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# NCERT Solutions for class 10th Maths Chapter 11 Constructions
### Exercise 11.1
In each of the following, give the justification of the construction also:
Question 1. Draw a line segment of length 7.6 cm and divide it in the ratio 5: 8. Measure the two parts.
Question 2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.
Question 3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.
Question 4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 11/2 times the corresponding sides of the isosceles triangles.
Question 5. Draw a triangle ABC with side BC= 6 cm, AB= 5 cm and LABC =60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Question 6. Draw a triangle ABC with side BC = 7 cm, LB = 45°, LA= 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of MBC.
Question 7. Draw a right triangle in which the sides other than the hypotenuse are of lengths 4cm and 3cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
### Exercise 11.2
In each of the following, given the justification of the construction.
Question 1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct a pair of tangents to the circle and measure their lengths.
Note: If centre of the circle is not given, then locate the centre first by taking two non parallel chords and then finding the point of intersection of their perpendicular bisectors.
Question 2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Question 3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter and on opposite sides of its centre, each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Question 4. Draw a pair of tangents to a circle of radius 5 cm, which are inclined to each other at an angle of 60°.
Question 5. Draw a line segment AB of length 8 cm. Taking A as the centre, draw a circle of radius 4 cm and taking B as the centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Question 6. Let ABC be a right triangle in which AB= 6 cm, BC= 8 cm and LB= 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Question 7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Related Articles:<|endoftext|>
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# Dilation
## Larger or smaller version of a figure that preserves its shape.
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Practice Dilation
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Dilations
When you dilate a line segment, how is the original line segment related to the image?
#### Watch This
https://www.youtube.com/watch?v=El7zOrCDzBs
#### Guidance
A transformation is a function that takes points in the plane as inputs and gives other points as outputs. You can think of a transformation as a rule that tells you how to create new points.
A dilation is an example of a transformation that moves each point along a ray through the point emanating from a fixed center point \begin{align*}P\end{align*}, multiplying the distance from the center point by a common scale factor, \begin{align*}k\end{align*}.
\begin{align*}\Delta ABC\end{align*} below has been dilated about point \begin{align*}P\end{align*} by a scale factor of 2. Notice that \begin{align*}P\end{align*}, \begin{align*}A\end{align*}, and \begin{align*}A^\prime\end{align*} are all collinear. Similarly, \begin{align*}P\end{align*}, \begin{align*}B\end{align*}, and \begin{align*}B^\prime\end{align*} are collinear and \begin{align*}P\end{align*}, \begin{align*}C\end{align*}, and \begin{align*}C^\prime\end{align*} are collinear. \begin{align*}PC=3\end{align*} and \begin{align*}PC^\prime=6\end{align*}. The scale factor of this dilation is 2 because \begin{align*}\frac{PC^\prime}{PC}=\frac{6}{3}=2\end{align*}. If you calculate \begin{align*}PA\end{align*}, \begin{align*}PA^\prime\end{align*}, \begin{align*}PB\end{align*} and \begin{align*}PB^{\prime}\end{align*} you will find that \begin{align*}\frac{PA^\prime}{PA}=\frac{PB^\prime}{PB}=2\end{align*} as well.
Note that a dilation is not a rigid transformation, because it does not preserve distance. In the dilation above, \begin{align*}\Delta A^\prime B^\prime C^\prime\end{align*} is larger than \begin{align*}\Delta ABC\end{align*}. Dilations do, however, preserve angles. A shape and its image after a dilation will be similar, meaning they will be the same shape but not necessarily the same size.
Example A
A shape is dilated by a scale factor of \begin{align*}\frac{1}{2}\end{align*}. How does the image relate to the original shape?
Solution: If the scale factor is less than 1, the image will be smaller than the original shape.
Example B
Dilate the line segment below about point \begin{align*}P\end{align*} by a scale factor of 3. Make at least two conjectures about how \begin{align*}\overline{AB}\end{align*} relates to \begin{align*}\overline{A^\prime B^\prime}\end{align*}.
Solution: To dilate the line segment, draw a ray starting at point \begin{align*}P\end{align*} through each end point. Use the grid lines to help you find points on these rays that are three times the distance from point \begin{align*}P\end{align*} as the original endpoints were.
Two conjectures you might make are that \begin{align*}A^\prime B^\prime=3AB\end{align*} or \begin{align*}\overline{A^\prime B^\prime} \ \| \ \overline{AB}\end{align*}
Example C
Show that \begin{align*}A^\prime B^\prime=3AB\end{align*} and \begin{align*}\overline{A^\prime B^\prime} \ \| \ \overline{AB}\end{align*} for the dilation in Example B.
Solution: To find the lengths of the segments, create right triangles with the segments as their hypotenuses.
Use the Pythagorean Theorem to find the lengths of the hypotenuses.
• \begin{align*}AB^2=4^2+2^2 \rightarrow AB^2=20 \rightarrow AB=\sqrt{20}=2\sqrt{5}\end{align*}
• \begin{align*}{A^\prime B^\prime}^2=12^2+6^2 \rightarrow {A^\prime B^\prime}^2=180 \rightarrow A^\prime B^\prime=\sqrt{180}=6\sqrt{5}\end{align*}
Therefore, \begin{align*}A^\prime B^\prime=3AB\end{align*}.
Two line segments are parallel if they have the same slope.
• Slope of \begin{align*}\overline{AB}: -\frac{1}{2}\end{align*}
• Slope of \begin{align*}\overline{A^\prime B^\prime}: -\frac{1}{2}\end{align*}
Therefore, \begin{align*}\overline{A^\prime B^\prime} \ \| \ \overline{AB}\end{align*}.
Concept Problem Revisited
When you dilate a line segment, the original line segment will always be parallel to (or on the same line as) the image. Also, if the length of the original line segment is \begin{align*}L\end{align*} and the scale factor is \begin{align*}k\end{align*}, the length of the image will be \begin{align*}kL\end{align*}.
#### Vocabulary
transformation is a function that takes points in the plane as inputs and gives other points as outputs.
rigid transformation preserves distance and angles.
dilation is an example of a transformation that moves each point along a ray through the point emanating from a fixed center point \begin{align*}P\end{align*}, multiplying the distance from the center point by a common scale factor, \begin{align*}k\end{align*}.
Informally, two shapes are similar if they are the same shape but not necessarily the same size.
#### Guided Practice
1. A shape is dilated by a scale factor of 1. How does the image relate to the original shape?
2. Dilate the line segment below about point \begin{align*}P\end{align*} by a scale factor of \begin{align*}\frac{1}{2}\end{align*}.
3. Using your answer to #2, show that \begin{align*}A^\prime B^\prime=\frac{1}{2}AB\end{align*} and \begin{align*}\overline{A^\prime B^\prime} \ \| \ \overline{AB}\end{align*}.
Answers:
1. If the scale factor is 1, the shape does not change size or move at all. The image will be equivalent to the original figure.
2. The distance from \begin{align*}P\end{align*} to \begin{align*}A^\prime\end{align*} should be half the distance from \begin{align*}P\end{align*} to \begin{align*}A\end{align*}. Similarly, the distance from \begin{align*}P\end{align*} to \begin{align*}B^\prime\end{align*} should be half the distance from \begin{align*}P\end{align*} to \begin{align*}B\end{align*}.
Notice that \begin{align*}A^\prime\end{align*} is the midpoint of \begin{align*}\overline{PA}\end{align*} and \begin{align*}B^\prime\end{align*} is the midpoint of \begin{align*}\overline{PB}\end{align*}.
3. Use the Pythagorean Theorem to compare the lengths of the two segments.
• \begin{align*}AB^2=4^2+2^2 \rightarrow AB^2=20 \rightarrow AB=\sqrt{20}=2 \sqrt{5}\end{align*}
• \begin{align*}A^\prime B^{\prime 2}=2^2+1^2 \rightarrow A^\prime B^{\prime 2}=5 \rightarrow A^\prime B^\prime=\sqrt{5}\end{align*}
Therefore, \begin{align*}A^\prime B^\prime=\frac{1}{2}AB\end{align*}.
Two line segments are parallel if they have the same slope.
• Slope of \begin{align*}\overline{AB}: \frac{1}{2}\end{align*}
• Slope of \begin{align*}\overline{A^\prime B^\prime}: \frac{1}{2}\end{align*}
Therefore, \begin{align*}\overline{A^\prime B^\prime} \ \| \ \overline{AB}\end{align*}.
#### Practice
1. Describe how to perform a dilation.
2. Explain why a dilation is not an example of a rigid transformation.
3. True or false: angle measures are preserved in a dilation.
4. A shape is dilated by a scale factor of \begin{align*}\frac{3}{2}\end{align*}. How does the image relate to the original shape?
5. In general, if \begin{align*}k > 1\end{align*} will the image be larger or smaller than the original figure?
6. In general, if \begin{align*}k < 1\end{align*} will the image be larger or smaller than the original figure?
7. Dilate the line segment below about point \begin{align*}P\end{align*} by a scale factor of 2.
8. Using your answer to #7, show that \begin{align*}A^\prime B^\prime=2AB\end{align*}.
9. Using your answer to #7, show that \begin{align*}\overline{A^\prime B^\prime} \ \| \ \overline{AB}\end{align*}.
10. If one of the points of your figure IS the center of dilation, what happens to that point when the dilation occurs?
11. Dilate the line segment below about point \begin{align*}P\end{align*} by a scale factor of \begin{align*}\frac{1}{4}\end{align*}.
12. Using your answer to #11, show that \begin{align*}A^\prime B^\prime=\frac{1}{4}AB\end{align*}.
13. Using your answer to #11, show that \begin{align*}\overline{A^\prime B^\prime} \ \| \ \overline{AB}\end{align*}.
You can perform dilations in Geogebra just like you can perform other transformations. Start by creating your figure and the point for your center of dilation. Then, select “Dilate an Object from Point by Factor”, then your figure, and then the center of dilation.
Enter the scale factor into the pop up window and your figure will be dilated.
14. Create a triangle in Geogebra and dilate it about the origin by a scale factor of 2.
15. Dilate the same triangle about a different point by a scale factor of 2.
16. Compare and contrast the two images from #13 and #14.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 6.1.
### Vocabulary Language: English
Dilation
Dilation
To reduce or enlarge a figure according to a scale factor is a dilation.
Quadrilateral
Quadrilateral
A quadrilateral is a closed figure with four sides and four vertices.
Ratio
Ratio
A ratio is a comparison of two quantities that can be written in fraction form, with a colon or with the word “to”.
Scale Factor
Scale Factor
A scale factor is a ratio of the scale to the original or actual dimension written in simplest form.
Transformation
Transformation
A transformation moves a figure in some way on the coordinate plane.
Vertex
Vertex
A vertex is a point of intersection of the lines or rays that form an angle.
Rigid Transformation
Rigid Transformation
A rigid transformation is a transformation that preserves distance and angles, it does not change the size or shape of the figure.
### Explore More
Sign in to explore more, including practice questions and solutions for Dilation.
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# Find the area of the triangle whose vertices are ${(at_1^2,2a{t_1}),(at_2^2,2a{t_2}){\text{ and }}(at_3^2,2a{t_3})}$
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Hint: Just remember the formula for the area of the triangle, and then by simply putting the values you will get the answer.
We know that,
The area of a triangle = $\dfrac{1}{2}[{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})$
Here we have,
$\dfrac{1}{2}[at_1^2(2a({t_2} - {t_3})) + at_2^2(2a({t_3} - {t_1})) + at_3^2(2a({t_1} - {t_2})]$
$\Rightarrow$ $\dfrac{1}{2}[at_1^2(2a{t_2} - 2a{t_3}) + at_2^2(2a{t_3} - 2a{t_1}) + at_3^2(2a{t_1} - 2a{t_2})]$
Taking $2{a^2}$ as common, we get,
$\dfrac{1}{2}(2{a^2})[t_1^2({t_2} - {t_3}) + t_2^2({t_3} - {t_1}) + t_3^2({t_1} - {t_2})]$
$\Rightarrow$ ${a^2}(t_1^2{t_2} - t_1^2{t_3} + t_2^2{t_3} - t_2^2{t_1} + t_3^2{t_1} - t_3^2{t_2})]$
$\Rightarrow$ ${a^2}[(t_1^2{t_2} - t_3^2{t_2}) + (t_2^2{t_3} - t_1^2{t_3}) + (t_3^2{t_1} - t_2^2{t_1})]$
$\Rightarrow$ ${a^2}[{t_2}(t_1^2 - t_3^2) + {t_3}(t_2^2 - t_1^2) + {t_1}(t_3^2 - t_2^2)]$
Now using identity ${a^2} - {b^2}$ = (a + b)(a - b)
${a^2}[{t_2}({t_1} + {t_3})({t_1} - {t_3}) + {t_3}({t_2} + {t_1})({t_2} - {t_1}) + {t_1}({t_3} + {t_2})({t_3} - {t_2})]$
So, this is the required answer.
Note: You should be careful while choosing the formula for finding the area for these kinds of questions. Since the vertices are given, we have to take the formula of area of the triangle in coordinate geometry.<|endoftext|>
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The magnetosphere is the interface between the interplanetary environment and the upper Earth’s atmosphere. It is a vast region evolving under the control of the Earth’s magnetic field. The magnetosphere is filled with a collision-free plasma, continuously out of thermodynamic equilibrium, turbulent, unstable, which undergoes almost daily global reconfigurations. The magnetosphere has only one visual manifestation, but of great aesthetic quality: the polar aurora.
1. The solar wind
Most of the Sun’s energy comes to us in the form of light, infrared, visible, ultraviolet, and in small and variable proportions, in X-rays. The Sun also sends us a tiny part of it in the form of a very diffuse wind of matter, composed of free electrons, protons, and a small proportion of helium nuclei. The components of the wind are electrically charged, it is said to be a plasma wind, or more simply a solar wind. Its density is very low, typically 10 particles per cubic centimetre, but it goes fast, between 300 and 800 kilometres per second at the level of the Earth’s orbit. That is, it travels from the Sun to the Earth’s orbit in just three days. This wind carries not only a plasma, but also the magnetic field of the solar corona. The power brought by the solar wind to the terrestrial environment has been estimated at 100 GW (gigawatts). By way of comparison, this is slightly more power than that produced in France by all power plants. This is therefore not much compared to the total solar energy received by the Earth every second (1017 watts), but more than enough to induce significant electromagnetic effects in the terrestrial environment.
When the solar wind reaches a planet without an atmosphere like Mercury or the Moon, it hits the surface directly, contributing in particular to the very slow erosion of the soil. With planets like Venus or Mars, it is the upper atmosphere that it erodes, taking with it some of its constituents. With the Earth, it is more complicated, because it has a magnetic field, and this has an effective action on the flow of the solar wind. Conversely, the action of the solar wind modifies the magnetic field and plasmas of the Earth’s environment, creating an interface region between the interplanetary environment traversed by the solar wind and the upper Earth atmosphere. This interface, explored from 1957 with the beginning of the space age, received a name in 1959: the magnetosphere.
2. The magnetosphere
This name does not mean that this region has the shape of a sphere, but simply that it is under the control of the Earth’s magnetic field, which disappears completely into interplanetary space. Figure 1 shows a cross-section of the Earth’s magnetosphere: the Earth is the small white dot in the middle, the solar wind comes from the left, the dark blue lines are the magnetic field lines, the light blue line is the magnetopause and the red line represents the shock wave, to which we will return later.
The magnetosphere is a dynamic object, which never seems to reach a state of equilibrium. It is subject to continuous variations in the solar wind’s thrust and magnetism, as well as a series of internal processes that destabilize it by causing sudden, almost daily reconfigurations.
The Earth’s magnetosphere is the subject of two types of approaches. The first is to describe the magnetosphere (an objective achieved by scientists) and to understand the physical mechanisms that govern it. The other approach, which is complementary, aims to predict its activity in order to warn economic operators (industries, electricity networks, transport) of the fluctuations it induces on our electromagnetic environment. This article focuses on the first approach; the second is presented in the article Space Meteorology and its Consequences on Earth.
The inner boundary of the magnetosphere, closest to the Earth, is the ionosphere, which is the high-altitude atmospheric layer (from 60 km) in which, mixed with neutral gas, ionized atoms, i.e. a plasma, appear. Unlike atoms in a neutral atmosphere, these electrically charged atoms make the ionosphere a good electrical conductor. In the lower part of the ionosphere, at altitudes below 400 km, neutral gases and plasma are still relatively dense, in the sense that collisions between the particles that constitute it (atoms, molecules, ions and electrons) are frequent. Collisions generate exchanges of pulse and energy between particles, and this promotes the rapid establishment of a local thermodynamic equilibrium throughout the ionosphere. With the basic thermodynamic quantities: density, velocity, temperature, pressure, and with electrical quantities such as charge density and current, we can describe all their physics.
In the magnetosphere, the density is significantly lower, and the plasma, of solar origin, is much more energetic. This cocktail has a property that makes it special: collisions between particles are extremely rare. A particle typically has to travel a hundred thousand kilometres before colliding with another particle. In scientific terms, the average free ride is about 100,000 kilometres. Since this is also the characteristic dimension of the magnetosphere, it is deduced that the plasma of the magnetosphere is made up of particles without collisions with each other. They say it’s a collision-free plasma.
This variety of plasma does not occur naturally on Earth. Collision-free plasmas are not in local thermodynamic equilibrium (link to article Thermodynamics). For example, when cold plasma of ionospheric origin is mixed with hot plasma of solar origin, we do not obtain a warm plasma, whose particles would have a thermal agitation rate that would be a kind of average between the thermal agitation rates of the ionosphere and the solar wind. On the contrary, there are still particles that move little (the cold ones) and others that move in all directions at high speeds (the hot ones), all together, in the same places. Mathematically, a temperature can always be defined, but this notion is very insufficient to give a precise idea of the state of magnetospheric plasma. This is extremely unintuitive and confusing for physicists accustomed to the fluids encountered in our daily environment and in laboratories.
The absence of collisions also makes magnetospheric plasma very special, without viscosity, thus becoming very easily turbulent. To figure out the viscosity, you can take liquid honey and lower it into a pipe. The flow will be very slow, and slowed down by the honey on the edge, which is almost immobilized on the walls. By doing the same experiment with water, whose viscosity is much lower than that of honey, the flow will be faster, but still not uniform. The water in the middle will flow much faster than the water near the wall (for water, we can see things better by observing the flow of rivers). It is also an effect of viscosity. With its lower viscosity, water has much more turbulent movements than honey. With the collision-free plasma of the magnetosphere, virtually viscosity-free, the turbulent effects are further amplified. This still prevents it from reaching the slightest state of equilibrium, even locally and briefly.
Finally, like any plasma, the plasma of the magnetosphere is formed of electric particles, its movement is controlled by the electromagnetic field, especially by the magnetic field of Earth origin. Conversely, since its charged particles are in motion, they are also sources of electromagnetic fields; this plasma therefore modifies the Earth’s electrical and magnetic environment.
If the magnetosphere and the solar wind did not exist, the magnetic field around the Earth would be a dipole magnetic field, like the one represented in Figure 2 by its field lines. It would extend infinitely in space, losing its intensity as it moves further and further away. Because of the solar wind, the magnetic field lines look as shown in Figure 1, and this representation shows several things: the magnetosphere has a very elongated shape. It extends 60,000 kilometres from Earth on the sun side (the day side), but hundreds of thousands of kilometres away on the opposite side (the night side). Then, with the topography of the field lines, several regions are distinguished, whose spatial probes have revealed distinct properties.
3. The regions of the magnetosphere
The outermost region of the magnetosphere is a shock wave through which the solar wind passes from a supersonic speed to a subsonic speed. Behind the shock wave is magnetogaine, a very turbulent region of plasma, and “noisy” because it is traversed by a wide variety of very low frequency electromagnetic waves. (However, it is not a noise audible to a human ear.) Then comes the actual limit of the magnetosphere: the magnetopause. In the magnetogaine, it is still the matter from the solar wind that flows, and the magnetic field is of solar origin. Behind the magnetopause, we find a plasma partly of solar origin, partly of terrestrial origin, with very different characteristics. Only a small part of the solar wind manages to enter the magnetosphere through the magnetopause. On average, the density of matter in the magnetosphere is lower than in the solar wind. It’s like a kind of bubble, with a complicated shape. The magnetic field of the magnetosphere is of terrestrial origin. By following the magnetic field lines, at least in one direction, we always end up reaching the Earth’s ionosphere. The most complex region of the magnetosphere is on the night side. There are the two lobes shown in Figure 1, to the north and south, almost devoid of plasma.
4. The activity of the magnetosphere
Between the two lobes, the region called the current layer is characterized by a weaker magnetic field, which changes direction. This is where most of the magnetosphere’s plasma accumulates. As plasma from the solar wind reaches the magnetosphere, the current layer fills up. This region is continuously traversed by an electric current that runs from east to west (perpendicular to the plane in Figure 1). It is also the most unstable region: when the plasma accumulates a little too much, the current layer becomes thinner (paradoxically), and the current intensifies. Then, the whole thing finally cracked.
How does the current layer crack? The researchers have discovered several mechanisms called “magnetic reconnection”, “interchange instability”, but this remains a very active and still controversial subject of study.
What’s going on? The current layer is emptied in a few tens of minutes. Some of the plasma goes to space, some to Earth. These reconfiguration episodes of the current layer where it suddenly empties are called magnetospheric sub-storms. Typically, one or more sub-storms, or pseudo-triggers, occur each day with varying intensities, releasing more or less plasma with more or less energy. The visible and spectacular manifestations of these sub-storms are the polar auroras, which we will discuss in section 6.
5. The acceleration of plasma in the magnetosphere
When we study the magnetosphere, we measure the energy of particles in electron volts (eV, read Energy). The particles of the solar wind enter the magnetosphere with a typical energy of 100 eV, i.e. as if they had been accelerated by a 100 volt drop in electrical potential. Particles in the ionosphere have an energy of 0.1 eV, but those escaping into the magnetosphere often gain an energy of a few eV. In the tail of the magnetosphere, particles of various energies, from 10 eV to 50 keV (kilo-eV), are found. Commonly, their energy is in the order of 100 eV. During sub-storms, they earn a few hundred eV. These acceleration mechanisms are diverse, associated with waves, changes in the magnetic field and the accompanying electric fields.
6. The polar auroras
The plasma that is heading towards the Earth following such a reconfiguration undergoes a new acceleration at an altitude of a few thousand kilometers. It is caused by electromagnetic waves and electrical structures generated during the reconfiguration of the magnetospheric tail. The best accelerated particles are electrons from the plasma layer, their energy reaches a few keV. When they arrive on the ionosphere (a medium denser than the magnetosphere, with collisions), they jostle a few dozen atoms and molecules before losing their kinetic energy and speed. These atoms are “excited” by the collision, and “de-excite” by emitting a photon. These photons have energies corresponding to the nature of the atoms encountered. For example, the typical green colour of auroras (Figure 3) is due to collisions with oxygen atoms. Other colours are also produced: red, purple, blue. The light thus produced is the only visible manifestation of the magnetosphere, but what a manifestation: a sublime spectacle. It’s the polar aurora.
The brightest auroras are all associated with the reconfiguration of the magnetosphere’s plasma layer, so they all occur during a substorm. Shortly before these episodes, we can also observe auroras, but they are less bright and very static. They often correspond to the end of the current layer filling period, when “leaks” begin to occur.
The solar wind, like any wind, also experiences storms. Several dozen of them occur each year, of varying intensities. These storms can affect the density and speed of the solar wind, but also the intensity and variations in the orientation of its magnetic field. And sometimes these storms compress the Earth’s magnetosphere. Once compressed, the magnetosphere enters a relaxation phase, during two or three days, during which its activity is stronger than usual. Substorms are stronger and more frequent, and auroras can be seen in places where it is uncommon to see them (in France, for example). These episodes are called magnetic storms. It is the purpose of space weather (see Space Weather and its Consequences on Earth) to predict magnetic storms that could have a significant influence on human activity.
The Encyclopedia of the Environment is published by Université Grenoble Alpes - www.univ-grenoble-alpes.fr
To cite this article: MOTTEZ Fabrice (2019), The magnetosphere: under the influence of the Earth and the Sun, Encyclopedia of the Environment, [online ISSN 2555-0950] url : https://www.encyclopedie-environnement.org/en/air-en/magnetosphere-under-influence-earth-and-sun/.
The articles of the Encyclopedia of the Environment are made available under the terms of the Creative Commons Attribution - No Commercial Use - No Modification 4.0 International license.<|endoftext|>
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# How Do You Calculate Mass Using Spring Constant and Frequency?
• SithsNGiggles
In summary: I found, in blue.In summary, the conversation discusses using a spring with a spring constant of 4 N/m to weigh items, assuming no friction. In part (a), a frequency of 0.8 Hz is given and the mass is calculated to be 0.158 kg. In part (b), a formula for the mass in terms of frequency is derived. In part (2), friction is added and the spring constant is unknown. Two reference weights of 1 kg and 2 kg are used to calibrate the setup, with measured frequencies of 1.1 Hz and 0.8 Hz, respectively. In part (a), the values for the spring constant k and damping constant c are found
SithsNGiggles
## Homework Statement
(1) Suppose you have a spring with spring constant 4 N/m. You want to use it to weigh items. Assume no friction. You place the mass on the spring and put it in motion.
a) You count and find that the frequency is 0.8 Hz (cycles per second). What is the mass?
b) Find a formula for the mass m given the frequency ω in Hz.
(2) Suppose we add possible friction to the previous situation. Further, suppose the spring constant is unknown, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz and for the 2 kg weight you measured 0.8 Hz.
a) Find k (spring constant) and c (damping constant).
b) Find a formula for the mass in terms of the frequency in Hz.
c) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know the mass of the unknown object is more than a kilogram.
## Homework Equations
##\omega=\sqrt{\frac{k}{m}}##
## The Attempt at a Solution
• I'm not sure if this is right, but for 1(a) I use the above formula and I get
##0.8=\sqrt{\frac{4}{m}}\\
m=6.25##
• For 1(b), here's what I did:
##\omega=\sqrt{\frac{k}{m}}\Rightarrow m=\frac{k}{\omega^2}##
But this seems too simple.
• For 2(a), I tried to find k using the formula under "Relevant equations," but I get two different values of k:
##1.1=\sqrt{\frac{k}{1}}\Rightarrow k=1.21\\
0.8=\sqrt{\frac{k}{2}}\Rightarrow k=1.28##
So I'm not really sure what's going on here.
Any input is appreciated, thanks!
SithsNGiggles said:
## Homework Statement
(1) Suppose you have a spring with spring constant 4 N/m. You want to use it to weigh items. Assume no friction. You place the mass on the spring and put it in motion.
a) You count and find that the frequency is 0.8 Hz (cycles per second). What is the mass?
b) Find a formula for the mass m given the frequency ω in Hz.
(2) Suppose we add possible friction to the previous situation. Further, suppose the spring constant is unknown, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz and for the 2 kg weight you measured 0.8 Hz.
a) Find k (spring constant) and c (damping constant).
b) Find a formula for the mass in terms of the frequency in Hz.
c) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know the mass of the unknown object is more than a kilogram.
## Homework Equations
##\omega=\sqrt{\frac{k}{m}}##
## The Attempt at a Solution
• I'm not sure if this is right, but for 1(a) I use the above formula and I get
##0.8=\sqrt{\frac{4}{m}}\\
m=6.25##
• For 1(b), here's what I did:
##\omega=\sqrt{\frac{k}{m}}\Rightarrow m=\frac{k}{\omega^2}##
But this seems too simple.
• For 2(a), I tried to find k using the formula under "Relevant equations," but I get two different values of k:
##1.1=\sqrt{\frac{k}{1}}\Rightarrow k=1.21\\
0.8=\sqrt{\frac{k}{2}}\Rightarrow k=1.28##
So I'm not really sure what's going on here.
Any input is appreciated, thanks!
Not right. For one the units of ω in your formula aren't Hz. They are radians/s. There is some difference between them. For the second part there is another formula for ω that include the effect of damping. Try to look it up!
SithsNGiggles said:
• I'm not sure if this is right, but for 1(a) I use the above formula and I get
##\color{red}{5.027}=\sqrt{\frac{4}{m}}##
##\color{red}{m=0.158}##
• For 1(b), here's what I did:
##x=\text{ frequency given in Hz}##
##2\pi x=\text{ given frequency in rad/s}##
##x=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\Rightarrow m=\frac{k}{(2\pi x)^2}##
• For 2(a), I tried to find k using the formula under "Relevant equations," but I get two different values of k:
##1.1=\sqrt{\frac{k}{1}}\Rightarrow k=1.21\\
0.8=\sqrt{\frac{k}{2}}\Rightarrow k=1.28##
So I'm not really sure what's going on here.
Ah, thanks, I've corrected the first part, in red. There doesn't appear to be any other formula for ω in my notes, though. I'll be on the lookout for it.
EDIT: also corrected 1(b)
Last edited:
## What is a spring system differential equation?
A spring system differential equation is a mathematical representation of the motion of a mass attached to a spring. It takes into account the properties of the spring, such as its stiffness, and the external forces acting on the mass.
## What is the equation for a spring system?
The equation for a spring system is typically written as m*d^2x/dt^2 + kx = F, where m is the mass of the object, x is its displacement from equilibrium, k is the spring constant, and F is the external force acting on the object.
## How do you solve a spring system differential equation?
To solve a spring system differential equation, you can use techniques from differential equations, such as separation of variables or the method of undetermined coefficients. You can also use the equation of motion approach to find the general solution.
## What factors influence the behavior of a spring system?
The behavior of a spring system is influenced by several factors, including the mass and stiffness of the spring, the initial conditions of the system, and any external forces acting on the system. Temperature and damping can also affect the behavior of a spring system.
## What are some real-life applications of spring system differential equations?
Spring system differential equations have many real-life applications, such as in modeling the motion of objects attached to springs, analyzing the vibrations of mechanical systems, and understanding the behavior of biological systems like the human cardiovascular system.
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# Midpoint Formula
In coordinate geometry, we know about the Cartesian plane and representation of point using the coordinate system. Thus many times we need to find the location of the midpoint between two points. Also, the Midpoint of a line segment may be required. This midpoint will work as the center point of a straight line. Sometimes we will need to find the number that is half of two particular numbers. In that similar manner, we use the midpoint formula in coordinate geometry to find the halfway number of two coordinates. In this article, the student will learn about the concept of midpoint and midpoint formula with examples. Let us begin learning!
Table of content
## Midpoint Formula
### Definition of Midpoint:
Midpoint formula is a mathematical equation that is used to locate the halfway point between two data points. Besides in geometry, the study of economics uses this calculation to find the coefficient of elasticity, etc. Therefore, we can say that it is used to calculate how consumer habits change as price, quantity demanded, and quantity supplied changes.
This equation is used to solve the elasticity of demand and supply in various scenarios of different models. The key characteristic of this equation is that it calculates the percentage changes based on the difference between the beginning and the ending values.
### Formula to Find the Midpoint:
To find the midpoint of the straight line in a graph, we use this midpoint formula that will enable us to find the coordinates of the endpoint of the given line. Suppose the endpoints of the line is $$(x_1, y_1)$$ and $$(x_2, y_2)$$ then the midpoint is given as:
The Midpoint Formula is given as,
$$(x,y) = [\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]$$
Where, $$x_1, x_2$$ are the coordinates of the x-axis. $$y_1, y_2$$ are the coordinates of the y-axis.
Similarly, if we want to find the midpoint of a segment in the 3-dimensional space, we can determine the midpoint coordinate using the formula:
$$(x,y,z) = [\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}]$$
Here in 3-D case coordinates of the point will be $$(x_1,y_1,z_1) and (x_2,y_2,z_2)$$
Also in statistics and other areas of science, we have a similar kind of formula. Let us take the case for calculating the elasticity. We will use the average percentage change in both quantity as well as price. This is popularly known as the midpoint method for elasticity. It is represented by the following equations:
$$\displaystyle\text{percent change in quantity}=\frac{Q_2- Q_1}{(Q_2+Q_1)\div{2}}\times{100}$$
i.e. $$\displaystyle\text{percent change in price}=\frac{P_2-P_1}{(P_2+P_1)\div{2}}\times{100}$$
## Solved Examples
Example-1: Find the midpoint of a line whose endpoints are (4, 5) and (6, 7)?
Solution:
Given, $$(x_1, y_1) = (4, 5)$$
And $$(x_2, y_2) = (6, 7)$$
According to the formula we can have the midpoint (x, y) as follows:
$$(x,y) = [\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]$$
i.e. $$(x,y) = [\frac{4+6}{2},\frac{5+7}{2}]$$
i.e. $$(x,y) = (5, 6)$$
Example-2: How do you solve a problem such the point (0,2) is the midpoint of (2,-3) and what point?
Solution:
The first point is (2,-3).
Let us assume that another point’s coordinate is (m,n).
Now,
(x,y) = $$[\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]$$
Putting known values, we get:
(0,2)= $$[\frac{2+m}{2}, \frac -3+n{2}]$$
i.e.
$$\frac{2+m}{2} = 0$$
And $$\frac{-3+n}{2} = 2$$
Solving we get
m= -2
and n=7
Thus the second point will have coordinate (-2,7)
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KUCKOO B
I get a different answer for first example.
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median 23 and
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# Large Printable Number Cards 1 20
PDF files are available
Updated: Nov 10, 2022
Author:
Editor:
Numbers with two digits are numbers that have only two values. So, the value is in units and tens. Two-digit numbers from 10-99. Meanwhile, the number before 10 is not a two-digit number, but a single digit.
Large Printable Number Cards 1 20
### Table of Images 👆
Printable Number Flash Card 1
## What are Two-Digit Numbers?
Numbers with two digits are numbers that have only two values. So, the value is in units and tens. Two-digit numbers from 10-99. Meanwhile, the number before 10 is not a two-digit number, but a single digit. So, the smallest number for a two-digit number is 10 and the largest number is 99.
The digits in the tens value can be from 1-9. An example is the numbers 75 and 28. In the concept of two-digit numbers, there is also place value. In two-digit numbers, the place value is ones and tens. An example is the number 25. It is a two-digit number consisting of 20 and 5. The number 20 is in the tens place and the number 5 is in the ones place.
### How to Write Two-Digit Numbers?
There are several ways that can be used to write these two-digit numbers. Two-digit numbers can also be written using words and expanded forms. Let's use an example to understand writing a two-digit number.
Two-digit number: 35
If written in the form of words it becomes: Thirty-five
If written in expanded form: 30 + 5
If it is written in words, the number 35 must be spelled according to its place value. Remember that the first number in a two-digit number is the tens, so it should be written as "tens." So the 3 in 35 is "thirty."
Meanwhile, the last digit in a two-digit number is units, so the 5 in 35 is simply written as "five." Now, do this to write for other two-digit numbers in words.
Then, what about writing the two-digit number in expanded form? Well, a two-digit number is a combination of two numbers. So, it can be said that a two-digit number is the sum of tens and ones.
Therefore, from the number 35 above, the expanded form is the sum of 30 + 5. Another example is 20 + 7 for the expanded form of the two-digit number 27.
### How to Teach Kids about Two-Digit Numbers?
Children, especially grade 2, have been introduced to the concept of two-digit numbers. So, how to introduce two-digit numbers for kids? There are a few easy steps to do this.
• First, ask the children to draw two boxes with the headings “place value.”
• After that, give the children a two-digit number. An example is 45. So, the number 4 must be placed in the tens column and the number 5 must be placed in the ones column.
• If the children are able to place the two-digit number in the correct place value, teach them to sum using two types of two-digit numbers. An example is the number 23 + 54 = 77. Now, the children have to create a place value column. Put the numbers 2 and 5 in the tens column. Then, numbers 3 and 4 are placed in the units column. Sum the numbers 2 and 5 and the numbers 3 and 4. The result is 77.
• The same way can also be done for the subtraction operations. For example, the numbers 57 - 32 = 25. So, there are numbers 5 and 3 in the tens column. Meanwhile, numbers 7 and 2 are placed in the units column. After that, subtract each number in that column. So, the number 5 minus 3 becomes 2 and the number 7 minus 2 becomes 5. The result is 25.
Printable Numbers 1 20
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## What are Two-Digit Number Activities?
After introducing two-digit numbers in the easiest way above, continue to do other numerical learning practices. So, there are four activities that children can do to understand these two-digit numbers.
The first way is to use number lines. So, ask the children to make a horizontal line on the paper or floor. Then, give the sum of the two-digit numbers. An example is the number 20 + 40 = 60.
So, the number 20 must be written on the number line. After that, make four jumps with a value of 10. The end result is 60 according to the results we have written above. The value 10 represents the number 40.
The second way is to use 10 blocks. So, download and print a two-digit addition template. An example is 27 + 52 = 79. Now, the children need to draw some blocks of the two types of two-digit numbers.
So, under the number 27 must be drawn 20 blocks and 7 lines. Meanwhile, under the number 52, 50 blocks and 2 lines are drawn. Then, all the blocks are added up and all the lines are also added up. The result is 79.
The third way is to use the break-apart method. This method has the concept of splitting two-digit numbers into tens and ones' forms. We give another example so you understand it.
For example, you ask the children to add up the numbers 65 and 41. So, the number 65 is written as 60 + 5 and the number 41 is written as 40 + 1. Then, all the tens digits are added up so 60 + 40 = 100. Then, all the numbers of the units are also added up so 5 + 1 = 6. Next, 100 + 6 + 106. How? Very easy, right?
The last way that is used as one of the two-digit number activities for children is to use the "vs." What's that? So, the tens digit must be added to the tens digit and the one's digit to the units digit. So, this is almost the same as the third method above. The difference is in the use of arrows to connect tens with tens and units with units.
Article written by Lestari Nur Aisah, last updated on Nov 10, 2022 and edited by Printablee Team.
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These large printable number cards are a handy resource for teaching and practicing numbers 1-20. Great for visual learners!
2. Gabriel
Love these printable number cards! They're so handy for teaching and practicing numbers 1-20. Great size and clarity – exactly what I needed.
3. Willow
These large printable number cards 1-20 are a practical teaching tool that enhances number recognition and counting skills in children, making learning math engaging and accessible.
4. Lance
The large printable number cards from 1 to 20 are a useful resource for educators and parents, providing a clear and easy way for children to learn and recognize numbers while also enhancing their number sense development.<|endoftext|>
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# How to Find Area of a Triangle: The Ultimate Guide
Table Contents:
## Introduction
Welcome, Asensio, to this comprehensive guide on finding the area of a triangle. Whether you are a student, a teacher, or just someone who wants to refresh their knowledge, you have come to the right place. In this article, we will cover everything you need to know about calculating the area of a triangle, including the different formulas, the steps involved, and some practical examples. So, let’s get started!
### The Importance of Knowing How to Find the Area of a Triangle
The area of a triangle is a fundamental concept in math and geometry. It plays a crucial role in many real-life applications, such as architecture, engineering, physics, and even art. For example, knowing how to find the area of a triangular roof can help an architect design a building that is both functional and aesthetically pleasing. Similarly, understanding the area of a triangle can help an engineer calculate the stress on a triangular-shaped bridge or beam. Therefore, mastering this skill can be valuable in various fields.
### What is a Triangle?
Before we dive into the details of finding the area of a triangle, let’s first define what a triangle is. A triangle is a 2D shape that has three sides, three angles, and three vertices (or corners). There are several types of triangles, depending on the length of their sides and the measure of their angles. Some common types are equilateral triangles, isosceles triangles, and scalene triangles. Regardless of their type, all triangles share some basic properties, such as the fact that the sum of their angles is always 180 degrees.
### What is the Area of a Triangle?
The area of a triangle is the amount of space that is enclosed by its three sides. It is usually measured in square units (such as square inches, square centimeters, or square meters). The formula for calculating the area of a triangle depends on the information available about its sides and angles. In general, there are three main formulas for finding the area of a triangle: the base-height formula, the side-angle-side formula, and the Heron’s formula. We will discuss each of them in detail in the following sections.
### Why are There Different Formulas for Finding the Area of a Triangle?
Good question, Asensio. The reason why there are multiple formulas for finding the area of a triangle is that triangles come in different shapes and sizes. Some triangles have a right angle (i.e., one angle that measures 90 degrees), while others don’t. Some triangles have equal sides, while others don’t. Some triangles have angles that are easy to measure, while others require more advanced techniques. Therefore, depending on the given information about a triangle, one formula may be more suitable than another.
### What Tools Do You Need to Find the Area of a Triangle?
To find the area of a triangle, you don’t need any fancy equipment or software. All you need is a pencil, a ruler, and some basic math skills. Of course, a calculator can also be helpful, especially if you are dealing with large or complex numbers. However, keep in mind that some formulas require more precision than others, so you may need to use more accurate tools or techniques, such as trigonometry or calculus.
### What Are the Steps to Find the Area of a Triangle?
The steps to find the area of a triangle depend on the formula you are using. However, in general, the process involves identifying the given information about the triangle (such as the length of the sides or the measure of the angles), applying the appropriate formula, and doing some basic arithmetic. Here is a general outline of the steps:
1. Determine the given information about the triangle (such as the length of the base and the height, or the length of two sides and the included angle).
2. Choose the formula that corresponds to the given information.
3. Substitute the values into the formula.
4. Solve for the area.
5. Round the answer to the desired level of precision and include the units (such as square inches or square centimeters).
### What are Some Practical Examples of Finding the Area of a Triangle?
Here are some real-life scenarios where you might need to find the area of a triangle:
• Calculating the area of a triangular-shaped garden bed to determine how much soil or mulch you need.
• Measuring the area of a triangular flag or banner to determine the amount of fabric or paint needed.
• Determining the area of a triangular roof to estimate the cost of shingles or tiles.
• Calculating the area of a triangular-shaped swimming pool to determine how much water it can hold.
• Measuring the area of a triangular-shaped piece of land to determine its value or potential use.
### What are Some Common Mistakes When Finding the Area of a Triangle?
When finding the area of a triangle, there are some common mistakes that you should avoid. Some of these include:
• Forgetting to convert the units of measurement (such as inches to feet or centimeters to meters).
• Using the wrong formula for the given information.
• Using the wrong value for the height (such as the slant height instead of the perpendicular height).
• Forgetting to square the base or the height in the formula.
• Rounding the answer too soon or too late.
## How to Find Area of a Triangle
### Formula 1: Base-Height Formula
The base-height formula, also known as the half-base times height formula, is the simplest and most commonly used formula for finding the area of a triangle. It is applicable to any type of triangle, as long as you know the length of the base and the height (i.e., the perpendicular distance from the base to the opposite vertex). Here is the formula:
Formula: Area = 1/2 x base x height
To use this formula, follow these steps:
1. Identify the base and the height of the triangle.
2. Multiply the base by the height.
3. Divide the product by 2.
4. Round the answer to the desired level of precision and include the units.
Here is an example:
Example 1: Find the area of a triangle whose base is 8 cm and height is 12 cm.
1. Base = 8 cm
2. Height = 12 cm
3. Area = 1/2 x 8 cm x 12 cm = 48 cm2
Therefore, the area of the triangle is 48 square centimeters.
### Formula 2: Side-Angle-Side Formula
The side-angle-side formula, also known as the SAS formula, is a bit more complex than the base-height formula, but it can be useful when you don’t know the height of the triangle. This formula requires you to know the length of two sides of the triangle and the angle between them (i.e., the included angle). Here is the formula:
Formula: Area = 1/2 x a x b x sin(C)
Where:
• a and b are the lengths of the two sides.
• C is the included angle (i.e., the angle between sides a and b).
• sin(C) is the sine of angle C.
To use this formula, follow these steps:
1. Identify the two sides and the included angle of the triangle.
2. Multiply the lengths of the two sides.
3. Multiply the product by the sine of the included angle.
4. Divide the result by 2.
5. Round the answer to the desired level of precision and include the units.
Here is an example:
Example 2: Find the area of a triangle whose sides are 6 cm, 8 cm, and 10 cm.
1. a = 6 cm, b = 8 cm, c = 10 cm (where c is the hypotenuse)
2. C = sin-1(a/c) = sin-1(6/10) = 36.87 degrees
3. Area = 1/2 x 6 cm x 8 cm x sin(36.87 degrees) = 14.70 cm2
Therefore, the area of the triangle is 14.70 square centimeters.
### Formula 3: Heron’s Formula
Heron’s formula is a more advanced formula for finding the area of a triangle, which is used when the lengths of all three sides are given. This formula derives its name from the ancient Greek mathematician Heron of Alexandria, who first presented it in his book “Metrica” in the first century AD. Here is the formula:
Formula: Area = √(s x (s-a) x (s-b) x (s-c))
Where:
• s = (a + b + c)/2 is the semiperimeter of the triangle (i.e., half of the perimeter).
• a, b, and c are the lengths of the three sides.
• √ is the square root function.
To use this formula, follow these steps:
1. Identify the lengths of all three sides of the triangle.
2. Calculate the semiperimeter by adding the three sides and dividing by 2.
3. Subtract each side from the semiperimeter to get three differences (s-a, s-b, and s-c).
4. Multiply the differences and the semiperimeter together.
5. Take the square root of the result.
6. Round the answer to the desired level of precision and include the units.
Here is an example:
Example 3: Find the area of a triangle whose sides are 7 cm, 8 cm, and 9 cm.
1. a = 7 cm, b = 8 cm, c = 9 cm
2. s = (a + b + c)/2 = (7 + 8 + 9)/2 = 12 cm
3. Area = √(12 cm x (12-7) cm x (12-8) cm x (12-9) cm) = 26.83 cm2
Therefore, the area of the triangle is 26.83 square centimeters.
## FAQs
### 1. How do I know which formula to use to find the area of a triangle?
The formula you should use to find the area of a triangle depends on the information you have about the triangle. If you know the length of the base and the height, use the base-height formula. If you know the lengths of two sides and the included angle, use the side-angle-side formula. If you know the lengths of all three sides, use Heron’s formula.
### 2. Is there a shortcut to find the area of an equilateral triangle?
Yes, there is. Since an equilateral triangle has three equal sides, you can use the base-height formula with the height being the square root of three divided by two times the length of one side. Here is the formula:
Formula: Area = (sqrt(3)/4) x a2
Where:
• a is the length of one side of the equilateral triangle.
• sqrt is the square root function.
### 3. Can I use the Pythagorean theorem to find the area of a right triangle?
No, the Pythagorean theorem only helps you find the length of the hypotenuse of a right triangle, not its area. To find the area of a right triangle, you need to use the base-height formula with the base being one of the legs and the height being the other leg.
### 4. What if I don’t know the height of a triangle?
If you don’t know the height of a triangle, you can use the side-angle-side formula or Heron’s formula to find the area. These formulas do not require the height explicitly, but they do require information about the lengths of the sides and angles.
### 5. Which units should I use to measure the sides and angles of a triangle?
You can use any unit of measurement for the sides and angles of a triangle, as long as you are consistent throughout your calculation. However, it is often helpful to use the same units for all measurements (such as centimeters or inches) and convert between units if necessary.
### 6. Do I need to round the answer when finding the area of a triangle?
Yes, you should round the answer to the desired level of precision, depending on the context of the problem. Most of the time, the answer should be rounded to two decimal places or fewer, but this may vary depending on the situation.
### 7. Can I use a calculator to find the area of a triangle?
Yes, you can use a calculator to find the area of a triangle. However, make sure you enter the correct values and use the correct formula for the given information. Also, be aware that some calculators may show the answer in scientific notation or fractions instead of decimal form.
## Conclusion
Congratulations, Asensio! You have reached the end of this comprehensive guide on finding the area of a triangle. We hope that you have found this article informative, engaging, and useful. We have covered everything from the basics of triangles to the different formulas for finding their area, as well as some examples, tips, and common mistakes to avoid. Now that you have learned the skill of finding the area of a triangle, you can apply it to a wide range of real-life problems and appreciate its importance in various fields. So, go ahead and put your knowledge into practice, and don’t forget to share this guide with your friends, colleagues, or anyone who might benefit from it.
## Closing Statement with Disclaimer
The information provided in this article is for educational and informational purposes only. It is not intended to be a substitute for professional advice, diagnosis, or treatment. Always seek the advice of a qualified professional with any questions you may have regarding a problem or situation. We make no representation or warranty of any kind, express or implied, regarding the accuracy, adequacy, validity, reliability,<|endoftext|>
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National Aeronautics and
Goddard Space Flight Center
Satellites provide observation capability for monitoring different fire characteristics such as fire susceptibility, active fires, burned areas, smoke, and trace gases. Specifically, the images on the front represent the detection of active fires and smoke associated with wildfires in Florida during June 1998.
Abnormally dry conditions combined with record high temperatures dried out woodland areas across the state, making conditions particularly ripe for wildfires. Fires visited 39 of Florida's 67 counties between May 25 and June 25 and burned more than 100,000 acres of land. Thick blankets of gray smoke were associated with burning eyes, scratchy throats, stuffed noses and heavy wheezing. Many communities were evacuated and major highways closed as firefighters battled to save forests, homes, and property.
The images on the front are National Oceanic and Atmospheric Administration (NOAA) Geostationary Operational Environmental Satellite (GOES)-8 false-color images of the Florida peninsula on June 6, 1998.
The image at left is a combination of visible and infrared (heat) data showing smoke plumes and hot spots (red areas) along Florida's east coast. Some of the smoke plumes actually spawned cumulus clouds as water vapor condensed on aerosol particles (called cloud condensation nuclei) in the smoke. This is evident off the coast of Cape Canaveral as westerly winds carried the smoke out over the Atlantic Ocean.
The image at right is a color-enhanced infrared image of the "hot spots." Infrared images from satellites allow scientists to see a snapshot of the amount of radiation being emitted from the surfaces of objects such as clouds, land, and the ocean. In this image, land surface areas emitting high amounts of infrared radiation because of the fires have been colored red and are easily detectable on the image. These images prove to be extremely valuable as smoke typically hampers efforts to correctly map wildfires from aircraft. The accuracy of these images aids firefighters in tracking and forecasting the movement of wildfires.
GOES-8 was launched into a geosynchronous orbit, which means it orbits the equator at a speed matching the Earth's rotation, allowing it to hover continuously over one position on the surface, at an altitude of 35,800 km.
Additional information can be found on the World Wide Web at:
NASA's Earth Science Enterprise
NASA's EOS Project Science Office
Images courtesy of NASA's EOS Terra (AM-1) Visualization Team and GOES Project Science Offices.
As wildfires become more common and destructive, "prescribed burning" is often mentioned as a wildfire prevention tool. Have your students research the following questions on the merits of prescribed burning (copious amounts of information can be found on the internet).
Does prescribed burning reduce the number, size and intensity of wildfires?
Are the effects of prescribed burning as a wildfire prevention tool measurable?
Does the use of prescribed burning eliminate the threat of wildfire?<|endoftext|>
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# Simple & Compound Interest
• Concepts:
Simple Interest, SI = PNR/100
Where P is Principal, R is rate of interest and N is the time period in years.
Let P is Principal, R is rate of interest and N is time period in years,
When interest is compounded annually,
Amount = P (1+(R/100))^N
When interest is compounded half yearly,
Amount = P (1+(R/2)/100)^2N
When interest is compounded quarterly,
Amount = P (1+(R/4)/100) ^4N
When interest is compounded annually, but time period in fraction say a b/c years
Amount = P*(1+R/100)^a * (1+bR/100c)
When rates are different for different years say R1 % , R2 % and R3 % for 1st,2nd and 3rd year respectively
Amount = P (1+ (R1/100)) *(1+ (R2/100)) * (1+ (R3/100))
Questions:
Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. Find the sum placed on simple interest.
Let Principal is P
P * 3 * 8/100 = 1/2(4000(110/100)^2-4000)
6P/25 = 1/2(40 * 121 -4000) = 1/2(4840-4000) = 1/2 (840) = 420
P = 25 * 420/6 = 25 * 70 = 1750
Find the difference between simple interest and compound on Rs. 1200 for one year at 10% per annum reckoned half-yearly?
P = 1200 N = 1 YEAR R= 10%
SI = 1200 * 1 * 10/100 = 120 RS
CI = 1200(1+(10/2)/100)^2 – 1200
= 1200 * 441/400 -1200
= 1200 * (441/400 – 1)
= 1200 * 41/400
= 123 RS
Difference = 3RS
The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the rate of interest per annum?
(15000(1+R/100) ^2-15000) – 15000 * 2 * R/100 = 96
15000((1+R/100) ^2-1 -2R/100) = 96
15000(1+2R/100+R^2/10000-1-2R/100) = 96
15000 * R^2/10000 = 96
3R^2 = 192
R^2 = 64
R = 8%
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. Find the sum.
Let P = 1 Rs
SI = 1 * 2 * 4/100 = 2/25
CI = 1 * (104/100)^2 – 1 = 676/625-1 = 51/625
CI-SI = 51/625 – 2/25
= 51-50/625 = 1/625
Hence for 1 Rs sum difference is 1/625 Rs
So for getting 1 Rs difference Sum will be 625 Rs.
The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. Find the period (in years)
Total Amount = 30000+4347 = 34347
34347 = 30000 * (1+7/100) ^N
(107/100)^N = 34347/30000 = 11449/10000
(107/100)^N = 11449/10000
N = 2 Years
Looks like your connection to MBAtious was lost, please wait while we try to reconnect.<|endoftext|>
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Paul's Online Notes
Home / Algebra / Common Graphs / Miscellaneous Functions
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### Section 4-5 : Miscellaneous Functions
The point of this section is to introduce you to some other functions that don’t really require the work to graph that the ones that we’ve looked at to this point in this chapter. For most of these all that we’ll need to do is evaluate the function as some $$x$$’s and then plot the points.
#### Constant Function
This is probably the easiest function that we’ll ever graph and yet it is one of the functions that tend to cause problems for students.
The most general form for the constant function is,
$f\left( x \right) = c$
where $$c$$ is some number.
Let’s take a look at $$f\left( x \right) = 4$$ so we can see what the graph of constant functions look like. Probably the biggest problem students have with these functions is that there are no $$x$$’s on the right side to plug into for evaluation. However, all that means is that there is no substitution to do. In other words, no matter what $$x$$ we plug into the function we will always get a value of 4 (or $$c$$ in the general case) out of the function.
So, every point has a $$y$$ coordinate of 4. This is exactly what defines a horizontal line. In fact, if we recall that $$f\left( x \right)$$ is nothing more than a fancy way of writing $$y$$ we can rewrite the function as,
$y = 4$
And this is exactly the equation of a horizontal line.
Here is the graph of this function.
#### Square Root
Next, we want to take a look at $$f\left( x \right) = \sqrt x$$. First, note that since we don’t want to get complex numbers out of a function evaluation we have to restrict the values of $$x$$ that we can plug in. We can only plug in value of $$x$$ in the range $$x \ge 0$$. This means that our graph will only exist in this range as well.
To get the graph we’ll just plug in some values of $$x$$ and then plot the points.
$$x$$ $$f(x)$$
0 0
1 1
4 2
9 3
The graph is then,
#### Absolute Value
We’ve dealt with this function several times already. It’s now time to graph it. First, let’s remind ourselves of the definition of the absolute value function.
$f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}x&{{\mbox{if }}x \ge 0}\\{ - x}&{{\mbox{if }}x < 0}\end{array}} \right.$
This is a piecewise function and we’ve seen how to graph these already. All that we need to do is get some points in both ranges and plot them.
Here are some function evaluations.
$$x$$ $$f(x)$$
0 0
1 1
-1 1
2 2
-2 2
Here is the graph of this function.
So, this is a “V” shaped graph.
#### Cubic Function
We’re not actually going to look at a general cubic polynomial here. We’ll do some of those in the next chapter. Here we are only going to look at $$f\left( x \right) = {x^3}$$. There really isn’t much to do here other than just plugging in some points and plotting.
$$x$$ $$f(x)$$
0 0
1 1
-1 -1
2 8
-2 -8
Here is the graph of this function.
We will need some of these in the next section so make sure that you can identify these when you see them and can sketch their graphs fairly quickly.
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# Math Expressions Grade 5 Student Activity Book Unit 1 Lesson 6 Answer Key Add and Subtract Like Mixed Numbers
This handy Math Expressions Grade 5 Student Activity Book Answer Key Unit 1 Lesson 6 Add and Subtract Like Mixed Numbers provides detailed solutions for the textbook questions.
## Math Expressions Grade 5 Student Activity Book Unit 1 Lesson 6 Add and Subtract Like Mixed Numbers Answer Key
Add. Write each total as a mixed number or a whole number.
Question 1.
1$$\frac{1}{9}$$ + 2$$\frac{1}{9}$$
The denominators of the fractions are the same. So, add the mixed fractions and simplify.
1$$\frac{1}{9}$$ + 2$$\frac{1}{9}$$
1 + $$\frac{1}{9}$$ + 2 + $$\frac{1}{9}$$
1 + 2 = 3
$$\frac{1}{9}$$ + $$\frac{1}{9}$$ = $$\frac{2}{9}$$
3 + $$\frac{2}{9}$$ = 3 $$\frac{2}{9}$$
So, 1$$\frac{1}{9}$$ + 2$$\frac{1}{9}$$ = 3 $$\frac{2}{9}$$
Question 2.
5$$\frac{3}{8}$$ + 2$$\frac{2}{8}$$
The denominators of the fractions are the same. So, add the mixed fractions and simplify.
5$$\frac{3}{8}$$ + 2$$\frac{2}{8}$$
5 + $$\frac{3}{8}$$ + 2 + $$\frac{2}{8}$$
5 + 2 = 7
$$\frac{3}{8}$$ + $$\frac{2}{8}$$ = $$\frac{5}{8}$$
7 + $$\frac{5}{8}$$ = 7$$\frac{5}{8}$$
Question 3.
1$$\frac{1}{6}$$ + 3$$\frac{4}{6}$$
The denominators of the fractions are the same. So, add the mixed fractions and simplify.
1$$\frac{1}{6}$$ + 3$$\frac{4}{6}$$
1 + $$\frac{1}{6}$$ + 3 + $$\frac{4}{6}$$
1 + 3 = 4
$$\frac{1}{6}$$ + $$\frac{4}{6}$$ = $$\frac{5}{6}$$
4 + $$\frac{5}{6}$$ = 4$$\frac{5}{6}$$
Question 4.
4$$\frac{1}{3}$$ + 6$$\frac{1}{3}$$
4$$\frac{1}{3}$$ + 6$$\frac{1}{3}$$
The denominators of the fractions are the same. So, add the mixed fractions and simplify.
4 + $$\frac{1}{3}$$ + 6 + $$\frac{1}{3}$$
4 + 6 = 10
$$\frac{1}{3}$$ + $$\frac{1}{3}$$ = $$\frac{2}{3}$$
10 + $$\frac{2}{3}$$ = 10$$\frac{2}{3}$$
Question 5.
2$$\frac{1}{4}$$ + 2$$\frac{3}{4}$$
The denominators of the fractions are the same. So, add the mixed fractions and simplify.
2$$\frac{1}{4}$$ + 2$$\frac{3}{4}$$
2 + $$\frac{1}{4}$$ + 2 + $$\frac{3}{4}$$
2 + 2 = 4
$$\frac{1}{4}$$ + $$\frac{3}{4}$$ = $$\frac{4}{4}$$ = 1
4 + 1 = 5
2$$\frac{1}{4}$$ + 2$$\frac{3}{4}$$ = 5
Question 6.
3$$\frac{2}{7}$$ + 2$$\frac{5}{7}$$
The denominators of the fractions are the same. So, add the mixed fractions and simplify.
3$$\frac{2}{7}$$ + 2$$\frac{5}{7}$$
3 + $$\frac{2}{7}$$ + 2 + $$\frac{5}{7}$$
3 + 2 = 5
$$\frac{2}{7}$$ + $$\frac{5}{7}$$ = $$\frac{7}{7}$$ = 1
5 + 1 = 6
Question 7.
1$$\frac{3}{5}$$ + 2$$\frac{4}{5}$$
The denominators of the fractions are the same. So, add the mixed fractions and simplify.
1$$\frac{3}{5}$$ + 2$$\frac{4}{5}$$
1 + $$\frac{3}{5}$$ + 2 + $$\frac{4}{5}$$
1 + 2 = 3
$$\frac{4}{5}$$ + $$\frac{3}{5}$$ = $$\frac{7}{5}$$ = 1$$\frac{2}{5}$$
3 + 1$$\frac{2}{5}$$ = 4$$\frac{2}{5}$$
Question 8.
40$$\frac{6}{7}$$ + 22$$\frac{5}{7}$$
The denominators of the fractions are the same. So, add the mixed fractions and simplify.
40$$\frac{6}{7}$$ + 22$$\frac{5}{7}$$
40 + $$\frac{6}{7}$$ + 22 + $$\frac{5}{7}$$
40 + 22 = 66
$$\frac{6}{7}$$ + $$\frac{5}{7}$$ = $$\frac{11}{7}$$ = 1$$\frac{4}{7}$$
66 + 1$$\frac{4}{7}$$ = 67$$\frac{4}{7}$$
Question 9.
6$$\frac{8}{9}$$ + 4$$\frac{7}{9}$$
The denominators of the fractions are the same. So, add the mixed fractions and simplify.
6$$\frac{8}{9}$$ + 4$$\frac{7}{9}$$
6 + $$\frac{8}{9}$$ + 4 + $$\frac{7}{9}$$
6 + 4 = 10
$$\frac{8}{9}$$ + $$\frac{7}{9}$$ = $$\frac{15}{9}$$ = 1$$\frac{6}{9}$$
10 + 1$$\frac{6}{9}$$ = 11$$\frac{2}{3}$$
Practice Subtracting Mixed Numbers
Subtract. Write the difference as a mixed number or a whole number.
