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https://edurev.in/studytube/PPT-Belts--Chains-Ropes/f355b541-4085-4340-a020-708abd4bc968_p | 1,632,085,535,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00546.warc.gz | 279,815,499 | 46,459 | Courses
# PPT: Belts, Chains & Ropes Notes | EduRev
## Mechanical Engineering : PPT: Belts, Chains & Ropes Notes | EduRev
``` Page 1
Belt, Rope and Chain Drives
Page 2
Belt, Rope and Chain Drives
The belts or ropes are used to transmit power from
one shaft to another by means of pulleys which
rotate at the same speed or at different speeds. The
amount of power transmitted depends upon the
following factors :
? Velocity of the belt.
? Tension under which the belt is placed on the
pulleys.
? Arc of contact between the belt and the smaller
pulley.
Page 3
Belt, Rope and Chain Drives
The belts or ropes are used to transmit power from
one shaft to another by means of pulleys which
rotate at the same speed or at different speeds. The
amount of power transmitted depends upon the
following factors :
? Velocity of the belt.
? Tension under which the belt is placed on the
pulleys.
? Arc of contact between the belt and the smaller
pulley.
Conditions under which the belt is used
? The shafts should be properly in line to insure
uniform tension across the belt section.
? The pulleys should not be too close together, in
order that the arc of contact on the smaller pulley
may be as large as possible.
? The pulleys should not be so far apart as to cause
the belt to weigh heavily on the shafts, thus
increasing the friction load on the bearings
Page 4
Belt, Rope and Chain Drives
The belts or ropes are used to transmit power from
one shaft to another by means of pulleys which
rotate at the same speed or at different speeds. The
amount of power transmitted depends upon the
following factors :
? Velocity of the belt.
? Tension under which the belt is placed on the
pulleys.
? Arc of contact between the belt and the smaller
pulley.
Conditions under which the belt is used
? The shafts should be properly in line to insure
uniform tension across the belt section.
? The pulleys should not be too close together, in
order that the arc of contact on the smaller pulley
may be as large as possible.
? The pulleys should not be so far apart as to cause
the belt to weigh heavily on the shafts, thus
increasing the friction load on the bearings
? A long belt tends to swing from side to side, causing
the belt to run out of the pulleys, which in turn
develops crooked spots in the belt.
? The tight side of the belt should be at the bottom, so
that whatever sag is present on the loose side will
increase the arc of contact at the pulleys.
? In order to obtain good results with flat belts, the
maximum distance between the shafts should not
exceed 10 metres and the minimum should not be
less than 3.5 times the diameter of the larger pulley.
Page 5
Belt, Rope and Chain Drives
The belts or ropes are used to transmit power from
one shaft to another by means of pulleys which
rotate at the same speed or at different speeds. The
amount of power transmitted depends upon the
following factors :
? Velocity of the belt.
? Tension under which the belt is placed on the
pulleys.
? Arc of contact between the belt and the smaller
pulley.
Conditions under which the belt is used
? The shafts should be properly in line to insure
uniform tension across the belt section.
? The pulleys should not be too close together, in
order that the arc of contact on the smaller pulley
may be as large as possible.
? The pulleys should not be so far apart as to cause
the belt to weigh heavily on the shafts, thus
increasing the friction load on the bearings
? A long belt tends to swing from side to side, causing
the belt to run out of the pulleys, which in turn
develops crooked spots in the belt.
? The tight side of the belt should be at the bottom, so
that whatever sag is present on the loose side will
increase the arc of contact at the pulleys.
? In order to obtain good results with flat belts, the
maximum distance between the shafts should not
exceed 10 metres and the minimum should not be
less than 3.5 times the diameter of the larger pulley.
Selection of a Belt Drive
? Speed of the driving and driven shafts,
? Speed reduction ratio,
? Power to be transmitted
? Centre distance between the shafts,
? Positive drive requirements
? Shafts layout,
? Space available
? Service conditions.
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!
## Theory of Machines (TOM)
94 videos|41 docs|28 tests
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,
,
,
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,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
; | 1,082 | 4,431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-39 | latest | en | 0.93594 |
https://math.stackexchange.com/questions/2900164/is-the-following-area-of-crescent-all-right | 1,652,925,302,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00585.warc.gz | 447,566,579 | 65,497 | # Is the following Area of Crescent all right?
In the figure below. There are two overlapping circles and the area of Crescent in Red that I have found is $A_{C} = \frac{\pi rw}{2}$, where $w$ is the shift from center $'X'$ in blue to $'X'$ in red.
Details: $$A_C = \frac{A_{elipse} - A_{circle}}{2} = \frac{[\pi r^2 + \pi r w] - \pi r^2}{2}$$
WLOG, assume both centers lie on the $x$-axis. You can use this diagram afterwards:
Since the area of the circle is $A=\pi r^2$, then the area of the crescent should be: $$A_{\text{crescent}}=\pi r^2-(2A_{\text{sector }EAF}-2A_{\triangle AEF})$$
This is because $A_{\text{sector }EAF}=A_{\text{sector }ECF}$, and so does their corresponding triangle. Since the area of a sector is $A=\frac12r^2 \theta$, with $\theta$ in radians, and the area of the triangle is $A=\frac12ab\sin C$. Then the area of the crescent can be re-written as: $$A=\pi r^2-\alpha r^2+r^2\sin\alpha\\ \implies A=r^2(\pi-\alpha+\sin \alpha)$$
• hmm...Thanks for your solution. On a sidenote, what software have you used for drawing your figures. It seems very elegant for technical writings. It seems your answer is correct. I will notify it as solution after solving it myself as well ^_^
– SJa
Aug 31, 2018 at 2:17
• This is just Geogebra online. Aug 31, 2018 at 2:20
Note that for $w=r$ we obtain
$$A_C=\frac{\pi r^2}{2}$$
which seems to be wrong.
Moreover how does the area for the ellipse come in the derivation?
For the general formula refer to Circle overlapping.
• hmm..thanks for notifying. I thought since both circles are of same radius, a slight shift would mean that they move as if formning ellipse. But I can see how you are right and I am wrong. Thanks
– SJa
Aug 31, 2018 at 2:16
• You are welcome! Bye
– user
Aug 31, 2018 at 7:05 | 559 | 1,774 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-21 | longest | en | 0.915035 |
https://bytes.com/topic/python/answers/45355-how-convert-string-list-tuple | 1,709,338,075,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475711.57/warc/CC-MAIN-20240301225031-20240302015031-00655.warc.gz | 151,226,857 | 14,831 | 473,239 Members | 1,453 Online
# how to convert string to list or tuple
a = "(1,2,3)"
I want convert a to tuple:(1,2,3),but tuple(a) return ('(', '1', ',',
'2', ',', '3', ')') not (1,2,3)
Jul 19 '05 #1
16 22966
On 5/26/05, flyaflya <fl******@gmail.com> wrote:
a = "(1,2,3)"
I want convert a to tuple:(1,2,3),but tuple(a) return ('(', '1', ',',
'2', ',', '3', ')') not (1,2,3)
Long answer - *don't* use eval unless you are in control of the source
of the string that you are evaluating.
--
Cheers,
Simon B,
si***@brunningonline.net,
http://www.brunningonline.net/simon/blog/
Jul 19 '05 #2
"flyaflya" <fl******@gmail.com> wrote:
a = "(1,2,3)"
I want convert a to tuple:(1,2,3),but tuple(a) return ('(', '1', ',',
'2', ',', '3', ')') not (1,2,3)
if you trust the source, use
eval(a)
if you don't trust it, you can use, say
tuple(int(x) for x in re.findall("\d+", a))
or, perhaps
tuple(int(x) for x in a[1:-1].split(","))
or some variation thereof.
(if you're using a version older than 2.4, add brackets inside
the tuple() call:
tuple([int(x) for x in a[1:-1].split(",")])
etc.
</F>
Jul 19 '05 #3
On Thu, 26 May 2005 19:53:38 +0800, flyaflya wrote:
a = "(1,2,3)"
I want convert a to tuple:(1,2,3),but tuple(a) return ('(', '1', ',',
'2', ',', '3', ')') not (1,2,3)
Others have already given some suggestions. Here are some others.
You didn't say where the input string a came from. Do you control
String_Tuple_To_Real_Tuple("(1,2,3)")
can you just create the tuple in the first place?
a = (1, 2, 3)
Second suggestion: if you know that the input string will ALWAYS be in the
form "(1,2,3)" then you can do this:
a = "(1,2,3)"
a = a[1:-1] # deletes leading and trailing parentheses
a = a.split(",") # creates a list ["1", "2", "3"] (items are strings)
a = [int(x) for x in a] # creates a list [1, 2, 3] (items are integers)
a = tuple(a) # coverts to a tuple
or as a one-liner:
a = "(1,2,3)"
a = tuple([int(x) for x in a[1:-1].split(",")])
Best of all, wrap your logic in a function definition with some
error-checking:
def String_Tuple_To_Real_Tuple(s):
"""Return a tuple of ints from a string that looks like a tuple."""
if not s:
return ()
if (s[0] == "(") and s[-1] == ")"):
s = s[1:-1]
else:
raise ValueError("Missing bracket(s) in string.")
return tuple([int(x) for x in s.split(",")])
Hope this helps,
--
Steven.
Jul 19 '05 #4
Simon Brunning wrote:
On 5/26/05, flyaflya <fl******@gmail.com> wrote:
a = "(1,2,3)"
I want convert a to tuple:(1,2,3),but tuple(a) return ('(', '1', ',',
'2', ',', '3', ')') not (1,2,3)
Long answer - *don't* use eval unless you are in control of the source
of the string that you are evaluating.
Or if you do use eval, don't give it access to any names.
import os
eval(raw_input(), {})
os.system("rm -rf *")
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<string>", line 0, in ?
NameError: name 'os' is not defined
Jul 19 '05 #5
Dan Bishop wrote:
Simon Brunning wrote:
[...]
Or if you do use eval, don't give it access to any names.
[...]
os.system("rm -rf *")
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<string>", line 0, in ?
NameError: name 'os' is not defined
Have you tried giving it the string '__import__("os").system("rm -rf *")'?
[Don't try that at home children!]
Even if you take steps to avoid that working by hiding the builtins, there
are still too many ways to do nasty things with eval for it ever to be
safe.
Jul 19 '05 #6
"Duncan Booth" <du**********@invalid.invalid> wrote in message
news:Xn*************************@127.0.0.1...
Dan Bishop wrote:
Simon Brunning wrote:
[...]
Or if you do use eval, don't give it access to any names.
[...]
os.system("rm -rf *")
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<string>", line 0, in ?
NameError: name 'os' is not defined
Have you tried giving it the string '__import__("os").system("rm -rf *")'?
[Don't try that at home children!]
Even if you take steps to avoid that working by hiding the builtins, there
are still too many ways to do nasty things with eval for it ever to be
safe.
There was a posting here Nov 5, 2003 by Huaiyu Zhu at IBM Almaden
that shows how to do eval type stuff safely. The basic notion is to use the
compiler and then check the ast to see if the result fits the straitjacket
you
want to put it into. Pass / Fail; trying to fix it up if it's "close" is
usually a
He gives an example, and there's a much more extensive set of working
code in the taBase.py module of PyFit that handles lists, tuples and
dicts which contain arbitrary literals including complex and arbitrarily
nested
lists, tuples and dicts.
------- code snippet starts here --------
def _safeEval(self, s):
"""
Evaluate strings that only contain the following structures:
const, tuple, list, dict
Taken from c.l.py newsgroup posting Nov 5, 2003 by Huaiyu Zhu at IBM
"""
#print "in _safeEval. input: '%s'" % s
node1 = compiler.parse(s)
# !!! special case of attempting to compile a lone string
if node1.doc is not None and len(node1.node.nodes) == 0:
#print "in _safeEval. string: '%s' found as docstring" %
node1.doc
return node1.doc
#print "in _safeEval. nodes: '%s'" % (node1,)
stmts = node1.node.nodes
assert len(stmts) == 1
node = compiler.parse(s).node.nodes[0]
nodes = node.getChildNodes()
assert len(nodes) == 1
result = self._safeAssemble(nodes[0])
#print "in _safeEval result: '%s'" % (result,)
return result
seq_types = {
compiler.ast.Tuple: tuple,
compiler.ast.List: list,
}
map_types = {
compiler.ast.Dict: dict,
}
oper_types = {
compiler.ast.Sub: operator.sub,
}
builtin_consts = {
"True": True,
"False": False,
"None": None,
}
def _safeAssemble(self, node):
""" Recursively assemble parsed ast node """
cls = node.__class__
if cls == compiler.ast.Const:
return node.value
elif cls in self.seq_types:
nodes = node.nodes
args = map(self._safeAssemble, nodes)
return self.seq_types[cls](args)
elif cls in self.map_types:
keys, values = zip(*node.items)
keys = map(self._safeAssemble, keys)
values = map(self._safeAssemble, values)
return self.map_types[cls](zip(keys, values))
elif cls in self.oper_types:
left = self._safeAssemble(node.left)
right = self._safeAssemble(node.right)
if type(left) == type(1.0j) or type(right) == type(1.0j):
return self.oper_types[cls](left, right)
else:
raise FitException, ("Parse001",)
elif cls == compiler.ast.Name:
result = self.builtin_consts.get(node.name, "?")
if result != "?":
return result
else:
raise FitException, ("Parse002", node.name)
else:
raise FitException, ("Parse003", cls)
------- end of code snippet -----------
John Roth
Jul 19 '05 #7
Duncan Booth wrote:
Dan Bishop wrote:
Or if you do use eval, don't give it access to any names. [snip] os.system("rm -rf *")
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<string>", line 0, in ?
NameError: name 'os' is not defined
Have you tried giving it the string '__import__("os").system("rm -rf *")'?
[Don't try that at home children!]
But you can try it at home if you set __builtins__ to something other
than the default:
py> eval("""__import__("os").system('echo "hello"')""",
dict(__builtins__=None))
Traceback (most recent call last):
File "<interactive input>", line 1, in ?
File "<string>", line 0, in ?
NameError: name '__import__' is not defined
If you're just doing work with constants, the lack of access to any
builtins is ok:
py> eval("(1,2,3)", dict(__builtins__=None))
(1, 2, 3)
I know there have been security holes in this technique before, but I
looked at the archives, and all the old ones I found have been patched.
(Or at least I wasn't able to reproduce them.)
STeVe
Jul 19 '05 #8
Steven Bethard wrote:
Have you tried giving it the string '__import__("os").system("rm -rf
*")'? [Don't try that at home children!]
But you can try it at home if you set __builtins__ to something other
than the default:
py> eval("""__import__("os").system('echo "hello"')""",
dict(__builtins__=None))
Traceback (most recent call last):
File "<interactive input>", line 1, in ?
File "<string>", line 0, in ?
NameError: name '__import__' is not defined
If you're just doing work with constants, the lack of access to any
builtins is ok:
py> eval("(1,2,3)", dict(__builtins__=None))
(1, 2, 3)
I know there have been security holes in this technique before, but I
looked at the archives, and all the old ones I found have been
patched.
(Or at least I wasn't able to reproduce them.)
I guess you are referring to things like this not working when you use eval
with an empty __builtins__:
eval('''[ cls for cls in {}.__class__.__bases__[0].__subclasses__()
if '_Printer' in `cls`
][0]._Printer__setup.func_globals['__builtins__']['__import__']''',
dict(__builtins__=None))
That gets blocked because func_globals is a 'restricted attribute', so I
can't get directly at __import__ that way, but what I can do is to access
any new style class you have defined and call any of its methods with
whatever arguments I wish.
Even with the big holes patched you are going to find it pretty hard to
write a safe program that uses eval on untrusted strings. The only way to
go is to filter the AST (or possibly the bytecode).
Jul 19 '05 #9
Duncan Booth wrote:
Steven Bethard wrote:
But you can try it at home if you set __builtins__ to something other
than the default:
py> eval("""__import__("os").system('echo "hello"')""",
dict(__builtins__=None))
Traceback (most recent call last):
File "<interactive input>", line 1, in ?
File "<string>", line 0, in ?
NameError: name '__import__' is not defined
[snip]
I know there have been security holes in this technique before, but I
looked at the archives, and all the old ones I found have been
patched.
(Or at least I wasn't able to reproduce them.)
I guess you are referring to things like this not working when you use eval
with an empty __builtins__:
eval('''[ cls for cls in {}.__class__.__bases__[0].__subclasses__()
if '_Printer' in `cls`
][0]._Printer__setup.func_globals['__builtins__']['__import__']''',
dict(__builtins__=None))
That gets blocked because func_globals is a 'restricted attribute', so I
can't get directly at __import__ that way
Among other things, yes, that's one of the big ones. func_globals is
inaccessible. Also, IIRC the file constructor is inaccessible.
but what I can do is to access
any new style class you have defined and call any of its methods with
whatever arguments I wish.
Any new style class that I've defined? Or just any one I pass in as
part of dict(__builtins__=None, ...)? If the former, could you
elaborate? If the latter, then yes, I can see the problem. However for
the case where all you pass in is dict(__builtins__=None), is there
still a risk? Note that in the OP's case, all that is necessary is
constant parsing, so no names need to be available.
STeVe
Jul 19 '05 #10
Steven Bethard wrote:
Duncan Booth wrote:
any new style class you have defined and call any of its methods with
whatever arguments I wish.
Any new style class that I've defined? Or just any one I pass in as
part of dict(__builtins__=None, ...)? If the former, could you
elaborate? If the latter, then yes, I can see the problem. However
for the case where all you pass in is dict(__builtins__=None), is
there still a risk? Note that in the OP's case, all that is necessary
is constant parsing, so no names need to be available.
Any new style class you have defined is accessible through
object.__subclasses__(), and as I showed object itself is always accessible
through {}.__class__.__bases__[0].
I'm assuming that the source code for your program is available. That means
I can find the name of an interesting class which has a method that does
something destructive, and call it.
e.g. Assuming that the MyDatabase class does something nasty to a file:
class MyDatabase(object): def __init__(self, filename):
self.filename = filename
def initialise(self):
print "Splat %s" % self.filename
eval('''[ cls for cls in {}.__class__.__bases__[0].__subclasses__()
if 'MyDatabase' in `cls`
][0]('importantfile').initialise()''', dict(__builtins__=None))
Splat importantfile
Jul 19 '05 #11
Duncan Booth wrote:
e.g. Assuming that the MyDatabase class does something nasty to a file:
class MyDatabase(object):
def __init__(self, filename):
self.filename = filename
def initialise(self):
print "Splat %s" % self.filename
eval('''[ cls for cls in {}.__class__.__bases__[0].__subclasses__()
if 'MyDatabase' in `cls`
][0]('importantfile').initialise()''', dict(__builtins__=None))
Splat importantfile
Interestingly, I don't seem to be able to create a file object as a
class attribute in restricted mode:
py> class C(object):
.... def __init__(self):
.... self.f = file('temp.txt', 'w')
....
py> eval('''[ cls for cls in {}.__class__.__bases__[0].__subclasses__()
if cls.__name__ == 'C'][0]().f.write("stuff")''', dict(__builtins__=None))
Traceback (most recent call last):
File "<interactive input>", line 1, in ?
File "<string>", line 0, in ?
AttributeError: 'C' object has no attribute 'f'
py> eval('''[ cls for cls in {}.__class__.__bases__[0].__subclasses__()
if cls.__name__ == 'C'][0]().__dict__''', dict(__builtins__=None))
{}
I don't get an error for calling the file constructor, but the f
attribute is never set AFAICT.
STeVe
Jul 19 '05 #12
Steven Bethard wrote:
Interestingly, I don't seem to be able to create a file object as a
class attribute in restricted mode:
py> class C(object):
... def __init__(self):
... self.f = file('temp.txt', 'w')
...
py> eval('''[ cls for cls in
{}.__class__.__bases__[0].__subclasses__() if cls.__name__ ==
'C'][0]().f.write("stuff")''', dict(__builtins__=None)) Traceback
(most recent call last):
File "<interactive input>", line 1, in ?
File "<string>", line 0, in ?
AttributeError: 'C' object has no attribute 'f'
py> eval('''[ cls for cls in
{}.__class__.__bases__[0].__subclasses__() if cls.__name__ ==
'C'][0]().__dict__''', dict(__builtins__=None)) {}
Weird. I copied and paste your class and eval exactly (apart from deleting
the ... prompts) and it worked exactly as expected: writing 'stuff' to
temp.txt. (Python 2.4)
Jul 19 '05 #13
flyaflya wrote:
a = "(1,2,3)"
I want convert a to tuple:(1,2,3),but tuple(a) return ('(', '1', ',',
'2', ',', '3', ')') not (1,2,3)
Probably a bit late... but there's always listquote - It's part of the
pythonutils module.
http://www.voidspace.org.uk/python/pythonutils.html
It will turn strings to lists, including nested lists.
Best Regards,
Fuzzy
http://www.voidspace.org.uk/python
Jul 19 '05 #14
Duncan Booth wrote:
Steven Bethard wrote:
Interestingly, I don't seem to be able to create a file object as a
class attribute in restricted mode:
py> class C(object):
... def __init__(self):
... self.f = file('temp.txt', 'w')
...
py> eval('''[ cls for cls in
{}.__class__.__bases__[0].__subclasses__() if cls.__name__ ==
'C'][0]().f.write("stuff")''', dict(__builtins__=None)) Traceback
(most recent call last):
File "<interactive input>", line 1, in ?
File "<string>", line 0, in ?
AttributeError: 'C' object has no attribute 'f'
py> eval('''[ cls for cls in
{}.__class__.__bases__[0].__subclasses__() if cls.__name__ ==
'C'][0]().__dict__''', dict(__builtins__=None)) {}
Weird. I copied and paste your class and eval exactly (apart from deleting
the ... prompts) and it worked exactly as expected: writing 'stuff' to
temp.txt. (Python 2.4)
So, I played around with this a little bit. If I start up a new
interpreter and type it in like above, I get the behavior you do. What
I had actually done (abbreviated) was:
py> class C(object):
.... pass
....
py> class C(object):
.... def __init__(self):
.... self.f = file('temp.txt', 'w')
....
py> eval('''[ cls for cls in {}.__class__.__bases__[0].__subclasses__()
if cls.__name__ == 'C'][0]().f.write("stuff")''', dict(__builtins__=None))
Traceback (most recent call last):
File "<interactive input>", line 1, in ?
File "<string>", line 0, in ?
AttributeError: 'C' object has no attribute 'f'
And the problem with this is that both __main__.C objects are now
subclasses of object:
py> eval('''[ cls for cls in {}.__class__.__bases__[0].__subclasses__()
if cls.__name__ == 'C']''', dict(__builtins__=None))
[<class '__main__.C'>, <class '__main__.C'>]
So I was getting the wrong __main__.C object. Sorry for the confusion!
Now, even using this technique, *your* code can't call the file constructor:
py> class C(object):
.... def __init__(self):
.... self.file = file
....
py> eval('''[ cls for cls in {}.__class__.__bases__[0].__subclasses__()
if cls.__name__ == 'C'][-1]().file("temp.txt", "w")''',
dict(__builtins__=None))
Traceback (most recent call last):
File "<interactive input>", line 1, in ?
File "<string>", line 0, in ?
IOError: file() constructor not accessible in restricted mode
But unless the person eval-ing your code *only* writes immaculate code I
can see that you can probably screw them. ;) I wonder why
__subclasses__ isn't a restricted attribute... Is it ever used for
something that isn't evil? ;)
STeVe
Jul 19 '05 #15
Steven Bethard schreef:
But unless the person eval-ing your code *only* writes immaculate code I
can see that you can probably screw them. ;) I wonder why
__subclasses__ isn't a restricted attribute... Is it ever used for
something that isn't evil? ;)
STeVe
Completely off topic, but I just cannot resist showing off.
Some time ago I used __subclasses__ in a way that is not evil. I think.
The details are described in the following thread:
A summary: I used __subclasses__ to apply the Chain-of-Responsibility
pattern to object creation. The code would appear to instantiate
an object of the root of a class hierarchy, but the actual object
that was created would be an instance of a subclass.
So to get back to your question: yes, there are non-evil
uses for __subclasses__. Weird perhaps, but non-evil.
Non-standard, sure . Too clever for my own good, very likely.
Regards,
Ruud
--
Ruud de Jong
'@'.join('.'.join(s) for s in (['ruud','de','jong'],['tiscali','nl']))
Jul 19 '05 #16
Ruud de Jong wrote:
Steven Bethard schreef:
But unless the person eval-ing your code *only* writes immaculate
code I can see that you can probably screw them. ;) I wonder why
__subclasses__ isn't a restricted attribute... Is it ever used for
something that isn't evil? ;)
STeVe
Completely off topic, but I just cannot resist showing off.
Some time ago I used __subclasses__ in a way that is not evil. I
think.
The details are described in the following thread:
ccb986c66cdc1/
A summary: I used __subclasses__ to apply the Chain-of-Responsibility
pattern to object creation. The code would appear to instantiate
an object of the root of a class hierarchy, but the actual object
that was created would be an instance of a subclass.
So to get back to your question: yes, there are non-evil
uses for __subclasses__. Weird perhaps, but non-evil.
Non-standard, sure . Too clever for my own good, very likely.
I've done almost exactly the same thing. The base class uses __subclasses__
to find the best matching subclass based on the factory parameters. In my
case I was retrieving files from the web, so I had a base Handler class and
created HtmlHandler, ImageHandler &c.
class Handler(object):
'''Class to process files'''
__map = {}
@classmethod
def _resolveClass(klass, isdir, name):
map = Handler.__map
if not map:
for c in klass.__subclasses__():
for ext in c.Extensions:
map['.'+ext.lower()] = c
if isdir:
klass = FolderHandler
else:
ext = os.path.splitext(name)[1].lower()
if ext not in map:
map[ext] = DefaultHandler
klass = map[ext]
return klass(name)
@classmethod
def fromPathname(klass, name, path, uri, db):
isdir = os.path.isdir(os.path.join(path, name))
obj = klass._resolveClass(isdir, name)
obj._initialize(name, path, uri, db)
return obj
@classmethod
def fromUrl(klass, uri, text, db=None):
... and so on ...
and then subclasses such as:
class ImageHandler(Handler):
Extensions = ('jpg', 'jpeg', 'gif', 'png')
type = 'Image'
class DefaultHandler(Handler):
Extensions = ('',)
type = 'Ignored'
This also contains the only code I think I've written with a class
definition in a for loop:
# General categories
EXTENSIONS = {
'js': 'javascript',
'php': 'php',
'doc': 'Word Document',
'ppt': 'Powerpoint',
'css': 'Stylesheet',
'swf': 'Flash',
'pdf': 'File',
'rtf': 'File',
'zip': 'File',
}
Classes = []
for ext in EXTENSIONS:
class GeneralHandler(Handler):
Extensions = (ext,)
type = EXTENSIONS[ext]
Classes.append(GeneralHandler)
Jul 19 '05 #17
This thread has been closed and replies have been disabled. Please start a new discussion.
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http://xrpp.iucr.org/Ab/ch4o2v0001/ | 1,558,359,723,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255944.3/warc/CC-MAIN-20190520121941-20190520143941-00311.warc.gz | 359,180,146 | 7,764 | International
Tables for
Crystallography
Volume A
Space-group symmetry
Edited by Th. Hahn
International Tables for Crystallography (2006). Vol. A, ch. 4.2, p. 61
https://doi.org/10.1107/97809553602060000508
## Chapter 4.2. Symbols for plane groups (two-dimensional space groups)
E. F. Bertauta
aLaboratoire de Cristallographie, CNRS, Grenoble, France
Chapters 4.1 , 4.2 and 4.3 contain extensive explanations and tabulations of the various types of space-group symbols. Chapter 4.2 presents a table with the short, full and extended symbols for the 17 plane groups, including symbols for the tetragonal c and the hexagonal h cell.
### 4.2.1. Arrangement of the tables
| top | pdf |
Comparative tables for the 17 plane groups first appeared in IT (1952). The classification of plane groups is discussed in Chapter 2.1 . Table 4.2.1.1 lists for each plane group its system, lattice symbol, point group and the plane-group number, followed by the short, full and extended Hermann–Mauguin symbols. Short symbols are included only where different from the full symbols. The next column contains the full symbol for another setting which corresponds to an interchange of the basis vectors a and b; it is only needed for the rectangular system. Multiple cells c and h for the square and the hexagonal system are introduced in the last column.
Table 4.2.1.1| top | pdf | Index of symbols for plane groups
System and lattice symbolPoint groupNo. of plane groupHermann–Mauguin symbolFull symbol for other settingMultiple cell
ShortFullExtended
Oblique 1 1 p1
p 2 2 p2
Rectangular
p, c
m 3 pm
4 pg
5 cm
2mm 6
7
8
9
Square
p
4 10
4mm 11
12
Hexagonal
p
3 13 p3 h3
3m 14
15
6 16 p6 h6
6mm 17 p6mm
### 4.2.2. Additional symmetry elements and extended symbols
| top | pdf |
Additional symmetry' elements are
(i) rotation points 2, 3 and 4, reproduced in the interior of the cell (cf. Table 4.1.2.1 and plane-group diagrams in Part 6 ); (ii) glide lines g which alternate with mirror lines m. In the extended plane-group symbols, only the additional glide lines g are listed: they are due either to c centring or to inclined' integral translations, as shown in Table 4.1.2.2 .
### 4.2.3. Multiple cells
| top | pdf |
The c cell in the square system is defined as follows: with centring points' at 0, 0; . It plays the same role as the three-dimensional C cell in the tetragonal system (cf. Section 4.3.4 ).
Likewise, the triple cell h in the hexagonal system is defined as follows: with centring points' at 0, 0; . It is the two-dimensional analogue of the three-dimensional H cell (cf. Chapter 1.2 and Section 4.3.5 ).
### 4.2.4. Group–subgroup relations
| top | pdf |
The following example illustrates the usefulness of multiple cells.
#### Example: p3m1 (14)
The symbol of this plane group, described by the triple cell h, is h31m, where the symmetry elements of the secondary and tertiary positions are interchanged. `Decentring' the h cell gives rise to maximal non-isomorphic k subgroups p31m of index [3], with lattice parameters (cf. Section 4.3.5 ).
### References
International Tables for X-ray Crystallography (1952). Vol. I, edited by N. F. M. Henry & K. Lonsdale. Birmingham: Kynoch Press. [Abbreviated as IT (1952).] | 920 | 3,269 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-22 | latest | en | 0.798429 |
https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0791.Custom-Sort-String/ | 1,726,421,086,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651632.84/warc/CC-MAIN-20240915152239-20240915182239-00528.warc.gz | 127,388,722 | 71,682 | 0791. Custom Sort String
# 791. Custom Sort String#
## 题目 #
`order` and `str` are strings composed of lowercase letters. In `order`, no letter occurs more than once.
