Solution - Blog Posts
Having everything provided to us has made us greedy and weak. Self sustainability will allow us to live naturally and successfully.
Just imagine if we could even get 50% of people to do something like this. Get outside, plant something, reconnect to Earth. Plus you will feel the satisfaction of providing for yourself!
Grandma Pizza - Pizza This grandma pizza in the style of Long Island, with its homemade savory sauce and cheese topping, is a fantastic substitute for takeout.
Pittsburgh Chipped Ham Barbecues - Pork
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PSY 490 Capstone Course in Psychology
PSY 490 Capstone Course in Psychology Version 9, ~~~~ Week 4 Quiz Not Included in this.
PSY 490 Week 1 DQ 1
PSY 490 Week 1 DQ 2
PSY 490 Week 1 Individual The Diverse Nature of Psychology Paper
PSY 490 Week 1 Summary
PSY 490 Week 2 DQ 1
PSY 490 Week 2 DQ 2
PSY 490 Week 2 Individual Portfolio Presentation
PSY 490 Week 2 Learning Team Psychological Issue Summary
PSY 490 Week 2 Summary
PSY 490 Week 3 DQ 1
PSY 490 Week 3 DQ 2
PSY 490 Week 3 Individual Ethics Awareness Inventory
PSY 490 Week 3 Summary
PSY 490 Week 4 DQ 1
PSY 490 Week 4 DQ 2
PSY 490 Week 4 Individual Pay it Forward
PSY 490 Week 4 Summary
PSY 490 Week 5 DQ 1
PSY 490 Week 5 DQ 2
PSY 490 Week 5 Learning Team Jeopardy Game Presentation
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PSY 205 Week 9 Assignment 2 Caring for a Loved One with Dementia NEW
Assignment 2: Caring for a Loved One with Dementia
Due Week 9 and worth 170 points Design an educational intervention to prepare new and potential caregivers of dementia patients for what to expect in caring for an older loved one. Note: Refer to Dementia Websites or other quality recourses to complete this assignment. Articles from Websites must be cited as articles. Example: Alzheimer’s Association (n.d.) 10 early signs and symptoms of Alzheimer’s . Write a two to three (2-3) page paper in which you: Identify a minimum of five (5) issues that are most important for caregivers to understand and address regarding the behavior of people with dementia. Propose at least three (3) interventions for dealing with the issues identified in Criterion 1 of this assignment. Support your proposal. Determine the possible outcomes of each intervention proposed in Criterion 2, including how effective the caregiver can expect them to be. Discuss the importance of support for the caregiver and suggest three to five (3-5) strategies for the caregiver to take care of him- or herself. Use at least four (4) quality resources in this assignment. Note: Wikipedia and similar Websites do not qualify as quality resources. Your assignment must follow these formatting requirements: Be typed, double spaced, using Times New Roman font (size 12), with one-inch margins on all sides; citations and references must follow APA or school-specific format. Check with your professor for any additional instructions. Include a cover page containing the title of the assignment, the student’s name, the professor’s name, the course title, and the date. The cover page and the reference page are not included in the required assignment page length. The specific course learning outcomes associated with this assignment are: Analyze the interaction between the environment and the individual as it affects stages of development such as infancy, adolescence, adulthood, and later life. Analyze contemporary concerns associated with lifespan development. Use technology and information resources to research issues in lifespan development.
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We have been given this one life, what much can we do with it apart from living as our hearts crave.
We have been given this one life and sadly you can’t live it all within the game of risking yet, confusingly those who risk it at most get out totall liberated.
@lifepath25
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I don’t have anything to hide My shadows are enlightened My words are spoken What are your secrets? Show me your ghosts the demons you try to drown every night in the liquid you call solution
How to pretend
Pretend you're into certain things and that you're expressly good at them. Display great fondness and commitment. When questioned about it, convince everyone, that you love those certain things because they are truly the greatest of all. Go on figurative and/or literary crusades. Organise everything in your life, so it would be interconnected with tose things. Have hobbies, studies, jobs, that relate to them. Have everybody decieved: trick the world, your co-workers, your best friend, your family, EVERYBODY, including yourself. And at the end of the day, when you have to face those certain things, you'll fail. Because you can't actually do anything that you said you could.