Question 10.
3$$\frac{2}{3}$$ – 1$$\frac{1}{3}$$
The denominators of the fractions are the same. So, subtract the mixed fractions and simplify.
3$$\frac{2}{3}$$ – 1$$\frac{1}{3}$$
3 + $$\frac{2}{3}$$ – 1 – $$\frac{1}{3}$$
3 – 1 = 2
$$\frac{2}{3}$$ – $$\frac{1}{3}$$ = $$\frac{1}{3}$$
2 + $$\frac{1}{3}$$ = 2$$\frac{1}{3}$$
Question 11.
5$$\frac{1}{6}$$ – 3$$\frac{5}{6}$$
The denominators of the fractions are the same. So, subtract the mixed fractions and simplify.
5$$\frac{1}{6}$$ – 3$$\frac{5}{6}$$
5 + $$\frac{1}{6}$$ – 3 – $$\frac{5}{6}$$
5 – 3 = 2
$$\frac{1}{6}$$ – $$\frac{5}{6}$$ = – $$\frac{4}{6}$$
2 – $$\frac{4}{6}$$ = 1$$\frac{1}{3}$$
Question 12.
2$$\frac{3}{4}$$ – 1$$\frac{3}{4}$$
The denominators of the fractions are the same. So, subtract the mixed fractions and simplify.
2$$\frac{3}{4}$$ – 1$$\frac{3}{4}$$
2 + $$\frac{3}{4}$$ – 1 – $$\frac{3}{4}$$
2 – 1 = 1
$$\frac{3}{4}$$ – $$\frac{3}{4}$$ = 0
2$$\frac{3}{4}$$ – 1$$\frac{3}{4}$$ = 1
Question 13.
4 – 2$$\frac{3}{8}$$
The denominators of the fractions are the same. So, subtract the mixed fractions and simplify.
4 – 2$$\frac{3}{8}$$
4 – 2 – $$\frac{3}{8}$$
4 – 2 = 2
2 – $$\frac{3}{8}$$ = 1$$\frac{5}{8}$$
Question 14.
15$$\frac{7}{9}$$ – 10$$\frac{5}{9}$$
The denominators of the fractions are the same. So, subtract the mixed fractions and simplify.
15$$\frac{7}{9}$$ – 10$$\frac{5}{9}$$
15 + $$\frac{7}{9}$$ – 10 – $$\frac{5}{9}$$
15 – 10 = 5
$$\frac{7}{9}$$ – $$\frac{5}{9}$$ = $$\frac{2}{9}$$
5 + $$\frac{2}{9}$$ = 5$$\frac{2}{9}$$
Question 15.
12$$\frac{2}{5}$$ – 8$$\frac{4}{5}$$
The denominators of the fractions are the same. So, subtract the mixed fractions and simplify.
12$$\frac{2}{5}$$ – 8$$\frac{4}{5}$$
12 + $$\frac{2}{5}$$ – 8 – $$\frac{4}{5}$$
12 – 8 = 4
$$\frac{2}{5}$$ – $$\frac{4}{5}$$ = –$$\frac{2}{5}$$
4 – $$\frac{2}{5}$$ = 3$$\frac{3}{5}$$
Solve Real World Problems
Write an equation. Then solve.
Question 16.
A bag contained 6$$\frac{2}{3}$$ cups of flour. Scout used 2$$\frac{1}{3}$$ cups to make some bread. How much flour was left in the bag after she made the bread?
Given,
A bag contained 6$$\frac{2}{3}$$ cups of flour.
Scout used 2$$\frac{1}{3}$$ cups to make some bread.
6$$\frac{2}{3}$$ – 2$$\frac{1}{3}$$
6 + $$\frac{2}{3}$$ – 2 – $$\frac{1}{3}$$
6 – 2 = 4
$$\frac{2}{3}$$ – $$\frac{1}{3}$$ = $$\frac{1}{3}$$
4 + $$\frac{1}{3}$$ = 4$$\frac{1}{3}$$
Question 17.
Tami and Ty were partners on a science project. Tami worked 4$$\frac{1}{4}$$ hours on the project, and Ty worked 1$$\frac{3}{4}$$ hours. How much time did they spend on the project in all?
Given,
Tami and Ty were partners on a science project.
Tami worked 4$$\frac{1}{4}$$ hours on the project, and Ty worked 1$$\frac{3}{4}$$ hours.
4$$\frac{1}{4}$$ + 1$$\frac{3}{4}$$
4 + $$\frac{1}{4}$$ + 1 + $$\frac{3}{4}$$
4 + 1 = 5
$$\frac{1}{4}$$ + $$\frac{3}{4}$$ = $$\frac{4}{4}$$ = 1
5 + 1 = 6
Solve Real World Problems
Question 18.
Maryl measures her height very precisely. Last month, she was 56$$\frac{7}{8}$$ inches tall. This month she is $$\frac{3}{8}$$ inch taller. How tall is she this month?
Given,
Maryl measures her height very precisely.
Last month, she was 56$$\frac{7}{8}$$ inches tall.
This month she is $$\frac{3}{8}$$ inch taller.
56$$\frac{7}{8}$$ + $$\frac{3}{8}$$
56 + $$\frac{7}{8}$$ + $$\frac{3}{8}$$
$$\frac{7}{8}$$ + $$\frac{3}{8}$$ = $$\frac{10}{8}$$ = 1$$\frac{2}{8}$$
56 + 1$$\frac{2}{8}$$ = 57$$\frac{1}{4}$$
Question 19.
Maryl (from Problem 18) measured her younger brother on January 1 of this year and found that he was 40$$\frac{3}{16}$$ inches tall. On January 1 of last year, he was 37$$\frac{15}{16}$$ inches tall. How much did he grow in a year’s time?
Given,
Maryl measured her younger brother on January 1 of this year and found that he was 40$$\frac{3}{16}$$ inches tall. On January 1 of last year, he was 37$$\frac{15}{16}$$ inches tall.
40$$\frac{3}{16}$$ – 37$$\frac{15}{16}$$
40 + $$\frac{3}{16}$$ – 37 + $$\frac{15}{16}$$
40 – 37 = 3
$$\frac{3}{16}$$ – $$\frac{15}{16}$$ = – $$\frac{12}{16}$$
3 – – $$\frac{12}{16}$$ = 2$$\frac{1}{4}$$
What’s the Error?
Dear Math Students,
This is a problem from my math homework. My teacher says my answer is not correct, but I can’t figure out what I did wrong. Can you help me find and fix my mistake?
5 $$\frac{7}{5}$$ – 5$$\frac{2}{5}$$ = $$\frac{5}{5}$$
$$\frac{5}{5}$$ – 3$$\frac{4}{5}$$ = -3$$\frac{1}{5}$$<|endoftext|>
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# Into Math Grade 4 Module 19 Lesson 1 Answer Key Identify Customary Measurement Benchmarks
We included HMH Into Math Grade 4 Answer Key PDF Module 19 Lesson 1 Identify Customary Measurement Benchmarks to make students experts in learning maths.
## HMH Into Math Grade 4 Module 19 Lesson 1 Answer Key Identify Customary Measurement Benchmarks
I Can select and use nonstandard units to measure lengths, weights, and liquid volumes.
Choose an object in your classroom. How would you describe the size of the object to a classmate?
Turn and Talk How could you describe the length of your object in terms of another object’s length?
Build Understanding
Question 1.
A. How could you use parts of your arm and hand as benchmark?
Parts of arms can be used as benchmark for foot and parts of hands can be used as a benchmark for inch.
B. If you have walked 20 minutes1 then you have likely walked about a mile. What benchmark could you use for 1 mile?
Connect to Vocabulary
A mile (mi) is a customary unit for measuring length. Liquid volume is the measure of the space a liquid occupies. Some customary units for measuring liquid volume are gallons (gal), quarts (qt), pints (pt), and cups (c).
The time taken to go from my home to school is 20 minutes by walk which can be used as benchmark for 1 mile.
Question 2.
Consider the objects shown and the liquid volume they can hold. Find at least one object in your classroom that would hold about 1 cup, 1 pint, 1 quart, and 1 gallon of liquid. Record your observations.
Question 3.
Consider the objects shown and their weights. Find at least one object in your classroom that would weigh about 1 ounce, 1 pound, or 1 ton. Record your observations.
Connect to Vocabulary
Weight is the heaviness of an object. Some customary units for measuring weight are ounces (oz), pounds (Ib), and tons (T).
A. What foods could you use as benchmarks to describe ounces and pounds?
A small chocolate weighs about 1 ounce and a pomegranate may weigh about 1 pound. So, these two foods may be used as benchmarks to describe ounces and pounds
B. What animal do you think weighs about 1 ton?
A lion will weigh about 1 ton.
Turn and Talk How can choosing certain objects as benchmarks affect precision when comparing relative sizes?
Check Understanding
Question 1.
The height of a desk could be used as a benchmark for which measurement unit?
inch
foot
yard
mile
Foot can be used as a benchmark for the height of a desk.
Use benchmarks to choose which unit would be best to measure each.
Question 2.
amount of liquid in a juice box
The amount of liquid in a juice box is measured in quarts.
Question 3.
weight of a car
The weight of a car is measured in Tons.
Question 4.
What object in your classroom weighs more than an ounce?
A duster will weight more than an ounce.
Question 5.
Construct Arguments Would you use tons to measure a dog’s weight? Which unit would you use to measure a puppy’s weight? Why?
We should not use tons to measure a dog’s weight. Pounds can be used as measuring unit for dog’s weight.
Puppy’s can be measured in pounds. Because a puppy can weigh nearly one to two pounds.
Question 6.
Which image represents a possible benchmark for a foot? Circle the answer.
A car length would be around 3 to 5 feet.
Attend to Precision Use benchmarks for length to choose the customary unit you would use to measure each.
Question 7.
height of a man __________
Height of a man is measured in feet.
Question 8.
width of an envelope ________
Width of an envelope is measured in inches.
Question 9.
length of a bus route ________
length of a bus route is measured in miles.
Question 10.
width of a house _________
Width of a house is measured in yards.
Complete the sentence. Write more or less.
Question 11.
A bath tub holds _______ water than a gallon.
A bath tub holds more water than a gallon.
Question 12.
A television weighs _________ than a ton.<|endoftext|>
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# Frequent question: What is the sample space when two dice are rolled?
Contents
The size of the sample space is the total number of possible outcomes. For example, when you roll 1 die, the sample space is 1, 2, 3, 4, 5, or 6. So the size of the sample space is 6.
## What is the sample space of two dice are rolled?
For an event of throwing of single die, sample space of die is given as: s = {1, 2, 3, 4, 5, 6}. Therefore, the number of favorable outcomes in the sample space = 6. Total number of possible outcomes of sample space = 36.
## What are the outcomes when two dice are thrown?
Since, a cubical dice has six faces and each face has a number that range from 1 to 6. So, the total number of outcomes when 1 dice is thrown is 6. Therefore, when two dice are thrown, the total number of outcomes will be the total combinations of the 6 numbers of one die with the 6 numbers of the other die.
## How many outcomes are there for adding the numbers of 12 dice?
A throw of twelve dice can result in 612 different outcomes, to all of which we attribute equal probabilities.
## When two sided dice are rolled There are 36 possible outcomes?
Every time you add an additional die, the number of possible outcomes is multiplied by 6: 2 dice 36, 3 dice 36*6 = 216 possible outcomes.
## What are the event of rolling a six-sided die?
A compound eventAn event with more than one outcome. is an event with more than one outcome. For example, in rolling one sixsided die, rolling an even number could occur with one of three outcomes: , , and . When we roll a sixsided die many times, we should not expect any outcome to happen more often than another.
## What is the probability of rolling a four with a six-sided die?
Two (6-sided) dice roll probability table
Roll a… Probability
4 3/36 (8.333%)
5 4/36 (11.111%)
6 5/36 (13.889%)
7 6/36 (16.667%)
## How do you calculate sample space?
The sample space is S = {H, T}. E = {H} is an event. Example 2 Tossing a die. The sample space is S = {1,2,3,4,5,6}.<|endoftext|>
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A convict is "a person found guilty of a crime and sentenced by a court" or "a person serving a sentence in prison". The word is sometimes abbreviated as "con". After a conviction, convicts often become prisoners in a prison. People convicted and sentenced but not sent to prison are not usually called "convicts". An ex-convict (or short: ex-con) is a person who has been let out of prison.
In past centuries many convicts were send to penal colonies. Many British convicts were sent to the Thirteen Colonies as cheap workers, but that stopped after the War of Independence. After this, convicts were transported to Australia in 1788, the very start of European settlement. They were used as cheap workers. Transportation was stopped in 1868. British convicts were also sent to Canada and India.
Related pages[change | change source]
References[change | change source]
- Webster's New World Dictionary of the American Language, p. 311 (2d Coll. Ed. 1978).
- Webster's New World Dictionary of the American Language, p. 292 (2d Coll. Ed. 1978).<|endoftext|>
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United States antitrust laws regulate the organization and conduct of business corporations on state and national levels to provide fair competition for the benefit of consumers. Why are they necessary?
The Federal Trade Commission (FTC) has the answer:
"Free and open markets are the foundation of a vibrant economy. Aggressive competition among sellers in an open marketplace gives consumers - both individuals and businesses - the benefits of lower prices, higher quality products and services, more choices, and greater innovation. The FTC's competition mission is to enforce the rules of the competitive marketplace - the antitrust laws. These laws promote vigorous competition and protect consumers from anticompetitive mergers in business practices. The FTC's Bureau of Competition, working in tandem with the Bureau of Economics, enforces the antitrust laws for the benefit of consumers."
The Sherman Antitrust Act, passed by Congress in 1890 under President Benjamin Harrison, was the first Federal act that outlawed interstate monopolistic business practices. It is considered a landmark decision because previous laws were limited to intrastate businesses.
In 1890 Utah, Oklahoma, New Mexico, Arizona, Alaska, and Hawaii were not even states. The Transcontinental Railroad that connected the eastern United States with the Pacific coast was in its infancy. That was then, this is now. Today there are 50 states, world travel is commonplace, and antitrust matters matter to every person on Earth.
Why? What do antitrust matters have to do with me? The answer is EVERYTHING.
The protections of antitrust laws focus on the unfairness of business monopolies on consumers. What about the unfairness of political monopolies on citizens? America broke free from the monopolistic centralized power of the British monarchy and won its independence after years of war and death. Why? To create a country based on the principles of individual freedom and liberty.
Every freedom articulated by our Founding Fathers was designed to protect the citizenry by promoting individualism and fairness. The Constitution and Bill of Rights are the greatest antitrust documents ever written providing:
● Freedom of religion
● Freedom of speech
● Freedom of the press
● Freedom of assembly
● Freedom to petition
● Right to keep and bear arms
● No quartering of soldiers
● Freedom from unreasonable searches and seizures
● Right to due process
● Right to speedy trial
● Freedom from cruel and unusual punishment
The United States Constitution protects citizens from anticompetitive mergers in government practices. The United States of America is the greatest experiment in individual freedom and liberty the world has ever known specifically because of its antitrust antimonopolistic infrastructure. The Constitution defines the checks and balances of power and the individual rights of its citizens.
The current battle for our national sovereignty is an antitrust struggle against global governance. Globalism must not be confused with fair global trade between independent nations. Globalism is the monopolistic anticompetitive political practice of concentrating world power in the hands of a few globalist elite rulers. It is a return to the feudal infrastructure of a paternalistic pyramid with the few elites at the top and the masses of enslaved people at its base.
In 1890 antitrust legislation was necessary to protect consumers from exploitive business practices that focused on goods and services in the interstate marketplace. In 2018 the competition is over the world of ideas. Political systems of national sovereignty supporting individual freedom are battling political systems supporting collectivist globalism.
Globalists are selling planetary governance. They are facilitating the sale by concentrating their power with mergers and acquisitions that limit the information the public is offered. Monopolistic information through censorship and partisan program content on the Internet, television, radio, films, and particularly in academic curricula indoctrinate the unsuspecting public toward collectivism and planetary governance.
Today's consumers need protection from exploitive political practices to provide fair competition for the benefit of consumers. The American public needs to become aware of how protective antitrust legislation that prevents monopolistic concentration of power affects them. Global governance is a worldwide concentration of power that eliminates individual freedom in the same way global monopolies on goods and services eliminate fair competition. Free and open markets are the foundation of a vibrant economy just like the free and open marketplace of ideas is the foundation of freedom.
The choice that Americans must make is between "free" stuff in a monopolistic collectivist political system of global governance that will enslave them, or freedom in an independent sovereign United States of America. The coordinated effort to delegitimize and overthrow President Donald Trump is the coordinated effort to delegitimize and overthrow our national sovereignty. Antitrust matters matter because monopolistic oppression is never good for the consumer whether they are buying goods, services, or their freedom.<|endoftext|>
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This book, subtitled “One Woman’s Courage in the Struggle for American Labor Union Rights” tells the story of labor activist Fannie Sellins.
Fannie Sellins was born Fanny Mooney in 1872. She married Charles Sellins, and after his relatively early death (when her youngest was just a baby) she had to support her four children. She went to work in a garment factory, one of the two sweatshops owned by the Marx & Haas Clothing Company. Girls as young as ten as well as grandmothers toiled there together, working ten- to fourteen-hour days, six days a week. Their pay averaged less than five dollars a week ($145 a week in today’s dollars). The building was stifling in summer, freezing in winter, and locked all day from the outside; this was a common practice in factories to prevent workers from taking unauthorized breaks and to reduce theft. [This same circumstance led to the tragic Triangle Shirtwaist Factory fire in New York, in 1911, in which 146 garment workers died.]
Fannie heard about the garment workers unions in Chicago and New York, and together with some other seamstresses organized Ladies’ Local 67 of the United Garment Workers of America in St. Louis. Fearing a strike, Marx & Haas nearly doubled workers’ wages and shortened the workday, but they did not improve conditions otherwise.
The air was filthy, and many of the employees contracted tuberculosis. When one tailor couldn’t make it up the six flights of stairs because of his illness, he was docked a week’s pay. To protest, Fannie and other union workers went on strike.
One month into the dispute, the local union leader died of tuberculosis, and Fannie became the new president. She traveled from city to city, speaking up to six times a day to all kinds of labor unions about the poor conditions at her factory and at other labor sites. She asked for support for the strikers. She also asked her audiences to buy only clothes with the “union-made” label inside them.
The book details the poor working conditions not only in the garment industry, but in the other places of work to which Fannie visited. Fannie was particularly affected by the West Virginia coal miners, who lived in abject poverty. Boys started in the mine as early as age six. The families lived in hovels with minimal food to eat and even without running water.
Fannie helped the miners of Colliers, West Virginia raise funds for a strike. Mine managers promptly evicted families from their houses – they now had to live in tents – and hauled in trainloads of strikebreakers. Fannie was arrested, although the first time, the judge released her with just a warning. Later she spent three months behind bars, during which time her health suffered greatly.
She moved on to help the coal miners in Western Pennsylvania organize, convincing thousands of miners to join the United Mine Workers of America (UMWA) and eventually to go on strike. The millionaire Lewis Hicks who ran fourteen notoriously bad mines went to the South to find strikebreakers, bringing up black sharecroppers by train. Hicks never told them they would be strikebreakers, or that they would be in danger because of it, just that they would be getting better wages. Hicks had the train doors locked from the outside [a favorite practice, it seems] so that union workers couldn’t get to the men.
When the train came to the mines, Fannie ran alongside it, yelling to the men inside through the windows not to break the strike and to support the union. She encouraged the men to climb out the train windows and join them. But the impasse was not broken until America entered World War I on April 6, 1917. The U.S. made a wage deal with coal mine operators to keep the mines working, and Hicks agreed to give workers a 50 percent pay raise, at least until the war ended.
Once again after the war, the UMWA initiated strikes. In August, 1919, Fannie was assigned to the Allegheny River Valley district to direct picketing by striking miners at Allegheny Coal and Coke Company. She was killed by sheriff’s deputies on August 26 who claimed that she led a “charging mob of men and women armed with clubs and bricks,” which was not true; in fact, Fannie rejected the idea that miners arm themselves for protection. Although there were a number of witnesses who gave sworn statements that the attack by the deputies was unprovoked, the local sheriff’s department refused to arrest them. When the case was reopened in 1923, again the deputies were acquitted. The author speculates that fears of communism – growing since the Bolshevik Revolution in Russia, colored the attitudes of the juries.
As the author writes:
“Today, both Fannie Sellins’s death and her passion for the welfare and rights of working people have been largely forgotten. But her name remains hallowed among union people in Western Pennsylvania, and her spirit lives on whenever someone stands up for the American ideals of equality and justice for all.”
At the end of the book, there is a glossary (words such as arbitration, lockout, and sweatshop), a detailed timeline of select events in the American labor struggle from 1877-1935, notes, sources, and a list of websites and books for further information.
This book for young readers is not a picture book, although the format is similar. Rather, it tells the story of Fannie Sellins with a great many photographs and reproductions of relevant documents.
Evaluation: This thoroughly researched book for ages nine and over includes an excellent selection of historical photos that brings the story to life in a way words could not, especially the depictions of poverty among mine workers. The story of Fannie Sellins’ belief in justice and personal sacrifices against the tenacious greed of factory owners is a lesson we still can learn today.
Published by Abrams Books for Young Readers, an imprint of ABRAMS, 2016<|endoftext|>
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## How many Tarot spreads are there?
More than you think!
Of course, there are as many different kinds of spreads, as there are readers – from drawing a single card, to three cards, to the standard ten-card Celtic Cross to huge spreads that use the whole deck. Here I’m asking more specifically once I choose a particular spread how many different results I can get from drawing different cards in a different order.
Start with a simple three card spread: We draw three cards and lay them out in order. How many different three-card arrangements can we get?
There are 78 choices for the first card, 77 choices for the second, and 76 choices for the third. 78 * 77 * 76 = 456,456 different layouts for a three-card spread.
That’s a lot! If you looked at one of these every minute of every hour of every day (no breaks!), it would take you almost 7 years to look at every single possibility.
Math section:
In math, we have a way of expressing this situation called permutations. For a three-card spread, you start with 78 cards and lay down 3 of them, and the order matters. That’s usually expressed as 78 permute 3, sometimes written 78 nPr 3, especially on calculators.
For calculating permutations, we use a math trick called factorials. A factorial is a way of saying “start with the number, then multiply it by all the numbers smaller than it down to 1.” Factorial is written as exclamation point: n! (pronounced “n factorial” or “n bang” if you want to hang with the kewl kids).
So 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1. (In more mathy language, n! = n * (n-1) * (n-2) * … * 3 * 2 * 1.) The thing about factorials is that they get big _really_ fast. Really, really fast. 10! = 3.6 million.
And factorials are really useful for calculating permutations. Here’s the trick: notice that 8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1, so 8! includes everything inside 10! except the 10 * 9. So if we take 10! divided by 8!, we can cancel out everything except that 9 and 10, and the answer is 90.
For the calculation above, we can say that the number of permutations possible for a three card spread drawn from 78 cards is 78! divided by 75! = 78 * 77 * 76 = 456,456. In general, to calculate how many ways there are to take n objects and choose permutations of r of them, we calculate n! / r!
Now let’s think about the standard Celtic Cross ten-card spread. We can calculate how many different spreads we can get by taking:
78 * 77 * 76 * 75 * 74 * 73 * 72 * 71 * 70 * 69
which is the same as 78 nPr 68 = 78! / 68!
which equals 4.56 times 10 ^ 18. That’s a really, really, really big number. It has nineteen digits. I’m not even going to try to write it all out. Let’s think about that some other ways.
There are four and a half _quintillion_ different possible Celtic Cross spreads. Quintillion, with a Q. (Come on, it’s a fun word!)
Suppose that a person started laying out a Celtic Cross spread, looking at it, shuffling the deck, and laying out a new one. If this person could lay out one spread a second, every second, from the moment of the Big Bang until now, that person would have seen only a tenth of the possible spreads so far in the lifetime of the known universe.
Or imagine every ten-card Tarot spread was represented by a grain of sand, and all that sand was piled into a giant cone. That cone’s tip would be twice as high as the roof of the Empire State Building, and its base would extend out to cover at least part of Grand Central Terminal and the same distance on the other side. This cone, 781 meters tall, would be almost as tall as the tallest building on earth, and have a base 1159 meters in radius – more than a mile across!
Two more geek comparisons for fun:
Suppose you were going to do a Celtic Cross spread for every cell in your body, plus all the microbes that live in and on your body. If you got together with about a thousand of your closest friends, all your cells together would use up all the available Tarot spreads.
The number of spreads is a few million times larger than the number of stars in our galaxy. Not just the stars you can see in the night sky, but all the stars in our galaxy. You’d need a few million galaxies for every star to have a corresponding Tarot spread. This is smaller than the estimated number of stars in the known universe, but not by a whole lot.
Note that none of this is counting reversed cards!
I got started on this by asking how likely it was to have five cards out of ten be Major Arcana, and calculating the answer was more complicated than I expected. If folks enjoy this, let me know, and I’ll continue with a series of two or three more posts, eventually answering the probabilities of the number of Major Arcana cards.
It took me a while to do these calculations, so if you cite them, please link back.
NB: All calculations assume that every card has an equal chance of being chosen – if that’s not true, then all bets are off.
Sources:
Size of a grain of sand. Note that sand comprises a range of sizes.
Estimate of number of cells in the body. Then I added in a factor of 100 to account for bacteria.
Estimate of the number of stars in the galaxy.
Beyond this place there be numbers:
4.56 x 10 ^ 18 spreads or grains of sand
4.56 x 10 ^ 18 times 2.51 x 10 ^ -10 m ^ 3 equals 1.1 x 10 ^ 9 m ^ 3, or a billion cubic meters
4.56 x 10 ^ 18 times 6 x 10 ^ -5 kg equals 3 x 10 ^12 kilos, or 3 trillion kilos
For the cone, Wikipedia states that dry sand has an angle of repose of 34 degrees.
Thus V = 1/3 pi r ^ 3 tan (34 deg) allows one to solve for the radius.
The rest is left as an exercise to the reader.
Literata is a Wiccan priestess and writer. She edited Crossing the River: An Anthology in Honor of Sacred Journeys, and her poetry, rituals, and nonfiction have appeared in works such as Mandragora, Unto Herself, and Anointed as well as multiple periodicals. Literata has presented rituals and workshops at Sacred Space conference, Fertile Ground Gathering, and other mid-Atlantic venues. Literata offers healing and divination services as well as customized life-cycle rituals. She is currently completing her doctoral dissertation in history with the support of her husband and four cats.
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### 6 Responses to How many Tarot spreads are there?
1. ToolTime says:
as many as you can create!
• Literata says:
In one sense, of course that’s true. You’ll see that in the sense I’m using it here, I mean how many different readings are possible within a single example of a spread. Maybe I should have said how many readings are possible, but I’m not sure everyone would use the terms the same way. We seem to use “spread” to mean both the arrangement of possible cards (imagine blank outlines that indicate where you will put cards down) and the actual layout of specific cards in a specific reading (“Her spread had a lot Major Arcana in it this time…”). Anybody want to suggest more specific terminology to differentiate those two?
• ToolTime says:
I’m a little lost outside of the scope of the statistical meaning of the post … the ten-card spread, with both positive & negative definitions of all the cards(78 in total), the possible outcomes for any given reading are 1:1.2 billion!
Gonna spell that, cause (for me) punctuation is hard to see in small print (guess I’m getting old) … that’s one in one-point-two BILLION!
and that’s a ten-card spread! My largest is a fifteen-card spread. Quite a card trick! Ehh?!?
Forgive me if I’m a little unsure of the scope of your post, as I was certain it was mathematical …. correct me if I’m wrong, as I enjoy the conversation!
• Literata says:
I mean that I’m using the word “spread” to mean both “the Celtic Cross spread” and “this Celtic Cross spread right here with card 1 this and card 2 that and card 3 the other…”. Of course there are as amny different ways to arrange the cards as the reader can invent, but within a given ten-card arrangement, there are a limited but very large number of possible readings.
The true number of possible readings with ten cards both upright and reversed is actually on the order of 4.6 sextillion – that’s ten to the 21st power. It’s a lot, lot bigger than billions. I’ll get to that – including reversals – in my next post.
2. Sara Mastros says:
I found this extremely fun; please continue. Also, let me know if you want any help with the calculating.
• Literata says:
Thank you! It’s odd – the most time-consuming part wasn’t actually the mathematically challenging bit of sorting through the probabilistic thinking but the effort to try different scales to find something to compare the number to that would make it comprehensible. More next week!<|endoftext|>
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A black king of New England, Nero Brewster, introduced a bill to gain his freedom to the General Assembly in 1779. It became law 234 years later, a fitting, if tardy, tribute to an extraordinary man.
Black King Nero had been born of royal lineage in Africa early in the 18th century. As a child, slavers captured him, took him to the American colonies and sold him to a wealthy tavern owner in Portsmouth, N.H. In that busy seacoast town he reclaimed his royal status as the acknowledged leader of the region’s black community.
For many years, Nero won the annual Negro Election and presided over an informal black government that mirrored the white polity.
Historians identified Nero as one of at least 31 elected black kings and governors throughout New England from about 1750 to 1850.
Portsmouth's African-Americans would elect Black King Nero every June, during Portsmouth’s 'Negroes Hallowday,' or 'Election Day.' The most important day of the year, it was held annually on the same day the white men of New England’s towns and cities gathered to vote for their leaders.
Black men elected their leaders not just in Portsmouth, but throughout the colonies. The Massachusetts towns of Salem and Lynn had 'Negroes Hallowday.' So did Newport and Narragansett in Rhode Island. So did Hartford, Norwich and Derby, Conn., and at least a half dozen other cities and towns.
In the royal colonies where the Crown appointed the governor – New Hampshire and Massachusetts – black men elected black kings.
In the charter colonies where white men could elect their own governor – Rhode Island and Connecticut – black men elected black governors. (For a list of known black governors of Connecticut, click here.)
For New England’s African slaves, Election Day meant more than a rare break from toil and a chance to have fun. They could express pride in their African heritage and rehearse for their role as free citizens.
And they could also elect black kings and governors to rule over the black social order as magistrate, spokesman, negotiator and leader.
Rum, Gingerbread and Election Cake
Election Day began with the Puritans as a jolly secular holiday. The colonists traveled to town in spring or early summer to elect their local leaders. Some had to travel quite far, and might stay overnight.
Important families hosted election-day celebrations, with rum and gingerbread and thick, fruit-studded election cakes.
The masters let the slaves take Election Day off or brought them into town. According to Salem minister William Bentley, African bondsmen and servants 'were too restless at home to be of any use till (the election holidays) were over.'
And so the Africans started organizing their own elections and their own celebrations, first in Newport and Hartford and then throughout New England.
In Connecticut, an enslaved African named London served as the first black governor known to history. London, who belonged to Capt. Thomas Seymour, was elected Connecticut governor in Hartford in 1755.
As Connecticut Colony's population grew, black residents began to elect leaders who lived nearby. Historians know of black governors in Derby, Durham, Farmington, New Haven, New London, Norwich and Seymour.
A Real Black King
Like Black King Nero, some of the black kings had royal African blood, and the Negroes Hallowday owes much to the Africans’ tradition of honoring their own royalty. Even before the Negro Election began in Lynn, African immigrants honored a prince of Africa named Pompey. Slavers had captured and sold him, but his master freed him when he grew too old to work.
Pompey then moved to a little glade near the Saugus River, where every year he hosted a holiday for the African bondsmen from nearby towns. He served as host, guest of honor and master of ceremonies. Women picked flowers to crown old King Pompey, and the men sat and talked about happier times on the Gambia River in West Africa.
A Monarch Held in Esteem
Black King Nero had dignity and stature. When he died in 1789, his obituary described him as, “A Monarch, who, while living, was held in reverential esteem by his subjects consequently, his death is greatly lamented.”
Similarly, black kings and governors throughout New England commanded respect -- because of their achievements, their strength or their ability.
Hartford’s Black Governor Peleg Nott, described as 'a first rate feller,' had traveled far, having driven a provision cart during the American Revolution. Black Governor Tobiah of Derby, Conn., was “a man of tact, courage and unusual intelligence.”
Black Governor Guy Watson of South Kingston fought in the 1st Rhode Island Regiment at the battles of Red Bank and Ticonderoga.
And black Governor Quosh Freeman of Derby, Conn., was a “man of herculean strength, a giant six-footer.”
Like all of the black kings and governors, Black King Nero’s rank reflected his master’s stataure. Col. William Brewster owned the Bell Tavern in Portsmouth, a well-known gathering place for patriots during the Revolutionary War.
Black King Nero embraced the revolutionary ideals of equality and freedom discussed so ardently within the tavern walls.
On each Election Day in June, Black King Nero dressed in his finest clothes to lead a procession of slaves and freedmen from Portsmouth and neighboring towns. His honor guard, perhaps decked out in feathers, flowers and ribbons, accompanied him.
The parade started with the crack of a gunshot. Down Middle Street they marched, and out Middle Road to a flat open area called Portsmouth Plains. They borrowed swords, guns and even horses from their masters for the festive parade.
Marchers made a happy racket with many African languages, more gunshots and music from tambourines, banjos, fiddles and drums. All the slaves wore their best clothes, often hand-me-downs from their masters. But they altered the outfits imaginatively with an African flair.
We can glean an idea of what their costumes looked like from a description of a runaway slave. He wore a Saxon blue jacket with bright green baize lining, slash sleeves and small metal buttons, a brown sleeveless jacket and scarlet breeches.
In Portsmouth and elsewhere, bondsmen deliberately played the fool in their gaudy Election Day costumes. White people looked on with enjoyment, belittling the slaves’ deportment and clothing as “fantastic.”
Today, historians note the white onlookers didn’t understand the black celebrants were making fun of their stiffness and pretensions. They also mocked the white people's obtuseness in not grasping they were laughingstocks.
Historians view their antics as a form of self-preservation. They didn’t want white people intimidated by their claim to participate in government.
The slaves' bitterness and contempt for their masters does show clearly in a written document signed by Black King Nero and 19 others in 1779.
Haranguing and Socializing
When the marchers reached the Plains, Nero and his opponent harangued each other as the crowd socialized and celebrated. Women, who couldn't vote, lobbied for their favorite candidates. The revelers reunited with friends and family who lived far away, and exchanged news and opinion about the white people they served.
More than a hundred took part, for in 1767 Portsmouth alone had 124 male and 63 female slaves, according to Brewster’s Rambles. (The total population of Portsmouth, one of the biggest cities in the colonies, reached 4,466 that year.)
After about three hours of electioneering and merriment, men voted and then declared Nero the black king. He subsequently announced his court: Viceroy Willie Clarkson, Sheriff Jock Odiorne and Deputy Sheriff Pharoah Shores. All came from Africa.
Who ran against Black King Nero, the acclaimed leader of Portsmouth’s African community? One wonders if it was ever Prince Whipple, the literate servant of William Whipple, who lived in the Moffatt-Ladd House.
Prince Whipple, born a prince of Africa, was sold into slavery as a child. He served as General Whipple’s bodyguard during the Revolutionary War, and he stood with the general at the Battle of Saratoga and at the signing of the Declaration of Independence.
According to legend, Prince Whipple accompanied George Washington during his famous crossing of the Delaware.
In the 1851 painting by Emanuel Gottlieb Leutze, some identify the black oarsman as Prince Whipple. A version of the painting now hangs in Marblehead’s Abbot Hall.
Regardless of whom he defeated at the polls, Black King Nero led the raucous procession back to his master's house for a victory party. As at the other Negroes Hallowdays, Nero’s master paid for the meal, though the black slaves and freedmen prepared it.
The black elections could get expensive for the masters. In Narragansett, R.I., the E.R. Potter’s slave John was elected governor. Potter, a state and federal legislator, told John that one of them would have to give up politics or the expenses would ruin them both. John stepped down.
The Black King and His Court
At his victory party in Portsmouth, Black King Nero served as host, master of ceremonies and guest of honor. He and his so-called Negro Court were toasted indoors, while games were played and athletic competitions held outside.
At the end of the day the celebrants moved to the slave quarters, where they.danced energetically to fiddle tunes of West African origin or influence.
Throughout the next year, Black King Nero served as leader, spokesman, arbiter and magistrate. He meted out judgments and punishments to slaves accused of petty crimes.
The shadow court system worked well for the white population as well as the black.
Trial and Punishment
In 1859, Portsmouth columnist C.W. Brewster described one trial in a typically condescending (for whites) account. A slave named Prince Jackson reportedly stole an ax. The black sheriff, Jock Odiorne, seized him and summoned the court.
Black King Nero sat for the examination in which evidence was submitted. He found Prince guilty and condemned him to 20 lashes on the bare back at the town pump.
The slaves gathered at the pump. Then, wrote Brewster,
...the Sheriff, after taking off his coat and tying up the convict to the pump, hands the whip to his deputy, Pharaoh Shores, addressing the company, "Gemmen, this way we s'port our government" - turning to his deputy - "Now, Pharaoh, pay on !" After the whipping was over, the sheriff dismissed the prisoner, telling him that the next time he is found this side Christian Shore, unless sent by his master, he will receive twenty lashes more.
Prince did not reform. Shortly afterward he was found guilty of larger thefts and taken to the white people's court.
“Here, we feel a just equality”
The festivities and satire of Negroes Hallowday belied a serious purpose. In the same way, the operation of the Negro Court throughout the year did not signify the slaves’ acceptance of their inferior status.
During the Revolutionary War, Black King Nero led at least one meeting of 20 black slaves to discuss their freedom. He may well have held more.
Inspired by revolutionary ideals and hatred of their miserable servitude, they drew up a petition demanding their freedom. They may have borrowed some of the language from the fervent patriots who talked of revolution at the Bell Tavern. They certainly took some language from the Declaration of Independence, which, one of the petition signers – Prince Whipple -- had witnessed.
The petition ensured that history remembered Black King Nero better than his master.
The men, seething with anger at their bondage, wrote a bold and bitter document. It drips with contempt for the pious Christians who enslaved them.
As in the Declaration, the signers list their grievances and denounce their enemies:
...thro' ignorance & brutish violence of their native countrymen and by similar designs of others, (who ought to have taught them better) & by the avarice of both, they, while but children, and incapable of self defense, whose infancy might have prompted protection, were seized, imprisoned, and transported from their native country, where (tho' ignorance and inchristianity prevailed) they were born free to a country, where (tho' knowledge, christianity and freedom, are their boast) they are compelled, and their unhappy posterity, to drag on their lives in miserable servitude. . . .
The Dignity of Human Nature
The bondsmen emphatically reject any idea that they are inferior to white people: 'here, we feel the passions and desires of men, tho' check'd by the rod of slavery! here, we feel a just equality!'
And they end with a prayer, 'that the name of SLAVE may no more be heard in a land gloriously contending for the sweets of freedom.'
They submitted the handwritten petition for freedom to the General Assembly in Exeter on Nov. 12, 1779. According to Black Portsmouth, the original manuscript has disappeared. No one knows whether the men signed their own names or used a mark.
It was read in the House of Representatives on April 25, 1780, and ordered published in the New Hampshire Gazette. The newspaper printed it that summer, with a disclaimer that it was intended “for the amusement “ of its readers.
The House then took up the petition on June 9, 1780, but decided to table it until “a more convenient opportunity.” No other legislative action would be taken on it for 234 years.
Black King Nero’s plea wasn’t the only petition in New England and it wasn’t the first. Two years earlier in 1777, eight Boston black men, mostly free, signed a petition to the Massachusetts Legislature to abolish slavery. A committee killed the petition, but the court abolished slavery in Massachusetts five years later.
In the spring of 1779, Connecticut black men in the towns of Stratford and Fairfield signed a petition asking the general assembly in Hartford if slavery wasn’t inconsistent with 'the present claims of the united States.'
And some Connecticut slaves petitioned for freedom because they’d became the property of the state after their Tory masters fled.
Six of the signers of the New Hampshire petition, including Prince Whipple, gained their freedom after the Revolution. Black King Nero died a slave, and possibly the last black king of Portsmouth. Historians haven’t unearthed any evidence that any other black kings were elected after he died.
In 2013, state Sen. Martha Fuller Clark of Portsmouth declared that Nero Brewster and the other 19 captive Africans deserved some closure. She filed a bill before the Legislature to posthumously free the remaining 15 slaves. The bill passed, and Gov. Maggie Hassan signed it into law. She granted King Nero Brewster his freedom, on Friday, June 7, 2013.
“The bill is now law,” she declared, as the room full of supporters for the bill broke into cheers.
Prince Whipple died in 1796 at age 46 and buried in Portsmouth’s North Cemetery. In 1908, local military veterans placed a stone on his grave.
In October 2003 a “Negro Burying Ground” was rediscovered under Portsmouth’s Chestnut Street.
Connecticut Colony paid for the first documented election cake (roughly £3), in Hartford, Conn. And the first known recipe appears in Amelia Simmons’ American Cookery in 1796, the first cookbook for an American market. For a recipe for Election Cake from the Boston Cooking-School Cook Book, (1911, orig. 1896), click here.
This story about black kings and governors was updated in 2018.<|endoftext|>
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Home > SAT > SAT Math Practice Online Test 32 Grid Ins Questions | SAT Online Classes AMBiPi
# SAT Math Practice Online Test 32 Grid Ins Questions | SAT Online Classes AMBiPi
Welcome to AMBiPi (Amans Maths Blogs). SAT (Scholastic Assessment Test) is a standard test, used for taking admission to undergraduate programs of universities or colleges in the United States. In this article, you will get SAT 2022 Math Practice Online Test 32 Grid Ins Questions with Answer Keys | SAT Online Classes AMBiPi.
### SAT 2022 Math Practice Online Test 32 Grid Ins Questions with Answer Keys
SAT Math Practice Online Test Question No 1:
When (x2 + 2x – 3)(2x + 5) – (x + 1)(x – 1)(x + 3) is expressed in the form ax3 + bx2 + cx + d, what is the value of a + b + c + d ?
Start by multiplying the terms together. To multiply (x2 + 2 x – 3)(2 x + 5), multiply each term in the left parenthesis by each term in the right parenthesis to get 2x3 + 5x2 + 4x2 + 10x – 6x – 15.
Combine like terms to get 2x3 + 9x2 + 4x – 15. To multiply (x + 1)(x – 1)(x + 3), multiply only two sets of parentheses first, then multiply that product by the remaining parenthesis. You may notice that (x + 1)( x – 1) is a common quadratic, which equals x2 – 1. Then, multiply x2 – 1 by (x + 3).
As you did before, multiply each term in the first parenthesis by each term in the second to get x3 + 3x2 – x – 3. Now, you can calculate (2x3 + 9x2 + 4x – 15) – (x3 + 3x2 – x – 3).
You can distribute the negative sign into the second parenthesis: (2x3 + 9x2 + 4x – 15) + (- x3 – 3x2 + x + 3). Now you can combine like terms to get x3 + 6x2 + 5x – 12. This is in the form ax3 + bx2 + cx + d , so a = 1, b = 6, c = 5, and d = -12. This means a + b + c + d = 1 + 6 + 5 + (-12) = 0.
SAT Math Practice Online Test Question No 2:
The Nile is a track and field athlete at North Sherahan High School. He hopes to qualify for the Olympic Games in his best field event, the javelin throw. He experiments with different javelin weights to build his arm strength and currently measures the results in feet.
During his preparations, Nile realizes that the upcoming Olympic qualifying competition will be judged in meters, rather than feet or yards. The Nile wants to make sure he can throw the javelin the minimum required distance so he can advance in the competition. If his current best throw is 60 yards, and one yard is approximately 0.9144 meters, how much further, to the nearest yard, must he throw to achieve the minimum required distance of 68.58 meters to qualify for the Olympics? (Disregard units when gridding your answer.)
Start by converting the qualifying distance of 68.58 meters into yards.
Set up a proportion of 1 yard/0.9144 meters = x yards/68.58 meters. Cross-multiply to get 0.9144 x = 68.58. Divide both sides by 0.9144 to find that the qualifying distance is 75 yards. If Nile’s current throw is 60 yards, he needs to throw 75 – 60 = 15 more yards.
SAT Math Practice Online Test Question No 3:
A bowl with 300 milliliters of water is placed under a hole where the rain gets in. If water drips into the bowl at a rate of 7 milliliters per minute, then how many milliliters of water will be in the bowl after 50 minutes?
Break the problem into Bite-Sized Pieces. First, determine how much water is added in 50 minutes. Rate = Amount/Time, so Rate × Time = Amount. 7 × 50 = 350 milliliters added over 50 minutes. Add this to the original 300 milliliters: 350 + 300 = 650.
SAT Math Practice Online Test Question No 4:
In one month, Rama and Siham ran for a total of 670 minutes. If Rama spent 60 fewer minutes running than Siham did, for how many minutes did Siham run?
To solve for Siham, translate from English to math and solve the system of equations.
Make the number of minutes Rama ran r and the number of minutes Siham ran s. If Rama and Siham ran for a total of 670 minutes, then the number of minutes Rama ran plus the number of minutes Siham ran equals 670, or r + s = 670.
For the second equation, Rama’s total amount of time was 60 minutes less than Sihams, so r = s – 60. Because you want to solve for s, you can substitute s – 60 for r in the first equation: (s – 60) + s = 670.
Combine like terms: 2 s – 60 = 670. Add 60 to both sides: 2 s = 730. Divide both sides by 2, and you get s = 365.
SAT Math Practice Online Test Question No 5:
Which f(x) = 1/[(x – 12)2 + 14(x – 12) + 49]
For what value of x is the function f above undefined?
For this function to be undefined, the denominator must be equal to 0. So, set the denominator equal to 0 and solve: (x – 12)2 + 14(x – 12) + 49 = 0 .
Start by FOILing the first parenthesis and distributing 14 in the second: x2 – 24x + 144 + 14x – 168 + 49 = 0 .
Combine like terms: x2 – 10x + 25 = 0. You might recognize this as a perfect square; it factors to (x – 5)(x – 5) = 0. Therefore, the value of x that makes function f undefined is 5.
SAT Math Practice Online Test Question No 6:
Marginal cost is the increase or decrease in the total cost a business will incur by producing one more unit of a product or serving one more customer. Marginal cost can be calculated using the equation M = C₂ – C₁/Q₂ – Q₁, where M is the marginal cost, C1 is the total cost for Q1 units, and C2 is the total cost for Q2 units.
At Carol’s Steakhouse, the total cost of serving 150 customers per day is \$900. Carol is interested in increasing her business but is concerned about the effect on marginal cost.
Carol successfully increases her business to 200 customers per day. However, her total cost for doing so is 50% greater than the expected \$1,600. What percent greater is the actual marginal cost than the expected marginal cost, to the nearest full percent? (Note: Ignore the percent sign when entering your answer. For example, if your answer is 326%, enter 326.)
Take this in Bite-Sized Pieces. If the actual cost is 50% greater than expected, and Carol expected the cost to be \$1,600 (as question 37 tells us), then the actual cost was 1,600 + 0.50(1,600) = \$2,400. This is the new value of C2 . Plug this value into the equation, using the same values as before for C1 , Q2 , and Q1 : M = 2,400 – 900/200 – 150 = 30.
To find the percent increase in marginal cost, use the equation percent change = difference/original x 100. So the percent change is (30 – 14)/14 x 100 = 114%.
SAT Math Practice Online Test Question No 7:
In the xy -plane, the point (2, 10) lies on the graph of the function f (x) = 2x2 + bx – 8. What is the value of b?
Because the point (2, 10) lies on the graph, you can make x = 2 and f (x) = 10 and solve for b : 10 = 2(2)2 + b (2) – 8; 10 = 8 + 2 b – 8; 10 = 2 b ; b = 5.
SAT Math Practice Online Test Question No 8:
A local frozen yogurt store views percentage rates of how many total viewers clicked on posts on its social media page. On the first 8 posts the store posted, the average (arithmetic mean) of the clicked-on percentage rates was 60%. What is the least value the page can receive for the 9th rating and still be able to have an average of at least 75% for the first 16 posts? (Disregard the % sign when gridding your answer.)
Average is total/number of things, which can be rearranged as average × number of things = total . If the desired average for 16 posts is 75%, then the sum of all the scores must be 75 × 16 = 1,200%.
For the first 8 posts, there was a total of 60 × 8 = 480%, leaving 1,200 – 480 = 720% for the last 8 posts. If you want the least value for the 9th post, then you want to assume that the 10th-16th posts to be as high as possible, or 100%.
This would give 100 × 7 = 700% total for the last 7 posts, making the least value possible for the 9th post 720 – 700 = 20%.
SAT Math Practice Online Test Question No 9:
In the figure above, point D is the center of the circle, line segments AB and BC are tangent to the circle at points A and C, respectively, and the segments intersect at point B as shown. If the circumference of the circle is 64, what is the length of major arc AC ?
Arc is proportionate to the central angle: central angle/360° = arc/circumference.
For a major arc, you need to use the central angle measure that’s greater than 180° (the central angle measure that’s less than 180° is the minor arc). Therefore, here you need to find angle ADC (the angle from the minor arc) and subtract that from 360° to get the angle for the major arc.
To get angle ADC, consider quadrilateral ABCD. A quadrilateral has 360°. Because AB and CB are tangents to the circle, angles DAB and DCB are each 90°, leaving 360 – 90 – 90 – 45 = 135° for angle ADC. The angle you need to use to find the major arc is therefore 360 – 135 = 225°. Insert this into the proportion: 225°/360° = x/64, where x is the arc.
Cross-multiply to get 14,440 = 360 x. Divide both sides by 360 to get x = 40.
SAT Math Practice Online Test Question No 10:
Mali is a landlocked country in western Africa. In 2015, its population was estimated to be 14.5 million. For the following 10 years, the population of Mali was projected to grow by 3 percent each year; this relationship can be modeled by the equation P = 14.5(r)y , where P is the population, in millions, y years after 2015.
What is the projected population of Mali in 2022, in millions, to the nearest tenth of a million?<|endoftext|>
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# Problem solving with multiplication and division (10x10)
Lesson
## Solving two step problems
Learn more about solving two step problems by watching this video.
Two step problems involve different parts to solve. When solving two step problems in mathematics it is helpful to break down the question into key parts. Sometimes it helps to underline the key ideas in the question.
When the question is about equal groups and we want to find a total we use multiplication to find the answer. If a question involves sharing something equally we use division to find the answer.
If you get stuck and don't know a strategy to solve the question, it helps to organise your thinking into an array diagram to represent the problem.
Here are some practice questions involving two steps. Some of them are about equals groups and finding totals and others involve sharing something equally.
Remember!
Use multiplication for equal groups and finding a total.
Use division for sharing into equal groups.
##### Question 1
There are $2$2 flower beds in Irene's backyard.
Each flower bed has $4$4 strawberry plants.
1. How many strawberry plants are there in total?
2. If each plant has $3$3 fruit growing on it, how many strawberries are there in total?
##### question 2
There are $5$5 strawberry plants in a backyard, each with $4$4 strawberries growing on them.
1. How many strawberries are there in total?
2. $2$2 friends pick the strawberries and want to equally share them.
How many strawberries will each friend get?
##### question 3
A group of $2$2 friends pool their money together and buy a big bag of marbles.
The bag has $22$22 marbles. Of these marbles, $20$20 of them are glass marbles and $2$2 are plastic marbles.
1. If the friends share the glass marbles amongst themselves equally, how many glass marbles will each of them get?
2. Gwen sells her bag of glass marbles to her brother. If each marble is sold for $£2$£2, how much does Gwen sell the bag for?<|endoftext|>
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# Important Questions for Class 8 Chapter 15-Introduction to Graph
The Important questions that we given in the PDF are based on each topic and chapter of the CBSE class 8th syllabus. Download your free PDF now and start practicing!
Important questions based on NCERT syllabus for Class 8 Chapter 15-Introduction to Graph:
Question 1: Observe the graph and answer the following questions
a. In which month is the TV viewership the maximum and for who?
b. Between which 2 months is the TV viewership not changing and for who?
c. In how many months is TV viewership more for boys than girls?
Solution:
From the graph, the analysis are as follows
a. TV viewership is maximum in March and it is for boys.
b. Between June and July the viewership is not changing for boys.
c. In 5 months- March, May, June, July and August the TV viewership is more for boys than girls.
Question 2:
a) I buy milk of different quantities at the rate of Rs.30 per litre. Which graph do I use to represent the data if I plot quantity of milk purchased on one axis and total cost of purchase on another axis.
b) What is a linear graph?
Solution:
a) I use a linear graph. As I purchase more milk, the money I spend on milk keeps increasing accordingly.
Hence the money spent on milk varies linearly with quantity of milk purchased.
b) If we plot points on a graph and on joining the plotted points we get a straight line. This graph is known as a linear graph.
Question 3: A bank gives 10% Simple Interest(S.I.) on deposits by senior citizens. Draw a graph to illustrate the relation between the sum deposited and simple interest earned. Find from your graph the following
a) The annual interest obtainable for an investment of Rs. 250.
b) The investment one has to make to get an annual simple interest of Rs. 70.
Solution:
Using the simple interest formula, for different values of the principal sum we get different values of annual interest as shown in table below.
The following graph is drawn with appropriate scale.
a) The annual interest for an investment of Rs.250 is Rs.25.
b) The investment one has to make to get an annual simple interest of Rs.70 is Rs.700.<|endoftext|>
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# Ch 25: Identifying & Working with Fractions
The brief videos in this chapter cover many aspects concerning fractions to help your third to fifth grader reinforce what they have been learning at school. The test at the end of the chapter can be especially useful to see how your child is doing and where they may need more help.
## Identifying & Working with Fractions - Chapter Summary
This chapter utilizes engaging videos to teach your student how to identify fractions, order them and use a number line. Students can also study the relationship between fractions, decimals and whole numbers, as well as strengthen their understanding of improper fractions and mixed numbers. Find out how well your student understands fractions by having them take the quizzes and chapter test.
## Chapter Lessons and Objectives
Lesson Objectives
What is a Fraction? - Definition and Types At the end of this lesson your student should be able to identify normal fractions, improper, mixed number and unlike fractions.
Identifying Fractions Students learn how to locate fractions on a number line, as well as how to view them as units, parts and divisions.
How to Raise and Reduce Fractions Instructors explain the procedure of raising and reducing fractions.
Relating Fractions and Decimals This lesson presents students with different examples of how fractions and decimals relate.
How to Find Least Common Denominators Your student will study various ways to identify least common denominators.
Comparing and Ordering Fractions This lesson provides an overview of how to figure out which fractions are greater than or less than others.
Using the Number Line to Compare Decimals, Fractions, and Whole Numbers Students gain an understanding of how number lines work and where various types of numbers fit on a number line.
Changing between Improper Fraction and Mixed Number Form Instructors show students the difference between improper fractions and mixed number form, as well as how to convert from one to the other.
8 Lessons in Chapter 25: Identifying & Working with Fractions
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
Test your knowledge of this chapter with a 30 question practice chapter exam.
Not Taken
Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
Not Taken
### Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.<|endoftext|>
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Also found in: Dictionary, Thesaurus, Medical, Legal, Acronyms, Idioms, Wikipedia.
race,one of the group of populations regarded as constituting humanity. The differences that have historically determined the classification into races are predominantly physical aspects of appearance that are generally hereditary. Genetically a race may be defined as a group with gene frequencies differing from those of the other groups in the human species (see heredityheredity,
transmission from generation to generation through the process of reproduction in plants and animals of factors which cause the offspring to resemble their parents. That like begets like has been a maxim since ancient times.
..... Click the link for more information. ; geneticsgenetics,
scientific study of the mechanism of heredity. While Gregor Mendel first presented his findings on the statistical laws governing the transmission of certain traits from generation to generation in 1856, it was not until the discovery and detailed study of the
..... Click the link for more information. ; genegene,
the structural unit of inheritance in living organisms. A gene is, in essence, a segment of DNA that has a particular purpose, i.e., that codes for (contains the chemical information necessary for the creation of) a specific enzyme or other protein.
..... Click the link for more information. ), but the genes responsible for the hereditary differences between the traditional races are extremely few when compared with the vast number of genes common to all human beings regardless of the race to which they belong. Many physical anthropologists now believe that, because there is as much genetic variation among the members of any given race as there is between the groups identified as different races, the concept of race is unscientific and unsound and racial categories are arbitrary designations. The term race is inappropriate when applied to national, religious, geographic, linguistic, or ethnic groups, nor can the physical appearances associated with race be equated with mental characteristics, such as intelligence, personality, or character.
All human groups belong to the same species (Homo sapiens) and are mutually fertile. Races arose as a result of mutationmutation,
in biology, a sudden, random change in a gene, or unit of hereditary material, that can alter an inheritable characteristic. Most mutations are not beneficial, since any change in the delicate balance of an organism having a high level of adaptation to its environment
..... Click the link for more information. , selection, and adaptational changes in human populations. The nature of genetic variation in human beings indicates there has been a common evolution for all races and that racial differentiation occurred relatively late in the history of Homo sapiens. Theories postulating the very early emergence of racial differentiation have been advanced (e.g., C. S. CoonCoon, Carleton Stevens,
1904–81, American anthropologist, archaeologist, and educator, b. Wakefield, Mass., grad. Harvard 1925, Ph.D. 1928. From 1925 to 1939 he was engaged in fieldwork and anthropological research in Arabia, the Balkans, and N Africa, where he discovered
..... Click the link for more information. , The Origin of Races, 1962), but they are now scientifically discredited.
Attempts at Classification
To classify humans on the basis of physical traits is difficult, for the coexistence of races through conquests, invasions, migrations, and mass deportations has produced a heterogeneous world population. Nevertheless, by limiting the criteria to such traits as skin pigmentation, color and form of hair, shape of head, stature, and form of nose, most anthropologists historically agreed on the existence of three relatively distinct groups: the Caucasoid, the Mongoloid, and the Negroid.
The Caucasoid, found in Europe, N Africa, and the Middle East to N India, is characterized as pale reddish white to olive brown in skin color, of medium to tall stature, with a long or broad head form. The hair is light blond to dark brown in color, of a fine texture, and straight or wavy. The color of the eyes is light blue to dark brown and the nose bridge is usually high.
The Mongoloid race, including most peoples of E Asia and the indigenous peoples of the Americas, has been described as saffron to yellow or reddish brown in skin color, of medium stature, with a broad head form. The hair is dark, straight, and coarse; body hair is sparse. The eyes are black to dark brown. The epicanthic fold, imparting an almond shape to the eye, is common, and the nose bridge is usually low or medium.
The Negroid race is characterized by brown to brown-black skin, usually a long head form, varying stature, and thick, everted lips. The hair is dark and coarse, usually kinky. The eyes are dark, the nose bridge low, and the nostrils broad. To the Negroid race belong the peoples of Africa south of the Sahara, the Pygmy groups of Indonesia, and the inhabitants of New Guinea and Melanesia.
Each of these broad groups can be divided into subgroups. General agreement is lacking as to the classification of such people as the aborigines of Australia, the Dravidian people of S India, the Polynesians, and the Ainu of N Japan within the traditional three race system. These exceptions highlight the problems associated with attempting to classify humanity into races and also challenge the validity of the notion of race when applied to human beings.
Race Classification and Racism
Attempts have been made to classify humans since the 17th cent., when scholars first began to separate types of flora and fauna. Johann Friedrich BlumenbachBlumenbach, Johann Friedrich
, 1752–1840, German naturalist and anthropologist. He introduced and developed the science of comparative anatomy in Germany. His De generis humani varietate nativa (1775; tr. On the Natural Varieties of Mankind, 1865, repr.
..... Click the link for more information. was the first to divide humanity according to skin color. In the 19th and early 20th cent., people such as Joseph Arthur GobineauGobineau, Joseph Arthur, comte de
, 1816–82, French diplomat and man of letters. The chief early French proponent of the theory of Nordic supremacy, he was antidemocratic and anti-Semitic.
..... Click the link for more information. and Houston Stewart Chamberlain, mainly interested in pressing forward the supposed superiority of their own kind of culture or nationality, began to attribute cultural and psychological values to race. This approach, called racism, culminated in the vicious racial doctrines and anti-Semitismanti-Semitism
, form of prejudice against Jews, ranging from antipathy to violent hatred. Before the 19th cent., anti-Semitism was largely religious and was expressed in the later Middle Ages by sporadic persecutions and expulsions—notably the expulsion from Spain under
..... Click the link for more information. of Nazi Germany and was used to justify slaveryslavery,
historicially, an institution based on a relationship of dominance and submission, whereby one person owns another and can exact from that person labor or other services.
..... Click the link for more information. and segregation (see integrationintegration,
in U.S. history, the goal of an organized movement to break down the barriers of discrimination and segregation separating African Americans from the rest of American society.
..... Click the link for more information. ) in the United States, apartheidapartheid
[Afrik.,=apartness], system of racial segregation peculiar to the Republic of South Africa, the legal basis of which was largely repealed in 1991–92. History
..... Click the link for more information. in the Republic of South Africa, and European imperialism and colonialism generally.
