a map (Arabian Peninsula, North Africa’s coastal and interior regions that connect with sub-Saharan Africa, and the Iberian Peninsula. Arabic literature, however, continued to develop after the 7th century in the eastern Mediterranean, Persia and Central Asia, and absorbed influences, instruments, and techniques from those regions, stringed instruments in particular flowing from Central Asia toward the west. Have students note the social class of origin of raï music, as well as its rural and urban roots before the 20th century. As an expression of the poorer classes, trace the connection with immigrants to France, Spain, and beyond that would help the music follow immigrants to Europe. Listen to the Cantigas, Andalus Ensemble performance, and Chaabi music from Morocco and Algeria.
3. Read the article “Women in Raï Music” (http://www.teachmideast.org/essays/37-culture/116-women-in-rai-music), watch and listen to the singers videos that follow. How do the videos illustrate what the essay described? Cheikha Rimitti is one of the most famous singers in the tradition, who paved the way for the fame of younger singers. The next few links introduce students to the evolving tradition and its growing popularity through radio, music cassettes, albums, and later music videos, which introduced the music to immigrant audiences in France and elsewhere.
4. The last part of the exploration illustrates the fusion of the music with European popular music, and its attainment of mass audiences in Europe and North Africa, as well as international audiences. The example is the music of Rachid Taha, particularly two famous songs—the song Ya Rayah fusing traditional and modern elements (see English translation of the lyrics) of pop performance and musical infectiousness. The other is the ambiguous phenomenon of Orientalism, “Rock the Casbah” by the Clash, which Rachid Taha claims to have inspired by giving the Clash a demo by his band Carte de Sejour (translation: ''Residence Permit''). Taha then covered the hit song in an ironic way, reclaiming its Algerian roots and Arabic lyrics in Rock El Casbah. (See links on Student Handout 6.7.1) Divide students into two groups to read the two New York Times articles on the crossover of Raï music to regional and international fame at “MUSIC - Shock the Casbah, Rock the French (And Vice Versa) - NYTimes.com.” http://query.nytimes.com/gst/fullpage.html?res=9B02E6D81F3DF930A25750C0A9639C8B63&pagewanted=all. “Arabic-Speaking Pop Stars Spread the Joy - New York Times.” http://www.nytimes.com/2002/02/06/arts/arabic-speaking-pop-stars-spread-the-joy.html .
5. Assessment: Students can write a reflective piece on the music, or conduct additional research into world music and describe its local and regional context and journey into the international arena. 6. Extension: Read the Wikipedia account of the influence of Rock the Casbah in the recent history of the Middle East, and discuss .
1. Activity B: Hip-Hop and Its Influence on the Arab Uprisings Distribute Student Handout 6.8.2 in electronic form, or project it in the classroom with a speaker setup for listening to the music and podcasts, and viewing the videos. This part of the lesson deals with the recent phenomenon of revolutions and social discontent against authoritarian regimes in Arab countries such as Tunisia, Egypt, Yemen, and Syria. In this case, the music style and substance being discussed involves an influence from outside the region having a major impact, in contrast to the first part that dealt with influence emanating from North Africa to Europe and beyond.
2. Begin by reading the NPR History Detectives dialogue to understand the origins of hip hop music in its urban setting in the Bronx, New York City during the 1970s. Students will list the different cultural influences from African American culture, Latino and Jamaican culture, and other influences. What was the impact of poverty on this type of music creation, and why did it become a global phenomenon?
3. Then trace the influence of this music and its social critique in the Arab uprisings of the early millennium. Notice that the movement had reached the Arab countries as well as many other countries, and did not emerge spontaneously, but had already become part of youth culture in these places. Note also, providing background for the students, that the youth population is very high as a percentage of total population, and many of them, though educated, cannot find jobs in order to live independently, marry, and fulfill the expectations of their generation. For a chart and explanation, see “Children and Youth in History | Arab Countries Youth Population Projection [Chart].” http://chnm.gmu.edu/cyh/primary-sources/424 .
4. Next, listen to the song by El General from Tunisia, and study the lyrics. Ask students to identify and list complaints issued by the artist against the government’s treatment of the people. Compare these lyrics’ spirit with those of rap artists working in a similar vein in the U.S., Africa and elsewhere around the world. The links are “The Rap That Sparked a Revolution: El General (Tunisia) | Hip Hop Diplomacy.” http://hiphopdiplomacy.org/2011/01/31/the-rap-that-sparked-a-revolution-el-general-tunisia/ and “Muftah » Revolutionary Arab Rap.” http://muftah.org/revolutionary-arab-rap/.
5. The next links relate to other hip hop artists whose songs became anthems of the Arab Awakening in Egypt and beyond. “The Rap Songs Of The Arab Spring : The Record : NPR.” http://www.npr.org/blogs/therecord/2011/06/09/137067390/the-rap-songs-of-the-arab-spring . Sample the five songs and compare; “Inside Tunisia’s Hip-Hop Revolution | SPIN | Profiles | Spotlight.” http://www.spin.com/articles/inside-tunisias-hip-hop-revolution/.
6. The next set of materials provides analysis of the phenomenon by a prominent blogger who was approached by major Italian Newspaper Corrierre Della Sera to write about it. “Revolutionary Arab Rap الراب العربي الثوري: My New Article in Italy’s Corriere Della Sera.” http://revolutionaryarabrap.blogspot.com/2012/02/my-new-article-in-italys-corriere-della.html . “Notizie Di Libri e Cultura Del Corriere Della Sera.” http://lettura.corriere.it/hip-hop-arab-youth-and-the-arab-awakening/ . Political scientists, sociologists, and historians have taken note of the phenomenon and have already held panels and conferences about the content, spirit and cultural influence of the music.
7. Extension: students can research hip hop music’s influence in other Mediterranean countries, and in other parts of the world.
Click below to view a document.<|endoftext|>
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