Modern day thinking requires us to rapidly make decisions, even the ones that will determine main matters in our lives. It is not hard to be drifted away to some unknown directions, that look tempting but in fact are alien to us. I believed, that I'd be a great engineer and could set up a huge company, that'll be providing me with a grand fortune but things don't work like this. Not all of us will find what we're looking for in popular careers or good-sounding ideologies.
What I have found though, is that prayer is a highly underrated element of happiness. Wherever I might go, it can always take me home from there. Always. So just be strong enough to kneel down and bring yourself before the Father. :)
Sangaku Sunday Bonus - what about the other problems?
In the sangaku series, we've solved two of the four problems on this tablet, the middle two, which I believe were the easiest to work on in terms of geometric arguments - we hardly ever used more than Pythagoras's theorem, though the second one needed some more advanced algebra to finish off.
Here's a quick look at the problems at each end of the tablet, and the main ideas I had to solve them.
On the far left, we have two circles tangent to one another (with centres A and B), inside a larger circle (with centre O) so that their diameters add up to the diameter of the largest. The radii of these three circles, respectively p, q and p+q, are known. The unknown is the radius r of the circle with centre C, which must be tangent to all three original circles (it has a twin on the right-hand side with the same radius).
This is quite quick to solve. Remember that tangent circles mean that the distances between centres is equal to the sum of the radii, e.g. AC = p+r, BC = q+r... Al-Kashi's theorem, which is a general version of Pythagoras's theorem, links the lengths of three sides of a triangle with one of the triangle's angles, and the triangles CAO and CAB have an angle in common, which yields the equation for r by isolating this angle in each application of Al-Kashi's theorem. The result is:
The problem on the far right seems to start in a similar fashion: two circles with fixed radii are offset by a fixed distance. A third circle has its diameter equal to the remainder of the diameter of one of the large circles: this radius can be calculated with little difficulty. What we want to do next is construct circles which are tangent to the two large ones, and the one previously constructed.
The radius of the circle with centre C1 can be obtained as above, but this method does not seem to extend to the subsequent circles, as O, D and C1 are no longer aligned, and there no longer appears to be a common angle in the triangles we want to work with. So I went for a parametric approach, understanding the curves that contain points that are equidistant from two circles. The red curve (which looks like a circle but isn't one) is the set of points at equal distance from the two largest circles, and we seek to intersect this with the set of points that are at equal distance from one large circle and the smaller one, the green curve. The intersection is equidistant from all three circles, so it is the centre of the circle we want to construct. Rotate and repeat for subsequent circles.
The general formulas are horrible and not worth showing, but this is another problem where I have been able to read the results on the tablet. The large circles have radii 61 and 72, and the offset is 23. The radii of the smaller circles, starting with the one in the middle and working outwards are:
17, 15.55, 12.292, 8.832 and 6.038 (I see 八, but I'll give the authors the benefit of the doubt as the top of the character 六 may have been erased by time)
The results with our exact formulas are:
17, 15.58, 12.795, 9.076 and 6.444
Rather close! As with the "three circles in a triangle", I do not know how the authors originally solved this problem.
Sangaku Saturday #14 - the grand finale!
We are only a few steps of algebra away from solving the "three circles in a triangle" problem we set in episode 7. This method will also yield general formulas for the solutions (first with height 1 and base b; for any height h and half-base k, set b=k/h and multiply the results by h).