See R. Benedict, Race: Science and Politics (rev. ed. 1943, repr. 1968); C. Lévi-Strauss, Race and History (1962); M. Mead et al., ed., Science and the Concept of Race (1968); S. M. Garn, ed., Readings on Race (2d ed. 1968) and Human Races (3d ed. 1971); J. C. King, The Biology of Race (1971); L. L. Cavalli-Sforza, The Origin and Differentiation of Human Races (1972); S. J. Gould, The Mismeasure of Man (1981); I. F. Haney Lopez, White by Law: The Legal Construction of Race (1996); A. Montagu, Man's Most Dangerous Myth: The Fallacy of Race (6th ed. 1998); G. M. Frederickson, Racism: A Short History (2002); M. Yudell, Race Unmasked: Biology and Race in the 20th Century (2014).
racea scientifically discredited term previously used to describe biologically distinct groups of persons who were alleged to have characteristics of an unalterable nature. The concept has been used in the English language since the 16th-century Its meaning has altered several times over the last 400 years in line with changing concepts about the nature of physical and cultural differences and, more importantly the ideological uses of the concept to justify relationships of superiority and exploitation. Banton in Racial Theories (1987) provides a comprehensive account of the different uses of the concept of race.
Social scientists now recognize that ‘race’ is exclusively a socially constructed categorization which specifies rules for identification of a given group. Many writers will not use the term except in inverted commas to distance the use of the word from its historical and biological connotations. It is preferable to refer to ETHNICITY or ETHNIC GROUPS. Despite the discredited nature of the concept of’race’, the idea still exerts a powerful influence in everyday language and ideology. See also RACE RELATIONS, RACISM, ETHNICITY, ETHNIC GROUP.
in biology, an ecologically or sometimes geographically related group of organisms within a species or subspecies.
The members of a race have similar morphological, physiological, and ecological characteristics and are distributed in a region that is part of the range of the species or subspecies. Various races are often found in the same locality, but they are differentiated by their living conditions (ecological race). Thus, many plant species include an alpine race, a xeromorphic race, and a race that requires shade. Among animals, there are seasonal races of crustaceans. Many races of parasites are distinguished by their functional adaptation (specialization) to certain plant or animal hosts (races based on hosts). In ichthyology the term “race” refers to local populations (schools) of fish.
Sometimes, geographic races are regarded as subspecies. The term “race” is also used with reference to breeds of domesticated animals.
What does it mean when you dream about a race?
Running a race may depict how the dreamer feels about his or her waking life (a hectic “rat race,” perhaps?), possibly indicating the dreamer should slow down or change his or her approach to life.
RACE(1) See race condition and RACE encoding.
(2) (Research And Development of Advanced Communications) A European program of telecommunications R&D introduced in 1987. Over the subsequent 10-year period, more than 100 projects were undertaken.
(3) (Random Access Card Equipment) An early magnetic card storage device from RCA that was used with its IBM-compatible Spectra 70 mainframes. The units read and wrote data on a deck of 4x18" cards with a magnetic recording surface. The card was released from the cartridge, passed down a raceway, wrapped around a read/write head and returned. Operating in the late 1960s, the machine jammed frequently, and an operator had to remain nearby to extricate and replace the damaged cards. See CRAM, Data Cell and racetrack memory.<|endoftext|>
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### Lesson 9 - Math Techniques for Adult Basic Education (ABE) Learners Illinois State Library
#### More Math Activities
Page 4 of 6
Many adult learners want to learn certain math skills like multiplication or fractions. They may need these math skills for specific purposes like securing better employment, helping their children with math assignments, or calculating their home budget.
The following are suggested strategies to build specific math skills.
#### Fractions (1/2, 1/3, 1/4)
One half
Fill a 1/2 cup measuring cup and empty into a 1-cup measuring cup. Talk with the student about how much is filled. Repeat and talk about how many 1/2-cup measuring cups it takes to fill the 1-cup.
Fold paper in 1/2 lengthwise and have the student find another way to fold it in 1/2.
One third
Use rectangular shapes for this activity, such as graham crackers. Discuss how many parts make up a whole cracker. Use the measuring cup activity and paper activity from the one half activity to demonstrate one third.
One fourth (or one quarter)
Repeat the activities in One half and One third using measuring cups, paper, or graham crackers to illustrate quarters.
#### Tools
Kit-Kat and Hershey's candy bars also divide well into fractions. Egg cartons are already divided into twelve sections and can be cut easily to further demonstrate fractions.
#### Adding and Subtracting 2-Digit Numbers without Borrowing
Place 2 dimes together, 3 pennies together, and 5 pennies together into three separate sets. Ask the learner to write the number in each set, tens and ones, and find out how many there are altogether. Stress that ones are added to ones and tens are added to tens.
Using dimes and pennies, show 47 cents. Ask the learner to write the number and to show the value that will be left if 6 pennies are taken away. Remind them that pennies are taken from pennies. Have them take the 6 pennies away.
Arrange 47 paper clips in 4 strings of 10 with 7 in the final column. Ask the learner how many will be left if 23 are taken away. Remove 2 groups of 10 and 3 loose paper clips.
Give the student a dime and 3 pennies and say: "Write the example that will tell how many are left if I take away 5 pennies. Help me find a way to take away 5 pennies." (The dime must be exchanged for 10 pennies.)
13 − 5 = 8
#### Multiplication Tables
Multiplication facts often present a problem for learners. One strategy the tutor can use to ease the fears of learners is to explain that there are really only 15 facts to learn (memorize). Start by showing your learner that any number multiplied by 1 is just the number itself.
Next, have the learner count by 2s showing the answers to 2x1, 2x2, 2x3, etc. Then have the learner do the same counting by 5s for 5x1, 5x2, 5x3, etc. The tutor should write out the problems and fill in the answers as the learner counts.
Finally, show the learner that multiplying by the greatly feared "9" is not what it appears. Start by showing your learner that if you put the problems in a row first. Then, start at 2x9 and count from 1 to 8. Last, start at 2x9 and count backwards from 8 to 1. You have your answers for the 9s. Also point out to learners that all the answers for each problem add up to 9.
First, Then, on left Last, on right Answer
1 x 9 = 0 9 9
2 X 9 = 1 8 18
3 x 9 = 2 7 27
4 x 9 = 3 6 36
5 x 9 = 4 5 45
6 x 9 = 5 4 54
7 x 9 = 6 3 63
8 x 9 = 7 2 72
9 x 9 = 8 1 81
Still another way to teach the 9s, particularly to learners who have tactile/kinesthetic learning strengths, is to use the hands. Try it and see!
1. Place both hands flat on the table.
2. Starting with the left little finger, number your fingers from 1-10.
3. For 9 x 1, go to finger #1 (left pinkie) and tuck it under.
4. Count all the fingers on the other side of that finger. Total is 9!
5. Try another! 9 x 7 = (tuck finger #7, right pointer finger)
6. How many fingers on the left of #7? 6-Right!
7. How many fingers on the right of #7? 3-Right! The answer is 63
8. GOOD! Now try a few! 9 x 4 =, 9 x 8 =
#### Probability
If a person tosses a coin, only 1 out of 2 sides can show when it lands, so the probability for either side is 1 out of 2 or 1/2. If a die with numbers 1-6 is rolled, the probability of any number showing is 1 out of 6 or 1/6. If there are 3 socks in a drawer - 2 red and 1 blue - the probability of blindly pulling out a red sock is twice as great as that for a blue sock. Red sock - 2 out of 3 or 2/3 and blue sock - 1 out of 3 or 1/3.
• You need two boxes, each containing 5 marbles (buttons, cards, socks) that are alike in color. Write A on one box and B on the other. In Box A, place 4 items of one color (say, red checkers) and 1 item of the other color (say, black) and in Box B, place 2 items of one color (say, red checkers) and 3 items of the other color (say, black).
• Using Box A, draw 1 checker. Use a graph to record each draw by coloring in a square in the proper column. Return the checker to the box after each draw.
• After about 15 draws, discuss which color was drawn more times (red) and why? Do the same with Box B. Compare the two graphs and discuss from which box did they draw more blacks and why? More red and why? Do the same thing again, only now don't replace the checker after the draw. Discuss how this changes the probability.
Show a die and ask which number will be on top when you roll it. Have the student roll the die 20-30 times and record the results on a graph.
Show the student the following graph and two dice. Have the student roll the dice several times, and after each roll, color in a square above the sum of the 2 numbers. After about 50 rolls, ask which sum was rolled the most times? (Probably 7) Which sums were rolled least? (Probably 2 and 12) Discuss why?
#### Place Value
To Reinforce Value of Hundreds Place:
Show the learner nine bundles of ten sticks and nine individual ones (Popsicle sticks, toothpicks, etc.). Make certain to have rubber bands on the table for this activity. Ask the learner to count the number of 10s and ones and record the number. Place one more stick with the 9 ones and ask how many sticks there are. Observe to see if the student bundles the 10 ones to make 1 ten. If not, ask: "What do we do when we have 10 ones?" (make a ten) Then ask how many 10s do we have now? (ten) We have 10 tens. What do we do when we have 10? (We bundle them) So we will bundle the 10 tens and we have? (100) One bundle of tens is 100. Continue this activity showing other sets of hundreds, tens, and ones from 100 to 999. Have the student record each number.
Hundreds Tens Ones
1 0 0
1 10
(Bundled)
0
1 10 100<|endoftext|>
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Forces Science KS2. Have fun with forces with links to free to use interactive teaching resources.
All About Forces Questions Interactive
A set of questions to test knowledge gained in 'All About Forces' and 'Parkworld Plot'
Forces In Action Interactive
Learn about forces in action as you experiment how gradients, weights, motion and resistance affect the movement of various objects with this fun science activity.
Friction Test Track Interactive
Click on the chequered flag to find out how the track surface and gradient of Freddy's test track affect how far the go cart travels.
Magnets and Springs Interactive
Magnets have north poles and south poles. These attract each other. But two north poles will repel each other, as will two south poles.
When a spring is stretched or squashed, it creates a force.
Parkworld Plot Interactive
Help the owner of Parkworld theme park find out which ride has been made unsafe by his competitor.
Water Rockets Video
Children from Links Primary School in London investigate water rockets, streamlining their rocket shape to try to make their rocket travel the furthest.<|endoftext|>
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Question
1. # The Radius Of A Circle Is Increased By 1%. Find How Much % Does Its Area Increases?
Every day, we use circles to help us understand the world around us. When you see a circular diagram in a math textbook, for example, it’s easy to understand how that information affects the world around you. But what about more abstract concepts? What about numbers that don’t have a real-world counterpart? In this blog post, we will explore just that—the radius of a circle is increased by 1%. Find out how much its area increases, and start using this information in your everyday life.
## What is Radius?
Radius is the distance from the center of a circle to its edge. The radius of a circle is increased by %. For example, if the radius is increased by .5, then the circumference (around the circle) will be increased by 50%.
The area of a circle can be calculated by multiplying its radius by its circumference. The area will increase by % for any increase in radius.
## How Radius is Increased?
Radius can be increased by a certain percentage by increasing the radius of a circle. When the radius is increased, the area of the circle also increases. This is because when circles are drawn with larger radii, their circumference (the distance around the circle) becomes larger than their diameter.
## Example:A Circle With a Radius of 4
The radius of a circle is increased by %. Find how much % does its area increases?
The radius of a circle is increased by %. The area of the circle will increase by the percentage amount.
## Conclusion
The radius of a circle is increased by 1%. Find how much % does its area increases? The radius of the circle becomes 2.5% larger due to this increase in percentage.
2. Are you curious to find out how much area a circle’s radius increases when the radius is increased by 1? This is a common problem faced by mathematicians, engineers, and architects. In this article, we are going to delve into the world of geometry to answer this question.
The formula for finding the area of a circle is pi multiplied by the square of its radius. To discover how much increase in area will occur when the radius of a circle is increased by 1, it is necessary to calculate what happens when pi multiplied by an increase of one squared results in an overall change in area. When we multiply these two values together and take into account that pi remains constant at 3.141592 regardless of any changes in the size or shape of the circle, we can determine that there will be an increase in total area equal to 9.
3. 😃 Welcome to the fascinating world of mathematics! Today, we are going to explore an interesting concept – how much does the area of a circle increase when its radius is increased by 1%?
Let’s begin by understanding the basics. The area of a circle is calculated by multiplying the square of its radius by pi. That is A = πr². So, when the radius of a circle is increased by 1%, its area will also increase by 1%.
But, it is important to note that this increase in the area of a circle is not linear. That is, the increase in the area of a circle is not proportional to the increase in its radius. Instead, the increase in the area of a circle is exponential. That is, when the radius of a circle increases by 1%, its area increases by more than 1%.
To calculate the exact percentage increase in the area of a circle when its radius is increased by 1%, we can use the formula A = πr² × (1 + r/100). This formula shows that the area of the circle increases by (1 + r/100) times its original area when the radius is increased by 1%.
For example, if the radius of a circle is 10 cm, the area of the circle would be π × 10² = 314.159 cm². Now, if the radius of the circle is increased by 1%, its area would be π × 11² × (1 + 10/100) = 354.948 cm². Thus, the area of the circle increases by 12.79%.
Hence, the area of a circle increases by more than 1% when its radius is increased by 1%. So the next time you come across this question, you know what to do!<|endoftext|>
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## Thursday, March 10, 2016
### 4 SIMPLE KEYS FOR INDUCTIVE PROOF
Whether mathematical induction seems routine or complicated, a few simple keys can make the process easier to follow.
1. Have a structure within which to organize your work.
1a. The left-hand column will be the same each time. Every teacher will have specific wording intended to describe the process. Additional comments like “Let n = 1” or “Assuming the base case is true for n = k, we will now show that the base case is true for k + 1 and therefore for all integers” will be redundant from problem to problem within the discrete classroom.
1b. By keeping the calculations aligned, it will be easier to follow the calculation steps.
2. Highlight specific information. Another organization tool is intended to identify what part of the Inductive Assumption (also called the Inductive Hypothesis) will be used to simplify calculations in the proof. Highlighting the terms can take several forms like underlining or enclosing in brackets.
3. Use the highlighted form. The Induction Step itself follows a predictable steps. The key is to employ strong algebra skills in order to manipulate the left side to look like a simplified right side!
Add (k + 1) to the 1 + 2 + 3 + … + k + (k + 1) = (k + 1)(k + 1 + 1) / 2
left side and replace
k with (k + 1) on the
right side
Replace the underlined k (k + 1) + (k + 1) = (k + 1)(k + 2)
section on the left. 2 2
Then do all that elementary Algebra on the left side…..
…..adding fractions k (k + 1) + 2(k + 1)
2
…..factor out common terms (k + 1)(k + 2) = (k + 1)(k + 2)
2 2
QED: the left side looks just
like the right side!
4. Recognize properties of exponents. 'Product' and 'Multiple' induction problems have unique applications. The objective is to pull out the divisor so multiplying it by anything else would prove divisibility. This strategy might require expanding an exponential and using the distributive property. Here’s an example:
Prove that 11^n - 4^n is ‘a multiple of 7’ (or is ‘divisible by 7’)
Initial Test Step Let n = 1
11^1 - 4^1 = 11 - 4 = 7 (true)
Assume true when 11^k - 4^k is a multiple of 7
n = k Therefore, 11^k - 4^k = 7x
and 11^k = 7x + 4^k
Prove true when 11^(k + 1) - 4^(k + 1) is a multiple of 7
n = k + 1
(11^k)(11^1) - (4^k)(4^1)
(7x + 4^k)(11^1) - (4^k)(4^1)
11(7x + 4^k) - (4^k)(4)
77x + [11(4^k) - 4(4^k)]
77x + (4^k)(11 - 4)
77x + (4^k)(7)
7 (11x + 4^k) QED
Often, classroom expectations include some description like “7 times any number is a multiple of 7.” Individual teachers will have a favored statement to use.
5. Partition integers. Sometimes it is useful to split up integers into the sum of two other numbers (called Partitioning - a favored activity in early elementary numeracy education). To use it requires recognizing that 9 can be written as (8 + 1) or (5 + 4). So when proving that something is divisible by or a multiple of 8 for example, split up 9 into (8 + 1) and use the distributive property. For example…
Given: 18^n -1 is divisible by 17
Initial Test Step Let n = 1
18 - 1 = 17 (true)
Assume true when 18^k -1 is divisible by 17
n = k
Prove true when 18^(k + 1) - 1 is divisible by 17
n = k + 1
(18^k)(18^1) - 1
18(18^k)-1
(17 + 1)(18^k) -1
17(18^k) + 1(18^k -1)
Now might come a lengthy description preferred by a teacher:
17(18^k) is obviously divisible by 17 AND
(18^k - 1) is assumed divisible by 17 AND the sum
of any multiples of 17 is divisible by 17. QED
6. Know how to use Pascal’s Triangle to expand higher power binomials. While it is possible to raise a binomial to a higher power through long hand distribution, Pascal will be a work saver. Be ready to recognize what is duplicated and look for ways to factor out the divisor. Watch this…
Prove that the sum of the cubes of any 3 consecutive, positive integers is divisible by 9.
Set up 3 consecutive, positive integers = x, x + 1, x + 2
Cubes = x^3, (x + 1)^3, (x + 2)^3
Base Case x^3 + (x + 1)^3 + (x + 2)^3 is divisible by 9
Initial Test Step Let x = 1
1^3 + 2^3 + 3^3 is divisible by 9 (true)
Assume true when
x = k k^3 + (k + 1)^3 + (k + 2)^3 is divisible by 9
Pascal Assistant: k^3 = k^3
(k + 1)^3 = (k^3 + 3k^2 + 3k + 1)
(k + 2)^3 = (k^3 + 6k^2 +12k + 8)
Prove true when
x = k + 1 (k + 1)^3 + (k + 2)^3 + (k + 3)^3 is divisible by 9
Pascal Assistant: (k + 3)^3 = (k^3 + 9k^2 + 81k + 27)
k^3 + (k + 1)^3 + (k + 2)^3 + 9k^2 + 81k +27
[k^3 + (k + 1)^3 + (k + 2)^3] + 9(k^2 + 9k + 3)
[assumed divisible by 9] obviously divisible
by 9
Well, proof by induction may never be ‘simple,’ but it IS predictable and relies on just a few well-known manipulations. Keep these keys in mind and the task will (eventually) become routine.<|endoftext|>
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###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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# The Cross Product of Vectors - Concept
Norm Prokup
###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Share
In Calculus we often use the cross product of vectors to find orthogonal vectors and the area of parallelograms in three dimensions. The cross product of vectors is found by identifying the 3x3 determinants, however we substitute one of the rows with symbols that represent unit vectors. Finding 3x3 determinants can be made easier if we understand how to simplify determinants.
One really important application of determinants is that it can be used to define the cross product of vectors so this brings us back to vectors. Let's review one idea though. A vector a 3D vector v whose components are a, b, c can be written in terms of the unit vectors i, j and k where i is the unit vector in the positive x direction, j is the unit vector in the positive y direction and k is the unit vector in the positive z direction so it's in alphabetic order just like xyz, ijk and this is how you would write vector v you'd write it as aah aah a times i plus b times j plus c times k right, the components are still easy to see.
Now let's suppose we have two vectors in this form in this i, j, k form then the cross product is defined v cross w equals the determinant and across the first row, I put the 3 unit vectors i, j and k and then I put across the second row the components of the first vector in the cross product v and across the last row I put the components of the second vector in the cross product w so that's really important the order that you put these two right? The first one in the cross product you put in the second row the second one you put in the third row. Let's try this in an example.
So I have two vectors v equals 4, -5, -3, w equals 2, 2, 1 I first have to put these in i j k form so that's really easy because all you have to do is just fill in the i-5j-3k same for this, 2i plus 2j plus 1k it's just k and then we cross w is the determinant i, j, k so you always put i, j, k across the top and then 4, -5, -3 those are the components of v and then the components of w 2, 2 and 1 and you expand this the way you would a normal determinant so I'll expand across the top row I have i times this minor -5, -3, 2, 1 then I have to subtract j times it's minor 4, -3, 2, 1 then I add k times its minor oops 4, -5, 2, 2 okay and then I just compute these two by two determinants so this one is going to be -5--6 so -5+6 which is 1 that gives me i times 1 minus j this will be 4--6, 4+6 10 so j times 10 plus k I have 8--10 8+10 is 18 so the answer is i-10j+18k and of course you can put that into regular component form if you like, v cross w is 1,-10, 18 so depending on what what form your teacher wants either of these is an acceptable answer.<|endoftext|>
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We’ve been looking at art all over the world this month, and today, I’m staying home in the United States of America. As with all places I have covered so far, it is hard to pick just one. This was the first artwork to pop in my head when I thought of the US thought, and I think it’s one of those images that sticks with you.
In the 1930s, a Great Depression engulfed the country. Many people lost their savings, their jobs, and their homes. Business, farms, banks, and factories failed. (Side Note: I originally wrote about the Stock Market crash of 1929 leading to the Great Depression, but it turns out it is much more complicated than that so I took it out. Watch this excellent Crash Course video to find out more.)
You would think that this period of devastation would lead to less art being produced, because people had to focus on things like survival, but the Franklin D. Roosevelt and Congress, in an effort to stimulate the economy and create jobs, created the New Deal. In the New Deal, there were many job programs to both help people support themselves and to help the country’s infrastructure.
Artists, sculptors, and photographers were hired to create art in public places, build buildings and bridges, and document the plight of the American people. Around 1,300 murals and 300 sculptures were commissioned in public spaces like courthouses, post offices, schools, federal buildings, and more.
“The New Deal arts projects provided work for jobless artists, but they also had a larger mission: to promote American art and culture and to give more Americans access to what President Franklin Roosevelt described as “an abundant life.” The projects saved thousands of artists from poverty and despair and enabled Americans all across the country to see an original painting for the first time, attend their first professional live theater, or take their first music or drawing class.” (Source: The National Archive: A New Deal for the Arts)
One program specifically was enacted by the Farm Security Administration. The FSA hired photographers to chronicle the lives of out-of-work farmers and migrant workers, and that was the assignment that led to our feature image today by Dorothea Lange. (It took me a minute to get there, but I got there! 🙂 )
Photographer Dorothea Lange worked for this program from 1935 to 1939. She took pictures and distributed them free to newspapers across the country. Her work “brought the plight of the poor and forgotten – particularly sharecroppers, displaced farm families, and migrant workers – to public attention” (Source: Wikipedia).
Lange’s most famous image is Migrant Mother. This is what she had to say about taking this photograph:
I saw and approached the hungry and desperate mother, as if drawn by a magnet. I do not remember how I explained my presence or my camera to her, but I do remember she asked me no questions. I made five exposures, working closer and closer from the same direction. I did not ask her name or her history. She told me her age, that she was thirty-two. She said that they had been living on frozen vegetables from the surrounding fields, and birds that the children killed. She had just sold the tires from her car to buy food. There she sat in that lean-to tent with her children huddled around her, and seemed to know that my pictures might help her, and so she helped me. There was a sort of equality about it.
Dorothea Lange, 1960, Popular Photography
Art Discussion Questions
- What’s going on? What do you see that makes you say that?
- What can you tell about these people based on what you see? What can’t you tell about the people in this photograph?
- What is the mood or feeling of this artwork?
- How do you feel about the people in this photograph? How do you think the artist felt about the people in this photograph?
- Why do you think this photograph was made?
Have students imagine that they are hired by the government today in 2014 to document life in America. What types of photographs would they take? What stories would they tell? How would they tell those stories through photography?
Additionally, try having students document the people in their lives with a camera. How can you tell the story of a person through picture only? What do pictures leave out?
Dorothea Lange Resources and Citations
- Biographical Information and Photos by Dorothea Lange
- More works by Dorothea Lange
- Museum of Contemporary Photography: Dorothea Lange, Migrant Mother, and the Documentary Tradition
- Crash Course – Great Depression
- Crash Course – New Deal
- The National Archive: A New Deal for the Arts, Exhibit
Remember you can get the whole month of Art Around the World posts as a PDF eBook at the end of the month by subscribing to my e-mail newsletter. Enter your e-mail to get that when it comes out! Also, check out my Facebook page. I’m sharing more art from around the world on there. 🙂
If you like this post, you may also like:<|endoftext|>
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We’ve been working on all sorts of subjects since my last post. We have been working a lot at the chalkboard with Dubard work evolving our sounds and words into their full sentence structure and continuing with naming their grammar parts. The girls have been choosing to do more math at their desks as opposed to the big table in the kitchen where they can spread out their work more easily but with added tables it makes it possible next to their desks too.
We finished up the Montessori Pink series language which was far easier than the work the girls are used to reading now but it was good practice in between reading out loud their reader sets. The pink boxes were based on sentences and matching the appropriate pictures.
The girls have been choosing to work more often in their workbooks choosing between 4-5 of their favorite ones that combine all subjects together. Some of the topics covered:
- Math with abacus work
- Sentence structure: the subject part and the action part.
- Finding the main idea.
- Problem solving
- Word problems in math
- Tens and ones and grouping
- number and number names.
- Dot to dots counting by 2s, 5s, 10s.
- Shapes and measurements
- Geometry solids.
- Learning about light
- Introduction to the moon and it’s lunar cycle.<|endoftext|>
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Hukawng Valley in Myanmar on the continent of Asia has one of the richest deposits of arthropods found in amber from the Cretaceous period, with 252 families found so far. The latest fossil found by a team from the College of Science at Oregon State University is a strange, parasitic wasp without wings.
The wasp, preserved in amber more than 100 million years old, belongs to no other family ever identified on Earth, although different parts of its anatomy looks similar to those of other insects. The wasp is amazingly well preserved and scientists speculate that it crawled along the ground at the base of trees trying to find other insects and a place to lay its eggs. It however eventually disappeared and is now extinct. This could have been due to pathogens or habitat loss, or maybe because it couldn’t fly.
After substantial debate, mentioning first one body part and then another, researchers eventually decided to create a new family for the specimen, called Aptenoperissidae. This new family is part of the larger Order of Hymenoptera, which includes modern wasps and bees. Within that family, this wasp was named Aptenoperissus burmanicus and it is now the only known specimen.
George Poinar, Jr., one of the world’s leading experts on plant and animal life forms found preserved in amber, did not know what he was looking at when first seeing the insect. Poinar, who is co-author on the study, explained that although it was obvious that it could give a painful sting and was robust and tough, it simply did not fit anywhere and a new family had to be created for it. It ultimately dying out created an evolutionary dead end for that family.
When various researchers and reviewers, each with their own frame of reference, looked at the fossil, many saw something different. If you focused on the antenna, it looked like an ant. Its strong hind legs looks like a grasshopper’s and the thick abdomen like a cockroach. The team finally decided it had to be some kind of Hymenoptera because the face looks mostly like a wasp.
The insect possibly burrowed into cavities to seek pupae of other insects into which to lay its eggs. It is female and has long legs that were used to help pull it out of the cavities. If this were the case, wings would have been a hindrance. Researchers also speculate that it may have attacked other beetles with its sharp and jagged stinger, while its strong hind legs gave it a strong leaping ability. It did have a cleaning device on the tip of its antenna that is typical of Hymenoptera.
The findings have been published in the journal Cretaceous Research.<|endoftext|>
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# 16.4: Green’s Theorem
Flux Form of Green’s Theorem
Flux Form of Green’s Theorem
16.4: Green’s Theorem
Page ID
2620
• Apply the circulation form of Green’s theorem.
• Apply the flux form of Green’s theorem.
• Calculate circulation and flux on more general regions.
In this section, we examine Green’s theorem, which is an extension of the Fundamental Theorem of Calculus to two dimensions. Green’s theorem has two forms: a circulation form and a flux form, both of which require region $$D$$ in the double integral to be simply connected. However, we will extend Green’s theorem to regions that are not simply connected.
Put simply, Green’s theorem relates a line integral around a simply closed plane curve $$C$$ and a double integral over the region enclosed by $$C$$. The theorem is useful because it allows us to translate difficult line integrals into more simple double integrals, or difficult double integrals into more simple line integrals.
## Extending the Fundamental Theorem of Calculus
Recall that the Fundamental Theorem of Calculus says that
$\int_a^b F′(x)\,dx=F(b)−F(a). \nonumber$
As a geometric statement, this equation says that the integral over the region below the graph of $$F′(x)$$ and above the line segment $$[a,b]$$ depends only on the value of $$F$$ at the endpoints $$a$$ and $$b$$ of that segment. Since the numbers $$a$$ and $$b$$ are the boundary of the line segment $$[a,b]$$, the theorem says we can calculate integral $$\int_a^b F′(x)\,dx$$ based on information about the boundary of line segment $$[a,b]$$ (Figure $$\PageIndex{1}$$). The same idea is true of the Fundamental Theorem for Line Integrals:
$\int_C \vecs \nabla f·d\vecs r=f(\vecs r(b))−f(\vecs r(a)). \nonumber$
When we have a potential function (an “antiderivative”), we can calculate the line integral based solely on information about the boundary of curve $$C$$.
Green’s theorem takes this idea and extends it to calculating double integrals. Green’s theorem says that we can calculate a double integral over region $$D$$ based solely on information about the boundary of $$D$$. Green’s theorem also says we can calculate a line integral over a simple closed curve $$C$$ based solely on information about the region that $$C$$ encloses. In particular, Green’s theorem connects a double integral over region $$D$$ to a line integral around the boundary of $$D$$.
## Circulation Form of Green’s Theorem
The first form of Green’s theorem that we examine is the circulation form. This form of the theorem relates the vector line integral over a simple, closed plane curve $$C$$ to a double integral over the region enclosed by $$C$$. Therefore, the circulation of a vector field along a simple closed curve can be transformed into a double integral and vice versa.
Let $$D$$ be an open, simply connected region with a boundary curve $$C$$ that is a piecewise smooth, simple closed curve oriented counterclockwise (Figure $$\PageIndex{2}$$). Let $$\vecs F=⟨P,Q⟩$$ be a vector field with component functions that have continuous partial derivatives on $$D$$. Then,
\begin{align} \oint_C \vecs F·d\vecs r =\oint_C P\,dx+Q\,dy \\[4pt] =\iint_D (Q_x−P_y)\,dA. \end{align} \nonumber
Notice that Green’s theorem can be used only for a two-dimensional vector field $$\vecs F$$. If $$\vecs F$$ is a three-dimensional field, then Green’s theorem does not apply. Since
$\displaystyle \int_C P\,dx+Q\,dy=\int_C \vecs F·\vecs T\,ds \nonumber$
this version of Green’s theorem is sometimes referred to as the tangential form of Green’s theorem.
The proof of Green’s theorem is rather technical, and beyond the scope of this text. Here we examine a proof of the theorem in the special case that $$D$$ is a rectangle. For now, notice that we can quickly confirm that the theorem is true for the special case in which $$\vecs F=⟨P,Q⟩$$ is conservative. In this case,
$\oint_C P\,dx+Q\,dy=0 \nonumber$
because the circulation is zero in conservative vector fields. $$\vecs F$$ satisfies the cross-partial condition, so $$P_y=Q_x$$. Therefore,
$\iint_D (Q_x−P_y)\,dA=\int_D 0\,dA=0=\oint_C P\,dx+Q\,dy \nonumber$
which confirms Green’s theorem in the case of conservative vector fields.
Let’s now prove that the circulation form of Green’s theorem is true when the region $$D$$ is a rectangle. Let $$D$$ be the rectangle $$[a,b]×[c,d]$$ oriented counterclockwise. Then, the boundary $$C$$ of $$D$$ consists of four piecewise smooth pieces $$C_1$$, $$C_2$$, $$C_3$$, and $$C_4$$ (Figure $$\PageIndex{3}$$). We parameterize each side of $$D$$ as follows:
$$C_1: \vecs r_1(t)=⟨t,c⟩$$, $$a≤t≤b$$
$$C_2: \vecs r_2(t)=⟨b,t⟩$$, $$c≤t≤d$$
$$−C_3: \vecs r_3(t)=⟨t,d⟩$$, $$a≤t≤b$$
$$−C_4: \vecs r_4(t)=⟨a,t⟩$$, $$c≤t≤d$$.
Then,
\begin{align*} \int_C \vecs F·d \vecs r &=\int_{C_1} \vecs F·d \vecs r+\int_{C_2} \vecs F·d \vecs r+\int_{C_3} \vecs F·d \vecs r+\int_{C_4} \vecs F·d \vecs r \\[4pt] &=\int_{C_1} \vecs F·d \vecs r+\int_{C_2} \vecs F·d \vecs r−\int_{−C_3} \vecs F·d \vecs r−\int_{−C_4} \vecs F·d \vecs r \\[4pt] &=\int_a^b \vecs F( \vecs r_1(t))· \vecs r_1′(t)\,dt+\int_c^d \vecs F( \vecs r_2(t))· \vecs r_2′(t)\,dt−\int_a^b \vecs F( \vecs r_3(t))· \vecs r_3′(t)\,dt−\int_c^d \vecs F( \vecs r_4(t))·\vecs r_4′(t)\,dt\\[4pt] &=\int_a^b P(t,c)\,dt+\int_c^dQ(b,t)\,dt−\int_a^bP(t,d)\,dt−\int_c^dQ(a,t)\,dt \\[4pt] &=\int_a^b(P(t,c)−P(t,d))\,dt+\int_c^d(Q(b,t)−Q(a,t))\,dt\\[4pt] &=−\int_a^b(P(t,d)−P(t,c))\,dt+\int_c^d(Q(b,t)−Q(a,t))\,dt. \end{align*}
By the Fundamental Theorem of Calculus,
$P(t,d)−P(t,c)=\int_c^d \dfrac{\partial}{\partial y}P(t,y)dy \nonumber$
and
$Q(b,t)−Q(a,t)=\int_a^b \dfrac{\partial}{\partial x} Q(x,t)\,dx. \nonumber$
Therefore,
$−\int_a^b(P(t,d)−P(t,c))\,dt+\int_c^d(Q(b,t)−Q(a,t))\,dt=−\int_a^b\int_c^d \dfrac{\partial}{\partial y} P(t,y)\,dy\,dt+\int_c^d\int_a^b \dfrac{\partial}{\partial x}Q(x,t)\,dx\,dt. \nonumber$
But,
\begin{align*} −\int_a^b\int_c^d \dfrac{\partial}{\partial y}P(t,y)\,dy\,dt+\int_c^d\int_a^b \dfrac{\partial}{\partial x}Q(x,t)\,dx\,dt &=−\int_a^b\int_c^d \dfrac{\partial}{\partial y}P(x,y)\,dy\,dx+\int_c^d\int_a^b \dfrac{\partial}{\partial x}Q(x,y)\,dx\,dy \\[4pt] &=\int_a^b\int_c^d(Q_x−P_y)\,dy\,dx\\[4pt] &=\iint_D(Q_x−P_y)\,dA. \end{align*}
Therefore, $$\displaystyle \int_C \vecs F\cdot d\vecs r=\iint_D(Q_x−P_y)\,dA$$ and we have proved Green’s theorem in the case of a rectangle.
$$\square$$
To prove Green’s theorem over a general region $$D$$, we can decompose $$D$$ into many tiny rectangles and use the proof that the theorem works over rectangles. The details are technical, however, and beyond the scope of this text.
Calculate the line integral
$\oint_C x^2ydx+(y−3)dy, \nonumber$
where $$C$$ is a rectangle with vertices $$(1,1)$$, $$(4,1)$$, $$(4,5)$$, and $$(1,5)$$ oriented counterclockwise.
Solution
Let $$\vecs F(x,y)=⟨P(x,y),Q(x,y)⟩=⟨x^2y,y−3⟩$$. Then, $$Q_x(x,y)=0$$ and $$P_y(x,y)=x^2$$. Therefore, $$Q_x−P_y=−x^2$$.
Let $$D$$ be the rectangular region enclosed by $$C$$ (Figure $$\PageIndex{4}$$). By Green’s theorem,
\begin{align*} \oint_C x^2ydx+(y−3)\,dy &=\iint_D (Q_x−P_y)\,dA \\[4pt] &=\iint_D −x^2 \,dA=\int_1^5\int_1^4−x^2\,dx\,dy \\[4pt] &=\int_1^5−21\,dy=−84.\end{align*}
Analysis
If we were to evaluate this line integral without using Green’s theorem, we would need to parameterize each side of the rectangle, break the line integral into four separate line integrals, and use the methods from the section titled Line Integrals to evaluate each integral. Furthermore, since the vector field here is not conservative, we cannot apply the Fundamental Theorem for Line Integrals. Green’s theorem makes the calculation much simpler.
Calculate the work done on a particle by force field
$\vecs F(x,y)=⟨y+\sin x,e^y−x⟩ \nonumber$
as the particle traverses circle $$x^2+y^2=4$$ exactly once in the counterclockwise direction, starting and ending at point $$(2,0)$$.
Solution
Let $$C$$ denote the circle and let $$D$$ be the disk enclosed by $$C$$. The work done on the particle is
$W=\oint_C (y+\sin x)\,dx+(e^y−x)\,dy. \nonumber$
As with Example $$\PageIndex{1}$$, this integral can be calculated using tools we have learned, but it is easier to use the double integral given by Green’s theorem (Figure $$\PageIndex{5}$$).
Let $$\vecs F(x,y)=⟨P(x,y),Q(x,y)⟩=⟨y+\sin x,e^y−x⟩$$. Then, $$Q_x=−1$$ and $$P_y=1$$. Therefore, $$Q_x−P_y=−2$$.
By Green’s theorem,
\begin{align*} W &=\oint_C(y+\sin(x))dx+(e^y−x)\,dy \\[4pt] &=\iint_D (Q_x−P_y)\,dA \\[4pt] &=\iint_D−2\,dA \\[4pt] &=−2(area(D))=−2\pi (2^2)=−8\pi. \end{align*}
Use Green’s theorem to calculate line integral
$\oint_C \sin(x^2)\,dx+(3x−y)\,dy. \nonumber$
where $$C$$ is a right triangle with vertices $$(−1,2)$$, $$(4,2)$$, and $$(4,5)$$ oriented counterclockwise.
Hint
Transform the line integral into a double integral.
$$\dfrac{45}{2}$$
In the preceding two examples, the double integral in Green’s theorem was easier to calculate than the line integral, so we used the theorem to calculate the line integral. In the next example, the double integral is more difficult to calculate than the line integral, so we use Green’s theorem to translate a double integral into a line integral.
Calculate the area enclosed by ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ (Figure $$\PageIndex{6}$$).
Solution
Let $$C$$ denote the ellipse and let $$D$$ be the region enclosed by $$C$$. Recall that ellipse $$C$$ can be parameterized by
• $$x=a\cos t$$,
• $$y=b \sin t$$,
• $$0≤t≤2\pi$$.
Calculating the area of $$D$$ is equivalent to computing double integral $$\iint_D \,dA$$. To calculate this integral without Green’s theorem, we would need to divide $$D$$ into two regions: the region above the x-axis and the region below. The area of the ellipse is
$\int_{−a}^a\int_0^{\sqrt{b^2−{(bx/a)}^2}} \,dy\,dx+\int_{−a}^{a} \int_{−\sqrt{b^2−{(bx/a)}^2}}^{0} \,dy\,dx. \nonumber$
These two integrals are not straightforward to calculate (although when we know the value of the first integral, we know the value of the second by symmetry). Instead of trying to calculate them, we use Green’s theorem to transform $$\iint_D \,dA$$ into a line integral around the boundary $$C$$.
Consider vector field
$F(x,y)=⟨P,Q⟩=⟨−\dfrac{y}{2},\dfrac{x}{2}⟩. \nonumber$
Then, $$Q_x=\dfrac{1}{2}$$ and $$P_y=−\dfrac{1}{2}$$, and therefore $$Q_x−P_y=1$$. Notice that $$\vecs F$$ was chosen to have the property that $$Q_x−P_y=1$$. Since this is the case, Green’s theorem transforms the line integral of $$\vecs F$$ over $$C$$ into the double integral of 1 over $$D$$.
By Green’s theorem,
\begin{align*} \iint_D \,dA &=\iint_D (Q_x−P_y)\,dA \\[4pt] &=\int_C \vecs F\cdot d\vecs r=\dfrac{1}{2}\int_C −y\,dx+x\,dy \\[4pt] &=\dfrac{1}{2}\int_0^{2\pi}−b \sin t(−a\sin t)+a(\cos t)b\cos t\,dt \\[4pt] &=\dfrac{1}{2}\int_0^{2\pi} ab \cos^2 t+ab \sin^2 t\,dt \\[4pt] &=\dfrac{1}{2}\int_0^{2\pi} ab\,dt =\pi ab. \end{align*}
Therefore, the area of the ellipse is $$\pi ab\;\text{units}^2$$.
In Example $$\PageIndex{3}$$, we used vector field $$\vecs F(x,y)=⟨P,Q⟩=⟨−\dfrac{y}{2},\dfrac{x}{2}⟩$$ to find the area of any ellipse. The logic of the previous example can be extended to derive a formula for the area of any region $$D$$. Let $$D$$ be any region with a boundary that is a simple closed curve $$C$$ oriented counterclockwise. If $$F(x,y)=⟨P,Q⟩=⟨−\dfrac{y}{2},\dfrac{x}{2}⟩$$, then $$Q_x−P_y=1$$. Therefore, by the same logic as in Example $$\PageIndex{3}$$,
$\text{area of} \; D=\iint_D dA=\dfrac{1}{2}\oint_C−ydx+xdy. \label{greenarea}$
It’s worth noting that if $$F=⟨P,Q⟩$$ is any vector field with $$Q_x−P_y=1$$, then the logic of the previous paragraph works. So. Equation \ref{greenarea} is not the only equation that uses a vector field’s mixed partials to get the area of a region.
Find the area of the region enclosed by the curve with parameterization $$r(t)=⟨\sin t\cos t,\sin t⟩$$, $$0≤t≤\pi$$.
Hint
Use Equation \ref{greenarea}.
$$\dfrac{4}{3}$$
## Flux Form of Green’s Theorem
The circulation form of Green’s theorem relates a double integral over region $$D$$ to line integral $$\oint_C \vecs F·\vecs Tds$$, where $$C$$ is the boundary of $$D$$. The flux form of Green’s theorem relates a double integral over region $$D$$ to the flux across boundary $$C$$. The flux of a fluid across a curve can be difficult to calculate using the flux line integral. This form of Green’s theorem allows us to translate a difficult flux integral into a double integral that is often easier to calculate.
Let $$D$$ be an open, simply connected region with a boundary curve $$C$$ that is a piecewise smooth, simple closed curve that is oriented counterclockwise (Figure $$\PageIndex{7}$$). Let $$\vecs F=⟨P,Q⟩$$ be a vector field with component functions that have continuous partial derivatives on an open region containing $$D$$. Then,
$\oint_C \vecs F·\vecs N\,ds=\iint_D P_x+Q_y\,dA. \label{GreenN}$
Because this form of Green’s theorem contains unit normal vector $$\vecs N$$, it is sometimes referred to as the normal form of Green’s theorem.
Recall that $$\displaystyle \oint_C \vecs F·\vecs N\,ds=\oint_C −Q\,dx+P\,dy$$. Let $$M=−Q$$ and $$N=P$$. By the circulation form of Green’s theorem,
\begin{align*} \oint_C−Q\,dx+P\,dy &=\oint_C M\,dx+N\,dy\\[4pt] &=\iint_D N_x−M_y \,dA\\[4pt] &=\iint_D P_x−{(−Q)}_y \,dA\\[4pt] &=\iint_D P_x+Q_y \,dA. \end{align*}
$$\square$$
Let $$C$$ be a circle of radius $$r$$ centered at the origin (Figure $$\PageIndex{8}$$) and let $$\vecs F(x,y)=⟨x,y⟩$$. Calculate the flux across $$C$$.
Solution
Let $$D$$ be the disk enclosed by $$C$$. The flux across $$C$$ is $$\displaystyle \oint_C \vecs F·\vecs N\,ds$$. We could evaluate this integral using tools we have learned, but Green’s theorem makes the calculation much more simple. Let $$P(x,y)=x$$ and $$Q(x,y)=y$$ so that $$\vecs F=⟨P,Q⟩$$. Note that $$P_x=1=Q_y$$, and therefore $$P_x+Q_y=2$$. By Green’s theorem,
$\int_C \vecs F\cdot\vecs N\,ds=\iint_D 2\,dA=2\iint_D \,dA. \nonumber$
Since $$\displaystyle \iint_D \,dA$$ is the area of the circle, $$\displaystyle \iint_D \,dA=\pi r^2$$. Therefore, the flux across $$C$$ is $$2\pi r^2$$.
Let $$S$$ be the triangle with vertices $$(0,0)$$, $$(1,0)$$, and $$(0,3)$$ oriented clockwise (Figure $$\PageIndex{9}$$). Calculate the flux of $$\vecs F(x,y)=⟨P(x,y),Q(x,y)⟩=⟨x^2+e^y,x+y⟩$$ across $$S$$.
Solution
To calculate the flux without Green’s theorem, we would need to break the flux integral into three line integrals, one integral for each side of the triangle. Using Green’s theorem to translate the flux line integral into a single double integral is much more simple.
Let $$D$$ be the region enclosed by $$S$$. Note that $$P_x=2x$$ and $$Q_y=1$$; therefore, $$P_x+Q_y=2x+1$$. Green’s theorem applies only to simple closed curves oriented counterclockwise, but we can still apply the theorem because $$\displaystyle \oint_C \vecs F·\vecs N\,ds=−\oint_{−S} \vecs F·\vecs N\,ds$$ and $$−S$$ is oriented counterclockwise. By Green’s theorem, the flux is
\begin{align*} \oint_C \vecs F·\vecs N\,ds &= \oint_{−S} \vecs F·\vecs N\,ds\\[4pt] &=−\iint_D (P_x+Q_y)\,dA \\[4pt] &=−\iint_D (2x+1)\,dA.\end{align*}
Notice that the top edge of the triangle is the line $$y=−3x+3$$. Therefore, in the iterated double integral, the $$y$$-values run from $$y=0$$ to $$y=−3x+3$$, and we have
\begin{align*} −\iint_D (2x+1)\,dA &= −\int_0^1\int_0^{−3x+3}(2x+1)\,dy\,dx \\[4pt] &=−\int_0^1(2x+1)(−3x+3)\,dx \\[4pt] &=−\int_0^1(−6x^2+3x+3)\,dx\\[4pt] &=−{[−2x^3+\dfrac{3x^2}{2}+3x]}_0^1 \\[4pt] &=−\dfrac{5}{2}. \end{align*}
Calculate the flux of $$\vecs F(x,y)=⟨x^3,y^3⟩$$ across a unit circle oriented counterclockwise.
Hint
Apply Green’s theorem and use polar coordinates.
$$\dfrac{3\pi}{2}$$
Water flows from a spring located at the origin. The velocity of the water is modeled by vector field $$\vecs v(x,y)=⟨5x+y,x+3y⟩$$ m/sec. Find the amount of water per second that flows across the rectangle with vertices $$(−1,−2)$$, $$(1,−2)$$, $$(1,3)$$,and $$(−1,3)$$, oriented counterclockwise (Figure $$\PageIndex{10}$$).
Solution
Let $$C$$ represent the given rectangle and let $$D$$ be the rectangular region enclosed by $$C$$. To find the amount of water flowing across $$C$$, we calculate flux $$\int_C \vecs v\cdot d\vecs r$$. Let $$P(x,y)=5x+y$$ and $$Q(x,y)=x+3y$$ so that $$\vecs v=⟨P,Q⟩$$. Then, $$P_x=5$$ and $$Q_y=3$$. By Green’s theorem,
\begin{align*} \int_C \vecs v\cdot d\vecs r &=\iint_D (P_x+Q_y)\,dA \\ &=\iint_D 8\,dA \\ &=8(area\space of\space D)=80. \end{align*}
Therefore, the water flux is 80 m2/sec.
Recall that if vector field $$\vecs F$$ is conservative, then $$\vecs F$$ does no work around closed curves—that is, the circulation of $$\vecs F$$ around a closed curve is zero. In fact, if the domain of $$\vecs F$$ is simply connected, then $$\vecs F$$ is conservative if and only if the circulation of $$\vecs F$$ around any closed curve is zero. If we replace “circulation of $$\vecs F$$” with “flux of $$\vecs F$$,” then we get a definition of a source-free vector field. The following statements are all equivalent ways of defining a source-free field $$\vecs F=⟨P,Q⟩$$ on a simply connected domain (note the similarities with properties of conservative vector fields):
1. The flux $$\displaystyle \oint_C \vecs F·\vecs N\,ds$$ across any closed curve $$C$$ is zero.
2. If $$C_1$$ and $$C_2$$ are curves in the domain of $$\vecs F$$ with the same starting points and endpoints, then $$\displaystyle \int_{C_1} \vecs F·\vecs N\,ds=\int_{C_2} \vecs F·\vecs N\,ds$$. In other words, flux is independent of path.
3. There is a stream function $$g(x,y)$$ for $$\vecs F$$. A stream function for $$\vecs F=⟨P,Q⟩$$ is a function g such that $$P=g_y$$ and $$Q=−g_x$$.Geometrically, $$\vecs F=\langle a,b\rangle$$ is tangential to the level curve of $$g$$ at $$(a,b)$$. Since the gradient of $$g$$ is perpendicular to the level curve of $$g$$ at $$(a,b)$$, stream function $$g$$ has the property $$\vecs F(a,b)\cdot\vecs\nabla g(a,b)=0$$ for any point $$(a,b)$$ in the domain of $$g$$. (Stream functions play the same role for source-free fields that potential functions play for conservative fields.)
4. $$P_x+Q_y=0$$
Verify that rotation vector field $$\vecs F(x,y)=⟨y,−x⟩$$ is source free, and find a stream function for $$\vecs F$$.
Solution
Note that the domain of $$\vecs F$$ is all of $$ℝ^2$$, which is simply connected. Therefore, to show that $$\vecs F$$ is source free, we can show any of items 1 through 4 from the previous list to be true. In this example, we show that item 4 is true. Let $$P(x,y)=y$$ and $$Q(x,y)=−x$$. Then $$P_x=0=Q_y$$, and therefore $$P_x+Q_y=0$$. Thus, $$\vecs F$$ is source free.
To find a stream function for $$\vecs F$$, proceed in the same manner as finding a potential function for a conservative field. Let $$g$$ be a stream function for $$\vecs F$$. Then $$g_y=y$$, which implies that
$$g(x,y)=\dfrac{y^2}{2}+h(x)$$.
Since $$−g_x=Q=−x$$, we have $$h′(x)=x$$. Therefore,
$$h(x)=\dfrac{x^2}{2}+C$$.
Letting $$C=0$$ gives stream function
$$g(x,y)=\dfrac{x^2}{2}+\dfrac{y^2}{2}$$.
To confirm that $$g$$ is a stream function for $$\vecs F$$, note that $$g_y=y=P$$ and $$−g_x=−x=Q$$.
Notice that source-free rotation vector field $$\vecs F(x,y)=⟨y,−x⟩$$ is perpendicular to conservative radial vector field $$\vecs \nabla g=⟨x,y⟩$$ (Figure $$\PageIndex{11}$$).
Find a stream function for vector field $$\vecs F(x,y)=⟨x \sin y,\cos y⟩$$.
Hint
Follow the outline provided in the previous example.
$$g(x,y)=−x\cos y$$
Vector fields that are both conservative and source free are important vector fields. One important feature of conservative and source-free vector fields on a simply connected domain is that any potential function $$f$$ of such a field satisfies Laplace’s equation $$f_{xx}+f_{yy}=0$$. Laplace’s equation is foundational in the field of partial differential equations because it models such phenomena as gravitational and magnetic potentials in space, and the velocity potential of an ideal fluid. A function that satisfies Laplace’s equation is called a harmonic function. Therefore any potential function of a conservative and source-free vector field is harmonic.
To see that any potential function of a conservative and source-free vector field on a simply connected domain is harmonic, let $$f$$ be such a potential function of vector field $$\vecs F=⟨P,Q⟩$$. Then, $$f_x=P$$ and $$f_x=Q$$ because $$\vecs \nabla f=\vecs F$$. Therefore, $$f_{xx}=P_x$$ and $$f_{yy}=Q_y$$. Since $$\vecs F$$ is source free, $$f_{xx}+f_{yy}=P_x+Q_y=0$$, and we have that $$f$$ is harmonic.
For vector field $$\vecs F(x,y)=⟨e^x\sin y,e^x\cos y⟩$$, verify that the field is both conservative and source free, find a potential function for $$\vecs F$$, and verify that the potential function is harmonic.
Solution
Let $$P(x,y)=e^x\sin y$$ and $$Q(x,y)=e^x \cos y$$. Notice that the domain of $$\vecs F$$ is all of two-space, which is simply connected. Therefore, we can check the cross-partials of $$\vecs F$$ to determine whether $$\vecs F$$ is conservative. Note that $$P_y=e^x \cos y=Q_x$$, so $$\vecs F$$ is conservative. Since $$P_x=e^x \sin y$$ and $$Q_y=e^x \sin y$$, $$P_x+Q_y=0$$ and the field is source free.
To find a potential function for $$\vecs F$$, let $$f$$ be a potential function. Then, $$\vecs \nabla f=\vecs F$$, so $$f_x(x,y)=e^x \sin y$$. Integrating this equation with respect to x gives $$f(x,y)=e^x \sin y+h(y)$$. Since $$f_y(x,y)=e^x \cos y$$, differentiating $$f$$ with respect to y gives $$e^x\cos y=e^x\cos y+h′(y)$$. Therefore, we can take $$h(y)=0$$, and $$f(x,y)=e^x\sin y$$ is a potential function for $$f$$.
To verify that $$f$$ is a harmonic function, note that $$f_{xx}(x,y)=\dfrac{\partial}{\partial x}(e^x\sin y)=e^x \sin y$$ and
$$f_{yy}(x,y)=\dfrac{\partial}{\partial x}(e^x\cos y)=−e^x\sin y$$. Therefore, $$f_{xx}+f_{yy}=0$$, and $$f$$ satisfies Laplace’s equation.
Is the function $$f(x,y)=e^{x+5y}$$ harmonic?
Hint
Determine whether the function satisfies Laplace’s equation.
No
## Green’s Theorem on General Regions
Green’s theorem, as stated, applies only to regions that are simply connected—that is, Green’s theorem as stated so far cannot handle regions with holes. Here, we extend Green’s theorem so that it does work on regions with finitely many holes (Figure $$\PageIndex{12}$$).
Before discussing extensions of Green’s theorem, we need to go over some terminology regarding the boundary of a region. Let $$D$$ be a region and let $$C$$ be a component of the boundary of $$D$$. We say that $$C$$ is positively oriented if, as we walk along $$C$$ in the direction of orientation, region $$D$$ is always on our left. Therefore, the counterclockwise orientation of the boundary of a disk is a positive orientation, for example. Curve $$C$$ is negatively oriented if, as we walk along $$C$$ in the direction of orientation, region $$D$$ is always on our right. The clockwise orientation of the boundary of a disk is a negative orientation, for example.
Let $$D$$ be a region with finitely many holes (so that $$D$$ has finitely many boundary curves), and denote the boundary of $$D$$ by $$\partial D$$ (Figure $$\PageIndex{13}$$). To extend Green’s theorem so it can handle $$D$$, we divide region $$D$$ into two regions, $$D_1$$ and $$D_2$$ (with respective boundaries $$\partial D_1$$ and $$\partial D_2$$), in such a way that $$D=D_1\cup D_2$$ and neither $$D_1$$ nor $$D_2$$ has any holes (Figure $$\PageIndex{13}$$).
Assume the boundary of $$D$$ is oriented as in the figure, with the inner holes given a negative orientation and the outer boundary given a positive orientation. The boundary of each simply connected region $$D_1$$ and $$D_2$$ is positively oriented. If $$\vecs F$$ is a vector field defined on $$D$$, then Green’s theorem says that
\begin{align} \oint_{\partial D} \vecs F·d\vecs{r} &=\oint_{\partial D_1}\vecs F·d\vecs{r}+\oint_{\partial D_2}\vecs F·d\vecs{r} \\ &=\iint_{D_1}Q_x−P_y\,dA+\iint_{D_2}Q_x−P_y\,dA \\ &=\iint_D (Q_x−P_y)\,dA.\end{align} \nonumber
Therefore, Green’s theorem still works on a region with holes.
To see how this works in practice, consider annulus $$D$$ in Figure $$\PageIndex{14}$$ and suppose that $$F=⟨P,Q⟩$$ is a vector field defined on this annulus. Region $$D$$ has a hole, so it is not simply connected. Orient the outer circle of the annulus counterclockwise and the inner circle clockwise (Figure $$\PageIndex{14}$$) so that, when we divide the region into $$D_1$$ and $$D_2$$, we are able to keep the region on our left as we walk along a path that traverses the boundary. Let $$D_1$$ be the upper half of the annulus and $$D_2$$ be the lower half. Neither of these regions has holes, so we have divided $$D$$ into two simply connected regions.
We label each piece of these new boundaries as $$P_i$$ for some $$i$$, as in Figure $$\PageIndex{14}$$. If we begin at $$P$$ and travel along the oriented boundary, the first segment is $$P_1$$, then $$P_2$$, $$P_3$$, and $$P_4$$. Now we have traversed $$D_1$$ and returned to $$P$$. Next, we start at $$P$$ again and traverse $$D_2$$. Since the first piece of the boundary is the same as $$P_4$$ in $$D_1$$, but oriented in the opposite direction, the first piece of $$D_2$$ is $$−P_4$$. Next, we have $$P_5$$, then $$−P_2$$, and finally $$P_6$$.
Figure $$\PageIndex{14}$$ shows a path that traverses the boundary of $$D$$. Notice that this path traverses the boundary of region $$D_1$$, returns to the starting point, and then traverses the boundary of region $$D_2$$. Furthermore, as we walk along the path, the region is always on our left. Notice that this traversal of the $$P_i$$ paths covers the entire boundary of region $$D$$. If we had only traversed one portion of the boundary of $$D$$, then we cannot apply Green’s theorem to $$D$$.
The boundary of the upper half of the annulus, therefore, is $$P_1\cup P_2\cup P_3\cup P_4$$ and the boundary of the lower half of the annulus is $$−P_4\cup P_5\cup −P_2\cup P_6$$. Then, Green’s theorem implies
\begin{align} \oint_{\partial D}\vecs F·d\vecs{r} &=\int_{P_1}\vecs F·d\vecs{r}+\int_{P_2}\vecs F·d\vecs{r}+\int_{P_3}\vecs F·d\vecs{r}+\int_{P_4}\vecs F·d\vecs{r}+\int_{−P_4}\vecs F·d\vecs{r}+\int_{P_5}\vecs F·d\vecs{r}+\int_{−P_2}\vecs F·d\vecs{r}+\int_{P_6}\vecs F·d\vecs{r} \\ &=\int_{P_1}\vecs F·d\vecs{r}+\int_{P_2}\vecs F·d\vecs{r}+\int_{P_3}\vecs F·d\vecs{r}+\int_{P_4}\vecs F·d\vecs{r}+\int_{P_4}\vecs F·d\vecs{r}+\int_{P_5}\vecs F·d\vecs{r}+\int_{−P_2}\vecs F·d\vecs{r}+\int_{P_6}\vecs F·d\vecs{r} \\ &=\int_{P_1}\vecs F·d\vecs{r}+\int_{P_3}\vecs F·d\vecs{r}+\int_{P_5}\vecs F·d\vecs{r}+\int_{P_6}\vecs F·d\vecs{r} \\ &=\oint_{\partial D_1}\vecs F·d\vecs{r}+\oint_{\partial D_2}\vecs F·d\vecs{r}\\ &=\iint_{D_1}(Q_x−P_y)\,dA+\iint_{D_2}(Q_x−P_y)\,dA \\ &=\iint_D(Q_x−P_y)\,dA. \end{align} \nonumber
Therefore, we arrive at the equation found in Green’s theorem—namely,
$\oint_{\partial D}\vecs F·d\vecs{r}=\iint_D (Q_x−P_y)\,dA. \nonumber$
The same logic implies that the flux form of Green’s theorem can also be extended to a region with finitely many holes:
$\oint_C F·N\,ds=\iint_D (P_x+Q_y)\,dA. \nonumber$
Calculate the integral
$\oint_{\partial D}(\sin x−\dfrac{y^3}{3})dx+(\dfrac{y^3}{3}+\sin y)dy, \nonumber$
where $$D$$ is the annulus given by the polar inequalities $$1≤r≤2$$, $$0≤\theta≤2\pi$$.
Solution
Although $$D$$ is not simply connected, we can use the extended form of Green’s theorem to calculate the integral. Since the integration occurs over an annulus, we convert to polar coordinates:
\begin{align*} \oint_{\partial D}(\sin x−\dfrac{y^3}{3})\,dx+(\dfrac{x^3}{3}+\sin y)\,dy &=\iint_D (Q_x−P_y)\,dA \\ &=\iint_D (x^2+y^2)\,dA\\ &=\int_0^{2\pi}\int_1^2 r^3\,drd\theta=\int_0^{2\pi} \dfrac{15}{4}\,d\theta \\ &=\dfrac{15\pi}{2}. \end{align*}
Let $$\vecs F=⟨P,Q⟩=⟨\dfrac{y}{x^2+y^2},-\dfrac{x}{x^2+y^2}⟩$$ and let $$C$$ be any simple closed curve in a plane oriented counterclockwise. What are the possible values of $$\oint_C \vecs F·d\vecs{r}$$?
Solution
We use the extended form of Green’s theorem to show that $$\oint_C \vecs F·d\vecs{r}$$ is either $$0$$ or $$−2\pi$$—that is, no matter how crazy curve $$C$$ is, the line integral of $$\vecs F$$ along $$C$$ can have only one of two possible values. We consider two cases: the case when $$C$$ encompasses the origin and the case when $$C$$ does not encompass the origin.