`order` was sorted in some custom order previously. We want to permute the characters of `str` so that they match the order that `order` was sorted. More specifically, if `x` occurs before `y` in `order`, then `x` should occur before `y` in the returned string.
Return any permutation of `str` (as a string) that satisfies this property.
``````Example:Input:
order = "cba"
str = "abcd"
Explanation:
"a", "b", "c" appear in order, so the order of "a", "b", "c" should be "c", "b", and "a".
Since "d" does not appear in order, it can be at any position in the returned string. "dcba", "cdba", "cbda" are also valid outputs.
``````
Note:
• `order` has length at most `26`, and no character is repeated in `order`.
• `str` has length at most `200`.
• `order` and `str` consist of lowercase letters only.
## 解题思路 #
• 题目只要求 T 中包含 S 的字符串有序，所以可以先将 T 中包含 S 的字符串排好序，然后再拼接上其他字符。S 字符串最长为 26 位，先将 S 中字符的下标向左偏移 30，并将偏移后的下标值存入字典中。再把 T 字符串按照字典中下标值进行排序。S 中出现的字符对应的下标经过处理以后变成了负数，S 中未出现的字符的下标还是正数。所以经过排序以后，S 中出现的字符按照原有顺序排列在前面，S 中未出现的字符依次排在后面。
## 代码 #
``````package leetcode
import "sort"
func customSortString(order string, str string) string {
magic := map[byte]int{}
for i := range order {
magic[order[i]] = i - 30
}
byteSlice := []byte(str)
sort.Slice(byteSlice, func(i, j int) bool {
return magic[byteSlice[i]] < magic[byteSlice[j]]
})
return string(byteSlice)
}
``````
Apr 8, 2023 | 566 | 1,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.507935 |
https://it.mathworks.com/matlabcentral/profile/authors/434782?s_tid=cody_local_to_profile | 1,607,029,529,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141732696.67/warc/CC-MAIN-20201203190021-20201203220021-00323.warc.gz | 337,887,990 | 24,437 | Community Profile
# Walter Roberson
##### Last seen: Today
48.708 total contributions since 2011
I do not do free private consulting. If you want to bring my attention to something, send a link to the MATLAB Answers location.
#### Walter Roberson's Badges
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Fimplicit not plotting properly
a=2; h=0/2; b=0; g=3/2; f=1/2; c=7; fun1=@(x,y) a.*x.^2+2*h.*x.*y+b*y.^2+2*g.*x+2*f.*y+c; fimplicit(fun1, [-10 10 -10 10]) Lo...
circa 11 ore ago | 1
How to do a rem while loop?
while true h = input('Enter the desired grid spacing: '); if isnumeric(h) && isscalar(h) && h > 0 && rem(h,1) == 0; br...
circa 11 ore ago | 0
The wk^3*exp(x*tau) is getting integrated to include a /tau^4 term -- the general pattern wk^n*exp(...*tau) integrates to someth...
circa 11 ore ago | 0
matlab error ode45
x0=[0,10,45]; %initial condtion tspan=[0,30];% time span u=10; odefun = @(t,y)reshape(rolling(t,y,u), [], 1); [t,x]=ode45(...
circa 12 ore ago | 0
vpasolve -> Empty sym: 0-by-1
I was seeing some oddities when I solved, so I tried solving two different ways: R=8.3144; T=298.15; F=96487; eps0= 8.854*10...
circa 18 ore ago | 0
What does [3, Inf] represents?
In addition to what the others say: that kind of variable would tend to be used as the size parameter for reading data, such as ...
circa 20 ore ago | 1
| accepted
For loop stores all zeros then the actual values of last iteration
Your statement finalsm = zeros(height(x),1); % 'height(x)' bc array changes size each iteration is throwing away all previ...
circa 21 ore ago | 0
| accepted
convert binary string into hex
lookup = containers.Map(cellstr(dec2bin(0:15,4)),cellstr(dec2hex(0:15))); binary_string = '0101010011101010'; output = cel...
circa 21 ore ago | 0
How to read text file backwards?
You can fopen() the file for reading and fseek() to a particular offset relative to the end of file. You will not be able to fs...
circa 21 ore ago | 0
ERROR Subscript indices must either be real positive integers or logicals.
for j=1:j_max+1 x=v((j-1:j+N0-2)*dt); j starts at 1. j-1 starts at 0. 0 multiplied by anything is 0. Therefore your first ind...
circa 21 ore ago | 0
Why do the lines not plot to completion
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circa 23 ore ago | 1
How Can I Pull A Specific Variable From The Determinant of Matrix?
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1 giorno ago | 0
How to combine the using of meshgrid and surface?
x = linspace(0,.1); y = linspace(.3, .4); [X, Y] = meshgrid(x, y); Z = X.^2 - 5.*Y.^3 + sin(5*2*pi*X); surf(X, Y, Z, 'edgeco...
1 giorno ago | 0
| accepted
Get rid of unnecessary space in output of full(A)
disp(num2str(full(A))) You can add a format specifier to num2str() to be more specific about widths.
1 giorno ago | 0
Merge data by date
Convert your data to a timetable() object, and use retime() to consolidate the information.
1 giorno ago | 0
| accepted
Can I pcolor base on wavelength, any colormap can do so?
When you pcolor(X,Y,I) then the I data will be used to decide which color information is used. By default, the numerically lowes...
1 giorno ago | 0
How to: write .xls into a dynamic folder name ?
folder = [Answer{1},' ',Answer{2},' ',num2str(Answer{3})]; if ~exist(folder, 'dir'); mkdir(folder); end filename = 'patient_...
1 giorno ago | 0
| accepted
How to run continuously
current_state = 0; writeDigitalPin(a,'D2',current_state); while true if readVoltage(a, 'A1') >= 2 if current_s...
1 giorno ago | 0
| accepted
how to solve the error "Not enough input arguments."
You are pressing the green Run button to run the code. Don't Do That. Go down to the command line and command alpham(18.35) al...
2 giorni ago | 0
[THIS QUESTION HAS BEEN REDACTED]
function [T_mp,rho,cp,k,alpha] = Material_Properties(MN) T_mp_vec = [1406 2192 693 2125]'; rho_vec = [19070 6100 7140 6570]'; ...
2 giorni ago | 0
| accepted
rotating a matrix without interpolation
When you rotate a matrix, then the only pixels that can be calculated without any interpolation are the ones whose coordinates a...
2 giorni ago | 1
Using size on a data file, etc... Could someone explain what each step here does exactly?
importdata() in some circumstances returns a struct that has a field named data that holds whatever importdata() decided was num...
2 giorni ago | 0
| accepted
Adding accents to the live scripts
That feature was not supported until R2019a.
2 giorni ago | 0
| accepted
Is there a way to remove the Column titles when using format longg?
disp(num2str(YourMatrix))
2 giorni ago | 1
| accepted
How to add a newline after a certain number of characters
while number of remaining characters > n remove and record the first n characters add them and newline to output end ...
2 giorni ago | 0
k=dlmread('11032020_2020-11-25_17-40-55_m1.ascii', ' ',[585 3 1457 4]) asks it to skip 3 columns and that the reading should pr...
2 giorni ago | 0
| accepted
Readtable function on Matlab
Unable to find or open 'Task3.xlsx' Task3.xlsx is not in your current directory, and not in any directory that is on the MATLAB...
2 giorni ago | 0
`integral` function: "Output of the function must be same size as input" error
integral() will call the given function with a vector of values, and the function must return one value for each element in the ...
2 giorni ago | 0
| accepted | 1,605 | 5,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-50 | latest | en | 0.760697 |
https://ainews.spxbot.com/2019/01/08/light-on-math-machine-learning-intuitive-guide-to-neural-style-transfer/ | 1,555,786,174,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529962.12/warc/CC-MAIN-20190420180854-20190420202854-00463.warc.gz | 334,547,562 | 14,539 | # Light on Math Machine Learning: Intuitive Guide to Neural Style Transfer
### Introduction
Neural style transfer (NST) is a very neat idea. NST builds on the key idea that,
it is possible to separate the style representation and content representations in a CNN, learnt during a computer vision task (e.g. image recognition task).
Following this concept, NST employs a pretrained convolution neural network (CNN) to transfer styles from a given image to another. This is done by defining a loss function that tries to minimise the differences between a content image, a style image and a generated image, which will be discussed in detail later. By the end of this tutorial you will be able to create very cool artwork like below.
This tutorial will be covering the following parts in the coming sections of the tutorial.
• Why neural style transfer and the high level architecture
• Defining inputs, outputs, losses and the optimiser for the neural style transfer network
• Defining an input pipeline to feed data to the network
• Training the network and saving the results
• Conclusion
The other articles of this series can be found below.
A B C D* E F G H I J K L* M N O P Q R S T U V W X Y Z
* denotes articles behind the Medium paywall
### Code
Note that I will be sharing only the most important code snippets in the article. You can get the full code as a Jupyter notebook here. The algorithm is implemented with TensorFlow.
### Why NST?
Deep neural networks have already surpassed human level performance in tasks such as object recognition and detection. However, deep networks were lagging far behind in tasks like generating artistic artefacts having high perceptual quality until recent times. Creating better quality art using machine learning techniques is imperative for reaching human-like capabilities, as well as opens up a new spectrum of possibilities. And with the advancement of computer hardware as well as the proliferation of deep learning, deep learning is right now being used to create art. For example, an AI generated art won’t be sold at an auction for a whopping \$432,500.
### High level architecture
As stated earlier, neural style transfer uses a pretrained convolution neural network. Then to define a loss function which blends two images seamlessly to create visually appealing art, NST defines the following inputs:
• A content image (c) — the image we want to transfer a style to
• A style image (s) — the image we want to transfer the style from
• An input (generated) image (g) — the image that contains the final result (the only trainable variable)
The architecture of the model as well as how the loss is computed is shown below. You do not need to develop a profound understanding of what is going on in the image below, as you will be seeing each component in detail in the next several sections to come. The idea is to give a high level understanding of the workflow taking place during style transfer.
High level architecture of NST model
You will be borrowing the VGG-16 weights from this webpage. You will need to download the vgg16_weights.npz file and place that in a folder called `vgg` in your project home directory (sorry I should have automated this, but I was lazy). You will only be needing the convolution and the pooling layers. Specifically, you will be loading the first `7` convolutional layers to be used as the NST network. You can do this using the `load_weights(...)` function given in the notebook.
Note: You are welcome to try more layers. But beware of the memory limitations of your CPU and GPU.
### Defining functions to build the style transfer network
Here you define several functions that will help you later to fully define the computational graph of the CNN given an input.
#### Creating TensorFlow variables
Here you load the loaded numpy arrays into TensorFlow variables. We will be creating following variables:
• content image (`tf.placeholder`)
• style image (`tf.placeholder`)
• generated image (`tf.Variable` and `trainable=True`)
• pretrained weights and biases (`tf.Variable` and `trainable=False`)
Make sure you leave the generated image trainable while keeping pretrained weights and biases frozen. Below we show two functions to define inputs and neural network weights.
#### Computing the VGG net output
Here you are computing the VGG net output by means of convolution and pooling operations. Note that you are replacing the `tf.nn.max_pool` with the `tf.nn.avg_pool` operation, as `tf.nn.avg_pool` gives better visually pleasing results during style transfer [1]. Feel free to experiment with `tf.nn.max_pool` by changing the operation in the function below.
### Loss functions
In this section we define two loss functions; the content loss function and the style loss function. The content loss function ensures that the activations of the higher layers are similar between the content image and the generated image. The style loss function makes sure that the correlation of activations in all the layers are similar between the style image and the generated image. We will be discussing the details below.
#### Content cost function
The content cost function is making sure that the content present in the content image is captured in the generated image. It has been found that CNNs capture information about content in the higher levels, where the lower levels are more focused on individual pixel values [1]. Therefore we use the top-most CNN layer to define the content loss function.
Let A^l_{ij}(I) be the activation of the l th layer, i th feature map and j th position obtained using the image I. Then the content loss is defined as,
The content loss
Essentially L_{content} captures the root mean squared error between the activations produced by the generated image and the content image. But why does minimising the difference between the activations of higher layers ensure the content of the content image is preserved?
#### Intuition behind content loss
If you visualise what is learnt by a neural network, there’s evidence that suggests that different feature maps in higher layers are activated in the presence of different objects. So if two images to have the same content, they should have similar activations in the higher layers.
We can define the content cost as follows.
#### Style loss function
Defining the style loss function requires more work. To extract the style information from the VGG network, we use all the layers of the CNN. Furthermore, style information is measured as the amount of correlation present between features maps in a given layer. Next, a loss is defined as the difference of correlation present between the feature maps computed by the generated image and the style image. Mathematically, the style loss is defined as,
w^l (chosen uniform in this tutorial) is a weight given to each layer during loss computation and M^l is an hyperparameter that depends on the size of the l th layer. If you would like to see the exact value, please refer to this paper. However in this implementation, you are not using M^l as that will be absorbed by another parameter when defining the final loss.
#### Intuition behind the style loss
Though the above equation system is a mouthful, the idea is relatively simple. The goal is to compute a style matrix (visualised below) for the generated image and the style image. Then the style loss is defined as the root mean square difference between the two style matrices.
Below you can see an illustration of how the style matrix is computed. The style matrix is essentially a Gram matrix, where the (i,j) th element of the style matrix is computed by computing the element wise multiplication of the i th and j th feature maps and summing across both width and height. In the figure, red cross denotes element wise multiplication and the red plus sign denotes summing across both width height of the feature maps.
You can compute the style loss as follows.
#### Why is it that style is captured in the Gram matrix?
It’s great that we know how to compute the style loss. But you still haven’t been shown “why the style loss is computed using the Gram matrix”. The Gram matrix essentially captures the “distribution of features” of a set of feature maps in a given layer. By trying to minimise the style loss between two images, you are essentially matching the distribution of features between the two images [3, 4].
Note: Personally, I don’t think the above question has been answered satisfactorily. For example [4] explains the similarities between the style loss and domain adaptation. But this relationship does not answer the above question.
So let me take a shot at explaining this a bit more intuitively. Say you have the following feature maps. For simplicity I assume only three feature maps, and two of them are completely inactive. You have one feature map set where the first feature map looks like a dog, and in the second feature map set, the first feature map looks like a dog upside down. Then if you try to manually compute content and style losses, you will get these values. This means that we haven’t lost style information between two feature map sets. However, the content is quite different.
Understanding style loss
#### Final loss
The final loss is defined as,
where α and β are user-defined hyperparameters. Here β has absorbed the M^l normalisation factor defined earlier. By controlling α and β you can control the amount of content and style injected to the generated image. You can also see a nice visualisation of different effects of different α and β values in the paper.
### Defining the optimiser
Next you use the Adam optimiser to optimise the loss of the network.
### Defining the input pipeline
Here you define the full input pipeline. `tf.data` provides a very easy to use and intuitive interface to implementing input pipelines. For most of the image manipulation tasks you can use the `tf.image` API, however the ability of `tf.image` to handle dynamic sized images is very limited. For example, if you want to dynamically crop and resize images it is better to do in the form of a generator as implemented below.
You have defined two input pipelines; one for content and one for style. The content input pipeline looks for `jpg` images that start with the word `content_`, where the style pipeline looks for images starting with `style_`.
### Defining the computational graph
Now you are ready to rock and roll! In this section you will be defining the full computational graph.
• Define iterators that provide inputs
• Define inputs and CNN variables
• Define the content, style and the total loss
• Define the optimisation operation
### Running style transfer
Time to run the computational graph and generate some artwork. The generated artwork will be saved to `data/gen_0`, `data/gen_1`, …, `data/gen_5`, etc. folders.
When you run the above code, you should be getting some neat art saved to your disk like below.
### Conclusion
In this tutorial, you learnt about neural style transfer. Neural style transfer allows to blend two images (one containing content and one containing style) together to create new art. You first went through why you need neural style transfer and an overview of the architecture of the method. Then you defined the specifics of the neural style transfer network with TensorFlow. Specifically, you defined several functions to define the variables/inputs, compute the VGG output, compute the losses and perform the optimisation. You next understood the two losses that allow us to achieve what we want; the content loss and the style loss in detail, and saw how they come together to define the final loss. Finally you ran the model and saw artwork generated by the model.
Code for this tutorial is available here. | 2,402 | 11,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-18 | longest | en | 0.920672 |
https://henmydiffi.web.app/1485.html | 1,723,089,770,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640719674.40/warc/CC-MAIN-20240808031539-20240808061539-00390.warc.gz | 229,120,740 | 6,135 | While supplying the nonlinear loads, harmonic currents flow in the transformer and to protect the windings and transformer oil from overheating, load on it needs to be reduced. Rating of transformers supplying harmonic rich loads. Factor k in europe, the transformer derating factor is calculated according to the formulae in bs 7821 part 4. Kfactor is a mathematical representation to characterize a transformers ability to withstand. Standard and custom designs through 10 mva, up to 35kv. K factor distribution transformers in north america are subject to minimum efficiency regulations in both the u. Transformers course to be more familiar with the contents of our new articles about the power and distribution transformers sizing calculations.
The k factor is a number calculated from the harmonic spectrum of the current in a load waveform such that, regardless of the actual spectrum, loads that require the same k factor have the same heating effect on the transformer. It is a k factor calculator developed by the new zealand company knac solutions. No load losses of a distribution transformer under harmonic loads can increase its operating cost. What is k factor and what does it mean regarding transformers. Calculating the prospective short circuit current 4. In this standard, the k factor is defined and its method of calculation shown. Calculating overcurrent protection device rating on the low voltage side of the transformer. Kfactor measures the heating effects of harmonics on transformer loading and losses.
A kfactor rated transformer is one which is used to deal with harmonic generating loads. A case study for selection and sizing of k rated isolation transformers in parallel redundant ups system. Pdf a study of kfactor power transformer characteristics by. Dry type distribution transformers solahd k factor transformers are designed to reduce the heating effects of harmonic currents created by loads like those shown in chart a. Power transformer kfactor understand the importance of power transformers k factor and follow these simple steps to verify transformer content and load.
K factor must be determined to calculate the right size transformer that is needed. As from the formula, while calculating the kfactor, the harmonic current is getting multiplied by the square of its number. These type of loads generate harmonic currents that cause transformers and system neutrals to overheat, which can destroy the transformer. The harmonic analysis module also offers the transformer kfactor ansiiec and factork bs calculation for the assessment of transformer rating with respect. K factor is a weighting of the harmonic load currents according to their effects on transformer heating, as derived from ansiieee c57. In short, k factor is a weighting of the harmonic load currents according to their effects on transformer heating, as derived from ansiieee c57. Short answer is the k factor conveys a transformers ability to serve varying degrees of nonlinear loads without exceeding the rated temperature rise limits. Krated transformer manufacturers mumbai india neelkanth power. All sheet metal workers know the importance of an accurate k factor or bend allowance when it comes to design. Only the fundamental oscillation is used in order to calculate the cosphi. It is important to have systems in place to monitor and analyze the power quality issues that can affect critical and expensive components in a power distribution system.
Miscellaneous documents power quality international. By downloading this spreadsheet you agree to the following terms. To increasedecrease the alternating volts in electric power, transformers are used. Rus bulletin 1724e301 guide for the evaluation of large. Underwriters laboratories recognized the problems with derating. K factor calculation depends upon the fundamental current and the harmonic current components, and it is an indicative value of the harmonic contents in the power system. The k factor transformer is an isolating transformer that can be loaded at 100% of load plus the harmonics load without overheating. It has two types, single phase and three phase transformer. Determination of transformer rating 52 temperature increases also reduce the power factor, productivity, efficiency, capacity and performance of the plant. Neel industrial systems k rated isolation transformer. Significance of k factor k factor of the electrical distribution network is the reason why transformers need to be derated. K factor transformers are specially assembled with a double sized neutral conductor, heavier gauge copper and either change the geometry of their conductors or use multiple conductors for the coils. Kfactor transformers are specially designed to accommodate harmonic.
I came across a community sponsored application that will aid any sheet metal designer. Significance of k factor in circuit breaker ratings powell industries, inc. The kfactor transformer power quality in electrical. Collection of formulas for umg measurement devices janitza. Alfa offers a complete range of new and reconditioned low voltage dry type transformers including general purpose, drive isolation, motorstarting autotransformers. The k rating number of the transformer 1, 4, 9, 20 is in indication of the amount of harmonic current the transformer is capable of handling. As the k factor of the transformer increases, generally the size and cost increase, low load efficiency k factor transformers are designed to withstand the extra heating and higher neutral currents caused by harmonics created by nonlinear loads such as vfd, dc power supplies and led lighting. Transformer capacity calculator three phase transformer.
Significance of k factor in circuit breaker ratings. An online three phase transformer calculator to find the current in different power units. A specific k factor rating indicates that a transformer can supply its rated kva load output to a load with a specified amount of harmonic content. Eaton s kfactor transformers address the problems caused by nonlinear loads and harmonics. K 40 k factor transformers are designed to be operated fully loaded with any harmonic load having a k factor equal to or less than its k rating. It is important to note, however, that these solutions are best only for those experts who have knowledge and experience with assembly transformers. K factor is the right company, iso9001 certified, reliable and. Calculating the prospective short circuit current across the low voltage side of the transformer. K factor is a unit measuring a transformer s ability to withstand the harmonics content of a system. Lft transformer loss factor which is the ratio of average transformer losses to peak transformer losses.
To download your copy of transformer calculations spreadsheet, click on. As long as the load k factor is lower than the k rating of the transformer, the full kva rating of the transformer can. In this paper, the formulations how to calculate the maximum permissible rms nonsinusoidal load current i max using the harmonic loss factors f hl and k factor were introduced. If you have some experience in assembling transformers, then this program is perfect for you. Find, read and cite all the research you need on researchgate. Mks mksh single phase kfactor dry type distribution. Equivalence table between k factor ul 1562 and over rating factor. Kfactor transformer, also known as k rated transformer, is designed for nonlinear or harmonic generating loads that a standard transformer could not adequately handle due to overheating. Calculate size of transformer and voltage due to starting of large size motor new.
A k factor transformer is designed to handle harmonic content in its load current without exceeding its operating temperature limits. Once the k factor of the load has been determined, it is a simple matter to specify a transformer with a higher k rating from the standard range of 4, 9, 20, 30, 40, 50. Professionals with experience in transformer assemblies use the computer program to calculate the number of coils and wire thickness. There are two widely used methods to calculate the kfactor. Are you looking for the best transformer calculator software for your next project. The k factor rating is an index of the transformer s ability to withstand harmonic content while operating within the temperature limits of its insulating system. Visit our website to view the publications listed below, download the software.
The derating measure for normal transformers is always taken to limit the temperature of windings or topoil. Today, we will explain other special cases for power and distribution transformers sizing calculations. Power and distribution transformers sizing calculations part seven. In the case of sinusoidal loading of the transformer the k factor 1. Three phase isolation transformers from 1 to 40kva. A study of k factor power transformer characteristics by modeling simulation. | 1,655 | 8,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-33 | latest | en | 0.905704 |
http://research.stlouisfed.org/fred2/series/PGD2USLBA621NUPN | 1,433,007,247,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207932182.89/warc/CC-MAIN-20150521113212-00232-ip-10-180-206-219.ec2.internal.warc.gz | 204,136,006 | 19,786 | # Purchasing Power Parity Converted GDP Per Capita Relative to the United States, average GEKS-CPDW, at current prices for Lebanon
2010: 30.47824 U.S.=100 (+ see more)
Annual, Not Seasonally Adjusted, PGD2USLBA621NUPN, Updated: 2012-09-17 11:21 AM CDT
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
This is the current per capita GDP expressed relative to the United State (US=100) in each year. More information is available at http://pwt.econ.upenn.edu/Documentation/append61.pdf
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: y2
Source: University of Pennsylvania
Release: Penn World Table 7.1
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(a) Purchasing Power Parity Converted GDP Per Capita Relative to the United States, average GEKS-CPDW, at current prices for Lebanon, U.S.=100, Not Seasonally Adjusted (PGD2USLBA621NUPN)
This is the current per capita GDP expressed relative to the United State (US=100) in each year. More information is available at http://pwt.econ.upenn.edu/Documentation/append61.pdf
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: y2
Purchasing Power Parity Converted GDP Per Capita Relative to the United States, average GEKS-CPDW, at current prices for Lebanon
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Name: Email: | 708 | 2,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2015-22 | latest | en | 0.759766 |
https://sinclairzxworld.com/viewtopic.php?f=5&p=29338 | 1,519,354,529,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814311.76/warc/CC-MAIN-20180223015726-20180223035726-00240.warc.gz | 764,248,513 | 7,172 | ## PEEK & POKE math (?? +256 ??)
Anything Sinclair ZX Basic related; history, development, tips - differences between BASIC on the ZX80 and ZX81
vwbz80a
Posts: 31
Joined: Wed Nov 29, 2017 2:35 pm
### PEEK & POKE math (?? +256 ??)
This must be really Obvious!!!!! But it is NOT computing between the ears. (or on paper).
Problem: why do you add 256 to the low byte decimal before multiplying by the high order decimal?
using a hex calculator there is no need for any additon.
and it is not that i can't get the right values to poke in, or peek at...........[here is where we loose everyone]....because i can.
It is that i just wish to use the simple calculator, manually. and follow the process. make sense ?
Andy Rea
Posts: 1313
Joined: Fri May 09, 2008 2:48 pm
Location: notts UK
### Re: PEEK & POKE math (?? +256 ??)
We are not adding 256 to the low byte... when we do something like
Code: Select all
``PRINT PEEK 16388 + 256 * PEEK 16389``
To display the current value of ramtop you have to remeber the order of processing.
the computer will do both PEEKS then it will multiple the second PEEK [highbyte] value by 256 and finally add on the first PEEK [lowbyte] value.
hth Andy
6 x ZX81, 1 x TS1500 , 1 x +3e, 1 x timex 2040 printer, 1 x timex 2020 cassette deck, siclair printer and some spectrum
vwbz80a
Posts: 31
Joined: Wed Nov 29, 2017 2:35 pm
### Re: PEEK & POKE math (?? +256 ??)
Andy Rea wrote:
Mon Jan 15, 2018 6:12 pm
We are not adding 256 to the low byte... when we do something like
Code: Select all
``PRINT PEEK 16388 + 256 * PEEK 16389``
To display the current value of ramtop you have to remeber the order of processing.
the computer will do both PEEKS then it will multiple the second PEEK [highbyte] value by 256 and finally add on the first PEEK [lowbyte] value.
hth Andy
thanks, Andy. | 548 | 1,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-09 | longest | en | 0.833447 |
https://www.numbersaplenty.com/1213698 | 1,701,472,024,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00638.warc.gz | 1,025,339,115 | 3,091 | Search a number
1213698 = 231773163
BaseRepresentation
bin100101000010100000010
32021122212210
410220110002
5302314243
642002550
713213323
oct4502402
92248783
101213698
11759962
124a6456
13336585
1423844a
1518e933
hex128502
1213698 has 32 divisors (see below), whose sum is σ = 2621376. Its totient is φ = 373248.
The previous prime is 1213673. The next prime is 1213721. The reversal of 1213698 is 8963121.
It is an unprimeable number.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 7365 + ... + 7527.
It is an arithmetic number, because the mean of its divisors is an integer number (81918).
21213698 is an apocalyptic number.
1213698 is an abundant number, since it is smaller than the sum of its proper divisors (1407678).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
1213698 is a wasteful number, since it uses less digits than its factorization.
1213698 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 258.
The product of its digits is 2592, while the sum is 30.
The square root of 1213698 is about 1101.6796267518. The cubic root of 1213698 is about 106.6686696672.
The spelling of 1213698 in words is "one million, two hundred thirteen thousand, six hundred ninety-eight". | 399 | 1,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-50 | latest | en | 0.861244 |
https://oeis.org/A318580 | 1,638,458,610,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362230.18/warc/CC-MAIN-20211202145130-20211202175130-00587.warc.gz | 499,835,978 | 4,514 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A318580 Expansion of e.g.f. exp(-1 + Product_{k>=1} 1/(1 - x^k)^k). 0
1, 1, 7, 55, 601, 7561, 116191, 1999327, 39267985, 850964401, 20332107991, 527930427751, 14838001344937, 447653776595065, 14440021169407471, 495398956418435791, 18012260306904120481, 691502230924473978337, 27948692251661337581095, 1185878351946613955122711 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS FORMULA E.g.f.: exp(-1 + exp(Sum_{k>=1} sigma_2(k)*x^k/k)). E.g.f.: A(x) = exp(B(x) - 1), where B(x) = o.g.f. of A000219. a(0) = 1; a(n) = Sum_{k=1..n} A000219(k)*k!*binomial(n-1,k-1)*a(n-k). MAPLE seq(n!*coeff(series(exp(-1+mul(1/(1-x^k)^k, k=1..100)), x=0, 20), x, n), n=0..19); # Paolo P. Lava, Jan 09 2019 MATHEMATICA nmax = 19; CoefficientList[Series[Exp[-1 + Product[1/(1 - x^k)^k, {k, 1, nmax}]], {x, 0, nmax}], x] Range[0, nmax]! nmax = 19; CoefficientList[Series[Exp[-1 + Exp[Sum[DivisorSigma[2, k] x^k/k, {k, 1, nmax}]]], {x, 0, nmax}], x] Range[0, nmax]! p[n_] := p[n] = Sum[DivisorSigma[2, k] p[n - k], {k, n}]/n; p[0] = 1; a[n_] := a[n] = Sum[p[k] k! Binomial[n - 1, k - 1] a[n - k], {k, n}]; a[0] = 1; Table[a[n], {n, 0, 19}] CROSSREFS Cf. A000219, A001157, A058892, A318250. Sequence in context: A083836 A326885 A159313 * A054910 A028562 A209668 Adjacent sequences: A318577 A318578 A318579 * A318581 A318582 A318583 KEYWORD nonn AUTHOR Ilya Gutkovskiy, Aug 29 2018 STATUS approved
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Last modified December 2 10:13 EST 2021. Contains 349437 sequences. (Running on oeis4.) | 836 | 2,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-49 | latest | en | 0.518445 |
https://atcoder.jp/contests/past202005-open/tasks/past202005_g | 1,627,090,107,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150067.87/warc/CC-MAIN-20210724001211-20210724031211-00506.warc.gz | 128,998,289 | 6,543 | Contest Duration: - (local time) (300 minutes) Back to Home
G - Gold General Goes on the Grid /
Time Limit: 2 sec / Memory Limit: 1024 MB
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• (x+1, y+1)
• (x, y+1)
• (x-1, y+1)
• (x+1, y)
• (x-1, y)
• (x, y-1)
### 制約
• 入力はすべて整数
• 1 \leq N \leq 800
• -200 \leq x_i, y_i, X, Y \leq 200
• (x_i, y_i) は相異なる。つまり、同じマスに 2 個以上の障害物はない。
• (0, 0) および (X, Y) には障害物はない
• (X, Y) \neq (0, 0)
### 入力
N X Y
x_1 y_1
:
x_N y_N
### 入力例 1
1 2 2
1 1
### 出力例 1
3
..G
.#.