Before we do that, it's worth noting what the sangaku tablet says. Now I don't read classical Japanese (the tablet dates back to 1854 according to wasan.jp), but I can read numbers, and fishing for these in the text at least allows me to understand the result. The authors of the sangaku consider an equilateral triangle whose sides measure 60: boxed text on the right: 三角面六尺, sankaku-men roku shaku (probably rosshaku), in which 尺, shaku, is the ten marker. In their writing of numbers, each level has its own marker: 尺 shaku for ten, 寸 sun for units, 分 fun for tenths and 厘 rin for hundredths (毛 mô for thousandths also appear, which I will ignore for brevity). Their results are as follows:
甲径三尺八寸八分六厘: diameter of the top (甲 kou) circle 38.86
乙径一尺六寸四分二厘: diameter of the side (乙 otsu) circle 16.42
反径一尺二寸四分二厘: diameter of the bottom (反 han) circle 12.42
I repeat that I don't know classical Japanese (or much modern Japanese for that matter), so my readings may be off, not to mention that these are the only parts of the tablet that I understand, but the results seem clear enough. Let's see how they hold up to our final proof.
1: to prove the equality
simply expand the expression on the right, taking into account that
(s+b)(s-b) = s²-b² = 1+b²-b² = 1.
2: the equation 2x²-(s-b)x-1 = 0 can be solved via the discriminant
As this is positive (which isn't obvious as s>b, but it can be proved), the solutions of the equation are
x+ is clearly positive, while it can be proved the x- is negative. Given that x is defined as the square root of 2p in the set-up of the equation, x- is discarded. This yields the formulas for the solution of the geometry problem we've been looking for:
3: in the equilateral triangle, s=2b. Moreover, the height is fixed at 1, so b can be determined exactly: by Pythagoras's theorem in SON,
Replacing b with this value in the formulas for p, q and r, we get
Now we can compare our results with the tablet, all we need to do is multiply these by the height of the equilateral triangle whose sides measure 60. The height is obtained with the same Pythagoras's theorem as above, this time knowing SN = 60 and ON = 30, and we get h = SO = 30*sqrt(3). Bearing in mind that p, q and r are radii, while the tablet gives the diameters, here are our results:
diameter of the top circle: 2hp = 45*sqrt(3)/2 = 38.97 approx.
diameter of the side circle: 2hr = 10*sqrt(3) = 17.32 approx.
diameter of the bottom circle: 2hq = 15*sqrt(3)/2 = 12.99 approx.
We notice that the sangaku is off by up to nearly a whole unit. Whether they used the same geometric reasoning as us isn't clear (I can't read the rest of the tablet and I don't know if the method is even described), but if they did, the difference could be explained by some approximations they may have used, such as the square root of 3. Bear in mind they didn't have calculators in Edo period Japan.
With that, thank you very much for following the Sangaku Weekends series, hoping that you found at least some of it interesting.
Sangaku Saturday #12
Having mentioned previously how mathematical schools were organised during the Edo period in Japan, we can briefly talk about how mathematicians of the time worked. This was a time of near-perfect isolation, but some information from the outside did reach Japanese scholars via the Dutch outpost near Nagasaki. In fact, a whole field of work became known as "Dutch studies" or rangaku.
One such example was Fujioka Yûichi (藤岡雄市, a.k.a. Arisada), a surveyor from Matsue. I have only been able to find extra information on him on Kotobank: lived 1820-1850, described first as a wasanka (practitioner of Japanese mathematics), who also worked in astronomy, geography and "Dutch studies". The Matsue City History Museum displays some of the tools he would have used in his day: ruler, compass and chain, and counting sticks to perform calculations on the fly.
No doubt that those who had access to European knowledge would have seen the calculus revolution that was going on at the time. Some instances of differential and integral calculus can be found in Japan, but the theory was never formalised, owing to the secretive and clannish culture of the day.
That said, let's have a look at where our "three circles in a triangle" problem stands.
The crucial step is to solve this equation,
and I suggested that we start with a test case, setting the sizes of the triangle SON as SO = h = 4 and ON = k = 3. Therefore, simply, the square root of h is 2, and h²+k² = 16+9 = 25 = 5², and our equation is
x = 1 is an obvious solution, because 32+64 = 96 = 48+48. This means we can deduce a solution to our problem:
Hooray! We did it!
What do you mean, "six"? The triangle is 4x3, that last radius makes the third circle way larger...