### Case 1: C Does Not Encompass the Origin
In this case, the region enclosed by $$C$$ is simply connected because the only hole in the domain of $$\vecs F$$ is at the origin. We showed in our discussion of cross-partials that $$\vecs F$$ satisfies the cross-partial condition. If we restrict the domain of $$\vecs F$$ just to $$C$$ and the region it encloses, then $$\vecs F$$ with this restricted domain is now defined on a simply connected domain. Since $$\vecs F$$ satisfies the cross-partial property on its restricted domain, the field $$\vecs F$$ is conservative on this simply connected region and hence the circulation $$\oint_C \vecs F·d\vecs{r}$$ is zero.
### Case 2: C Does Encompass the Origin
In this case, the region enclosed by $$C$$ is not simply connected because this region contains a hole at the origin. Let $$C_1$$ be a circle of radius a centered at the origin so that $$C_1$$ is entirely inside the region enclosed by $$C$$ (Figure $$\PageIndex{15}$$). Give $$C_1$$ a clockwise orientation.
Let $$D$$ be the region between $$C_1$$ and $$C$$, and $$C$$ is orientated counterclockwise. By the extended version of Green’s theorem,
\begin{align*} \int_C \vecs F·d\vecs{r}+\int_{C_1}\vecs F·d\vecs{r} &=\iint_D Qx_−P_y \,dA \\[4pt] &=\iint_D−\dfrac{y^2−x^2}{{(x^2+y^2)}^2}+\dfrac{y^2−x^2}{{(x^2+y^2)}^2}dA \\[4pt] &=0, \end{align*}
and therefore
$\int_C \vecs F·d\vecs{r}=−\int_{C_1} \vecs F·d\vecs{r}. \nonumber$
Since $$C_1$$ is a specific curve, we can evaluate $$\int_{C_1}\vecs F·d\vecs{r}$$. Let
$x=a\cos t, \;\; y=a\sin t, \;\; 0≤t≤2\pi \nonumber$
be a parameterization of $$C_1$$. Then,
\begin{align*} \int_{C_1}\vecs F·d\vecs{r} &=\int_0^{2\pi} F(r(t))·r′(t)dt \\[4pt] &=\int_0^{2\pi} ⟨−\dfrac{\sin(t)}{a},−\dfrac{\cos(t)}{a}⟩·⟨−a\sin(t),−a\cos(t)⟩dt \\[4pt] &=\int_0^{2\pi}{\sin}^2(t)+{\cos}^2(t)dt \\[4pt] &=\int_0^{2\pi}dt=2\pi. \end{align*}
Therefore, $$\int_C F·ds=−2\pi$$.
Calculate integral $$\oint_{\partial D}\vecs F·d\vecs{r}$$, where $$D$$ is the annulus given by the polar inequalities $$2≤r≤5$$, $$0≤\theta≤2\pi$$, and $$F(x,y)=⟨x^3,5x+e^y\sin y⟩$$.
Hint
Use the extended version of Green’s theorem.
$$105\pi$$
Imagine you are a doctor who has just received a magnetic resonance image of your patient’s brain. The brain has a tumor (Figure $$\PageIndex{16}$$). How large is the tumor? To be precise, what is the area of the red region? The red cross-section of the tumor has an irregular shape, and therefore it is unlikely that you would be able to find a set of equations or inequalities for the region and then be able to calculate its area by conventional means. You could approximate the area by chopping the region into tiny squares (a Riemann sum approach), but this method always gives an answer with some error.
Instead of trying to measure the area of the region directly, we can use a device called a rolling planimeter to calculate the area of the region exactly, simply by measuring its boundary. In this project you investigate how a planimeter works, and you use Green’s theorem to show the device calculates area correctly.
A rolling planimeter is a device that measures the area of a planar region by tracing out the boundary of that region (Figure $$\PageIndex{17}$$). To measure the area of a region, we simply run the tracer of the planimeter around the boundary of the region. The planimeter measures the number of turns through which the wheel rotates as we trace the boundary; the area of the shape is proportional to this number of wheel turns. We can derive the precise proportionality equation using Green’s theorem. As the tracer moves around the boundary of the region, the tracer arm rotates and the roller moves back and forth (but does not rotate).
Let $$C$$ denote the boundary of region $$D$$, the area to be calculated. As the tracer traverses curve $$C$$, assume the roller moves along the y-axis (since the roller does not rotate, one can assume it moves along a straight line). Use the coordinates $$(x,y)$$ to represent points on boundary $$C$$, and coordinates $$(0,Y)$$ to represent the position of the pivot. As the planimeter traces $$C$$, the pivot moves along the y-axis while the tracer arm rotates on the pivot.
Watch a short animation of a planimeter in action.
Begin the analysis by considering the motion of the tracer as it moves from point $$(x,y)$$ counterclockwise to point $$(x+dx,y+dy)$$ that is close to $$(x,y)$$ (Figure $$\PageIndex{18}$$). The pivot also moves, from point $$(0,Y)$$ to nearby point $$(0,Y+dY)$$. How much does the wheel turn as a result of this motion? To answer this question, break the motion into two parts. First, roll the pivot along the y-axis from $$(0,Y)$$ to $$(0,Y+dY)$$ without rotating the tracer arm. The tracer arm then ends up at point $$(x,y+dY)$$ while maintaining a constant angle $$\phi$$ with the x-axis. Second, rotate the tracer arm by an angle $$d\theta$$ without moving the roller. Now the tracer is at point $$(x+dx,y+dy)$$. Let ll be the distance from the pivot to the wheel and let L be the distance from the pivot to the tracer (the length of the tracer arm).
1. Explain why the total distance through which the wheel rolls the small motion just described is $$\sin \phi dY+ld\theta=\dfrac{x}{L}dY+ld\theta$$.
2. Show that $$\oint_C d\theta=0$$.
3. Use step 2 to show that the total rolling distance of the wheel as the tracer traverses curve $$C$$ is
Total wheel roll $$=\dfrac{1}{L}\oint_C xdY$$.
Now that you have an equation for the total rolling distance of the wheel, connect this equation to Green’s theorem to calculate area $$D$$ enclosed by $$C$$.
4. Show that $$x^2+(y−Y)^2=L^2$$.
5. Assume the orientation of the planimeter is as shown in Figure $$\PageIndex{18}$$. Explain why $$Y≤y$$, and use this inequality to show there is a unique value of $$Y$$ for each point $$(x,y)$$: $$Y=y=\sqrt{L^2−x^2}$$.
6. Use step 5 to show that $$dY=dy+\dfrac{x}{L^2−x^2}dx.$$
7. Use Green’s theorem to show that $$\displaystyle \oint_C \dfrac{x}{L^2−x^2}dx=0$$.
8. Use step 7 to show that the total wheel roll is
$\text{Total wheel roll}\quad =\quad 1L\oint_C x\,dy. \nonumber$
It took a bit of work, but this equation says that the variable of integration Y in step 3 can be replaced with y.
9. Use Green’s theorem to show that the area of $$D$$ is $$\oint_C xdy$$. The logic is similar to the logic used to show that the area of $$\displaystyle D=12\oint_C −y\,dx+x\,dy$$.
10. Conclude that the area of $$D$$ equals the length of the tracer arm multiplied by the total rolling distance of the wheel.
You now know how a planimeter works and you have used Green’s theorem to justify that it works. To calculate the area of a planar region $$D$$, use a planimeter to trace the boundary of the region. The area of the region is the length of the tracer arm multiplied by the distance the wheel rolled.
## Key Concepts
• Green’s theorem relates the integral over a connected region to an integral over the boundary of the region. Green’s theorem is a version of the Fundamental Theorem of Calculus in one higher dimension.
• Green’s Theorem comes in two forms: a circulation form and a flux form. In the circulation form, the integrand is $$\vecs F·\vecs T$$. In the flux form, the integrand is $$\vecs F·\vecs N$$.
• Green’s theorem can be used to transform a difficult line integral into an easier double integral, or to transform a difficult double integral into an easier line integral.
• A vector field is source free if it has a stream function. The flux of a source-free vector field across a closed curve is zero, just as the circulation of a conservative vector field across a closed curve is zero.
## Key Equations
• Green’s theorem, circulation form
$$\displaystyle ∮_C P\,dx+Q\,dy=∬_D Q_x−P_y\,dA$$, where $$C$$ is the boundary of $$D$$
• Green’s theorem, flux form
$$\displaystyle ∮_C\vecs F·\vecs N\,ds=∬_D P_x+Q_y\,dA$$, where $$C$$ is the boundary of $$D$$
• Green’s theorem, extended version
$$\displaystyle ∮_{\partial D}\vecs F·d\vecs{r}=∬_D Q_x−P_y\,dA$$
## Glossary
Green’s theorem
relates the integral over a connected region to an integral over the boundary of the region
stream function
if $$\vecs F=⟨P,Q⟩$$ is a source-free vector field, then stream function $$g$$ is a function such that $$P=g_y$$ and $$Q=−g_x$$
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Matrix Transformations
## Matrix Transformations
Matrix transformations are performed through matrix multiplication of a point matrix by a transformation matrix. The transformation matrix affects the point matrix, creating a new point matrix.
[T][P]=[N] where:
• [T] is the transformation matrix
• [P] is the point matrix being transformed
• [N] is the new transformed point matrix
The matrices are multiplied in this order in order to obtain a new point matrix and not a larger matrix.
A point matrix, P, takes the form of an n x 1 though generally, 2 x 1 and 3 x 1 matrices are generally used as they model 2D and 3D systems respectivey. The value P1,1 is the x co-ordinate and the value P2,1 is the y co-ordinate. In a 3D point matrix the value of P3,1 is the z co-ordinate.
For example:
=
is the form Therefore, the co-ordinates of this point matrix are: (4,3,1)
A transformation matrix is a square matrix which has an affect on a point matrix so as to transform it's position on it's co-ordinate axis when they are multiplied. The no. of rows of the point matrix determines the dimensions of the transformation matrix as the no. of columns of the transformation matrix must equal the no. of rows of the point matrix in order to allow matrix multiplication to occur as previously discussed in Basic Matrix Arithmetic.
For example: This transformation matrix will affect a reflection in the line y=x for the point matrix, as shown in the example below:
Transformation Matrix Point MatrixMultiplicationNew Point Matrix
= =
The transformation matrix has caused the point matrix to reflect in the line y=x creating the co-ordinate (3,4) from the co-ordinate (4,3).
A list of some 2D transformation matrices and their affects can be found in the section 2D Matrix Rotations and Reflections.
The point matrix denoting the co-ordinates (4,3) will be used as an example throughout this page in order to emphasise the changes that are taking placing in the point matrices under transformation matrices.
## 2D Matrix Rotations and Reflections
Matrix rotation and reflection is achieved through matrix multiplication with specific matrices which achieve certain rotations or reflections. This is effectively plotting the co-ordinates of the previous matrix and then matrix multiplying it and ending up with new co-ordinates of the matrix at another point on the graph.
There are a few rules which need to be followed when rotating or reflecting a matrix:
1. The transformation matrix is always the leading matrix in the multiplication [T][P]=[N] where:
• [T] is the transformation matrix
• [P] is the point matrix
• [N] is the new matrix
2. A point matrix is a n x 1 vector as it describes a point on a graph
3. The no. of rows of the transformation matrix must equal the no. of columns of the point matrix.
The matrices listed below all perform different rotations/reflections:
Transformation MatrixAffect of Transformation Matrix
on Point Matrix
Example
This transformation matrix is the identity matrix. When multiplying by this matrix, the point matrix is unaffected and the new matrix is exactly the same as the point matrix.
= =
This transformation matrix creates a reflection in the x-axis. When multiplying by this matrix, the x co-ordinate remains unchanged, but the y co-ordinate changes sign.
= =
This transformation matrix creates a reflection in the y-axis. When multiplying by this matrix, the y co-ordiante remains unchanged, but the x co-ordinate changes sign.
= =
This transformation matrix creates a rotation of 180 degrees. When multiplying by this matrix, the point matrix is rotated 180 degrees around (0,0). This changes the sign of both the x and y co-ordinates.
= =
This transformation matrix creates a reflection in the line y=x. When multiplying by this matrix, the x co-ordinate becomes the y co-ordinate and the y-ordinate becomes the x co-ordinate.
= =
This transformation matrix rotates the point matrix 90 degrees clockwise. When multiplying by this matrix, the point matrix is rotated 90 degrees clockwise around (0,0).
= =
This transformation matrix rotates the point matrix 90 degrees anti-clockwise. When multiplying by this matrix, the point matrix is rotated 90 degrees anti-clockwise around (0,0).
= =
This transformation matrix creates a reflection in the line y=-x. When multiplying by this matrix, the point matrix is reflected in the line y=-x changing the signs of both co-ordiantes and swapping their values.
= =
## Matrix Enlargements
The enlargement of a matrix can be achieved through matrix multiplication of a transformation matrix followed by a point matrix: [T][P]=[N]
The section 2D Reflections and Rotations shows how transforming with the identity matrix, doesn't appear to alter the co-ordiantes of the point matrix. This is due to:
• the new x co-ordinate being constructed of the old x co-ordinate multiplied by one added to the old y co-ordiante multiplied by zero
• the new y co-ordinate being constructed of the old y co-ordinate multiplied by one added to the old x co-ordiante multiplied by zero
eg.
= =
However, if the one's in this transformation matrix, were not ones and were another integer number, then the value of the new point matrix would be an enlargement of that integer number, on the original matrix. This is because the new x co-ordiante is the old one multiplied by T1,1 and the new y-co-ordinateis the old one multiplied by T2,2. This means that the identity matrix really performs an enlargment of scale factor 1.
Below are some examples of transformation matrices which enlarge point matrices:
Transformation MatrixAffect of Transformation Matrix
on Point Matrix
Example
This transformation matrix is the identity matrix multiplied by the scalar 6. When multiplying by this matrix, the point matrix is enlarged by a factor of 6 in the x and y directions.
= =
This transformation matrix is the identity matrix but T1,1 has been larged by a factor of 7 and T2,2 has been enlarged by a factor of 0. When multiplying by this matrix, the x co-ordinate is enlarged by a factor of 7, whilst the y co-ordinate is enlarged by a factor of 0.
= =
This transformation matrix is the identity matrix but T1,1 has been larged by a factor of a and T2,2 has been enlarged by a factor of b. When multiplying by this matrix, the x co-ordinate is enlarged by a factor of a, whilst the y co-ordinate is enlarged by a factor of b.
= =
This transformation matrix creates a rotation and an enlargement. When multiplying by this matrix, the point matrix is rotated 90 degrees anticlockwise around (0,0), whilst the x xo-ordinate of the new point matrix is enlarged by a factor of -5 and the y co-ordinate of the new point matrix is enlarged by a factor of 7.
= =
The table above shows that it is the position and sign of elements in a rotation or reflection transformation matrix that decides the rotation or reflection of the point matrix and not the value of the elements as enlarging the value of an element in a rotation matrix only enlarges the rotation. Enlarging a point matrix by a factor in either the x or y direction requires that the non-zero elements of the transformation matrix be enlarged by that factor.<|endoftext|>
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The Origin of Fossils
Contrary to what many scientific texts report, an orderly succession of evolutionary fossils does not exist in the most studied canyon in all the world—the Grand Canyon. In fact, there are many fossils on the surface of the canyon, but we find very few hiking through the canyon. Furthermore, the evolutionary fossil succession does not exist in any canyon in the world—a little known fact discussed in the Fossil Model. Fossils did not form covered by miles of sediment for millions of years. Find the answers to all of your questions about fossil formation in the Fossil Model, which explains the true origin of fossils; answers like, how wood turns into a stone instead of decomposing or how soft tissue jellyfish and feathers became rock are typical questions answered in Chapter 11, the Fossil Model.<|endoftext|>
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# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 4/5 - 3/4 = 1/20 = 0.05
Spelled result in words is one twentieth.
### How do you solve fractions step by step?
1. Subtract: 4/5 - 3/4 = 4 · 4/5 · 4 - 3 · 5/4 · 5 = 16/20 - 15/20 = 16 - 15/20 = 1/20
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(5, 4) = 20. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 5 × 4 = 20. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - four fifths minus three quarters = one twentieth.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Unknown number
I think the number - its sixth is 3 smaller than its third.
• Cupcakes 2
Susi has 25 cupcakes. She gives 4/5. How much does she have left?
• Brick
Isosceles scale has on one side all brick and second weight 1 kg and 1/4 of brick. The balance is in equilibrium. What is the weight of a brick?
• Trees
From the total number of trees in the orchard, there are two-fifths pearls and apples are three eighty. The rest of the trees are 9 ceremonial. How many trees are in the set?
• Trip
On the trip drank 3/10 of pupils tea, 2/5 cola, 1/4 mineral water and remaining 3 juice. How many students were on the trip?
• A man
A man spends 5/9 of his money on rent, and 5/16 of the remainder on electricity. If the final balance remaining is 550 find how much was spent on rent
• On a farm
On a farm, there are 90 rabbits, 700 sheep, 300 cattle and 500 pigs. What percentage of the total number of animals are rabbits?
• Two pumps together
The first pump will fill the tank itself in 3 hours and the second one in 6 hours. How many hours will the tank be full if both pumps are worked at the same time?
• One sixth
How many sixths are two thirds?
• Class 8.A
Three quarters of class 8.A went skiing. Of those who remained at home one third was ill and the remaining six were on math olympic. How many students have class 8.A?
• Sales tax
A sales tax on a Php 10,800 appliance is Php.1,620. What is the rate of sales tax?
• PC disks
Peter has 45 disks in three colors. One-fifth of the disks are blue, red are twice more than the white. How much is blue, red and white disks?<|endoftext|>
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Notes: Determinant
Table of contents Determinants of the Third Order Value of a Determinant Minors and Cofactors Expansion of a Determinant in terms of the Elements of any Row or Column Properties of Determinants Multiplication of Determinants: Special Determinants: System of Linear Equations (Cramer's Rule) :
If the equations a1x + b1 = 0 and a2x + b2 = 0 are satisfied by the same value of x, then a1b2 – a2b1 = 0.
The expression a1b2 – a2b1 is called a determinant of the second order, and is denoted by:
A determinant of the second order consists of two rows and two columns.
## Determinants of the Third Order
Consider the system of equations:
• a1x + b1y + c1 = 0
• a2x + b2y + c2 = 0
• a3x + b3y + c3 = 0
If these equations are satisfied by the same values of x and y, then on eliminating x and y we get:
a1(b2c3 – b3c2) + b1(c2a3 – c3a2) + c1(a2b3 – a3b2) = 0
The expression on the left is called a determinant of the third order, and is denoted by:
A determinant of the third order consists of three rows and three columns.
## Value of a Determinant
The value of a determinant of order three is given by:
D = a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2)
Using the Sarrus diagram to find the value of a third-order determinant:
Example 1: Find the value:
Solution:
= (27 + 42) – 2 (–36 –12) + 3 (28 – 6) = 231
Alternatively, Using Sarrus Diagram,
= (27 + 24 + 84) – (18 – 42 – 72)= 135 – (18 – 114) = 231
## Minors and Cofactors
The minor of a given element of a determinant is the determinant obtained by deleting the row and the column in which the given element stands.
For example, the minor of a1 in
Hence a determinant of order three will have “9 minors”. If Mij represents the minor of the element belonging to ith row and jth column then the cofactor of that element is given by : Cij = (–1)i + j . Mij.
## Expansion of a Determinant in terms of the Elements of any Row or Column
The sum of the products of the elements of any row (or column) of a matrix with their corresponding cofactors always equals the value of the determinant of the matrix.
This determinant $DD$ can be expressed in any of the following six forms:
1. $a1A1+b1B1+c1C1a_1 A_1 + b_1 B_1 + c_1 C_1$
2. $a1A1+a2A2+a3A3a_1 A_1 + a_2 A_2 + a_3 A_3$
3. $a2A2+b2B2+c2C2a_2 A_2 + b_2 B_2 + c_2 C_2$
4. $b1B1+b2B2+b3B3$
5. $a3A3+b3B3+c3C3a_3 A_3 + b_3 B_3 + c_3 C_3$
6. $c1C1+c2C2+c3C3c_1 C_1 + c_2 C_2 + c_3 C_3$
Here, Ai, Bi, and $C_i$Ci (for $i=1,2,3i = 1, 2, 3$) denote the cofactors of ai, $b_i$bi, and $c_i$ci respectively.
Additionally, the sum of the products of the elements of any row (or column) with the cofactors of another row (or column) is always equal to zero. Consequently, we have:
$a2A1+b2B1+c2C1=0a_2 A_1 + b_2 B_1 + c_2 C_1 = 0$, $b1A1+b2A2+b3A3=0b_1 A_1 + b_2 A_2 + b_3 A_3 = 0$ and so forth.
In these equations, $A_i$Ai, Bi, and Ci (for $i=1,2,3i = 1, 2, 3$) are again the cofactors of ai, $b_i$bi, and $c_i$ci respectively.
## Properties of Determinants
(a) The value of a determinant remains unaltered, if the rows & columns are inter-changed.
(b) If any two rows (or columns) of a determinant be interchanged, the value of determinant is changed in sign only. e.g.
(c) If all the elements of a row (or column) are zero, then the value of the determinant is zero.
(d) If all the elements of any row (or column) are multiplied by the same number, then the determinant is multiplied by that number.
(e) If all the elements of a row (or column) are proportional (or identical) to the element of any other row, then the determinant vanishes, i.e. its value is zero.
Example: Prove that:
(f) If each element of any row (or column) is expressed as a sum of two (or more) terms, then the determinant can be expressed as the sum of two (or more) determinants.
(g) Row - column operation : The value of a determinant remains unaltered under a column (Ci ) operation of the form Ci + aCj + bCk (j, k is not equal to i ) or row (Ri ) operation of the form Ri → Ri + aRj + bRk (j, k ≠ i). In other words, the value of a determinant is not altered by adding the elements of any row (or column) to the same multiples of the corresponding elements of any other row (or column).
(h) Factor theorem : If the elements of a determinant D are rational integral functions of x and two rows (or columns) become identical when x = a then (x – a) is a factor of D. Note that if r rows become identical when a is substituted for x, then (x – a)r–1 is a factor of D.
## Multiplication of Determinants:
Similarly two determinants of order three are multiplied. (a) Here we have multiplied row by column. We can also multiply row by row, column by row and column by column. (b) If D1 is the determinant formed by replacing the elements of determinant D of order n by their corresponding cofactors then D1 = Dn–1
Example: Let a & b be the roots of equation ax2 + bx + c = 0 and Sn = an + bn for n ≥ 1. Evaluate the value of the determinant
## System of Linear Equations (Cramer's Rule) :
### (a) Equations involving two variables :
(i) Consistent Equations : Definite & unique solution (Intersecting lines)
(ii) Inconsistent Equations : No solution (Parallel lines)
(iii) Dependent Equations : Infinite solutions (Identical lines)
### (b) Equations Involving Three variables :
Note :
(i) If D ≠ 0 and atleast one of D1 , D2 , D3 ≠ 0, then the given system of equations is consistent and has unique non trivial solution.
(ii) If D ≠ 0 & D1 = D2 = D3 = 0, then the given system of equations is consistent and has trivial solution only.
(iii) If D = 0 but atleast one of D1 , D2 , D3 is not zero then the equations are inconsistent and have no solution.
(iv) If D = D1 = D2 = D3 = 0, then the given system of equations may have infinite or no solution.
### (c) Homogeneous system of linear equations :
Applications of Determinants:
The document Notes: Determinant | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
All you need of JEE at this link: JEE
## Mathematics (Maths) for JEE Main & Advanced
209 videos|443 docs|143 tests
## FAQs on Notes: Determinant - Mathematics (Maths) for JEE Main & Advanced
1. How do you find the value of a determinant?
Ans. The value of a determinant is found by multiplying the elements of a row (or column) by their respective minors and adding them up.
2. What are minors and cofactors of a determinant?
Ans. Minors of a determinant are the determinants formed by removing the row and column of a specific element. Cofactors are the minors multiplied by -1 raised to the power of the sum of the row and column numbers.
3. How can a determinant be expanded in terms of the elements of any row or column?
Ans. A determinant can be expanded by choosing a row or column and multiplying each element by its cofactor, then summing them up to get the value of the determinant.
4. What are some properties of determinants?
Ans. Some properties of determinants include linearity, scalar multiplication, the determinant of the identity matrix, the determinant of a transpose, and the determinant of a product of matrices.
5. How does Cramer's Rule relate to determinants in solving a system of linear equations?
Ans. Cramer's Rule uses determinants to solve a system of linear equations by expressing the solutions in terms of ratios of determinants.
## Mathematics (Maths) for JEE Main & Advanced
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The word learning is used in many different contexts and its definition varies based on the context and the author. It can get quite confusing.
Let’s look at a few definitions…
Learning is the act of acquiring new or modifying and reinforcing existing knowledge, behaviors, skills, values, or preferences which may lead to a potential change in synthesizing information, depth of the knowledge, attitude or behavior relative to the type and range of experience.
the acquisition of knowledge or skills through study, experience, or being taught.
1. the act or experience of one that learns2. knowledge or skill acquired by instruction or study3. modification of a behavioral tendency by experience
While the list here is not exhaustive, these definitions focuses on the act of acquiring skill and knowledge through various means such as instruction, experience, study and other various methods. These definitions view skills and knowledge as a thing to be acquired or possessed.
In the learning and development industry, various experts define learning differently.
A persisting change in human performance or performance potential…[which] must come about as a result of the learner’s experience and interaction with the world.
A process that leads to change, which occurs as a result of experience and increases the potential of improved performance and future learning.
A change in human disposition or capability that persists over a period of time and is not simply ascribable to processes of growth.
While not an exhaustive list, the commonality across these expert definitions is that learning is a process, rather than an acquisition. Learning is seen as a journey or an experience, in which the learner’s goal is to enhance one’s performance, capabilities and character.
While the two sets of definitions are similar in many ways, there is one subtle distinction. The first set of standard definition looks at learning as a thing to have. While the second set of definitions by industry experts, focuses on the learners becoming through experience.
Why is this distinction important?
When we approach learning as a thing to have, there is a greater emphasis on the learner to collect certificates, produce handwritten notes, read x number of books/articles, commit words to memory, etc. Because having is tangible, there is documented evidence that the learner has possess a set of skills and knowledge. This form of learning does not necessarily transform the behaviour of the learner because their primary motivation is to acquire learning evidence.
Eric Fromm, a distinguished humanist psychologist and philosopher, described a different kind of learner in his book called, To Have or To Be?. He calls them the being learner.
instead of being passive receptacles of words and ideas, they listen, they hear and most important, they receive and they respond in an active productive way. What they listen to stimulates their own thinking processes. New questions, new ideas, new perspectives arise in their minds…
Being learners focuses on seeking answers, to help solve problems of their own or of others. The affects of a being learners is not as tangible as obtaining evidence, but it is certainly observable. What is observed, is that each learner has been affected and has become different after the learning experience than they were before it.
To be aware of this distinction, between having and being, will help you identify your learner’s motivation. Also, it will help to shape your programs and its methods that will better engage your leaners as well as effectively transfer the required skill, knowledge and behaviour.
In our daily lives, both types of learning is required. We cannot escape society’s expectation to have learning evidence. After all, a certificate are required for job applications, for university entrance, etc. It’s the way our world is shaped.
Though, we must not forget that the ultimate pursuit of learning is to give ourselves the opportunity to live up to our highest potential and become the best of ourselves.<|endoftext|>
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SC.3.E.5 Humans continue to explore Earth's place in space. Gravity and energy influence the formation of galaxies, including our own Milky Way Galaxy, stars, the Solar System, and Earth. Humankind's need to explore continues to lead to the development of knowledge and understanding of our Solar System.
SC.3.E.5.1 Explain that stars can be different; some are smaller, some are larger, and some appear brighter than others; all except the Sun are so far away that they look like points of light.
SC.3.E.5.2 Identify the Sun as a star that emits energy; some of it in the form of light.
SC.3.E.5.3 Recognize that the Sun appears large and bright because it is the closest star to Earth.
SC.3.E.5.4 Explore the Law of Gravity by demonstrating that gravity is a force that can be overcome.
SC.3.E.5.5 Investigate that the number of stars that can be seen through telescopes is dramatically greater than those seen by the unaided eye.
SC.3.E.6 Humans continue to explore the composition and structure of the surface of Earth. External sources of energy have continuously altered the features of Earth by means of both constructive and destructive forces. All life, including human civilization, is dependent on Earth's water and natural resources.
SC.3.E.6.1 Demonstrate that radiant energy from the Sun can heat objects and when the Sun is not present, heat may be lost.
SC.3.L Life Science
SC.3.L.14 Organization and Development of Living Organisms
A All plants and animals, including humans, are alike in some ways and different in others.
SC.3.L.14.2 Investigate and describe how plants respond to stimuli (heat, light, gravity), such as the way plant stems grow toward light and their roots grow downward in response to gravity.
SC.3.L.15 Diversity and Evolution of Living Organisms
A Earth is home to a great diversity of living things, but changes in the environment can affect their survival.
B Individuals of the same kind often differ in their characteristics and sometimes the differences give individuals an advantage in surviving and reproducing.
SC.3.L.15.1 Classify animals into major groups (mammals, birds, reptiles, amphibians, fish, arthropods, vertebrates and invertebrates, those having live births and those which lay eggs) according to their physical characteristics and behaviors.
SC.3.L.15.2 Classify flowering and nonflowering plants into major groups such as those that produce seeds, or those like ferns and mosses that produce spores, according to their physical characteristics.
A Plants and animals, including humans, interact with and depend upon each other and their environment to satisfy their basic needs.
A Scientific inquiry is a multifaceted activity; The processes of science include the formulation of scientifically investigable questions, construction of investigations into those questions, the collection of appropriate data, the evaluation of the meaning of those data, and the communication of this evaluation.
B The processes of science frequently do not correspond to the traditional portrayal of "the scientific method."
C Scientific argumentation is a necessary part of scientific inquiry and plays an important role in the generation and validation of scientific knowledge.
D Scientific knowledge is based on observation and inference; it is important to recognize that these are very different things. Not only does science require creativity in its methods and processes, but also in its questions and explanations.
SC.3.N.1.1 Raise questions about the natural world, investigate them individually and in teams through free exploration and systematic investigations, and generate appropriate explanations based on those explorations.
SC.3.N.1.2 Compare the observations made by different groups using the same tools and seek reasons to explain the differences across groups.
SC.3.N.1.3 Keep records as appropriate, such as pictorial, written, or simple charts and graphs, of investigations conducted.
SC.3.N.1.4 Recognize the importance of communication among scientists.
SC.3.N.1.5 Recognize that scientists question, discuss, and check each others' evidence and explanations.
SC.3.N.3.3 Recognize that all models are approximations of natural phenomena; as such, they do not perfectly account for all observations.
SC.3.P Physical Science
SC.3.P.8 Properties of Matter
A All objects and substances in the world are made of matter. Matter has two fundamental properties: matter takes up space and matter has mass.
B Objects and substances can be classified by their physical and chemical properties. Mass is the amount of matter (or "stuff") in an object. Weight, on the other hand, is the measure of force of attraction (gravitational force) between an object and Earth.
SC.3.P.8.1 Measure and compare temperatures of various samples of solids and liquids.
SC.3.P.8.2 Measure and compare the mass and volume of solids and liquids.<|endoftext|>
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# Evaluation of $I=\int \frac{1}{\sin (x-a) \sin(x-b) \sin(x-c)}.dx$
Evaluate the given integral:
$$I=\int \frac{1}{\sin (x-a) \sin(x-b) \sin(x-c)}.dx$$
When it was $I=\int \frac{1}{\sin (x-a) \sin(x-b)}.dx$, I solved it by multiplying and dividing by $\sin (b-a)$. But I am not getting how to proceed here..
Could someone help me with this?
• It would be helpful to me to know if you really need a primitive or if you are just interested in the integral computed over some interval. A brute-force approach is just to use partial fraction decomposition, since $\frac{1}{\sin(z)}$ is a meromorphic function with known singularities and residues. – Jack D'Aurizio Aug 4 '16 at 18:10
• Moreover, we may remove a parameter by a suitable substitution like $x=t+b$, so you problem is equivalent to finding $$\int_{?}\frac{dx}{\sin x}\cdot\frac{1}{\sin(x-u)\sin(x+v)}.$$ – Jack D'Aurizio Aug 4 '16 at 18:12
I'd like to expand on @Kajelad explanation. The system can be solved as follows.
We want to perform the decomposition
$$\frac{1}{\sin(x-a)\sin(x-b)\sin(x-c)} = \frac{A}{\sin(x-a)} + \frac{B}{\sin(x-b)} + \frac{C}{\sin(x-c)}$$
or by rewriting we obtain
$$1 = A\sin(x-b)\sin(x-c) + B\sin(x-a)\sin(x-c) + C\sin(x-a)\sin(x-b).$$
As we want this to hold for all $x$ we may select a few values to see if we get a solution. Naturally, the equation should hold if $x=a$ hence we obtain
$$1 = A\sin(a-b)\sin(a-c) + B\sin(0)\sin(a-c) + C\sin(0)\sin(a-b)$$
but as $\sin(0) = 0$ then
$$A = \frac{1}{\sin(a-b)\sin(a-c)}.$$
Similarly, by selecting $x=b$ and $x=c$ we obtain
$$B = \frac{1}{\sin(b-a)\sin(b-c)}$$
$$C = \frac{1}{\sin(c-a)\sin(c-b)}.$$
Through brute work and using the odd property of $\sin$ and trigonometric formulas these should verify that the decomposition is justified. If so, then the solution is given by
$$\int \left(A\csc(x-a) + B\csc(x-b) + C\csc(x-c)\right) dx$$ $$= \frac{\ln(\tan(x-a))}{\sin(a-b)\sin(a-c)} + \frac{\ln(\tan(x-b))}{\sin(b-a)\sin(b-c)} + \frac{\ln(\tan(x-c))}{\sin(c-a)\sin(c-b)} + C$$
I looked at the partial feaction decomposition method. It is not particularly elegant, but it reduces the calculus problem to a linear algebra one. It is likely a more efficient method for this integral exists, but I am not sure what it would be.
$$\int\frac{dx}{\sin(x-a)\sin(x-b)\sin(x-c)}$$
$$=\int\left(\frac{A}{\sin(x-a)}+\frac{B}{\sin(x-b)}+\frac{C}{\sin(x-c)}\right)dx$$
If $A$, $B$, and $C$ are constants, we can use the identity $\int\csc x\ dx=\ln\left(\tan\frac x2\right)+k.$
$$=A\ln\left(\tan\frac{x-a}2\right)+B\ln\left(\tan\frac{x-b}2\right)+C\ln\left(\tan\frac{x-c}2\right)+k$$
In order to find these constants, we can rewrite $\sin(x-a)=\sin x\cos a-\sin a\cos x.$
$$=\int\frac{\begin{array}A(\sin x\cos b-\sin b\cos x)(\sin x\cos c-\sin c\cos x)\\+B(\sin x\cos a-\sin a\cos x)(\sin x\cos c-\sin c\cos x)\\+C(\sin x\cos b-\sin b\cos x)(\sin x\cos c-\sin c\cos x)\end{array}}{(\sin x\cos a-\sin a\cos x)(\sin x\cos b-\sin b\cos x)(\sin x\cos c-\sin c\cos x)}dx$$
Rearranging, and rewiring some sines using the sum identity:
$$=\int\frac{\begin{array}(A\cos b\cos c+B\cos a\cos c+C\cos a\cos\ b)\sin^2x\\-(A\sin(b+c)+B\sin(a+c)+C\sin(a+b))\sin x\cos x\\+(A\sin b\sin c+B\sin a\sin c+C\sin a\sin b)\cos^2x\end{array}}{(\sin x\cos a-\sin a\cos x)(\sin x\cos b-\sin b\cos x)(\sin x\cos c-\sin c\cos x)}dx$$
The double angle identity can be used to show that $$\alpha\sin^2x+\beta\sin x\cos x+\gamma\cos^2x=\frac12(\alpha+\gamma)+\frac12(-\alpha+\gamma)\cos 2x+\beta\sin 2x,$$ which is equal to $1$ for all $x$ if and only if $\alpha=1,\beta=0,\gamma=1$. This gives a linear system of equations.
$$A\cos b\cos c+B\cos a\cos c+C\cos a\cos b=1$$ $$A\sin b\sin c+B\sin a\sin c+C\sin a\sin b=1$$ $$A\sin(b+c)+B\sin(a+c)+C\sin(a+b)=0$$
Rewriting as a matrix:
$$\begin{bmatrix}\cos b\cos c & \cos a\cos c & \cos a\cos b\\ \sin b\sin c & \sin a\sin c & \sin a\sin b\\ \sin(b+c) & \sin(a+c) & \sin(a+b)\\\end{bmatrix}\begin{bmatrix}A\\B\\C\\\end{bmatrix}=\begin{bmatrix}1\\1\\0\\\end{bmatrix}$$
For specific values of $a$, $b$, and $c$, this equation above can be solved by row reduction. Technically, the inverse of the $3\times3$ matrix above can be computed to find the general solution, but it would be a rather cumbersome process.<|endoftext|>
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Getting around in a public place isn’t just about not walking into other people or avoiding collisions with pillars and park benches. When we walk, or just stand around in public, there are all kinds of unwritten rules we follow. And Stanford’s Jackrabbot is a robot designed to learn these rules and to obey them.
Most of what we do in public, we do without thinking. For instance, when you stop to do something, you instinctively move over to the side of the sidewalk to get out of the way of others (or you should, anyway). And when you meet someone coming in the other direction, you both move around each other. You don’t just stop and wait for the other person to do something, like a robot might do–in fact, if you did, it would seem aggressive.
And, unlike on the road, there’s no rule book. Everything is done according to etiquette learned subconsciously, over years. The Jackrabbot is attempting to model these rules, effectively teaching itself how to behave around people.
The goal is to eventually build robots that can work alongside humans “in dynamic crowded environments such as terminals, malls, or campuses,” says Stanford’s project team, at the Computational Vision and Geometry Lab. As more jobs are taken over by robots, their ability to get along with people will become more important. As part of its training, the Jackrabbot is delivering mail, food, tools, and goods on the Stanford campus.
The training, which will teach the robot when to do things like taking its turn when waiting in line to, say, get to a busy exit, should help future robots blend into crowded areas. But the technology could also help elsewhere. Driverless cars, for example, could learn better how to interact with human drivers on the roads, perhaps becoming a little less meek in situations where people currently take advantage of autonomous cars’ excessive politeness. And smooth crowd-navigation could also help visually-impaired people to better navigate public spaces, improving on the classic white stick.
The Jackrabbot is designed to be unthreatening. “We tried to make it cute, so that people will be happy to share their space with this robot,” Stanford’s Alex Alahi told the Bay Area’s KCBS news.
[All Images: via Stanford Computational Geometry & Vision Lab]
Have something to say about this article? You can email us and let us know. If it’s interesting and thoughtful, we may publish your response.<|endoftext|>
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Elementary Algebra Review for Pretest
1. Graph –3.2 on the number line provided.
2. Graph on the number line provided .
3. True or false:
(a) –3 < –7 (b) 5 ≤ 5 (c) 9 ≥ –9 (d) –5 > –3 (e) –2 ≥ –2
4. Which symbol, > or <, makes a true statement when placed in the blank between the given pairs of numbers?
5. Arrange the numbers –3.04, –3.1, –3.024, and from smallest to largest.
6. Arrange the numbers from smallest to largest.
7. Evaluate:
8. Divide:
10. Subtract: –3.02 –(–4.5)
11. Multiply:
12. Divide:
13. Evaluate:
14. Evaluate:
15. Evaluate:
16. Simplify: 5 – 2[–3 – 2(4 – 6)]
17. Simplify:
18. Simplify:
19. Subtract and simplify if possible:
20. Combine like terms:
21. Simplify: 3(–4m + 7) – (5m – 6)
22. Solve :
23. Solve: 5.2 – 3s = s – 3.5
24. Solve: 4v – (v – 2) = 7 – 2(3v – 4)
25. Write an expression to represent the expression "5 less than the product of 3 and a number". Let
n re present the number .
26. Write an expression to represent the expression "the quotient of 3 more than a number and
double the number". Let x represent the number.
27. Simplify:
28. Evaluate 2x^3 – 3x^2 + 5x – 9 for x = –2.
29. Add: (3x^2 – 5x + 2) + (–x^2 + 7x – 8)
30. Subtract: (–2x^2 + 4x – 7) – (8x – 10)
31. Multiply: –5t(4t^2 – 3t)
32. Reduce:
33. Express each of the following as the product of prime factors: (a) 42 (b) 48
34. Find the LCM for 42 and 48.
35. What number is 45% of 62.4?
36. What percent of 36 is 4.5?
37. 240 is 160% of what number?
38. The formula for the area of a circle is A = π *r^2 . What is the area of a circle that has
39. The sum of triple a number and 8 is –13. What is the number?
40. The product of 7 and a number is 2 less than the number. What is the number?
41. If a recipe for sugar cookies requires cups of flour to make 36 cookies, how much flour
would be needed to make 60 of the cookies?
42. At a particular college the ratio of men to women is 35 to 45. If there are 9135 women at the
college, how many men are there at the college?
(Each answer is followed by the name of the topic that the problem is most closely associated with. If you get a wrong answer and want additional information on how to do the problem, look for this topic in the index of a p realgebra textbook .)
(plotting real numbers on a number line)
(plotting real numbers on a number line)
3. (a) False (b) True (c) True (d) False (e) True
(ordering real numbers, less than, greater than)
4. (a) > (b) > (c) > (d) < (e) <
(ordering real numbers, less than, greater than)
5. , –3.1, –3.04, –3.024 (ordering real numbers, less than, greater than)
(ordering real numbers, less than, greater than)
7. 5 ( absolute value )
10. 1.48 (addition of signed decimals )
( multiplication of signed fractions)
12. 80 ( division of signed decimals)
13. –25 (exponents, order of operations)
(exponents, multiplication of signed fractions)
15. 9.61 (exponents, multiplication of signed decimals)
16. 3 (order of operations)
17. –24 (order of operations)
18. –57 (order of operations)
(combining like terms)
20. –x^2 – 4x + 2 (combining like terms)
21. –17m + 27 (distributive property, combining like terms)
22. 18 ( linear equations with one variable )
23. 2.175 (linear equations with one variable)
(linear equations with one variable)
25. 3n – 5 (translating English into algebra )
(translating English into algebra)
(properties of exponents)
28. –47 (evaluating polynomials)
29. 2x^2 + 2x – 6 (adding polynomials)
30. –2x^2 – 4x + 3 (subtracting polynomials)
31. –20t^3 + 15t^2 (multiplying polynomials)
(reducing rational expressions)
( factoring integers into primes)
34. 336 (least common multiple)
35. 28.08 (percentages)
36. 12.5% (percentages)
37. 150 (percentages)
38. 113.0 in2 (evaluating formulas)
39. –7 (applications of linear equations with one variable)
(applications of linear equations with one variable)
cups (ratio and proportion)
42. 7105 men (ratio and proportion)
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Heavy metals have been a topic of concern in recent years, which is quite justified among toxic metals such as mercury, lead, and aluminum. But did you know other metals can actually be beneficial to the body? Silver, copper, and brass all possess a special quality that makes them as hard-working as they are good-looking: they destroy bacteria.
How? Termed the “Oligodynamic Effect”, the process actually isn’t well understood, but it’s thought that, in simple terms, the metal ions bind with proteins in bacteria cells, a fatal attraction which renders the bacteria inactive. And the process is potent: one study tested water contaminated with E.coli and Salmonella in brass, silver and copper pots. Within four hours, the copper pot had killed off all the Salmonella in its water; silver and brass followed suit shortly after, at eight and twelve hours, respectively. Thus, we find the reason why the copper kettle and the silver spoon have been used throughout history: their usage helped to prevent food poisoning.
Indeed, even the ancients were aware of this effect. Ancient Egyptians were using copper more than 4,000 years ago to sterilize wounds and drinking water, and Aztecs used copper to treat skin conditions.
Considering further applications of the oligodynamic effect, the next logical place would be on common spaces we all share and dread, especially during cold and flu season: door handles. One study tested hospital doorknobs and found those made of brass and copper exhibited an antibacterial effect, with the copper having partially sterilized itself within just 15 minutes. Amazingly, copper was also found effective against the MRSA superbug; one study showed a copper surface at room temperature neutralized 100 million MRSA bacteria cells within 90 minutes.
Considering hospitals are hotbeds of infectious germs, where patients are especially vulnerable to MRSA, it would make sense for designers to use copper in hospitals as much as possible. Copper door handles, light switches, faucets, and toilet seats could significantly cut back on sickness and death rates, a hallmark of a more evolved society. And in fact, this has already been put into practice at the Harburg Asklepios Clinic in Germany, a 774-bed hospital, in a major installation of copper touch surfaces in 2014.
Unfortunately, stainless steel and aluminum door handles are cheaper and therefore much more commonly used. Likewise, most flatware today is made with stainless steel instead of silver — and they are brimming with bacteria. In the same aforementioned study that tested MRSA on copper, MRSA was also tested on steel and aluminum surfaces, and on those surfaces, the number of bacteria actually increased.
Without the oligodynamic effect to work for you on these types of door handles and other surfaces, you are quite right to spray and wipe them often when sick people have been in the building, and to use a non-alcohol and non-triclosan based hand sanitizer when in any public setting. Flatware — forks, spoons and knives — may be automatically sanitized in a dishwasher, but if you are washing them by hand, further measures have to be taken, such as soaking them in a bleach solution.
But those who don’t have any actual silver in their silverware can benefit from the oligodynamic effect in another way, which also allows avoiding use of bleach or triclosan. Modern applications of oligodynamic metals are now found in hydrosol form — namely, colloidal silver. This product has seen a dramatic uptick in popularity in recent years, and in fact, this article assignment was a serendipitous reminder that the author needed to stop at the local health food store and pick up a bottle.
Colloidal silver is proving to be a worthy addition to the home medicine cabinet, with numerous applications topically as well as internally. Users report relief from everything from sinus infections to acne to cold and flu and even stronger viral infections such as EBV (Epstein-Barr Virus). In a world of resistant bacteria that have evolved from the overuse of pharmaceutical antibiotics and antibacterial soap products, a high quality colloidal silver product seems to be the answer we are looking for, and we can expect to see it used in more products to come.
Additional new oligodynamic products on the market include copper mugs and water bottles. Amazon.com is flooded with listings for these; a search of “copper mug” returns over 6,000 results. And while this metal might look fancy, it’s actually relatively inexpensive; you can get a copper mug for as low as ten dollars. And it’ll give you something fresh to talk about during coffee breaks!<|endoftext|>
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# Find the point B if M is the midpoint of the line segment joining points A and B.
A(−1, −8), M(6, −1) Using these two points I have to figure out what B is. I'm not terribly good at math and I would love to know how to do this.
### 3 Answers by Expert Tutors
Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
0
Here is a quick way:
You go from A to M in x-direction in 7 units. You go another 7 units to reach B. So, the x-coordinate of B is 6+7 = 13. In the same way, the y-coordinate of B is -1+7 = 6.
Aisha S. | Qualified and Caring EducatorQualified and Caring Educator
0
You can also use a graph paper and draw the two lines using X, Y coordinates.
Christine M. | Patient and Experienced Math TutorPatient and Experienced Math Tutor
5.0 5.0 (179 lesson ratings) (179)
0
First, you will need to know the midpoint formula:
M=((x1+x2)/2 , (y1+y2)/2)
Let us call A = (x1, y1). This means that x1=-1 and y1=-8.
We know that the midpoint, M, is (6, -1).
B will be the unknown point, (x2, y2).
Therefore, we must plug in what we know:
(6, -1) = ((-1+x2)/2 , (-8+y2)/2).
From here, we will create two equations. One will help us to solve for x2, and the other will help us to solve for y2. We will create the two equations by setting the x value on the left hand side equal to the x value on the right hand side:
6 = (-1+x2)/2 {multiplying both sides by 2 gives us}
12 = -1 + x2 {adding 1 to both sides gives us the value for x2}
x2 = 13
Now we will need to find y2 using the same method (setting the value for y on the left hand side equal to the value for y on the right hand side).
-1 = (-8+y2)/2 {multiplying both sides by 2 gives us}
-2 = -8 + y2 {adding 8 to both sides gives us the value for y2}
y2 = 6
Therefore, we know that point B is (13, 6). That is the answer.
To check your answer, plug A and B back into the midpoint formula:
M = ((-1 + 13)/2 , (-8 + 6)/2)
M = ((12/2) , (-2/2))
M = (6,-1)
which is correct!<|endoftext|>
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Through pattern recognition, your digital footprints caused by frequent visits to the e-commerce sites, keep a track of your online search to offer you results based on historical your search preferences. Thus probability is bright that the holiday package you were just browsing online may come with a buy option at different web pages that you visit later. Such is the power of Technology!
Leveraging Robotic Vision
Computer Vision and Robotic Vision (also called Robot Vision) are the two most impressive inventions in the field of AI. Many perceive more often than usual, these two entities to be the same. However, there are major differences between these two futuristic vision technologies. Robotic vision powers robotics while computer vision is the technology behind image recognition.
Robot Vision at its core is a combination of multitudes of cameras, computer algorithms and other hardware components that work together to provide visual insights to the robot or machine they cater to. It is the Robot Vision that helps any robot to accomplish complex tasks requiring a visual understanding of its surroundings. For instance, it is the Robot Vision techniques that guide a machine or a robotic arm to pick an object and place it somewhere as per requirements. Imagine a scenario where sensors and cameras detect an object placed on an elevated height which is subsequently lifted up by the robotic arm deploying complex Robot Vision algorithms. For object detection the Robot is teamed with the ordinary 2D cameras if the situation is more complex like the robotic arm has to mount its wheels on a moving vehicle, advanced 3D Stereo Cameras are used. The process of robotic vision works in two simple steps:
1. Imaging: In this process, the robot uses vision technology for scanning or “seeing” objects. The robot can scan two-dimensional things including barcode scanning and line scanning, in addition to the X-ray imaging and 3D imaging for inspection purposes.
2. Image Processing: Image processing follows Imaging process. After object detection, the robot processes it or starts to think about it. For instance, the robot finds the presence of an obstruction and detects edges, discovers and manipulates objects according to its programming, counts pixels, and recognizes patterns.
Talking in the crux, a Robot would apparently be blind if there was no Robot Vision technology to show the way. Essentially, robot vision is a sophisticated technology helping an automated robot, to identify things, navigate, find objects, and inspect things along its way. Robot vision deploys a series of carefully-calibrated algorithms, temperature detection sensors that come with a varying range of complexity and application.
This innovative and path-breaking technology has the potential to cut operating costs to create practical application solutions for all types of automation. Robots working side by side, with robotic vision technology, would not collide with each other performing a series of different tasks like measuring, reading barcodes and scanners, inspecting engine parts, surface inspection and packaging, checking the orientation of components and pieces and inspecting for defects among others.
The Potential of Computer Vision
Computer Vision is a component of Artificial Intelligence aiming to render advanced visual capabilities to computers through camera hardware and complex algorithms. Computer Vision primarily deals with image recognition. The method begins with the initial extraction of useful information from the different videos and digital images. This information is further processed and analysed usually with the help of webcams, by acquiring high-dimensional data from the real world. This data is finally processed to generate symbolic information powering the computer to take important decisions. Further, Computer vision is categorised into various subfields which include object pose estimation, image restoration object recognition, video tracking and event detection.
The latest advances in Computer Vision have enabled nearly 300 million visually impaired people around the world to see what the sighted often take for granted, any number of the nearly 2 billion photos which are uploaded onto Facebook on a daily basis. Computer vision is additionally deployed by consumer offerings, social media platforms, law enforcement agencies, and industrial production houses.
Computer vision and Robotic vision have already entered our lives and businesses in ways many of us may not be aware of improving the quality of our lives with each passing day.<|endoftext|>
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Phylogeny describes the ancestral relationships between organisms. The connections often take the form of a tree, with tips, branches, nodes, and roots, called the phylogenetic tree.
Specifically, the tips of the tree represent extant—or living—taxa, and the branches denote evolutionary changes between ancestors and descendants, such as the change in the DNA sequence or the evolution of a characteristic like feathers.
Species that share an immediate common ancestor, sister taxa, are their closest relatives and share nodes—points where branches meet, like lizards and birds, and rodents and humans. A basal node, the root, corresponds to the most recent common ancestor of all organisms in the tree, the Metaspriggina in this case.<|endoftext|>
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# Box Method Multiplication Worksheets for 2 Digit Numbers
We are here with box method multiplication worksheet to download and print to start a sweet mathematical journey!
Your kid had learnt one-digit and two-digit multiplication. Now, it’s time to learn multi-digit multiplication. Your child may find it bit difficult at first but, we are here to help your child with the easiest thing possible to help him learn and understand it in a better way.
your child needs to upgrade his multiplication skills from one-digit to multi-digit multiplication because he has to apply this in real-life chores and he should learn and be prepared for his future problems.
The easiest and most convenient method of performing multi-digit multiplication is with the help of box method multiplication.
## What is Box Method Multiplication?
Also known as the Grid method multiplication is an initial approach to solve multi-digit multiplication for numbers more than 10. This method is often introduced to children during their elementary years. Hence, also known as the Grammar School method.
So, let’s get started.
To teach your kid the Box method you have to follow the following steps:
STEP 1: Draw a two-by-two grid-like shown in the above worksheet. Draw two lines vertically and two horizontally.
STEP 2: Now, you need to expand the given numbers and write the numbers vertically and horizontally outside the boxes. Like for example let me show to by solving the first problem of worksheet 2.
So, the number 33 will be 30+3, and 68 will be 60+8. You have to write 30 and 3 vertically and 60 and 8 horizontally outside the grids. You can also change the position of the numbers but in the same pattern.
STEP 3: For the third step you have to perform the basic step and that is multiplication. Teach your kid to ignore zeros for a while just multiply 3×6 rather than 30×60. This will be a lot easier procedure for them.
So, now they have to just add the zeros behind the numbers. Here, we ignored two zeros so we will add two zeros after 3×6=12. By adding zero it will be 1200.
STEP 4: For the next step you have to multiply same term i.e., 30 with the another number beside 60 i.e., 8 and answer will be 8×6=48 and by adding a zero that we ignored earlier it will be 480.
Repeat step 4 for the following number in a vertical grid that is 3. Multiply it by 60 to get the answer which is 3×6=12 and adding the zero the answer will be 120.
Again, multiply 3 by 8 and the answer will be 3×8=24.
STEP 5: For the last and the final step you just simply add all the numbers inside the boxes which you get after multiplication.
Like for our problem 1200+240+120+24= 1584
## 2-Digit Box Method Multiplication Worksheets
Use the box method to work out these multiplication questions.
### 2-Digit Box Method Multiplication Worksheet: 9
This simple and easy method will make your kid understand every concept of multi-digit multiplication. This will not only boost his confidence when he will be able to perform and apply this method in actual life but will also help him to stay ahead in his academics.
Remember to make your kid focus properly while solving these problems. To do this you can offer them sweets as they complete their problems. One sweet for each problem.
These box method multiplication worksheets will make the activity more fun and he will always love to study and learn new things. Our printables will help your kid to solve them and learn this method. You can download these worksheets very conveniently. We will keep working to prepare more exciting and fun worksheets for your kids.
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# Solving Equations With Like Terms Worksheet
Solving Equations With Like Terms Worksheet - Research is showing coloring in the classroom activates the brain!students will solve 15 equations where they will have to combine. Like terms are terms that contain the same. Web distributive property and combining like terms worksheets are a great resource for students to implement the understanding of solving algebraic expressions. X + 1 + 3x = 29. Click “show answer” underneath the. Web this series of lessons and worksheets teach the steps for simplifying equations by matching or combining terms that are alike until there are no more steps that can be.
Your first step should be to combine any like terms. Like terms are terms that have the same variables. Worksheets are algebra work combining like terms and solving, solving equationscombinin. Simplify each equation by combining like terms, then solve. Like terms are terms that have the same variables raised to.
X + 1 + 3x = 29. These free equations and word problems worksheets will help your students practice. Fall in love with math. The terms of an expression are the parts to be added or subtracted. 1) 4 n − 2n = 4 2) −12 = 2 + 5v + 2v 3) 3 = x + 3 − 5x 4) x + 3 − 3 = −6 5) −12 = 3 − 2k − 3k 6) −1 = −3r.
## Solving Linear Equations Worksheet Pdf
Solving Equations With Like Terms Worksheet - Simplify each equation by combining like terms, then solve. Web this series of lessons and worksheets teach the steps for simplifying equations by matching or combining terms that are alike until there are no more steps that can be. 1) 4 n − 2n = 4 2) −12 = 2 + 5v + 2v 3) 3 = x + 3 − 5x 4) x + 3 − 3 = −6 5) −12 = 3 − 2k − 3k 6) −1 = −3r. Like terms are terms that have the same variables. Solve for x in each equation below. Web distributive property and combining like terms worksheets are a great resource for students to implement the understanding of solving algebraic expressions. Your first step should be to combine any like terms. Your first step should be to combine any like terms. Worksheets are algebra work combining like terms and solving, solving equationscombinin. Research is showing coloring in the classroom activates the brain!students will solve 15 equations where they will have to combine.
1) 4 n − 2n = 4 2) −12 = 2 + 5v + 2v 3) 3 = x + 3 − 5x 4) x + 3 − 3 = −6 5) −12 = 3 − 2k − 3k 6) −1 = −3r. Web tlw review solving equations with variables on both sides. Web solving equations—combining like terms #2. Inverse operations to get the variable all by itself on one side of. Solve for x in each equation below.
Fall in love with math. Web tlw review solving equations with variables on both sides. Click “show answer” underneath the. Inverse operations to get the variable all by itself on one side of.
Solve equations involving like terms across the equals sign. Like terms are terms that have the same variables. Web combining like terms and solving worksheet.
Like terms are terms that have the same variables raised to. Like terms are terms that contain the same. Your first step should be to combine any like terms.
## Your First Step Should Be To Combine Any Like Terms.
Your first step should be to combine any like terms. Inverse operations to get the variable all by itself on one side of. Click “show answer” underneath the. Web solving equations—combining like terms #1.
## Like Terms Are Terms That Have The Same Variables Raised To.
Research is showing coloring in the classroom activates the brain!students will solve 15 equations where they will have to combine. Solve for x in each equation below. X + 1 + 3x = 29. These free equations and word problems worksheets will help your students practice.
## Worksheets Are Algebra Work Combining Like Terms And Solving, Solving Equationscombinin.
Web solving equations—combining like terms #2. Your first step should be to combine any like terms. Like terms are terms that contain the same. The terms of an expression are the parts to be added or subtracted.
## X = 6 + 4X + X = 46.
1) 4 n − 2n = 4 2) −12 = 2 + 5v + 2v 3) 3 = x + 3 − 5x 4) x + 3 − 3 = −6 5) −12 = 3 − 2k − 3k 6) −1 = −3r. Combining like terms and solving equations worksheet. Web tlw review solving equations with variables on both sides. Like terms are terms that have the same variables.<|endoftext|>
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The remoras //, sometimes called suckerfish, are a family (Echeneidae) of ray-finned fish in the order Perciformes. They grow to 7–75 cm (2.8 in–2 ft 5.5 in) long. Their distinctive first dorsal fins take the form of a modified oval, sucker-like organ with slat-like structures that open and close to create suction and take a firm hold against the skin of larger marine animals. The disk is made up of stout, flexible membranes that can be raised and lowered to generate suction. By sliding backward, the remora can increase the suction, or it can release itself by swimming forward. Remoras sometimes attach to small boats, and have been observed attaching to divers as well. They swim well on their own, with a sinuous, or curved, motion.
|Common remora, Remora remora|
Remora front dorsal fins have evolved to enable them to adhere by suction to smooth surfaces and they spend their lives clinging to a host animal such as a whale, turtle, shark or ray. It is probably a mutualistic arrangement as the remora can move around on the host, removing ectoparasites and loose flakes of skin, while benefiting from the protection provided by the host and the constant flow of water across its gills. Although it was initially believed that remoras fed off particulate matter from the host's meals, this has been shown to be false; in reality, their diets are composed primarily of host feces.
Remoras are tropical open-ocean dwellers, but are occasionally found in temperate or coastal waters if they have attached to large fish that have wandered into these areas. In the mid-Atlantic Ocean, spawning usually takes place in June and July; in the Mediterranean Sea, it occurs in August and September. The sucking disc begins to show when the young fish are about 1 cm (0.4 in) long. When the remora reaches about 3 cm (1.2 in), the disc is fully formed and the remora can then attach to other animals. The remora's lower jaw projects beyond the upper, and the animal lacks a swim bladder.
Some remoras associate with specific host species. They are commonly found attached to sharks, manta rays, whales, turtles, and dugongs, hence the common names "sharksucker" and "whalesucker". Smaller remoras also fasten onto fish such as tuna and swordfish, and some small remoras travel in the mouths or gills of large manta rays, ocean sunfish, swordfish and sailfish.
Remoras, like many other fishes, have two different modes of ventilation. Ram ventilation is the process in which at higher speeds, the remora uses the force of the water moving past it to create movement of fluid in the gills. Alternatively, at lower speeds the remora will use a form of active ventilation, in which the fish actively moves fluid through its gills. In order to use active ventilation, a fish must actively use energy to move the fluid; however, determining this energy cost is normally complicated due to the movement of the fish when using either method. As a result, the remora has proved invaluable in finding this cost difference (since they will stick to a shark or tube, and hence remain stationary despite the movement or lack thereof of water). Experimental data from studies on remora found that the associated cost for active ventilation created a 3.7–5.1% increased energy consumption in order to maintain the same quantity of fluid flow the fish obtained by using ram ventilation.