S..
ここで、S はマス (0, 0) を、G はマス (X, Y) = (2, 2) を、# は障害物のあるマスを表します。
### 入力例 2
1 2 2
2 1
### 出力例 2
2
### 入力例 3
5 -2 3
1 1
-1 1
0 1
-2 1
-3 1
### 出力例 3
6
Score : 6 points
### Warning
Do not make any mention of this problem until June 6, 2020, 6:00 p.m. JST. In case of violation, compensation may be demanded. After the examination, you can reveal your total score and grade to others, but nothing more (for example, which problems you solved).
### Problem Statement
We have an infinite two-dimensional grid. Snuke is initially standing at the square (0,0) in this grid. There are N squares that contain an obstacle, and Snuke cannot enter those squares. The i-th obstacle is at the square (x_i, y_i). When Snuke is at the square (x, y), he can get to one of the following six squares in one move:
• (x+1, y+1)
• (x, y+1)
• (x-1, y+1)
• (x+1, y)
• (x-1, y)
• (x, y-1)
Print the minimum number of moves Snuke needs to reach the square (X, Y). If this square is unreachable, print -1.
### Constraints
• All values in input are integers.
• 1 \leq N \leq 800
• -200 \leq x_i, y_i, X, Y \leq 200
• The pairs (x_i, y_i) are distinct. That is, no square contains two or more obstacles.
• The squares (0, 0) and (X, Y) do not contain an obstacle.
• (X, Y) \neq (0, 0)
### Input
Input is given from Standard Input in the following format:
N X Y
x_1 y_1
:
x_N y_N
### Output
Print an integer as specified in Problem Statement.
### Sample Input 1
1 2 2
1 1
### Sample Output 1
3
Shown below is a part of the grid:
..G
.#.
S..
Here, S, G, and # stand for the square (0, 0), (X, Y) = (2, 2), and a square with an obstacle.
The minimum number of moves needed to reach (X, Y) = (2, 2) is three. One way to achieve it is (0, 0) \to (0, 1) \to (1, 2) \to (2, 2).
### Sample Input 2
1 2 2
2 1
### Sample Output 2
2
### Sample Input 3
5 -2 3
1 1
-1 1
0 1
-2 1
-3 1
### Sample Output 3
6 | 1,063 | 2,509 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-31 | latest | en | 0.325554 |
https://www.collegemaitrisedelacathedrale.fr/2020-Jun-23/22.html | 1,638,025,682,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358189.36/warc/CC-MAIN-20211127133237-20211127163237-00490.warc.gz | 805,692,035 | 6,514 | # exterior algebra
• ### What is the exterior algebra of textbf R 2
2017-7-7 · The exterior algebra Λ R 2 is a real vector space of dimension 4 with basis 1 e 1 e 2 e 1 ∧ e 2. So its every element is a unique linear combination of these basis elements say a 1 ⋅ 1 a 2 e 1 a 3 e 2 a 4 e 1 ∧ e 2 for real numbers a 1 a 2 a 3 a 4 which can be chosen arbitrarily.
• ### Exterior Algebra -- from Wolfram MathWorld
Exterior Algebra. Exterior algebra is the algebra of the wedge product also called an alternating algebra or Grassmann algebra. The study of exterior algebra is also called Ausdehnungslehre or extensions calculus. Exterior algebras are graded algebras.
• ### Exterior AlgebraMichigan State University
2007-5-14 · The Algebra of the Exterior Product also called an Alternating Algebra or Grassmann Algebra.The study of exterior algebra is also called Ausdehnungslehre and Extensions Calculus.Exterior algebras are Graded Algebras. In particular the exterior algebra of a Vector Space is the Direct Sum over in the natural numbers of the Vector Spaces of alternating -forms on that Vector Space.
• ### The Exterior Algebra and Central Notions in Mathematics
2015-3-10 · The exterior algebra originated in the workof Hermann Grassmann (1809–1877) in his bookAusdehnungslehrefrom 1844 and the thoroughlyrevised 1862 version which now exists in an Englishtranslation from 2000. Grassmann workedas a professor at the gymnasium in Stettin thenGermany. Partly because Grassmann was an originalthinker and maybe partly because his
• ### Exterior algebraInfogalactic the planetary knowledge core
2020-5-13 · The exterior algebra or Grassmann algebra after Hermann Grassmann is the algebraic system whose product is the exterior product. The exterior algebra provides an algebraic setting in which to answer geometric questions. For instance blades have a concrete geometric interpretation and objects in the exterior algebra can be manipulated
• ### Exterior Algebras ScienceDirect
Exterior algebra is an important tool for studying endomorphisms over E. In particular in the same way as vectors of E are employed to construct vectors of Λ pE we address the following question. Select 11Λ2E Algebra. Book chapter Full text access.
• ### math/0001161v1 The exterior algebra and `Spin of an
2000-1-28 · Abstract A well-known result of Kostant gives a description of the G-module structure for the exterior algebra of Lie algebra . We give a generalization of this result for the isotropy representations of symmetric spaces. If is a Z_2-grading of a simple Lie algebra we explicitly describe a -module such that the exterior algebra of is the
• ### 4 Exterior algebraPeople
2020-4-19 · 4 Exterior algebra 4.1 Lines and 2-vectors The time has come now to develop some new linear algebra in order to handle the space of lines in a projective space P(V). In the projective plane we have seen that duality can deal with this but lines in higher dimensional spaces behave differently.
• ### The exterior algebra of a vector space.
The exterior algebra of a vector space. If is a vector space we define a -linear map to be a map where there are copies of which is linear in each factor. That is. We define a -linear map to be totally antisymmetric if for all vectors and all . Note that it follows that and if is a permutation of letters then.
• ### Exterior algebraWikiMili The Best Wikipedia Reader
2020-4-1 · The exterior algebra over the complex numbers is the archetypal example of a superalgebra which plays a fundamental role in physical theories pertaining to fermions and supersymmetry. A single element of the exterior algebra is called a supernumber 21 or Grassmann number.
• ### Exterior algebra Math Wiki Fandom
2021-7-12 · Exterior algebra is a type of algebra characterized by the Wedge product and the Anti-wedge product. A∧B is the wedge product of A and B which is a simple bivector or 2-blade
• ### Exterior algebraen.LinkFang
The exterior algebra Λ(V) of a vector space V over a field K is defined as the quotient algebra of the tensor algebra T(V) by the two-sided ideal I generated by all elements of the form x ⊗ x for x ∈ V (i.e. all tensors that can be expressed as the tensor product of a vector in V by itself). The ideal I contains the ideal J generated by elements of the form ( displaystyle xotimes y y
• ### Tensor Exterior and Symmetric AlgebrasThe Rising Sea
2021-6-28 · A graded R-algebra is an R-algebra Awhich is also a graded ring in such a way that the image of the structural morphism R−→ Ais contained in A 0. Equivalently Ais a graded ring and a R-algebra and all the graded pieces A d d≥ 0 are R-submodules. A morphism of graded R-algebras is an R-algebra morphism which preserves degree.
• ### Wedge Product -- from Wolfram MathWorld
Wedge Product. The wedge product is the product in an exterior algebra. If and are differential k -forms of degrees and respectively then. (Spivak 1999 p. 203) where and are constants. The exterior algebra is generated by elements of degree one and so the wedge product can be defined using a basis for when the indices are distinct and
• ### Exterior Algebra SpringerLink
2017-11-7 · Exterior Algebra. In super case there are two possible definitions leading to nonisomorphic algebras (which are however isomorphic as vector spaces). First definition. For a ℤ 2graded vector spaceV = V0 ⊕ V1 define Λ ( V ) = T ( V )/ I where T ( V) is the tensor algebra of V
• ### Exterior algebraInfogalactic the planetary knowledge core
2020-5-13 · The exterior algebra or Grassmann algebra after Hermann Grassmann is the algebraic system whose product is the exterior product. The exterior algebra provides an algebraic setting in which to answer geometric questions. For instance blades have a concrete geometric interpretation and objects in the exterior algebra can be manipulated
• ### Tensor Spaces and Exterior AlgebraAMS
2019-2-12 · Exterior Algebra and its Applications 77 §1. Definition of exterior algebra and its properties 77 §2. Applications to determinants 83 §3. Inner (interior) products of exterior algebras 88 §4. Applications to geometry 91 Exercises 97 . VI CONTENTS Chapter IV. Algebraic Systems with Bilinear Multiplication. Lie
• ### The exterior algebra Mathematics for Physics
The exterior algebra (AKA Grassmann algebra alternating algebra) is the tensor algebra modulo the relation v ∧ v ≡ 0 and can be written as ΛV ≡ ΣΛkV = R ⊕ Λ1V ⊕ Λ2V ⊕ ⋯ ⊕ ΛnV where n is the dimension of V (since ΛkV automatically vanishes for k > n).
• ### exterior algebra Arithmetic variety
Tag Archives exterior algebra Post navigation Trace is the derivative of determinant. Posted on June 5 2020 by rohanjoshi. 0. A question I always had when learning linear algebra is "what does the trace of a matrix mean conceptually " For example the determinant of a matrix is roughly speaking the factor by which the matrix expands
• ### What is the exterior algebra of textbf R 2
2017-7-7 · The exterior algebra Λ R 2 is a real vector space of dimension 4 with basis 1 e 1 e 2 e 1 ∧ e 2. So its every element is a unique linear combination of these basis elements say a 1 ⋅ 1 a 2 e 1 a 3 e 2 a 4 e 1 ∧ e 2 for real numbers a 1 a 2 a 3 a 4 which can be chosen arbitrarily.
• ### Exterior algebra Math Wiki Fandom
2021-7-12 · Exterior algebra is a type of algebra characterized by the Wedge product and the Anti-wedge product. A∧B is the wedge product of A and B which is a simple bivector or 2-blade
• ### exterior algebraPlanetMath
2020-2-9 · Λ(V) the exterior algebra of V we are referring to the isomorphism class of all such models. It is also common to identify Vwith its image ι(V) and to write vrather than
• ### Exterior algebra Math Wiki Fandom
2021-7-12 · Exterior algebra is a type of algebra characterized by the Wedge product and the Anti-wedge product. A∧B is the wedge product of A and B which is a simple bivector or 2-blade
• ### math/0001161v1 The exterior algebra and `Spin of an
2000-1-28 · Abstract A well-known result of Kostant gives a description of the G-module structure for the exterior algebra of Lie algebra . We give a generalization of this result for the isotropy representations of symmetric spaces. If is a Z_2-grading of a simple Lie algebra we explicitly describe a -module such that the exterior algebra of is the
• ### The exterior algebra of a vector space.
The exterior algebra of a vector space. If is a vector space we define a -linear map to be a map where there are copies of which is linear in each factor. That is. We define a -linear map to be totally antisymmetric if for all vectors and all . Note that it follows that and if is a permutation of letters then.
• ### Tensor Spaces and Exterior AlgebraAMS
2019-2-12 · Exterior Algebra and its Applications 77 §1. Definition of exterior algebra and its properties 77 §2. Applications to determinants 83 §3. Inner (interior) products of exterior algebras 88 §4. Applications to geometry 91 Exercises 97 . VI CONTENTS Chapter IV. Algebraic Systems with Bilinear Multiplication. Lie
• ### Exterior Algebra Physics Forums
2014-9-12 · General Math Calculus Differential Equations Topology and Analysis Linear and Abstract Algebra Differential Geometry Set Theory Logic Probability Statistics MATLAB Maple Mathematica LaTeX Hot Threads
• ### Exterior algebraEncyclopedia of Mathematics
2020-6-5 · The exterior algebra for M is defined as the direct sum ∧ M = ⊕ r ≥ 0 ∧ r M where ∧ 0 M = A with the naturally introduced multiplication. In the case of a finite-dimensional vector space this definition and the original definition are identical. The exterior algebra of a module is employed in the theory of modules over a principal
next: skylar grey | 2,386 | 9,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-49 | latest | en | 0.843779 |
https://brainly.my/tugasan/136661 | 1,485,036,597,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281226.52/warc/CC-MAIN-20170116095121-00469-ip-10-171-10-70.ec2.internal.warc.gz | 795,579,298 | 10,001 | # A local furniture store is advertising a discount of 30% off of their selection of sofas. Pauline wants to buy a sofa that has an original cost of \$450. What is the sale price of the sofa?
1
dari lu9nnewfikashahma
## Jawapan
2016-03-13T11:24:29+08:00
450 * 100-30/100 =
450 * 70/100 =
31500/100 =
\$315 | 103 | 308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-04 | latest | en | 0.810336 |
https://www.instructables.com/id/How-To-Count-In-Binary-With-Nothing-But-Your-Hands/ | 1,571,605,048,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986718918.77/warc/CC-MAIN-20191020183709-20191020211209-00441.warc.gz | 944,337,187 | 19,807 | # How to Count in Binary With Nothing But Your Hands
5,474
4
3
Materials:
- A hand.
Here are Tara's Three Rules for counting in binary on your hands.
1. When a finger is down, that is '0'.
2. When a finger is up, that is '1.'
3. Do not count your thumb.
In this picture, all fingers are down. So this means zero.
You maybe wondering why the thumb is out.
The answer is that it does not matter where the thumb is. I do not count the thumb when counting in binary. When we get to number 10, you will see why I leave the thumb out.
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: One
This is 1.
Put the index finger up.
1 finger is up, so this is the binary number 1.
## Step 2: 10
Read the fingers from left to right. The middle finger is up, so that is 1. The index finger is down, so that is zero.
This makes the number 10.
This number is why I leave my thumb out when counting in binary. I started doing that when a teacher thought I was flicking her off. She was not impressed when I said, "I am not flicking you off, I am counting in binary.
## Step 3: 11
Read the fingers from left to right.
You get the number 11.
## Step 4: 100
This is hard to do because my ring finger is not flexible.
## Step 5: 101
Ring up, middle down, ring up is 101 in binary.
## Step 6: 110
Ring up, middle up, index down is
110.
## Step 7: 111
Ring up, middle up, and index up means the number is 111.
## Step 8: 1000 Through 1111
Here are the next 8 binary numbers. Can you read them?
If you can, congratulations! You have learned to count up in binary!
## Recommendations
• ### Lamps Class
9,733 Enrolled
## 3 Discussions
By excluding the thumb, and leaving it to stick out instead, you can only count half as high. When you account for both hands, you can only count a quarter as high as you could by using your thumbs. If you are worried about how people will take 0010000100, you can use the simple expedient of reversing your hands, facing your palm away from you, and beginning at the pinky instead of the thumb.
how we know we read correct figures as u mentioned last 8 | 574 | 2,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-43 | latest | en | 0.949693 |
http://www.conceptispuzzles.com/forum/tm.aspx?m=37320 | 1,603,579,848,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107885059.50/warc/CC-MAIN-20201024223210-20201025013210-00606.warc.gz | 142,917,882 | 8,943 |
## Need help with Kakuro
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Need help with Kakuro - 4/20/2019 9:20:33 AM
JohnWK
Posts: 2
Joined: 4/20/2019
From: Australia
Status: offline
Who can help me solve this Kakuro? It is number 15 from the free hard starters. Have been looking at it for ages but can’t see the way out.
Thanks,
John
Attachment (1)
< Message edited by JohnWK -- 4/25/2019 1:35:54 AM >
RE: Need help with Kakuro - 4/25/2019 1:25:29 AM
Ahlyis
Posts: 208
Joined: 12/30/2009
From: United States
Status: offline
Bad logic on my part. Post edited. Let me look some more...
< Message edited by Ahlyis -- 4/25/2019 1:28:46 AM >
RE: Need help with Kakuro - 4/25/2019 7:27:45 PM
Ahlyis
Posts: 208
Joined: 12/30/2009
From: United States
Status: offline
First, a minor piece.
You can eliminate the 2's from the last row, columns 5 and 6. No matter which way they go, the 2 will have to be in either column 1 or 3.
Now, for something better.
The top of column 3 is 26 down. The bottom 2 digits cannot be 3 and 5. Not just because it would make it impossible to determine the order between them and the 3-5 in column 2 (which I do not accept as valid logic for an elimination). But if you put in the 3 and 5, in either order, that puts a 1 at the third position. Tracing that out, you can fill in a ton of numbers, but you eventually end up with row 5, 33 across, having a 1, 2, 3, 4 and 5. There is no way to make the 33 with all of those. Therefore you can eliminate the 3 and 5 as possibles for the bottom 2 positions of the 26 down in column 3.
That leaves you with either 2-4 or 2-6 and whichever it is, the 2 must be the bottom number.
Knowing that the 4th position must be 4 or 6 eliminates the 3 as a possible starting number for row 4, 32 across.
Hope that helps.
RE: Need help with Kakuro - 4/25/2019 7:48:31 PM
Ahlyis
Posts: 208
Joined: 12/30/2009
From: United States
Status: offline
With the previous in place. Take a look at row 4, 17 across. If you put in the 9 you can fill in a bunch of numbers and eventually end up with an impossible situation at the intersection of row 5 and column 5 (or possibly elsewhere).
I'll stop with that unless you ask for more. :)
RE: Need help with Kakuro - 4/26/2019 9:28:04 AM
JohnWK
Posts: 2
Joined: 4/20/2019
From: Australia
Status: offline
Hi Ahlyis, thanks for your help in pointing me in the right direction!
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| 946 | 3,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-45 | latest | en | 0.886044 |
https://openubiproject.org/2022/08/16/how-much-does-a-5-gallon-bucket-of-sap-weigh/ | 1,679,388,533,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943637.3/warc/CC-MAIN-20230321064400-20230321094400-00574.warc.gz | 513,138,949 | 18,445 | # How much does a 5 gallon bucket of sap weigh?
## How much does a 5 gallon bucket of sap weigh?
Maple syrup must meet exacting standards for purity. High quality pure maple syrup can be made only by the evaporation of pure maple sap, and by weight may contain no less than 66 percent sugar (Brix). In Vermont and New Hampshire the minimum sugar content is 66.9%.
The “Rule of 86” is used to calculate the gallons of sap needed to produce one gallon of syrup. It states that the number of gallons of sap you need to produce one gallon of syrup is equal to 86 divided by the percent sugar. Rule of 86 Gallons of sap to produce one gallon of syrup = 86 / % sugar content in sap.
The volume of sap produced during one season varies from 10-20 gallons per tap, depending on the tree, weather conditions, length of the sap season, and method of collecting sap.
Rule of 86 Gallons of sap to produce one gallon of syrup = 86 / % sugar content in sap. For example, you would need 43 gallons of sap with 2% sugar content to produce one gallon of syrup. Why is maple syrup different colors? The color in maple syrup results from a browning reaction that occurs during the latter stages of evaporation.
Should you be plugging maple tap holes at the end of the season? Nope! No need for you to plug maple tap holes with twigs or anything else. Trees know how to heal their wounds all on their own.
Some people enjoy drinking sap fresh from the tree, while others prefer to boil it for a brief period to kill any bacteria or yeast. Since it is certainly possible for harmful bacteria to be found in sap, the cautious solution is to pasteurize it before drinking.
between 66°
The unit of measure most often used for syrup density is Brix – one Brix is equal to about 1% sugar content. The correct density for maple syrup is between 66° and 68° Brix, with some local jurisdictions that have strict maple laws requiring a narrower range.
The weighted average was \$2.87 per tap or \$11.48 per gallon (assumes four taps required to produce a gallon of syrup). The average annual investment cost for a plastic tubing system ranged from \$7.90 for a 500-tap operation to \$6.03 for a 10,000-tap system.
40 gallons
Usually about 40 gallons of sap are required to produce one gallon of finished syrup. Actually this figure can vary from 20 to 60 gallons or more depending primarily on sap sugar content. A large amount of water must be evaporated from the sap to produce the finished syrup of 66 to 67 percent sugar.
Maple syrup has a density of 1.37 grams per milliliter, heavier than water, milk and corn syrup but not as dense as honey. When the temperature of the syrup reaches 7 degrees above the boiling point of water, it has reached the correct thickness.
Prices. The average U.S. price per gallon for maple syrup in 2017 was \$35, down \$1.70 from 2016. The average price per gallon in Vermont was \$30, and 80 percent of the sales were bulk.
Five gallons of sap weigh 55 pounds.
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# 1. NUMERAL SYSTEM
A numeral system is how we represent a number. The value of a number depends on the digit we use and its position.
For example, in decimal system (the most widely used system), we use ten figures for representing a number (0,1,….,8,9) and 89 has not the same value as 98, despite the fact that we use the same figures for representing both numbers.
Ideally, a numeral system will:
• Represent all the numbers (rational, integers, etc.)
• Use just one representation for each number.
• Reflect the algebraic and arithmetic structure of the numbers.
Source:
wikipedia
Some ancient numeral systems:
picture source
http://math.tutorvista.com/number-system.html
PLEASE, DO ACTIVITY 1 BEFORE CONTINUING
Short url: https://clilstore.eu/cs/4981 | 196 | 806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-22 | latest | en | 0.790699 |
https://astronomy.stackexchange.com/questions/57028/unreasonable-ratios-in-astronomy-as-ways-of-remembering/57029 | 1,719,047,417,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862310.25/warc/CC-MAIN-20240622081408-20240622111408-00117.warc.gz | 94,244,698 | 39,326 | # Unreasonable ratios in astronomy as ways of remembering
In mathematics there are approximations to pi such as:
$$\pi\approx\dfrac{355}{113}$$
that have extraordinary accuracy and are useful ways of remembering or being able to calculate a value close to pi that may be used in calculations.
I noticed the the ratio of the diameter of the moon to the earth seems unreasonably close to $$\frac{3}{11}$$.
Is my observation correct?
And are there other useful ways of remembering approximate metrics of various heavenly bodies with surprising economy?
• 1 year is $\simeq \pi \times 10^7$ seconds and see xkcd.com/1047 for other examples of numerology. Commented Feb 26 at 12:51
• Fun fact: No rational approximation to π is really useful, since the number of digits to be remembered will never be smaller than the number of decimals in π to which it is accurate. For instance, 355/113 involves six digits, but is only correct to 5th decimal in π.
– pela
Commented Feb 26 at 14:18
• @pela is that always true? It seems to hold for the first few terms in the continued fraction, but I can't think of why it would hold in general. Commented Feb 26 at 15:31
• can't be true for all irrational numbers eg consider 1 + pi/10^9, the rational number 1/1 involves two digits, but is accurate to the 8th decimal place. @pela Commented Feb 26 at 18:33
• The error in a good rational approximation (eg from the continued fraction) $p/q$ is $<1/q^2$. See en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem 355/113 is ~3.14159292, so it gives you pi (~3.14159265) to 7 digits. Commented Feb 26 at 19:14
The angular size of the Moon and the Sun, as viewed from the Earth is pretty much exactly $$1/1$$ with slight variations due to the eccentricity of the Moons orbit. Another coincidence is that the acceleration of 1 $$g$$ at the Earth's surface is almost exactly equal to $$1 c/$$year where $$c$$ is the speed of light. In terms of masses I find the following chain of ratios useful to get a grip on the size of things: $$80$$ Moons in the Earth, $$300$$ Earths in Jupiter, and $$1000$$ Jupiters in the Sun. And, while not uniquely astronomical, I've remembered since high school that there are $$\pi\times10^7$$ seconds in a year (to within half a percent error!).
It's also worth bearing in mind that a lot of astronomical units are ratios to begin with (technically, all units are, but typically not with respect to a relevant baseline). Thus any time something is given in AUs or years it's a ratio to the radius and period of the Earth's orbit around the Sun.
• FWIW, 1 g = 9.80665 m/s2 ~= 1.0323 ly/y^2, using Julian years of 365.25 days. Commented Feb 26 at 19:41
• I learned that last example as “π seconds in a nanocentury”. Commented Feb 27 at 12:38
• @gidds I mean, that works, but nanocentury definitely feels like a cursed unit though. Like kilo Watt hours or km per megaparsec per second. Commented Feb 27 at 14:27
• @PM2Ring Rocket engine specific impulse is often represented in seconds (as a somewhat misleading shorthand for "pound-force-seconds-per-pound-mass"), which differs by a factor of g from the other commonly used unit, meters per second (really Newton-seconds per kilogram, but it's the same dimension). For back-of-the-envelope conversions, g=10 is a good approximation, and likewise 10N thrust against 1 kg of rocket gives you about 1 g of acceleration. Commented Feb 28 at 7:03
1 km/s is roughly 1 pc per million years. This is roughly a consequence of there being $$\pi \times 10^7$$ seconds in a year and roughly $$\pi$$ light years in a parsec. Useful.
Since the Sun is 8 light minutes away, and the speed of light is a foot per nanosecond, the diameter of the Earth’s orbit is a trillion feet.
• This is only useful to Americans though…
– pela
Commented Feb 27 at 11:29
• Diameter of earth's orbit is ~1000 light seconds. Commented Feb 27 at 13:03
• Similarly, the diameter of the earth's orbit is $3 \times 10^{11}$ meters if you work in metric - the diameter of the orbit is $10^3$ seconds and the speed of light is $3 \times 10^8$ m/s. Commented Feb 27 at 14:34
• @pela Geocentric => Heliocentric => Americocentric Commented Feb 27 at 16:30
• The americans should just change their basic length unit to nanolightsecond. This would please the science nerds, and wouldn't change much for people accustomed to feet. Commented Feb 28 at 1:18 | 1,166 | 4,383 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-26 | latest | en | 0.932232 |
https://math.stackexchange.com/questions/388530/proving-an-identity-with-a-combinatorial-proof-binomnk-binomkr-binom | 1,560,955,764,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999000.76/warc/CC-MAIN-20190619143832-20190619165832-00018.warc.gz | 520,096,497 | 37,404 | # Proving an identity with a combinatorial proof: $\binom{n}{k}\binom{k}{r}=\binom{n}{r}\binom{n-r}{k-r}$
For any integers $$n$$, $$k$$, $$r$$ where $$n\geq k\geq r \geq 0$$, give a combinatorial proof of the following identity: \begin{align} \binom{n}{k}\binom{k}{r}=\binom{n}{r}\binom{n-r}{k-r} . \end{align}
The problem is that I can't come up with a good counting argument of what exactly the two sides are doing. The left hand side is quite mysterious, and the right hand side is apparently choosing $$r$$ items and then choosing $$k-r$$ items from the remaining, which should be equivalent to $$\binom{n}{k}$$ but somehow isn't. How should I approach this problem?
• I think the reason the RHS is not $\binom{n}{k}$ is that in choosing $r$ items and then $k-r$ items, you can distinguish between the items chosen in the first batch and those chosen in the second. – Alex Becker May 11 '13 at 14:46
Think of the problem of how to choose a sports team team consisting of $k$ people, and then to choose $r$ people from that team to play in a particular match (leaving $k-r$ people on the sidelines). How many ways are there to do this if you have $n$ total people from which to choose?
We can make these choices in two different ways. First of all, how many ways are there to pick the entire team first, and then pick the players for the match from the team? Secondly, how many ways are there to pick just the players for the match first, and then fill out the rest of the team from the people you didn't pick?
In more general terms, we you can think of this identity as saying that if we make a selection from a set of $n$ things so that $k$ of them have a property $1$ and $r$ of them have properties $1$ and $2$, it doesn't matter the order in which we assign the properties.
• Good way of explaining this. That is, let $A$ be a set with $n$ elements. You are counting the pairs $(B, C)$, where $B \subseteq C \subseteq A$, with $\lvert B \rvert = r$, and $\lvert C \rvert = k$, in two ways, either $C$ first (LHS), or $B$ first (RHS). – Andreas Caranti May 11 '13 at 14:51
• @AndreasCaranti Thank you! It's probably my favourite combinatorial proof! I'm never quite sure how generally to phrase a proof to be both understandable and to make sure it applies to all related situations, but the subset terminology is nice. – Tom Oldfield May 11 '13 at 14:56
• I'm not failing to get the identity. How can we formalize this? How can the left hand side be formulated into the size of a set $S$? – ithisa May 11 '13 at 15:00
• @EricDong, look at my comment, it is the set of all pairs $(B, C)$ etc. – Andreas Caranti May 11 '13 at 15:01
You have $$n$$ children. Give $$k$$ of them plates, and on $$r$$ of those plates, put a cookie. That gives you the left hand side.
Now, with the same $$n$$ children, give $$r$$ of them a plate with a cookie. Of the $$n-r$$ children that don't have a plate with a cookie, give $$k-r$$ of them empty plates. This gives you the right hand side.
Choose a $$k$$-member team (from $$n$$ people) and then choose an $$r$$-member starting line up. That gives you the LHS. Now do the same thing in a different way. Choose the $$r$$-member starting line up first then choose the $$k-r$$ member substitution from the remaining $$n-r$$ people. This gives you the RHS. | 918 | 3,302 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-26 | latest | en | 0.932619 |
https://gateoverflow.in/371891/Gate-cse-2022-question-45 | 1,656,217,935,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037089.4/warc/CC-MAIN-20220626040948-20220626070948-00704.warc.gz | 323,787,816 | 28,223 | 3,197 views
Consider routing table of an organization’s router shown below:
$$\begin{array} {|l|l|l|} \hline \text{Subnet Number} & \text{Subnet Mask} & \text{Next Hop} \\\hline 12.20.164.0 & 255.255.252.0 & \text{R1} \\\hline 12.20.170.0 & 255.255.254.0 & \text{R2} \\\hline 12.20.168.0 & 255.255.254.0 & \text{Interface 0} \\\hline 12.20.166.0 & 255.255.254.0 & \text{Interface 1} \\\hline \text{default} & & \text{R3} \\\hline \end{array}$$
Which of the following prefixes in $\text{CIDR}$ notation can be collectively used to correctly aggregate all of the subnets in the routing table?
1. $12.20.164.0/20$
2. $12.20.164.0/22$
3. $12.20.164.0/21$
4. $12.20.168.0/22$
@Arjun Sir
Can anyone say Why C And D can't collectively used here!? I think B and C are generating same sort of lowest and highest address.
### Subscribe to GO Classes for GATE CSE 2022
GATE official initial key has given B,D as answers.