Okay, looking back at how the problem was formulated, one has to admit that this is a solution: the third circle is tangent to the first two, and to two sides of the triangle SNN' - you just need to extend the side NN' to see it.
But evidently, we're not done.
Sangaku Sunday #10
On the historical front, we previously established that mathematics didn't stop during the Edo period. Accountants and engineers were still in demand, but these weren't necessarily the people who were making sangaku tablets. The problems weren't always practical, and often, the solutions were incomplete, as they didn't say how the problems were solved.
There was another type of person who used mathematics at the time: people who regarded mathematics as a field in which all possibilities should be explored. Today, these would be called researchers, but in Edo-period Japan, they probably regarded mathematics more as an art form.
As in many other art forms (Hiroshige's Okazaki from The 53 Stations of the Tôkaidô series as an example), wasan mathematics organised into schools with masters and apprentices. This would have consequences on how mathematics advanced during this time, but besides that, wasan schools were on the look-out for promising talents. In this light, sangaku appear as an illustration of particular school's abilities with solved or unsolved problems to bait potential recruits, who would prove their worth by presenting their solutions.
Speaking which, we now continue to present our solution to the "three circles in a triangle" problem.
Recall that we are looking for two expressions of the length CN.
1: Knowing that ON = b and OQ = 2*sqrt(qr), it is immediate that QN is the subtraction of the two. Moreover, CQ = r, so by using Pythagoras's theorem in the right triangle CQN, we get
2: We get a second expression by using a cascade of right triangles to reach CN "from above". Working backwards, in the right triangle CRN, we known that CR = r, but RN is unknown, and we would need it to conclude with Pythagoras's theorem. We can get RN if we know SR, given that SN = SR+RN is known by using Pythagoras's theorem in the right triangle SON, with SO = 1 and ON = b. But again, in the right triangle CRS, we do not know CS, but (counter-but!) we could get CS by using the right triangle PCS, where PC and PS are both easy to calculate. We've reached a point where we can start calculating, so let's work forward from there.
Step 1: CPS. PCQO is a rectangle, so PC = OQ and PS = SO-OP = SO-CQ = 1-r, therefore
Step 2: CRS. Knowing CR = r, we deduce
At this point, we can note that 2r-4qr = 2r(1-2q) = 2r*2p, using the first relation between p and q obtained in the first post on this problem. So SR² = 1-4pr.
Step 3a: SON. Knowing SO = 1 and ON = b, we have SN² = 1+b².
Step 3b: CRN. From SN and SR, we deduce
so, using Pythagoras's theorem one more time:
Conclusion. At the end of this lengthy (but elementary) process, we can write CN² = CN² with different expressions either side, and get the final equation for our problem:
Note that 2*(p+q) = 1, and divide by 2 to get the announced result.
Sangaku Saturday #8
Having established that sangaku were, in part, a form of advertisement for the local mathematicians, we can look at the target demographic. Who were the mathematicians of the Edo period? What did they work on and how?
The obvious answer is that the people in the Edo period who used mathematics were the ones who needed mathematics. As far back as the time when the capital was in Kashihara, in the early 8th century, evidence of mathematical references has been uncovered (link to a Mainichi Shinbun article, with thanks to @todayintokyo for the hat tip). All kinds of government jobs - accounting, such as determining taxes, customs, or engineering... - needed some form of mathematics. Examples above: 8th-century luggage labels and coins at the Heijô-kyô Museum in Nara, and an Edo-period ruler used for surveying shown at Matsue's local history museum.
As such, reference books for practical mathematics have existed for a long time, and continued to be published to pass on knowledge to the next generation. But sangaku are different: they are problems, not handbooks.
More on that soon. Below the cut is the solution to our latest puzzle.
Recall that SON is a right triangle with SO = 1 and ON = b. These are set values, and our unknowns are the radii p, q and r of the circles with centres A, B and C. While these are unknown, we assume that this configuration is possible to get equations, which we can then solve.