Other research into the remora's physiology came about as a result of studies across multiple taxa, or using the remora as an out-group for certain evolutionary studies. Concerning the latter case, remoras were used as an outgroup when investigating tetrodotoxin resistance in remoras, pufferfish, and related species, finding remoras (specifically Echeneis naucrates) had a resistance of 6.1–5.5×10−8 M.
Use for fishingEdit
Some cultures use remoras to catch turtles. A cord or rope is fastened to the remora's tail, and when a turtle is sighted, the fish is released from the boat; it usually heads directly for the turtle and fastens itself to the turtle's shell, and then both remora and turtle are hauled in. Smaller turtles can be pulled completely into the boat by this method, while larger ones are hauled within harpooning range. This practice has been reported throughout the Indian Ocean, especially from eastern Africa near Zanzibar and Mozambique, and from northern Australia near Cape York and Torres Strait.
Similar reports come from Japan and from the Americas. Some of the first records of the "fishing fish" in the Western literature come from the accounts of the second voyage of Christopher Columbus. However, Leo Wiener considers the Columbus accounts to be apocryphal: what was taken for accounts of the Americas may have been, in fact, notes Columbus derived from accounts of the East Indies, his desired destination.
In ancient times, the remora was believed to stop a ship from sailing. In Latin, remora means "delay", while the genus name Echeneis comes from Greek εχειν, echein ("to hold") and ναυς, naus ("a ship"). In a notable account by Pliny the Elder, the remora is blamed for the defeat of Mark Antony at the Battle of Actium and, indirectly, for the death of Caligula. A modern version of the story is given by Jorge Luis Borges in Book of Imaginary Beings (1957).
- Friedman, Matt, et al. "An early fossil remora (Echeneoidea) reveals the evolutionary assembly of the adhesion disc." Proc. R. Soc. B 280.1766 (2013): 20131200.
- Froese, Rainer, and Daniel Pauly, eds. (2013). "Echeneidae" in FishBase. April 2013 version.
- "Echeneidae". Integrated Taxonomic Information System. Retrieved 20 March 2006.
- "Sharksucker fish's strange disc explained". Natural History Museum. 28 January 2013. Archived from the original on 1 February 2013. Retrieved 5 February 2013.
- Beer, Amy-Jane. Derek Hall. (2012). The Illustrated World Encyclopedia of Marine Fish & Sea Creatures. Leicestershire: Lorenz Books. p. 235. ISBN 978-0-7548-2290-5.
- Jackson, John (30 November 2012). "How does the Remora develop its sucker?". National History Museum. Retrieved 2 January 2016.
- Williams, E. H.; Mignucci-Giannoni, A. A.; Bunkley-Williams, L.; Bonde, R. K.; Self-Sullivan, C.; Preen, A.; Cockcroft, V. G. (2003). "Echeneid-sirenian associations, with information on sharksucker diet". Journal of Fish Biology. 63 (5): 1176. CiteSeerX 10.1.1.619.4020. doi:10.1046/j.1095-8649.2003.00236.x.
- Willmer, Pat; Stone, Graham; Johnston, Ian (2009-03-12). Environmental Physiology of Animals. John Wiley & Sons. ISBN 9781444309225.
- Steffensen, J. F.; Lomholt, J. P. (1983-03-01). "Energetic cost of active branchial ventilation in the sharksucker, Echeneis naucrates". Journal of Experimental Biology. 103 (1): 185–192. ISSN 0022-0949. PMID 6854201.
- Kidokoro, Yoshiaki; Grinnell, Alan D.; Eaton, Douglas C. (1974). "Tetrodotoxin sensitivity of muscle action potentials in pufferfishes and related fishes". Journal of Comparative Physiology. 89: 59. doi:10.1007/BF00696163.
- Gudger, E. W. (1919). "On the Use of the Sucking-Fish for Catching Fish and Turtles: Studies in Echeneis or Remora, II., Part 1". The American Naturalist. 53 (627): 289–311. doi:10.1086/279716. JSTOR 2455925.
- Gudger, E. W. (1919). "On the Use of the Sucking-Fish for Catching Fish and Turtles: Studies in Echeneis or Remora, II., Part 2". The American Naturalist. 53 (628): 446–467. doi:10.1086/279724. JSTOR 2456185.
- MacGillivray, John (1852). Narrative of the Voyage of H.M.S. Rattlesnake, Commanded By the Late Captain Owen Stanley, R.N., F.R.S. Etc. During the Years 1846–1850. 2. London: Lords Commissioners of the Admiralty. (Dr. Gudger's accounts are more authoritative, but this source is noted as an early account that Gudger appears to have missed.)
- Wiener, Leo (1921). "Once more the sucking-fish". The American Naturalist. 55 (637): 165–174. doi:10.1086/279802. JSTOR 2456418.
- Pliny the Elder. "Book 32, Chapter 1". Natural History. (cited in Gudger, E. W. (1930). "Some old time figures of the shipholder, Echeneis or Remora, holding the ship". Isis. 13 (2): 340–352. doi:10.1086/346461. JSTOR 224651.)
|Wikimedia Commons has media related to Echeneidae.|<|endoftext|>
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# A retailer agreed to take 5000 ball point pen . How ever he found that 12% are faulty . What was the percentage decrement.
By Emma
A retailer agreed to take 5000 ball point pen . How ever he found that 12% are faulty . What was the percentage decrement.
### 2 thoughts on “A retailer agreed to take 5000 ball point pen . How ever he found that 12% are faulty . What was the percentage decrement.”
Step-by-step explanation:
Given Damaged articles = 20%
Left articles =20kg
Required = the original weight =?
Solutions
Let the original weight be Xkg
20% of x
20% x X
20/100
## 88%
Step-by-step explanation:
12% of pen is faulty.
So, 12% of 5000 = 12 / 100 * 5000
= 600
600 pens are faulty
Now remaining pens will be 5000 – 600 = 4400
4400 pens are in good condition
therefore, in %
4400 / 5000 * 100 = 88%
thus the pens’ value is reduced to 88%
### Quicker solution:
Out of 100%, 12% were faulty.
thus, % decrement = 100 – 12 = 88%<|endoftext|>
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Mathematics » Introducing Integers » Multiply and Divide Integers
# Multiplying Integers
## Multiplying Integers
Since multiplication is mathematical shorthand for repeated addition, our counter model can easily be applied to show multiplication of integers. Let’s look at this concrete model to see what patterns we notice. We will use the same examples that we used for addition and subtraction.
We remember that $$a·b$$ means add $$a,\phantom{\rule{0.2em}{0ex}}b$$ times. Here, we are using the model shown in the figure below just to help us discover the pattern.
Now consider what it means to multiply $$5$$ by $$-3.$$ It means subtract $$5,3$$ times. Looking at subtraction as taking away, it means to take away $$5,3$$ times. But there is nothing to take away, so we start by adding neutral pairs as shown in the figure below.
In both cases, we started with $$\mathbf{\text{15}}$$ neutral pairs. In the case on the left, we took away $$\mathbf{\text{5}},\mathbf{\text{3}}$$ times and the result was $$-\mathbf{\text{15}}.$$ To multiply $$\left(-5\right)\left(-3\right),$$ we took away $$-\mathbf{\text{5}},\mathbf{\text{3}}$$ times and the result was $$\mathbf{\text{15}}.$$ So we found that
$$\begin{array}{ccc}5·3=15\hfill & & -5\left(3\right)=-15\hfill \\ 5\left(-3\right)=-15\hfill & & \left(-5\right)\left(-3\right)=15\hfill \end{array}$$
Notice that for multiplication of two signed numbers, when the signs are the same, the product is positive, and when the signs are different, the product is negative.
### Multiplication of Signed Numbers
The sign of the product of two numbers depends on their signs.
Same signsProduct
•Two positives
•Two negatives
Positive
Positive
Different signsProduct
•Positive • negative
•Negative • positive
Negative
Negative
## Example
Multiply each of the following:
1. $$−9·3$$
2. $$\phantom{\rule{0.2em}{0ex}}-2\left(-5\right)\phantom{\rule{1em}{0ex}}$$
3. $$\phantom{\rule{0.2em}{0ex}}4\left(-8\right)\phantom{\rule{1em}{0ex}}$$
4. $$\phantom{\rule{0.2em}{0ex}}7·6$$
### Solution
$$–9\cdot 3$$ Multiply, noting that the signs are different and so the product is negative. $$–27$$
$$–2\left(–5\right)$$ Multiply, noting that the signs are the same and so the product is positive. $$10$$
$$4\left(–8\right)$$ Multiply, noting that the signs are different and so the product is negative. $$–32$$
$$7\cdot 6$$ The signs are the same, so the product is positive. $$42$$
When we multiply a number by $$1,$$ the result is the same number. What happens when we multiply a number by$$-1?$$ Let’s multiply a positive number and then a negative number by $$-1$$ to see what we get.
$$\begin{array}{ccc}\hfill -1·4\hfill & \phantom{\rule{2em}{0ex}}& \hfill -1\left(-3\right)\hfill \\ \hfill -4\hfill & \phantom{\rule{2em}{0ex}}& \hfill 3\hfill \\ \hfill -4\phantom{\rule{0.2em}{0ex}}\text{is the opposite of}\phantom{\rule{0.2em}{0ex}}\mathbf{\text{4}}\hfill & \phantom{\rule{2em}{0ex}}& \hfill \mathbf{\text{3}}\phantom{\rule{0.2em}{0ex}}\text{is the opposite of}\phantom{\rule{0.2em}{0ex}}-3\hfill \end{array}$$
Each time we multiply a number by $$-1,$$ we get its opposite.
### Multiplication by -1
Multiplying a number by $$-1$$ gives its opposite.
$$-1a=-a$$
## Example
Multiply each of the following:
1. $$\phantom{\rule{0.2em}{0ex}}-1·7\phantom{\rule{1em}{0ex}}$$
2. $$\phantom{\rule{0.2em}{0ex}}-1\left(-11\right)$$
### Solution
The signs are different, so the product will be negative. $$-1\cdot 7$$ Notice that −7 is the opposite of 7. $$-7$$
The signs are the same, so the product will be positive. $$-1\left(-11\right)$$ Notice that 11 is the opposite of −11. $$11$$<|endoftext|>
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Why is Perspective Taking so important?
Perspective Taking is so important for children to learn because they need perspective taking skills to relate to others, to make others feel comfortable around us, and to know how to influence others in a positive way.
To help children develop and improve their perspective taking skills, we need to first explain what perspective taking is, then provide many examples of situations which would require good perspective taking, in order to have a successful social experience.
In other words, define it and practice it.
Perspective Taking is understanding something from someone else’s point of view.
*Note: You do not have to agree with someone, to understand their perspective. Understanding someone’s perspective means that you can cognitively understand why they are doing, thinking or saying something; but it does not mean that you have to agree with it.
For example; if a student can take the perspective of their teacher, they may understand that she gives homework because she wants her students to practice their academic skills. This doesn’t mean the student agrees that they should be given homework.
I have created a fun and interactive Perspective Taking activity set to help my students learn and practice this important skill.
These 4 differentiated SOCIAL SCENARIO activities help children who struggle to understand the importance of considering other’s perspectives.
With these activities, children will practice perspective taking using a wide variety of social scenarios which will help them to be more thoughtful and make better decisions in real life social situations.
This activity includes 6 Behavior Charts (1 Color K-2nd, 1 Color 3rd-5th, 1 B/W K-2nd, 1 B/W 3rd-5th, 2 Editable)
Here is an example of a K-2nd grade behavior chart in color.
Here is an example of a K-2nd grade behavior ~challenge chart.
This chart is great to use with students who are ready for a little extra discussion.
The “It Depends” column allows students to discucss situations where there isn’t one right answer, but… they have to “defend” their choice to put a card in the “it depends” column.
I LOVE options!
Here are two behavior charts that your 3rd-5th graders may like…
And here is the challenge behavior chart…
Here is a sample of some of the scenario cards…
There are 24 Positive Behavior Scenario Cards (can be used with k-5) and 24 Negative Behavior Scenario Cards (can be used with k-5)
There is also a page of EDITABLE cards for you to write your own scenarios to individualize to specific students and situations.
Discussion Cards for k-2nd and 3rd-5th
This activity includes 18 Discussion Cards appropriate for k-2nd grade/ability,18 Discussion Cards appropriate for 3rd-5th grade/ability
and 1 page of EDITABLE cards for you to write your own scenarios to individualize to your students.
Thought Bubble and Speech Bubble Visuals (To be used with any of the scenario cards)
Students listen to scenario cards and then, write in the thought bubble, what they think the person in the scenario was thinking. Next, in the speech bubble, they write what they think would be a better way the person in the scenario could have said what he/she said.
This is a great way to help students who are sensitive to criticism and benefit from learning skills by discussing someone else, vs. talking about their own behavior.
Here is a sample of activity 4 Question Cards…
This activity includes 18 Question Cards appropriate for k-2nd grade/ability, 18 Question Cards appropriate for 3rd-5th grade/ability
and 1 page of EDITABLE cards for you to individualize to your students.
A Thumbs Up/Thumbs Down Visual to use with the activity. Kids love using the Thumbs Up/Thumbs Down cards. You can use them as is or you can cut them out and glue them to a popsicle stick for students to hold up.
You can also use these handy visuals to help teach the concept and to put on your bulletin board 🙂
Who Can Teach Perspective Taking Skills?
Social Workers/Psychologists: Can teach children who struggle to understand other’s perspectives and have difficulty interacting with others in social situations such as their classroom peers, sports groups, outside of school activities.
Special Educators: Can teach students on their caseload who struggle to understand other’s perspectives. They tend to think only their point of view is important or correct and tend to get into arguments with peers and adults and tend to say offensive things to others due to a lack of perspective taking.
Regular Education Teachers: Can teach students who struggle to understand other’s perspectives and may be having social challenges in the classroom during group projects or on the playground engaging in group games and activities.
SLP’s: Can teach students on their case load for social pragmatics. Great for students who need to increase and improve their language and behaviors in social situations.
These activities lead to great discussions and really encourage children to think about others in a wide variety of social situations.
They are great to use in social groups, friendship groups and lunch bunches. They can be used as a whole class lesson too!
I created these activities because my students were struggling to understand a variety of everyday social situations and how to handle them. The practice and discussion is very helpful for them.
I hope it can be helpful for your students as well!
Could you use some visuals to help you teach your students important perspective taking terms? Grab this set of 5 beautiful posters by filling out the form below.
Thanks So Much and Happy Teaching!
Cindy ~Socially Skilled Kids<|endoftext|>
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# Electric Field Due To a Uniformly Charged Ring - JEE
## What is the Significance of the Axis of a Charged Ring?
Last updated date: 19th Mar 2023
Total views: 62.4k
Views today: 0.42k
An electric field gets associated with any charged point in space. Such an electric field is often quantified using terms such as the electric field strength and intensity. The electric field can be present in any object irrespective of how the charge is distributed.
For instance, an electric field gets created along with the spherical, linear, and planar distribution of charges. In all the above cases, the electric field has a magnitude and a direction. Likewise, when a ring is charged either uniformly or non-uniformly, an electric field gets created along the axis of the ring, the magnitude, and direction of which gets determined by the charge on the ring itself.
In this article, you will learn about the axis of a uniformly charged ring and the electric field due to the ring. It is significant to understand the nature and magnitude of the electric field at various points along the axis to understand the force it would exert on any unit positive charge kept nearby.
## Electric Field due to Arc of Ring
An electric field around any charge distribution can be found by creating an element out of infinitesimal point charges. In the case of a uniformly charged ring, the electric field on the axis of a ring, which is uniformly charged, can be found by superimposing the electric fields of an infinitesimal number of charged points. The ring is then treated as an element to derive the electric field of a uniformly charged disc.
### Deriving the Circular Ring Formula:
Consider the following figure as a charged ring whose axis is subjected to an electric field of varying intensity from the centre of the charged ring.
Electric Field at P due to an Infinitesimal Charge dq on a Charged Ring
Let’s now derive the equation to find the electric field along the axis at a distance of x from the centre of the charged ring. Here, $r = \sqrt{x^{2} + a^{2}}$ is the distance of point p from the arc element dq.
According to the principle of superposition, the total electric field at point p (along the axis of the charged ring) is the vector sum of individual electric fields due to all the point charges.
According to Gauss Law, the electric field caused by a single point charge is as follows:
$\overrightarrow{E} = \dfrac {1}{4~\pi ~ \epsilon_0} \times {\dfrac {q}{r^2}}~\hat {r}$
The electric field at point p due to the small point charge dq which is at a radius of a from the centre of the charged ring can be written as:
$\overrightarrow{dE} = \dfrac {1}{4~\pi ~ \epsilon_0} \times {\dfrac {dq}{r^2}}~\hat {r}$
It is important to note that since there is a corresponding piece of point charge on the opposite side of the ring, the y components of the electric field will get nulled throughout.
Hence, only the x component of the electric field will be significant in deriving the total electric field at point p due to a charged ring.
However, we can see that the tiny x component forms an angle Θ at point p along the axis of the charged ring. Hence the electric field equation will be adjusted while considering this angle and hence becomes:
$dE_{x} = \dfrac {1}{4~\pi ~ \epsilon_0} \times {\dfrac {q}{r^2}}~\cos \Theta$
Since the axis forms a right angle with the distance from dq to point p and Θ is unknown, we replace r and $\cos \Theta$ with the known distances x and a. The above equation thus becomes:
$dE_{x} = \dfrac {1}{4~\pi ~ \epsilon_0} \times {\dfrac {dq}{\left({x^2}+{a^2}\right)}}~{\dfrac {x}{\sqrt {\left({x^2}+{a^2}\right)}}}$
Upon simplification,
$dE_{x} = \dfrac {1}{4~\pi ~ \epsilon_0} \times {\dfrac {x~dq}{\left({x^2}+{a^2}\right)^{\dfrac {3}{2}}}}$
Now, by integrating the above equation with respect to dq, the total electric field at point p on the axis of a charged ring is given by the following equation:
$E_{x}=\dfrac {1}{4~\pi ~ \epsilon_0} \times {\dfrac {x~Q}{\left({x^2}+{a^2}\right)^{\dfrac {3}{2}}}}$
Note: The above equation holds good only for finding the electric field on any point on the x-axis.
Let us now draw a graph that closely represents the relationship between the electric field along the axis of a charged ring and the distance from the centre of the charged ring.
The electric field is zero at the centre and increases to a maximum on either side of the ring, and then gradually falls back to zero as x approaches infinity. The magnitude of the electric field will be the same due to symmetry and uniformity in the distribution of charge, and the closest graphical representation or the graph of a uniformly charged ring is drawn below:
Electric Field Intensity along the Axis of a Charged Ring
## Key Points for JEE Exams
• When the point p is very far from the centre of the charged ring i.e., x >> a, then the electric field equation is the same as that due to a point charge, as a is very insignificant such that it becomes zero. Hence, the electric field equation when x >> a is
$E_{x} = (Q/4\pi \epsilon _{o}x^{2} )$
• When the point p is at the centre of the ring, x = 0. Hence the electric field at the centre of a charged ring is zero which is in conformance with symmetry and uniformity.
• The maximum possible electric field intensity can be derived using dE/dx = 0. In this case, it is observed that the maximum electric field strength occurs when $\pm a\sqrt{2}$ and is given by the following equation:
$E_{max} = Q/6\sqrt{3}\pi \epsilon a^{2}$
• The magnetic field due to the ring is $B=2\left( R+x \right)\mu lx$.
## Conclusion
The charge distribution along the axis of an electrically charged ring will be symmetric on either side of the ring, and, hence, the electric field will be in a direction that is along the axis of the charged ring. The resultant electric field at the centre of the ring will be zero, and it will increase to a maximum at a distance of $a/\sqrt{2}$ on either side of the charged ring. Then, at an infinite distance from the centre, the electric field becomes zero again.
Competitive Exams after 12th Science
## FAQs on Electric Field Due To a Uniformly Charged Ring - JEE
1. Does a ring charge behave like a point charge? If so, when does this happen?
Yes. A charged ring will behave like a point charge when the distance from the point p to the centre of the charged ring exceeds the radius of the charge distribution to a greater extent. In that case, the radius is negligible and the charge distribution starts to act as a single point charge i.e., at such a far away distance, a ring charge is perceived as a point charge.
The electric field intensity in such a case will be $E_{x} = \dfrac {Q}{4~\pi ~ \epsilon _{o}~{x^2}}$.
2. What is a uniformly charged ring?
A uniformly charged ring is the one in which the charge is distributed uniformly along the circumference of the ring. The uniformly charged ring plays an important role in understanding electric field generation.<|endoftext|>
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Shunt DC Motor
In the shunt motor, the field winding is connected in parallel or in shunt with the armature winding. [Figure 10-285] The resistance in the field winding is high. Since the field winding is connected directly across the power supply, the current through the field is constant. The field current does not vary with motor speed, as in the series motor and, therefore, the torque of the shunt motor will vary only with the current through the armature. The torque developed at starting is less than that developed by a series motor of equal size.
The speed of the shunt motor varies very little with changes in load. When all load is removed, it assumes a speed slightly higher than the loaded speed. This motor is particularly suitable for use when constant speed is desired and when high starting torque is not needed.
Compound DC Motor
The compound motor is a combination of the series and shunt motors. There are two windings in the field: a shunt winding and a series winding. A schematic of a compound motor is shown in Figure 10-286. The shunt winding is composed of many turns of fine wire and is connected in parallel with the armature winding. The series winding consists of a few turns of large wire and is connected in series with the armature winding. The starting torque is higher than in the shunt motor but lower than in the series motor. Variation of speed with load is less than in a series wound motor but greater than in a shunt motor. The compound motor is used whenever the combined characteristics of the series and shunt motors are desired.
Like the compound generator, the compound motor has both series and shunt field windings. The series winding may either aid the shunt wind (cumulative compound) or oppose the shunt winding (differential compound). The starting and load characteristics of the cumulative compound motor are somewhere between those of the series and those of the shunt motor.
Because of the series field, the cumulative compound motor has a higher starting torque than a shunt motor. Cumulative compound motors are used in driving machines, which are subject to sudden changes in load. They are also used where a high starting torque is desired, but a series motor cannot be used easily.
In the differential compound motor, an increase in load creates an increase in current and a decrease in total flux in this type of motor. These two tend to offset each other and the result is a practically constant speed. However, since an increase in load tends to decrease the field strength, the speed characteristic becomes unstable. Rarely is this type of motor used in aircraft systems.
A graph of the variation in speed with changes of load of the various types of DC motors is shown in Figure 10-287.
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# Stat 1510: Statistical Thinking and Concepts Scatterplots and Correlation.
## Presentation on theme: "Stat 1510: Statistical Thinking and Concepts Scatterplots and Correlation."— Presentation transcript:
Stat 1510: Statistical Thinking and Concepts Scatterplots and Correlation
Agenda 2 Explanatory and Response Variables Displaying Relationships: Scatterplots Interpreting Scatterplots Adding Categorical Variables to Scatterplots Measuring Linear Association: Correlation Facts About Correlation
Objectives 3 Define explanatory and response variables Construct and interpret scatterplots Add categorical variables to scatterplots Calculate and interpret correlation Describe facts about correlation
4 The most useful graph for displaying the relationship between two quantitative variables is a scatterplot. Scatterplot A scatterplot shows the relationship between two quantitative variables measured on the same individuals. The values of one variable appear on the horizontal axis, and the values of the other variable appear on the vertical axis. Each individual in the data appears as a point on the graph. 1.Decide which variable should go on each axis. If a distinction exists, plot the explanatory variable on the x-axis and the response variable on the y-axis. 2.Label and scale your axes. 3.Plot individual data values. 1.Decide which variable should go on each axis. If a distinction exists, plot the explanatory variable on the x-axis and the response variable on the y-axis. 2.Label and scale your axes. 3.Plot individual data values. How to Make a Scatterplot 4
5 Scatterplot Example: Make a scatterplot of the relationship between body weight and pack weight for a group of hikers. Body weight (lb)120187109103131165158116 Backpack weight (lb)2630262429353128
6 Interpreting Scatterplots To interpret a scatterplot, follow the basic strategy of data analysis discussed earlier. Look for patterns and important departures from those patterns. As in any graph of data, look for the overall pattern and for striking departures from that pattern. You can describe the overall pattern of a scatterplot by the direction, form, and strength of the relationship. An important kind of departure is an outlier, an individual value that falls outside the overall pattern of the relationship. As in any graph of data, look for the overall pattern and for striking departures from that pattern. You can describe the overall pattern of a scatterplot by the direction, form, and strength of the relationship. An important kind of departure is an outlier, an individual value that falls outside the overall pattern of the relationship. How to Examine a Scatterplot
Interpreting Scatterplots 7 Two variables have a positive association when above-average values of one tend to accompany above-average values of the other, and when below-average values also tend to occur together. Two variables have a negative association when above-average values of one tend to accompany below-average values of the other. Two variables have a positive association when above-average values of one tend to accompany above-average values of the other, and when below-average values also tend to occur together. Two variables have a negative association when above-average values of one tend to accompany below-average values of the other. Direction Form Strength There is one possible outlier, the hiker with the body weight of 187 pounds seems to be carrying relatively less weight than are the other group members. There is a moderately strong, positive, linear relationship between body weight and pack weight. It appears that lighter hikers are carrying lighter backpacks.
Adding Categorical Variables 8 Consider the relationship between mean SAT verbal score and percent of high-school grads taking SAT for each state. To add a categorical variable, use a different plot color or symbol for each category. Southern states highlighted
Measuring Linear Association 9 A scatterplot displays the strength, direction, and form of the relationship between two quantitative variables. The correlation r measures the strength of the linear relationship between two quantitative variables. r is always a number between -1 and 1. r > 0 indicates a positive association. r < 0 indicates a negative association. Values of r near 0 indicate a very weak linear relationship. The strength of the linear relationship increases as r moves away from 0 toward -1 or 1. The extreme values r = -1 and r = 1 occur only in the case of a perfect linear relationship. The correlation r measures the strength of the linear relationship between two quantitative variables. r is always a number between -1 and 1. r > 0 indicates a positive association. r < 0 indicates a negative association. Values of r near 0 indicate a very weak linear relationship. The strength of the linear relationship increases as r moves away from 0 toward -1 or 1. The extreme values r = -1 and r = 1 occur only in the case of a perfect linear relationship.
Correlation 10
Facts About Correlation 1. Correlation makes no distinction between explanatory and response variables. 2. r has no units and does not change when we change the units of measurement of x, y, or both. 3. Positive r indicates positive association between the variables, and negative r indicates negative association. 4. The correlation r is always a number between -1 and 1. Cautions: Correlation requires that both variables be quantitative. Correlation does not describe curved relationships between variables, no matter how strong the relationship is. Correlation is not resistant. r is strongly affected by a few outlying observations. Correlation is not a complete summary of two-variable data. Cautions: Correlation requires that both variables be quantitative. Correlation does not describe curved relationships between variables, no matter how strong the relationship is. Correlation is not resistant. r is strongly affected by a few outlying observations. Correlation is not a complete summary of two-variable data. 11
Correlation Practice 12 For each graph, estimate the correlation r and interpret it in context.
13 Case Study Per Capita Gross Domestic Product and Average Life Expectancy for Countries in Western Europe
14 Case Study CountryPer Capita GDP (x)Life Expectancy (y) Austria21.477.48 Belgium23.277.53 Finland20.077.32 France22.778.63 Germany20.877.17 Ireland18.676.39 Italy21.578.51 Netherlands22.078.15 Switzerland23.878.99 United Kingdom21.277.37
15 Case Study xy 21.477.48-0.078-0.3450.027 23.277.531.097-0.282-0.309 20.077.32-0.992-0.5460.542 22.778.630.7701.1020.849 20.877.17-0.470-0.7350.345 18.676.39-1.906-1.7163.271 21.578.51-0.0130.951-0.012 22.078.150.3130.4980.156 23.878.991.4891.5552.315 21.277.37-0.209-0.4830.101 = 21.52 = 77.754 sum = 7.285 s x =1.532s y =0.795
16 Case Study<|endoftext|>
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There are four pairs of sinuses (named for the skull bones in which they are located):
- Frontal sinuses: The right and left frontal sinuses are located in the center of the forehead (frontal bone) just above each eye.
- Maxillary sinuses: These are the largest of the sinuses and are located behind the cheekbones near the maxillae, or upper jaws.
- Sphenoid sinuses: The sphenoid sinuses are located in the sphenoid bone near the optic nerve and the pituitary gland on the side of the skull.
- Ethmoid sinuses: The ethmoid sinuses are located in the ethmoid bone, which separates the nasal cavity from the brain. These sinuses are not single sacs but a collection of six to 12 small air cells that open independently into the nasal cavity. They are divided into front, middle, and rear groups.
Like the nasal cavity, the sinuses are all lined with mucus. The mucus secretions produced in the sinuses are continually being swept into the nose by the hair-like structures (called ‘cilia’) on the surface of the respiratory membrane.
When they aren’t moistening the air we breathe through our noses, the hollow sinuses act to lighten the bones of the skull. Sinuses also serve as sound-resonance chambers for speech.
The paired and often asymmetrical (not perfectly mirrored) sinuses are small, or rudimentary, at birth but grow as the skull grows. They are fairly well developed by age seven or eight, but don’t reach their maximum size until after puberty. In adults, the sinuses vary considerably in size and shape.
Sinuses often get infections. Sinusitis is inflammation of a sinus caused by a bacterial infection that can follow a viral infection. This causes pus and mucus to accumulate in the sinus. Symptoms can include fever, headache, stuffy nose, and impaired sense of smell.<|endoftext|>
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Write out a list of contractions without the apostrophe For example: weve and cant. Have your child put elbow macaroni in the place of the missing apostrophes. Be sure to check for accuracy.
The week is almost over. Hang in there! You’re doing an awesome job. Tonight, during dinner, have everybody go around and say one thing that is great about each other.
Brainstorm a verb for every letter in your name. Have your child do the same. For example: Mike: Move, Illustrate, Kick and Exercise.
Write base words “hop, stop, pat, trip, and shop” on paper. Ask your child to add the suffix “ing” to the words. Remember to follow the rule: you must have double consonants before adding “ing”.
This week will focus on honing some writing skills. When learning to become a writer not only is it important to read other authors and to practice writing, but it’s essential to get feedback from someone who is reading your work. Keep that in mind as you work on writing skills. One specific activity you can do is remind your child, “when 2 vowels go walking, the 1st does the talking.” Help your child hunt for words that follow this pattern in books. Example: Juice, Paint, Goal, etc.
Before reading a book, ask your child to scan the pages for question marks and exclamations. Have your child practice reading these sentences aloud, and while reading have them practice changing the intonation of their voice.
Practice punctuation. Write 5 sentences and have your child fill in the correct punctuation: period, question mark, or exclamation point.<|endoftext|>
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Pauli matrices and the complex number matrix representation
The three spin Pauli matrices are: $\sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
According to problem 2.2.3 (Mathematical methods for physicists, Arfken Weber & Harris, Seventh Edition) Complex numbers, a + ib, with a and b real, may be represented by (or are isomorphic with) 2 × 2 matrices as follows:
$a + ib = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}$
This implies that: $i = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, -i = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
And therefore according to the above proposition. $\sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = i \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = i(-i) = 1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},$
By exchanging the rows in $\sigma_1$, we can also write it as the identity matrix. Therefore according to this we get $\sigma_1 = \sigma_2$ which to me is quite surprising. They have the property that $\sigma_1^2 = \sigma_2^2 = I$. One can also prove that $\sigma_1 \sigma_2 = i \sigma_3$. Since $\sigma_1 = \sigma_2 = I$, $\sigma_1 \sigma_2 = I$. The RHS becomes:
$i\sigma_3 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$
The problem arises when I try to prove the identity $\sigma_i \sigma_j + \sigma_j \sigma_i = 2\delta_{ij}I_2$ in Problem 2.2.11(c). This clearly does not hold since $\sigma_2 \sigma_3 + \sigma_3 \sigma_2$ $= I \sigma_3 + \sigma_3 I \neq 0$
Where am I making a mistake here ?
• Real 2x2 matrices. – NickD Oct 16 '17 at 20:46
• Yes where a and b are real – 256ABC Oct 16 '17 at 20:47
• I would look at the minimal polynomial $f_j(x)= x^2+b_j x+c_j$ of each matrix to obtain $\mathbb{R}[\sigma_j] \cong \mathbb{R}[x]/(f_j(x))$. The ring $\mathbb{R}[\sigma_1,\sigma_2,\sigma_3]$ is a non-commutative subring of $M_2(\mathbb{C})$ (the ring of $2\times 2$ complex matrices) – reuns Oct 16 '17 at 20:51
• Exchanging rows in $\sigma_1$ gives a completely different matrix. – Aretino Oct 16 '17 at 21:15
• The point is that the Pauli matrices are NOT of this form. – NickD Oct 16 '17 at 21:17
The mapping $a +ib \Rightarrow \begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ where $a,b \in \mathbb{R}$ does not map any complex number to a Pauli matrix. The Pauli matrices are not of the right form. So saying that $1$ maps to $\sigma_2$ (or even worse, that it is equal to it) is meaningless.
• But you can bring $\sigma_1$ to this form by just interchanging the rows right ? – 256ABC Oct 16 '17 at 21:44<|endoftext|>
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Or download our app "Guided Lessons by Education.com" on your device's app store.
L.1.1.J Worksheets, Workbooks, Lesson Plans, and Games
Our Take:In year two, students are taught to independently write different types of sentences. These types include declarative (a statement of fact or opinion), interrogative (a question), imperative (a command or request), and exclamatory (a statement that shows an emotional response and often ends with an exclamation point).
These worksheets and lesson plans can help students practise this Common Core State Standards skill.
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No exercises found for this common core node.<|endoftext|>
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Children begin learning to read usually between the ages of three to five. Some take to it quickly; others need more help. But what is going on in your little one’s brain to allow them to make sense from a meaningless jumble of shapes?
The first thing a baby does is learn to talk. More specifically, it begins to learn the language you speak to them. They begin to communicate with you from the moment of birth, as they listen to your sounds and mimic them. Gradually, the sounds start to make sense and children can master complex language within a few years.
As children learn words, they also learn that words are made up of different sounds or syllables. This is known as phonic awareness. This helps when they start to understand the phonic sounds in writing.
When a child first sees letters and numbers, they make very little sense. However, once they are shown the letters of the alphabet, and are taught their accompanying sounds, a change happens. They will begin by learning individual letters such as A, B and C.
From here on, your child’s memory is going to be their most important tool. Memorising letters and sounds takes some time but gradually your little one will understand them and be able to take in how simple words are made up. Constant practice is needed, so take the time to point to things, say their name and then spell it out phonetically.
Of course, as English is not a totally phonetic language, it’s not quite that simple! We know that letters can take on different sounds depending on the word they are in, so it may be a case of explaining that letters are not always what they seem but not to get too bogged down in explanations.
Next comes the ability to recognise individual words, although this may still be largely to do with their shape. They may for example recognise the word ‘take’ but get confused by ‘talk’.
Next, they begin to recognise letter sequences in similar words, such as lake, bake, fake and take. This is called chunking. This means they can take whole sounds at a time rather than having to recognise all the letters individually. This will lead to the ability to increase their vocabulary and to being able to read sentences, then paragraphs. Soon, they’ll be coming home with a ‘chapter book’ from school and they will truly be flying.
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Manage your Families Locations, email preferences, listings and events in your profile.<|endoftext|>
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The fossil record is frustratingly incomplete and the earliest fossils date back approximately 3.8 billion years. Philip Donoghue, one of the co-authors of the published study, describes a “second record of life” that can be used to supplement the fossil record, which exists inside the genomes of all living organisms found on modern Earth. Here, the research that allowed scientists to revise their estimate of the date of life’s emergence on Earth relied upon the use of so-called “molecular clocks.” A molecular clock is established by counting the number of mutations in the genomic DNA of several species with the supposition that this number is proportional to how far back in time their lineages diverged from a common ancestor.
To that end, the research team behind this recent study combed the genetic material of 102 separate organisms for changes in the DNA code of 29 distinct genes. The data generated from this exhaustive search allowed them to develop a timeline from which all significant clusters of life originated on Earth from bacteria to advanced multi-cellular organisms. As stated earlier, the “molecular clocks” have inferred that LUCA sprung into existence far earlier than the existing fossil record suggested; right before the late heavy bombardment.
The idea that life originated during this cataclysmic period was unexpected and remains controversial to say the least. This would place LUCA’s emergence just after a major event in the geological history of Earth occurred: when another planet named Theia collided with the early Earth, breaking off large chunks that would coalesce into the moon. Such a collision would have certainly destroyed any life that may have been present at that time. And many scientists believe that the environmental conditions on Earth after that impact would have been far too volatile to have supported the emergence of life.
It is highly unlikely that any fossil record of LUCA will be found given that the earliest life would consist of microscopic cells, the artifacts of which are difficult to identify and validate. Therefore, any “molecular clock” evidence will lack corroboration from the fossil record and likely to remain a subject of much contention. Barring Q’s intervention, the precise date that life first arose on Earth will continue to fuel much debate in the coming years.<|endoftext|>
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Hutchinson-Gilford Progeria Syndrome (HGPS)
Progeria is a rare and fatal congenital disorder marked by physical symptoms resembling that of aging but having early onset in childhood. The word “Progeria” is derived from the Greek word “Progerus” meaning “prematurely old”. It has a reported incidence of about 1 in 4 – 8 million newborns. It affects both sexes equally and all races.
Point mutation in the LMNA gene that encodes the protein Lamina A which is a part of the building blocks of nuclear envelope. Hence, the production of abnormal Lamina A proteins leads to premature death of cells. It is not inherited but it is almost always a chance occurrence that is extremely rare. Progeria is related with these diseases:
- Werner’s syndrome
- Cockayne’s syndrome
- Xeroderma pigmentosum
Signs and Symptoms
Children with progeria generally appear normal at birth. By 12 months, signs and symptoms, such as skin changes and hair loss, begin to appear.
- Growth failure during the first year of life
- A narrowed face and beaked nose
- Narrow, shrunken or wrinkled face
- Alopecia (baldness, loss of eyebrows and eyelashes)
- Scleroderma like skin conditions
- Dwarfism and mental retardation
- Large head for size of face (macrocephaly)
- Small jaw (micrognathia)
- Nail and dental abnormalities
- Prominent scalp veins
- Skeletal defects with limited range of motion
- Atherosclerosis and cardiac problems
- Myocardial infarction (MI or heart attack)
- Stroke (sudden death of brain cells in a localized area due to inadequate blood flow)
- On the basis of above signs and symptoms
- Blood test revealing low level of High Density Lipoprotein (HDL)
- Confirmed through genetic tests
- Earlier the diagnosis is made, treatment can be commenced to ease the signs and symptoms and prevent the possible complications like heart attack and stroke.
Progeria is an incurable disease. Most treatment focuses on reducing possible complications. These are:
- Artery bypass surgery or angioplasty.
- Low dose aspirin.
- High calorie diet.
- Growth hormone therapy.
- Physical & occupational therapy to minimize joint stiffness and improve mobility.
- Extraction of milk teeth to provide enough space for prematurely erupting adult teeth.
- Infants who feed poorly may benefit from a feeding tube and a syringe.
Average age of 13 years with a range of about 8 – 21 years (death due to atherosclerosis)
Progeria in Nepal
No cases of progeria has been identified in Nepal by this date.
Where to donate for Progeria?
Progeria Research Foundation is a trusted foundation that conducts campaign to collect funds and aims to find the cure and effective treatment for this disease.
Progeria as depicted in Movie “PAA”
Amitabh Bachchan plays the 13 year old character named Auro who is suffering from Progeria. This disease rapidly accelerates his aging and towards the end of the movie by the time the character hits his teens he dies.
Main sources of information:
Progeria Research Foundation<|endoftext|>
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In this lecture, what I want to do is talk about the velocity of the robot's end-effector. So the last couple of lectures, we have looked at concepts like forward kinematics and forward kinematics is the relationship between the joint angles and the robot's end-effector pose, so I put in the joint angles and I get out the end-effector pose.
The inverse kinematics is sort of the opposite. I say what I want the end-effector pose to be and what comes out of the algorithm of the joint angles that I need in order to achieve that end-effector pose.
The problem, what we are going to talk about in this particular lecture is velocity transform. If the joints move at this particular velocity, what is the velocity of the robot's end-effector and you note that I have written the velocity of the joints as q-dot, so dot represents the time rate of change. It is a common notation in differential calculus and the pose of the robot's end-effector I have been using this symbol, psi E, to represent the pose of the end-effector and we have talked about how that is combination of position and orientation.
So, in this particular case, what we are talking about now is cosine dot. We are talking about the rate of change of a pose and that is an interesting and not necessarily intuitive thing; so first of all we are going to talk about what does the rate of change of a pose mean, what kind of mathematical object do we use to represent that and then what is the relationship between that thing, the psi dot, and the rate of change of the joint angles of the robot itself.
So, that is what this lecture is about. It is about understanding relationship between velocity of the joints and the rate of change of pose.
We will learn about the relationship, in 2D, between the velocity of the joints and the velocity of the end-effector — the velocity kinematics. This relationship is described by a Jacobian matrix which also provides information about how easily the end-effector can move in different Cartesian directions.
This content assumes high school level mathematics and requires an understanding of undergraduate-level mathematics; for example, linear algebra - matrices, vectors, complex numbers, vector calculus and MATLAB programming.<|endoftext|>
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# Permutation and combination? In your own words, explain the difference between a permutation and a combination. Also, give an example of a problem whose solution is a permutation and an example of a problem whose solution is a combination.?
See explanation.
Permutation of a set is a sequence in which the order is important (it means that for example ##135## and ##153## are two different permutations of a set ##{1,3,5}##)
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Permutation and combination? In your own words, explain the difference between a permutation and a combination. Also, give an example of a problem whose solution is a permutation and an example of a problem whose solution is a combination.?
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Combination of a set is a subset in which the order is not important. This means that for example in a set ##{1,2,3,4,5,6}## you can find a 3-element subset ##{2,3,5}## and if you write it in decreasing order (as ##{5,3,2}##) it is still the same combination but written in a different way.
A problem including permutations would require ordering some elements. It can be a problem like:
Jack has a set of four books. Each is marked with a number 1,2,3 and 4. If he puts them randomly on a shelf, what is the probability that they are ordered either increasingly or decreasingly
Solution
To calculate the probability we have to calculate all possibilities first.There are 4 books and to find all possibilities of putting them on a shelf we have to calculate number of permutations of four element set:
##bar(bar(Omega))=P_4=4! =1*2*3*4=24##
The orders which are mentioned in the task are ##1234## (increasing order) and ##4321## (decreasing order), so
##bar(bar(A))=2##
Finally we can calculate the probability:
##P(A)=bar(bar(A))/bar(bar(Omega))=2/24=1/12##
Answer: The probability is ##1/12##
A problem which includes combinations would deal with choosing a subset from a larger set. For example:
Before a science contest a team has to be chosen from a class. Assuming that there are 15 boys and 10 girls in how many ways can a team be choosen if it has to contain 2 girls and two boys ?
Solution:
In this task the order is not important. No matter if you choose Jack first and Ann next or the other way Ann first and next Jack, the team consists of the same people. So you have to use combinations to solve this task.
To calculate the number we can assume that we choose girls first, then boys, so the number would be calculated as:
##n=C””_10^2*C””_15^2=(10!)/(2!*8!)(15!)/(2!*13!)##
##n=15750##
Answer There are ##15750## ways to choose the team.
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# What are sequences in math?
An arithmetic sequence is a list of numbers with a definite pattern. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence.
The constant difference in all pairs of consecutive or successive numbers in a sequence is called the common difference, denoted by the letter d. We use the common difference to go from one term to another. How? Take the current term and add the common difference to get to the next term, and so on. That is how the terms in the sequence are generated.
• If the common difference between consecutive terms is positive, we say that the sequence is increasing.
• On the other hand, when the difference is negative we say that the sequence is decreasing.
## Illustrative Examples of Increasing and Decreasing Arithmetic Sequences
Here are two examples of arithmetic sequences. Observe their common differences.
With this basic idea in mind, you can now solve basic arithmetic sequence problems.
### Examples of How to Apply the Concept of Arithmetic Sequence
Example 1: Find the next term in the sequence below.
First, find the common difference of each pair of consecutive numbers.
Since the common difference is 8 or written as d=8, we can find the next term after 31 by adding 8 to it. Therefore, we have 31 + 8 = 39.
Example 2: Find the next term in the sequence below.
Observe that the sequence is decreasing. We expect to have a common difference that is negative in value.
To get to the next term, we will add this common difference of d=-7 to the last term in the sequence. Therefore, 10 + left( { – 7}
ight) = 3.
Example 3: Find the next three terms in the sequence below.
Be careful here. Don’t assume that if the terms in the sequence are all negative numbers, it is a decreasing sequence. Remember, it is decreasing whenever the common difference is negative. So let’s find the common difference by taking each term and subtracting it by the term that comes before it.
The common difference here is positive four left( { + ,4}
ight) which makes this an increasing arithmetic sequence. We can obtain the next three terms by adding the last term by this common difference. Whatever is the result, add again by 4, and do it one more time.
Here’s the calculation:
The next three terms in the sequence are shown in red.
Example 4: Find the seventh term (7th) in the sequence below.
Sometimes you may encounter a problem in an arithmetic sequence that involves fractions. So be ready to use your previous knowledge on how to add or subtract fractions.
Also, always make sure that you understand what the question is asking so that you can have the correct strategy to approach the problem.
In this example, we are asked to find the seventh term, not simply the next term. It is a good practice to write all the terms in the sequence and label them, if possible.
Now we have a clear understanding of how to work this out. Find the common difference, and use this to find the seventh term.
Finding the common difference,
Then we find the 7th term by adding the common difference starting with the 4th term, and so on. Here’s the complete calculation.
Therefore, the seventh term of the sequence is zero (0). We can write the final answer as,
Example 5: Find the color{red}{35^{th}} term in the arithmetic sequence 3, 9, 15, 21, …
You can solve this problem by listing the successive terms using the common difference. This method is tedious because you will have to keep adding the common difference (which is 6) thirty-five times starting with the last term in the sequence.
You don’t have to do this because it is cumbersome. And not only that, it is easy to commit a careless error during the repetitive addition process.
If you decide to find the color{red}{35^{th}} term of the sequence using this “successive addition” method, your solution will look similar below. The “dot dot dot” means that there are calculations there but not shown as it can easily occupy the entire page.
• You might also be interested in:
• Arithmetic Sequence Formula
• More Practice Problems with the Arithmetic Sequence Formula
See also: How to write numbers: a guide to using them correctly
## Number Sequences (solutions, examples, videos)
Related Topics: More Lessons for Arithmetic Math Worksheets How to Find The Next Term In A Number Sequence?
A number sequence is a list of numbers arranged in a row. Let us look at two examples below.
(i) 4, 6, 1, 10, 14, 5, …
(ii) 4, 7, 10, 13, ….
Number sequence (i) is a list of numbers without order or pattern. You cannot tell what number comes after 5.
Number sequence (ii) has a pattern. Do you observe that each number is obtained by adding 3 to the preceding number (i.e. the number just before it)?
In this section, we will only study number sequences with patterns .
Some other examples of number sequences are:
Number Sequence Pattern 3, 6, 9, 12, … add 3 12, 17, 22, 27, … add 5 70, 65, 60, 55, … subtract 5 15, 19, 23, 27, … add 4 81, 27, 9, 3, … divide by 3
How to Complete Missing Terms In A Number Sequence?
• Each of the number in the sequence is called a term.
• In order to find the missing terms in a number sequence, we must first find the pattern of the number sequence.
• Example :
• Find the missing term in the following sequence:
• 8, ______, 16, ______, 24, 28, 32
• Solution:
To find the pattern, look closely at 24, 28 and 32. Each term in the number sequence is formed by adding 4 to the preceding number. So, the missing terms are 8 + 4 =12 and 16 + 4 = 20. Check that the pattern is correct for the whole sequence from 8 to 32.
1. Example :
2. What is the value of n in the following number sequence?
3. 16, 21, n, 31, 36
4. Solution:
5. We find that the number pattern of the sequence is “add 5” to the preceding number. So, n = 21 + 5 = 26
How to find the next term in a number sequence? The following video shows some examples of how to determine the next term in a number sequence. Examples: Find the next number 1. 1, 8, 15, 22, … 2. 1, 8, 64, 512, … 3. 1, 8, 27, 64, ….
4. 1, 8, 16, 15, …
• Show Step-by-step Solutions
The following diagrams give the formulas for Arithmetic Sequence and Geometric Sequence. Scroll down the page for examples and solutions. How to find the nth Term of an Arithmetic Sequence Example:
7, 9, 11, 13, 15, …
• Show Step-by-step Solutions
How to find the nth Term of a Geometric Sequence? Example:
5, 10, 20, 40, …
• Show Step-by-step Solutions
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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## Describing Sequences | Number Patterns
A sequence is an ordered list of items, usually numbers. Each item which makes up a sequence is called a “term”.
Sequences can have interesting patterns. Here we examine some types of patterns and how they are formed.
Examples:
• (1; 4; 7; 10; 13; 16; 19; 22; 25; ldots)
• There is difference of ( ext{3}) between successive terms.
• The pattern is continued by adding ( ext{3}) to the previous term.
1. (13; 8; 3; -2; -7; -12; -17; -22; ldots)
There is a difference of (- ext{5}) between successive terms.
The pattern is continued by adding (- ext{5}) to (i.e. subtracting ( ext{5}) from) the previous term.
1. (2; 4; 8; 16; 32; 64; 128; 256; ldots)
2. This sequence has a factor of ( ext{2}) between successive terms.
3. The pattern is continued by multiplying the previous term by 2.
• (3; -9; 27; -81; 243; -729; 2187; ldots)
• This sequence has a factor of (- ext{3}) between successive terms.
• The pattern is continued by multiplying the previous term by (- ext{3}).
1. (9; 3; 1; frac{1}{3}; frac{1}{9}; frac{1}{27}; ldots)
2. This sequence has a factor of (frac{1}{3}) between successive terms.
3. The pattern is continued by multiplying the previous term by (frac{1}{3}) which is equivalent to dividing the previous term by 3.
Some learners may see example 3 as (2^{1}; 2^{2}; 2^{3}; ldots) and see a pattern with the powers. You may choose to discuss this in class as a precursor to geometric series which will be introduced in Grade 12.
## Sequences
A sequence is a set of ordered numbers. For example, the sequence 2, 4, 6, 8, … has 2 as its first term, 4 as its second, etc. The nth term in a sequence is usually called sn. The terms of a sequence may be arbitrary, or they may be defined by a formula, such as sn = 2n.
In general, n starts at 1 for sequences, but there are times when it is convenient for n to start at 0, in which case the first term is s0. If we add up the first n terms of a sequence we get a partial sum, usually referred to as Sn (i.e., with a capital letter).
This device cannot display Java animations. The above is a substitute static image
See About the calculus applets for operating instructions.
### 1. Arithmetic sequence
The applet shows the sequence defined by sn = 2 + 3(n – 1). This is called an arithmetic sequence and each term of the sequence is found by adding a constant amount (e.g., 3 in this example) to the preceeding element.
The general formula for an arithmetic sequence is sn = s1 + d(n – 1), where s1 is the first term and d is the common difference (i.e., the amount added to get the next term).
The partial sum of the first 10 terms is shown in the upper left corner of the graph, and you can change the number of terms by moving the max n slider or typing in the max n input box.
One of the issues that we are concerned with when working with sequences is what happens to the values of the terms when n heads to infinity.
In other words, does
have a value, or does sn head off to infinity or jump around as n gets big? If there is a limit, we say that the sequence converges or is convergent.
If this limit does not exist, the sequence diverges or is divergent. Obviously the arithmetic sequence diverges, because the terms keep getting bigger.
Select the second example from the drop down menu, showing a geometric sequence defined by
sn = 2n In a geometric sequence each term is a constant multiple of the previous term (the multiple here is 2). The general form of a geometric sequence is
sn = s1rn – 1 where r is the common ratio (i.e., the amount that each term is multiplied by to get the next term). Obviously, r = 1 and r = 0 are not useful cases (both just give a constant value for all terms). It is clear from the graph that the example sequence is divergent, because the terms keep getting bigger.
### 3. Another geometric sequence
Select the third example, showing another geometric sequence with a common ratio of 1/2. Does this one converge? The terms get closer and closer to zero, so this sequence does converge. Geometric sequences converge if the common ratio is between 0 and 1, and diverge if the common ratio is greater than 1.
### 4. Alternating geometric sequence
Select the fourth example, showing another geometric sequence with a negative common ratio. Note that the terms alternate on the positive and negative side of the axis. This sequence also converges towards 0, so we can extend our knowledge of geometric sequence convergence to say that the sequence converges if |r| < 1.
### Explore
You can experiment with your own sequences by typing in a rule, using n as the variable.
## Sequences – Sequences – AQA – GCSE Maths Revision – AQA – BBC Bitesize
• Number sequences are sets of numbers that follow a pattern or a rule.
• If the rule is to add or subtract a number each time, it is called an arithmetic sequence.
• If the rule is to multiply or divide by a number each time, it is called a geometric sequence.
• Each number in a sequence is called a term.
• A sequence which increases or decreases by the same amount each time is called a linear sequence.
The term to term rule of a sequence describes how to get from one term to the next.
### Example 1
Write down the term to term rule and then work out the next two terms in the following sequence.
3, 7, 11, 15, …
Firstly, work out the difference in the terms.
This sequence is going up by four each time, so add 4 on to the last term to find the next term in the sequence.
3, 7, 11, 15, 19, 23, …
To work out the term to term rule, give the starting number of the sequence and then describe the pattern of the numbers.
The first number is 3. The term to term rule is 'add 4'.
Once the first term and term to term rule are known, all the terms in the sequence can be found.
### Example 2
Write down the term to term rule and then work out the next two terms in the following sequence.
-1, -0.5, 0, 0.5, …
The first term is -1. The term to term rule is 'add 0.5'.
Question
What is the term to term rule and the next two terms of the sequence: 17, 14, 11, 8, …?
To work out the term to term rule, give the first term and then the pattern. The first term is 17, and the pattern is to subtract 3 each time, so the term to term rule is 'start at 17 and subtract 3'.
The next two terms of the sequence are 5 and 2, giving the sequence as:
Question
What are the next three terms of a sequence that has a first term of 1, where the term to term rule is multiply by 2?
The first term is given as 1. Each number that follows is double the number before.
## Sequences | Brilliant Math & Science Wiki
A sequence is an ordered set with members called terms.
Usually, the terms are numbers. A sequence can have infinite terms.
An example of a sequence is
1,2,3,4,5,6,7,8,… .1,2,3,4,5,6,7,8,dots.1,2,3,4,5,6,7,8,….
There are different types of sequences. For example, an arithmetic sequence is when the difference between any two consecutive terms in the sequence is the same. So,
5,14,23,32,41,505, 14, 23, 32, 41,505,14,23,32,41,50
is an arithmetic sequence with common difference 999, first term 555, and number of terms 6.6.6.
Another type of sequence is a geometric sequence. This is when the ratio of any two consecutive terms in the sequence is the same. For example,
2,6,18,54,1622, 6, 18, 54, 1622,6,18,54,162
is a geometric sequence with common ratio 333, first term 222, and number of terms 5.5.5.
In a sequence, it is conventional to use the following variables:
• aaa is the first term in the sequence.
• nnn is the number of terms in the sequence.
• Tn{ T }_{ n }Tn is the nthn^ ext{th}nth term in the sequence.
• Sn{ S }_{ n }Sn is the sum of the first nnn terms of the sequence.
• ddd is the common difference between any two consecutive terms (arithmetic sequences only).
• rrr is the common ratio between any two consecutive terms (geometric sequence only).
For example, if a series starts with 111 and has a common difference of 1,1,1, we have Sn=n(n+1)2.{ S }_{ n }= dfrac{n(n + 1)}{2}.Sn=2n(n+1).
Similarly, for the series of squares 12,22,32,…,n2,1^2,2^2,3^2,dots,n^2,12,22,32,…,n2, we have Sn=n(n+1)(2n+1)6.{ S }_{ n } = dfrac{n(n + 1)(2n + 1)}{6}.Sn=6n(n+1)(2n+1).
For the series of cubes 13,23,33,…,n3,1^3,2^3,3^3,dots,n^3,13,23,33,…,n3, we have Sn=(n(n+1)2)2.{ S }_{ n } = left(dfrac{n(n+1)}{2}
ight)^2.Sn=(2n(n+1))2.
Some special types of sequences can be found in Arithmetic Progressions, Geometric Progressions, and Harmonic Progression.<|endoftext|>
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Medical waste must be properly handled and disposed of to prevent the unnecessary spread of infectious diseases.
What is Medical Waste
Medical waste is any type of material that is itself or is contaminated by (or even potentially contaminated by) infectious materials.
This type of waste includes items such as blood and other bodily fluids, samples taken in a laboratory or used during medical research, and includes any items/tools/materials that come (or may have the potential to come) into contact with these items (such as bandages, bed linens, needles, swabs, etc.)
Different Types of Medical Waste
Because the general definition of medical waste is broad, and different organizations typically produce certain kinds of medical waste, different categories exist to help distinguish between different types of waste.
Categories are formed based on how dangerous waste items are, what type of container certain materials should be stored in, and what type of institution might produce this waste.
Not all medical waste is considered dangerous, so the law has created the term “regulated medical waste” to classify potentially harmful waste. These are the types of waste most commonly regulated by states:
- Contaminated sharps
- Contaminated animal carcasses, body parts and bedding
- Isolation waste
- Pathological waste
- Human blood and blood products
- Cultures and stocks of infectious agents
Stages in the Medical Waste Disposal Process
Organizations such as hospitals might have a hard time properly managing medical waste because they produce so much medical waste at such a rapid pace and it falls into so many different categories.
One of the best ways to combat improper medical waste disposal and reduce risk is to create a clear, thought out plan and educate all team members on proper disposal.
Understanding the medical waste disposal process full cycle will help you make better decisions for your disposal plan.
Step 1: Collection and Sorting
The first step in proper medical waste disposal process is collecting and sorting medical waste before it gets muddled and possibly thrown in with the normal garbage.
Medical waste containers come in different colors to easily denote which type of waste goes in them. They come in different sizes and shapes for convenience and are made of different materials for extra protection against some items like sharps.
Audit your building and strategically place medical waste containers in areas that make disposal quick and easy to reduce the risk of improper disposal.
Step 2: Storage
Medical waste should be disposed of regularly. Biohazardous waste must legally be disposed of every 30 days at a maximum (unless it’s stored in below freezing temperatures).
Whether or not your medical waste is a biohazard, regularly disposing of your waste will keep your work space clean, risk-free, compliant, and safe.
Even though medical waste may not be ready for immediate disposal at this point, it’s important to store materials in the proper waste container (hard shelled for sharps, leak proof for infectious, etc.) until they are ready to be disposed of.
Step 3: Transportation
Transporting medical waste involves hauling it to your vehicle and subsequently driving it to a facility for final disposal.
Whether you gather and transport your own waste to a nearby facility (which is uncommon) or you have a service provider picking up your waste from your location, it’s imperative that your waste be properly contained to avoid contamination during transport.
Ensure your waste is in the proper container, properly labeled, thoroughly sealed, and make sure no container is overflowing or able to spill.
Step 4: Final Disposal
The most common means of destruction of medical waste is incineration. Materials can also be treated via radiation, thermally, or chemically, depending on the type of material.
This process should be completed by professionals, and not performed in house. Medical waste disposal services make the process easy—coming to you to pick up and transport your materials and safely disposing of them for you.
Find a Safe and Reliable Medical Waste Disposal Service Provider Near You
One of the easiest ways to manage the medical waste you produce is to use a collection service. Medical waste disposal professionals can stop by your location as often as you need to pick up your waste and safely dispose of it for you.
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Shred Nations partners with all types of destruction service providers across North America. Call us at (800) 747-3365 or fill out form to get free quotes from medical waste disposal service providers in your area.<|endoftext|>
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# User:IssaRice/Linear algebra/Change of basis example in two dimensions
This example comes from this video. To make it easier to go back and forth between this page and the video, the notation on this page tries to follow that of the video (where the discussion overlaps), but we distinguish between matrices and vectors.
We are working in $\mathbf R^2$, the plane. To be slightly pedantic, we will distinguish between matrices and vectors: $\begin{bmatrix}1 \\ 0\end{bmatrix} \in \mathbf R^{2,1}$ and $(1,0) \in \mathbf R^2$. If $v$ is a vector and $\beta$ is a basis, we write $[\vec{\mathbf v}]^\beta \in \mathbf R^{n,1}$ for the "column vector" (n-by-1 matrix) and $[\vec{\mathbf v}]_\beta \in \mathbf R^{1,n}$ for the "row vector" (1-by-n matrix). It's usually not necessary to be this pedantic, but here the whole point of the discussion is to understand how coordinate systems and translation between coordinate systems works, so it's worthwhile to be pedantic.
Jennifer's basis vectors: $\vec{\mathbf b}_1 := (2,1)$ and $\vec{\mathbf b}_2 := (-1, 1)$.
To Jennifer, $\vec{\mathbf b}_1$ looks like $(1, 0)$ and $\vec{\mathbf b}_2$ looks like $(0, 1)$.
If Jennifer says "$(-1, 2)$", to us (in the standard basis) this is the vector $-1\vec{\mathbf b}_1 + 2\vec{\mathbf b}_2 = -1 (2, 1) + 2(-1, 1) = (-4, 1)$.