I strongly feel this question should be challenged to give marks for all
I have prepared a draft to submit. Please find it here – here
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Some Binary Maths-
$255 = 11111111\\ 254 = 11111110\\ 252 = 11111100$
$255.255.254.0 = /23 \\ 255.255.252.0 = /22$
$\color{brown}{\textbf{Network1: }} 12.20.164.0/22 = \color{blue}{12.20.1010}\color{red}{01}00.0$
$\color{brown}{\textbf{Network2: }} 12.20.170.0/23 = \color{blue}{12.20.1010}\color{red}{101}0.0$
$\color{brown}{\textbf{Network3: }} 12.20.168.0/23 = \color{blue}{12.20.1010}\color{red}{100}0.0$
$\color{brown}{\textbf{Network4: }} 12.20.166.0/23 = \color{blue}{12.20.1010}\color{red}{011}0.0$
Ignore above color coding (why some part in $\color{blue}{\text{blue}}$ and some in $\color{red}{\text{red}}$). This will clear in a minute.
Now, We have these 4 networks. How can we identify all of these networks with some common name?
-Ok, You can say all of these start with $12$.
But can you be more specific? –
All of these start with $12.20$
But can you be as specific as you can?-
We can say all of these networks start with $\color{blue}{12.20.1010}$.
Now we say, we have supernet. And anything that start with $\color{blue}{12.20.1010}$ is part of my supernet.
This is used to simplify the routing table so that it does not have to maintain multiple entries. 1 entry will work here instead of 4 different entries.
We want initial 20 bits to be $\color{violet}{\textbf{fixed}}$ as $\color{blue}{12.20.1010}$ and remaining bits to has be $\color{red}{\textbf{free}}$.
The option(s) which has $\color{blue}{12.20.1010}$ as initial bits (and rest bits as free) is/are correct option(s)
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Option A : $12.20.164.0/20 = \color{blue}{12.20.1010}0100.0$
it has initial 20 bits $\color{violet}{\textbf{fixed}}$ as $\color{blue}{12.20.1010}$. Hence correct option.
Option B : $12.20.164.0/22 = \color{blue}{12.20.1010}\color{red}{01}00.0$.
It has initial 22 bits as fixed. Hence wrong option.
Option C : $12.20.164.0/21 = \color{blue}{12.20.1010}\color{red}{0}100.0$.
It has initial 21 bits as fixed. Hence wrong option.
Option D : $12.20.168.0/22 = \color{blue}{12.20.1010}\color{red}{10}00.0$.
It has initial 22 bits as fixed. Hence wrong option.
Options like $0.0.0.0/0$ or $12.0.0.0/8$ or $12.20.0.0/16$ or $12.20.10\text{xxxxxx}.0/18$ are also correct.
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Actually option A contains a lot many ip addresses which doesn’t belong to any of the 4 networks. But that is ok, because it atleast contains all of the 4 networks, there could be some more ip addresses or networks.
Infact $0.0.0.0/0$ this supernet contains everything in world hence can be used for route aggregation but its not most specific.
Number of IP addresses that option A contains = $2^{\text{free bits}} = 2^{12}$
Number of IP addresses all networks contains combinedly $= \underbrace{2^{10} }_\color{brown}{\textbf{Network1}} +\underbrace{2^9 }_\color{brown}{\textbf{Network2}} +\underbrace{2^9 }_\color{brown}{\textbf{Network3}} +\underbrace{2^9 }_\color{brown}{\textbf{Network4}} = 5\times2^9$
Video explanation for route aggregation – Route aggregation or Supernetting | Easiest explanation
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Edit: GATE official initial key has given B,D as answer.
Possible Reason: “collectively” word in the question. obviously B and D collectively are more precise than A.
Explanation for B and D – $\color{brown}{\textbf{Network4}}$ is already a part of $\color{brown}{\textbf{Network1}}$ hence we can just ignore $\color{brown}{\textbf{Network4}}$ for supernetting purpose.
Network1 (and Network4) is directly given by option B.
Network2 and 3 has $\color{blue}{12.20.1010}\color{red}{10}$ as common and can be represented by option D.
In this way, we do not have to aggregate extra ips and combinedly (B,D) are more precise than A.
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
by
221 277 348
I dont think the given answer key is going to change. There is no confusion regarding the solution to the problem, but the following two highlighted words are surely confusing. They could have at least used “precisely” than “correctly” and highlighted the “collectively” part.
collectively used to correctly aggregate
@Arjun sir,
Please give the above article a read. It is not the wording of the question.
We cannot use IP addresses not assigned to us for aggregation as it would cause packets to drop. If everyone started doing this, the internet would become extremely slow.
Also sir, if you could give some idea regarding the fate of question 39(minimum spanning tree), it would relieve our anxieties a bit.
Thank you
So you think b,d has to be correct right ?, I strongly feel that it has to be they won't change the keys and especially gives MTA, I dont know why there is so much buzz about this, Also in the document they donot followed the rules of aggregation, I mean why they are being given as the rules if we donot have to follow them. Also option a cannot be correct ever because it is one of the host id's it can never be the subnet id if it will be .160 at last instead of 164, then It will be correct, because then we can have .160 as one of the subnet id, we cannot make .164 in last part by only /20. But idk why they complicated it so much..It is a simple aggregation based question just given in table format and some mixed language, according to the rules b and d will be correct and also I dont see them changing the keys, Also your link says the same about it.
Also the mst question hate to say it, I am gonna lose marks but they won't change it..
@aspirant22
@N.S. Same here.
12.20.164.0/22 → 12.20.10100100.00000000
12.20.170.0/23 → 12.20.10101010.00000000
12.20.168.0/23 → 12.20.10101000.00000000
12.20.166.0/23 → 12.20.10100110.00000000
12.20.170.0 and 12.20.168.0 can be combined to 12.20.10101000.0 (12.20.168.0/22).
12.20.166.0/23 comes inside 12.20.164.0/22. So, 12.20.164.0/22 can be used for route aggregation of both.
12.20.168.0/22 and 12.20.164.0/22 can’t be combined further.
So, answer is option B and D.
Question is asking which prefixes can be collectively used to correctly aggregate all of the subnets in the routing table. Not which options independently does it.
Also, Option A contains IP address 12.20.175.1 which doesn’t belong to the specified organization.
A is the answer by combining the last common bits by the principle of supernetting.
by
3 5 8
### 1 comment
It is not, a can never be correct, How can you make .164 in 3 octet by just /20. It is one of the host id's it can never be a subnet id, Also the routing table stores the subnet or network id's.
You are blindly applying the rules, but it is not the case you have to think carefully in this question..
i think this is right .
by
5 12 34 | 2,406 | 7,764 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-27 | latest | en | 0.733586 |
https://docs.microsoft.com/en-us/dotnet/api/system.windows.rect.op_equality?view=netframework-4.8 | 1,596,671,188,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735989.10/warc/CC-MAIN-20200805212258-20200806002258-00113.warc.gz | 285,324,290 | 8,758 | Rect.Equality(Rect, Rect) Operator
Definition
Compares two rectangles for exact equality.
``````public:
static bool operator ==(System::Windows::Rect rect1, System::Windows::Rect rect2);``````
``public static bool operator == (System.Windows.Rect rect1, System.Windows.Rect rect2);``
``static member ( = ) : System.Windows.Rect * System.Windows.Rect -> bool``
``Public Shared Operator == (rect1 As Rect, rect2 As Rect) As Boolean``
Parameters
rect1
Rect
The first rectangle to compare.
rect2
Rect
The second rectangle to compare.
Returns
Boolean
`true` if the rectangles have the same Location and Size values; otherwise, `false`.
Examples
The following example shows how to use the Equality operator to determine if two rectangles are exactly equal.
``````private Boolean overloadedEqualityOperatorExample()
{
// Initialize new rectangle.
Rect myRectangle = new Rect();
// The Location property specifies the coordinates of the upper left-hand
// corner of the rectangle.
myRectangle.Location = new Point(10, 5);
// Set the Size property of the rectangle with a width of 200
// and a height of 50.
myRectangle.Size = new Size(200, 50);
// Create second rectangle to compare to the first.
Rect myRectangle2 = new Rect();
myRectangle2.Location = new Point(0, 0);
myRectangle2.Size = new Size(200, 50);
// Check if the two Rects are exactly equal using the overloaded equality operator.
// areEqual is false because although the size of each rectangle is the same,
// the locations are different.
bool areEqual = (myRectangle == myRectangle2);
// Returns false.
return areEqual;
}
``````
``````Private Function overloadedEqualityOperatorExample() As Boolean
' Initialize new rectangle.
Dim myRectangle As New Rect()
' The Location property specifies the coordinates of the upper left-hand
' corner of the rectangle.
myRectangle.Location = New Point(10, 5)
' Set the Size property of the rectangle with a width of 200
' and a height of 50.
myRectangle.Size = New Size(200, 50)
' Create second rectangle to compare to the first.
Dim myRectangle2 As New Rect()
myRectangle2.Location = New Point(0, 0)
myRectangle2.Size = New Size(200, 50)
' Check if the two Rects are exactly equal using the overloaded equality operator.
' areEqual is false because although the size of each rectangle is the same,
' the locations are different.
Dim areEqual As Boolean = (myRectangle = myRectangle2)
' Returns false.
Return areEqual
End Function
``````
Remarks
This operation tests for object equality.
In this comparison, two instances of Double.NaN are considered equal.
Note
A rectangle's position and dimensions are described by Double values. Because Double values can lose precision when operated upon, a comparison between two values that are logically equal might fail.
The equivalent method for this operator is Rect.Equals(Rect, Rect) | 641 | 2,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-34 | latest | en | 0.709592 |
http://nlab.mathforge.org/nlab/show/Jacobian | 1,368,966,362,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2013-20/segments/1368697503739/warc/CC-MAIN-20130516094503-00037-ip-10-60-113-184.ec2.internal.warc.gz | 187,497,311 | 8,253 | # nLab Jacobian
### Context
#### Differential geometry
differential geometry
synthetic differential geometry
# Contents
For Jacobian in the sense of Jacobian variety (of an algebraic curve), see there (also more general intermediate Jacobians).
## Definition
If $f:{ℝ}^{n}\to {ℝ}^{m}$ is a ${C}^{1}$-differentiable map, between Cartesian spaces, its Jacobian matrix is the $\left(m×n\right)$ matrix
$J\left(f\right)\in {\mathrm{Mat}}_{m×n}\left({C}^{0}\left(ℝ,ℝ\right)\right)$J(f) \in Mat_{m \times n}(C^0(\mathbb{R}, \mathbb{R}))
$J\left(f{\right)}_{j}^{i}:=\frac{\partial {f}^{i}}{\partial {x}^{j}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}i=1,\dots ,m;j=1,\dots ,n,$J(f)^i_j := \frac{\partial f^i}{\partial x^j},\,\,\,\,\,\,\,i=1,\ldots,m; j = 1,\ldots,n,
where $x=\left({x}^{1},\dots ,{x}^{n}\right)$. Here the convention is that the upper index is a row index and the lower index is the column index; in particular ${R}^{n}$ is the space of real column vectors of length $n$.
In more general situation, if $f=\left({f}^{1}\left(x\right),\dots ,{f}^{m}\left(x\right)\right)$ is differentiable at a point $x$ (and possibly defined only in a neighborhood of $x$), we define the Jacobian ${J}_{p}f$ of map $f$ at point $x$ as a matrix with real values $\left({J}_{p}f{\right)}_{j}^{i}=\frac{\partial {f}^{i}}{\partial {x}^{j}}{\mid }_{x}$.
That is, the Jacobian is the matrix which describes the total derivative?.
If $n=m$ the Jacobian matrix is a square matrix, hence its determinant $\mathrm{det}\left(J\left(f\right)\right)$ is defined and called the Jacobian of $f$ (possibly only at a point). Sometimes one refers to Jacobian matrix rather ambigously by Jacobian as well.
## Properties
The chain rule may be phrased by saying that the Jacobian matrix of the composition ${R}^{n}\stackrel{f}{\to }{R}^{m}\stackrel{g}{\to }{R}^{r}$ is the matrix product of the Jacobian matrices of $g$ and of $f$ (at appropriate points).
If $g:M\to N$ is a ${C}^{1}$-map of ${C}^{1}$-manifolds, then the tangent map $Tg:TM\to TN$ defined point by point abstractly by $\left({T}_{p}g\right)\left({X}_{p}\right)\left(f\right)={X}_{p}\left(f\circ g\right)$, for $p\in M$, can in local coordinates be represented by a Jacobian matrix. Namely, if $\left(U,\varphi \right)\ni p$ and $\left(V,\psi \right)\ni g\left(p\right)$ are charts and ${X}_{p}=\sum {X}^{i}\frac{\partial }{\partial {x}^{i}}{\mid }_{p}$ (i.e. ${X}_{p}\left(f\right)={\sum }_{i}{X}_{p}^{i}\frac{\partial \left(f\circ {\varphi }^{-1}\right)}{\partial {x}^{i}}{\mid }_{\varphi \left(p\right)}$ for all germs $f\in {ℱ}_{p}$), then
$\left({T}_{p}g\right)\left({X}_{p}\right)=\sum _{i,j}{J}_{p}\left(\psi \circ g\circ {\varphi }^{-1}{\right)}_{i}^{j}{X}_{p}^{i}\frac{\partial }{\partial {y}^{j}}{\mid }_{g\left(p\right)}$(T_p g)(X_p) = \sum_{i,j} J_p(\psi \circ g\circ\phi^{-1})_i^j X^i_p \frac{\partial}{\partial y^j}|_{g(p)}
Revised on October 15, 2012 22:01:11 by Zoran Škoda (161.53.130.104) | 1,135 | 3,165 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 34, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2013-20 | latest | en | 0.726038 |
http://blog.leanote.com/post/icontofig/1fa992ed3790 | 1,623,574,184,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487607143.30/warc/CC-MAIN-20210613071347-20210613101347-00593.warc.gz | 8,182,099 | 6,665 | 二分图最大匹配KM算法之BFS版本模板
```#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 405;
const int INF = 0x3f3f3f3f
ll n, a[maxn],b[maxn],c[maxn],p[maxn];
ll w[maxn][maxn];
ll lx[maxn] , ly[maxn];
ll slack[maxn];
bool visy[maxn];
ll pre[maxn];
void bfs( ll k ){
ll x , y = 0 , yy = 0 , delta;
memset( pre , 0 , sizeof(pre) );
for(int i = 1 ; i <= n ; i++ ) slack[i] = INF;
while(1){
x = linker[y]; delta = INF; visy[y] = true;
for(int i = 1 ; i <= n ;i++ ){
if( !visy[i] ){
if( slack[i] > lx[x] + ly[i] - w[x][i] ){
slack[i] = lx[x] + ly[i] - w[x][i];
pre[i] = y;
}
if( slack[i] < delta ) delta = slack[i] , yy = i ;
}
}
for(int i = 0 ; i <= n ; i++ ){
if( visy[i] ) lx[linker[i]] -= delta , ly[i] += delta;
else slack[i] -= delta;
}
y = yy ;
if( linker[y] == -1 ) break;
}
while( y ) linker[y] = linker[pre[y]] , y = pre[y];
}
ll KM(){
memset( lx , 0 ,sizeof(lx) );
memset( ly , 0 ,sizeof(ly) );
for(int i = 1 ; i <= n ; i++ ){
memset( visy , false , sizeof(visy) );
bfs(i);
}
ll res = 0 ;
for(int i = 1 ; i <= n ; i++ ){
if( linker[i] != -1 ){
res += w[linker[i]][i] ;
}
}
return res;
}``` | 446 | 1,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-25 | latest | en | 0.145945 |
https://risingentropy.com/2018/11/ | 1,618,854,232,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038916163.70/warc/CC-MAIN-20210419173508-20210419203508-00550.warc.gz | 581,764,732 | 65,791 | # Fractals and Epicycles
There is no bilaterally-symmetrical, nor eccentrically-periodic curve used in any branch of astrophysics or observational astronomy which could not be smoothly plotted as the resultant motion of a point turning within a constellation of epicycles, finite in number, revolving around a fixed deferent.
Norwood Russell Hanson, “The Mathematical Power of Epicyclical Astronomy”
A friend recently showed me this image…
…and thus I was drawn into the world of epicycles and fractals.
Epicycles were first used by the Greeks to reconcile observational data of the motions of the planets with the theory that all bodies orbit the Earth in perfect circles. It was found that epicycles allowed astronomers to retain their belief in perfectly circular orbits, as well as the centrality of Earth. The cost of this, however, was a system with many adjustable parameters (as many parameters as there were epicycles).
There’s a somewhat common trope about adding on endless epicycles to a theory, the idea being that by being overly flexible and accommodating of data you lose epistemic credibility. This happens to fit perfectly with my most recent posts on model selection and overfitting! The epicycle view of the solar system is one that is able to explain virtually any observational data. (There’s a pretty cool reason for this that has to do with the properties of Fourier series, but I won’t go into it.) The cost of this is a massive model with many parameters. The heliocentric model of the solar system, coupled with the Newtonian theory of gravity, turns out to be able to match all the same data with far fewer adjustable parameters. So by all of the model selection criteria we went over, it makes sense to switch over from one to the other.
Of course, it is not the case that we should have been able to tell a priori that an epicycle model of the planets’ motions was a bad idea. “Every planet orbits Earth on at most one epicycle”, for instance, is a perfectly reasonable scientific hypothesis… it just so happened that it didn’t fit the data. And adding epicycles to improve the fit to data is also not bad scientific practice, so long as you aren’t ignoring other equally good models with fewer parameters.)
Okay, enough blabbing. On to the pretty pictures! I was fascinated by the Hilbert curve drawn above, so I decided to write up a program of my own that generates custom fractal images from epicycles. Here are some gifs I created for your enjoyment:
## Negative doubling of angular velocity
(Each arm rotates in the opposite direction of the previous arm, and at twice its angular velocity. The length of each arm is half that of the previous.)
## Negative trebling
Here’s a still frame of the final product for N = 20 epicycles:
## ωn ~ (n+1) 2n
(or, the Fractal Frog)
## ωn ~ 2n, rn ~ 1/n2
And here’s a still frame of N = 20:
(All animations were built using Processing.py, which I highly recommend for quick and easy construction of visualizations.)
# How to Learn From Data, Part 2: Evaluating Models
I ended the last post by saying that the solution to the problem of overfitting all relied on the concept of a model. So what is a model? Quite simply, a model is just a set of functions, each of which is a candidate for the true distribution that generates the data.
Why use a model? If we think about a model as just a set of functions, this seems kinda abstract and strange. But I think it can be made intuitive, and that in fact we reason in terms of models all the time. Think about how physicists formulate their theories. Laws of physics have two parts: First they specify the types of functional relationships there are in the world. And second, they specify the value of particular parameters in those functions. This is exactly how a model works!
Take Newton’s theory of gravity. The fundamental claim of this theory is that there is a certain functional relationship between the quantities a (the acceleration of an object), M (the mass of a nearby object), and r (the distance between the two objects): a ~ M/r2. To make this relationship precise, we need to include some constant parameter G that says exactly what this proportionality is: a = GM/r2.
So we start off with a huge set of probability distributions over observations of the accelerations of particles (one for each value of G), and then we gather data to see which value of G is most likely to be right. Einstein’s theory of gravity was another different model, specifying a different functional relationship between a, M, and r and involving its own parameters. So to compare Einstein’s theory of gravity to Newton’s is to compare different models of the phenomenon of gravitation.
When you start to think about it, the concept of a model is an enormous one. Models are ubiquitous, you can’t escape from them. Many fancy methods in machine learning are really just glorified model selection. For instance, a neural network is a model: the architecture of the network specifies a particular functional relationship between the input and the output, and the strengths of connections between the different neurons are the parameters of the model.
So. What are some methods for assessing predictive accuracy of models? The first one we’ll talk about is a super beautiful and clever technique called…
## Cross Validation
The fundamental idea of cross validation is that you can approximate how well a model will do on the next data point by evaluating how the model would have done at predicting a subset of your data, if all it had access to was the rest of your data.
I’ll illustrate this with a simple example. Suppose you have a data set of four points, each of which is a tuple of two real numbers. Your two models are T1: that the relationship is linear, and T2: that the relationship is quadratic.
What we can do is go through each data point, selecting each in order, train the model on the non-selected data points, and then evaluate how well this trained model fits the selected data point. (By training the model, I mean something like “finding the curve within the model that best fits the non-selected data points.”) Here’s a sketch of what this looks like:
There are a bunch of different ways of doing cross validation. We just left out one data point at a time, but we could have left out two points at a time, or three, or any k less than the total number of points N. This is called leave-k-out cross validation. If we partition our data by choosing a fraction of it for testing, then we get what’s called n-fold cross validation (where 1/n is the fraction of the data that is isolated for testing).
We also have some other choices besides how to partition the data. For instance, we can choose to train our model via a Likelihoodist procedure or a Bayesian procedure. And we can also choose to test our model in various ways, by using different metrics for evaluating the distance between the testing set and the trained model. In leave-one-out cross validation (LOOCV), a pretty popular method, both the training and testing procedures are Likelihoodist.
Now, there’s a practical problem with cross-validation, which is that it can take a long time to compute. If we do LOOCV with N data points and look at all ways of leaving out one data point, then we end up doing N optimization procedures (one for each time you train your model), each of which can be pretty slow. But putting aside practical issues, for thinking about theoretical rationality, this is a super beautiful technique.
Next we’ll move on to…
## Bayesian Model Selection!
We had Bayesian procedures for evaluating individual functions. Now we can go full Bayes and apply it to models! For each model, we assess the posterior probability of the model given the data using Bayes’ rule as usual:
$Pr(M | D) = \frac{Pr(D | M)}{Pr(D)} Pr(M)$
Now… what is the prior Pr(M)? Again, it’s unspecified. Maybe there’s a good prior that solves overfitting, but it’s not immediately obvious what exactly it would be.
But there’s something really cool here. In this equation, there’s a solution to overfitting that pops up not out of the prior, but out of the LIKELIHOOD! I explain this here. The general idea is that models that are prone to overfitting get that way because their space of functions is very large. But if the space of functions is large, then the average prior on each function in the model must be small. So larger models have, on average, smaller values of the term Pr(f | M), and correspondingly (via $Pr(D | M) = \sum_{f \in M} Pr(D | f, M) Pr(f | M)$) get a weaker update from evidence.
So even without saying anything about the prior, we can see that Bayesian model selection provides a potential antidote to overfitting. But just like before, we have the practical problem that computing Pr(M | D) is in general very hard. Usually evaluating Pr(D | M) involves calculating a really complicated many-dimensional integral, and calculating many-dimensional integrals can be very computationally expensive.
Might we be able to find some simple unbiased estimator of the posterior Pr(M | D)? It turns out that yes, we can. Take another look at the equation above.
$Pr(M | D) = \frac{Pr(D | M)}{Pr(D)} Pr(M)$
Since $Pr(D)$ is a constant across models, we can ignore it when comparing models. So for our purposes, we can attempt to maximize the equation:
$Pr(D | M) Pr(M) = \sum_{f \in M} Pr(D | f, M) Pr(f | M) Pr(M)$
If we assume that P(D | f) is twice differentiable in the model parameters and that Pr(f) behaves roughly linearly around the maximum likelihood function, we can approximate this as:
$Pr(D | M) Pr(M) \approx \frac {Pr(D | f^*(D))} {\sqrt{N}}^k Pr(M)$
f* is the maximum likelihood function within the model, and k is the the number of parameters of the model (e.g. k = 2 for a linear model and k = 3 for a quadratic model).
\We can make this a little neater by taking a logarithm and defining a new quantity: the Bayesian Information Criterion.
$argmax_M [ Pr(M | D) ] \\~\\ = argmax_M [ Pr(D | M) Pr(M) ] \\~\\ \approx argmax_M [ \frac {Pr(D | f^*(D))} {\sqrt{N}}^k Pr(M)] \\~\\ = argmax_M [ \log(Pr(M)) + \log (Pr(D | f^*(D)) - \frac{k}{2} \log(N) ] \\~\\ = argmin_M [ BIC - \log Pr(M) ]$
$BIC = \frac{k}{2} log(N) - \log \left( Pr(D | f^*(D)) \right)$
Thus, to maximize the posterior probability of a model M is roughly the same as to minimize the quantity BIC – log Pr(M).
If we assume that our prior over models is constant (that is, that all models are equally probable at the outset), then we just minimize BIC.
## Bayesian Information Criterion
Notice how incredibly simple this expression is to evaluate! If all we’re doing is minimizing BIC, then we only need to find the maximum likelihood function f*(D), assess the value of Pr(D | f*), and then penalize this quantity with the factor k/2 log(N)! This penalty scales in proportion to the model complexity, and thus helps us avoid overfitting.
We can think about this as a way to make precise the claims above about Bayesian Model Selection penalizing overfitting in the likelihood. Remember that minimizing BIC made sense when we assumed a uniform prior over models (and therefore when our prior doesn’t penalize overfitting). So even when we don’t penalize overfitting in the prior, we still end up getting a penalty! This penalty must come from the likelihood.
Some more interesting facts about BIC:
• It is approximately equal to the minimum description length criterion (which I haven’t discussed here)
• It is only valid when N >> k. So it’s not good for small data sets or large models.
• It the truth is contained within your set of models, then BIC will select the truth with probability 1 as N goes to infinity.
Okay, stepping back. We sort of have two distinct clusters of approaches to model selection. There’s Bayesian Model Selection and the Bayesian Information Criterion, and then there’s Cross Validation. Both sound really nice. But how do they compare? It turns out that in general they give qualitatively different answers. And in general, the answers you get using cross validation tend to be more predictively accurate that those that you get from BIC/BMS.
A natural question to ask at this point is: BIC was a nice simple approximation to BMS. Is there a corresponding nice simple approximation to cross validation? Well first we must ask: which cross validation? Remember that there was a plethora of different forms of cross validation, each corresponding to slightly different criterion for evaluating fits. We can’t assume that all these methods give the same answer.
Let’s choose leave-one-out cross validation. It turns out that yes, there is a nice approximation that is asymptotically equivalent to LOOCV! This is called the Akaike information criterion.
## Akaike Information Criterion
First, let’s define AIC:
$AIC = k - \log \left( Pr(D | f^*(D)) \right)$
Like always, k is the number of parameters in M and f* is chosen by the Likelihoodist approach described in the last post.
What you get by minimizing this quantity is asymptotically equivalent to what you get by doing leave-one-out cross validation. Compare this to BIC:
$BIC = \frac{k}{2} \log (N) - \log \left( Pr(D | f^*(D) ) \right)$
There’s a qualitative difference in how the parameter penalty is weighted! BIC is going to have a WAY higher complexity penalty than AIC. This means that AIC should in general choose less simple models than BIC.
Now, we’ve already seen one reason why AIC is good: it’s a super simple approximation of LOOCV and LOOCV is good. But wait, there’s more! AIC can be derived as an unbiased estimator of the Kullback-Leibler divergence DKL.
What is DKL? It’s a measure of the information theoretic distance of a model from truth. For instance, if the true generating distribution for a process is f, and our model of this process is g, then DKL tells us how much information we lose by using g to represent f. Formally, DKL is:
$D_{KL}(f, g) = \int { f(x) \log \left(\frac{f(x)} {g(x)} \right) dx }$
Notice that to actually calculate this quantity, you have to already know what the true distribution f is. So that’s unfeasible. BUT! AIC gives you an unbiased estimate of its value! (The proof of this is complicated, and makes the assumptions that N >> k, and that the model is not far from the truth.)
Thus, AIC gives you an unbiased estimate of which model is closest to the truth. Even if the truth is not actually contained within your set of models, what you’ll end up with is the closest model to it. And correspondingly, you’ll end up getting a theory of the process that is very similar to how the process actually works.
There’s another version of AIC that works well for smaller sample sizes and larger models called AICc.
$AICc = AIC + \frac {k (k + 1)} {N - (k + 1)}$
And interestingly, AIC can be derived as an approximation to Bayesian Model Selection with a particular prior over models. (Remember earlier when we showed that $argmax_M \left( Pr(M | D) \right) \approx argmin_M \left( BIC - \log(Pr(M)) \right)$? Well, just set $\log Pr(M) = BIC - AIC = k \left( \frac{1}{2} \log N - 1 \right)$, and you get the desired result.) Interestingly, this prior ends up rewarding theories for having lots of parameters! This looks pretty bad… it seems like AIC is what you get when you take Bayesian Model Selection and then try to choose a prior that favors overfitting theories. But the practical use and theoretical virtues of AIC warrant, in my opinion, taking a closer look at what’s going on here. Perhaps what’s going on is that the likelihood term Pr(D | M) is actually doing too much to avoid overfitting, so in the end what we need in our prior is one that avoids underfitting! Regardless, we can think of AIC as specifying the unique good prior on models that optimizes for predictive accuracy.
There’s a lot more to be said from here. This is only a short wade into the waters of statistical inference and model selection. But I think this is a good place to stop, and I hope that I’ve given you a sense of the philosophical richness of the various different frameworks for inference I’ve presented here.
# How to Learn From Data, Part I: Evaluating Simple Hypotheses
Here’s a general question of great importance: How should we adjust our model of the world in response to data? This post is a tour of a few of the big ideas that humans have come up with to address this question. I find learning about this topic immensely rewarding, and so I shall attempt to share it! What I love most of all is thinking about how all of this applies to real world examples of inference from data, so I encourage you to constantly apply what I say to situations that you find personally interesting.
First of all, we want a formal description of what we mean by data. Let’s describe our data quite simply as a set: $D = \{(x_1, y_1), (x_2, y_2),..., (x_N, y_N)\}$, where each $x_i$ is a value that you choose (your independent variable) and each $y_i$ is the resulting value that you measure (the dependent variable). Each of these variables can be pretty much anything – real numbers, n-tuples of real numbers, integers, colors, whatever you want. The goal here is to embrace generality, so as to have a framework that applies to many kinds of inference.
Now, suppose that you have a certain theory about the underlying relationship between the x variables and the y variables. This theory might take the form of a simple function: $T: y = f(x)$ We interpret T as making a prediction of a particular value of the dependent variable for each value of the independent variable. Maybe the data is the temperatures of regions at various altitudes, and our theory T says that one over the temperature (1/T) is some particular linear function of the altitude.
What we want is a notion of how good of a theory T is, given our data. Intuitively, we might think about doing this by simply assessing the distance between each data point $y_n$ and the predicted value of y at that point: $f(x_n)$, using some metric, then adding them all up. But there are a whole bunch of distance metrics available to us. Which one should we use? Perhaps the taxicab measure comes to mind ($\sum_{n=1}^N {|y_n - f(x_n)|}$), or the sum of the squares of the differences ($SOS = \sum_{n=1}^N {(y_n - f(x_n))^2}$). We want a good theoretical justification for why any one of these metrics should be preferred over any other, since in general they lead to different conclusions. We’ll see just such justifications shortly. Keep the equation for SOS in mind, as it will turn up repeatedly ahead.
Now here’s a problem: If we want a probabilistic evaluation of T, then we have to face the fact that it makes a deterministic prediction. Our theory seems to predict with 100% probability that at $x_n$ the observed $y_n$ will be precisely $f(x_n)$. If it’s even slightly off of this, then our theory will be probabilistically disconfirmed to zero.