1: The two circles with centres B and C are tangent to a same line, so we can just re-use the very first result from this series, so
2: Also recalling what we said in that first problem about tangent circles, we know that
Moreover, PA = AO - OP = AO - CQ = (p+2*q) - r. Thus, using Pythagoras's theorem in the right triangle APC, we get a new expression for PC:
since 2(p+q)=1 (the first relation). Equating the two expressions we now have of PC², we solve the equation for r:
again using the first relation to write 2q-1 = -2p.
It only remains to find a third equation for p to solve the problem.
Sangaku Sunday #6
We are about to solve our first sangaku problem, as seen on the tablet shown above from Miminashi-yamaguchi-jinja in Kashihara.
First, we should conclude our discussion: what are sangaku for? There's the religious function, as an offering, and this offering was put on display for all to see, though not all fully understood the problems and their solutions. But a few people would understand, and these would have been the mathematicians of the time. When they visited a new town, they would typically stop at a temple or shrine for some prayers, and they would see the sangaku, a sample of what the local mathematicians were capable of. Whether the problems were solved or open, the visitor could take up the challenges and find the authors to discuss.
And this is where everything lined up: the local school of mathematics would have someone new to talk to, possibly to impress or be impressed by, and maybe even recruit. With the Japanese-style mathematics of the time, called wasan, being considered something of an art form, there would be masters and apprentices, and the sangaku was therefore a means to perpetuate the art.
Now, what about that configuration of circles, second from right on the tablet?
Recall that we had a formula for the radii of three circles which are pairwise tangent and all tangent to the same line. Calling the radii p, q, r, s and t for the circles of centres A, B, C, D and E respectively, we have
for the circles with centres A, B and C (our previous problem), and adapting this formula to two other systems of three circles, we get
for the circles with centres A, C and D, and
for the circles with centres B, C and E. Add these together, and use the first relation on the right-hand side, we get a rather elegant relation between all five radii:
Of course, we can get formulas for s and t,
r having been calculated previously using just p and q, which were our starting radii.
For example, setting p=4 and q=3, we get, approximately, r=0.86, s=0.4 and t=0.37 (this is the configuration shown in the figure, not necessarily the one on the tablet - I will be able to make remarks about that on another example).
Sangaku Saturday #4
In the previous info post, we went over the debate on the religious aspect of sangaku, and the fact that the absence of prayers on these tablets was more puzzling to some than the mathematics. As such, the tablets are not ema prayer tablets, but donations, which usually don't feature prayers on them. Case in point, some consecrated sake and French wine seen at Meiji-jingû in 2016.
Beyond wishing for good fortune and health, such donations serve two very worldly purposes: to contribute to the life and prestige of the shrine or temple (having a famous contributor makes the shrine famous by association), and to advertise the donor in return, as their name is on display. See this large torii at Fushimi Inari Taisha paid for by TV Asahi (テレビ朝日).
With that in mind, Meijizen's cynical comment from 1673 that sangaku aim "to celebrate the mathematical genius of their authors" may not far from the truth. The authors of sangaku are looking to gain notoriety through the publicity that the shrine or temple provides. But was the bemused Meijizen the target audience?
More on that in a couple of weeks. Below the cut is the solution to last week's problem.
The solution to the first problem (below the cut in this post) is the key. Name K, L and M the intersections of the three circles with the horizontal line. Then, by using that previous result,
Indeed, as in that problem, we can construct three right triangles, ABH, ACI and BCJ and apply Pythagoras's theorem in each.
Now, it suffices to note that KL = KM + LM, so
or, dividing by 2*squareroot(pqr), we get the desired result:
Inverting and squaring this yields the formula for r:
This gives us the means to construct this figure on paper using a compass and a marked ruler. Having chosen two radii p and q and constructed the two large circles (remember that AB=p+q) and a line tangent to both, placing M and C is done after calculating the lengths IK=CM=r and IC=KM=2*sqrt(pr).
Sangaku Sunday #2
As the tags in a reblog by @todayintokyo indicated, I waffled about what we'll do in this series in the first post without really defining its main object!