We can also write the above calculation as $\begin{bmatrix}\uparrow & \uparrow \\ \vec{\mathbf b}_1 & \vec{\mathbf b}_2 \\ \downarrow & \downarrow\end{bmatrix}\begin{bmatrix}-1 \\ 2\end{bmatrix} = \begin{bmatrix}2 & -1 \\ 1 & 1\end{bmatrix} \begin{bmatrix}-1 \\ 2\end{bmatrix} = \begin{bmatrix}-4 \\ 1\end{bmatrix}$.
Notice that $\begin{bmatrix}\uparrow & \uparrow \\ \vec{\mathbf b}_1 & \vec{\mathbf b}_2 \\ \downarrow & \downarrow\end{bmatrix} \begin{bmatrix}\uparrow \\ \vec{\mathbf e}_1 \\ \downarrow\end{bmatrix} = \begin{bmatrix}\uparrow \\ \vec{\mathbf b}_1 \\ \downarrow\end{bmatrix}$ and $\begin{bmatrix}\uparrow & \uparrow \\ \vec{\mathbf b}_1 & \vec{\mathbf b}_2 \\ \downarrow & \downarrow\end{bmatrix} \begin{bmatrix}\uparrow \\ \vec{\mathbf e}_2 \\ \downarrow\end{bmatrix} = \begin{bmatrix}\uparrow \\ \vec{\mathbf b}_2 \\ \downarrow\end{bmatrix}$, i.e., this matrix transforms our (standard) basis vectors into Jennifer's basis vectors.
How can we write this using change of basis notation? When Jennifer says "$(-1, 2)$", this is the vector $\vec{\mathbf v} \in \mathbf R^2$ such that, when written in Jennifer's coordinate system, it has coordinates $(-1,2)$. In other words, it is the vector $v$ such that $[\vec{\mathbf v}]^{(\vec{\mathbf b}_1, \vec{\mathbf b}_2)} = \begin{bmatrix}-1 \\ 2\end{bmatrix}$. To find out what this vector means in our coordinate system, we must compute $[\vec{\mathbf v}]^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}$.
We can write $[I]_{(\vec{\mathbf b}_1, \vec{\mathbf b}_2)}^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)} [\vec{\mathbf v}]^{(\vec{\mathbf b}_1, \vec{\mathbf b}_2)} = [\vec{\mathbf v}]^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}$. What is the meaning of the matrix $[I]_{(\vec{\mathbf b}_1, \vec{\mathbf b}_2)}^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}$? The notation means that the columns of the matrix are Jennifer's basis vectors written using our coordinate system. It takes coordinates written in Jennifer's system and translates it into our coordinates. In other words, it translates from Jennifer's language to our language. But geometrically, it transforms our grid into Jennifer's grid. Aren't these two opposites? This is a point made in the video. The idea is to think of the matrix as transforming our misconception of what Jennifer is saying into what she is actually saying. When Jennifer says "$(1,0)$" she actually means $\vec{\mathbf b}_1$, which is the mapping $\vec{\mathbf e}_1 \mapsto \vec{\mathbf b}_1$, which, geometrically, is transforming our basis vector $\vec{\mathbf e}_1$ into Jennifer's basis vector $\vec{\mathbf b}_1$.
Now consider the linear transformation $T : \mathbf R^2 \to \mathbf R^2$ defined by $T\vec{\mathbf e}_1 := \vec{\mathbf b}_1$ and $T\vec{\mathbf e}_2 := \vec{\mathbf b}_2$. Since $(\vec{\mathbf e}_1, \vec{\mathbf e}_2)$ is a basis of $\mathbf R^2$, there is exactly one such linear transformation, i.e., our specification is well-defined. We can check that $T$ is the map $(x,y)\mapsto (2x-y, x+y)$. What is the matrix of $T$? We can look at where it takes the standard basis vectors to see that the first column is $\vec{\mathbf b}_1$ and the second column is $\vec{\mathbf b}_1$, i.e., we have $[T]_{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)} = \begin{bmatrix}\uparrow & \uparrow \\ \vec{\mathbf b}_1 & \vec{\mathbf b}_2 \\ \downarrow & \downarrow\end{bmatrix} = \begin{bmatrix}2 & -1 \\ 1 & 1\end{bmatrix} = [I]_{(\vec{\mathbf b}_1, \vec{\mathbf b}_2)}^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}$.
We should also verify that $[I]_{(\vec{\mathbf b}_1, \vec{\mathbf b}_2)}^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)} = [T]_{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}$. On the one hand, the $k$th column of $[I]_{(\vec{\mathbf b}_1, \vec{\mathbf b}_2)}^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}$ is $[I\vec{\mathbf b}_k]^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)} = [\vec{\mathbf b}_k]^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}$. On the other hand, the $k$th column of $[T]_{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}$ is $[T\vec{\mathbf e}_k]^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)} = [\vec{\mathbf b}_k]^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}$ (the $\vec{\mathbf e}_k$ comes from the basis in the subscript). So the two matrices are indeed equal.
To summarize, we can write the same equation in multiple ways:
Equation Description
$-1\vec{\mathbf b}_1 + 2\vec{\mathbf b}_2 = -1 (2, 1) + 2(-1, 1) = (-4, 1)$ Linear combination of Jennifer's basis vectors
$-1\begin{bmatrix}\uparrow \\ \vec{\mathbf b}_1 \\ \downarrow\end{bmatrix} + 2\begin{bmatrix}\uparrow \\ \vec{\mathbf b}_2 \\ \downarrow\end{bmatrix} = -1 \begin{bmatrix}2 \\ 1\end{bmatrix} + 2\begin{bmatrix}-1 \\ 1\end{bmatrix} = \begin{bmatrix}-4 \\ 1\end{bmatrix}$ Linear combination of Jennifer's basis vectors, written using column vectors
$\begin{bmatrix}\uparrow & \uparrow \\ \vec{\mathbf b}_1 & \vec{\mathbf b}_2 \\ \downarrow & \downarrow\end{bmatrix}\begin{bmatrix}-1 \\ 2\end{bmatrix} = \begin{bmatrix}2 & -1 \\ 1 & 1\end{bmatrix} \begin{bmatrix}-1 \\ 2\end{bmatrix} = \begin{bmatrix}-4 \\ 1\end{bmatrix}$ Matrix multiplication
$[I]_{(\vec{\mathbf b}_1, \vec{\mathbf b}_2)}^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)} [\vec{\mathbf v}]^{(\vec{\mathbf b}_1, \vec{\mathbf b}_2)} = [\vec{\mathbf v}]^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}$ Change of coordinate equation
$T(-1\vec{\mathbf e}_1 + 2\vec{\mathbf e}_2) = -1\vec{\mathbf b}_1 + 2\vec{\mathbf b}_2$ Application of a linear transformation to a vector
$[T]_{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)} \begin{bmatrix}-1 \\ 2\end{bmatrix} = [T]_{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)}^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)} [(-1,2)]^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)} = [T(-1,2)]^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)} = [(-4,1)]^{(\vec{\mathbf e}_1, \vec{\mathbf e}_2)} = \begin{bmatrix}-4 \\ 1\end{bmatrix}$ Matrix of linear transformation in standard coordinates<|endoftext|>
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In 2010, the gene controlling pea flower colour – first studied by Gregor Mendel 150 years earlier – was identified by researchers from New Zealand and overseas.
Mendel’s experiments on peas
Key principles ofwere developed from Gregor Mendel’s ground-breaking experiments on inheritance in the 1860s. Mendel cross-bred pure lines of pea plants (Pisum sativum) and recorded the inheritance of visible traits, such as flower colour, stem length, pod shape and pod colour.
Mendel’s traits in the 21st century
In 2010, a team of researchers from Plant & Food Research in New Zealand, along with collaborators in the UK, France and the US, identified thethat controls flower colour in peas, one of the 7 traits that Mendel followed.
Purple flowers accumulatepigment molecules, whereas white flowers do not. For this reason, the researchers speculated that the gene they were looking for was likely to regulate anthocyanin .
Finding the gene for pea flower colour
The researchers knew approximately where the gene was in the pea– its – because of genetic markers close by. Unfortunately, there wasn’t much information available about the pea genes in this region, and a full genome sequence wasn’t available. By comparing pea genes with genes in a similar region in plants with fully sequenced genomes, like barrel clover and petunia, they found a gene that was likely to affect anthocyanin pigment production.
A gene controlling pigment manufacture
When the researchers looked at the switch on expression of genes. The pea plants with this version of the gene don’t make anthocyanin pigments and have white flowers.sequence of the same gene in white-flowering pea plants. they noticed a difference in just 1 . This single base change, from G to A, stops the gene being made into protein, so it can’t
Confirming the gene function
To test the gene’s function, the researchers expressed the gene transiently in pea petals. They did this using– firing copies of the gene construct into pea petals from a . In areas of the petal where the gene was expressed, the flowers accumulated anthocyanin pigment and were coloured purple.
Identifying Mendel’s other pea genes
This is the 4th of the Mendel’s peagenes to be identified. Researchers have already identified the genes behind plant size, pea seed colour and seed shape.
Anthocyanins: important pigment molecules
Anthocyanin is responsible for red, blue and purple colours seen in many plants. Anthocyanin pigments are antioxidants – they mop up DNA-damaging free radicals that are released during , particularly during high light. This minimises damage to the plant’s DNA.
If we eat plants containing anthocyanin, the pigment molecules seem to make our own systems more effective. This is part of the reason why having a mixed diet of colourful fruits and vegetables is good for our health. This is also why plant breeders are interested in breeding fruits that contain more anthocyanin pigments. Find out more about this in resources on breeding red-fleshed apples.
The full article on the Identification of Mendel’s white flower character is freely available at PLoS One
Video of Roger Hellens explaining the discovery of the gene controlling pea flower colour.<|endoftext|>
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Teaching and Group Activities for Understanding
Day 1 Teaching
Write the number sentence 3 + ☐ = 8. Children discuss. Did they count on or use their fingers? Hold up a strip of 8 cubes and beneath it a strip of 3 cubes. How many more cubes to make 8? Fill in the number sentence 3 + 5 = 8. Repeat for other bonds to 8. Record the bonds to 8. Choose an addition; cover one number, ask children to find it.
-- Find the matching number to make 8.
-- Order number bonds to 8; find matching number bond partners.
Day 2 Teaching
Write 9 + 0, 3 + 6, 7 + 2, 4 + 5, 8 + 1 and show that these are all bonds to 9. Can we write them differently? Show that we can write addition either way round. 3 + 6 = 9 and 6 + 3 = 9. Rehearse using these bonds to write subtraction facts, e.g. 9 – 4 = 5.
-- Making number bonds to 9 with cubes in a feely bag.
-- Play a game creating number bond number sentences with a partner.
Day 3 Teaching
What is double 3? Explain that when we double a number we actually add it to itself, e.g. 4+4, 6+6 etc. Ask children to work in pairs to show doubles to 6 on their fingers.
Draw a 3 by 2 grid on IWB. Fill with numbers 1 to 6. Children copy and write doubles.
Use the investigation suggestions in ‘Double Trouble’ from NRICH for today’s group activity.
Or, use these activities:
-- Fish for doubles, collecting as many fish as they can.
-- Quick-fire game of doubles and halves.<|endoftext|>
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Treaty of Paris
TREATY OF PARIS
The agreement in 1783 between the United States and Great Britain known as the Treaty of Paris formally ended the struggle for American independence. The British "acknowledged" their former colonial subjects as "free, sovereign and independent." Both sides opted for political reconciliation and commercial cooperation rather than continuous hostilities and competition.
In October 1781 Continental troops forced General Charles Cornwallis to surrender his troops in the wake of the Battle of Yorktown. The British decline spurred by this defeat was exacerbated by other defeats to the French and the Spanish on other continents and compounded by accumulating debt. Subsequently, the political situation changed. In March 1782 King George III installed a new cabinet. Its leaders secretly negotiated with senior American diplomats authorized by the Continental Congress. Five men were commissioned. John Adams, John Jay, and Benjamin Franklin (who, being pro-French, initially objected to the talks) conducted the bargaining; Henry Laurens was captured and held by the British; and Thomas Jefferson remained in America until after the deal was sealed. Jefferson was more inclined than the others toward the French perspective, so his absence facilitated an Anglo-American agreement.
On 30 November 1782 the peace treaty was initialed in Paris. It ended the Revolutionary War by February 1783. On 15 April 1783 the preliminary Articles of Peace were ratified by the United States. On 6 August 1783 Great Britain did the same. On 3 September 1783 the Definitive Treaty of Paris (merely adding procedural details) was signed by American and British representatives. On 14 January 1784 this treaty was ratified by the United States and went into formal effect. On 9 April 1784 Britain followed suit.
American and British diplomats sidetracked the ambitious French, although the Americans had explicitly promised in 1778 not to sign a separate treaty. Britain had an interest in making concessions to the United States; doing so positioned the Americans as a potential ally, which aroused the ire of the French. The separate British-U.S. arrangement minimized gains for the French and their Spanish allies. The British exchanged with them territories in the Caribbean, West Africa, and the Mediterranean but maintained their fortress of Gibraltar. The Anglo-Saxon powers totally overlooked the interests of indigenous and racial populations.
As a result of the Treaty of Paris, the British ceded—without compensation—vast territories they possessed to the United States, whose boundaries were set in the Great Lakes and along the Mississippi River and thirty-one degrees north latitude, although New Orleans was excluded. This transfer of sovereignty doubled the size of the original colonies, primarily at the expense of native tribes. The terms, however, compared poorly with American aspirations upon independence in 1776 and what the Continental Congress had stipulated in 1779. Canada remained British. The Mississippi River itself and its navigation did not become exclusively American. Spain regained Florida. The French continued to possess vast territories beyond the Mississippi until the Louisiana Purchase of 1803. American diplomats secured much, but their ability to maneuver amid the conflict of their interests with those of the British, French, and Spanish was limited.
Both American and British sailors were authorized to navigate the Mississippi River. U.S. citizens retained their previous fishing rights to rich British waters such as the Grand Banks and all other banks of Newfoundland as well as the Gulf of St. Lawrence. Americans were also permitted to dry and cure their catch on unsettled beaches in Labrador and Nova Scotia.
The United States pledged that its Congress would "earnestly recommend" to state and local authorities the restoration of property confiscated from British Loyalists during the war, prohibit future expropriation, release the Loyalists from confinement, and halt their persecution. These commitments had a weak legal basis and were rarely observed. Both sides promised that creditors would recover their prewar debts, but implementation was imperfect.
Brecher, Frank W. Securing American Independence: John Jay and the French Alliance. Westport, Conn.: Praeger, 2003.
Stockley, Andrew. Britain and France at the Birth of America: The European Powers and the Peace Negotiations of 1782–1783. Exeter, U.K.: University of Exeter Press, 2001.
Itai Nartzizenfield Sneh
"Treaty of Paris." Encyclopedia of the New American Nation. . Encyclopedia.com. (May 20, 2019). https://www.encyclopedia.com/history/encyclopedias-almanacs-transcripts-and-maps/treaty-paris
"Treaty of Paris." Encyclopedia of the New American Nation. . Retrieved May 20, 2019 from Encyclopedia.com: https://www.encyclopedia.com/history/encyclopedias-almanacs-transcripts-and-maps/treaty-paris
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Study Materials
# NCERT Solutions for Class 10th Mathematics
Page 1 of 4
## Chapter 1. Real Numbers
### Exercise 1.1
1. Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Sol:
(1) 135 and 225
a = 225, b = 135 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
225 = 135 ×1 + 90
135 = 90 ×1 + 45
90 = 45 × 2 + 0 {when we get r=0, our computing get stopped}
b = 45 {b is HCF}
Hence: HCF = 45
Sol:
(ii) 196 and 38220
a = 38220, b = 196 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}
b = 196 {b is HCF}
Hence: HCF = 196
Sol:
(iii) 867 and 255
a = 867, b = 255 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}
b = 196 {b is HCF}
Hence: HCF = 196
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Sol:
Let a is the positive odd integer
Where b = 6,
When we divide a by 6 we get reminder 0, 1, 2, 3, 4 and 5, {r < b}
Here a is odd number then reminder will be also odd one.
We get reminders 1, 3, 5
Using Euclid’s division algorithm
So we get
a = 6q + 1, 6q+3 and 6q+5
3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Sol:
Maximum number of columns = HCF (616, 32)
a = 616, b = 32 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
616 = 32 ×19 + 8 {when we get r=0, our computing get stopped}
32 = 8 × 4 + 0
b = 8 {b is HCF}
HCF = 8
Hence: Maximum number of columns = 8
4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]
Solution :
To Show :
a2 = 3m or 3m + 1
a = bq + r
Let a be any positive integer, where b = 3 and r = 0, 1, 2 because 0 ≤ r < 3
Then a = 3q + r for some integer q ≥ 0
Therefore, a = 3q + 0 or 3q + 1 or 3q + 2
Now we have;
a2 = (3q + 0)2 or (3q + 1)2 or (3q +2)2
a2 = 9q2 or 9q2 + 6q + 1 or 9q2 + 12q + 4
a2 = 9q2 or 9q2 + 6q + 1 or 9q2 + 12q + 3 + 1
a2 = 3(3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1
Let m = (3q2) or (3q2 + 2q) or (3q2 + 4q + 1)
Then we get;
a2 = 3m or 3m + 1 or 3m + 1
5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let , a is any positive integer
By using Euclid’s division lemma;
a = bq + r where; 0 ≤ r < b
Putting b = 9
a = 9q + r where; 0 ≤ r < 9
when r = 0
a = 9q + 0 = 9q
a3 = (9q)3 = 9(81q3) or 9m where m = 81q3
when r = 1
a = 9q + 1
a3 = (9q + 1)3 = 9(81q3 + 27q2 + 3q) + 1
= 9m + 1 where m = 81q3 + 27q2 + 3q
when r = 2
a = 9q + 2
a3 = (9q + 2)3 = 9(81q3 + 54q2 + 12q) + 8
= 9m + 2 where m = 81q3 + 54q2 + 12q
⇒ The End
Page 1 of 4
Chapter Contents:<|endoftext|>
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Battle of Jackson, Mississippi
The Battle of Jackson, fought on May 14, 1863, in Jackson, Mississippi, was part of the Vicksburg Campaign in the American Civil War. Union commander Maj. Gen. Ulysses S. Grant and the Army of the Tennessee defeated elements of the Confederate Department of the West, commanded by General Joseph E. Johnston, seizing the city, cutting supply lines, and opening the path to the west and the Siege of Vicksburg.
On May 9, Gen. Johnston received a dispatch from the Confederate Secretary of War directing him to "proceed at once to Mississippi and take chief command of the forces in the field." As he arrived in Jackson on May 13, from Middle Tennessee, he learned that two army corps from the Union Army of the Tennessee—the XV, under Maj. Gen. William Tecumseh Sherman, and the XVII, under Maj. Gen. James B. McPherson—were advancing on Jackson, intending to cut the city and the railroads off from Vicksburg, Mississippi which was a major port on the Mississippi River. These corps, under the overall command of Grant, had crossed the Mississippi River south of Vicksburg and driven northeast toward Jackson. The railroad connections were to be cut to isolate the Vicksburg garrison. And if the Confederate troops in Jackson were defeated, they would be unable to threaten Grant's flank or rear during his eventual assault on Vicksburg.
Johnston consulted with the local commander, Brig. Gen. John Gregg, and learned that only about 6,000 troops were available to defend the town. Johnston ordered the evacuation of Jackson, but Gregg was to defend Jackson until the evacuation was completed. By 10 a.m., both Union army corps were near Jackson and had engaged the enemy. Rain, Confederate resistance, and poor defenses prevented heavy fighting until around 11 a.m., when Union forces attacked in numbers and slowly but surely pushed the enemy back. In mid-afternoon, Johnston informed Gregg that the evacuation was complete and that he should disengage and follow.
Soon after, Union troops entered Jackson and had a celebration in the Bowman House, hosted by Grant, who had been traveling with Sherman's corps. They then burned part of the town and cut the railroad connections with Vicksburg. Johnston's evacuation of Jackson was premature because he could, by late on May 14, have had 11,000 troops at his disposal and by the morning of May 15, another 4,000. The fall of the former Mississippi state capital was a blow to Confederate morale.
General Sherman appointed Brig. Gen. Joseph A. Mower to the position of military governor of Jackson and ordered him to destroy all facilities that could benefit the war effort. With the discovery of a large supply of rum, it was impossible for Mower's Brigade to keep order among the mass of soldiers and camp followers, and many acts of pillage took place. Grant left Jackson on the afternoon of May 15 and proceeded to Clinton, Mississippi. On the morning of May 16 he sent orders for Sherman to move out of Jackson as soon as the destruction was complete. Sherman marched almost immediately, clearing the city by 10 a.m.. By nightfall on May 16, Sherman's corps reached Bolton, Mississippi, and the Confederacy had reoccupied what remained of Jackson. Jackson had been destroyed as a transportation center and the war industries were crushed. But more importantly the Confederate concentration of men and materials aimed at saving Vicksburg were scattered. Sherman would later lead an expedition against Jackson following the fall of Vicksburg to clear Johnston's relief force from the area.<|endoftext|>
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# What is the weight of a 70 kg body on the surface of a planet whose mass is ${{\dfrac{1}{7}}^{th}}$ that of the earth and radius is half of the earth?A. 20 kgB. 40 kgC. 70 kgD. 140 kg
Last updated date: 19th Sep 2024
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Hint: Weight of a body on any planet is given by mg, where g is the gravitational acceleration on the planet. So, to find the weight of this body on another planet we need to find the gravitational acceleration g’ on the surface of that planet arising due to the gravitational attraction. Which depends on the mass and the radius of the planet.
Gravitational attraction force, which is essentially the weight measured on a planet, exerted by the planet on a body of mass m is given as:
$\overset{\to }{\mathop{F}}\,=\dfrac{GMm}{{{R}^{2}}}$
Where,
M is the mass of the planet
m is the mass of body
R is the radius of the planet
G is the gravitational constant
Since we know that weight is also represented mathematically as mg.
Let us take the mass of earth as M and radius as R. Therefore, the acceleration experienced by the object of mass m on earth will be g:
$g=\dfrac{GM}{{{R}^{2}}}$
According to the question:
Mass of that planet M’ $=\dfrac{M}{7}$ where M is the mass of earth
Radius of that planet R’ $=\dfrac{R}{2}$ where R is the Radius of earth
Now we can put these values in the above given equation as;
\begin{align} & g'=\dfrac{GM'}{{{R}^{'2}}} \\ & \Rightarrow g'=\dfrac{4GM}{7{{R}^{2}}} \\ & \Rightarrow g'=\dfrac{4}{7}g \\ \end{align}
As we know that mg = 70kg is the weight on earth. Multiplying m on both sides:
\begin{align} & mg'=\dfrac{4}{7}mg \\ & \Rightarrow mg'=\dfrac{4}{7}70kg=40kg \\ \end{align}
So, the weight of a 70 kg body on the given planet will be 40kg. Option B.
Note: Gravitational force is a conservative force, which means that work done by this force doesn’t depend on the path followed. Gravitational force’s effect decreases with increase in distance, and increases with increase in mass.<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Induction and Factors
## Proving divisibility using properties of integers and inductive proofs.
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Progress
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Induction and Factors
Proof by induction is common in mathematics. In this lesson we will use it to prove different kinds of hypotheses and to practice the use of the process in different situations. Specifically, we will be applying the principles of induction to prove various statements regarding factors and factorability.
### Watch This
Khan Academy: Proof by Induction
### Guidance
There are two properties of integers and their factors that will be useful for the proofs in this lesson.
Property 1: If a is a factor of b, and a is a factor of c, then a is a factor of the sum b + c.
For example, the set of numbers a = 3, b = 6, and c = 9 satisfies the hypothesis because 3 is a factor of 6 and 3 is a factor of 9. Property 1 states that therefore 3 is a factor of 9 + 6 = 15, which we know is true because 3 × 5 = 15. We can prove that this property is true for all integers if we think about what the term factor means. If a is a factor of b, then there exists some integer M such that aM = b. Similarly if a is a factor of c, then there exists some integer N such that aN = c. So we can write the sum b + c as aM + aN. We know
b + c = aM + aN = a(M + N).
Because we can write the sum as a product of a and another number, a is a factor of the sum b + c.
Property 2: If a is a factor of b and b is a factor of c, then a is a factor of c.
We can prove this property in a similar manner. If a is a factor of b, then there exists some integer M such that aM = b. If b is a factor of c, then there exists some integer N such that bN = c. We can write c in the following manner:
c = bN = (aM)N = a(MN)
Therefore a is a factor of c because we can write c as the product of a and another integer, MN.
As noted above, these properties of integers are useful for proving statements about integers and factors via induction.
#### Example A
Prove that 3 is a factor of 4n - 1 for all positive integers n.
Solution:
Proof by induction:
1. The base case: if n = 1, then 4n - 1 = 4 - 1 = 3. 3 is a factor of itself because 3 × 1 = 3 .
If the base case does not convince you, you can always test out additional values of n. For example, if n = 2, 42 - 1 = 16 - 1 = 15 = 5 × 3.
2. The inductive hypothesis: assume that 3 is a factor of 4k - 1.
3. The inductive step: show that 3 is a factor of 4k+1 - 1.
If 3 is a factor of 4k - 1, then there exists some integer M such that 3M = 4k - 1. We can write 4k+1 - 1 in a manner that allows us to use the inductive hypothesis:
4k+1 - 1
4(4k-1) + 3 Factor out 4, but add 3 in order to keep the value of 4k+1-1
4(3M) + 3 Substitute: 3M = 4k - 1
Note that the substitution is the same as using property 2 above: if 3 is a factor of 4k - 1 , then 3 is a factor of 4(4k - 1). Using the substitution simply makes the fact a bit more obvious.
This last step proves that 3 is a factor of 4k+1 - 1 , by property 1 above.
The technique of rewriting the k + 1 term can also be used to prove statements about polynomials and factors.
#### Example B
Prove that x - y is a factor of xn - yn for all positive integers n.
Note: Since we are talking about polynomials that are factorable now, not integers, then we say that if x - y is a factor of xn - yn, then there exists a polynomial P such that P(x - y) = xn - yn.
Solution:
Proof by induction:
1. The base case: If n = 1, we have xn - yn = x - y , and x - y is a factor of itself, as x - y = 1(x - y).
As we did above, we can also check n = 2 in order to convince ourselves. If n = 2, we have x2 - y2 = (x - y)(x + y), so x - y is clearly a factor.
2. The inductive hypothesis: assume that x - y is a factor of xk - yk.
3. Show that x - y is a factor of xk+1 - yk+1.
From the inductive step, we know that there is some polynomial P such that P(x - y) = xk - yk. We can rewrite xk+1 - yk+1 in a manner that allows us to use the inductive hypothesis:
xk+1 - yk+1
=xk+1 - xyk + xyk - yk+1
=x(xk - yk) + yk(x - y)
=x(P(x - y)) + yk(x - y)
=Px(x - y) + y'k-1(x - y)
Again, by property 1 above, this shows that x - y is a factor of xk+1 - yk+1. Therefore we have shown that x - y is a factor of xn - yn for all positive integers n.
#### Example C
a) Without adding, determine if 7 a factor of 49 + 70.
b) Consider the sum 23 + 54 = 77. Is 7 a factor of 77? What does this tell you about the first factor property in the lesson?
Solution:
a) Use Property 1 from the lesson: If a is a factor of b, and a is a factor of c, then a is a factor of the sum b + c.
7 is a factor of 49, since \begin{align*}7 \times 7 = 49\end{align*}
7 is a factor of 70, since \begin{align*}7 \times 10 = 70\end{align*}
Therefore 7 is a factor of 119, since \begin{align*}49 + 70 = 119\end{align*}
b) This is a test of the converse of Property 1, which would be "If a number is a factor of the sum, then it is a factor of the factors of the sum"
7 is a factor of the sum: 77
7 is not a factor of 23 or 54
This tells us that the converse of the property is not necessarily true.
-->
### Guided Practice
1) Prove that 9n - 1 is divisible by 8 for all positive integers n.
2) Prove that xn - 1 is divisible by x - 1 for all positive integers n.
3) Prove that n2 - n is even for all positive integers n.
4) Prove that 52n-1 + 1 is divisible by 6 for all positive integers n.
Answers
1) Prove that 9n - 1 is divisible by 8 for all positive integers n.
1. Base case: If n = 1, 9n - 1 = 9 - 1 = 8 = 8(1)
2. Inductive hypothesis: Assume that 9k - 1 is divisible by 8.
3. Inductive step: Show that 9k+1 - 1 is divisible by 8.
9k - 1 divisible by 8 \begin{align*}\Rightarrow\end{align*} 8W = (9k - 1) for some integer W
9k + 1 - 1 = 9 (9k - 1) + 8 = 9 (8W) + 8, which is divisible by 8
2) Prove that xn - 1 is divisible by x - 1 for all positive integers n.
1. Base case: If n = 1, xk - 1 = x - 1 = (x - 1)(1)
2. Inductive hypothesis: Assume that xk - 1 is divisible by x - 1
3. Inductive step: Show that xk+1 - 1 is divisible by x - 1.
xk - 1 divisible by x - 1 \begin{align*}\Rightarrow\end{align*} P(x - 1) = (x k - 1) for some polynomial P
xk + 1 - 1 = x(xk - 1) + (x - 1) = Px(x - 1) + (x - 1),which is divisible by x - 1
3) Prove that n2 - n is even for all positive integers n
1. Base case: If n = 1, 12 - 1 = 1 - 1 = 0 = 2 × 0
2. Inductive hypothesis: Assume that k2 - k is even
3. Inductive step: Show that (k + 1)2 - (k + 1) is even.
If k2 - k is even, then k2 - k = 2 M for some integer M
(k + 1)2 - (k + 1) = k2 + 2k + 1 - k - 1 = k2 - k + 2k = 2M + 2k = 2(M + k) which is even because M + K is an integer.
4) Prove that 52n - 1 + 1 is divisible by 6 for all positive integers n.
1. Base case: If n = 1, 51 + 1 = 5 + 1 = 6 = 6(1)
2. Inductive hypothesis: Assume that 52k-1 + 1 is divisible by 6.
3. Inductive step: Show that 52(k + 1) - 1 + 1 is divisible by 6.
If 52k - 1 + 1 is divisible by 6, then 52k - 1 + 1 = 6M for some integer M.
52(k + 1) - 1 + 1 = 52k + 1 + 1 = 52 (52k - 1 + 1) - 24 = 52 (6M) - 24 which is divisible by 6.
### Explore More
1. Without adding, determine if 7 a factor of 49 + 70
2. Consider the sum \begin{align*}23 + 54 = 77\end{align*} Is 7 a factor of 77? What does this tell you about the first factor property in the lesson?
3. Prove that any positive integer n > 1 a) is prime or, b) can be represented as a product of prime factors.
4. Found within set "J" are all positive integers, from the number 1 to 2n. Prove that there are two numbers, one that is a factor of another, from any (n = 1) numbers chosen from set "J"
5. Prove that \begin{align*}(x^n + \frac{1}{x^n})\end{align*} is also an integer for any positive integer n if the following is an integer: \begin{align*}(x + \frac{1}{x})\end{align*}
6. Prove the formula \begin{align*}n_{k + m} = n_{k-1} u_m + n_k n_{m+1}\end{align*}for the sequence of Fibonacci numbers: \begin{align*}n_1 = 1, n_2 = 1, u_{k+1}= n_k + n_{k-1}, k = 2, 3...\end{align*}
Prove the following identities.
1. \begin{align*}1^2 + 2^2 + 3^2 + ... +n^2 =\frac{ n(n + 1)(2n + 1)}{6}\end{align*}
2. \begin{align*}1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n + 1)^2}{4}\end{align*}
3. \begin{align*}1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + n(n + 1)(n + 2) = \end{align*} \begin{align*}\frac{n(n + 1)(n + 2) (n + 3)}{4}\end{align*}
4. \begin{align*}1 x \cdot 1! + 2 x \cdot 2! + ... + n \cdot n! = (n + 1)! - 1\end{align*}
5. \begin{align*}n^2 \frac{(n + 1)^2}{4} = 1^3 + 2^3 + 3^3 + ... + n^3\end{align*}
6. \begin{align*}n (n + 1) \frac{(2n + 1)}{6} = 1^2 + 2^2 + 3^2 + ... + n^2\end{align*}
Prove the following divisibilities.
1. Prove that \begin{align*}n \frac{(n+1)}{2}\end{align*} is a factor of \begin{align*}1 + 2 + 3 +... + n\end{align*} for all positive integers \begin{align*}n\end{align*}
2. Prove that 3 is a factor of \begin{align*}n^3 + 2n\end{align*} for all positive integers n.
3. Prove that \begin{align*}n^2 \frac{(n + 1)^2}{4}\end{align*} is a factor of \begin{align*}1^3 + 2^3 + 3^3 + ... + n^3\end{align*}
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 7.7.
### Vocabulary Language: English
factor
factor
Factors are the numbers being multiplied to equal a product. To factor means to rewrite a mathematical expression as a product of factors.
induction
induction
Induction is a method of mathematical proof typically used to establish that a given statement is true for all positive integers.
Integer
Integer
The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
proof
proof
A proof is a series of true statements leading to the acceptance of truth of a more complex statement.
### Explore More
Sign in to explore more, including practice questions and solutions for Induction and Factors.
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The uvea is the part of the eye made up of the iris, the ciliary body and the choroid. The choroid is the middle layer or vascular tunic of the eye located between the sclera, which is the fibrous protective outer coat (the white of the eye) and the retina, which is the light sensitive surface within the eye.
The iris and the ciliary body together form the anterior uveal tract. The iris is the pigmented or colored membrane behind the cornea (clear outer surface of the eye). It is responsible for the color of the eye. The ciliary body is a thickened extension of the choroid and connects the choroid and the iris. The ciliary body produces a fluid called the aqueous humor that provides important nutrients to the eye and that maintains intra-ocular pressure (IOP). The ciliary body contains the suspensory ligament and ciliary muscles which support the lens and control its shape and hence its ability to focus images.
Uveitis is an inflammation of one or more of the structures making up the uvea. If all three structures are involved, ciliary body, iris and choroid, is the inflammation is called true uveitis or pan-uveitis. If only the ciliary body and the iris are inflamed it is called iridocyclitis or anterior uveitis, while inflammation of the choroid is choroiditis or posterior uveitis.
What causes uveitis?
There are many potential causes of uveitis. Sometimes the true cause is never discovered. Common causes are:
- Infection - viral, bacterial, parasitic or fungal
- Metabolic disease
- Diabetes mellitus
- High blood pressure
- Immune mediated - particularly autoimmune disease where the dog produces antibodies against its own tissues
- Trauma to the eye
- Lens damage resulting in the leakage of lens protein, and tumors
What are the clinical signs of uveitis?
"The usual signs of uveitis are severe pain with an intense reddening of the visible parts of the eye."
The usual signs of uveitis are severe pain with an intense reddening of the visible parts of the eye. The eye is usually kept shut and most pets avoid bright lights. Cloudiness of the eye may be noticed. Sometimes there is bleeding into the eye. There may be excessive tearing.
How is uveitis diagnosed?
Many of the signs of uveitis are similar to glaucoma. With uveitis, intraocular pressure (IOP) is reduced (low) whereas with glaucoma it is elevated (high). Measurement of IOP is often performed to differentiate between the two conditions and is a simple, painless procedure. A complete and thorough physical examination of the pet must be performed since generalized illnesses can have uveitis as one of their signs. Often there is a color change of the iris, which may be permanent. Special diagnostic procedures such as ultrasound may be used to examine the eye.
What is the treatment of uveitis?
Treatment is initially aimed at reducing inflammation and providing pain relief. Treatment of uveitis due to trauma can involve repair of any corneal tears or removal of a foreign body in the eye. This may involve referral to a specialist. A combination of drops or ointments combined with tablets may be required.
In order to assess the initial response to treatment, your veterinarian will need to examine your dog frequently.
What is the prognosis of uveitis?
When properly treated, most cases of uveitis begin to improve within twenty-four hours. If the eye is very cloudy or if hemorrhage has occurred, this may take a few more days to clear.
Complications are more common after very severe or recurrent cases of uveitis. These can involve the development of synechiae and glaucoma. Synechiae are adhesions between the lens and the iris. Glaucoma is an increase in IOP. Both complications may need specialist treatment. Severe uveitis can result in irreversible blindness.<|endoftext|>
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# If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB - Mathematics
Sum
If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB
#### Solution
Since, AC is a diameter line, so angle in semi-circle formed is 90°.
i.e., ∠ABC = 90°
In ∆ABC,
∠CAB + ∠ABC + ∠BCA = 180° .....[Angle sum property]
⇒ ∠CAB + ∠BCA = 180°-90° = 90° ......(i)
Since, diameter of a circle is perpendicular to the tangent.
i.e., CA ⊥ AT
∴ ∠CAT = 90°
⇒ ∠CAB + ∠BAT =90° .......(ii)
From equation (i) and (ii),
∠CAB + ∠ACB = ∠CAB + BAT
⇒ ∠ACB = ∠BAT
Concept: Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles
Is there an error in this question or solution?
#### APPEARS IN
NCERT Mathematics Exemplar Class 10
Chapter 9 Circles
Exercise 9.4 | Q 4 | Page 110
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Example: Finding 3 Consecutive Integers by Forming an Equation in One Variable
Post to:
Explanation:
Consecutive integer
Consecutive integers are integers that follow each other in order. They have a difference of 1 between every two numbers. (Our solved example in mathguru.com uses this concept)
In a set of consecutive integers, the mean and the median are equal.
If n is an integer, then n, n+1, and n+2 will be consecutive integers.
Examples:
1, 2, 3, 4, 5
-3, -2, -1, 0, 1, 2
1004, 1005, 1006
http://simple.wikipedia.org/wiki/Consecutive_integer
# Equation
An equation is a mathematical statement that asserts the equality of two expressions. Equations consist of the expressions that have to be equal on opposite sides of an equal sign as in
## Properties
If an equation in algebra is known to be true, the following operations may be used to produce another true equation:
1. Any real number can be added to both sides.
2. Any real number can be subtracted from both sides.
3. Any real number can be multiplied to both sides.
4. Any non-zero real number can divide both sides.
5. Some functions can be applied to both sides. Caution must be exercised to ensure that the operation does not cause missing or extraneous solutions. For example, the equation y*x=x has 2 solutions: y=1 and x=0. Dividing both sides by x "simplifies" the equation to y=1, but the second solution is lost.
http://en.wikipedia.org/wiki/Equation
# Equation solving
In mathematics, to solve an equation is to find what values (numbers, functions, sets, etc.) fulfill a condition stated in the form of an equation (two expressions related by equality). These expressions contain one or more unknowns, which are free variables for which values are sought that cause the condition to be fulfilled.
A solution of the equation is an assignment of expressions to the unknowns that satisfies the equation; in other words, expressions such that, when they are substituted for the unknowns, the equation becomes a tautology (a provably true statement). (Our solved example in mathguru.com uses this concept)
http://en.wikipedia.org/wiki/Equation_solving<|endoftext|>
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Instructional designers today are not only tasked with traditional design projects but are expected to design with a more human-centered approach. The design thinking process can help designers achieve that goal. It consists of five stages that aren’t necessary linear but require the designer to go back and forth in order to create the best learning experience. The five stages are empathize, define, ideate, prototype and test.
Designers connect with learners and gain more insight in the empathize and define stages by interviewing learners, observing them, and creating empathy maps and learner personas. Once designers interpret the data from these stages, they define a problem statement and are ready to ideate. In the ideation stage, designers look at every possible angle for their problem statement.
Equipped with a multitude of ideas, designers then move to the prototype stage. In this stage, the design team produces a number of inexpensive, scaled down versions of the solution. The goal is to determine which is the best possible solution for the design challenge at hand. It is recommended to share prototypes within the team and, if possible, with a wider audience. In this experimental phase, designers accept, improve, and re-examine or reject solutions based on the users’ experience. Through this process, the design team will be able to predict how a real user would behave, think and feel when interacting with the learning solution. In this stage, designers move from an abstract idea to a more tangible product.
Types of Prototyping
Prototyping doesn’t have to be time-consuming, expensive or difficult. Different levels of fidelity (detail, functionality or interactivity) allow designers to come up with solutions quickly. Here are some ideas.
Low-fidelity prototypes don’t have a lot of detail, images or colors. Instead, they use placeholders for images and text but show the flow and functionality of a solution. You can create this type of prototype on paper; it can be a simple visualization and doesn’t require expensive software. Storyboarding, sketching and card-sorting are three inexpensive and simple ways to create low-fidelity prototypes.
Low-fidelity prototyping is quick and inexpensive and allows for instant changes. It helps designers develop an overall view of the end solution with minimal time and effort. On the other hand, low-fidelity prototyping has a lack of realism and may lack validity.
High-fidelity prototypes look and function closer to the finished product. It could involve creating a 3D model with moveable parts or using a prototype tool to build out the experience on a phone or desktop computer. This prototype has most design assets and components developed and integrated. High-fidelity prototypes are more engaging for the users to test, and it is easier for designers to judge if the solution will work. On the other hand, they take longer to create, and users might focus more on small details instead of the content the prototype is presenting. It takes longer to update high-fidelity prototypes as well, and sometimes, it is difficult for a designer to make changes after hours have been invested in building the prototype.
Guidelines for Prototyping
You might want to start with low-fidelity prototypes and then move to high-fidelity prototyping at a later stage in the design process. Don’t overthink this process; simply start building a prototype. It will help you think about the learning solution in a different way and potentially give you different insights. Don’t spend so much time, money and resources on the prototype that you become attached to it, in case you have to make changes later. Never lose sight of the issue you are trying to test with your prototype, and always build with the user in mind.<|endoftext|>
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Congratulations - you have completed . You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%%
Your answers are highlighted below.
What was the main difference between Federalists and Anti-Federalists?
Federalists wanted independence from Britain, but Anti-Federalists wanted to remain as British colonies.
Anti-Federalists wanted independence from Britain, but Federalists wanted to remain as British colonies.
Federalists wanted a weak national government and strong state governments, but Anti-Federalists wanted a strong national government and weak state governments.
Anti-Federalists wanted a weak national government and strong state governments, but Federalists wanted a strong national government and weak state governments.
Question 1 Explanation:
The fight between the Federalists and Anti-Federalists was essentially the fight over whether the Articles of Confederation would remain the core of the American government (the will of the Anti-Federalists) or if the Articles would be scrapped for a new arrangement that involved a much stronger federal government (the will of the Federalists).
What was the main difference between the Virginia Plan and the New Jersey Plan brought forth at the Constitutional Convention?
The Virginia Plan called for legislative representatives to be proportional to the number of citizens in each state, while the New Jersey Plan called for a single representative for each state regardless of population.
The New Jersey Plan called for the abolition of slavery while the Virginia Plan called for legalized slavery in all 13 states.
The New Jersey Plan called for an electoral college system to elect the president, while the Virginia Plan called for a popular election.
The Virginia Plan and New Jersey plan were virtually identical; each state just wanted credit for the idea.
Question 2 Explanation:
The dispute over the two plans would end in compromise. The bicameral legislature laid out in the United States Constitution is a mix of each style of representation: The House of Representatives has representatives in number proportional to each state’s population, and the Senate has two representatives for each state regardless of population.
Why did Anti-Federalists, such as George Mason, push for the inclusion of a bill of rights in the United States Constitution.
Anti-Federalists wanted to ensure the Constitution would not gain enough state approval to replace the Articles of Confederation.
Anti-Federalists were worried that without a guarantee of certain rights, the federal government could overstep their bounds and encroach upon its citizens’ freedoms.
Anti-Federalists wanted to mirror the constitutions of other successful governments.
Anti-Federalists wanted to give the federal government more power.
Question 3 Explanation:
Although its creation was largely a nod to the American Anti-Federalist movement, the Bill of Rights would go on to become one of the most important aspects of the United States Constitution. By clearly defining the essential rights of citizens rather than simply spelling out the powers of government, the United States government was established with the freedoms of its people at the forefront.
Which of the following best describes the type of government created by the Articles of Confederation and the United States Constitution?
Question 4 Explanation:
The United States is a republic where citizens elect representatives to make governmental decisions on their behalf.
How did the political power of the American federal government change from The Articles of Confederation to The Constitution?
The federal government’s power was greatly increased under the Constitution.
The federal government was weakened under the Constitution.
State governments were given more power under the Constitution.
The federal government lost the power to tax under the Constitution.
Question 5 Explanation:
The Articles of Confederation created a government where states held the majority of the power to govern themselves. The federal government existed primarily as a means to unify the states for military purposes should the need arise. In contrast, the Constitution gave the federal government significantly more power to do things like tax, regulate trade, and settle disputes between states.
Which of the following best describes the election of George Washington as America’s first president:
The American public elected George Washington in a landslide victory.
George Washington was elected unanimously by the Electoral College.
George Washington narrowly defeated John Adams to become the first President.
George Washington was not elected to the Presidency; he was given this role due to his service and leadership during the Revolutionary War.
Question 6 Explanation:
The first presidential election was held in January of 1789. George Washington was reluctant to get involved, but Alexander Hamilton and others helped convinced him to run. In 1789 only Pennsylvania and Maryland held elections in order to choose their presidential electors. State legislatures chose the electors in all of the other states. Washington received all 69 of the electoral votes, and John Adams was elected as Vice President.
Who did George Washington name as the first Secretary of the Treasury?
Question 7 Explanation:
Washington was impressed with Alexander Hamilton’s business exploits as well as his sound understanding of economics. Without Hamilton’s leadership, America would have had a difficult time navigating the financial hardships associated with starting a new nation.
What tactic did Alexander Hamilton use to get the Southern states to agree to his plan for the federal government to pay off the states’ war debts?
Hamilton threatened to withhold future federal funds from states that did not go along with his repayment plan.
States that did not support his plan would be considered outside the realm of American military protection.
Hamilton agreed to move the nation’s capital into the South.
Hamilton was able to gather enough votes in Congress to overrule the Southern states’ objections.
Question 8 Explanation:
Washington, D.C. remains America’s capital to this day thanks to this deal struck by Hamilton. By using land from Virginia and Maryland to build the new capital, the Southern states felt they had a true stake in the new nation despite paying for war debts that were mainly accumulated by the Northern states.
What was the main source of revenue for the new American government in the 1790s?
Tariffs on imports
Donations from the states
Question 9 Explanation:
Nearly all of the money raised by the early American government was through taxes on goods imported from Europe. Hamilton’s hope was that the tariffs would help America raise money in the short term while motivating Americans to develop their own industry for long term growth.
Which of the following best describes Hamilton’s argument in favor of establishing a national bank:
A national bank would enable the government to issue loans to struggling Americans.
A national bank would provide a safe place to keep the money that the United States had acquired from Britain after winning the Revolutionary War.
The United States would lose business to the banks of other nations if they did not create one of their own.
A national bank was necessary in order to stabilize and improve the nation's credit.
Question 10 Explanation:
Hamilton stated that, "The tendency of a national bank is to increase public and private credit. The former gives power to the state for the protection of its rights and interests, and the latter facilitates and extends the operations of commerce amongst individuals."
What brought about the end of the Whiskey Rebellion?
President George Washington asserted the Presidency's powers to enforce federal law, by force when necessary.
The whiskey tax was repealed.
Western Pennsylvania began the process of creating a new government independent of the United States.
The federal government raised taxes further on bartered goods.
Question 11 Explanation:
Pennsylvania farmers had the right to protest the whiskey tax that they felt unfairly targeted them and their largely barter-based economy. However, when farmers refused to pay the tax, they were in violation of the law. When it became clear that the taxes were not being paid and the protests were turning violent, Washington sent in the army to successfully end the rebellion. Washington’s response to the 1794 rebellion was one of the first major tests for the young republic.
How did President George Washington’s involve America in the French Revolution?
He supplied the French rebels with weapons, but decided not to commit American soldiers to the conflict.
He couldn’t spare money or supplies for the French rebels, but he did commit American soldiers to the conflict.
He sent supplies and soldiers to France in order to help the French rebels overthrow their government.
He decided to remain neutral in the conflict.
Question 12 Explanation:
Despite the fact that France provided America with military and financial support during the Revolutionary War, Washington issued a formal Proclamation of Neutrality in regards to the French Revolution. America was facing its own challenges with Native Americans and a struggling economy, so Washington did not want to get the nation involved in European affairs.
Why did George Washington only serve two terms as president?
He lost reelection to his vice president, John Adams.
The Constitution barred presidents from serving a third term.
He did not seek reelection to a third term.
He was too sick to remain in office.
Question 13 Explanation:
While he was under no obligation to do so, Washington refused to seek a third term (despite many who urged him to maintain his post). This established the tradition of a maximum of two terms for a president. The only president to serve more than two terms was Franklin D. Roosevelt. Washington's farewell address warned against political parties and foreign entanglements.
Which of the following best describes the ideological differences between the Federalist Party and the Democratic-Republican Party?
Federalists were in favor of a strong national government and a loose interpretation of the Constitution. while the Democratic-Republicans favored stronger state governments and a stricter interpretation of the Constitution.
Federalists were focused on protecting farmers and the working class from government overreach, but Democratic-Republicans were more concerned with bolstering the nation’s industries and supporting the wishes of the wealthy.
Federalists wanted to expand international trade, while Democratic-Republicans wanted to use tariffs to protect American business.
None of the above.
Question 14 Explanation:
The divide between the Federalists and Democratic-Republicans was an extension of the fight that had happened over the Articles of Confederation and Constitution years earlier. The divide would lead to a presidential election in 1796 where the American people would face a choice between candidates of rival parties with vastly different visions for the new nation.
What was the result of the election of 1796?
Federalist John Adams was elected president and Democratic-Republican Thomas Jefferson was elected vice president.
Federalist John Adams was elected president and fellow Federalist Charles Pinckney was elected vice president.
Democratic-Republican Thomas Jefferson was elected president and Federalist John Adams was elected vice president.
Democratic-Republican Thomas Jefferson was elected president and fellow Democratic-Republican Aaron Burr was elected vice president.
Question 15 Explanation:
Adams won the presidency with 71 electoral votes. According to the Constitution, the second-place vote-getter in the Electoral College was awarded the vice presidency. Following these rules, Thomas Jefferson became vice president, resulting in an executive branch made up of both parties.
What was the XYZ Affair?
A successful covert attack on the United States by France.
The Adams administration’s failed attempt to declare English the national language of the United States.
A British plan to trade with the United States that avoided the steep import tariffs.
The failed attempt by French government agents to extort a bribe from the United States.
Question 16 Explanation:
In 1797, France sent three agents to meet with an American delegation to demand a bribe. In exchange for a cash payment and a hefty loan, the French would stop capturing American trade ships bound for Britain. The United States refused the bribe request and launched into a “quasi-war” with France that lasted until 1800.
What powers were given to the president with the passage of the Alien Acts of 1798?
The president could investigate UFO sightings and negotiate with extraterrestrial civilizations.
The president could imprison and deport non-citizens who were deemed dangerous or who were from a hostile nation.
The president could change immigration policies without the approval of Congress.
None of the above.
Question 17 Explanation:
The Alien Acts were a direct response to the growing tensions with France prior to the turn of the 19th century. American citizens were worried about how immigrants who were not legal citizens would act if war were to break out with France or another European nation.
Why did the Sedition Act of 1798 face staunch opposition from the American public?
The law encroached on American citizen’s rights to free speech and assembly.
The law could not effectively be enforced.
The law was too weak to achieve its goals.
The law was unclear and hard to understand.
Question 18 Explanation:
The Sedition Act made it illegal to speak against the United States government either aloud or in print. The Bill of Rights guaranteed the right to these freedoms, therefore the law was deemed unconstitutional. Some states used their constitutional right to quickly nullify the law prior to its eventual repeal.
How did the passage of the Naturalization Act of 1798 affect the path to American citizenship?
It decreased the amount of time an immigrant had to be a resident of the United States before becoming a citizen.
It increased the amount of time an immigrant had to be a resident of the United States before becoming a citizen.
It created a series of committees that were responsible for screening applicants seeking citizenship.
It removed legal citizenship from the majority of Americans.
Question 19 Explanation:
This Act increased the residence period necessary for immigrants to become naturalized citizens in the United States from 5 to 14 years. It was promoted as a way to protect national security, but was more likely an effort to decrease the number of voters who disagreed with the Federalist political party.
What were the political aims of the Naturalization, Alien, and Sedition Acts?
To make the United States more democratic in it’s political practices.
To help immigrants who were living in the United States.
To bolster relationships between the Federalist and Democratic-Republican political parties.
To strengthen the Federalist Party at the expense of the Democratic-Republican Party.
Question 20 Explanation:
These laws were clear attempts to make the federal government and the Federalist Party stronger by eliminating the voices of their adversaries and overstepping their constitutional limits of power. The laws proved so unpopular that many point to their passage as the beginning of the end for the Federalist Party.
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 20 questions to complete.<|endoftext|>
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Modulus and Argument
The Argand Diagram
Definition
An Argand diagram has a horizontal axis, referred to as the real axis, and a vertical axis, referred to as the imaginary axis.
A complex number $z = a + bi$ is plotted at coordinates $(a,b)$, as $a$ is the real part of the complex number, and $b$ the imaginary part.
Worked Example
Example 1
Plot the following complex numbers on an Argand diagram.
\begin{align} z_1 &= 3+i \\ z_2 &= -2-4i \\ z_3 &=-1+3i \\ z_4 &= -2i \end{align}
Modulus and Argument
Definition
Any complex number $z$ can be represented by a point on an Argand diagram. We can join this point to the origin with a line segment. The length of the line segment is called the modulus of the complex number and is denoted $\lvert z \rvert$. The angle measured from the positive real axis to the line segment is called the argument of the complex number, denoted $arg(z)$ and often labelled $\theta$. The modulus and argument can be calculated using trigonometry.
The modulus of a complex number $z = a + b i$ is $\lvert z \rvert = \sqrt{a^2+b^2}$
When calculating the argument of a complex number, there is a choice to be made between taking values in the range $[-\pi,\pi]$ or the range $[0,\pi]$. Both are equivalent and equally valid. On this page we will use the convention $-\pi \lt \theta \lt \pi$.
The 'naive' way of calculating the angle to a point $(a,b)$ is to use $\arctan(\frac{b}{a})$ but, since $\arctan$ only takes values in the range $[-\frac{\pi}{2},\frac{\pi}{2}]$, this will give the wrong result for coordinates with negative $x$-component. You can fix this by adding or subtracting $\pi$, depending on which quadrant of the Argand diagram the point lies in.
• First quadrant: $\theta = \arctan \left(\dfrac{b}{a}\right)$.
• Second quadrant: $\theta = \arctan \left(\dfrac{b}{a}\right) + \pi$.
• Third quadrant: $\theta = \arctan \left(\dfrac{b}{a}\right) -\pi$.
• Fourth quadrant: $\theta = \arctan \left(\dfrac{b}{a}\right)$.
It's a good idea to draw an Argand diagram of the complex number when making the decision about which formula to use.
Note: be careful to watch out for the case when $a=0$, i.e. the complex number has no real part. In this case, the $\arctan$ method doesn't work, but the argument is either $\frac{\pi}{2}$ or $-\frac{\pi}{2}$ for numbers with positive and negative imaginary parts, respectively.
Example
$z_1=1+i$ has the argument $\arg z_1 = \arctan \left(\dfrac{1}{1}\right) = \arctan (1) = \dfrac{\pi}{4}.$
However, the same calculation for $z_2=-1-i$ gives $\arctan \left(\frac{-1}{-1}\right) = \arctan (1) = \dfrac{\pi}{4}$, the same number!
If we draw $z_2$ on an Argand diagram, we can see that it falls in the third quadrant, so the argument should be between $-\frac{\pi}{2}$ and $-\pi$. We must subtract $\pi$ to correct this and therefore get $\arg z_2 = -\dfrac{3\pi}{4}$.
Worked Examples
Example 1
Find the modulus and argument of the complex number $z = 3+2i$.
Solution
\begin{align} \lvert z \rvert &= \sqrt{3^2+2^2}\\ &=\sqrt{9+4}\\ &=\sqrt{13} \end{align}
As the complex number lies in the first quadrant of the Argand diagram, we can use $\arctan \frac{2}{3}$ without modification to find the argument.
\begin{align} \arg z &= \arctan \left(\frac{2}{3}\right) \\ &=0.59 \text{ radians (to 2 d.p.)} \end{align}
Example 2
Find the modulus and argument of the complex number $z=4i$.
Solution
\begin{align} \lvert z \rvert &= \sqrt{0^2+4^2}\\ &=\sqrt{16}\\ &=4 \end{align}
The simplest way to find the argument is to look at an Argand diagram and plot the point $(0,4)$. The point lies on the positive vertical axis, so $\arg z = \frac{\pi}{2}$
Example 3
Find the modulus and argument of the complex number $z = -2+5i$.
Solution
\begin{align} \lvert z \rvert &= \sqrt{(-2)^2+5^2}\\ &=\sqrt{4+25}\\ &=\sqrt{29} \end{align}
As $z$ is in the second quadrant of the Argand diagram, we need to add $\pi$ to the result obtained from $\arctan \left(\frac{5}{-2}\right)$.
\begin{align} \arg z &= \arctan \left(\frac{5}{-2}\right) + \pi \\ &=-1.19 + \pi \\ &= 1.95 \text{ radians (to 2 d.p.)} \end{align}
Example 4
Find the modulus and argument of the complex number $z = -4-3i$.
Solution
\begin{align} \lvert z \rvert &= \sqrt{(-4)^2+(-3)^2}\\ &=\sqrt{16+9}\\ &=\sqrt{25}\\ &=5 \end{align}
As $z$ lies in the third quadrant of the Argand diagram, we need to subtract $\pi$ from the result of $\arctan \left(\frac{-3}{-4}\right)$.
\begin{align} \arg z &= \arctan \left(\frac{-3}{-4}\right) - \pi\\ &= \arctan \left(\frac{3}{4}\right) - \pi\\ &= 0.64 - \pi \\ &= -2.50 \text{ radians (to 2 d.p.)} \end{align}
Note: Alternatively, the answer could have been given in the range $0 \lt \theta \lt 2\pi$, where we would have added $\pi$, instead of subtracting it, and got an answer of $\arg z = 3.67$ radians (to 2 d.p.)
Example 5
Find the modulus and argument of the complex number $z = 1-4i$.
Solution
\begin{align} \lvert z \rvert &= \sqrt{1^2+(-4)^2}\\ &=\sqrt{1+16}\\ &=\sqrt{17} \end{align}
As $z$ lies in the fourth quadrant of the Argand diagram, we don't need to modify the result of $\arctan \left(\frac{-4}{1}\right)$ to find the argument.
\begin{align} \arg z &= \arctan \left(\frac{-4}{1}\right)\\ &= \arctan \left(-4\right) \\ &= -1.33 \text{ radians (to 2 d.p.)} \end{align}
Video Example
Prof. Robin Johnson draws the complex numbers $z=-1-i$ and $z=-4+3i$ on an Argand diagram, and finds their modulus and argument.
Workbook
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.
More Support
You can get one-to-one support from Maths-Aid.<|endoftext|>
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This we week we wrapped up some aspects of Renaissance exploration by thinking about why exactly Europe experienced such a huge burst of exploration activity in the mid-late 15th century.
Most of us might tend to think that the key to the increased activity was the advent of new technology. That is, Europeans discovered new tools that would help them sail the seas, and so now they could make the attempts to find new lands that lack of technology made previously impossible.
In his book Pathfinders, historian Felipe-Fernandez Armesto discounts this notion. Very few technological advances took place in the decades leading up to the great expansion of exploration. One thing did change significantly, however, and that was their desire to explore. Quite simply, they wanted to go, whereas before they did not. Exploration resulted from a belief that mankind should take great risks to find out more about the world. While they made some technological advances as a result of their sailing, things continued more or less as they had been from 1450 until the discovery of how to measure longitude in late 18th century. Belief, not technology, spurred on exploration.
What happened? The Renaissance shifted the emphasis from orienting one’s life from “top to bottom,” as the Medievals viewed life and thought, to a more “side to side” perspective that focused on the knowable, observable, and measurable. Whether this shift indicated that the Renaissance tried to “improve upon God’s handiwork of creation,” as the great Umberto Eco stated, or that, “the people of the Renaissance had a renewed sense of humanity’s responsibility and stewardship of creation,” as the great art appreciator Sister Wendy postulated, is a question I want the students to consider.
So, what building cathedrals was to medievals, exploration was to the Renaissance. Both capture the spirit of the times, and show the values of each time and place. A good question for us to consider is, “What values does our society pursue?” Do cultures need to dream, to risk, to reach beyond themselves to function well?
We then went on to discuss the controversial political philosopher Nicolo Machiavelli. To help set the groundwork for understanding him, I asked the students a few questions:
1. Martin Luther supposedly said, “I would rather be ruled by a wise Turk than a foolish Christian” (though many now believe that Luther never said this exactly, though he said other things like it). If we agree with Luther, this assumes that what we want from political governance differs from what we want from our spiritual leaders. If we followed Machiavelli, for example, we would not put moral character or spiritual guidance at the top of our list for qualities we look for in political leadership. Consider these two alternatives for president
- A solid Christian in belief and morals, but possessing little political experience, imagination, or intelligence,
- A shrewd, intelligent, and experienced leader with respect from the international community, but who does not consider himself a Christian.