We can solve this problem by modifying our theory T to not just be a theory of the data, but also a theory of error. In other words, we expect that the value we get will not be exactly the predicted value, and give some account of how on average observation should differ from theory.
$T: y = f(x) + \epsilon$, where $\epsilon$ is some random variable drawn from a probability distribution $P_E$.
This error distribution can be whatever we please – Gaussian, exponential, Poisson, whatever. For simplicity let’s say that we know the error is normal (drawn from a Gaussian distribution) with a known standard deviation σ.
$T: y = f(x) + \epsilon, \epsilon \sim N(0, \sigma)$
A note on notation here: $\epsilon \sim N(\mu, \sigma)$ denotes a random variable drawn from a Gaussian distribution centered around $\mu$ with a standard deviation of $\sigma$.
This gives us a sensible notion of the probability of obtaining some value of y from a chosen x given the theory T.
$Pr(y | x, T) \sim N(f(x), \sigma) \\~\\ Pr(y | x, T) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2 \sigma^2} (y - f(x))^2}$
Nice! This is a crucial step towards figuring out how to evaluate theories: we’ve developed a formalism for describing precisely how likely a data point is, given a particular theory (which, remember, so far is just a function from values of independent variables to values of dependent variables, coupled with a theory of how the error in your observations works).
Let’s try to extend this to our entire data set D. We want to assess the probability of the particular values of the dependent variables, given the chosen values of the dependent variables and the function. We’ll call this the probability of D given f.
$Pr(D | f) = Pr(x_1, y_1, x_2, y_2,..., x_N, y_N | T) \\~\\ Pr(D | f) = Pr(y_1, y_2,...,y_N | x_1, x_2,..., x_N, T)$
(It’s okay to move the values of the independent variable x across the conditional bar because of our assumption that we know beforehand what values x will take on.)
But now we run into a problem: we can’t really do anything with this expression without knowing how the data points depend upon each other. We’d like to break it into individual terms, each of which can be evaluated by the above expression for $Pr(y | x, T)$. But that would require an assumption that our data points are independent. In general, we cannot grant this assumption. But what we can do is expand our theory once more to include a theory of the dependencies in our data points. For simplicity of explication, let’s proceed with the assumption that each of our observations is independent of the others. Said another way, we assume that given T, $x_n$ screens off all other variables from $y_n$). Combined with our assumption of normal error, this gives us a nice simple reduction.
$Pr(D | f) = Pr(y_1, y_2,...,y_N | x_1, x_2,..., x_N, T) \\~\\ Pr(D | f) = Pr(y_1 | x_1, T) Pr(y_2 | x_2, T) ... Pr(y_N | x_N, T) \\~\\ Pr(D | f) = \prod_{n=1}^N { \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2 \sigma^2} (y_n - f(x_n))^2} } \\~\\ Pr(D | f) = (2 \pi \sigma^2)^{-N/2} e^{-\frac{1}{2 \sigma^2} \sum_{n=1}^N {(y_n - f(x_n))^2}} \\~\\ Pr(D | f) = (2 \pi \sigma^2)^{-N/2} e^{-\frac{SOS(f, D)}{2 \sigma^2} }$
We see an interesting connection arise here between Pr(D | f) and the sum of squares evaluation of fit. More will be said of this in just a moment.
But first, let’s take a step back. Our ultimate goal here is to find a criterion for theory evaluation. And here we’ve arrived at an expression that looks like it might be right for that role! Perhaps we want to say that our criterion for evaluating theories is just maximization of Pr(D | f). This makes some intuitive sense… a good theory is one which predicts the observed data. If general relativity or quantum mechanics predicted that the sun should orbit the earth, or that atoms should be unstable, then we wouldn’t be very in favor of it.
And so we get the Likelihoodism Thesis!
Likelihoodism: The best theory f* given some data D is that which maximizes Pr(D | f). Formally: $f^*(D) = argmax_f [ Pr(D | f) ]$
Now, since logarithms are monotonic, we can express this in a more familiar form:
$argmax_f [Pr(D | F)] = argmax_f \left[ e^{-\frac{SOS(f, D)}{2 \sigma^2} } \right] = argmax_f \left[-\frac{SOS(f, D)}{2 \sigma^2} \right] = argmin_f [ SOS(f, D) ]$
Thus the best theory according to Likelihoodism is that which minimizes the sum of squares! People minimize sums of squares all the time, and most of the time they don’t realize that it can be derived from the Likelihoodism Thesis and the assumptions of Gaussian error and independent data. If our assumptions had been different, then we would have found a different expression, and SOS might no longer be appropriate! For instance, the taxicab metric arises as the correct metric if we assume exponential error rather than normal error. I encourage you to see why this is for yourself. It’s a fun exercise to see how different assumptions about the data give rise to different metrics for evaluating the distance between theory and observation. In general, you should now be able to take any theory of error and assess what the proper metric for “degree of fit of theory to data” is, assuming that degree of fit is evaluated by maximizing Pr(D | f).
Now, there’s a well-known problem with the Likelihoodism thesis. This is that the function f that minimizes SOS(f, D) for some data D is typically going to be a ridiculously overcomplicated function that perfectly fits each data points, but does terribly on future data. The function will miss the underlying trend in the data for the noise in the observations, and as a result fail to get predictive accuracy.
This is the problem of overfitting. Likelihoodism will always prefer theories that overfit to those that don’t, and as a result will fail to identify underlying patterns in data and do a terrible job at predicting future data.
How do we solve this? We need a replacement for the Likelihoodism thesis. Here’s a suggestion: we might say that the problem stems from the fact that the Likelihoodist procedure recommends us to find a function that makes the data most probable, rather than finding a function that is made most probable by the data. From this suggestion we get the Bayesian thesis:
Bayesianism: The best theory f given some data is that which maximizes Pr(f | D). $f^*(D) = argmax_f [ Pr(f | D) ]$
Now, what is this Pr(f | D) term? It’s an expression that we haven’t seen so far. How do we evaluate it? We simply use the theorem that the thesis is named for: Bayes’ rule!
$Pr(f | D) = \frac{Pr(D | f)}{Pr(D)} Pr(f)$
This famous theorem is a simple deductive consequence of the probability axioms and the definition of conditional probability. And it happens to be exactly what we need here. Notice that the right-hand side consists of the term Pr(D | f), which we already know how to calculate. And since we’re ultimately only interested in varying f to find the function that maximizes this expression, we can ignore the constant term in the denominator.
$argmax_f [ Pr(f | D) ] = argmax_f [ Pr(D | f) Pr(f) ] \\~\\ argmax_f [ Pr(f | D) ] = argmax_f [ \log Pr(D | f) + \log Pr(f) ] \\~\\ argmax_f [ Pr(f | D) ] = argmax_f [ -\frac{SOS(f, D)}{2 \sigma^2} + \log Pr(f) ] \\~\\ argmax_f [ Pr(f | D) ] = argmin_f [ SOS(f, D) - 2 \sigma^2 \log Pr(f) ]$
The last steps we get just by substituting in what we calculated before. The 2σ² in the first term comes from the fact that the exponent of our Gaussian is (y – f(x))² / 2σ². We ignored it before because it was just a constant, but since we now have another term in the expression being maximized, we have to keep track of it again.
Notice that what we get is just what we had initially (a sum of squares) plus an additional term involving a mysterious Pr(f). What is this Pr(f)? It’s our prior distribution over theories. Because of the negative sign in front of the second term, a larger value of Pr(f) gives a smaller value for the expression that we are minimizing. Similarly, the more closely our function follows the data, the smaller the SOS term becomes. So what we get is a balance between fitting the data and having a high prior. A theory that fits the data perfectly can still end up with a bad evaluation, as long as it has a low enough prior. And a theory that fits the data poorly can end up with a great evaluation if it has a high enough prior.
(Those that are familiar with machine learning might notice that this feels similar to regularization. They’re right! It turns out that different regularization techniques just end up corresponding to different Bayesian priors! L2 regularization corresponds to a Gaussian prior over parameters, and L1 regularization corresponds to an exponential prior over parameters. And so on. Different prior assumptions about the process generating the data lead naturally to different regularization techniques.)
What this means is that if we are trying to solve overfitting, then the Bayesianism thesis provides us a hopeful ray of light. Maybe we can find some perfect prior Pr(f) that penalizes complex overfitting theories just enough to give us the best prediction. But unsurprisingly, finding such a prior is no easy matter. And importantly, the Bayesian thesis as I’ve described it gives us no explicit prescription for what this prior should be, making it insufficient for a full account of inference. What Bayesianism does is open up a space of possible methods of theory evaluation, inside which (we hope) there might be a solution to our problem.
Okay, let’s take another big step back. How do we progress from here? Well, let’s think for a moment about what our ultimate goal is. We want to say that overfitting is bad, and that therefore some priors are better than others insofar as they prevent overfitting. But what is the standard we’re using to determine that overfitting is bad? What’s wrong with an overfitting theory?
Here’s a plausible answer: The problem with overfitting is that while it maximizes descriptive accuracy, it leads to poor predictive accuracy! I.e., theories that overfit do a great job at describing past data, but tend to do a poor job of matching future data.
Let’s take this idea and run with it. Perhaps all that we truly care about in theory selection is predictive accuracy. I’ll give a name to this thesis:
Predictivism: The ultimate standard for theory evaluation is predictive accuracy. I.e., the best theory is that which does the best at predicting future data.
Predictivism is a different type of thesis than Bayesianism and Likelihoodism. While each of those gave precise prescriptions for how to calculate the best theory, Predictivism does not. If we want an explicit algorithm for computing the best theory, then we need to say what exactly “predictive accuracy” means. This means that as far as we know so far, Predictivism could be equivalent to Likelihoodism or to Bayesianism. It all depends on what method we end up using to determine what theory has the greatest predictive accuracy. Of course, we have good reasons to suspect that Likelihoodism is not identical to Predictivism (Likelihoodism overfits, and we suspect that Predictivism will not) or to Bayesianism (Bayesianism does not prescribe a particular prior, so it gives no unique prescription for the best theory). But what exactly Predictivism does say depends greatly on how exactly we formalize the notion of predictive accuracy.
Another difference between Predictivism and Likelihoodism/Bayesianism is that predictive accuracy is about future data. While it’s in principle possible to find an analytic solution to P(D | f) or P(f | D), Predictivism seems to require us to compute terms involving future data! But these are impossible to specify, because we don’t know what the future data will be!
Can we somehow approximate what the best theory is according to Predictivism, by taking our best guess at what the future data will be? Maybe we can try to take the expected value of something like $Pr(y_{N+1} | x_{N+1}, T, D)$, where $(x_{N+1}, y_{N+1})$ is a future data point. But there are two big problems with this.
First, in taking an expected value, you must use the distribution of outcomes given by your own theory. But then your evaluation picks up an inevitable bias! You can’t rely on your own theories to assess how good your theories are.
And second, by the assumptions we’ve included in our theories so far, the future data will be independent of the past data! This is a really big problem. We want to use our past data to give some sense of how well we’ll do on future data. But if our data is independent, then our assessment of how well T will do on the next data point will have to be independent of the past data as well! After all, no matter what value $y_{N+1}$ has, $Pr(y_{N+1} | x_{N+1}, T, D) = Pr(y_{N+1} | x_{N+1}, T).$
So is there any way out of this? Surprisingly, yes! If we get creative, we can find ways to approximate this probability with some minimal assumptions about the data. These methods all end up relying on a new concept that we haven’t yet discussed: the concept of a model. The next post will go into more detail on how this works. Stay tuned!
# The Match Problem
This puzzle has been popping up on my Facebook feed recently. Try to see how high you can get before reading on!
(…)
Now, would you be surprised if I told you that with a little interpretive freedom and creativity, you can get numbers larger than the number of atoms in the observable universe? How about if I told you that you can get to numbers large enough that they break set theory? Let me demonstrate for you.
You can move the bottom match in the 5 up to make it a 9. Then you can rotate the bottom left vertical match in the 0 to make it a nine. This gives 998.
Can we do better? Sure! Take the top and bottom matches in the central zero and move them to squeeze in another one, giving you 51,118.
So far we’ve been working purely in base 10. Let’s try to do better by moving to a more exotic number system.
Hexadecimal is a base-16 number system, the digits of which are written as 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F. For instance, we can construct F88:
F88 is only 3976, which is not better than our best so far (51,118). If we’re allowed to interpret a square of matches as a D instead of a zero, we can do slightly better (though to do so we have to be allowed to throw out a toothpick, or place it directly on top of another one):
FDB is 4059, which is still not better than our current best. To get better, we need to get more clever.
# Exponentiation!
Our first strategy is just to shift two matches in the first digit down so as to make it into a 9:
We can interpret this as 98 = 43,046,721. This is our best so far! But we can do better by applying Knuth’s up-arrow notation.
This is 518, which is almost 3.8 trillion, 100 thousand times better than 98! But we’re not done yet! If we use a caret (^) to represent exponentiation, we can get even higher!
5118 is 5.4 nonillion, a number with 31 decimal digits long. We could try to do better than this by squeezing in a caret between the 5 and the 1, making 5118 (a number with 83 decimal digits) but this becomes pretty messy and gross.
Alright, so we got up to 83 digit long numbers. Can we get any better? Yep!
# Tetration
Tetration is the level above exponentiation. The nth tetration of a is defined as follows:
Just as multiplication is repeated addition and exponentiation is repeated multiplication, tetration is repeated exponentiation! Now we get into numbers whose size is truly impossible to grasp. Let’s shift the top and bottom matches on the middle 0:
We can write this equivalently as:
How big is this number? There’s really no meaningful way to describe it. We’ve gone far beyond any quantities that can be made physical sense of, like the number of cubic nanometers in the universe, which is a measly 10107. But we’re not done yet!
# Busy Beavers!
The busy beaver numbers are a sequence of numbers that arise from the properties of Turing machines. Here’s a brief description: The nth busy beaver number B(N) is the greatest number of ones that a finitely-running N state Turing machine can print on a tape which starts all zero.
Did you think tetration is big? Well, busy beaver numbers are unimaginably larger. In fact, the busy beaver sequence grows larger than any computable function! There’s a neat proof of this that involves the uncomputability of the halting problem. To skim over it, it can be shown that were we to have an upper bound on the busy beaver sequence, then we could find a deterministic algorithm for solving the halting problem in a finite amount of time, which we know is impossible. And if any computable function F(N) grew faster than B(N), then we could find an upper bound on the busy beaver sequence. Thus, it must be the case that no computable function grows as fast as B(N)!
We can exploit the absurd growth rate of the busy beaver sequence if we are allowed the interpretative freedom of assuming parentheses, so that Bn = B(n).
Let’s think for a minute about how large B(118) must be. So far, the only values of n for which B(n) is known are B(1), B(2), B(3), and B(4). After this the values grow out of control. A lower bound on B(6) is . For B(7), we have as a lower bound . B(8) almost certainly beats our current record. And B(118) is unthinkably large.
We can get even higher than the busy beaver numbers with the maximum shifts function S(N), defined as the number of steps that the longest-finitely-running N state Turing machine takes before halting. This function is guaranteed to be larger than B(N) for all N. Using S(N), and taking the same liberties as above with respect to parentheses, we can get an insanely high value:
This is S(1118), and while it’s undoubtedly larger than B(118), there’s no way to get any intuitive grasp on how much larger. But wait, there’s more! We can get even larger by recursively nesting a busy beaver function within a maximum shifts function:
We interpret this as S(B(9)). Why is this larger than S(1118)? Well, B(9) is some enormous number, certainly larger than 1118, so S(B(9)) is certainly greater than S(1118).
Now, are we finally done? Have we reached the peak yet? No! It’s time for the largest solution of them all.
The reason that the Busy Beaver numbers and Maximum Shift function are so big is because of the uncomputability of the halting problem. But if we consider Turing machines that have an oracle for the halting problem (call these meta Turing machines), we get a new meta-halting problem: when do these meta Turing machines halt? From the meta-halting problem comes an associated new sequence of Busy Beaver numbers, which grows uncomputable faster than the original Busy Beaver sequence. Then we can equip Turing machines with an oracle for the meta-halting problem, generating a meta-meta-Busy Beaver sequence.
Thus we get a hierarchy of Busy Beaver functions, which, following the notation used by Scott Aaronson here, can be described with Bn(x). Each Bn grows uncomputably faster than the previous Bn-1. There’s a similar hierarchy for the maximum shifts function, and each S_n is going to be an upper bound on each Sn-1.
So we can exploit this hierarchy to create an unimaginably large number (whose actual value is almost certainly independent of the axioms of set theory): Move around the top and bottom matches on the 0 to give S a subscript of 11. Then we get the 11th-up-in-the-hierarchy maximum shifts function S11 applied to 118: S11(118).
It’s a little gross-looking, but I think it works! I challenge anybody to try to come up with a better solution. 🙂
# Gödel’s Second Incompleteness Theorem: Explained in Words of Only One Syllable
Somebody recently referred me to a 1994 paper by George Boolos in which he writes out a description of Gödel’s Second Incompleteness Theorem, using only words of one syllable. I love it so much that I’m going to copy the whole thing here in this post. Enjoy!
First of all, when I say “proved”, what I will mean is “proved with the aid of the whole of math”. Now then: two plus two is four, as you well know. And, of course, it can be proved that two plus two is four (proved, that is, with the aid of the whole of math, as I said, though in the case of two plus two, of course we do not need the whole of math to prove that it is four). And, as may not be quite so clear, it can be proved that it can be proved that two plus two is four, as well. And it can be proved that it can be proved that it can be proved that two plus two is four. And so on. In fact, if a claim can be proved, then it can be proved that the claim can be proved. And that too can be proved.
Now, two plus two is not five. And it can be proved that two plus two is not five. And it can be proved that it can be proved that two plus two is not five, and so on.
Thus: it can be proved that two plus two is not five. Can it be proved as well that two plus two is five? It would be a real blow to math, to say the least, if it could. If it could be proved that two plus two is five, then it could be proved that five is not five, and then there would be no claim that could not be proved, and math would be a lot of bunk.
So, we now want to ask, can it be proved that it can’t be proved that two plus two is five? Here’s the shock: no, it can’t. Or, to hedge a bit: if it can be proved that it can’t be proved that two plus two is five, then it can be proved as well that two plus two is five, and math is a lot of bunk. In fact, if math is not a lot of bunk, then no claim of the form “claim X can’t be proved” can be proved.
So, if math is not a lot of bunk, then, though it can’t be proved that two plus two is five, it can’t be proved that it can’t be proved that two plus two is five.
By the way, in case you’d like to know: yes, it can be proved that if it can be proved that it can’t be proved that two plus two is five, then it can be proved that two plus two is five.
George Boolos, Mind, Vol. 103, January 1994, pp. 1 – 3 | 9,810 | 41,166 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 94, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-17 | longest | en | 0.95725 |
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### What is Proposition and Logical Deduction in elitmus?
Proposition is one of the important part in Elitmus pH test, in verbal section contains more than three questions are asked from proposition part. Proposition is easiest one in verbal section. In this post you find more details about Proposition and logical deduction questions and important rules for handling those questions.
Proposition
In Logic, any categorical statement is termed as the Proposition.
A Proposition is a statement that asserts that either a part of, or the whole of, one set of objects - the set identified by the subject term in the sentence expressing that statement - either is included in, or is excluded from, another set - the set identified by the predicate term in that sentence.
The standard form of a proposition is:
Quantifier + Subject + Copula + Predicate
Thus, the proposition consists of four parts:
1) Quantifier: The words 'all', 'no' and 'some' are called quantifiers because they specify a quantity 'All' and 'no' are universal quantifiers because they refer to every object in a certain set, while the quantifier 'some' is a particular quantifier because it refers to at least one existing object in a certain set.
2) Subject (denoted by 'S'): The subject is that about which something is said.
3) Predicate (denoted by 'P'): The predicate is the part of the proposition denoting that which is affirmed or denied about the subject.
4) Copula: The copula is that part of the proposition which denotes the relation between the subject and the predicate.
Four-Fold Classification of Propositions:
A proposition is said to have a universal quantity if it begins with a universal quantifier and a particular quantity if it begins with a particular quantifier. Besides, propositions which assert something about the inclusion of the whole or a part of one set in the other are said to have affirmative quality, while those which deny the inclusion of the whole or a part of one set in the other are said to have a negative quality. Also, a term is distributed in a proposition if it refers to all members of the set of objects denoted by that term. Otherwise, it is said to be undistributed. Based on the above facts, propositions can be classified into four types:
1) Universal Affirmative Proposition (denoted by A): It distributes only the subject i.e. the predicate is not interchangeable with the subject while maintaining the validity of the proposition.
2) Universal Negative Proposition (denoted by E): It distributes both the subject and the predicate i.e. an entire class of predicate term is denied to the entire class of the subject term, as in the proposition.
3) Particular Affirmative Proposition (denoted by I): It distributes neither the subject nor the predicate.
4) Particular Negative Proposition (denoted by O): It distributes only the predicate. e.g., Some animals are not wild. Here, the subject term 'animals' is used only for a part of its class and hence is undistributed while the predicate term 'wild' is denied in entirety to the subject term and hence is distributed. These facts can be summarized as follows:
Statement Form Quantity Quality Distributed (A): All S is P. Universal Affirmative S only (E): No S is P. Universal Negative Both S and P (I): Some S is P. Particular Affirmative Neither S nor P (O): Some S is not P Particular Negative P only
Logical Deduction:
The phenomenon of deriving a conclusion from a single proposition or a set of given propositions, is known as logical deduction. The given propositions are also referred to as the premises.
Two Inferential Processes of Deduction:
I. Immediate Deductive Inference:
Here, conclusion is deduced from one of the given propositions, by any of the three ways -conversion, obversion and contraposition.
1) Conversion: The Conversion proceeds with interchanging the subject term and the predicate term i.e. the subject term of the premise becomes the predicate term of the conclusion and the predicate term of the premise becomes the subject of the conclusion.
The given proposition is called converted, whereas the conclusion drawn from it is called its converse.
Table of Valid Conversions
Converted Converse A: All S is P Ex. All pins are tops. I: Some P is S Some tops are pins. E: No S is P. Ex. No fish is whale. E: No P is S. No whale is fish. I: Some S is P. Ex. Some boys are poets. I: Some P is S. Some poets are boys. O: Some S is not P. No valid conversion
Note that in a conversion, the quality remains the same and the quantity may change.
2) Obversion: In obversion, we change the quality of the proposition and replace the predicate term by its complement.
Table of Valid Obversions
Obverted Obverse A: All birds are mammals. E: No birds are non-mammals. E: No poets are singers. A: All poets are non-singers. I: Some nurses are doctors. O: Some nurses are not non-doctors. O: some politicians are not statesmen. I: Some politicians are non-statesmen
Contraposition: To obtain the contra positive of a statement, we first replace the subject and predicate terms in the proposition and then exchange both these terms with their complements.
Table of Valid Contrapositions
Proposition Contra positive A: All birds are mammals. A: All non-mammals are non-birds. I: Some birds are mammals. I: Some non-mammals are non-birds.
Note: The valid converse, obverse or contra positive of a given proposition always logically follows from the proposition.
II. Mediate Deductive Inference (SYLLOGISM): First introduced by Aristotle, a Syllogism is a deductive argument in which conclusion has to be drawn from two propositions referred to as the premises.
Example:
1. All lotus are flowers.
2. All flowers are beautiful.
3. All lotus are beautiful.
Clearly, the propositions 1 and 2 are the premises and the proposition 3, which follows from the first two propositions, is called the conclusion.
Term: In Logic, a term is a word or a combination of words, which by itself can be used as a subject or predicate of a proposition.
Syllogism is concerned with three terms:
1. Major Term: It is the predicate of the conclusion and is denoted by P (first letter of 'Predicate').
2. Minor Term: It is the subject of the conclusion and is denoted by S (first letter of 'Subject').
3. Middle Term: It is the term common to both the premises and is denoted by M (first letter of 'Middle'). | 1,422 | 6,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-26 | longest | en | 0.919489 |
https://www.java-forums.org/new-java/60912-i-am-having-issues-run-code-print.html | 1,493,373,651,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122886.86/warc/CC-MAIN-20170423031202-00230-ip-10-145-167-34.ec2.internal.warc.gz | 941,008,697 | 4,834 | # I am having issues to run this code ?
• 06-23-2012, 10:19 PM
dantzig
I am having issues to run this code ?
the idea is we want generate the Key stream for encryption and decryption, we will process with the following steps
21 24 2 5 9 15 18 1 16 12 27 4 10 19 7 20 23 26 13 3 28 25 22 8 6 14 11 17
Step 1: Swap the A joker (27) with the card following it. Thus, we swap 27 and 4.
21 24 2 5 9 15 18 1 16 12 4 27 10 19 7 20 23 26 13 3 28 25 22 8 6 14 11 17
Step 2: Move the B joker (28) down two places.
21 24 2 5 9 15 18 1 16 12 4 27 10 19 7 20 23 26 13 3 25 22 28 8 6 14 11 17
Step 3: Do the triple cut. Everything above the first joker (27, in this case) goes to the bottom of the deck. Everything below the second joker (28) goes to the top.
8 6 14 11 17 27 10 19 7 20 23 26 13 3 25 22 28 21 24 2 5 9 15 18 1 16 12 4
Step 4: The value of the bottom card is 4, so we take the first four cards of the deck and place them right before the 4.
17 27 10 19 7 20 23 26 13 3 25 22 28 21 24 2 5 9 15 18 1 16 12 8 6 14 11 4
Step 5: The top card is 17. So the keystream value is the 18th card, i.e., 9.
Thank you.
public class Deck {
public Deck(int[] deck)
{
Set s = new HashSet();
for (int i = 1; i<29 ; i++)
{
}
for (int i = 0 ; i<deck.length ; i++)
{
if (!s.contains(deck[i]));
System.out.println("The deck doesn't contains the integers from 1 to 28");
}
}
{
int[] array = new int[deck.size()];
for(int i = 0;i < array.length;i++)
array[i] = deck.get(i);
return array;
}
}
int[] b = toIntArray(deck);
for (int i = 0 ; i<b.length ;i++)
{
if (b[i]== 27)
{
int temp = b[i];
b[i]=b[i+1] ;
b[i] = temp ;
}
}
int[]c = b ;
for (int i = 0 ; i <c.length ; i++)
{
if (c[i] == 28)
{
int temp1 = c[i];
c[i]= c[i+1];
c[i]= temp1 ;
int temp2 = c[i+1] ;
c[i+1]=c[i+2];
c[i+2]= temp2 ;
}
}
int pos = 0 ;
int[] temp = new int[deck.size()] ;
if (deck.indexOf(27) <deck.indexOf(28) ){
// copy back to the front of middle
for (int i = deck.indexOf(28)+1 ; i<deck.size() ;i++){
temp[pos]= deck.get(i) ;
pos++ ;
}
// copy middle section to the middle of temp
for (int i =deck.indexOf(27); i<deck.indexOf(28) ; i++){
temp[pos]=deck.get(i);
}
// Copy first section to temp array
for (int i = 0; i < deck.indexOf(27); i++)
{
temp[pos] = deck.get(i);
pos++;
}
//Copy temp array back to original array
for (pos = 0; pos <deck.size(); pos++)
{
deck.set(pos, temp[pos]);
}
}
//if the joker 28 is after 27 in the deck
else{
// copy back to the front of middle
for (int i = deck.indexOf(27)+1 ; i<deck.size() ;i++){
temp[pos]= deck.get(i) ;
pos++ ;
}
// copy middle section to the middle of temp
for (int i =deck.indexOf(28); i<deck.indexOf(27) ; i++){
temp[pos]=deck.get(i);
}
// Copy first section to temp array
for (int i = 0; i < deck.indexOf(28); i++)
{
temp[pos] = deck.get(i);
pos++;
}
//Copy temp array back to original array
for (pos = 0; pos <deck.size(); pos++)
{
deck.set(pos, temp[pos]);
}
int t1 = deck.get(0);
int t2 = deck.get(1);
int t3 = deck.get(2);
int t4 = deck.get(3);
for (int i = 4 ; i<deck.size()-1 ; i++){
deck.set(i-4, deck.get(i)) ;
}
deck.set(23, t1);
deck.set(24, t2);
deck.set(25, t3);
deck.set(26, t4);
}
return deck;
}
public int getKeystreamValue(){
int x = deck.get(0) + 1 ;
return deck.get(x);
}
}
• 06-23-2012, 10:25 PM
Fubarable
Re: I am having issues to run this code ?
OK, you've posted your assignment, you've posted some unformatted code, and have mentioned that you are having "issues" but I think that for us to be able to help you, you need to tell us more.
• First and foremost, tell us the details of the problems you're having.
• If the code doesn't compile, please post any and all error messages, and indicate which lines in the code are the source of these errors.
• Likewise for any runtime exceptions -- please post the full exception stacktrace and indicate which lines are causing the errors.
• Consider editing your post above and adding [code] [/code] tags around your pasted code so that it will retain its formatting and be easier to read.
Best of luck and welcome to the java-forums. org!
• 06-23-2012, 10:33 PM
dantzig
Re: I am having issues to run this code ?
Thank you.
Let me send it again with the main method.
• 06-23-2012, 10:47 PM
dantzig
Re: I am having issues to run this code ?