Sangaku are wooden tablets on display at Shintô shrines or Buddhist temples in Japan, featuring geometry problems and their solutions, usually without proof. They started appearing in the Edo period, a particular time for the Japanese people and Japanese scientists. The votive role of these tablets has been debated as far back as the Edo period, as indicated by Meijizen who wrote in 1673:
"There appears to be a trend these days, of mathematical problems on display at shrines. If they were true votive tablets (ema), they should contain a prayer of some sort. Lacking that, one wonders what they are for, other than to celebrate the mathematical genius of their authors. Their meaning eludes me."
I feel the debate on their religious role is overrated. If you look at some food offerings at shrines today, I don't think you'll find a prayer on the bottle of tea or pack of rice, as the prayer is made at the time of offering. It likely is the same for sangaku tablets, which went on display with other offerings. But, as Meijizen hinted, they did have another purpose.
Until we expand on that, below the cut is the solution of last weekend's problem.
Place the point H on the line between A and C1 so that the distance between A and C1 is equal to r2. As the lines (AC1) and (BC2) are both perpendicular to the line (AB), they are parallel, and since AH=BC2=r2, HABC2 is a parallelogram with two right angles: it's a rectangle.
So the length we want, AB, is equal to HC2. The triangle HC1C2 has a right angle at the vertex H, so we can use Pythagoras's theorem:
HC1² + HC2² = C1C2²
In this equality, two lengths are known: C1C2=r1+r2, and
HC1 = AC1-AH = r1-r2 (assuming r1>r2, if not just switch the roles of r1 and r2)
Thus, HC2² = (r1+r2)²-(r1-r2)² = 4 r1 r2 after expanding both expressions (e.g. (r1+r2)² = (r1+r2)x(r1+r2) = r1² + 2 r1 r2 + r2²).
Taking the square root yields the result.
How To Cure Sleep Deprivation Naturally
Did you sleep well last night? If not, you may want to read this...
Sleep apnea or sleep deprivation is a common problem of not getting enough sleep. This can happen due to your work pressure, misbalanced life or any kind of stress. And many people purposely avoid sleeping, because they think that sleep is a waste of time.
It is so common nowadays that we neglect sleep apnea as much as we can. But if you are sleep deprived for a long time, you will face some serious medical problems. If you think that there is nothing wrong with your sleeping habits, check out this symptom of sleep deprivations...
Moodiness
Fatigue
Irritability
Depressed mood
Difficulty learning new concepts
Forgetfulness
Inability to concentrate or a "fuzzy" head
Lack of motivation
Clumsiness
Increased appetite and carbohydrate cravings
Weight gain
Is this enough to make you understand the importance of sleep? Now, if you are suffering from some or all of these problems, you need to take this seriously. There are things you can do at home to treat your sleep apnea, without the help of any pills.
Sleep deprivation solution...
A comfy Bed - Make sure that the bed you spend your night in, is comfortable for you. A stiff bed can harm your back and too much soft mattress can do the same. So, choose a medium soft mattress for your bed.
Follow A Routine - Make a fixed time for you to sleep and waking up. This will balance your sleep time and give you the energy for the day.
Exercise daily - Working out regularly will help you maintain your physique and make you fall asleep as soon as you hit the bed. Do some yoga, aerobics or any physical activity that you prefer.
Avoid Sleep debt - Sleep debt is a word to describe a number of hours you lack sleep. For example, if you need 8 hours of sleep and you only sleep for 6, then you have the sleep debt of 2 hours. If this continues for 5 days, then there are 10 hours of debt. And you will know that you have paid back your sleep debt when you wake up feeling refreshed.
Put That Coffee Down - Don’t consume caffeine before going to the bed, caffeine was made to make you feel energized. So, this will only make you stay wide awake.
Complete Darkness - Make your bedroom as dark as possible to avoid any disturbance in your sleep time. If you can’t do this, use an eye mask.
Follow these steps and it will be enough to make you sleep like a baby. But if you still are awake at night, you should seek some professional help for this.