“See,” Machiavelli might argue, “Contrary to your instincts, religion is not most important in politics.” Machiavelli encourages rulers not to be hostile to religion, but believes that politics operates independently from it.
2. Can politics have a redemptive effect on humanity? St. Augustine argued that politics, as it dealt with the ordering of earthly relationships, could not by definition help lead one to God. Other theologians disagree with Augustine, but if you agree with him, then one opens the door for politics to have different rules than “normal” life. For example, we have no problem admitting that trying to bluff in poker is not a sin, however much one tries to deceive others in the game. Poker is not “normal” life. When we play poker, we enter into an agreed upon alternate reality.
Politics functioned in a similar kind of alternate reality, according to Machiavelli. There are times when we expect our leaders to lie or disseminate false information, especially about military operations. Most of us would not only expect it, we might even admire the tactic should it prove successful and give our country a greater measure of safety. Whether we agree or not, if we understand these questions we can understand where Machiavelli came from with some of his ideas.
In a famous phrase intended as jibe against Plato, Machiavelli urges us not to seek out “imagined republics.” Like the Renaissance in general he sought guidance from what he saw in front of him, a consummate political realist. For example. . .
1. It would be best if you (the ruler) were perfect. But you’re not, so you will have faults and vices. First, seek to turn your faults to your advantage if you can. If you lack consistency of character, perhaps this could mean that your enemies will fear your unpredictability. Failing that, make sure you avoid vices that will directly effect your ability to rule. Much better for you to run around with women, for example, than to steal from the public till. God can forgive all sin, people will probably forgive the former, but not the latter. Above all, power is your guiding star. Do what you needed to do to maintain and keep power, for without that, nothing else matters (from a political perspective).
2. Should a ruler prefer to be loved or feared? Again, ideally the answer is, “both.” But very few can achieve this. Since nearly all of us must choose one or the other, Machiavelli writes,
Because this is to be asserted in general of men, that they are ungrateful, fickle, false, cowardly, covetous, and as long as you succeed they are yours entirely; they will offer you their blood, property, life and children, as is said above, when the need is far distant; but when it approaches they turn against you. And that prince who, relying entirely on their promises, has neglected other precautions, is ruined; because friendships that are obtained by payments, and not by greatness or nobility of mind, may indeed be earned, but they are not secured, and in time of need cannot be relied upon; and men have less scruple in offending one who is beloved than one who is feared, for love is preserved by the link of obligation which, owing to the baseness of men, is broken at every opportunity for their advantage; but fear preserves you by a dread of punishment which never fails.
Some of you may remember the controversy Surgeon General C. Everett Koop created when he allowed for contraceptive education in public schools. He stood against abortion and as a Christian privately supported abstinence, but as a public servant he believed in that if teens did have sex, they should use contraceptives, as it would protect them from disease and reduce teen pregnancy and abortions. Some Christians applauded this stance. Others believed that Koop did not just take sin into account with his policy, he gave it the victory. One can level the same charge against Machiavelli. Koop and Machiavelli both, though in different ways and to different degrees, touch on the dilemma between public service and personal belief and practice.
Few Christians would want to go as far as Machiavelli did, yet many would probably find themselves agreeing with some of his assumptions. Drawing the line appropriately will require great wisdom, and the students did a great job discussing some of these tough questions this past week.<|endoftext|>
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# Multiplying a Binomial by a Trinomial
In this video, we are going to look at how to multiply a binomial by a trinomial.
For example:
To multiply $(x+6)(x^2+4x-5)$ we have to distribute each term in the binomial to each term in the trinomial.
When we distribute the x to the terms in the trinomial, we get $x^3$, $4x^2$, and$-5x$. When we distribute the 6 to the trinomial, we get $6x^2$,$24x$, and $-30$. So now we are left with
$x^3+4x^2-5x+6x^2+24x-30$
From here, we want to combine like terms, to give us a final answer of
$x^3+10x^2+16x-30$
## Video-Lesson Transcript
Let’s go over multiplying a binomial by a trinomial.
It’s a two-term expression times a three-term expression.
For example:
$(x + 6) (x^2 + 4x - 5)$
Similar in multiplying a binomial by a binomial, we have to distribute each term on the first expression into each term of the second expression.
$x^3 + 4x^2 - 5x + 6x^2 + 24x - 30$
Then combine like terms
$x^3 + 10x^2 + 19x - 30$
The answer would still be the same even if we write the expressions like this:
$(x^2 + 4x - 5) (x + 6)$
$x^3 + 6x^2 + 4x^2 + 24x - 5x - 30$
If you take a closer look, this is the same as our answer above, they are just in different order.
So when we combine like terms, our answer is
$x^3 + 10x^2 + 19x - 30$
Both cases will give us the same answer.<|endoftext|>
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ABOUT THE TUTORIAL
This tutorial will give you a quick start to SQL. It covers most of the topics required for a basic understanding of SQL and to get a feel of how it works.
SQL is a database computer language designed for the retrieval and management of data in a relational database. SQL stands for Structured Query Language.
SQL is a language to operate databases; it includes database creation, deletion, fetching rows, modifying rows, etc. SQL is an ANSI (American National Standards Institute) standard language, but there are many different versions of the SQL language.
What is SQL?
SQL is Structured Query Language, which is a computer language for storing, manipulating and retrieving data stored in a relational database.
SQL is the standard language for Relational Database System. All the Relational Database Management Systems (RDMS) like MySQL, MS Access, Oracle, Sybase, Informix, Postgres and SQL Server use SQL as their standard database language.
Also, they are using different dialects, such as:
- MS SQL Server using T-SQL,
- Oracle using PL/SQL,
- MS Access version of SQL is called JET SQL (native format) etc
SQL is an interface between a human and a database.
Why SQL required?
SQL is one of the most important inventions in the software engineering world, equally important to the invention of Internet. Most of the Web applications today store their data in Relational Databases and SQL is the ubiquitous language that is used to and many more things.
- Create the tables and relationships between them.
- query data
- update data
- delete data
What is Database?
A database is a collection of information that is organized so that it can easily be accessed, managed, and updated.
An SQL database is a relational database that makes uses of the Structured Query Language (SQL) to query or communicate with the database.
A database is simply a collection of organized information, usually as a set of related lists of similar entries. The data is often organized so that it is easily accessible.
- The following are examples of databases that we use often:
– address book – dictionary – telephone book
A cylindrical structure is used to display the image of a database.
A database consists of a number of interrelated tables.
Each table has a number of records which are used to represent real world objects. –
- For example, the police may have a record for each criminal that has ever been arrested (i.e., the “rap”-sheet)
Each record has a number of fields which are data items used to specify a characteristic of the record.
Examples of fields are:
– name – employee number – address – prior convictions etc…
What is RDMS?
RDBMS stands for Relational Database Management System. RDBMS is the basis for SQL, and for all modern database systems like MS SQL Server, IBM DB2, Oracle, MySQL, and Microsoft Access.
A Relational database management system (RDBMS) is a database management system (DBMS) that is based on the relational model as introduced by E. F. Codd.
What is Table?
The data in an RDBMS is stored in database objects which are called as tables. This table is basically a collection of related data entries and it consists of numerous columns and rows.
Remember, a table is the most common and simplest form of data storage in a relational database. The following program is an example of a employee table −
| Id | name | dept | salary |
| 100 | Thomas| Sales | 5000 |
| 200 | Jason | Technology | 5500 |
| 300 | Mayla | Technology | 7000 |
| 400 | Nisha | Marketing | 9500 |
| 500 | Randy | Technology | 6000 |
5 rows in set (0.00 sec)
What is a field?
Every table is broken up into smaller entities called fields. The fields in the CUSTOMERS table consist of ID, NAME, AGE, ADDRESS and SALARY.
A field is a column in a table that is designed to maintain specific information about every record in the table.
What is a Record or a Row?
A record is also called as a row of data is each individual entry that exists in a table. For example, there are 5 records in the above EMPLOYEE table. Following is a single row of data or record in the EMPLOYEE table −
| id | name | dept | salary |
| 100 | Thomas | Sales | 5000 |
A record is a horizontal entity in a table.
What is a column?
A column is a vertical entity in a table that contains all information associated with a specific field in a table.
For example, a column in the EMPLOYEE table is DEPT, which represents location description and would be as shown below –
| dept |
| Sales |
| Technology |
| Technology |
| Marketing |
| Technology |
What is a NULL value?
A NULL value in a table is a value in a field that appears to be blank, which means a field with a NULL value is a field with no value.
It is very important to understand that a NULL value is different than a zero value or a field that contains spaces. A field with a NULL value is the one that has been left blank during a record creation.
Constraints are the rules enforced on data columns on a table. These are used to limit the type of data that can go into a table. This ensures the accuracy and reliability of the data in the database.
Constraints can either be column level or table level. Column level constraints are applied only to one column whereas; table level constraints are applied to the entire table.
Following are some of the most commonly used constraints available in SQL −
- NOT NULL Constraint− Ensures that a column cannot have a NULL value.
- DEFAULT Constraint− Provides a default value for a column when none is specified.
- UNIQUE Constraint− Ensures that all the values in a column are different.
- PRIMARY Key− Uniquely identifies each row/record in a database table.
- FOREIGN Key− Uniquely identifies a row/record in any another database table.
- CHECK Constraint− The CHECK constraint ensures that all values in a column satisfy certain conditions.
- INDEX− Used to create and retrieve data from the database very quickly.
The following categories of data integrity exist with each RDBMS −
- Entity Integrity −There are no duplicate rows in a table.
- Domain Integrity −Enforces valid entries for a given column by restricting the type, the format, or the range of values.
- Referential integrity −Rows cannot be deleted, which are used by other records.
- User-Defined Integrity −Enforces some specific business rules that do not fall into entity, domain or referential integrity.
Difference between RDBMS AND DBMS
BASIS FOR COMPARISON
Database Management System
Relational Database Management System
Data is stored in the navigational model.
Data is stored in the relational model (in tables).
Does not present
RDBMS uses normalization to reduce or eliminate redundancy.
Modification in data is complex.
Modification in data is easy and simple.
Consumes more time.
Faster as compared to DBMS.
Schema-based constraints and data dependencies
Employed in RDBMS.
Keys and indexes
Does not used.
To establish relationship keys and indexes are used in RDBMS.
Inefficient, error prone and insecure.
Efficient and secure.
Supported by RDBMS.
Dbase, Microsoft Access, LibreOffice Base, FoxPro.
SQL server, Oracle, mysql, MariaDB, SQLite.
MySQL is an open source SQL database, which is developed by a Swedish company – MySQL AB. MySQL is pronounced as "my ess-que-ell," in contrast with SQL, pronounced "sequel."
MySQL is supporting many different platforms including Microsoft Windows, the major Linux distributions, UNIX, and Mac OS X.
MySQL has free and paid versions, depending on its usage (non-commercial/commercial) and features. MySQL comes with a very fast, multi-threaded, multi-user and robust SQL database server.
- Development of MySQL by Michael Widenius & David Axmark beginning in 1994.
- First internal release on 23rdMay 1995.
- Windows Version was released on the 8thJanuary 1998 for Windows 95 and NT.
- Version 3.23: beta from June 2000, production release January 2001.
- Version 4.0: beta from August 2002, production release March 2003 (unions).
- Version 4.01: beta from August 2003, Jyoti adopts MySQL for database tracking.
- Version 4.1: beta from June 2004, production release October 2004.
- Version 5.0: beta from March 2005, production release October 2005.
- Sun Microsystems acquired MySQL AB on the 26thFebruary 2008.
- Version 5.1: production release 27thNovember 2008.
- High Performance.
- High Availability.
- Scalability and Flexibility Run anything.
- Robust Transactional Support.
- Web and Data Warehouse Strengths.
- Strong Data Protection.
- Comprehensive Application Development.
- Management Ease.
- Open Source Freedom and 24 x 7 Support.
- Lowest Total Cost of Ownership.
MS SQL Server
MS SQL Server is a Relational Database Management System developed by Microsoft Inc. Its primary query languages are −
- ANSI SQL
- 1987 - Sybase releases SQL Server for UNIX.
- 1988 - Microsoft, Sybase, and Aston-Tate port SQL Server to OS/2.
- 1989 - Microsoft, Sybase, and Aston-Tate release SQL Server 1.0 for OS/2.
- 1990 - SQL Server 1.1 is released with support for Windows 3.0 clients.
- Aston - Tate drops out of SQL Server development.
- 2000 - Microsoft releases SQL Server 2000.
- 2001 - Microsoft releases XML for SQL Server Web Release 1 (download).
- 2002 - Microsoft releases SQLXML 2.0 (renamed from XML for SQL Server).
- 2002 - Microsoft releases SQLXML 3.0.
- 2005 - Microsoft releases SQL Server 2005 on November 7th, 2005.
- High Performance
- High Availability
- Database mirroring
- Database snapshots
- CLR integration
- Service Broker
- DDL triggers
- Ranking functions
- Row version-based isolation levels
- XML integration
- Database Mail
It is a very large multi-user based database management system. Oracle is a relational database management system developed by 'Oracle Corporation'.
Oracle works to efficiently manage its resources, a database of information among the multiple clients requesting and sending data in the network.
It is an excellent database server choice for client/server computing. Oracle supports all major operating systems for both clients and servers, including MSDOS, NetWare, UnixWare, OS/2 and most UNIX flavors.
Oracle began in 1977 and celebrating its 32 wonderful years in the industry (from 1977 to 2009).
- 1977 - Larry Ellison, Bob Miner and Ed Oates founded Software Development Laboratories to undertake development work.
- 1979 - Version 2.0 of Oracle was released and it became first commercial relational database and first SQL database. The company changed its name to Relational Software Inc. (RSI).
- 1981 - RSI started developing tools for Oracle.
- 1982 - RSI was renamed to Oracle Corporation.
- 1983 - Oracle released version 3.0, rewritten in C language and ran on multiple platforms.
- 1984 - Oracle version 4.0 was released. It contained features like concurrency control - multi-version read consistency, etc.
- 1985 - Oracle version 4.0 was released. It contained features like concurrency control - multi-version read consistency, etc.
- 2007 - Oracle released Oracle11g. The new version focused on better partitioning, easy migration, etc.
- Read Consistency
- Locking Mechanisms
- Quiesce Database
- Self-managing database
- Resource Manager
- Data Warehousing
- Materialized views
- Bitmap indexes
- Table compression
- Parallel Execution
- Analytic SQL
- Data mining
This is one of the most popular Microsoft products. Microsoft Access is an entry-level database management software. MS Access database is not only inexpensive but also a powerful database for small-scale projects.
MS Access uses the Jet database engine, which utilizes a specific SQL language dialect (sometimes referred to as Jet SQL).
MS Access comes with the professional edition of MS Office package. MS Access has easyto-use intuitive graphical interface.
- 1992 - Access version 1.0 was released.
- 1993 - Access 1.1 released to improve compatibility with inclusion the Access Basic programming language.
- The most significant transition was from Access 97 to Access 2000.
2007 - Access 2007, a new database format was introduced ACCDB which supports complex data types such as multi valued and attachment fields.
- Users can create tables, queries, forms and reports and connect them together with macros.
- Option of importing and exporting the data to many formats including Excel, Outlook, ASCII, dBase, Paradox, FoxPro, SQL Server, Oracle, ODBC, etc.
- There is also the Jet Database format (MDB or ACCDB in Access 2007), which can contain the application and data in one file. This makes it very convenient to distribute the entire application to another user, who can run it in disconnected environments.
- Microsoft Access offers parameterized queries. These queries and Access tables can be referenced from other programs like VB6 and .NET through DAO or ADO.
- The desktop editions of Microsoft SQL Server can be used with Access as an alternative to the Jet Database Engine.
Microsoft Access is a file server-based database. Unlike the client-server relational database management systems (RDBMS), Microsoft Access does not implement database triggers, stored procedures or transaction logging.<|endoftext|>
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A calendar era is the year numbering system used by a calendar. For example, the Gregorian calendar numbers its years in the Western Christian era (the Coptic Orthodox and Ethiopian Orthodox churches have their own Christian eras). The instant, date, or year from which time is marked is called the epoch of the era. There are many different calendar eras such as Saka Era.
In antiquity, regnal years were counted from the accession of a monarch. This makes the Chronology of the ancient Near East very difficult to reconstruct, based on disparate and scattered king lists, such as the Sumerian King List and the Babylonian Canon of Kings. In East Asia, reckoning by era names chosen by ruling monarchs ceased in the 20th century except for Japan, where they are still used.
Ancient dating systems
For over a thousand years, ancient Assyria used a system of eponyms to identify each year. Each year at the Akitu festival (celebrating the Mesopotamian new year), one of a small group of high officials (including the king in later periods) would be chosen by lot to serve as the limmu for the year, which meant that he would preside over the Akitu festival and the year would bear his name. The earliest attested limmu eponyms are from the Assyrian trading colony at Karum Kanesh in Anatolia, dating to the very beginning of the 2nd millennium BC, and they continued in use until the end of the Neo-Assyrian Period, ca. 612 BC.
Assyrian scribes compiled limmu lists, including an unbroken sequence of almost 250 eponyms from the early 1st millennium BC. This is an invaluable chronological aid, because a solar eclipse was recorded as having taken place in the limmu of Bur-Sagale, governor of Guzana. Astronomers have identified this eclipse as one that took place on 15 June, 763 BC, which has allowed absolute dates of 892 to 648 BC to be assigned to that sequence of eponyms. This list of absolute dates has allowed many of the events of the Neo-Assyrian Period to be dated to a specific year, avoiding the chronological debates that characterize earlier periods of Mesopotamian history.
Among the ancient Greek historians and scholars, a common method of indicating the passage of years was based on the Olympic Games, first held in 776 BC. The Olympic Games provided the various independent city-states with a mutually recognizable system of dates. Olympiad dating was not used in everyday life. This system was in use from the 3rd century BC. The modern Olympic Games (or Summer Olympic Games beginning 1896) do not continue the four year periods from ancient Greece: the 669th Olympiad would have begun in the summer of 1897, but the modern Olympics were first held in 1896.:769
Another common system was the indiction cycle (15 indictions made up an agricultural tax cycle in Roman Egypt, an indiction being a year in duration). Documents and events began to be dated by the year of the cycle (e.g., "fifth indiction", "tenth indiction") in the 4th century, and this system was used long after the tax ceased to be collected. It was used in Gaul, in Egypt until the Islamic conquest, and in the Eastern Roman Empire until its conquest in 1453.
The rule for computing the indiction from the AD year number, which he had just invented, was stated by Dionysius Exiguus: add 3 and divide by 15; the remainder is the indiction, with 0 understood to be the fifteenth indiction.:770 Thus 2001 was the ninth indiction. The beginning of the year for the indiction varied.:769–71
The Seleucid era was used in much of the Middle East from the 4th century BC to the 6th century AD, and continued until the 10th century AD among Oriental Christians. The era is computed from the epoch 312 BC: in August of that year Seleucus I Nicator captured Babylon and began his reign over the Asian portions of Alexander the Great's empire. Thus depending on whether the calendar year is taken as starting on 1 Tishri or on 1 Nisan (respectively the start of the Jewish civil and ecclesiastical years) the Seleucid era begins either in 311 BC (the Jewish reckoning) or in 312 BC (the Greek reckoning: October–September).
An early and common practice was Roman 'consular' dating. This involved naming both consules ordinarii who had taken up this office on January 1 (since 153 BC) of the relevant civil year.:6 Sometimes one or both consuls might not be appointed until November or December of the previous year, and news of the appointment may not have reached parts of the Roman empire for several months into the current year; thus we find the occasional inscription where the year is defined as "after the consulate" of a pair of consuls.
The use of consular dating ended in AD 541 when the emperor Justinian I discontinued appointing consuls. The last consul nominated was Anicius Faustus Albinus Basilius. Soon afterwards, imperial regnal dating was adopted in its place.
Dating from the founding of Rome
Another method of dating, rarely used, was anno urbis conditae (Latin: "in the year of the founded city" (abbreviated AUC), where "city" meant Rome). (It is often incorrectly given that AUC stands for ab urbe condita, which is the title of Titus Livius's history of Rome.)
Several epochs were in use by Roman historians. Modern historians usually adopt the epoch of Varro, which we place in 753 BC.
The system was introduced by Marcus Terentius Varro in the 1st century BC. The first day of its year was Founder's Day (April 21), although most modern historians assume that it coincides with the modern historical year (January 1 to December 31). It was rarely used in the Roman calendar and in the early Julian calendar – naming the two consuls that held office in a particular year was dominant. AD 2019 is thus approximately the same as AUC 2772 (2019 + 753).
About AD 400, the Iberian historian Orosius used the AUC era. Pope Boniface IV (about AD 600) may have been the first to use both the AUC era and the Anno Domini era (he put AD 607 = AUC 1360).
Regnal years of Roman emperors
Another system that is less commonly found than might be thought was the use of the regnal year of the Roman emperor. At first, Augustus indicated the year of his reign by counting how many times he had held the office of consul, and how many times the Roman Senate had granted him Tribunican powers, carefully observing the fiction that his powers came from these offices granted to him, rather than from his own person or the many legions under his control. His successors followed his practice until the memory of the Roman Republic faded (about AD 200), when they began to use their regnal year openly.
Dating from the Roman conquest
Some regions of the Roman Empire dated their calendars from the date of Roman conquest, or the establishment of Roman rule.
The Spanish era counted the years from 38 BC, probably the date of a new tax imposed by the Roman Republic on the subdued population of Iberia. The date marked the establishment of Roman rule in Spain and was used in official documents in Portugal, Aragon, Valencia, and in Castile, into the 14th century. This system of calibrating years fell to disuse in 1381 and was replaced by today's Anno Domini.
Throughout the Roman and Byzantine periods, the Decapolis and other Hellenized cities of Syria and Palestine used the Pompeian era, counting dates from the Roman general Pompey's conquest of the region in 63 BC.
A different form of calendar was used to track longer periods of time, and for the inscription of calendar dates (i.e., identifying when one event occurred in relation to others). This form, known as the Long Count, is based upon the number of elapsed days since a mythological starting-point. According to the calibration between the Long Count and Western calendars accepted by the great majority of Maya researchers (known as the GMT correlation), this starting-point is equivalent to August 11, 3114 BC in the proleptic Gregorian calendar or 6 September in the Julian calendar (−3113 astronomical).
Other dating systems
A great many local systems or eras were also important, for example the year from the foundation of one particular city, the regnal year of the neighboring Persian emperor, and eventually even the year of the reigning Caliph.
Late Antiquity and Middle Ages
- The Etos Kosmou of the Byzantine Calendar places Creation at the beginning of its year 1, namely 5509 BC. Its first known use occurred in the 7th century AD, although its precursors were developed about AD 400. The year 7509 of this era began in September 2000.
- The Era of Martyrs or Era of Diocletian is reckoned from the beginning of the reign of Roman Emperor Diocletian; the first year of this era was 284/5. It was not the custom to use regnal years in Rome, but it was the custom in Roman Egypt, which the emperor ruled through a prefect (the king of Egypt). The year number changed on the first day of the Egyptian month Thoth (29 August three years out of four, 30 August the year before a Roman leap year.) Diocletian abolished the special status of Egypt, which thereafter followed the normal Roman calendar: consular years beginning on 1 January. This era was used in the Easter tables prepared in Alexandria long after the abdication of Diocletian, even though Diocletian was a notorious persecutor of Christians. The Era of Diocletian was retained by the Coptic Church and used for general purposes, but by 643 the name had been changed to Era of the Martyrs.:766–7
- The Incarnation Era is used by Ethiopia. Its epoch is August 29, AD 8 in the Julian calendar.
- The Armenian calendar has its era fixed at AD 552.
Dionysian "Common Era"
The era based on the Incarnation of Christ was introduced by Dionysius Exiguus in 525 and is in continued use with various reforms and derivations. The distinction between the Incarnation being the conception or the Nativity of Jesus was not drawn until the late ninth century.:881 The beginning of the numbered year varied from place to place: when, in 1600, Scotland adopted January 1 as the date the year number changes, this was already the case in much of continental Europe. England adopted this practice in 1752.:7
- A.D. (or AD) – for the Latin Anno Domini, meaning "in the year of (our) Lord". This is the dominant or Western Christian Era; AD is used in the Gregorian calendar. Anno Salutis, meaning "in the year of salvation" is identical. Originally intended to number years from the Incarnation of Jesus, according to modern thinking the calculation was a few years off. Years preceding AD 1 are numbered using the BC era, avoiding zero or negative numbers. AD was also used in the medieval Julian calendar, but the first day of the year was either March 1, Easter, March 25, September 1, or December 25, not January 1. To distinguish between the Julian and Gregorian calendars, O.S. and N.S. were often added to the date, especially during the 17th and 18th centuries, when both calendars were in common use. Old Style (O.S.) was used for the Julian calendar and for years not beginning on January 1. New Style (N.S.) was used for the Gregorian calendar and for Julian calendar years beginning on January 1. Many countries switched to using January 1 as the start of the numbered year at the same time as they switched from the Julian calendar to the Gregorian calendar, but others switched earlier or later.
- B.C. (or BC) – meaning "Before Christ". Used for years before AD 1, counting backwards so the year n BC is n years before AD 1. Thus there is no year 0.
- C.E. (or CE) and B.C.E. (or BCE) – meaning "Common Era" and "Before the Common Era", numerically equivalent to AD and BC, respectively (in writing, "AD" precedes the year number, but "CE" follows the year: AD 1 = 1 CE.) The Latin equivalent vulgaris aera was used as early as 1615 by Johannes Kepler. The English abbreviations C.E. and B.C.E. were introduced in the 19th century by Jewish intellectuals, wishing to avoid the abbreviation for dominus "lord" in implicit reference to Christ. By the later 20th century, the abbreviations had come into wider usage by authors who wished to emphasize secularism.
- Astronomical year numbering equates its year 0 with 1 BC, and counts negative years from 2 BC backward (−1 backward), so 100 BC is −99.
- The human era, also named Holocene era, proposed by Cesare Emiliani adds 10,000 to AD years, so that AD 1 would be the year 10,001.
- Anno Lucis of Freemasonry adds 4000 years to the AD year.
- A.H. (or AH) – for the Latinized Anno Hegirae, meaning "in the year of the Hijra", Muhammad's flight from Mecca to Medina in September 622, which occurred in its first year, used in the Islamic calendar. Since the Islamic calendar is a purely lunar calendar of about 354 days, its year count increases faster than that of solar and lunisolar calendars.
- A.H.S. (or AHS) is used by the Iranian calendar to denote the number of solar years since the Hijra. The year beginning at the vernal equinox equals the number of the Gregorian year beginning at the preceding January 1 minus 621.
- Hindu calendar, counting from the start of the Kali Yuga, with its epoch on February 18, 3102 BC Julian (January 23, 3102 BC Gregorian), based on Aryabhata (6th century).
- Vikrama Samvat, 56-57 BC, introduced about the 12th century.
- S.E. or (SE) – for the Saka Era, used in some Hindu calendars and in the Indian national calendar, with an epoch near the vernal equinox of year 78 (its year 0); its usage spread to Southeast Asia before year 1000. This era is also used (together with the Gregorian calendar) in the Indian national calendar, the official civil calendar used in communiques issued by the Government of India.
The Hindu Saka Era influences the calendars of southeast Asian indianized kingdoms.
- B.E. – for the Buddhist Era, introduced by Vajiravudh in 1912, which has an epoch (origin) of 544 BC. This year is called year 1 in Sri Lanka and Burma, but year 0 in Thailand, Laos and Cambodia. Thus the year 2500 B.E. occurred in 1956 in the former countries, but in 1957 in the latter. In Thailand in 1888 King Chulalongkorn decreed a National Thai Era, dating from the founding of Bangkok on April 6, 1782. In 1912 New Year's Day was shifted to April 1. In 1941 Prime Minister Phibunsongkhram decided to count the years since 543 BC. This is the Thai solar calendar using the Thai Buddhist Era aligned to the western solar calendar.
- BE for Burmese Era – from Burmese calendar originally with an epochal year 0 date of 22 March 638; from which derived CS for Chula Sakarat era; variously known as LE Lesser Era; ME Minor Era – the Major or Great Era being the Saka Era of the Indian national calendar
- B.E. of the Bahá'í calendar is below.
- B.E. – The Bahá'í calendar dates from the year of the declaration of the Báb. Years are counted in the Bahá'í Era (BE), which starts its year 1 from March 21, 1844.
- A.M. (or AM) – for the Latin Anno Mundi, meaning "in the year of the world", has its epoch in the year 3761 BC. This was first used to number the years of the modern Hebrew calendar in 1178 by Maimonides. Precursors with epochs one or two years later were used since the 3rd century, all based on the Seder Olam Rabba of the 2nd century. The year beginning in the northern autumn of 2000 was 5761 AM).
- The Zoroastrian calendar used regnal years since the reform by Ardeshir I, but after the fall of the Sassanid empire, the ascension of the last Sassanid ruler, Yazdegerd III of Persia, crowned June 16, 632, continued to be used as the reference year, abbreviated Y.Z. or "Yazdegerd era".
- The Republican Era of the French Republican Calendar was dated from September 22, 1792, the day of the proclamation of the French First Republic. It was used in Revolutionary France from October 24, 1793 (on the Gregorian calendar) to December 31, 1805.
- The Positivist calendar of 1844 takes 1789 as its epoch.
- The Republican era is used by the Republic of China since 1912, which is the first year of the republic. Coincidentally, this is the same as the Juche era used in North Korea, the year of the birth of its founder Kim Il-Sung.
- The Era Fascista 'Fascist Era' was instituted by the Italian Fascists and used Roman numerals to denote the number of years since the March on Rome in 1922. Therefore, 1934, for example, was XII E.F. (era fascista). This era was abolished with the fall of fascism in Italy on July 25, 1943, but restored in the northern part of the country during the Italian Social Republic. The Gregorian calendar remained in simultaneous use and a double numbering was adopted: the year of the Common era was presented in Arabic numerals and the year of the fascist era in Roman numerals. The year of the Fascist calendar began on October 29, so, for example, October 27, 1933 was XI E.F. but October 30, 1933 was XII E.F.
- China traditionally reckoned by the regnal year of its emperors, see Chinese era name. Most Chinese do not assign numbers to the years of the Chinese calendar, but the few who do, like expatriate Chinese, use a continuous count of years from the reign of the legendary Yellow Emperor, using 2698 BC as year 1. Western writers begin this count at either 2637 BC or 2697 BC (see Chinese calendar). Thus, the Chinese years 4637, 4697, or 4698 began in early 2000.
- In Korea, from 1952 until 1961 years were numbered via Dangi years, where 2333 BC was regarded as the first such year.
- The Assyrian calendar, introduced in the 1950s, has its era fixed at 4750 BC.
- The Japanese calendar dates from the accession of the current Emperor of Japan. The current emperor took the throne in early 1989, which became Heisei 1, which was until then Shōwa 64 (for its first seven days).
- The United States government sometimes uses a calendar of the era of its Independence, fixed on July 4, 1776, together with the Anno Domini civil calendar. For instance, its Constitution is dated "the Seventeenth Day of September in the Year of our Lord one thousand seven hundred and Eighty seven and of the Independence of the United States of America the Twelfth." Presidential proclamations are also dated in this way.
- A.D. – "After Dianetics". In Scientology, years are numbered relative to the first publication of the book Dianetics: The Modern Science of Mental Health (1950).
- Y.O.L.D. – In the Discordian calendar, the standard designation for the year number is YOLD (Year of Our Lady of Discord). The calendar begins counting from January 1, 1166 BC in the Discordian year 0, ostensibly the date of origin of the Curse of Greyface. An alternate designation, A.D.D. has been occasionally seen (Anno Domina Discordia, a Latin translation of YOLD, but presumably also a play on attention deficit disorder).
- e.v. – Era vulgaris. (From Latin, meaning "common era", usually stylized in lowercase.) The Thelemic calendar is used by some Thelemites to designate a number of years since Aleister Crowley's inauguration of the so-called Aeon of Horus, which occurred on March 20, 1904, and coincides with both the Thelemic new year and a holiday known as the Equinox of the Gods. The abbreviation "A.N.", for Aerae Novae ("New Era" in Latin), is sometimes also used.
- B.P. – for Before Present, specifically, the number of radiocarbon years before 1950.
- HE – for counting elapsed years of the Holocene from near the beginning of the Neolithic revolution of the Holocene epoch, specifically by adding exactly 10,000 years to AD (Anno Domini) or CE (Common Era) years, and subtracting BC/BCE years from 10001.
- Julian day number – for counting days, not years, its era fixed at noon January 1, 4713 BC in the proleptic Julian calendar. This equals November 24, 4714 BC in the proleptic Gregorian calendar. From noon of this day to noon of the next day was day 0. Multiples of 7 are Mondays. Negative values can also be used. Apart from the choice of the zero point and name, this Julian day and Julian date are not related to the Julian calendar. It does not count years, so, strictly speaking, it has no era, but it does have an epoch. Today (noon-to-noon UTC) the value is 2458622.
- Unix time – for counting elapsed seconds since the Unix epoch set at 00:00:00 or midnight UTC of January 1, 1970, though there are problems with Unix implementation of Coordinated Universal Time (UTC).
- "CDLI: The Old and Middle Assyrian limmu officials". Cuneiform Digital Library Initiative. Retrieved 18 May 2016.
- Millard, Alan (1994). The Eponyms of the Assyrian Empire, 910-612 BC (State Archives of Assyria Studies, Vol. 2). Helsinki: Neo-Assyrian Text Corpus Project. ISBN 978-9514567155.
- Blackburn, B. & Holford-Strevens, L. (1999, 2003). The Oxford Companion to the Year: an exploration of calendar customs and time-reckoning (corrected printing). Oxford University Press. ISBN 978-0-19-214231-3.
- Nautical Almanac Office of the United States Naval Observatory and Her Majesty's Nautical Almanac Office. (2000)The Nautical Almanac for the year 2001. Washington, DC: Government Printing Office and London, England, UK: The Stationery Office. p. B4.
- Gedaliah ibn Jechia the Spaniard, Shalshelet Ha-Kabbalah, Jerusalem 1962, p. 271 (Hebrew)
- Associated Press Stylebook. (2007). New York: Basic Books. p. 6. ISBN 978-0-465-00489-8. "Because the full phrase would read in the year of the Lord 96, the abbreviation A.D. goes before the figure for the year: A.D. 96."
- A 1635 English edition of that book has the title page in English – so far, the earliest-found usage of "Vulgar Era" in English. The English phrase "common Era" appears at least as early as 1708. In Latin, "Common Era" is written as Vulgaris Aera. It also occasionally appears as æra vulgaris, aera vulgaris, anni vulgaris, vulgaris aera Christiana, and anni vulgatae nostrae aerae Christianas.
- Use of "C.E." and "B.C.E.": Morris Jacob Raphall. Post-Biblical History of The Jews (1856). Explicit use of "b.c.e." for "before the common era": Max Stern, Lemaʼan Yilmedu: A Second Hebrew Reader for Jewish Schools and Private Instruction (1881), p. 37.
- John Sumser, The Conflict Between Secular and Religious Narratives in the United States: Wittgenstein, Social Construction, and Communication (2016), p. 69.
- Cesare, E. (1993). [Correspondence]. Nature, 336, 716.
- U.S. Constitution
- The White House: Presidential Actions
- Sappell, J., & Welkos, R. W. (1990, June 28). Costly Strategy Continues to Turn Out Bestsellers. Los Angeles Times.<|endoftext|>
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People who live farther from the equator have larger eye sockets than their tropical counterparts, a new study finds. And as people inhabited higher and higher latitudes, eye socket size grew along with the northerly or southerly extent of their migrations.
"It's never been shown before that latitude and vision are related in this way in humans," says Robert Barton, an evolutionary anthropologist from Durham University in England, who was not involved in the research.
The experimental design was quite simple. The researchers studied a museum collection of 73 healthy 200-year-old skulls from 12 different populations. For each skull, they lined an eye socket (which anthropologists call the "orbit") with laboratory film and filled it with small glass beads to measure the volume. Then they charted the location where each skull was dug up, and calculated average light levels for each locale, based on its latitude and longitude. It turned out that skull orbits obtained from higher latitudes held more glass beads: Those from nearer to the equator held 22.5 milliliters, on average, whereas the ones from inhabitants living more than 40 degrees north or south of the equator held closer to 27 milliliters.
"Days are shorter as you move away from the equator, and the duration of dawn and dusk increases," explains Eiluned Pearce, a University of Oxford doctoral student who co-authored the paper with her advisor, evolutionary anthropologist Robin Dunbar. Because of the diminished light, Pearce and Dunbar hypothesized that humans at higher latitudes would need to develop specialized visual systems to continue to be able to spot predators and recognize social cues from other humans.
The implications of the report, published in the February issue of Biology Letters, are not completely surprising. A few years ago, ecologist Robert Thomas of Cardiff University in Wales found that songbirds with larger eyes sing at lower daily light levels (such as at dawn and dusk) than do birds with smaller eyes—presumably because animals with larger eyes have larger pupils which funnel more light into the eye, thereby increasing the length of activities associated with daylight. "We all imagine people as a bit different from animals and maybe not subject to the evolutionary circumstances," Thomas says, "but here's a fantastic example of the evolutionary process working on human societies that live in different places."
However, it is risky to use eye-socket volume as a proxy for eyeball size, cautions physical anthropologist Christopher Kirk from the University of Texas at Austin. The actual eyeball takes up less than one third of the volume of the socket—the rest of it is filled with fat, muscle and connective tissue. One alternate explanation is that humans at higher latitudes require larger eye sockets to accommodate larger fat pads to act as insulation in colder weather. Other factors such as chewing muscles, the shape of the nasal cavity and the size of the frontal lobe also influence eye-socket size and shape. For these reasons and more, it would be best to measure eyeballs themselves, Kirk says.
And that is exactly what Pearce is working on now. With the help of a network of researchers, she is collecting and analyzing MRI scans from all over the world, in order to measure the eyeball sizes of modern populations and to determine how exactly they vary with latitude. At this point, it is not clear whether the eye-socket volume is genetically heritable or if it develops over an individual's lifetime. "This seems to be the first time anyone has ever really thought about this, so this paper was really about putting that hypothesis out there," Pearce says. There's still a lot of work to be done and many questions to be answered.
"For some reason, orbit size increases with latitude," Kirk says. "That's a phenomenon that needs to be explained."<|endoftext|>
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To safeguard the future survival of our planet and ocean, ambitious climate objectives and a reduction in greenhouse gas emissions are needed, said marine biologists Prof Dr Hans Otto Pörtner and Prof. Dr. Ulf Riebesell, ahead of the Paris World Climate Conference at the breakfast meeting of the German Climate Consortium (DKK) and the German Marine Research Consortium (KDM)
Pörtner, a biologist at the Alfred Wegener Institute, Helmholtz Centre for Polar and Marine Research and the newly elected Co Chair of IPCC Working Group II, described the projected risks for the oceans brought about by climate change: global warming, rising sea levels and ocean acidification. In addition to the tropical coral reefs, the sea ice regions in the Arctic are also considered to be the more vulnerable ecosystems.
Presently, the critical change thresholds for organisms and ecosystems and the resultant risks are analysed and represented in temperature. According to Pörtner, the human-induced warming of global temperatures must be limited to 1.5 degrees Celsius, rather than 2 degree Celsius.
He added that in the case of coral reefs, fifty percent of them can be preserved if the increase in temperature is limited to 1.2 degree Celsius—however, this figure does not take into account the effects of ocean acidification.
Ocean acidification: Oceans changed by greenhouse gases One of the greatest climate risks for the oceans is acidification: 24 million tons of carbon dioxide is absorbed into the ocean every day. This works out to about one-third of the pre-industrial carbon dioxide released, thus mitigating the effects of climate change. Today, the carbon dioxide uptake of the ocean averages 28 percent higher than in pre-industrial times. If the emissions are left unchecked, the ocean’s acidity will more than double by the end of this century. The more acidic the oceans become, the less carbon dioxide they can absorb from the atmosphere.
"The rate of projected ocean acidification is unprecedented in Earth's history," asserted Riebesell, Professor of Biological Oceanography at GEOMAR Helmholtz Centre for Ocean Research Kiel.
"Above all, calcifying organisms are among the losers of ocean acidification, in addition to corals, mussels, snails, sea urchins, starfish and many calcifying plankton."
GEOMAR field experiment in Norway: Winners and losers of acidification
Riebesell had first presented the results of a field experiment conducted in spring 2015 in the Norwegian Raunefjord, south of Bergen. It highlighted the effects of ocean acidification within several mesocosms (closed experimental systems set up to simulate biological, chemical and physical processes) in the fjord over the course of several months. It was discovered that some species like the winged snails and calcareous algae would not be able to survive the effects of ocean acidification, unlike the picoplankton, micro-organisms at the bottom of the food chain.
Hence, Riebesell concluded that small changes in the ecosystem could have huge consequences and revolutionise not only the food web in the ocean, but also affect aquaculture and fisheries.
Climate change and ocean acidification
The problem of ocean acidification is the result of carbon dioxide emissions by human activities. The interaction of environmental factors on the marine organisms, as well as man-made changes (like eutrophication and pollution) leads to complex changes in the ecosystem, migration of species and an overall decrease in biodiversity. Even now, we are just starting to understand the implications of such changes in our environment. Source: http://www.geomar.de<|endoftext|>
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# MOREFB - Editorial
Author: Amit Pandey
Tester: Mahbubul Hasan and Sunny Aggarwal
Editorialist: Balajiganapathi Senthilnathan
Russian Translator: Sergey Kulik
Mandarian Translator: Minako Kojima
Medium-Hard
### PREREQUISITES:
FFT, Divide and conquer
### PROBLEM:
Given a set S with N elements and a number K, find \sum_{s \subset S \land |s| = k} Fibo(sum(s))
### SHORT EXPLANATION
Since this is modulo a prime for which square root of 5 exists, we can calculate the n^th fibonacci number modulo P as: a(b^n - c^n) mod P for the correct values of a, b and c. Then finding the answer boils down to finding the coefficient of x^{(N - K)} in the equation (x + b^{a_1}) * (x + b^{a_2}) * .. * (x + b^{a_N}). The multiplication can be done using FFT.
### EXPLANATION:
Note that mod = 99991 is a prime number. There is a closed form for finding the n^{th} fibonacci number:
F_n = \frac{\sqrt{5}}{5}((\frac{1 + \sqrt{5}}{2})^{n} - (\frac{1 - \sqrt{5}}{2})^{n})
Now since we are taking everything modulo mod, we can calculate each of the term modulo mod.
So we will get an equation of the form:
F_n = a(b^n - c^n)
Where a, b and c are the corresponding values from the above equation modulo mod.
Now let us use this information to calculate the coefficient of b.
For example, if S = {1, 2} and K = 1, then the answer is F_1 + F_2 = a((b^1 + b^2) - (c^1 + c^2)). Let us call the part with b g(b). We will discuss how to calculate g(b), calculating for the c part is similar.
Let us see the multiplication of (x + b^1)(x + b^2). It comes out to be x^2 + (b^1 + b^2) x + b^3. Note that the coefficient of x is the answer we require. Also, if K were 2, then the answer would have been F_3 whose g(b) would be the coefficient of x^0. Seeing this pattern we can generalize the idea.
The g(b) for any K is the coefficient of x^{N - K} in the polynomial (x + b^{a_1}) * (x + b^{a_2}) * .. * (x + b^{a_N}).
So, how do we calculate this value? A naive implementation of the multiplication will TLE as it will take O(N^2). However we can use FFT (the tutorial is in Russian you may have to translate it first) or Karatsuba’s algorithm to speed up the multiplication.
O(N logN logN)
### AUTHOR’S AND TESTER’S SOLUTIONS:
6 Likes
Please could any one explain this editorial more. I really want to learn this problem.
3 Likes
I just wanted to point out that for the given constraints a simpler algorithm for polynomial multiplication, namely [Karatuba][1], suffices (which has a runtime of O(N^{\log_2(3)})). This is cause the implementation of FFT is fairly involved and calls for maintaining complex values, whereas Karatsuba only requires integers and in essence is a simple divide-and-conquer algorithm.
Also for those curious, the closed form for F_n mod 99991 ends up being:
F_n = \frac{a^n - b^n}{a - b}, a = 55048, b = 44944.
This is because the characteristic polynomial for F_n, i.e. x^2 - x - 1, has integer roots mod 99991 (since \sqrt(5) = 10104 \in Z_{99991})
[1]: https://en.wikipedia.org/wiki/Karatsuba_algorithm
3 Likes
The prime modulo 99991 is not suitable for use with NTT(Number Theoretic Transform), so we will have to resort to using FFT with complex doubles. To avoid rounding errors, after each polynomial multiplication, round the coefficients to the nearest integers. If using double still results in rounding errors, as it did in my case, replacing it with long double will eliminate the problem. This was my first time at solving an FFT problem and it feels nice to have learnt something new!
2 Likes
We can implement the same strategy via matrix multiplication can’t we ?
Would be interesting to know if it can be solved for 100pts with matrix multiplication.
I solved 2 subtasks(40pts) with a DP solution and matrix exponentiation, if you’re interested: http://www.codechef.com/viewsolution/7180786
Why can’t the editorial be more clear, for those who are very new to such topics, finds many difficulty in learning them as we cannot understand anything by seeing someone’s code. The main idea of editorials is for those who are willing to learn which is not sustained here(seems the editorial can be understandable by those who have already a idea about the topic.)
19 Likes
Interesting thing to know about Fibonacci numbers is that when we take mod with 99991 the Fibonacci cycle repeats itself after length of 33330 which can be verified here http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibmaths.html#section6.1
1 Like
Can anyone please explain the last part i.e. Also, if K were 2, then the answer would have been F3 whose g(b) would be the coefficient of x0
@bbackspace, can you explain your approach a little bit, specially that lucas number part.
I propose an alternate solution:
We can prove by induction that the matrix:
[1 1]
[1 0]
raised to the n-th power gives
[F_n+1 F_n ]
[F_n F_n-1]
Let’s call the matrix M, for short.
We should easily figure out that
F_(a+b) is the top-right (or bottom-left) element of M^a * M^b
Then how can we use this to out advantage?
Well, if we dive deeper into this statement, we can argue that F_sum(V), over any V, is the top-right cell of the matrix product(M^V[i]).
Also, from M^a + M^b we can translate easily to F_a + F_b (we just take the element I recalled);
Let’s summarize:
• M^a + M^b <-> F_a + F_b
• M^a * M^b <-> F_(a+b)
then SUM of F_(SUM of elements) can be translated to SUM of PRODUCT of corresponding matrices.
Let’s continue.
If we take all subsequences of cardinality K, then FIBOSUM(k) is the sum of all PRODUCT(M^v[i]), for each v subsequence of V of cardinality K. If we further analyze this, by the rule of Viete, it turns out to be the (N-K)th term of the polynom P(X) = PRODUCT(X + M^V[i]), for i = 1,n. Basically, we have to find out the (matrix) coefficients of the polynom P(X), which can be done with divide et impera and FFT or Karatsuba (gave me TLE).
Why would this be better? Well, it’s not. For this case at least. However, it does work in any case, no matter how the modulo is chosen.
4 Likes
I waited two days for this editorial hoping to find a good FFT explanation relevant to this question, and examples.
Why TL is so huge? Karatsuba version passed in 0.8 sec.
Hey, how did you calculate a and b? I don’t know what concept is used here. Please give me a link , if you have so I can read in detail.
Editorial is not clear. Please explain the FFT part.
I learned fft from a very well written article, which is very friendly and not so hectic:
After understanding this, you can try reading some more papers and more easily feel for it.
Google is our friend on this one!
@filmwalams you might have to read up on solving linear recurrences. Here’s a good link for the Fibonacci sequence specifically: http://gozips.uakron.edu/~crm23/fibonacci/fibonacci.htm. The idea is that the characteristic polynomial factors nicely mod 99991:
$(x - 55048)(x - 44944) = x^2 - 99992x + 2474077312 = x^2 - x - 1 (mod 99991)$.
Yes, it happens after Pisano period.
You may see the editorials of following problem for more details : https://www.hackerrank.com/challenges/emma-and-sum-of-products/editorial
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You may see the editorials of following problem for more details about FFT : https://www.hackerrank.com/challenges/emma-and-sum-of-products/editorial
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//<|endoftext|>
| 4.46875 |
328 |
Calculate the mean deviation from the median of the following data:
Question:
Calculate the mean deviation from the median of the following data:
Solution:
Given the frequency distribution
Now we have to find the mean deviation from the median
Let us make a table of the given data and append other columns after calculations
Now, here N=20, which is even.
Now, here $\mathrm{N}=20$, which is even.
Here median class $=\frac{\mathrm{N}}{2}=10^{\text {th }}$ term
This observation lie in the class interval 12-18, so median can be written as,
$\mathrm{M}=\mathrm{l}+\frac{\frac{\mathrm{N}}{2}-\mathrm{cf}}{\mathrm{f}} \times \mathrm{h}$
Here $I=12, c f=9, f=3, h=6$ and $N=20$, substituting these values, the above equation becomes,
$M=12+\frac{\frac{20}{2}-9}{3} \times 6$
$\Rightarrow M=12+\frac{10-9}{3} \times 6$
$\Rightarrow \mathrm{M}=12+\frac{1 \times 6}{3}$
$\Rightarrow M=12+2=14$
Hence Mean Deviation becomes,
$M . D=\frac{\sum f_{i} d_{i}}{\sum f_{i}}=\frac{140}{20}=7$
Therefore, the mean deviation about the median of the distribution is 7<|endoftext|>
| 4.5625 |
13,994 |
Chronology of discoveries of water on Mars
To date, interplanetary spacecraft have provided abundant evidence of water on Mars, dating back to the Mariner 9 mission, which arrived at Mars in 1971. This article provides a mission by mission breakdown of the discoveries they have made. For a more comprehensive description of evidence for water on Mars today, and the history of water on that planet, see Water on Mars.
Mariner 9 imaging revealed the first direct evidence of water in the form of river beds, canyons (including the Valles Marineris, a system of canyons over about 4,020 kilometres (2,500 mi) long), evidence of water erosion and deposition, weather fronts, fogs, and more. The findings from the Mariner 9 missions underpinned the later Viking program. The enormous Valles Marineris canyon system is named after Mariner 9 in honor of its achievements.
Warrego Valles, as seen by Mariner 9. This image suggests that rain/snow was necessary to form this kind of branched network of channels.
By discovering many geological forms that are typically formed from large amounts of water, Viking orbiters caused a revolution in our ideas about water on Mars. Huge river valleys were found in many areas. They showed that floods of water broke through dams, carved deep valleys, eroded grooves into bedrock, and traveled thousands of kilometers. Large areas in the southern hemisphere contained branched valley networks, suggesting that rain once fell. The flanks of some volcanoes are believed to have been exposed to rainfall because they resemble those occurring on Hawaiian volcanoes. Many craters look as if the impactor fell into mud. When they were formed, ice in the soil may have melted, turned the ground into mud, then the mud flowed across the surface. Normally, material from an impact goes up, then down. It does not flow across the surface, going around obstacles, as it does on some Martian craters. Regions, called "chaotic terrain", seemed to have quickly lost great volumes of water which caused large channels to form downstream. The amount of water involved was almost unthinkable—estimates for some channel flows run to ten thousand times the flow of the Mississippi River. Underground volcanism may have melted frozen ice; the water then flowed away and the ground just collapsed to leave chaotic terrain.
The images below, some of the best from the Viking Orbiters, are mosaics of many small, high resolution images. Click on the images for more detail. Some of the pictures are labeled with place names.
Streamlined Islands seen by Viking showed that large floods occurred on Mars. Image is located in Lunae Palus quadrangle.
The ejecta from Arandas crater acts like mud. It moves around small craters (indicated by arrows), instead of just falling down on them. Craters like this suggest that large amounts of frozen water were melted when the impact crater was produced. Image is located in Mare Acidalium quadrangle and was taken by Viking Orbiter.
Branched channels in Thaumasia quadrangle, as seen by Viking Orbiter. Networks of channels like this are strong evidence for rain on Mars in the past.
The branched channels seen by Viking from orbit strongly suggested that it rained on Mars in the past. Image is located in Margaritifer Sinus quadrangle.
Results from Viking lander experiments strongly suggest the presence of water in the present and in the past of Mars. All samples heated in the gas chromatograph-mass spectrometer (GSMS) gave off water. However, the way the samples were handled prohibited an exact measurement of the amount of water. But, it was around 1%. General chemical analysis suggested the surface had been exposed to water in the past. Some chemicals in the soil contained sulfur and chlorine that were like those remaining after sea water evaporates. Sulfur was more concentrated in the crust on top of the soil, than in the bulk soil beneath. So it was concluded that the upper crust was cemented together with sulfates that were transported to the surface dissolved in water. This process is common on Earth's deserts. The sulfur may be present as sulfates of sodium, magnesium, calcium, or iron. A sulfide of iron is also possible. Using results from the chemical measurements, mineral models suggest that the soil could be a mixture of about 90% iron-rich clay, about 10% magnesium sulfate (kieserite?), about 5% carbonate (calcite), and about 5% iron oxides (hematite, magnetite, goethite?). These minerals are typical weathering products of mafic igneous rocks. The presence of clay, magnesium sulfate, kieserite, calcite, hematite, and goethite strongly suggest that water was once in the area. Sulfate contains chemically bound water, hence its presence suggests water was around in the past. Viking 2 found similar group of minerals. Because Viking 2 was much farther north, pictures it took in the winter showed frost.
Photo of the Viking 2 lander taken by the Mars Reconnaissance Orbiter in December 2006.
Mars Global SurveyorEdit
The Mars Global Surveyor's Thermal Emission Spectrometer (TES) is an instrument able to detect mineral composition on Mars. Mineral composition gives information on the presence or absence of water in ancient times. TES identified a large (30,000 square-kilometer) area (in the Nili Fossae formation) that contained the mineral olivine. It is thought that the ancient impact that created the Isidis basin resulted in faults that exposed the olivine. Olivine is present in many mafic volcanic rocks; in the presence of water it weathers into minerals such as goethite, chlorite, smectite, maghemite, and hematite. The discovery of olivine is strong evidence that parts of Mars have been extremely dry for a long time. Olivine was also discovered in many other small outcrops within 60 degrees north and south of the equator. Olivine has been found in the SNC (shergottite, nakhlite, and chassigny) meteorites that are generally accepted to have come from Mars. Later studies have found that olivine-rich rocks cover more than 113,000 square kilometers of the Martian surface. That is 11 times larger than the five volcanoes on the Big Island of Hawaii.
Below are some examples of gullies that were photographed by Mars Global Surveyor.
Group of gullies on north wall of crater that lies west of the crater Newton (41.3047 degrees south latitude, 192.89 east longitide). Image is located in the Phaethontis quadrangle.
Gullies on one wall of Kaiser crater. Gullies usually are found in only one wall of a crater. Location is Noachis quadrangle.
Many places on Mars show dark streaks on steep slopes, such as crater walls. Dark slope streaks have been studied since the Mariner and Viking missions. It seems that streaks start out being dark, then they become lighter with age. Often they originate with a small narrow spot, then widen and extend downhill for hundreds of meters. Streaks do not seem to be associated with any particular layer of material because they do not always start at a common level along a slope. Although many of the streaks appear very dark, they are only 10% or less darker than the surrounding surface. Mars Global Surveyor found that new streaks have formed in less than one year on Mars.
Several ideas have been advanced to explain the streaks. Some involve water, or even the growth of organisms. The generally accepted explanation for the streaks is that they are formed from the avalanching of a thin layer of bright dust that is covering a darker surface. Bright dust settles on all Martian surfaces after a period of time.
Dark streaks can be seen in the images below, as seen from Mars Global Surveyor.
Tikhonravov crater floor in Arabia quadrangle. Click on image to see dark slope streaks and layers.
Dark streaks in Diacria quadrangle.
Some parts of Mars show inverted relief. This occurs when materials are deposited on the floor of a stream then become resistant to erosion, perhaps by cementation. Later the area may be buried. Eventually erosion removes the covering layer. The former streams become visible since they are resistant to erosion. Mars Global Surveyor found several examples of this process. Many inverted streams have been discovered in various regions of mars, especially in the Medusae Fossae Formation, Miyamoto Crater, and the Juventae Plateau.
The image below shows one example.
Pathfinder found temperatures varied on a diurnal cycle. It was coldest just before sunrise (about −78 Celsius) and warmest just after Mars noon (about −8 Celsius). These extremes occurred near the ground which both warmed up and cooled down fastest. At this location, the highest temperature never reached the freezing point of water (0 °C), so Mars Pathfinder confirmed that where it landed it is too cold for liquid water to exist. However, water could exist as a liquid if it were mixed with various salts.
Surface pressures varied diurnally over a 0.2 millibar range, but showed 2 daily minima and two daily maxima. The average daily pressure decreased from about 6.75 millibars to a low of just under 6.7 millbars, corresponding to when the maximum amount of carbon dioxide had condensed on the south pole. The pressure on the Earth is generally close to 1000 millibars, so the pressure on Mars is very low. The pressures measured by Pathfinder would not permit water or ice to exist on the surface. But, if ice were insulated with a layer of soil, it could last a long time.
Other observations were consistent with water being present in the past. Some of the rocks at the Mars Pathfinder site leaned against each other in a manner geologists term imbricated. It is believed strong flood waters in the past pushed the rocks around until they faced away from the flow. Some pebbles were rounded, perhaps from being tumbled in a stream. Parts of the ground are crusty, maybe due to cementing by a fluid containing minerals.
There was evidence of clouds and maybe fog.
In July 2003, at a conference in California, it was announced that the Gamma Ray Spectrometer (GRS) on board the Mars Odyssey had discovered huge amounts of water over vast areas of Mars. Mars has enough ice just beneath the surface to fill Lake Michigan twice. In both hemispheres, from 55 degrees latitude to the poles, Mars has a high density of ice just under the surface; one kilogram of soil contains about 500 g of water ice. But, close to the equator, there is only 2 to 10% of water in the soil. Scientists believe that much of this water is locked up in the chemical structure of minerals, such as clay and sulfates. Previous studies with infrared spectroscopes have provided evidence of small amounts of chemically or physically bound water. The Viking landers detected low levels of chemically bound water in the Martian soil. It is believed that although the upper surface only contains a percent or so of water, ice may lie just a few feet deeper. Some areas, Arabia Terra, Amazonis quadrangle, and Elysium quadrangle contain large amounts of water. Analysis of the data suggest that the southern hemisphere may have a layered structure. Both of the poles showed buried ice, but the north pole had none close to it because it was covered over by seasonal carbon dioxide (dry ice). When the measurements were gathered, it was winter at the north pole so carbon dioxide had frozen on top of the water ice. There may be much more water further below the surface; the instruments aboard the Mars Odyssey are only able to study the top meter or so of soil. If all holes in the soil were filled by water, this would correspond to a global layer of water 0.5 to 1.5 km deep.
The Phoenix lander confirmed the initial findings of the Mars Odyssey. It found ice a few inches below the surface and the ice is at least 8 inches deep. When the ice is exposed to the Martian atmosphere it slowly sublimates. In fact, some of the ice was exposed by the landing rockets of the craft.
Thousands of images returned from Odyssey support the idea that Mars once had great amounts of water flowing across its surface. Some pictures show patterns of branching valleys. Others show layers that may have formed under lakes. Deltas have been identified.
For many years researchers believed that glaciers existed under a layer of insulating rocks. Lineated deposits are one example of these probable rock-covered glaciers. They are found on the floors of some channels. Their surfaces have ridged and grooved materials that deflect around obstacles. Some glaciers on the Earth show such features. Lineated floor deposits may be related to lobate debris aprons, which have been proven to contain large amounts of ice by orbiting radar.
Channels near Warrego Valles. These branched channels are strong evidence for flowing water on Mars, perhaps during a much warmer period.
Erosion features in Ares Vallis – the streamined shape was probably formed by running water.
Delta in Lunae Palus quadrangle.
Delta in Margaritifer Sinus quadrangle.
Much of the surface of Mars is covered by a thick smooth mantle that is thought to be a mixture of ice and dust. This ice-rich mantle, a few yards thick, smooths the land, but in places it displays a bumpy texture, resembling the surface of a basketball. The low density of craters on the mantle means it is relatively young.
Changes in Mars's orbit and tilt cause significant changes in the distribution of water ice. During certain climate periods water vapor leaves polar ice and enters the atmosphere. The water returns to the ground at lower latitudes as deposits of frost or snow mixed generously with dust. The atmosphere of Mars contains a great deal of fine dust particles. Water vapor condenses on the particles, then they fall down to the ground due to the additional weight of the water coating. When ice at the top of the mantling layer returns to the atmosphere, it leaves behind dust, which insulates the remaining ice.
Dao Vallis begins near a large volcano, called Hadriaca Patera, so it is thought to have received water when hot magma melted huge amounts of ice in the frozen ground. The partially circular depressions on the left side of the channel in the image above suggests that groundwater sapping also contributed water.
In some areas large river valleys begin with a landscape feature called "chaos" or chaotic terrain." It is thought that the ground collapsed, as huge amounts of water were suddenly released. Examples of chaotic terrain, as imaged by THEMIS, are shown below.
The Phoenix lander confirmed the existence of large amounts of water ice in the northern regions of Mars. This finding was predicted by theory. and was measured from orbit by the Mars Odyssey instruments. On June 19, 2008, NASA announced that dice-sized clumps of bright material in the "Dodo-Goldilocks" trench, dug by the robotic arm, had vaporized over the course of four days, strongly implying that the bright clumps were composed of water ice which sublimated following exposure. Even though dry ice also sublimates under the conditions present, it would do so at a rate much faster than observed.