That is the error
Exception in thread "main" java.lang.ClassCastException: [I cannot be cast to java.lang.Integer
at WhitespaceString.tripleCut(WhitespaceString.java:1 39)
at WhitespaceString.main(WhitespaceString.java:52)
import java.util.Arrays;
import java.util.Iterator;
import java.util.Scanner;
import java.io.File;
import java.lang.String;
public class WhitespaceString{
private static int[] c;
public static void main(String[] args){
//21 24 2 5 9 15 18 1 16 12 27 4 10 19 7 20 23 26
// 13 3 28 25 22 8 6 14 11 17
int []a = toIntArray(deck) ;
int x = getKeystreamValue();
System.out.println("The value of key stream is"+ x);
for (int i = 0 ; i<l.size() ; i++){
System.out.println(l.get(i));
}
}
int[] array = new int[deck.size()];
for(int i = 0;i < array.length;i++){
array[i] = deck.get(i);
System.out.println(array[i]) ;
}
return array;
}
int[] b = toIntArray(deck);
for (int i = 0 ; i<b.length ;i++)
{
if (b[i]== 27)
{
int temp = b[i];
b[i]=b[i+1] ;
b[i] = temp ;
}
}
int[]c = b ;
for (int i = 0 ; i <c.length ; i++)
{
if (c[i] == 28)
{
int temp1 = c[i];
c[i]= c[i+1];
c[i]= temp1 ;
int temp2 = c[i+1] ;
c[i+1]=c[i+2];
c[i+2]= temp2 ;
}
}
int pos = 0 ;
int[] temp = new int[deck.size()] ;
if (deck.indexOf(27) <deck.indexOf(28) ){
// copy back to the front of middle
for (int i = deck.indexOf(28)+1 ; i<deck.size() ;i++){
temp[pos]= deck.get(i) ;
pos++ ;
}
// copy middle section to the middle of temp
for (int i =deck.indexOf(27); i<deck.indexOf(28) ; i++){
temp[pos]=deck.get(i);
}
// Copy first section to temp array
for (int i = 0; i < deck.indexOf(27); i++)
{
temp[pos] = deck.get(i);
pos++;
}
//Copy temp array back to original array
for (pos = 0; pos <deck.size(); pos++)
{
deck.set(pos, temp[pos]);
}
}
//if the joker 28 is after 27 in the deck
else{
// copy back to the front of middle
for (int i = deck.indexOf(27)+1 ; i<deck.size() ;i++){
temp[pos]= deck.get(i) ;
pos++ ;
}
// copy middle section to the middle of temp
for (int i =deck.indexOf(28); i<deck.indexOf(27) ; i++){
temp[pos]=deck.get(i);
}
// Copy first section to temp array
for (int i = 0; i < deck.indexOf(28); i++)
{
temp[pos] = deck.get(i);
pos++;
}
//Copy temp array back to original array
for (pos = 0; pos <deck.size(); pos++)
{
deck.set(pos, temp[pos]);
}
int t1 = deck.get(0);
int t2 = deck.get(1);
int t3 = deck.get(2);
int t4 = deck.get(3);
for (int i = 4 ; i<deck.size()-1 ; i++){
deck.set(i-4, deck.get(i)) ;
}
deck.set(23, t1);
deck.set(24, t2);
deck.set(25, t3);
deck.set(26, t4);
}
return deck;
}
public static int getKeystreamValue(){
int x = deck.get(0) + 1 ;
return deck.get(x);
}
}
• 06-23-2012, 11:14 PM
Fubarable
Re: I am having issues to run this code ?
Again, please edit your post above and add the code tags as described in my first post. Please indicate which line is causing the exception, again as I have requested previously.
• 06-24-2012, 01:53 AM
Ronin
Re: I am having issues to run this code ?
dantzig,
As I have free time and want the refresher, I have taken the liberty of debugging your program. There are multiple errors within your code but you can address these one at a time. The first issue is to get your code to compile.
Comment out or remove the code below and take it from there.
Code:
`deck = new LinkedList(Arrays.asList(c));`
Regards. | 2,518 | 7,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-17 | longest | en | 0.535829 |
https://nl.mathworks.com/matlabcentral/profile/authors/3536192?page=2 | 1,680,213,953,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949387.98/warc/CC-MAIN-20230330194843-20230330224843-00148.warc.gz | 469,550,638 | 5,219 | how to generate uniformly random varaibles with inequality constraints
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ongeveer 10 jaar ago | 0 answers | 0 | 1,892 | 6,529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-14 | latest | en | 0.733781 |
https://www.csplib.org/Problems/prob054/models/queens.pi.html | 1,659,889,727,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00797.warc.gz | 658,787,774 | 3,412 | /*
N-Queens problem in Picat.
Model created by Hakan Kjellerstrand, hakank@gmail.com
*/
% Licenced under CC-BY-4.0 : http://creativecommons.org/licenses/by/4.0/
import util.
import cp.
main => go.
%
% Note that $Q[I] is needed here. % queens3(N, Q) => Q=new_list(N), Q :: 1..N, all_different(Q), all_different([$Q[I]-I : I in 1..N]),
all_different([$Q[I]+I : I in 1..N]), solve([ff],Q). queens3(N) => queens3_all(N, Solutions), % writeln(Solutions), writeln(len=Solutions.length). % generate all solutions via solve_all (don't work right now) queens3_all(N, Solutions) => Q=new_list(N), Q :: 1..N, all_different(Q), all_different([$Q[I]-I : I in 1..N]),
all_different([$Q[I]+I : I in 1..N]), % This yield "Unknown procedure solve/2". Solutions = solve_all([ff],Q). % This works: % Solutions = findall(Q,$solve($Q)). % Using all_distinct instead queens3b(N, Q) => Q=new_list(N), Q :: 1..N, all_distinct(Q), all_distinct([$Q[I]-I : I in 1..N]),
all_distinct([$Q[I]+I : I in 1..N]), solve([ff],Q). % alternative approaches queens4(N, Q) => Q = new_list(N), Q :: 1..N, foreach(A in [-1,0,1]) all_different([$Q[I]+I*A : I in 1..N])
end,
solve([ff],Q).
% Decomposition of alldifferent
all_different_me(L) =>
Len = length(L),
foreach(I in 1..Len, J in I+1..Len) L[I] #!= L[J] end.
% Using all_different_me (my decomposition)
queens5(N, Q) =>
Q=new_list(N),
Q :: 1..N,
all_different_me(Q),
all_different_me([$Q[I]-I : I in 1..N]), all_different_me([$Q[I]+I : I in 1..N]),
solve([ff],Q).
go =>
queens3(8,Q),
writeln(Q),
N2 = 12,
queens3_all(8, Solutions),
% writeln(Solutions),
Len=Solutions.length,
writef("N:%w %w solutions.\n%w\n", N2, Len, Solutions).
go2 =>
foreach(N in 2..14)
statistics(backtracks, Backtracks),
statistics(runtime, [_, _Runtime1]),
queens3_all(N, Solutions),
Len=Solutions.length,
statistics(backtracks, Backtracks2),
B = Backtracks2 - Backtracks,
Backtracks := Backtracks2,
statistics(runtime, [_, Runtime2]),
writef("N:%3d %10d solutions (%d backtracks, %d millis)\n", N, Len, B, Runtime2)
end.
%
% Times per Picat v 0.1-beta 10:
% queens3 : 6.7s (2 backtracks)
% queens3b: 10.83s (0 backtracks)
% queens4 : 4.25s (2 backtracks)
% queens5 : 6.86s (2 backtracks)
%
go3 =>
Ps = [queens3=1000, queens3b=1000, queens4=1000,queens5=1000],
foreach(P=N in Ps)
printf("%w(%d)\n", P, N),
time2(once(call(P,N,Q))),
writeln(Q),
nl
end.
% Using permutations/1. Very slow.
go4 =>
N = 8,
C = 0,
foreach(P in permutations(1..N))
if check4(P) then
% writeln(P),
C := C +1
end
end,
writeln(sols=C),
nl.
go5 =>
N=100,
queens3(N,Q),
writeln(Q),
nl.
go6 =>
N = 10000,
println("timing queens4(10000,Q)"),
time2(queens4(N,_Q)),
nl.
check4(P) =>
N = length(P),
foreach(I in 1..N, J in I+1..N)
P[I]-I != P[J]-J,
P[I]+I != P[J]+J
end. | 966 | 2,751 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-33 | latest | en | 0.647271 |
https://numberworld.info/15760 | 1,642,972,736,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00561.warc.gz | 461,918,993 | 3,902 | # Number 15760
### Properties of number 15760
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 5 * 197
Divisors:
1, 2, 4, 5, 8, 10, 16, 20, 40, 80, 197, 394, 788, 985, 1576, 1970, 3152, 3940, 7880, 15760
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
3d90
Base 32:
fcg
sin(15760)
0.97997642695433
cos(15760)
-0.19911354201516
tan(15760)
-4.9216965206701
ln(15760)
9.6652303634119
lg(15760)
4.1975562131535
sqrt(15760)
125.5388386118
Square(15760)
### Number Look Up
Look Up
15760 which is pronounced (fifteen thousand seven hundred sixty) is a unique figure. The cross sum of 15760 is 19. If you factorisate the number 15760 you will get these result 2 * 2 * 2 * 2 * 5 * 197. 15760 has 20 divisors ( 1, 2, 4, 5, 8, 10, 16, 20, 40, 80, 197, 394, 788, 985, 1576, 1970, 3152, 3940, 7880, 15760 ) whith a sum of 36828. The number 15760 is not a prime number. 15760 is not a fibonacci number. The number 15760 is not a Bell Number. The figure 15760 is not a Catalan Number. The convertion of 15760 to base 2 (Binary) is 11110110010000. The convertion of 15760 to base 3 (Ternary) is 210121201. The convertion of 15760 to base 4 (Quaternary) is 3312100. The convertion of 15760 to base 5 (Quintal) is 1001020. The convertion of 15760 to base 8 (Octal) is 36620. The convertion of 15760 to base 16 (Hexadecimal) is 3d90. The convertion of 15760 to base 32 is fcg. The sine of 15760 is 0.97997642695433. The cosine of 15760 is -0.19911354201516. The tangent of the number 15760 is -4.9216965206701. The square root of 15760 is 125.5388386118.
If you square 15760 you will get the following result 248377600. The natural logarithm of 15760 is 9.6652303634119 and the decimal logarithm is 4.1975562131535. that 15760 is impressive number! | 736 | 1,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-05 | latest | en | 0.745591 |
https://number.academy/39563 | 1,720,900,761,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00420.warc.gz | 389,482,738 | 11,483 | # Number 39563 facts
The odd number 39,563 is spelled 🔊, and written in words: thirty-nine thousand, five hundred and sixty-three. The ordinal number 39563rd is said 🔊 and written as: thirty-nine thousand, five hundred and sixty-third. The meaning of the number 39563 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 39563. What is 39563 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 39563.
## Interesting facts about the number 39563
### Asteroids
• (39563) 1992 RB is asteroid number 39563. It was discovered by R. H. McNaught from Siding Spring Observatory on 9/2/1992.
## What is 39,563 in other units
The decimal (Arabic) number 39563 converted to a Roman number is (X)(X)(X)(IX)DLXIII. Roman and decimal number conversions.
#### Length conversion
39563 kilometers (km) equals to 24584 miles (mi).
39563 miles (mi) equals to 63671 kilometers (km).
39563 meters (m) equals to 129799 feet (ft).
39563 feet (ft) equals 12059 meters (m).
#### Time conversion
(hours, minutes, seconds, days, weeks)
39563 seconds equals to 10 hours, 59 minutes, 23 seconds
39563 minutes equals to 3 weeks, 6 days, 11 hours, 23 minutes
### Zip codes 39563
• Zip code 39563 Moss Point, Mississippi, Jackson, USA a map
### Codes and images of the number 39563
Number 39563 morse code: ...-- ----. ..... -.... ...--
Sign language for number 39563:
Number 39563 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Share in social networks
#### Is Prime?
The number 39563 is a prime number. The closest prime numbers are 39551, 39569.
#### Factorization and factors (dividers)
The prime factors of 39563
Prime numbers have no prime factors smaller than themselves.
The factors of 39563 are 1, 39563.
Total factors 2.
Sum of factors 39564 (1).
#### Prime factor tree
39563 is a prime number.
#### Powers
The second power of 395632 is 1.565.230.969.
The third power of 395633 is 61.925.232.826.547.
#### Roots
The square root √39563 is 198,9045.
The cube root of 339563 is 34,074519.
#### Logarithms
The natural logarithm of No. ln 39563 = loge 39563 = 10,58565.
The logarithm to base 10 of No. log10 39563 = 4,597289.
The Napierian logarithm of No. log1/e 39563 = -10,58565.
### Trigonometric functions
The cosine of 39563 is -0,602862.
The sine of 39563 is -0,797846.
The tangent of 39563 is 1,323431.
## Number 39563 in Computer Science
Code typeCode value
39563 Number of bytes38.6KB
Unix timeUnix time 39563 is equal to Thursday Jan. 1, 1970, 10:59:23 a.m. GMT
IPv4, IPv6Number 39563 internet address in dotted format v4 0.0.154.139, v6 ::9a8b
39563 Decimal = 1001101010001011 Binary
39563 Decimal = 2000021022 Ternary
39563 Decimal = 115213 Octal
39563 Decimal = 9A8B Hexadecimal (0x9a8b hex)
39563 BASE64Mzk1NjM=
39563 MD5dc10fce584f2cdf09d6690e0f2883227
39563 SHA1f8ee08f747647a06c9dc31e0c1698d8820cc1139
39563 SHA224c7a9308df1361a38b9506dd0f8dbe921131b1e89cb95b20e4529e639
39563 SHA384cf8664148736fd7d00c0c0ce2636aba6c30e8373d7415f7e220f2c2a65699dde4b779306572b608ab0486bc726bd2ff6
More SHA codes related to the number 39563 ...
If you know something interesting about the 39563 number that you did not find on this page, do not hesitate to write us here.
## Numerology 39563
### Character frequency in the number 39563
Character (importance) frequency for numerology.
Character: Frequency: 3 2 9 1 5 1 6 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 39563, the numbers 3+9+5+6+3 = 2+6 = 8 are added and the meaning of the number 8 is sought.
## № 39,563 in other languages
How to say or write the number thirty-nine thousand, five hundred and sixty-three in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 39.563) treinta y nueve mil quinientos sesenta y tres German: 🔊 (Nummer 39.563) neununddreißigtausendfünfhundertdreiundsechzig French: 🔊 (nombre 39 563) trente-neuf mille cinq cent soixante-trois Portuguese: 🔊 (número 39 563) trinta e nove mil, quinhentos e sessenta e três Hindi: 🔊 (संख्या 39 563) उनतालीस हज़ार, पाँच सौ, तिरेसठ Chinese: 🔊 (数 39 563) 三万九千五百六十三 Arabian: 🔊 (عدد 39,563) تسعة و ثلاثون ألفاً و خمسمائة و ثلاثة و ستون Czech: 🔊 (číslo 39 563) třicet devět tisíc pětset šedesát tři Korean: 🔊 (번호 39,563) 삼만 구천오백육십삼 Danish: 🔊 (nummer 39 563) niogtredivetusinde og femhundrede og treogtreds Hebrew: (מספר 39,563) שלושים ותשעה אלף חמש מאות שישים ושלוש Dutch: 🔊 (nummer 39 563) negenendertigduizendvijfhonderddrieënzestig Japanese: 🔊 (数 39,563) 三万九千五百六十三 Indonesian: 🔊 (jumlah 39.563) tiga puluh sembilan ribu lima ratus enam puluh tiga Italian: 🔊 (numero 39 563) trentanovemilacinquecentosessantatré Norwegian: 🔊 (nummer 39 563) trettini tusen fem hundre og sekstitre Polish: 🔊 (liczba 39 563) trzydzieści dziewięć tysięcy pięćset sześćdziesiąt trzy Russian: 🔊 (номер 39 563) тридцать девять тысяч пятьсот шестьдесят три Turkish: 🔊 (numara 39,563) otuzdokuzbinbeşyüzaltmışüç Thai: 🔊 (จำนวน 39 563) สามหมื่นเก้าพันห้าร้อยหกสิบสาม Ukrainian: 🔊 (номер 39 563) тридцять дев'ять тисяч п'ятсот шістдесят три Vietnamese: 🔊 (con số 39.563) ba mươi chín nghìn năm trăm sáu mươi ba Other languages ...
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Six a squared plus five a minus six? Find answers now! No. 1 Questions & Answers Place. More questions about Science & Mathematics, Mathematics
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of a number and 2 two . plus. a number two . ... of a number and two a number . minus. 2 two . ... Five less than six times a number divided by twice the ...
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2 + x + 2 Subtraction difference minus subtracted from less than ... Five less than six times a ... TRANSLATE WORD SENTENCES INTO ALGEBRAIC ...
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may be read "x plus six, the quantity squared" or "the quantity x plus six, squared." See: ... 3. the quantity of x minus three times x plus four.
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Accuplacer Elementary Algebra Study Guide for Screen Readers . ... Three squared will equal nine. ... Five plus one equals six and four minus two equals two. | 230 | 937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2016-44 | longest | en | 0.809755 |
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# 1.
3 Stretches
Once again – let’s first look back where we’ve been….
\$ #2
First, we saw how translations occur when we add (or subtract) numbers.
## For example, a function \$ # 2 can be
transformed to \$ # 3 2 2
## 2 units Graph horizontally And vertically
translates 3 units right translates 2 units up
3 units (opposite direction of the sign) (same direction* of the sign)
## To horizontally translate the graph 3 To vertically translate the graph 2
units right, replace “#” with “# 3” units up, replace “\$” with “\$ 2”
2
So, becomes \$ 2 # 3 Simplifies to
When we think of vertical translations this way, we can treat it “the same” as horizontal!
*That is, the opposite direction of the sign in the equation: \$ ( # → / % .
## We can similarly think of reflections with replacements.
2 horiz. reflection \$ # 4 2
For example, given the graph of \$ # 4 …
## To horizontally reflect, To vertically reflect,
replace “#” with “ #” replace“\$” with “ \$” vertical
reflection
## We can optionally simplify to: Now divide both sides by 1
2 to isolate “\$”!
\$ 1 # 4
2 2 2
\$ 1 # 4 \$ # 4
2
\$ # 4
%
Replacing “\$” with “ \$” is identical to “making the entire right side negative”
\$ ( # \$ ( #
\$ 5 ( # \$ ( # 5
## Vertical translation 5 units 5 units
- up, if 5 6 0 - down, if 5 7 0
/ %
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and schools throughout the 2021-22 School Year
1.3 Stretches
Exploration #1
The graph of ( # # 4 1 is on the right
## 1 On the same grid, construct a new graph of \$ < #
by multiplying all of the \$-coordinates by 2.
\$ ( #
Be sure to transform all points indicated ()
2 Is the resulting stretch
horizontal or vertical?
3 How could this transformation be
described with a mapping rule?
## 4 Describe the location and coordinates of any
invariant point(s). (Points that don’t change)
5 Use your graphing calculator to confirm the equation of < # is: < # 2 # 4 1
Exploration #2
The graph of ( # # 4 1 is again on the right
## 1 On the same grid, construct a new graph of \$ 5 #
\$ ( #
by multiplying all #-coordinates by 2.
2 Is the resulting stretch
horizontal or vertical?
3 How could this transformation be
described with a mapping rule?
4 Describe the location and coordinates
of any invariant point(s).
5 Use your graphing calculator to confirm the equation of 5 # is: 5 # 1⁄2 # 4 1
Exploration #3
## 1 Use your graphing calculator to help sketch the graphs
of \$ 2 # 2 4 and \$ 0.5 # 2 4 on the same grid.
Graph and together to compare, then graph and .
Be sure to indicate the points corresponding to those indicated ()
## 2 Explain the effect of the “2” and “0.5”
in terms of stretching.
## 3 Describe the location and coordinates of
any invariant point(s).
Page |28
Chapter 1 – Transformations
Exploration #4
## 1 Use your graphing calculator to help sketch the graphs
of \$ 2# 2 4 and \$ 0.5# 2 4 on the same grid.
Be sure to indicate the points corresponding to those indicated ()
## 2 Explain the effect of the “2” and “0.5”
in terms of stretching.
## The graph of a function % , transformed to -% , -%
is vertically stretched about the #-axis by a factor of -. %
!, -
## All points on the graph of \$ 0( # !,
are “-” times further from the #-axis
#-intercepts
% are invariant
%
Invariant Point
is on #-axis For example, the graph of a function ( # # 4 1
can be vertically stretched by a factor of 2,
4, 1
4, 2 Giving an equation \$ # 4 1
All pts move 2 times
Which can be simplified to: \$ 2 # 4 2
further from #-axis
% % J
The graph of a function % , transformed to % J ,
is horizontally stretched about the #-axis by a factor of /J.
1
All points are transformed #, \$ → #, \$
I Reciprocal
## All points on the graph of \$ ( I# ,! ,!
J
are “ /J” times further from the \$-axis
\$-intercept is
invariant
## For example, the graph of a function ( # # 4 1
Invariant Point %
is on \$-axis can be horizontally stretched by a factor of 2,
% Giving an equation ⁄
## 8, 1 4, 1 Which can be simplified to: \$ 1/2 # 8 2
All pts move 2 times
further from \$-axis
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and schools throughout the 2021-22 School Year
1.3 Stretches
In the transformation \$ ( # →\$ 0( I#
## The vertical stretch is “-” The horizontal stretch is “ /J”
(straightforward!) (reciprocal!)
Another way to look at this is to treat both the vertical and horizontal stretches the same.
So if, for example, a function \$ ( # is horizontally stretched by a factor of 3 and vertically
stretched as a factor of 4, we “could” ....
1
Replace # with #
3
\$ ( # So here both the horizontal and vertical
1 stretches are reciprocals in the equation
and, similarly… Replace \$ with \$
4
Note that this can be simplified to:
Which is how we would
\$ ( #
“normally” view this!
## Given the graph of \$ ( # on the right,
1
(a) Construct a mapping rule to sketch the graph of < # ( #
2
%
## (c) Compare the domain and range of \$ ( # and \$ < # .
Describe which is affected, and how.
\$ ( # D: \$ < # D:
R: R:
## Given the graph of \$ ( # on the right,
(a) Construct a mapping rule to sketch the graph
of < # ( 3#
## (b) State the location of any invariant points.
%
(c) Compare the domain and range of \$ ( #
and \$ < # . Describe which is affected, and
how.
\$ ( # D: \$ < # D:
R: R:
Page |30
Chapter 1 – Transformations
## Worked The graph of % C & is shown on the right. %
Example \$ < # is obtained by vertically stretching the graph of ( # about the
line \$ 0, by a factor of 3, and reflecting the graph about the #-axis.
\$ ℎ # is obtained by horizontally stretching the graph of
( # about the line # 0, by a factor of 2.
(a) State an equation for and sketch the graph of \$ < # , on the same
grid. Be sure to indicate the new location of all indicated points. ()
(b) State an equation and sketch the graph of \$ ℎ # , on the same grid.
## Sol: (a) Equation in terms of ( # : ' %
,
%
Vertical reflection Vertical stretch
Equation in terms of #: \$ 3 #2 6# 5
Simplifies to: G &
For the equation, multiply all \$-coordinates by 3. Invariant points are on the
Mapping rule: All points , → , #-axis, at 1, 0 and 5, 0 . ,
3, 4 → 3, 12
%
2, 3 → 2, 9 and so on…
## (b) Equation in terms of ( # : . %
Horizontal stretch (reciprocal) %
1 2
1
Equation in terms of #: \$ # 6 # 5
2 2
%
Simplifies to: &
For the equation, multiply all #-coordinates by 2. Invariant point is on
Mapping rule: All points , → , the \$-axis, at 0, 5 .
, C,
3, 4 → 6, 4
2, 3 → 4, 3 and so on…
## A function ( # # 5 8 has a range of \$ 8, \$ ∈ ℝ . A function \$ < # is obtained by reflecting the
graph of \$ ( # about the line \$ 0, and applying a vertical stretch by a factor of 1/4.
(a) Determine an equation for \$ < # , both in terms of ( # and in terms of #.
## (b) State the range of < # .
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and schools throughout the 2021-22 School Year
1.3 Stretches
## The graph of % , is shown. The graph of \$ < # is
obtained by vertically stretching the graph of ( # about the line \$ 0 by a
factor of 3, and reflecting it about the line \$ 0.
## (a) Sketch the graph of \$ < # . Be sure to indicate
the new location of the vertex, any # and
\$-intercepts, and any other key points. Mapping Rule:
## (b) Determine an equation for \$ < # ,
In terms In terms
of % : of :
## (c) State the location and coordinates
of any invariant point(s).
## (d) State the range of both \$ ( # and \$ < # .
\$ ( # R: \$ < # R: %
## The graph of % , is again shown below.
The graph of \$ ℎ # is obtained by horizontally stretching
the graph of ( # about the line # 0 by a factor of 3.
## (e) Sketch the graph of \$ ℎ # . Be sure to indicate the
new location of the vertex, and any # and \$-intercepts.
Mapping Rule:
## (f) Determine an equation for \$ ℎ # ,
In terms In terms
of % : of :
## (g) State the location and coordinates
of any invariant point(s).
%
## The graph of % C is shown on the right.
A new function < # is defined < # ( 3# . %
(a) Sketch the graph of \$ < # , on the same grid.
Mapping Rule:
## (b) Determine an equation for
\$ < # , in terms of # .
Page |32
Chapter 1 – Transformations
Worked For each pair of graphs below, the graph of \$ < # is obtained by stretching the graph of ( # .
Example For each, determine an equation for \$ < # .
'
'
(a) (b)
!, C
% G
!,
## Sol: (a) (b)
“start” point for ( # Then on < # it’s at
is at # 8 !, C \$ 36
## …and for < # it’s # 2 !,
\$-intercept for ( # is \$ 12
The start point (as with all other points) is stretched
horizontally, as it moves closer to the \$-axis.
Also note that the invariant point is on the \$-axis The \$-intercept (as with all other points) is stretched
vertically, as it moves further from the \$-axis.
8 K horiz. stretch 2 horiz. stretch 2L8 Also note that the invariant points are on the #-axis
12 K vert. stretch 36
horizontal stretch factor
vert. stretch 36 L 12
So, equation is: ' %
vertical stretch factor
## Class Example 1.36 Determining the Stretch from a Graph
For each pair of graphs below, the graph of \$ < # is obtained by stretching the graph of ( # .
Determine an equation for \$ < # , in terms of ( # .
(a) (b)
%
% '
'
## Mapping Rule: Mapping Rule:
Equation: Equation:
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and schools throughout the 2021-22 School Year
1.3 Stretches
## Class Example 1.37 Determining the Stretch from a Graph
For each graph below, the graph of \$ < # is obtained by stretching the graph of ( # . Points indicated ()
have integer coordinates. Determine an equation for < # for each, and identify all indicated characteristics.
(a)
'
% G
Equation in terms of % :
Equation in terms of :
-intercept of ' :
(b)
% C
'
Equation in terms of % :
Equation in terms of :
-intercept of % :
(use an algebraic process)
Page |34
1.3 Practice Questions
1. Given each graph of \$ ( # below, sketch each transformed function, and provide all indicated characteristics.
1
(a) \$ ( # (b) \$ 4( #
3 i Transformation
% in Words:
ii Mapping Rule:
%
i Transformation iii Range of ( # :
in Words:
ii Mapping Rule:
… of transformed
function:
iii Domain of ( # :
… Domain of iv Coordinates of
transformed invariant point(s):
function:
iv Coordinates of
invariant point(s):
1
(c) \$ ( 3#
2
i Mapping Rule:
ii Domain of ( # :
Range of ( # :
## iii Domain of transformed function:
%
Range of transformed function:
## 2. Determine an equation, as specified, when the function ( # 2# 2 4# 12 is:
(a) Vertically stretched by a factor of 2 (b) Horizontally stretched by a factor of 2
i Equation in i Equation in
terms of ( # : terms of ( # :
ii Equation in ii Equation in
terms of #: terms of #:
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and schools throughout the 2021-22 School Year
1.3 Stretches
## 3. The graph of % G is shown below.
The graph of \$ < # is obtained by stretching the graph of ( # about the line \$ 0 by a factor of 2.
(a) Sketch the graph of \$ < # . Be sure to indicate the new location of the
vertex, any # and \$-intercepts, and any other key points.
Mapping Rule:
## (b) Determine an equation for \$ < # ,
i In terms of % : ii In terms of :
## (c) State the location and coordinates
of any invariant point(s).
%
## 4. A function \$ ( # has a range of 8, ∞ , a domain of ∞, 4 , an #-intercept 3, 0 and a
\$-intercept 0, 5 . Determine these characteristics for the following transformed functions.
1
(a) \$ 5( # i Domain: (b) \$ ( # i Domain:
4
ii Range: ii Range:
## iii #-intercept: iii #-intercept:
iv \$-intercept: iv \$-intercept:
1. (a) i Horiz. str, factor of 3 (b) i Vertical stretch
%
ii #, \$ → 3#, \$ factor of 4
iii ( # : 3, 4 ii #, \$ → #, 4\$
transformed:
9, 4 % iii ( # : 3, 4
transformed:
iv 0, 2 on \$-axis 12, 16
1 1
(c) i #, \$ → #, \$ iv 1, 0 and 5, 0
3 2 on #-axis
ii Domain: 9, 12
Range: 6, 6
2. (a) i % ii G
iii Domain: 3, 4 %
(b) i % ii
Range: 3, 3
Page |36
Chapter 1 – Transformations
## 5. The graph of % G is shown below.
The graph of \$ < # is obtained by stretching the graph of ( # about the line # 0 by a factor of 1/4.
(a) Sketch the graph of \$ < # . Indicate the new location of key points.
Mapping Rule:
## (b) Determine an equation for \$ < # ;
%
i In terms of ( # : ii In terms of #:
## (c) State the location and coordinates
of any invariant point(s).
(d) State the domain of: i \$ ( # ii \$ < #
## (f) The point B 10, 4 is on the graph of \$ ( # . Determine the coordinates
of the corresponding point to B on the graph of \$ < # .
6. Below is the graph of * C . The #-intercepts have integer coordinates, and the
\$-intercept is 0, 48 . The two points indicated on the graph have coordinates rounded to the nearest tenth.
The graph of \$ < # is obtained by stretching the graph of 1 # about the line # 0 by a factor of 1/2.
(a) Sketch the graph of \$ < # . Mapping *
Rule:
(b) Determine an equation for \$ < # ; . C, C&. +
i In terms of p # : ii In terms of #:
## (c) State the location and coordinates of any invariant point(s).
. , .&
3. (a) #, \$ → #, 2\$ , C 4. (a) i ∞, 4 (b) i ∞, 16
(b) i < # 2( #
2
ii ∞, 40 ii 8, ∞
ii < # # 3 16 !, +
(c) On the #-axis; 7, 0 include all iii 3, 0 iii 12, 0
and 1, 0 indicated ()
iv 0, 25 iv 0, 5
(d) i ∞, 8 ii ∞, 16 points!
(e) i 0, 3.5 ii 0, 7
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1.3 Stretches
## 7. The graph of % is shown on the right.
C
1
The graph of \$ < # is defined as < # ( #
6
(a) Sketch the graph of \$ < # . Be sure to indicate the new location
Mapping of the vertex and intercepts.
Rule:
## (c) State the location and coordinates
of any invariant point(s).
(d) State the range of:
i \$ ( # ii \$ < #
## 8. The graph of % is shown on the right.
1
The graph of \$ < # is defined as < # ( #
3
(a) Sketch the graph of \$ < # .
Be sure to indicate the new location of the vertex and intercepts.
Mapping Rule:
## (d) The point B 4, 5 is on the graph of \$ ( # .
Determine the coordinates of the corresponding
point to B on the graph of \$ < # .
1 '
5. (a) #, \$ → #, \$ (c) On the \$-axis; 0, 2 . &,
4
(b) i < # ( 4# (d) i 8, ∞ ii 2, ∞
## ii < # 2 4# 8 2 (e) i 6, 0 ii 3/2, 0
Optionally simplify: ' G (f) E becomes: 2.5, 4 ,
1 !. G, C&. +
6. (a) #, \$ → #, \$ ' (b) i < # 1 2#
2
Mult all #-coords by 1/2 …. ii < # 2# 4 2# 2 2# 6
New #-intercepts are …. Optionally simplify: < # 2 # 2 2 # 1 2 # 3
2, 0 , 1, 0 , and 3, 0 ' G
## (c) On the \$-axis…. 0, 48
. , .&
Page |38
Chapter 1 – Transformations
## 9. The graph of % G is shown on the right.
The graph of < # is obtained by reflecting the graph of ( #
about the #-axis, and vertically stretching by a factor of 2.
(a) Sketch the graph of \$ < # .
Mapping Rule:
## (d) State the range of:
i \$ ( # ii \$ < # Be sure to indicate the new location of all key points.
## 10. The graph of % is shown on the right.
The graph of \$ < # is obtained by reflecting the
graph of ( # about the \$-axis, and horizontally
stretching the graph by a factor of 1/2.
Mapping Rule:
i \$ ( # ii \$ < #
## (e) The point B 14, 1 is on the graph of \$ ( # . Determine the
coordinates of the corresponding point to B on the graph of \$ < # .
Step-by-step solutions, along with videos and additional practice exams, can be found at math30-1edge.com
1
7. (a) #, \$ → #, \$ 8. (a) #, \$ → 3#, \$ '
6
1 ' 1 2
1
(b) < # # 6 2 (b) < # # 2 # 3
2 3 3
simplify to…
(c) on #-axis… 2, 0 and 6, 2 1 2 2
10, 0 < # # # 3
9 3
(d) i 12, ∞ ii 2, ∞ Vertex: 3, 4
(c) on \$-axis… 0, 3 (d) 12, 5
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1.3 Stretches
11. For each pair of graphs below, the graph of \$ < # is obtained by stretching the graph of ( # . For
each, (i) determine a mapping rule, (ii) an equation for \$ < # , in terms of ( # .
(a) (b)
%
'
'
i Mapping Rule:
i Mapping Rule:
ii Equation:
ii Equation:
(c)
i Mapping Rule:
% ii Equation:
## iii ALSO: #-intercepts
of \$ < # :
(exact rational values)
## Note: Both a horizontal and
' vertical stretch have been applied:
12. The graph of \$ < # is obtained by stretching the graph of ( # . The domain of ( # is ∞, 5 and the
range is 20, 15 . The domain of \$ < # is ∞, 5 and the range is 16, 12 .
Determine an equation for \$ < # , in terms of ( # .
9. (a) #, \$ → #, 2\$ 2, 16 1
10. (a) #, \$ → #, \$
2
2
(b) < # # 2 16 ' 0, 2 3
(b) < # 2# 2 3
(c) on #-axis… 6, 0 and simplify to…
2, 0 3.5, 0
< # 2 # 1 3
1, 3
(d) i 8, ∞ ii ∞, 16 (c) on \$-axis… 0, 2
' 3
(d) i 2, ∞ ii ∞, 1 (e) 7, 1
Page |40
Chapter 1 – Transformations
13. The graph of \$ < # is obtained by stretching the graph of ( # . The domain of ( # is 8, ∞ and the
range is 9, 6 . The domain of \$ < # is 2, ∞ and the range is 6, 4 .
Determine an equation for \$ < # , in terms of ( # .
## 14. The graph of \$ < # is obtained by stretching the graph
1 % '
of ( # # 2 2 # 4
4
(a) State the mapping rule that describes the
transformation from \$ ( # to \$ < #
## (b) Determine an equation for \$ < #
i In terms ii In terms
of % : of
## (c) The point B 6, 32 is on the graph of \$ ( # . Determine the
coordinates of the corresponding point to B on the graph of \$ < # .
2 3 3 4
11. (a) i #, \$ → #, 2\$ (b) i #, \$ → #, \$ (c) i #, \$ → #, \$ 12. < # ( #
3 4 2 5
ii \$ 2( # 3 3 4
ii \$ ( # ii \$ ( # iii 5.25, 0 , 1.5, 0 and 6, 0
2 2 3
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and schools throughout the 2021-22 School Year
1.3 Stretches
## 15. The graph of \$ < # is obtained by stretching the graph
2
of ( # 2# 12# 10 The vertices and intercepts of both
graphs all have integer coordinates.
(a) State the mapping rule that describes the
transformation from \$ ( # to \$ < #
## (b) Determine an equation for \$ < #
i In terms ii In terms
of % : of
%
'
## 16. The graph of \$ < # is obtained by stretching the graph
of ( # # 9 2
(a) State the mapping rule that describes the
transformation from \$ ( # to \$ < #
'
%
(b) Determine an equation for \$ < #
i In terms ii In terms
of % : of
## (c) Determine the #-intercept of \$ < #
2 3 2
13. < # ( 4# 14. (a) #, \$ → #, 3\$ (b) i < # 3( # ii < # # 2 # 4 (c) 6, 96
3 4
Page |42
Chapter 1 – Transformations
## 17. The graph of \$ < # is obtained by stretching and reflecting the
%
graph of ( # # 2 8# 12
The vertices and intercepts of both graphs all have integer coordinates.
(a) State the mapping rule that describes the
transformation from \$ ( # to \$ < #
## (b) Determine an equation for \$ < #
i In terms ii In terms
of % : of
'
## 18. The graph of \$ < # on the right is obtained by horizontally
and vertically stretching the graph of \$ ( # .
(a) State the mapping rule that describes the
transformation from \$ ( # to \$ < #
%
'
## (c) Determine the domain of \$ ( # , if the domain of
\$ < # is 3.12, 3.12 .
1 1 2
1
15. (a) #, \$ → 2#, \$ (b) i < # ( # ii < # 2 # 12 # 10 Simplify to: ' C !
2 2 2
2 3 3
16. (a) #, \$ → #, \$ (b) i < # ( # ii < # # 9 2 Simplify to: ' C
3 2 2
(c) 10/3,0
Copyright © RTD Learning. Open Use Access provided to all Alberta teachers Page |43
and schools throughout the 2021-22 School Year
1.3 Stretches
19. For each of the following, the graph of \$ < # is obtained by stretching and / or reflecting the
graph of \$ ( # . Provide a mapping rule to describe the transformations for each.
(a) ( # 6 # 5 2 8 (b) ( # 16# 2 8# 4 (c) ( # 2 # 3 ? 4
< # 1.5 # 5 2 2 < # # 2 2# 4 < # 2 3# 3 ? 4
## (d) ( # 3# 5 6 (e) ( # 6# 8 (f) ( # 2# 2 3# 5
< # 9 # 5 18 < # 2# 8 < # 8# 2 6# 5
20. Describe what happens to the graph of \$ ( # when the following changes are made to its equation,
and provide a mapping rule.
1
(a) Replace # with 5# (b) Replace \$ with \$ (c) Replace \$ with 3\$
2
2
(d) Replace # with # and replace \$ with \$ (e) Replace # with # 1 and replace \$ with \$
3
1
(f) Replace # with # and replace \$ with \$ 4 (g) Replace \$ with 5\$ and replace # with # 2
5
3 3 3 2
17. (a) #, \$ → #, \$ (b) i < # ( # ii < # # 6# 9
4 4 4
3 3 3 5
18. (a) #, \$ → #, \$ (b) < # ( # (c) 5.2, 5.2
5 2 2 3
Page |44
Chapter 1 – Transformations
## 21. Given the function ( # 3 # 4 # 2 ,
(a) State the zeros of \$ ( # (b) State the zeros of \$ 4( #
## (c) State the zeros of \$ ( 2# (d) State the zeros of \$ ( #
1
(e) State the zeros of \$ ( # (f) State the zeros of \$ ( #
3
## 22. The graph of \$ < # on the right is obtained by applying one
stretch and one reflection to the graph of ( # # 2 4# 5.
The vertices of both graphs have integer coordinates.
(a) Describe the transformation from \$ ( # to \$ < # ,
(i) in words and (ii) with a mapping rule.
'
%
(b) Determine an equation for \$ < # , (i) in terms of ( #
and (ii) in terms of #.
(c) Determine the \$-intercept
of \$ < # .
1 1
19. (a) #, \$ → #, \$ (b) #, \$ → 4#, \$ (c) #, \$ →
#, \$
4 3
1
(d) #, \$ → #, 3\$ (e) #, \$ → 3#, \$ (f) #, \$ → #, \$
2
1
20. (a) Horizontal stretch about the \$-axis (or line # 0 , factor of 1/5 #, \$ → #, \$
5
(b) Vertical stretch about the #-axis (or line \$ 0), factor of 2 #, \$ → #, 2\$
1
(c) Vertical stretch about \$ 0, factor of 1/3, and reflection about \$ 0
#, \$ → #, \$
3
3
(d) Horiz. stretch about # 0, factor of 3/2, and reflection about #-axis #, \$ → #, \$
2
(e) Vert. reflection about \$ 0, horizontal translation 1 unit right #, \$ → # 1, \$
(f) Horiz. stretch about # 0, factor of 5, vertical translation 4 units down #, \$ → 5#, \$ 4
1
(g) Vert. stretch about #-axis, factor of 1/5, refl. about #-axis, horiz. translation 2 units right #, \$ → # 2, \$
5
Copyright © RTD Learning. Open Use Access provided to all Alberta teachers Page |45
and schools throughout the 2021-22 School Year
1.3 Stretches
23. The graph of \$ ( # is reflected in the line # 0, stretched about the line \$ 0 by a factor
Exam
Style of 1⁄4, and stretched about the \$-axis by a factor of 3 to create the graph of \$ < # .
Point B 3, 12 on the graph of \$ ( # is transformed to which point on the graph of < # ?
A. 1, 3
B. 1, 3
C. 9, 3
D. 9, 3
## 24. The graph of \$ ( # is shown on the right.
Exam
Style W
The graphs of \$ ( # and \$ 2( # Y
intersect at: %
V X Z [
A. Point R
B. Point S U
C. Points P and Q
D. Points R and T
## 25. The graph of \$ ( # is shown on the right.
Exam
Style 1
The graph of \$ < # is defined as \$ ( # .
2
The graph of \$ < # will contain the point:
A. 4, 2
B. 1, 2
C. 3, 4 %
D. 3, 4
21. (a) 4 and 2 (b) 4 and 2 (c) 2 and 1 (d) 4 and 2 (e) 2 and 4 (f) 12 and 6
1
22. (a) i vertical stretch about the #-axis, factor of 1/3, plus a horiz. reflection about the \$-axis. ii #, \$ → #, \$
9 2 N O 3
1 5
(b) i < # ( # ii < # ?
#
?
#
? (c) 0,
3 3
Page |46 | 7,638 | 22,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-14 | latest | en | 0.792933 |
https://mimikersting.com/tag/comprehension-exercises-for-grade-1/ | 1,600,795,338,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206329.28/warc/CC-MAIN-20200922161302-20200922191302-00083.warc.gz | 539,168,739 | 21,010 | # Comprehension Exercises For Grade 1
## Letter C Worksheets For KindergartenLetter C Worksheets For Kindergarten
Published at Sunday, May 24th 2020, 04:42:26 AM by Pierretta Chauvin. математика класс. Veroyatno, luchshiy sposob ubeditsya, chto klass idet v nogu s tem, chto on izuchayet, - eto ustanovit bazovyy protsent dlya klassa, kotoryy idet v nogu s materialom, i togo, kotoryy ne uspevayet, i togo, kotoryy tozhe legko provesti vremya s nim. Naprimer, yesli klass v tselom pravilno otvechayet po krayney mere na 70% voprosov, kotoryye im zadayut, uchitel mozhet reshit, chto eto oznachayet, chto oni znayut material. Kak tolko oni naberut bolshe, chem, skazhem, 90% pravilnykh otvetov, uchitel mozhet reshit, chto eto oznachayet, chto on poluchayet dostatochno, chtoby oni mogli pereyti k sleduyushchey teme. Eto budet oznachat, chto klass ot 70% do 90% popadayet v tsel, klass nizhe 70% dolzhen zamedlitsya i peresmotret, a klass vyshe 90% gotov dvigatsya dalshe.
## Initial Sounds WorksheetInitial Sounds Worksheet
Published at Wednesday, May 13th 2020, 15:05:02 PM by Tihana Nurdiana. математика класс. Odnoy iz nepriyatnykh problem dlya mnogikh studentov-matematikov yavlyayetsya tot fakt, chto vse klassy matematiki v starshikh klassakh zavisyat ot formuly. Nekotorym uchenikam slozhno zapominat formuly. Drugiye uchashchiyesya mogut bez truda zapominat formuly, no ne ponimayut, kak polzovatsya formuloy, ili ne ponimayut, chto oznachayet rezultat formuly. Yesli vy popadayete v lyubuyu iz etikh kategoriy, togda budet slozhno dobitsya uspekha v klasse matematiki. K schastyu, yest neskolko sovetov i priyemov, kotoryye mogut pomoch. Formuly delyatsya na dve osnovnyye kategorii: 1) Vychislitelnyye navyki. Eto tip formul, naiboleye chasto vstrechayushchiysya v algebre, i yego osobenno trudno ponyat, potomu chto simvoly chasto ne imeyut znacheniya. Zapomnit, chto a ^ 2 - b ^ 2 = (a + b) (a - b) mozhet bytlegko, no eto mozhet ne imet dlya vas absolyutno nikakogo znacheniya i, sledovatelno, ne budet imet dlya vas osoboy tsennosti. Tekhniki, kotoryye my obsudim, osobenno vazhny dlya etogo tipa formul. 2) Ochevidnoye primeneniye v realnoy zhizni. Oni chashche vstrechayutsya v geometrii i trigonometrii. Ikh legche ponyat, no metody takzhe mogut pomoch v etom.
## Free Printable Letter B WorksheetsFree Printable Letter B Worksheets
Published at Friday, January 10th 2020, 10:04:23 AM by Robinette Durand. математика класс. Vsego za dve nedeli ya byl udivlen potryasayushchimi rezul tatami. Sara, dav yey neskol ko minut poigrat s ney, nachala izuchat matematiku. Ona skazala mne, chto oshiblas , kogda skazala, chto matematika - eto ne veselo. Cherez mesyats ona udivila menya vysokoy otsenkoy po matematike. YA byl tak schastliv, potomu chto ona byla pervoy v svoyem klasse matematiki. Dazhe yeye uchitel byl udivlen sluchivshimsya. YA prodolzhil etu strategiyu i slava Bogu, chto ona preuspela v shkole, osobenno po matematike. Etot neveroyatnyy opyt pobudil menya vklyuchit knigu v svoyu strategiyu obucheniya. I ya ne oshibsya v svoikh ozhidaniyakh, ya takzhe uvidel uluchsheniya v moikh uchenikakh. Na samom dele 96% iz nikh uluchshili matematiku i polyubili yeye. Tereza byla tak rada rezul tatam i khotela by podelit sya svoyey udivitel noy knigoy so vsemi uchitelyami i roditelyami, u kotorykh tozhe yest takaya zhe problema. Matematika pri pravil nom vzglyade obladayet ne tol ko istinoy, no i vysshey krasotoy - krasotoy, kholodnoy i surovoy, kak krasota skul ptury.
## Letter J Coloring SheetLetter J Coloring Sheet
Published at Friday, May 29th 2020, 16:26:02 PM. математика класс By Georgette Guillon. Dostupno yeshche mnogo rabochikh listov. Oni razlichayutsya po slozhnosti zanyatiya v zavisimosti ot vozrasta i urovnya obucheniya rebenka. Drugiye zanyatiya dlya detskogo sada vklyuchayut v sebya golovolomki. Sushchestvuyut takzhe rabochiye listy, kotoryye rasskazyvayut detyam o nekotorykh bibleyskikh personazhakh i o tom, kak pomoc soobshchestvu. Eti meropriyatiya ne ogranichivayutsya individualnoy rabotoy. Sushchestvuyut takzhe rabochiye listy, trebuyushchiye gruppovykh usiliy dlya vypolneniya zadachi. Tak deti uchatsya rabotat v komande. Eto ochen effektivnoye gruppovoye zanyatiye, pomogayushcheye im osoznat vazhnost sotrudnichestva i komandnoy raboty.
## Digraph Worksheets For KindergartenDigraph Worksheets For Kindergarten
Published at Friday, May 29th 2020, 16:07:59 PM. математика класс By Laycie Thomas. Ladno, eto proiskhodit kazhdyy god, poetomu mnogiye ucheniki pozvolili sebe otstavat v klasse matematiki, i oni prosto ne mogut poluchit pomoshch, v kotoroy oni nuzhdayutsya, dostatochno bystro, chtoby naverstat upushchennoye, poka klass prodolzhayet svoy put. Zatem sleduyushcheye, chto oni uznayut, - den poslednego ekzamena uzhe na podkhode. Pozhaluysta, ne razvalivaytes, i nikakaya matematika ne vkhodit v chislo predmetov, po kotorym ya rekomenduyu zubrezhku ili ne spat vsyu noch, chtoby proyti final, potomu chto eto trebuyet slishkom bol shoy kontsentratsii, tochnosti i logiki - myshleniya pravogo polushariya.
### Free Printable Letter AFree Printable Letter A
Published at Friday, May 29th 2020, 15:57:39 PM. математика класс By Fantine Pierre. No v kakikh imenno situatsiyakh ucheniku sleduyet propustit urok matematiki? Na etot vopros mozhet byt slozhno otvetit, potomu chto resheniye o tom, gotov li uchenik propustit tselyy god po matematike, zavisit ot sposobnosti opredelit uroven komforta etogo uchenika s matematicheskimi kontseptsiyami za ves god na osnove vsego odnogo provedennogo testa. v nachale goda. Yesli, gipoteticheski govorya, u uchenikov prosto «udachnyy den, i im udalos otvetit na neskol ko voprosov, v kotorykh on ili ona ne byl polnost yu uveren, uchenik mog by v konechnom itoge popast v boleye prodvinutyy klass, chem on ili ona mogli by ne byt gotovym k. I naoborot, u umnogo uchenika mozhet byt plokhoy den, i v itoge on potratit god na izucheniye togo, chto on ili ona deystvitel no uzhe znayet.
#### F Tracing WorksheetF Tracing Worksheet
Published at Thursday, May 28th 2020, 15:56:33 PM. математика класс By Icha Nurhayati. Poymite, chto naiboleye vazhnoye znacheniye formul sostoit v tom, chto oni konstatiruyut otnosheniya, kotoryye sushchestvuyut vsegda - VSEGDA ISTINNY. Eto oznachayet, chto oni VSEGDA mogut pomoch vam! Odnako, chtoby formuly pomogli vam, oni dolzhny imet dlya vas znacheniye. Vychislitelnyye formuly trebuyut znachitelno bolshe usiliy dlya sozdaniya etogo znacheniya. My podrobno obsudim eto v № 3. Zapominayte formuly po mere ikh poyavleniya. Vy mozhete sdelat domashneye zadaniye, prosto vzglyanuv na formulu, no ne zhdite nochi pered ekzamenom, chtoby popytatsya zapomnit mnogiye formuly. Rabota s formulami po mere ikh postupleniya i nemedlennoye sokhraneniye ikh v dolgovremennoy pamyati garantiruyet, chto eta informatsiya budet usvoyena navsegda, a ne tolko dlya testa.
##### Letter Sound Games PrintableLetter Sound Games Printable
Published at Thursday, May 28th 2020, 15:23:38 PM. математика класс By Favor Pelletier. No v kakikh imenno situatsiyakh ucheniku sleduyet propustit urok matematiki? Na etot vopros mozhet byt slozhno otvetit, potomu chto resheniye o tom, gotov li uchenik propustit tselyy god po matematike, zavisit ot sposobnosti opredelit uroven komforta etogo uchenika s matematicheskimi kontseptsiyami za ves god na osnove vsego odnogo provedennogo testa. v nachale goda. Yesli, gipoteticheski govorya, u uchenikov prosto «udachnyy den, i im udalos otvetit na neskol ko voprosov, v kotorykh on ili ona ne byl polnost yu uveren, uchenik mog by v konechnom itoge popast v boleye prodvinutyy klass, chem on ili ona mogli by ne byt gotovym k. I naoborot, u umnogo uchenika mozhet byt plokhoy den, i v itoge on potratit god na izucheniye togo, chto on ili ona deystvitel no uzhe znayet.
###### Letter X Pictures For KindergartenLetter X Pictures For Kindergarten
Published at Thursday, May 28th 2020, 15:23:28 PM. математика класс By Sydnee Georges. Vsego za dve nedeli ya byl udivlen potryasayushchimi rezul tatami. Sara, dav yey neskol ko minut poigrat s ney, nachala izuchat matematiku. Ona skazala mne, chto oshiblas , kogda skazala, chto matematika - eto ne veselo. Cherez mesyats ona udivila menya vysokoy otsenkoy po matematike. YA byl tak schastliv, potomu chto ona byla pervoy v svoyem klasse matematiki. Dazhe yeye uchitel byl udivlen sluchivshimsya. YA prodolzhil etu strategiyu i slava Bogu, chto ona preuspela v shkole, osobenno po matematike. Etot neveroyatnyy opyt pobudil menya vklyuchit knigu v svoyu strategiyu obucheniya. I ya ne oshibsya v svoikh ozhidaniyakh, ya takzhe uvidel uluchsheniya v moikh uchenikakh. Na samom dele 96% iz nikh uluchshili matematiku i polyubili yeye. Tereza byla tak rada rezul tatam i khotela by podelit sya svoyey udivitel noy knigoy so vsemi uchitelyami i roditelyami, u kotorykh tozhe yest takaya zhe problema. Matematika pri pravil nom vzglyade obladayet ne tol ko istinoy, no i vysshey krasotoy - krasotoy, kholodnoy i surovoy, kak krasota skul ptury.
## Lowercase Letter PracticeLowercase Letter Practice
Published at Thursday, May 28th 2020, 15:17:34 PM. математика класс By Ninette Lacroix. Vsegda chitayte formuly kak zakonchennyye mysli ili predlozheniya. Formuly vsegda dolzhny imet znak ravenstva. Takim obrazom, formula dlya rasstoyaniya NE rt. Formula dlya rasstoyaniya d = rt. Vsegda proveryayte pravil nost formul. Naprimer: raznitsa dvukh polnykh kvadratov mozhet vyglyadet kak 25-4. Ochevidno, znacheniye etogo chisla ravno 21, no yesli my poluchim eto zhe znacheniye s pomoshchyu formuly, my proverim formulu. Po formule 25-4 mozhno bylo by zapisat kak 5 ^ 2-2 ^ 2 i razlozhit na mnozhiteli kak (5 + 2) (5-2) = 7 (3) = 21. Takim obrazom, formula proverena. Znayte, kak oboznachat otvety po formule i chto oni oznachayut. Bud te konkretny pri chtenii formul. Poskol ku vse formuly ploshchadi nachinayutsya s A =, boleye polezno dobavit dopolnitelnuyu informatsiyu. V A = (pi) r ^ 2 luchshe vsego chitat eto tak: «Ploshchad kruga ravna pi, umnozhennomu na kvadrat radiusa kruga». Obyazatelno pomnite, chto ploshchad vsegda izmeryayetsya v kvadratnykh yedinitsakh.
Any content, trademark/s, or other material that might be found on this site that is not this site property remains the copyright of its respective owner/s. | 3,722 | 10,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-40 | latest | en | 0.137687 |
https://git.chrisbeckstrom.com/chris/tidal-autotechno/commit/385895799410374e71f686541e0a9cb5d09cedc6.diff | 1,657,157,003,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-27/segments/1656104683020.92/warc/CC-MAIN-20220707002618-20220707032618-00432.warc.gz | 319,824,196 | 3,442 | diff --git a/auto-techno.py b/auto-techno.py index 4ea03a7..196946a 100644 --- a/auto-techno.py +++ b/auto-techno.py @@ -21,7 +21,7 @@ lowestBpm = 92 highestBpm = 135 # force everything to happen within 1 bar -oneBar = 1 +oneBar = 0 # stack name # this doesn't matter, can be any string @@ -60,8 +60,8 @@ percNames = ["kick", "hh", "sd", "cp", "ohh","ride","cymbal"] noteRange = [12,24,36] # define the shortest and longest sequence you want # eg if shortest is 7, there will be no melodic sequences shorter than 7 -shortestMelodic = 4 -longestMelodic = 16 +shortestMelodic = 3 +longestMelodic = 10 # for melodic lines, what divisions do you want to choose from? # eg 4 = quarter notes, 8 = eighth notes @@ -141,6 +141,7 @@ bpm = "setcps ("+str(random.randrange(lowestBpm, highestBpm, 1))+"/60/4)" ####################### FUNCTIONS ################### + def generateBinary(busyness): """ busy-ness (sic) @@ -323,6 +324,26 @@ def fills(): fills = str(random.choice(fillChoices))+" "+fills return "whenmod "+str(random.choice(whenmods))+" ("+fills.rstrip()+")" +def sometimesBeQuiet(): + """ + Generate a gain pattern so that sometimes a track + is silent, and other times it isn't + """ + #def genOnOff(limit,busyness,struct): + # decide whether to do this at all + if (random.randint(0,10) > 5): + # for now, 8 bar phrases + # (# gain "[0 1 0 0 1]/8") + # how many cycles to make? + quietCycles = random.choice([2,3,4]) + quietBinary = genOnOff(quietCycles,2,0) + # split 001001 into 0 0 1 0 0 1 + quietBinary = list(quietBinary) + finalQuietBinary = "" + for i in quietBinary: + finalQuietBinary = i + " " + finalQuietBinary + print(indent+indent+"\$ degradeBy \"<"+str(finalQuietBinary).rstrip()+">/8\"") + ########################### LISTS ########################## # a collection of typical-ish kick drum patterns @@ -341,9 +362,11 @@ print("do") #print("") # create "let" statements +print(indent+"let fourToTheFloor = 0") +print(indent+" -- control density per track") for i in percNames: #print("let "+i+"M = \"["+genOnOff(4,5,0)+"]\"") - print(indent+"let "+i+"M = \"[1111]\"") + print(indent+" "+i+"M = \"[1111]\"") #print("") @@ -382,7 +405,7 @@ print(indent+indent+"\$ stack [") # this is an empty track to make dealing with commas easier # just 4 to the floor in case you want it print(indent+indent+"-- four to the floor") -print(indent+indent+"(#gain 1) \$ struct \"[t*4]\" \$ n \"0\" # midichan 0,") +print(indent+indent+"degradeBy fourToTheFloor \$ struct \"[t*4]\" \$ n \"0\" # midichan 0,") # generate rhythm tracks # set the starting midi channel @@ -401,6 +424,10 @@ for i in percNames: # alter how dense the pattern is, by altering "let nnM = "[1111] print(indent+indent+"mask "+i+"M") + # sometimes be quiet + if (i != "kick"): + sometimesBeQuiet() + #### VARIATIONS ##### # insert some funStuff, maybe if (random.randint(0,10) > 3): @@ -415,7 +442,7 @@ for i in percNames: ##### THE RHYTHMS ##### # how many bars to create? - rhythmLengthChoices = [1,1,1,2] + rhythmLengthChoices = [1,1,1,1,1,1,2,2] rhythmLength = random.choice(rhythmLengthChoices) # for the number of bars, do this: @@ -424,7 +451,7 @@ for i in percNames: # choose between random, type-specific (bd,hh,etc), # rhythm library, or euclidean rhythms - rChoice = random.choice([0,1,2,3]) + rChoice = random.choice([0,1,1,1,2,3]) rhythm = "" # placeholder for the rhythm if (rChoice == 0): # if choice = 0, choose hardcoded rhythms @@ -496,8 +523,11 @@ for i in range(melodyCount): print(indent+indent+"\$ struct \""+genOnOff(16,(busyness),1)+"\"") print(indent+indent+"\$ "+genNotes(1)) else: - genOnOff(32,(busyness)) - print(indent+indent+"\$ struct \""+genNotes(1))+"\"" + # how many 16th notes? + thisLength = random.randint(3,16) + #genOnOff(8,(busyness)) + print(indent+indent+"\$ struct \""+genOnOff(thisLength,(busyness),1)+"\"") + print(indent+indent+"\$ "+genNotes(0)) print(indent+indent+"# midichan "+str(melodicMidiChan)+" + n (-42)") melodicMidiChan = melodicMidiChan + 1 diff --git a/deleteme.tidal b/deleteme.tidal index 9d44029..428a644 100644 --- a/deleteme.tidal +++ b/deleteme.tidal @@ -1,63 +1,74 @@ -p "techno" \$ struct "t*4" \$ n "0" # sunvox - -hush - - --- seed: -413486887982277853 --- generated at 1633532891.3 +-- raw input: 268982558670766555 +-- hex seed: 0x3BB9E364458C1DB +-- decimal seed: 268982558670766555 +-- generated at 1633539511.09 +-- Auto-Techno Generator v0.1 do - let kickM = "[1111]" - let hhM = "[1111]" - let sdM = "[1111]" - let cpM = "[1111]" - let ohhM = "[1111]" - let rideM = "[1111]" - let cymbalM = "[1111]" + let fourToTheFloor = 0 + -- control density per track + kickM = "[1111]" + hhM = "[1111]" + sdM = "[1111]" + cpM = "[1111]" + ohhM = "[1111]" + rideM = "[1111]" + cymbalM = "[1111]" p "techno" - \$ whenmod 8 7 (stut 4 1 "1/16") - \$ whenmod 32 31 (rev) + -- fills + \$ whenmod 32 30 (rev) + \$ whenmod 32 31 (stut 2 1 "0.125") + \$ whenmod 32 30 (scramble 16) + \$ whenmod 16 15 (rev) + \$ whenmod 64 62 (rev) \$ stack [ -- four to the floor - (#gain 0) \$ struct "[t*4]" \$ n "0" # midichan 0, - -- kick ----------------- + degradeBy fourToTheFloor \$ struct "[t*4]" \$ n "0" # midichan 0, + -------------- kick --------------- mask kickM - \$ every 3 ((0.1875 ~>)) - \$ every 10 ((0.1875 ~>)) - \$ every 7 (rev) + \$ every 6 ((0.125 ~>)) \$ (0.125 ~>) - \$ struct "<[t*4]>" - \$ n 0 # midichan 0, - -- hh ----------------- + \$ struct "<[{0101011}%16]>" + \$ n 1 # midichan 0, + -------------- hh --------------- mask hhM - \$ every 16 ((0.1875 ~>)) + \$ every 16 (scramble 8) + \$ every 15 ((0.125 ~>)) + \$ struct "<[1*16]>" + \$ n 3 # midichan 1, + -------------- sd --------------- + mask sdM + \$ every 15 (stut 5 1 "0.0625") \$ (0.125 ~>) - \$ struct "<{000101}%16 {100000001}%16>" - \$ n 2 # midichan 1, - -- ride ----------------- + \$ struct "<[t(15,16)]>" + \$ n 2 # midichan 2, + -------------- cp --------------- + mask cpM + \$ degradeBy "[0 0 1]/8" + \$ struct "<[{0000100}%16]>" + \$ n 3 # midichan 3, + -------------- ohh --------------- + mask ohhM + \$ struct "<[tt*2]*4 [[0001]]*4>" + \$ n 0 # midichan 4, + -------------- ride --------------- mask rideM - \$ every 3 (stut 2 1 "1/8") - \$ (0.125 ~>) - \$ struct "<[t(2,8)] [tttt*2?]>" - \$ n 3 # midichan 5, - -- cymbal ----------------- - mask cymbalM - \$ every 4 (scramble 8) - \$ every 12 ((0.1875 ~>)) + \$ every 10 (scramble 8) + \$ every 5 (rev) + \$ every 13 (scramble 8) \$ (0.1875 ~>) - \$ struct "<[01]*2>" - \$ n 1 # midichan 6, + \$ struct "<[t(4,16)]>" + \$ n 0 # midichan 5, + -------------- cymbal --------------- + mask cymbalM + \$ every 8 (rev) + \$ every 7 (rev) + \$ every 12 (stut 5 1 "0.0625") + \$ struct "<[{10011000}%16]>" + \$ n 3 # midichan 6, ----------------------- degradeBy 0 - \$ struct "{1011111111011101}%16" - \$ n "{23 21 20 3 6 32 10 19 9 9 0 0 10 0 2 11 }%16" + \$ struct "[{0010100000000000}%16]" + \$ n "[{18 6 8 7 19 35 8 24 11 7 7 8 3 23 8 33 }%16]" # midichan 7 + n (-42) ] # sunvox - - -p "techno" -\$ every 16 ((0.1875 ~>)) -\$ (0.125 ~>) --- \$ struct "<{000101}%16 {100000001}%16>" -\$ struct "t*4" -\$ n 2 # midichan 1, | 2,550 | 7,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-27 | latest | en | 0.608946 |
https://longroom.com/discussion/868167/chinese-math-exam-question-has-the-internet-completely-puzzled-and-we-can-see-why | 1,571,343,767,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986676227.57/warc/CC-MAIN-20191017200101-20191017223601-00194.warc.gz | 570,268,706 | 12,161 | # Chinese math exam question has the internet completely puzzled — and we can see why
Rare | 1/31/2018 | Zuri Davis
Level 4
Click For Photo: https://coxrare.files.wordpress.com/2018/01/1-31-18-test.jpg?w=1200&h=627&crop=1
It’s a nightmare of a question for anyone who dreads math.