On July 31, 2008, NASA announced that Phoenix confirmed the presence of water ice on Mars. During the initial heating cycle of a new sample, the Thermal and Evolved-Gas Analyzer's (TEGA) mass spectrometer detected water vapor when the sample temperature reached 0 °C. Liquid water cannot exist on the surface of Mars with its present low atmospheric pressure, except at the lowest elevations for short periods.
Results published in the journal Science after the mission ended reported that chloride, bicarbonate, magnesium, sodium potassium, calcium, and possibly sulfate were detected in the samples. Perchlorate (ClO4), a strong oxidizer, was confirmed to be in the soil. The chemical when mixed with water can greatly lower freezing points, in a manner similar to how salt is applied to roads to melt ice. Perchlorate may be allowing small amounts of liquid water to form on Mars today. Gullies, which are common in certain areas of Mars, may have formed from perchlorate melting ice and causing water to erode soil on steep slopes.
Additionally, during 2008 and early 2009, a debate emerged within NASA over the presence of 'blobs' which appeared on photos of the vehicle's landing struts, which have been variously described as being either water droplets or 'clumps of frost'. Due to the lack of consensus within the Phoenix science project, the issue had not been raised in any NASA news conferences. One scientist's view poised that the lander's thrusters splashed a pocket of brine from just below the Martian surface onto the landing strut during the vehicle's landing. The salts would then have absorbed water vapor from the air, which would have explained how they appeared to grow in size during the first 44 Martian days before slowly evaporating as Mars temperature dropped. Some images even suggest that some of the droplets darkened, then moved and merged; this is strong physical evidence that they were liquid.
For about as far as the camera can see, the land is flat, but shaped into polygons between 2–3 meters in diameter and are bounded by troughs that are 20 cm to 50 cm deep. These shapes are due to ice in the soil expanding and contracting due to major temperature changes.
Comparison between polygons photographed by Phoenix on Mars...
... and as photographed (in false color) from Mars orbit...
The microscope showed that the soil on top of the polygons is composed of flat particles (probably a type of clay) and rounded particles. Clay is a mineral that forms from other minerals when water is available. So, finding clay proves the existence of past water. Ice is present a few inches below the surface in the middle of the polygons, and along its edges, the ice is at least 8 inches deep. When the ice is exposed to the Martian atmosphere it slowly sublimates.
Snow was observed to fall from cirrus clouds. The clouds formed at a level in the atmosphere that was around −65 °C, so the clouds would have to be composed of water-ice, rather than carbon dioxide-ice (dry ice) because the temperature for forming carbon dioxide ice is much lower—less than −120 °C. As a result of mission observations, it is now believed that water ice (snow) would have accumulated later in the year at this location. The highest temperature measured during the mission was −19.6 °C, while the coldest was −97.7 °C. So, in this region the temperature remained far below the freezing point (0°) of water. Bear in mind that the mission took place in the heat of the Martian summer.
Interpretation of the data transmitted from the craft was published in the journal Science. As per the peer reviewed data the site had a wetter and warmer climate in the recent past. Finding calcium carbonate in the Martian soil leads scientists to believe that the site had been wet or damp in the geological past. During seasonal or longer period diurnal cycles water may have been present as thin films. The tilt or obliquity of Mars changes far more than the Earth; hence times of higher humidity are probable. The data also confirms the presence of the chemical perchlorate. Perchlorate makes up a few tenths of a percent of the soil samples. Perchlorate is used as food by some bacteria on Earth. Another paper claims that the previously detected snow could lead to a buildup of water ice.
Mars Exploration RoversEdit
The Mars Rovers Spirit and Opportunity found a great deal of evidence for past water on Mars. Designed to last only three months, both were still operating after more than six years. Spirit got trapped in a sand pit in 2006, with NASA officially cutting with the rover in 2011. Opportunity lost contact with NASA on June 10 2018 and its mission was declared complete on February 13 2019.
The Spirit rover landed in what was thought to be a huge lake bed. However, the lake bed had been covered over with lava flows, so evidence of past water was initially hard to detect. As the mission progressed and the Rover continued to move along the surface more and more clues to past water were found.
On March 5, 2004, NASA announced that Spirit had found hints of water history on Mars in a rock dubbed "Humphrey". Raymond Arvidson, the McDonnell University Professor and chair of Earth and planetary sciences at Washington University in St. Louis, reported during a NASA press conference: "If we found this rock on Earth, we would say it is a volcanic rock that had a little fluid moving through it." In contrast to the rocks found by the twin rover Opportunity, this one was formed from magma and then acquired bright material in small crevices, which look like crystallized minerals. If this interpretation holds true, the minerals were most likely dissolved in water, which was either carried inside the rock or interacted with it at a later stage, after it formed.
By Sol 390 (Mid-February 2005), as Spirit was advancing towards "Larry's Lookout", by driving up the hill in reverse, it investigated some targets along the way, including the soil target, "Paso Robles", which contained the highest amount of salt found on the red planet. The soil also contained a high amount of phosphorus in its composition, however not nearly as high as another rock sampled by Spirit, "Wishstone". Squyres said of the discovery, "We're still trying to work out what this means, but clearly, with this much salt around, water had a hand here".
As Spirit traveled with a dead wheel in December 2007, pulling the dead wheel behind, the wheel scraped off the upper layer of the martian soil, uncovering a patch of ground that scientists say shows evidence of a past environment that would have been perfect for microbial life. It is similar to areas on Earth where water or steam from hot springs came into contact with volcanic rocks. On Earth, these are locations that tend to teem with bacteria, said rover chief scientist Steve Squyres. "We're really excited about this," he told a meeting of the American Geophysical Union (AGU). The area is extremely rich in silica – the main ingredient of window glass. The researchers have now concluded that the bright material must have been produced in one of two ways. One: hot-spring deposits produced when water dissolved silica at one location and then carried it to another (i.e. a geyser). Two: acidic steam rising through cracks in rocks stripped them of their mineral components, leaving silica behind. "The important thing is that whether it is one hypothesis or the other, the implications for the former habitability of Mars are pretty much the same," Squyres explained to BBC News. Hot water provides an environment in which microbes can thrive and the precipitation of that silica entombs and preserves them. Squyres added, "You can go to hot springs and you can go to fumaroles and at either place on Earth it is teeming with life – microbial life.
Opportunity rover was directed to a site that had displayed large amounts of hematite from orbit. Hematite often forms from water. When Opportunity landed, layered rocks and marble-like hematite concretions ("blueberries") were easily visible. In its years of continuous operation, Opportunity sent back much evidence that a wide area on Mars was soaked in liquid water.
During a press conference in March 2006, mission scientists discussed their conclusions about the bedrock, and the evidence for the presence of liquid water during their formation. They presented the following reasoning to explain the small, elongated voids in the rock visible on the surface and after grinding into it (see last two images below). These voids are consistent with features known to geologists as "vugs". These are formed when crystals form inside a rock matrix and are later removed through erosive processes, leaving behind voids. Some of the features in this picture are "disk-like", which is consistent with certain types of crystals, notably sulfate minerals. Additionally, mission members presented first data from the Mössbauer spectrometer taken at the bedrock site. The iron spectrum obtained from the rock El Capitan shows strong evidence for the mineral jarosite. This mineral contains hydroxide ions, which indicates the presence of water when the minerals were formed. Mini-TES data from the same rock showed that it consists of a considerable amount of sulfates. Sulfates also contain water.
Mars Reconnaissance OrbiterEdit
The Mars Reconnaissance Orbiter's HiRISE instrument has taken many images that strongly suggest that Mars has had a rich history of water-related processes. A major discovery was finding evidence of hot springs. These may have contained life and may now contain well-preserved fossils of life.
Research, in the January 2010 issue of Icarus, described strong evidence for sustained precipitation in the area around Valles Marineris. The types of minerals there are associated with water. Also, the high density of small branching channels indicates a great deal of precipitation because they are similar to stream channels on the Earth.
Channels near the rim of Ius Chasma, as seen by HiRISE. The pattern and high density of these channels support precipitation as the source of the water. Location is Coprates quadrangle.
Channels in Candor plateau, as seen by HiRISE. Location is Coprates quadrangle. Click on image to see many small, branched channels which are strong evidence for sustained precipitation.
Some places on Mars show inverted relief. In these locations, a stream bed appears as a raised feature, instead of a depression. The inverted former stream channels may be caused by the deposition of large rocks or due to cementation of loose materials. In either case erosion would erode the surrounding land and consequently leave the old channel as a raised ridge because the ridge will be more resistant to erosion. Images below, taken with HiRISE show sinuous ridges that are old channels that have become inverted.
In an article published in January 2010, a large group of scientists endorsed the idea of searching for life in Miyamoto Crater because of inverted stream channels and minerals that indicated the past presence of water.
Inverted Channel with many branches in Syrtis Major quadrangle.
Using data from Mars Global Surveyor, Mars Odyssey and the Mars Reconnaissance Orbiter, scientists have found widespread deposits of chloride minerals. Usually chlorides are the last minerals to come out of solution. A picture below shows some deposits within the Phaethontis quadrangle. Evidence suggests that the deposits were formed from the evaporation of mineral-enriched waters. Lakes may have been scattered over large areas of the Martian surface. Carbonates, sulfates, and silica should precipitate out ahead of them. Sulfates and silica have been discovered by the Mars Rovers. Places with chloride minerals may have once held various life forms. Furthermore, such areas should preserve traces of ancient life.
Rocks on Mars have been found to frequently occur as layers, called strata, in many different places. Columbus Crater is one of many craters that contain layers. Rock can form layers in a variety of ways. Volcanoes, wind, or water can produce layers. Many places on Mars show rocks arranged in layers. Scientists are happy about finding layers on Mars since layers may have formed under large bodies of water. Sometimes the layers display different colors. Light-toned rocks on Mars have been associated with hydrated minerals like sulfates. The Mars Rover Opportunity examined such layers close-up with several instruments. Some layers are probably made up of fine particles because they seem to break up into fine dust. In contrast, other layers break up into large boulders so they are probably much harder. Basalt, a volcanic rock, is thought to form layers composed of boulders. Basalt has been identified all over Mars. Instruments on orbiting spacecraft have detected clay (also called phyllosilicates) in some layers. Scientists are excited about finding hydrated minerals such as sulfates and clays on Mars because they are usually formed in the presence of water. Places that contain clays and/or other hydrated minerals would be good places to look for evidence of life.
Below are a few of the many examples of layers that have been studied with HiRISE.
Layers in west slope of Asimov Crater. Location is Noachis quadrangle.
Close-up of layers in west slope of Asimov Crater. Shadows show the overhang. Some of the layers are much more resistant to erosion, so they stick out. Location is Noachis quadrangle.
Much of the surface of Mars is covered by a thick smooth mantle that is thought to be a mixture of ice and dust. This ice-rich mantle, a few yards thick, smoothes the land. But in places it displays a bumpy texture, resembling the surface of a basketball. Because there are few craters on this mantle, the mantle is relatively young. The images below, all taken with HiRISE, show a variety of views of this smooth mantle.
Dissected Mantle with layers. Location is Noachis quadrangle.
Changes in Mars's orbit and tilt cause significant changes in the distribution of water ice from polar regions down to latitudes equivalent to Texas. During certain climate periods water vapor leaves polar ice and enters the atmosphere. The water returns to the ground at lower latitudes as deposits of frost or snow mixed generously with dust. The atmosphere of Mars contains a great deal of fine dust particles. Water vapor condenses on the particles, then they fall down to the ground due to the additional weight of the water coating. When ice at the top of the mantling layer goes back into the atmosphere, it leaves behind dust, which insulates the remaining ice.
HiRISE has carried out many observations of gullies that are assumed to have been caused by recent flows of liquid water. Many gullies are imaged over and over to see if any changes occur. Some repeat observations of gullies have displayed changes that some scientists argue were caused by liquid water over the period of just a few years. Others say the flows were merely dry flows. These were first discovered by the Mars Global Surveyor.
Below are some of the many hundreds of gullies that have been studied with HiRISE.
Crater wall inside Mariner Crater showing a large group of gullies.
Gullies in Green Crater.
Close-up of gullies in Green Crater. Image located in Argyre quadrangle.
Scalloped Terrain at Peneus Patera. Scalloped terrain is quite common in some areas of Mars.
Maunder Crater. The overhang is part of the degraded south (toward bottom) wall of crater. The scale bar is 500 meters long.
Asimov Crater. Bottom of picture shows southeastern wall of crater. Top of picture is edge of mound that fills most of the crater.
Gullies on mound in Asimov Crater. Location is Noachis quadrangle.
Close-up of gullies in crater, as seen by HiRISE under the HiWish program. Location is Phaethontis quadrangle.
Close-up of gullies in trough, as seen by HiRISE under the HiWish program. These are some of the smaller gullies visible on Mars. Location is Phaethontis quadrangle.
Gullies in Phaethontis quadrangle. Notice how channels curve around obstacles.
Gullies with branches in Phaethontis quadrangle.
Gullies near Newton Crater, as seen by HiRISE, under the HiWish program. Place where there was an old glacier is labeled. Image from Phaethontis quadrangle.
Of interest from the days of the Viking Orbiters are piles of material surrounding cliffs. These deposits of rock debris are called lobate debris aprons (LDAs). These features have a convex topography and a gentle slope from cliffs or escarpments; this suggests flow away from the steep source cliff. In addition, lobate debris aprons can show surface lineations just as rock glaciers on the Earth. Recently[when?], research with the Shallow Radar on the Mars Reconnaissance Orbiter has provided strong evidence that the LDAs in Hellas Planitia and in mid northern latitudes are glaciers that are covered with a thin layer of rocks. Radar from the Mars Reconnaissance Orbiter gave a strong reflection from the top and base of LDAs, meaning that pure water ice made up the bulk of the formation (between the two reflections). Based on the experiments of the Phoenix lander and the studies of the Mars Odyssey from orbit, frozen water is now known to exist at just under the surface of Mars in the far north and south (high latitudes). The discovery of water ice in LDAs demonstrates that water is found at even lower latitudes. Future colonists on Mars will be able to tap into these ice deposits, instead of having to travel to much higher latitudes. Another major advantage of LDAs over other sources of Martian water is that they can easily detected and mapped from orbit. Lobate debris aprons are shown below from the Phlegra Montes, which are at a latitude of 38.2 degrees north. The Phoenix lander set down at about 68 degrees north latitude, so the discovery of water ice in LDAs greatly expands the range of easily available on Mars. It is far easier to land a spaceship near the equator of Mars, so the closer water is available to the equator the better it will be for future colonists.
Below are examples of lobate debris aprons that were studied with HiRISE.
Close-up of surface of a lobate debris apron. Note the lines that are common in rock glaciers on the Earth. Image located in Hellas quadrangle.
View of lobate debris apron along a slope. Image located in Arcadia quadrangle.
Place where a lobate debris apron begins. Note stripes which indicate movement. Image located in Ismenius Lacus quadrangle.
Research, reported in the journal Science in September 2009, demonstrated that some new craters on Mars show exposed, pure, water ice. After a time, the ice disappears, evaporating into the atmosphere. The ice is only a few feet deep. The ice was confirmed with the Compact Imaging Spectrometer (CRISM) on board the Mars Reconnaissance Orbiter (MRO). The ice was found in five locations. Three of the locations are in the Cebrenia quadrangle. These locations are 55.57° N, 150.62° E; 43.28° N, 176.9° E; and 45° N, 164.5° E. Two others are in the Diacria quadrangle: 46.7° N, 176.8° E and 46.33° N, 176.9° E. This discovery proves that future colonists on Mars will be able to obtain water from a wide variety of locations. The ice can be dug up, melted, then taken apart to provide fresh oxygen and hydrogen for rocket fuel. Hydrogen is the powerful fuel used by the space shuttle main engines.
In 2012, the NASA's rover Curiosity discovered solid evidence for an ancient streambed that used to flow through the rocks. The rover discovered conglomerates, which are rocks made up of sand and gravel. After studying pictures of these rocks, scientists concluded that shape and size of the pebbles that make up the conglomerates signify that they were eroded by water, perhaps several billion years ago. Satellites used to capture evidence of existing channels, which could indicate running water, but did not prove it. This was the first solid major evidence that support these satellite images.
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5,103 |
# Circles, Chords and Tangents - Mathematics Form 3 Notes
## Length of an Arc
• The Arc length as shown in the figure above is given by;
θ x 2πr. where r is the radius of the circle
360
Example
Find the length of an arc subtended by an angle of 2500 at the centre of the circle of radius 14 cm.
Solution
Length of an arc = θ x 2πr
360
= 250 x 2 x 22 x 14 = 61.11 cm
360 7
Example
The length of an arc of a circle is 11.0 cm.Find the radius of the circle if an arc subtended an angle of 900 at the centre.
Solution
Arc length = θ x 2πr but θ = 900
360
Therefore 11 = 90 x 2 x 22 x r
360 7
r = 7.0 cm
Example
Find the angle subtended at the centre of a circle by an arc of 20 cm, if the circumference of the circle is 60 cm.
Solution
= θ x 2πr = 20
360
But 2πr = 60 cm
Therefore, θ x 60 = 20
360
θ = 20 x 360
60
θ = 120
0
## Chords
• Chord of a circle: A line segment which joins two points on a circle.
• Diameter: a chord which passes through the center of the circle.
• Radius: the distance from the center of the circle to the circumference of the circle
### Perpendicular Bisector of a Chord
• A perpendicular drawn from the centre of the circle to a chord bisects the chord.
• Note;
• Perperndicular drawn from the centre of the circle to chord bisects the cord (divides it into two equal parts)
• A straight line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.
Example
The radius of a circle centre O is 13 cm.Find the perpendicular distance from O to the chord, if AB is 24 cm.
Solution
OC bisects chord AB at C
Therefore, AC =12 cm
In ∆AOC ,OC2 = AO2 AC2
= 132 - 122 = 25
Therefore, OM = √25 = 5 cm
### Parallel Chords
• Any chord passing through the midpoints of all parallel chords of a circle is a diameter
Example
In the figure below CD and AB are parallel chords of a circle and 2 cm apart. If CD = 8 cm and AB= 10 cm, find the radius of the circle
Solution
• Draw the perpendicular bisector of the chords to cut them at K and L .
• Join OD and OC
• In triangle ODL,
• DL = 4 cm and KC =5 cm
• Let OK = x cm
• Therefore ( x + 2)2 + 42 = r2
In triangle OCK;
• x2 + 52 =r2
• Therefore ( x + 2)2 + 42 = x2 + 52
• x2 + 4x + 20) + 42 = x2 + 52
• 4x + 20 = 25
• 4x = 5
• x =
Using the equation x2 + 52 = r2
r2 = (5/4)2 52
= 25/16 + 25
425/16
r =√(425/16)
= 5.154 cm
Example
A hemispherical pot is used for a hanging basket. The width of the surface of the soil is 30cm. The maximum depth of the soil is 10 cm. Find the radius of the pot
Solution
c2 = a2 + b2
r2 = 152 + (r − 10)2
r2 = 225 + r2 − 20r + 100
20r = 325
r = 325/20
r = 16.25 cm
### Intersecting Chords
In general DE = EB or DE ×EC = EB × AE
AE EC
Example
In the example above AB and CD are two chords that intersect in a circle at Given that AE = 4 cm, CE =5 cm and DE = 3 cm, find AB.
Solution
Let EB = x cm
4
× x = 5 × 3
4x = 15
x = 3.75 cm
Since AB = AE + EB
AB = 4 + 3.75
= 7.75 cm
### Equal Chords.
• Angles subtended at the centre of a circle by equal chords are equals
• If chords are equal they are equidistant from the centre of the circle
### Secant
• A chord that is produced outside a circle is called a secant
BC = CD OR BC × CA = CD × EC
EC CA
Example
Find the value of AT in the figure below. AR = 4 cm, RD = 5 cm and TC = 9 cm.
Solution
AC x AT = AO x AR
(x + 9) x = (5 + 4) 4
x
2 + 9x = 36
x
2 + 9x − 36 = 0
(x + 12) (x − 3) = 0
Therefore, x = − 12 or x = 3
x can only be a positive number not negative hence x = 3 cm
## Tangent and Secant
#### Tangent
• A line which touches a circle at exactly one point is called a tangent line and the point where it touches the circle is called the point of contact
#### Secant
• A line which intersects the circle in two distinct points is called a secant line (usually referred to as a secant).
In the figures below, A shows a secant while B shows a tangent
### Construction of a Tangent
• Draw a circle of any radius and centre O.
• Join O to any point P on the circumference
• Produce OP to a point P outside the circle
• Construct a perpendicular line SP through point P
• The line is a tangent to the circle at P as shown below.
Note;
• The radius and tangent are perpendicular at the point of contact.
• Through any point on a circle , only one tangent can be drawn
• A perpendicular to a tangent at the point of contact passes thought the centre of the circle.
Example
In the figure below PT = 15 cm and PO = 17 cm, calculate the length of PQ.
Solution
OT2 = OP2 PT2
= 172 - 152
= √64
OT = 8 cm
### Properties of Tangents to a Circle from an External Point
If two tangents are drawn to a circle from an external point
• They are equal
• They subtend equal angles at the centre
• The line joining the centre of the circle to the external point bisects the angle between the tangents
Example
The figure below represents a circle centre O and radius 5 cm. The tangents PT is 12 cm long. Find:
1. OP
2. Angle TPT'
Solution
1. Join O to P
OP2 = OC2 PC2 (pythagoras theorem)
OP
2 = 52 + 122
= 25 + 144
= √169
Therefore, OP = 13 cm
2. <TPT' = 2TPO( PO bisect < TPT')
<OTP = 900
∆TPO is a right angled at T
cos <TPO= 12/13 = 0.9231
Therefore, <TPO = 22.62 0
Hence <TPT1 = 22.62 0 ×2
= 45.240
### Tangent Problem
• The common-tangent problem is named for the single tangent segment that’s tangent to two circles.
• Your goal is to find the length of the tangent.
• These problems are a bit involved, but they should cause you little difficulty if you use the straightforward three-step solution method that follows.
• The following example involves a common external tangent (where the tangent lies on the same side of both circles).
• You might also see a common-tangent problem that involves a common internal tangent (where the tangent lies between the circles). No worries: The solution technique is the same for both.
Example
Given the radius of circle A is 4 cm and the radius of circle Z is 14 cm and the distance between the two circles is 8 cm. Find the length of the common tangent.
Here’s how to solve it:
1. Draw the segment connecting the centers of the two circles and draw the two radii to the points of tangency (if these segments haven’t already been drawn for you).
Draw line AZ and radii AB and ZY.
The following figure shows this step. Note that the given distance of 8 cm between the circles is the distance between the outsides of the circles along the segment that connects their centers.
2. From the center of the smaller circle, draw a segment parallel to the common tangent till it hits the radius of the larger circle (or the extension of the radius in a common-internal-tangent problem).
You end up with a right triangle and a rectangle; one of the rectangle’s sides is the common tangent. The above figure illustrates this step.
3. You now have a right triangle and a rectangle and can finish the problem with the Pythagorean Theorem and the simple fact that opposite sides of a rectangle are congruent.
The triangle’s hypotenuse is made up of the radius of circle A, the segment between the circles, and the radius of circle Z. Their lengths add up to 4 + 8 + 14 = 26. You can see that the width of the rectangle equals the radius of circle A, which is 4; because opposite sides of a rectangle are congruent, you can then tell that one of the triangle’s legs is the radius of circle Z minus 4, or 14 – 4 = 10.
You now know two sides of the triangle, and if you find the third side, that’ll give you the length of the common tangent.
You get the third side with the Pythagorean Theorem:
x2 + 10= 262
x2 + 100 = 676
x2 = 576
x=24
(Of course, if you recognize that the right triangle is in the 5 : 12 : 13 family, you can multiply 12 by 2 to get 24 instead of using the Pythagorean Theorem.)Because opposite sides of a rectangle are congruent, BY is also 24, and you’re done.
Now look back at the last figure and note where the right angles are and how the right triangle and the rectangle are situated; then make sure you heed the following tip and warning.
Note the location of the hypotenuse. In a common-tangent problem, the segment connecting the centers of the circles is always the hypotenuse of a right triangle. The common tangent is always the side of a rectangle, not a hypotenuse.
In a common-tangent problem, the segment connecting the centers of the circles is never one side of a right angle. Don’t make this common mistake.
### How to Construct a Common Exterior Tangent Line to Two Circles
• In this lesson you will learn how to construct a common exterior tangent line to two circles in a plane such that no one is located inside the other using a ruler and a compass.
Problem 1
For two given circles in a plane such that no one is located inside the other, to construct the common exterior tangent line using a ruler and a compass.
Solution
We are given two circles in a plane such that no one is located inside the other (Figure 1a).
We need to construct the common exterior tangent line to the circles using a ruler and a compass.
First, let us analyze the problem and make a sketch (Figures 1 a and 1b). Let AB be the common tangent line to the circles we are searching for.
Let us connect the tangent point A of the first circle with its center P and the tangent point B of the second circle with its center Q (Figure 1 a and 1 b).
Then the radii PA and QB are both perpendicular to the tangent line AB (lesson A tangent line to a circle is perpendicular to the radius drawn to the tangent point under the topic Circles and their properties ).
Hence, theradii PA and QB are parallel.
Figure 1a. To the Problem 1
Figure 1b. To the solution of the Problem 1
Figure 1c. To the construction step 3
Next, let us draw the straight line segment CQ parallel to AB through the point Q till the intersection with the radius PA at the point C (Figure 1b). Then the straight line CQ is parallel to AB.
Hence, the quadrilateral CABQ is a parallelogram (moreover, it is a rectangle) and has the opposite sides QB and CA congruent. The point C divides the radius PA in two segments of the length r2 (CA) and r1 - r2 (PC).
It is clear from this analysis that the straight line QC is the tangent line to the circle of the radius r1 - r2 with the center at the point P (shown in red in Figure 1b).
It implies that the procedure of constructing the common exterior tangent line to two circles should be as follows:
1. draw the auxiliary circle of the radius r1 - r2 at the center of the larger circle (shown in red in Figure 1b);
2. construct the tangent line to this auxiliary circle from the center of the smaller circle (shown in red in Figure 1b). In this way you will get the tangent point C on the auxiliary circle of the radius r1 - r2;
3. draw the straight line from the point P to the point C and continue it in the same direction till the intersection with the larger circle (shown in blue in Figure 1 b). The intersection point A is the tangent point of the common tangent line and the larger circle. Figure 1 c reminds you how to perform this step.
4. draw the straight line QB parallel to PA till the intersection with the smaller circle (shown in blue in Figure 1b). The intersection point B is the tangent point of the common tangent line and the smaller circle;
5. the required common tangent line is uniquely defined by its two points A and B.
Note that all these operations 1 - 4 can be done using a ruler and a compass. The problem is solved.
Problem 2
Find the length of the common exterior tangent segment to two given circles in a plane, if they have the radii r1 and r2 and the distance between their centers is d.
No one of the two circles is located inside the other.
Solution
Let us use the Figure 1b from the solution to the previous Problem 1 .
This Figure is relevant to the Problem 2. It is copied and reproduced in the Figure below on the right for your convenience.
It is clear from the solution of the Problem 1 above that the common exterior tangent segment |AB| is congruent to the side |CQ| of the quadrilateral (rectangle) CABQ.
From the other side, the segment CQ is the leg of the right-angled triangle DELTAPCQ. This triangle has the hypotenuse's measure d and the other leg's measure
r1 r2 . Therefore, the length of the common exterior tangent segment |AB| is equal to
|AB| = √(
d− (r r2)2)
Note that the solvability condition for this problem is d > r1 r2.
It coincides with the condition that no one of the two circles lies inside the other.
Example 1
Find the length of the common exterior tangent segment to two given circles in a plane, if their radii are 6 cm and 3 cm and the distance between their centers is 5 cm.
Solution
Use the formula (1 ) derived in the solution of the Problem 2.
According to this formula, the length of the common exterior tangent segment to the two given circles is equal to
√[52 − (6 − 3)2] = √(52 − 32) = √(25 − 9)
= 4 cm
The length of the common exterior tangent segment to the two given circles is 4 cm
### Contact of Circles
• Two circle are said to touch each other at a point if they have a common tangent at that point.
• Point T is shown by the red dot.
Note;
• The centers of the two circles and their point of contact lie on a straight line
• When two circles touch each other internally, the distance between the centers is equal to the difference of the radii i.e. PQ= TP − TQ
• When two circles touch each other externally, the distance between the centers is equal to the sum of the radii i.e. OR =TO +TR
### Transverse (Interior) Common Tangents
In the figure below P and Q are centres of two circles with radiir, and r, respectively. Given that r> r2, construct the transverse common tangents to both circles.
Procedure
1. With centre P, construct a circle whose radius PR is equal to r+ r2
2. Join P to Q and bisect PQ to get point C.
3. With centre C and radius PC, draw arcs to cut the circle whose radius is r1 + r2, at R and S. Join to R and Q to S.
4. Draw the lines PR and PS to cut the circle whose radius is r1 at M and N respectively.
5. Draw line QX parallel to PS and line QY parallel to PR.
6. Draw lines MY and NX. These are the required transverse (interior) common tangents.
Note:
PR = r1 + r2, (construction).
PM =r1 (given)
∴ RM = PR − PM (r1 + r2) − r1
∴ RM QY But RM is parallel to QY (construction)
∴ MRQY is a parallelogram (opposite sides equal and parallel).
QR is tangent to circle radius PR (construction).
∠PRO = 90° (radius ⟂ tangent).
But ∠YQR = 90° (interior ∠s of a parallelogram).
∴ MRQY is a rectangle.
∴ ∠PMY = ∠QYM = 90°.
### Alternate Segment Theorem
The angle which the chord makes with the tangent is equal to the angle subtended by the same chord in the alternate segment of the circle.
Angle a = Angle b
Note;
The blue line represents the angle which the chord CD makes with the tangent PQ which is equal to the angle b which is subtended by the chord in the alternate segment of the circle.
Illustrations
• Angle s = Angle t
• Angle a = Ange b we use the alternate segment theorm
### Tangent-secant Segment Length Theorem
If a tangent segment and secant segment are drawn to a circle from an external point, then the square of the length of the tangent equals the product of the length of the secant with the length of its external segment.
(TV)
2 = TW.TX
Example
In the figure above ,TW=1 0 cm and XW = 4 cm. find TV
Solution
(TV)2 = TW.TX
(TV)
2 = 10 ×6 ( tx = tw − xw)
= 16
TV = 4 cm
## Circles and Triangles
### Inscribed Circle
• Construct any triangle ABC.
• Construct the bisectors of the three angles
• The bisectors will meet at point I
• Construct a perpendicular from O to meet one of the sides at M
• With the centre I and radius IM draw a circle
• The circle will touch the three sides of the triangle ABC
• Such a circle is called an inscribed circle or in circle.
• The centre of an inscribed circle is called the incentre
### Circumscribed Circle
• Construct any triangle ABC.
• Construct perpendicular bisectors of AB , BC, and AC to meet at point O.
• With O as the centre and using OB as radius, draw a circle
• The circle will pass through the vertices A , B and C as shown in the figure below
### Escribed Circle
• Construct any triangle ABC.
• Extend line BA and BC
• Construct the perpendicular bisectors of the two external angles produced
• Let the perpendicular bisectors meet at O
• With O as the centre draw the circle which will touch all the external sides of the triangle
Note;
• Centre O is called the ex-centre
• AO and CO are called external bisectors.
## Past KCSE Questions on the Topic.
1. The figure below represents a circle a diameter 28 cm with a sector subtending an angle of 750 at the centre.
Find the area of the shaded segment to 4 significant figures
1. ∠PST
2. The figure below represents a rectangle PQRS inscribed in a circle centre 0 and radius 17 cm. PQ = 16 cm.
Calculate
1. The length PS of the rectangle
2. The angle POS
3. The area of the shaded region
3. In the figure below, BT is a tangent to the circle at B. AXCT and BXD are straight lines. AX = 6 cm, CT = 8 cm, BX = 4.8 cm and XD = 5 cm.
Find the length of
1. XC
2. BT
4. The figure below shows two circles each of radius 7 cm, with centers at X and Y. The circles touch each other at point Q.
Given that ∠AXD = ∠BYC = 1200 and lines AB, XQY and DC are parallel, calculate the area of:
1. Minor sector XAQD (Take π 22/7)
2. The trapezium XABY
5. The figure below shows a circle, centre, O of radius 7 cm. TP and TQ are tangents to the circle at points P and Q respectively. OT =25 cm.
Calculate the length of the chord PQ
6. The figure below shows a circle centre O and a point Q which is outside the circle
Using a ruler and a pair of compasses, only locate a point on the circle such that angle OPQ = 90o
7. In the figure below, PQR is an equilateral triangle of side 6 cm. Arcs QR, PR and PQ arcs of circles with centers at P, Q and R respectively.
Calculate the area of the shaded region to 4 significant figures
8. In the figure below AB is a diameter of the circle. Chord PQ intersects AB at N. A tangent to the circle at B meets PQ produced at R.
Given that PN = 14 cm, NB = 4 cm and BR = 7.5 cm, calculate the length of:
1. NR
2. AN
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Question
# In the following figure of an isosceles triangle, find the value of $'x'$?(All dimensions are in centimeter)
Hint: Notice that there are two right-angled triangles in the triangle $\Delta ABC$. They have a common side. Use Pythagoras’ theorem to find the measure of that side. Then use that value while applying Pythagoras’ theorem in another triangle to obtain the value of $'x'$.
Let’s analyse the question through the diagram first. Here we have an isosceles triangle $\Delta ABC$, which has an altitude $BE$ on the side $AC$. The altitude divides the $\Delta ABC$ into two right-angled triangles $\Delta BEC$ and $\Delta BEA$. Using all this information we need to find the value of $'x'$
If we notice $\Delta BEC$ in the figure, we can see that $x$ is the hypotenuse of this triangle.
Now we can use Pythagoras’ theorem that states that, “the sum of the squares of the sides of a right triangle is equal to the square on the hypotenuse”
Using Pythagoras’ theorem in $\Delta BEA$, we get:
$\Rightarrow A{B^2} = B{E^2} + A{E^2}$
Now we can easily substitute the given values in the above equation
$\Rightarrow {9^2} = B{E^2} + {7^2}$
$\Rightarrow B{E^2} = 81 - 49 = 32$
$\Rightarrow BE = \sqrt {32} = 4\sqrt 2 cm$
Similarly, again using Pythagoras’ theorem in $\Delta BEC$, we can see that:
$\Rightarrow B{C^2} = B{E^2} + E{C^2}$
Let’s substitute the already known values in the above equation
$\Rightarrow B{C^2} = {\left( {4\sqrt 2 } \right)^2} + {2^2} = 32 + 4 = 36$
$\Rightarrow BC = x = 6 cm$
Hence, we got the value of $'x'$ as $6 cm$.
Note: Try to first understand all the given information in the diagram given in the question. Be careful while taking square root or squaring both sides in equations. An alternative approach to the problem is to apply Pythagoras’ theorem in both the right-angled triangles and equate for the value of $BE$. This will give you an equation with one unknown, i.e. $'x'$.<|endoftext|>
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NCERT Solutions for Class 11th Biology Chapter 13 Photosynthesis in Higher Plants
National Council of Educational Research and Training (NCERT) Book Solutions for class 11th
Chapter: Chapter 13 – Photosynthesis in Higher Plants
Class 11th Biology Chapter 13 Photosynthesis in Higher Plants NCERT Solution is given below.
Question 1: By looking at a plant externally can you tell whether a plant is C3 or C4? Why and how?
Answer One cannot distinguish whether a plant is C3 or C4 by observing its leaves and other morphological features externally. Unlike C3 plants, the leaves of C4 plants have a special anatomy called Kranz anatomy and this difference can only be observed at the cellular level. For example, although wheat and maize are grasses, wheat is a C3 plant, while maize is a C4 plant.
Question 2: By looking at which internal structure of a plant can you tell whether a plant is C3 or C4? Explain.
Answer The leaves of C4 plants have a special anatomy called Kranz anatomy. This makes them different from C3 plants. Special cells, known as bundle-sheath cells, surround the vascular bundles. These cells have a large number of chloroplasts. They are thick-walled and have no intercellular spaces. They are also impervious to gaseous exchange. All these anatomical features help prevent photorespiration in C4 plants, thereby increasing their ability to photosynthesise.
Question 3: Even though a very few cells in a C4 plant carry out the biosynthetic – Calvin pathway, yet they are highly productive. Can you discuss why?
Answer The productivity of a plant is measured by the rate at which it photosynthesises. The amount of carbon dioxide present in a plant is directly proportional to the rate of photosynthesis. C4 plants have a mechanism for increasing the concentration of carbon dioxide. In C4 plants, the Calvin cycle occurs in the bundle-sheath cells. The C4 compound (malic acid) from the mesophyll cells is broken down in the bundle-
sheath cells. As a result, CO2 is released. The increase in CO2 ensures that the enzyme RuBisCo does not act as an oxygenase, but as a carboxylase. This prevents photorespiration and increases the rate of photosynthesis. Thus, C4 plants are highly productive.
Question 4: RuBisCo is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCo carries out more carboxylation in C4 plants?
Answer The enzyme RuBisCo is absent from the mesophyll cells of C4 plants. It is present in the bundle-sheath cells surrounding the vascular bundles. In C4 plants, the Calvin cycle occurs in the bundle-sheath cells. The primary CO2 acceptor in the mesophyll cells is phosphoenol pyruvate – a three-carbon compound. It is converted into the four-carbon compound oxaloacetic acid (OAA). OAA is further converted into malic acid. Malic acid is transported to the bundle-sheath cells, where it undergoes decarboxylation and CO2 fixation occurs by the Calvin cycle. This prevents the enzyme RuBisCo from acting as an oxygenase.
Question 5: Suppose there were plants that had a high concentration of Chlorophyll-b, but lacked chlorophyll-a, would it carry out photosynthesis? Then why do plants have chlorophyll-b and other accessory pigments?
Answer Chlorophyll-a molecules act as antenna molecules. They get excited by absorbing light and emit electrons during cyclic and non-cyclic photophosphorylations. They form the reaction centres for both photosystems I and II. Chlorophyll-b and other photosynthetic pigments such as carotenoids and xanthophylls act as accessory pigments. Their role is to absorb energy and transfer it to chlorophyll-a. Carotenoids and xanthophylls also protect the chlorophyll molecule from photo-oxidation. Therefore, chlorophyll-a is essential for photosynthesis.
If any plant were to lack chlorophyll-a and contain a high concentration of chlorophyll-b, then this plant would not undergo photosynthesis.
Question 6: Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable?
Answer Since leaves require light to perform photosynthesis, the colour of a leaf kept in the dark changes from a darker to a lighter shade of green. Sometimes, it also turns yellow. The production of the chlorophyll pigment essential for photosynthesis is directly proportional to the amount of light available. In the absence of light, the production of chlorophyll-a molecules stops and they get broken slowly. This changes the colour of the leaf gradually to light green. During this process, the xanthophyll and carotenoid pigments become predominant, causing the leaf to become yellow. These pigments are more stable as light is not essential for their production. They are always present in plants.
Question 7: Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
Answer Light is a limiting factor for photosynthesis. Leaves get lesser light for photosynthesis when they are in shade. Therefore, the leaves or plants in shade perform lesser photosynthesis as compared to the leaves or plants kept in sunlight.
In order to increase the rate of photosynthesis, the leaves present in shade have more chlorophyll pigments. This increase in chlorophyll content increases the amount of light absorbed by the leaves, which in turn increases the rate of photosynthesis. Therefore, the leaves or plants in shade are greener than the leaves or plants kept in the sun.
Question 8: Figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions:
(a) At which point/s (A, B or C) in the curve is light a limiting factor?
(b) What could be the limiting factor/s in region A?
(c) What do C and D represent on the curve?
(a) Generally, light is not a limiting factor. It becomes a limiting factor for plants growing in shade or under tree canopies. In the given graph, light is a limiting factor at the point where photosynthesis is the minimum. The least value for photosynthesis is in region A. Hence, light is a limiting factor in this region.
(b) Light is a limiting factor in region A. Water, temperature, and the concentration of carbon dioxide could also be limiting factors in this region.
(c) Point D represents the optimum point and gives the light intensity at which the maximum photosynthesis is recorded. The rate of photosynthesis remains constant after this point, even though the intensity of light.
Question 9: Give comparison between the following:
(a) C3 and C4 pathways
(b) Cyclic and non-cyclic photophosphorylation
(c) Anatomy of leaf in C3 and C4 plants
Answer (a) C3 and C4 pathways
(c) Anatomy of the leaves in C3 and C4 plants
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# How to Find the Height of a Triangle
The height of a triangle can be useful in a variety of geometry calculations, like determining the area of a triangle or the volume of a triangular prism. While it may seem difficult at first to find the height of a triangle, especially with non-right and non-equilateral triangles, it is not hard with the correct process.
In this guide, you’ll learn everything you need to know to be able to find the height of any triangle you might come across in your studies.
## Figure Out What You Know
Depending on which pieces of information you have about the triangle you’re dealing with, there are different strategies for determining its height. Based on the information you know, you’ll be able to select a technique for finding the height of the triangle that is guaranteed to work.
• If you have the area and the base length of the triangle, you can work backwards using the area formula.
• If the triangle is equilateral and you know the side lengths, you can use the Pythagorean theorem to find the height because the height of an equilateral triangle divides it into two identical right triangles.
• If you have the lengths of all three sides but no angles, you can use Heron’s formula to find the area and work backwards to find the height with the area formula.
• If you have the lengths of one side and an angle on either side of it, you can use the trigonometry formula to find the height, where c is the side and A is the angle.
## Finding Height Given Area and Base Length
The area formula for a triangle is , where A is the area, b is the base length, and h is the height perpendicular to the base. If you know A and b, you can find h by plugging in A and b into the area formula and solving.
## Finding Height of an Equilateral Triangle With Side Length
Equilateral triangles have the same length on all three sides and the same interior angle (60 degrees) throughout. Because of this, the height of an equilateral triangle divides the triangle into perfect halves.
Use the Pythagorean theorem on one half to find the missing height, since you know two of the three side lengths in that half.
## Finding Height With Three Sides
Heron’s formula allows you to find the area of a triangle given three side lengths.
Find the value s using this formula: s = (a + b + c)/2
Then, the area is √(s(s-a)(s-b)(s-c)) With this area and the bottom side as the base length, follow the “Finding Height Given Area and Base Length” instructions.
## Finding Height With One Side and One Angle
Assuming that the angle is on either side of the side you know the length of, the height is equal to , where c is the side length and A is the angle.
## Still Can’t Find It?
Determining the height of a triangle without at least knowing the base is impossible. However, you can rotate the triangle so that any side is the base and find the height relative to that base.<|endoftext|>
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# How to Use the 68 95 and 99.7 Rule Calculator
How to Use the 68 95 and 99.7 Rule Calculator
Probability and statistics play a crucial role in various fields, including finance, science, and social sciences. One fundamental concept in statistics is the normal distribution. The normal distribution, also known as the bell curve, is often used to analyze and interpret data. To understand the distribution better, statisticians developed the 68 95 and 99.7 rule, which provides valuable insights into the probability of events occurring within certain ranges. In this article, we will explore how to use the 68 95 and 99.7 rule calculator and answer some frequently asked questions.
Understanding the 68 95 and 99.7 Rule:
Before diving into the calculator, it’s essential to grasp the concept of the 68 95 and 99.7 rule. The rule states that for a normal distribution:
– Approximately 68% of the data falls within one standard deviation of the mean.
– Approximately 95% of the data falls within two standard deviations of the mean.
– Approximately 99.7% of the data falls within three standard deviations of the mean.
This rule helps us understand the likelihood of certain events occurring within different standard deviation ranges. By using the 68 95 and 99.7 rule calculator, you can easily determine the probabilities associated with specific data sets.
Using the 68 95 and 99.7 Rule Calculator:
To use the 68 95 and 99.7 rule calculator, follow these steps:
1. Input the mean: Start by entering the mean value of your data set into the calculator. The mean represents the average value around which your data is centered.
2. Input the standard deviation: Next, input the standard deviation, which measures the spread of your data. It tells you how much your data deviates from the mean.
3. Select the desired range: Choose the range you want to calculate probabilities for. For example, if you want to determine the probability of an event occurring within two standard deviations of the mean, select the “2” option.
4. Calculate the probability: Once you’ve entered all the necessary information, click on the “Calculate” button. The calculator will provide you with the probability associated with the selected range.
5. Interpret the results: The calculator will display the probability as a decimal or percentage. For instance, if the probability is 0.68, this means there is a 68% chance that the event will occur within the selected range.
Q1. Why is the 68 95 and 99.7 rule important?
A1. The 68 95 and 99.7 rule helps us understand the distribution of data and provides valuable insights into the probability of events occurring within specific ranges. It is widely used in statistical analysis and decision-making processes.
Q2. What if my data is not normally distributed?
A2. The 68 95 and 99.7 rule is specifically designed for data that follows a normal distribution. If your data does not meet this assumption, the rule may not accurately represent the probabilities. In such cases, alternative statistical methods should be used.
Q3. How can I interpret the probabilities obtained from the calculator?
A3. The probabilities obtained from the calculator represent the likelihood of an event occurring within the selected range. For example, a probability of 0.95 indicates a 95% chance of the event falling within the specified range.<|endoftext|>
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# How do you describe a system of equations?
## How do you describe a system of equations?
A “system” of equations is a set or collection of equations that you deal with all together at once. Linear equations (ones that graph as straight lines) are simpler than non-linear equations, and the simplest linear system is one with two equations and two variables.
## What are the three types of solutions to a system of linear equations?
There are three possible outcomes for a system of linear equations: one unique solution, infinitely many solutions, and no solution.
## What are the applications of rational functions in real life situation?
Rational expressions and rational equations can be useful tools for representing real life situations and for finding answers to real problems. In particular, they are quite good for describing distance-speed-time questions, and modeling multi-person work problems.
## How do you teach linear equations?
There are many ways to teach about linear equations, right. You can (a) use the old t-table approach; (b) you can draw a line and then figure out its slope and y-intercept; or you can (c) first explain the slope-intercept formula and then explain how the equation aligns with the formula.
## What are systems of linear equations used for in real life?
Almost any situation where there is an unknown quantity can be represented by a linear equation, like figuring out income over time, calculating mileage rates, or predicting profit. Many people use linear equations every day, even if they do the calculations in their head without drawing a line graph.
## How do you explain an equation to a child?
What Is an Equation? An equation is a mathematical sentence that has two equal sides separated by an equal sign. 4 + 6 = 10 is an example of an equation. We can see on the left side of the equal sign, 4 + 6, and on the right hand side of the equal sign, 10.
## What is a system of a linear equation?
A system of linear equations is just a set of two or more linear equations. In two variables (x and y) , the graph of a system of two equations is a pair of lines in the plane. The lines intersect at infinitely many points. (The two equations represent the same line.)
## What are the features of rational equation?
A rational equation is an equation containing at least one fraction whose numerator and denominator are polynomials, \frac{P(x)}{Q(x)}. Q(x)P(x). These fractions may be on one or both sides of the equation.
## Why are systems of linear equations important?
Linear equations are an important tool in science and many everyday applications. They allow scientist to describe relationships between two variables in the physical world, make predictions, calculate rates, and make conversions, among other things. Graphing linear equations helps make trends visible.
## What determines a rational function?
A rational function is defined as the quotient of polynomials in which the denominator has a degree of at least 1 . In other words, there must be a variable in the denominator. The general form of a rational function is p(x)q(x) , where p(x) and q(x) are polynomials and q(x)≠0 .
## What does write an equation mean?
An equation is a mathematical sentence containing an equals sign. It tells us that two expressions mean the same thing, or represent the same number. An equation can contain variables and constants. Using equations, we can express math facts in short, easy-to-remember forms and solve problems quickly.
## What makes a system linear?
As a rule of thumb, a system is linear, if the operations on the input signal are all linear and no signal-independent terms are contained. What are linear operations? Scaling of the input signal: y(t)=ax(t) Time-shifting the input signal y(t)=x(t−a) Scaling the argument of the signal y(t)=x(at)
## How do you introduce an equation?
In many cases, an equation contains one or more variables. These are still written by placing each expression on either side of an equals sign (= ). For example, the equation x+3=5 x + 3 = 5 , read “x plus three equals five”, asserts that the expression x+3 is equal to the value 5.<|endoftext|>
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Mike 111 is a popular Researcher on the Egyptsearch Forum: Egypt. Below is a piece he did on the Yamassee War.
The Yamasee War
The Yamasee Indians were part of the Muskhogean language group. Their traditional homelands lay in present-day northern Florida and southern Georgia. The advent of the Spanish in the late 16th century forced the Yamasee to migrate north into what would become South Carolina. Relations between the tribe and English settlers in that region were generally positive during the latter half of the 17th century.
Not surprisingly, problems between the races developed. The continuing influx of white settlers put pressure on Indian agricultural and hunting lands. The relationship was further complicated in that the tribe had become dependent on English firearms and other manufactured items, and had incurred a large debt, typically payable in deerskins. White fur traders acted on their displeasure by enslaving a number of Yamasee women and children to cover portions of the outstanding debt.
In the spring of 1715, the Yamasee formed a confederation with other tribes and struck at the white settlements in South Carolina. Several hundred settlers were killed, homes burned and livestock slaughtered. The frontier regions were emptied; some fled to the relative safety of North Carolina and others pushed on to even more secure Virginia. Charleston also received large numbers of frightened settlers.
At the height of the fighting, it appeared that the tribal confederation's overwhelming numerical superiority would end in the white settlements' complete destruction in the region. This would have been a virtual certainty if the confederacy had successfully drawn the Cherokee into their cause. Instead, the Cherokee gave in to the lure of English weapons and other goods, and chose to aid the Carolinians. In a further stroke of good fortune, the besieged settlers also managed to gain support from Virginia ~ez_mdash~ an event not assured in this age of intense colonial rivalries.
The tide turned against the Yamasee, who were slowly pushed south through Georgia back into their ancestral lands in northern Florida. There, the tribe was virtually annihilated by protracted warfare with the Creeks, but some members were absorbed by the Seminole.
The Yamasee War took a heavy toll in South Carolina. Such terror had been instilled in the minds of the frontiersmen that it would take nearly 10 years for resettlement to occur in many areas. The warfare also brought a sharp change to the region's economy. Originally, farming had been the settlers' primary occupation, but the livestock supply had been so drastically depleted that many farms disappeared. In their absence, enterprising South Carolinians turned to the forests as a source of naval stores (tar, pitch and turpentine) and soon developed a lucrative trade with England. Later, the economy would develop rice and indigo as its primary products.<|endoftext|>
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This series offers an overview of the various peripheral functions that support effective use of MCUs. In this session, we look at how the MCU uses serial communication to communicate with external peripherals.
Parallel Connection vs. Serial Connection
As we've discussed before, the MCU functions as the "brain" of the embedded device, whereas external peripherals units as the "hands and feet". This means that the MCU must communicate with each of these units. Consider, for example, how we might connect up a sensor. The MCU's built-in GPIO (general-purpose input/output) port, explained in the first session of this series, could be used to connect eight signal lines with the sensor, enabling the sensor to send eight bits of data at a time—using up eight GPIO data pins. This type of transfer is referred to as parallel, as one full byte of data is transferred at a single time along parallel lines. (See Figure 1, left) But using eight lines to connect to a single sensor is usually a waste of resources. Is there a way to accomplish the same thing using fewer lines?
Of course there is. We can send the bits one at a time over a single data line. Because the bits are now sent in series, we call this serial communication. (See Figure 1, right) Note, however, that the MCU uses parallel communication for its internal processing and communication. This means that the MCU must also handle the necessary conversions: "serial-to-parallel" conversion of data received from the sensor, and "parallel-to-serial" conversion of data being sent out to the sensor. On the RX63N, these conversions are handled by the SCI (serial communications interface).
To sum up: assume that we wish to send a single text character, which requires that we send an eight- bit char-type value. If we use parallel transfer, we need one data line for each bit. With serial parallel, we send the eight data bits one after another, over the same single line.
Clearly, serial communication requires fewer pins and wires. In today's world, most MCU-to-peripheral connections are serial. Serial mode is used not just for communication with switches and from ON/OFF sensors, but also by the GPIO to output software-generated motor-drive signals, LED flashing signals, and more.
Built-in UART Enables Easy Serial Communication
Serial communication can be implemented in numerous ways, in accordance with varying electrical characteristics and the requirements of differing protocols. The simplest implementation—requiring only a single wire—is referred to as "start-stop synchronous communication." This mode is often used when communicating with wireless LAN modules and drive monitors.
In start-stop synchronous communication, data is sent in character units1. For control purposes, a start bit is placed at the beginning of each set, and a stop bit is placed at the end. (See Figure 2) This eliminates the need to control transfer timing through use of a separate clock signal line—as required in other common serial modes, such as I2C ("I squared C") and SPI (serial parallel interface). To help ensure transfer accuracy, start/stop data sets can also include a parity bit2.
Start-stop synchronous communication is handled by a component called a "UART" (universal asynchronous receiver/transmitter). The RX63N's built-in SCI includes a UART for this purpose.
2. A bit appended to each (7- or 8-bit) data set. If using even parity, the parity bit is set so that the total number of 1s (in most implementations) is even: if the data value has an odd number of 1s, the parity bit is set to 1; if the data value has an even number of 1s, the parity bit is set to 0. The receiving side will therefore know an error has occurred if it receives a data set with an odd number of 1s. Similarly, if using odd parity, the parity bit is set so that the number of 1s will always be odd.
Start-stop synchronous communication can be implemented as either full duplex or half duplex. With full duplex, two lines are used: one for communication from the MCU to the peripheral, and the other for communication going the other way. This means that communication can go in both directions at the same time. With half-duplex, a single line is used, so that communication can only go one way at a time. The direction is changed by operating a switch. (See Figure 3)
Trying It Out: Communicating with a Computer
Once again, we turn to the GR-SAKURA. This time, we will run a sample program that will enable communication between the GR-SAKURA and a terminal emulator on your personal computer. Specifically, the program will operate as follows: When the terminal emulator sends a "?" character, the GR-SAKURA board will return the character string "GR-SAKURA". Note that, in order to use this program, your computer must be running a terminal emulator. If you are using a Windows machine, you can install and use the free TeraTerm emulator, or use another emulator of your choice. On Mac machines, you can use the built-in "Terminal" program.
Note that the sample program makes use of the Serial class, which is available from the GR-SAKURA's Sakura library. This class supports full-duplex start-stop synchronous communication.
Line 8 of the code begins the serial communication, via USB. Line 13 causes the GR-SAKURA to read the character data from the computer and determine whether it is a "?" character. If so, the GR-SAKURA then returns "GR-SAKURA" to the computer. During execution, processing within the RX63N (the MCU) proceeds as follows: The SCI receives each serial data value from the computer and converts it into parallel form; the CPU checks whether the value is "?"; if it is, the CPU returns the "GR-SAKURA" string via the SCI, which converts the parallel string data into serial form and outputs the result to the computer.
This very simple program is sufficient to send character information in both directions between the computer and the GR-SAKURA. Although our example sends only the predefined string "GR-SAKURA," we encourage you to try expanding the program so that the return string varies according to whatever conditions you decide to set.
/*GR-SAKURA Sketch Template Version: V1.08*/
#define SPEED 9600 //Set transfer speed to 9600 bps
Serial.begin( SPEED, SCI_USB0); //Set serial channel speed. Set serial transfer to use USB port.
if( Serial.read() == '?') //Check if incoming character is "?"
Serial.println( "GR-SAKURA"); //If so, return string "GR-SAKURA"
*Text following "//" is a comment, and does not affect the execution of the program.
*The purpose of this program is for understanding the principle. It is not sufficient for a full implementation.
If you have any problems using the TeraTerm emulator, feel free to visit the following (Japanese-based) support website.
In our next session, we will look at interrupts. We look forward to seeing you then.<|endoftext|>
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Circumference Proof - Geometry Made Easy
# Proof that C=2πr
May 2016 - Proving that the formula for the circumference of a circle is always true is harder than just figuring out that it is true. Now we need to show that C=2πr is always true for every possible circle.
Here is how Archimedes proved it. Draw any circle. Make a point anywhere on the circumference of the purple circle. Use that point as the center of a blue circle with the same radius as the purple circle. The edge of the blue circle should touch the center of the purple circle.
Draw the line segment connecting the centers of the two circles. That's the radius of the both circles. Now draw the line connecting the center of the blue circle to where it crosses the purple circle on both sides, and complete the triangles. You should have two equilateral triangles whose sides are equal to the radius of the purple circle.
Now extend all of the radius lines so they become diameter lines, all the way across the circle, and finish drawing all of the triangles to connect them. You've got six equilateral triangles now, that make an orange hexagon. So the perimeter of your hexagon is the same as six times the radius of your circle. But your circumference is a little bigger than the perimeter of your hexagon, because the shortest distance between two points is always a straight line. This shows you that the circumference of the purple circle has to be more than 6r, so if C=2πr then π (pi) has to be a little bigger than 3, which it is.
Now let's try to get a little closer to the real value of π. Suppose we make our triangles narrower, so that instead of drawing a hexagon, we draw a dodecagon - a shape with twelve sides? We can do that by drawing more points halfway between the points of the hexagon. If we do that, we'll see that that the perimeter of the dodecagon is a little bigger, and closer to being a circle. We have twelve congruent isosceles triangles. Each triangle has two sides that are as long as the radius of the circle, and a third side that we want to know the length of, in order to figure out the perimeter of the dodecagon.
It's pretty hard to figure out the perimeter of the dodecagon. Start by drawing a red line from the center of the purple circle to one of the points of the dodecagon (B). We know that the line AC is the same as the radius of the circle - let's say the radius is 4. It's also the hypotenuse of the bluish right triangle ADC. We also know that the line AD is the same as half the radius of the circle (remember the hexagon was made of equilateral triangles), so AD = 2. Because the bluish triangle is a right triangle, we can use the Pythagorean Theorem to tell us that the third side, CD, is the square root of 12. (4 squared = 2 squared + 12).
Or you can get the same proof from this video
We also know that the red line CB is the same as the radius of the circle, so that's also 4. So the distance from point B to point D must be the radius (4) minus the length of CD, or the square root of 12. BD = 4 - the square root of 12. Now look at the little orange triangle ABD. That's also a right triangle, and now we know two of its sides. AD = 2, and BD = 4 - the square root of 12. We can use the Pythagorean Theorem again to calculate that the green line AB must be 2.07, so the whole perimeter is 12 x 2.07 = 24.84. The circumference of the purple circle has to be more than 24.72, or more than 6.21 r. If C = 2πr, then π has to be a little bigger than 3.1.
The more sides we draw on our polygon, the closer we will get to the real value of pi (3.14159 etc.). Using a polygon with 96 sides, Archimedes was able to calculate that π was a little bigger than 3.1408, which is pretty close.
## More GeometryMore about MathQuatr.us home
Professor Carr
Karen Eva Carr, PhD.
Assoc. Professor Emerita, History
Portland State University
Professor Carr holds a B.A. with high honors from Cornell University in classics and archaeology, and her M.A. and PhD. from the University of Michigan in Classical Art and Archaeology. She has excavated in Scotland, Cyprus, Greece, Israel, and Tunisia, and she has been teaching history to university students for a very long time.
Professor Carr's PSU page
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• Christian Persecution
• Christian Empire
Happy New Year! Welcome back! Get ready for Martin Luther King day with these articles about medieval Africa, slavery, the Civil War, emancipation, the civil rights movement, and Martin Luther King Jr. himself. More about King here...<|endoftext|>
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February 6, 2016
Show . . . and Tell
Bringing tangible symbols into the classroom can often provide memorable metaphors for the students. For example, in this lesson you may wish to develop the theme of the perils of idolatry, or putting anything above God. To talk about it theoretically is good. To illustrate it with tangible symbols might prove to be more effective.
To do this, you could bring twenty-first-century “idols” to display as props. You could work your way down a long line of items (such as an iPod, a TV, a picture of an athlete, a plate of junk food, designer clothes, etc.) while engaging them in a conversation about how each one might be “worshipped” (i.e., placed above God), and how they could set healthy boundaries in their use.
When they encounter these common items during the week, they could then well remember the discussion in Sabbath School and be more cognizant of the temptation of worshipping the “idol.”<|endoftext|>
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Protection of biological diversity in Iceland aims at strengthening and preserving for the future those species which have from its earliest days created Icelandic nature and have thrived in the country for millennia, and to prevent the extinction of species through human activities.
The UN Convention on Biological Diversity (CBD) was adopted at the 1992 Rio de Janeiro Earth Summit. Iceland signed the Convention at the conference and it entered into force in the country in 1994. The Convention has three objectives: the conservation of biological diversity; the sustainable use of the components of biological diversity; and the fair and equitable sharing of the benefits of genetic resources.
States are to incorporate various provisions of the Convention into national law, which has to a large extent been done through the Nature Conservation Act, the Act on the Icelandic Institute of Natural History and Natural History Centres, and several other Acts. The Government of Iceland adopted a biological diversity strategy in 2008 and a corresponding action plan in 2010 on the proposal of the Minister for the Environment. Work is underway on updating the strategy to take into consideration the 2020 objectives of the Convention.
See further information about biological diversity in Iceland at the website of the Icelandic Institute of Natural History.<|endoftext|>
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