At least one test in China has everyone scratching their heads after seemingly asking students to solve a problem but clearly not giving them enough information, reports the BBC.
#### Ship - Sheep - Goats - Ship - Captain
“If a ship had 26 sheep and 10 goats onboard, how old is the ship’s captain?” read a confusing question, which was intended for 11-year-olds in the fifth grade.
#### Others - Discussion - Weight - Ship - Cargo
Others got sidetracked by a discussion on the weight of the theoretical ship’s imaginary cargo.
How on earth can 26 sheep and 10 goats can weigh 7700 kgs!!! Dey mean to say a sheep or goat weighs more den 200 kgs!!!
At least... | 240 | 949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-43 | latest | en | 0.923082 |
https://nedcod.com/recursive-algorithms-competitive-programming/ | 1,657,044,460,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104597905.85/warc/CC-MAIN-20220705174927-20220705204927-00641.warc.gz | 460,498,236 | 10,206 | Recursion often provides an elegant way to implement an algorithm. In this section, we discuss recursive algorithms that systematically go through candidate solutions to a problem. First, we focus on generating subsets and permutations and then discuss the more general backtracking technique.
• Generating Subsets
Our first application of recursion is generating all subsets of a set of n elements. For example, the subsets of {1, 2, 3} are ∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, and {1, 2, 3}.
The following recursive function search can be used to generate the subsets. The function maintains a vector
vector<int> subset;
that will contain the elements of each subset. The search begins when the function is called with parameter
void search(int k) {
if (k == n+1) {
// process subset
} else {
// include k in the subset
subset.push_back(k);
search(k+1);
subset.pop_back();
// don’t include k in the subset
search(k+1);
}
}
When the function search is called with parameter k, it decides whether to include the element k in the subset or not, and in both cases, then calls itself with parameter k +1. Then, if k = n +1, the function notices that all elements have been processed and a subset has been generated.
Illustrates the generation of subsets when n = 3. At each function call, either the upper branch (k is included in the subset) or the lower branch (k is not included in the subset) is chosen.
• Generating Permutations
Next we consider the problem of generating all permutations of a set of n elements. For example, the permutations of {1, 2, 3} are (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), and (3, 2, 1). Again, we can use recursion to perform the search. The following function search maintains a vector
vector<int> permutation;
that will contain each permutation, and an array bool chosen[n+1]; which indicates for each element if it has been included in the permutation. The search begins when the function is called without parameters.
void search() {
if (permutation.size() == n) {
// process permutation
} else {
for (int i = 1; i <= n; i++) {
if (chosen[i]) continue;
chosen[i] = true;
permutation.push_back(i);
search();
chosen[i] = false;
permutation.pop_back();
}
}
}
Each function call appends a new element to permutation and records that it has been included in chosen. If the size of permutation equals the size of the set, a permutation has been generated.
Note that the C++ standard library also has the function next_permutation that can be used to generate permutations. The function is given a permutation, and it produces the next permutation in lexicographic order.
The following code goes through the permutations of {1, 2, . . . , n}:
for (int i = 1; i <= n; i++) {
permutation.push_back(i);
}
do {
// process permutation
} while (next_permutation(permutation.begin(),
permutation.end()));
• Backtracking
A backtracking algorithm begins with an empty solution and extends the solution step by step. The search recursively goes through all different ways how a solution can be constructed.
As an example, consider the problem of calculating the number of ways n queens can be placed on an n × n chessboard so that no two queens attack each other. For example, shows the two possible solutions for n = 4.
The problem can be solved using backtracking by placing queens on the board row by row. More precisely, exactly one queen will be placed on each row so that no queen attacks any of the queens placed before. A solution has been found when all n queens have been placed on the board.
For example, shows some partial solutions generated by the backtracking algorithm when n = 4. At the bottom level, the three first configurations are illegal, because the queens attack each other. However, the fourth configuration is valid, and it can be extended to a complete solution by placing two more queens on the board.
There is only one way to place the two remaining queens.
The algorithm can be implemented as follows:
void search(int y) {
if (y == n) {
count++;
return;
}
for (int x = 0; x < n; x++) {
if (col[x] || diag1[x+y] || diag2[x-y+n-1]) continue;
col[x] = diag1[x+y] = diag2[x-y+n-1] = 1;
search(y+1);
col[x] = diag1[x+y] = diag2[x-y+n-1] = 0;
}
}
The search begins by calling search(0). The size of the board is n, and the code calculates the number of solutions to count. The code assumes that the rows and columns of the board are numbered from 0 to n − 1. When search is called with parameter y, it places a queen on row y and then calls itself with parameter y+1. Then, if y = n, a solution has been found, and the value of count is increased by one.
The array col keeps track of the columns that contain a queen, and the arrays diag1 and diag2 keep track of the diagonals. It is not allowed to add another queen to a column or diagonal that already contains a queen.
Shows the numbering of columns and diagonals of the 4 × 4 board. The above backtracking algorithm tells us that there are 92 ways to place 8 queens on the 8 × 8 board. When n increases, the search quickly becomes slow, because
the number of solutions grows exponentially. For example, it takes already about a minute on a modern computer to calculate that there are 14772512 ways to place 16 queens on the 16 × 16 board.
In fact, nobody knows an efficient way to count the number of queen combinations for larger values of n. Currently, the largest value of n for which the result is known is 27: there are 234907967154122528 combinations in this case. This was discovered in 2016 by a group of researchers who used a cluster of computers to calculate the
result. | 1,360 | 5,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2022-27 | latest | en | 0.824825 |
https://backtoweb.info/?post=1740 | 1,632,833,327,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060803.2/warc/CC-MAIN-20210928122846-20210928152846-00293.warc.gz | 177,526,697 | 5,223 | What number goes in 28 and 35-11-60c
Sequence of numbers
"Intelligence is what an intelligence test measures".
We all know these tests, which are supposed to measure our intelligence. Today we know that the intelligence of a person consists of many different components, which only together give a reasonably reliable statement about the intelligence. However, intelligence tests only ever measure a few of these components.
Test items that measure understanding of numbers are widespread. These are also very popular in application tests. It's often about that Continuation of a sequence of numbers.
Example: Determine the next member of the sequence3, 5, 7, ...
Not difficult, many will think: the next link is 9because this is the sequence of odd numbers.
However, this is not the only possibility, as it could also be a sequence of prime numbers, with the next being 11 would be (and not 9, since 9 is not a prime number).
If you think about it further, other possible sequelae could also be found, since a sequence of only 3 numbers cannot yet be determined clearly enough.
In this article a few tricks are to be shown how one can get on the track of such sequences of numbers and thereby maybe become a little more 'intelligent' in the next test.
But first we want to introduce and describe some well-known numerical sequences from mathematics.
1st sequence of natural numbers:
1, 2, 3, 4, .... next number: 5
2nd sequence of even numbers:
2, 4, 6, 8, .... next number: 10
3rd sequence of odd numbers:
1, 3, 5, 7, .... next number: 9
4th sequence of square numbers:
1, 4, 9, 16, ... next number: 25
5. Sequence of prime numbers:
1, 2, 3, 5, 7, ... next number: 11
6. Distance numbers:
1, 2, 4, 7, 11, 16, next number: 22
here the distances between the numbers increase by 1
7. Fibonacci numbers:
1, 1, 2, 3, 5, 8, ... next number: 13
In the famous Fibonacci sequence, the following numbers result from the
Sum of the two previous numbers. The interesting thing is that the quotient is
two consecutive numbers closer and closer to the value 1.61 .. which
the measure of the 'golden section' is used as a measured value in art and architecture
for a division corresponding to the ideal of beauty.
8. Triangle numbers:
1, 3, 6, 10, 15, ... next number: 21
The nth number results from the sum of the numbers from 1 to n.
That means the 5th number is 1 + 2 + 3 + 4 +5 = 15.
(These also correspond to the binomial coefficient n over 2)
9. Tetrahedral numbers:
1, 4, 10, 20, ... next number: 35
The nth number results from the sum of the first n triangular numbers
(These also correspond to the binomial coefficient n over 3)
10. Faculties:
1, 2, 6, 24, 120, ... next number: 720
The nth number results from the multiplication of the first n numbers
E.g. the 5th number is 1 x 2 x 3 x 4 x 5 = 120.
A algorithm to determine the next number in a sequence of numbers assumes that the numbers in the sequence are derived from the
basic mathematical arithmetic operations (+, -, *, :). In addition, at least 3 terms of the sequence must be known.
Step 1: Find the difference between two consecutive terms and write this under the two numbers
Repeat this step until all differences are equal.
Step 2: If all the differences are equal, add the differences determined for the last numbers to the last number of
Follow and get the next number. (I.e. we reverse the steps we performed in step one)
example: What is the name of the next link in the sequence: 6, 8, 12, 18, 26, 36
Find the differences: 2 4 6 8 10 Repeat this step
2 2 2 2 differences are constant
Find the next term: 36 + 10 + 2 = 48
If one does not obtain a constant sequence of differences in the way described above, one can try to obtain a constant sequence by dividing two consecutive terms:
Example: What is the name of the next link in the sequence: 2, 4, 12, 48, ...
We divide 2 3 4 now we apply subtraction:
1 1 we reverse the steps:
The next term is: 1 + 4 = 5, 5 * 48 = 240
The basic arithmetic operations alternate with some sequences of numbers. Example: 2, 5, 10, 13, 26, ...
Here 3 is alternately added and then multiplied by 2. Here you can apply the algorithm described above to every 2nd element.
But there are also other sequences of numbers for which the algorithm does not help. One such is, for example, the sequence of prime numbers shown above (No. 5). In the following of this type, however, only a certain basic mathematical knowledge helps to recognize relationships.
Nevertheless, most number sequences from so-called intelligence tests can be solved with the algorithm shown above. | 1,190 | 4,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2021-39 | latest | en | 0.961114 |
https://www.jwfsanctuary.club/variation-types/shapes/ | 1,719,061,962,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862396.85/warc/CC-MAIN-20240622112740-20240622142740-00756.warc.gz | 749,823,396 | 50,151 | Shapes
Variations that ignore the input and generate specific shapes. Often called “blurs” because they create a blur effect when used on the same transform with other variations.
blur
Circle with a bright center.
Type: 2D blur
blur_circle
Circle with even coloring.
Type: 2D blur
Author: Joel and Michael Faber
Date: 15 Mar 2009
Square blur followed by circlize. Same result as circleblur. An older version had a “hole” parameter which didn’t work correctly and was later removed.
http://sourceforge.net/p/apo-plugins/code/HEAD/tree/
https://sourceforge.net/projects/apo-plugins/files/apo-plugins/80810/
blur_heart
Creates a heart from two ellipses.
Type: 2D blur
Author: Luca G (dark-beam)
Date: 14 Nov 2011
Keep |a|≤|b| to keep the dimple below the zero point; if it goes below, the dimple will have some fill in it.
https://www.deviantart.com/dark-beam/art/Blur-heart-hearty-apoplugins-269015914
http://www.mathematische-basteleien.de/heart.htm
http://www.deviantart.com/art/BCs-BDs-Gnarly-Blur-Heart-Script-284934920
blur3D / pre_blur3D
Three dimensional Gaussian blur.
Type: 3D blur
chrysanthemum
Chrysanthemum flower shaped curve.
Type: 2D blur
Author: Jesus Sosa
Date: 1 Feb 2018
http://paulbourke.net/geometry/chrysanthemum/
circleblur
Circle with even coloring.
Type: 2D blur
Author: Anton Liasotskiy (zy0rg)
Date: 9 Jan 2013
http://sourceforge.net/p/apo-plugins/code/HEAD/tree/
https://sourceforge.net/projects/apo-plugins/files/apo-plugins/80810/
gaussian_blur
Fuzzy circle with a bright center, made from a Gaussian distribution.
Type: 2D blur
nBlur
Polygon shaped blur
Type: 2D blur
Author: FractalDesire
Date: 21 Dec 2010
http://zweezwyy.deviantart.com/art/nBlur-a-useful-tool-207495126
https://www.deviantart.com/fractaldesire/art/nBlur-plugin-190401515
pie
A circle with wedges missing, like pieces of pie.
Type: 2D blur
Author: Joel Faber
Date: 16 Sep 2007
http://sourceforge.net/projects/apo-plugins/files/apo-plugins/80810/
http://joelfaber.deviantart.com/art/A-Recipe-for-Plastic-Pie-35295850
pie_fl
Fluid version of pie, allows fractional value for slices.
Type: 2D blur
Author: Fred E (morphapoph)
Date: 8 Sep 2010
Same as pie, but relaxes the restriction that slices is an integer.
https://www.deviantart.com/morphapoph/art/Apo-Anim-friendly-Plugins-178559281
pie3D
Three dimensional version of pie.
Type: 3D blur
Author: Andreas Maschke (thargor6)
Date: 21 Nov 2011
pre_blur
Pre version of gaussian_blur.
Type: 2D blur
primitives_wf
Blur with selectable two or three dimensional shape.
Type: 2D or 3D blur (depends on shape parameter)
Author: Andreas Maschke (thargor6)
Date: 12 Oct 2013
sineblur
A circle with a shading effect.
Type: 2D blur
Author: Anton Liasotskiy (zy0rg)
Date: 9 Jan 2013
The appearance is greatly influnced by the background; the default of 1 works well for dark backgrounds, but a somewhat higher value (like 20) is better for light backgrounds. Very high values will have low density in the center. Values less than 1 will produce a smaller circle with a fuzzy edge.
https://www.deviantart.com/zy0rg/art/Blur-Package-347648919
sphereblur
3D version of sineblur
Type: 3D blur
Author: Anton Liasotskiy (zy0rg)
Date: 9 Jan 2013
http://zy0rg.deviantart.com/art/Sphereblur-687844594
square
Square shaped blur
Type: 2D blur
Author: Antonio Intieri (gygrazok)
http://lu-kout.deviantart.com/art/Apophysis-Plugin-Pack-1-v0-4-59907275
Cube shaped blur
Type: 3D blur
starblur
Star shaped blur
Type: 2D blur
Author: Anton Liasotskiy (zy0rg)
Date: 9 Jan 2013
Negative values are allowed for both parameters, and can generate interesting shapes.
https://www.deviantart.com/zy0rg/art/Blur-Package-347648919
superShape3d
General 3D shape generator using two superformula instances.
Type: 3D blur
Name: David Young (Sc0t0ma)
Date: 19 Oct 2008
https://www.deviantart.com/sc0t0ma/art/SuperShape3d-101195889
http://paulbourke.net/geometry/supershape/
triangle
Triangular blur (3D)
Type: 3D blur
Author: Jesus Sosa
Date: 7 May 2020
waveblur_wf
Creates waves, like ripples in a pond.
Type: 3D blur
Author: Andreas Maschke
Date: 30 Jan 2015
xheart_blur_wf
Heart shaped blur
Type: 2D blur
Author: Andreas Maschke
Date: 30 Aug 2014
Based on xheart (which is not a blur).
zblur
Gaussian blur for the z axis only (no effect on x or y).
Type: 3D blur
This information has all been created by Rick Sidwell as a guide to the more popular variations used in fractal flames, and very generously allowed me to reproduce it here. Not all of the variations are included with JWildfire, but a great many are, so it is worthwhile learning about them.
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# Round Function
#### Joe Caverly
While TCC does not have a round function, the PSHELL command and/or function can be used to round a number;
Code:
``````echo %@pshell[[math]::Round(22/7)]
3``````
..or...
Code:
``````pshell /s "\$a=22/7"
echo %@pshell[\$a]
3.14285714285714
echo %@pshell[[math]::Round(\$a)]
3``````
Joe
The .NET [math]::Round Method offers additional conventions. Take a look at the Microsoft documentation for more, including;
Rounding Away From Zero
Round to the nearest, or banker's rounding
...and other examples.
Joe
For rounding to the nearest integer the easiest way is to add 0.5 and modulus divide by 1. So, for instance:
set myval=3.49999
echo %@eval[(%myval+.5)\1]
TCC also has the @CEILING and @FLOOR functions.
This also works:
echo %@eval[3.14=0]
3
echo %@eval[3.57=0]
4
This also works:
echo %@eval[3.14=0]
3
echo %@eval[3.57=0]
4
Good point!
That probably should be mentioned expressly in the Help. When I first saw the post I thought "That looks like a programming error--he's comparing 3.14 to 0?? In a math function, not a logic function?"
That probably should be mentioned expressly in the Help.
I see it documented at the bottom of the @EVAL help page.... the part that starts with "if displayformat is i.a, then..."
I see it documented at the bottom of the @EVAL help page.... the part that starts with "if displayformat is i.a, then..."
I did see that--but it's not clear that setting it to 0 will cause a simple "round to nearest integer".
I doubt that most people would think "Why not use @eval and set the precision to 0?" Like the original poster, most people would be thinking "Where is the @ROUND[] function? Maybe TCC uses a different name?"
I did see that--but it's not clear that setting it to 0 will cause a simple "round to nearest integer".
I doubt that most people would think "Why not use @eval and set the precision to 0?" Like the original poster, most people would be thinking "Where is the @ROUND[] function? Maybe TCC uses a different name?"
Yeah? Well *I* thought of it. Maybe it's because I'm an ancient FORTRAN programmer. :-)
Yeah? Well *I* thought of it. Maybe it's because I'm an ancient FORTRAN programmer. :-)
That seems more intuitive to me than launching an instance of PowerShell. Which is kind of like mounting a military invasion of Russia to get a cup of tea.
[FOX] Ultimate Translator
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SMKMERBAUMIRISARAWAK
PART1:MASTERINGTHEDEFINITION
PHYSICSTERM
Physicalquantity
Basequantity
Derivedquantity
Scalarquantity
Vectorquantity
Distance
Displacement
Speed
DEFINITION
FORM4CHAPTER1INTRODUCTIONTOPHYSICS
Isaquantitythatcanbemeasured.
Isaquantitythatcannotbedefinedinothertermsofphysicalquantities.
EXAMPLE/QUESTION
Isaquantitythatcanbedefinedinothertermsofphysicalquantitiesbyeithermultiplication
ordivisionorboth.
Isaquantitywithmagnitudeonly.
Example:distance,speed,time,mass,temperature
Isaquantitywithbothmagnitudeanddirection.
Example:displacement,velocity,acceleration,force
FORM4CHAPTER2FORCESANDMOTION
Isthetotallengthbetweentwopoints.
Measuretheactuallength.
Isthetotallengthbetweentwopointsatcertaindirection.
Measuretheshortcutlength
Distance
Istherateofchangeofdistance.
Speed=
Time taken
Unitofspeedisms1.
Velocity
Istherateofchangeofdisplacement.
Velocity,v=
Displacement
Time taken
Unitofvelocityisms1.
Acceleration
Istherateofchangeofvelocity.
Acceleration,a=
Change of velocity
Time taken
Unitofaccelerationisms2.
Anda=
Inertia
Momentum
Principleofconservationof
momentum
Impulse
v-u
No formula as it is not a physical quantity that can be
measured.
inertia.
Momentum,p=mvm=mass,v=velocity
Unitofmomentumiskgms1
Statesthatinclosedsystem,totalmomentumbeforecollisionisequaltothetotal Elastic:m1u1+m2u2=m1v1+m1v2
momentumafterthecollisionsuchthatthetotalmomentumisconserved.
Inelastic:m1u1+m2u2=(m1+m1)v
Explosion:0=m1v1+m1v2
Isthechangeofmomentum.
Impulse=FtF=force,t=time
Impulse=mvmu
Unitofimpulseiskgms1
Isthetendencyofobjecttoresistthesuddenchangeactingonthesystem
OR
Isthetendencyofobjecttoremainatrestifrestorcontinuetomovewithuniform
velocityinstraightlineifmovingunlessexternalforceactingonit.
Istheproductofmassandvelocity.
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PHYSICSTERM
Impulsiveforce
DEFINITION
Istherateofchangeofmomentum.
EXAMPLE/QUESTION
Impulsiveforce,F=
mv - mu
Unitofimpulsiveforceiskgms2
Shorterimpacttime,biggerimpulsiveforce.
Force
Istheproductofmassandacceleration.
Force,F=mam=mass
a=acceleration
Unitofforceiskgms2orNewtonorN
Newtonsfirstlawofmotion Statesthatabodycontinuesinstateofrestoruniformvelocityinstraightlineunless Alwaysreferredtoinertiacondition!!
acteduponbyanexternalforce.
Newtonssecondlawof
States that the rate of change of momentum of moving object is directly Alwaysreferredtocollisionsystems!!
motion
proportionaltoandinthesamedirectionastheforceactingonit.
Newtonsthirdlawof
States that if one body exerts a force on another, there is an equal but opposite Alwaysreferredtoactionandreaction!!
motion
forcecalledreactionexertedonthefirstbodybythesecond.
(Ihitballaction)
(Theballcausesmyhandpainreaction)
Energy
Istheabilityofdoingwork.
Workdone
Istheproductofforceanddisplacementwhichisparalleltothedisplacementofthe Workdone,W=Fs[noangle]
object.
Workdone,W=Fscos[ifangle]
F=force
s=displacement
UnitofworkdoneisJouleorJ
Potentialenergy
Istheenergypossessedbytheobjectduetoitspositionorlocation.
Potentialenergy,Ep=mgh
m=mass
g=gravityvalue=10ms2
h=height
UnitofpotentialenergyisJoule
Kineticenergy
Istheenergypossessedbytheobjectduetoitsconditionofmovement.
Kineticenergy,Ek=mv2
m=mass
v=velocity
UnitofkineticenergyisJoule
Principleofconservationof Statesthatinaclosedsystem,theenergycannotbecreatedordestroyedbutitcan
energy
be changed from one form to another form that is the total energy is being
conserved.
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PHYSICSTERM
Power
DEFINITION
EXAMPLE/QUESTION
Istherateofworkdone.
Power,P=
WorkDone
Time
OR
Power,P=
Energy
Time
UnitofpowerisWattorW
Efficiency
Istheratioofoutputworkdonetotheinputenergysuppliedbythesystem.
Hookeslaw
Statesthattheextensionofspringisdirectlyproportionaltotheforceactingonit Force,F=kx
suchthattheelasticlimitisnotexceeded.
k=forceconstant
x=extensionofspring
Fistotheweightofobjectwhereweight=massx10
Is the ability of an object to resume to its original state once the applied force is
removed.
Istheforceperunitlengthofextension.
k=F/x
UnitofspringconstantisNm1
FORM4CHAPTER3FORCESANDPRESSURE
Istheforceactingnormallytothesurfaceperunitarea.
Pressure,p=F/A[ifsolid]
F=force
A=area
UnitofpressureisNm2orPascalorPa
OR
Pressure,p=hg[ifliquid]
h=depth
=densityofliquid
g=gravityvalue=10ms2
UnitofpressureisNm2orPascalorPa
OR
Pressure,p=76cmHg+unbalance[mercury]
Isthepressureexertedbytheatmosphereonthesurfaceoftheearth.
Elasticity
Springconstant
Pressure
Atmosphericpressure
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Efficiency,e=
OutputWork
x100%
InputEnergy
PHYSICSTERM
Pascalsprinciple
DEFINITION
States that in enclosed system, the applied pressure will be transmitted equally to
everypartofthefluidandalsotothewallofthecontainer.
EXAMPLE/QUESTION
F1 F2
=
A1 A2
AndA1d1=A2d2
A=area
d=distancemoved
Achimedesprinciple
Statesthatwhenanobjectisimmersedpartiallyorwhollyintoafluid,theweightof Buoyantforce,F=Vg
water being displaced due buoyant force is equal to the weight of object being =densityofliquid
immersed.
V=Immersedvolumeoftheobject
g=gravityvalue=10ms2
UnitofbuoyantforceisNewtonorN
Bernoullisprinciple
Statesthataregionwhereexperienceshighairspeedwillhaslowairpressureand Highspeedlowpressure
viceversa.
Lowspeedhighpressure
FORM4CHAPTER4HEAT
Thermalequilibrium
Isaconditionwheretwoobjectsincontacthavethesametemperatureandthereis
nonettransferofheatbetweentwoobjects.
Heat
Isaformofenergy.
Temperature
Isthedegreeofhotnessofanobject.
Specificheatcapacity
Istheamountofheatrequiredtoincreasethetemperatureof1kgobjectby1C Heat,H=mc
withoutchangeinphysicalstate.
m=mass,c=specificheatcapacity,=riseintemperature
UnitofheatisJouleorJ
Lowspecificheatcapacityfastergettinghot
Highspecificheatcapacityslowergettinghot
Waterisagoodcoolingagentasithashighspecificheat
capacity
Meltingpoint
Isthemaximumtemperaturepointthatcanbesustainedbytheobjectbeforethe Highmeltingpointcanwithstandhightemperature
objectstartstomelt.
beforeitgetsmelt
Boilingpoint
Isthemaximumtemperaturepointthatcanbesustainedbytheobjectbeforethe
objectstartstoboil.
Specificlatentheatoffusion Is the amount of heat required to change the 1 kg object physically from solid to Heat,H=mLf
liquidwithoutthechangeintemperature.
m=mass
Lf=specificlatentheatoffusion
Specificlatentheatof
vapourisation
UnitofheatisJouleorJ
Is the amount of heat required to change the 1 kg object physically from liquid to Heat,H=mLv
steamwithoutthechangeintemperature.
m=mass
Lv=specificlatentheatofvapourisation
UnitofheatisJouleorJ
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PHYSICSTERM
Boyleslaw
Charleslaw
DEFINITION
Statesthatforafixedmassofgas,thepressureofgasisinverselyproportionaltoits
volumesuchthatthetemperatureofgasiskeptconstant.
Statesthatforafixedmassofgas,thevolumeofgasisdirectlyproportionaltoits
absolutetemperaturesuchthatthepressureofgasiskeptconstant.
EXAMPLE/QUESTION
P1V1=P2V2
P=pressure,V=volume
V1 V 2
=
T1 T2
V=volume,T=temperaturemustbeinunitKelvin
T=(+273)K
Pressurelaw
Snellslaw
Mirror
Lens
Refractiveindex
Statesthatforafixedmassofgas,thepressureofgasisdirectlyproportionaltoits P1 P2
=
absolutetemperaturesuchthatthevolumeofgasiskeptconstant.
T1 T2
P=pressure,T=temperaturemustbeinunitKelvin
T=(+273)K
FORM4CHAPTER5LIGHT
Statesthattheangleofincidence,angleofreflectionandthenormaltothesurface
all lie in the same plane such that the angle of incidence is equal to the angle of
reflection.
Isanobjectwithonlyonesideoffocus/viewwhereitwillreflecttheincidentray.
Isanobjectwithtwosidesoffocus/viewwhereitallowstherefractionoflight.
sin i
Istheratioofsineofincidenceangletothesineofrefractedangle.
Refractiveindex,n=
sin r
imustbeinairandrinmedium
Apparentdepth
Realdepth
Isthedistanceofthevirtualimagefromthesurfaceofthewater.
Isthedistanceoftherealobjectfromthesurfaceofthewater.
Criticalangle
1
Is defined as the angle of incidence in the denser medium when the angle of
Refractiveindex,n=
refractioninthelessdensemediumis90.
sin c
c=criticalangle
Statesthatwhentheangleofincidenceisfurtherincreasesothatitisgreaterthan Example:mirage,opticalfibre
thecriticalangle,thenthelightisnolongerrefractedbutitisreflectedinternally.
Isapointwherealltherayswillfocusat.
IsthedistancebetweenthecentreofthelenswiththeprincipleF.
Isthereciprocalofthefocallengthofalens.
Poweroflens,P=1/f
f=focallength(mustconverttounitmetre)
UnitofpoweroflensisDioptreorD
Formulaoflens: 1 + 1 = 1
Totalinternalreflection
Focalpoint
Focallength,f
Poweroflens
Refractiveindex,n=
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Real Depth
Apparent Depth
u=objectdistance,v=imagedistance,f=focallength
Imagemagnification,m=v/u
PHYSICSTERM
Wave
Transversewave
DEFINITION
EXAMPLE/QUESTION
FORM5CHAPTER1WAVES
Isthevibrationoroscillationofparticlewhichtransfersenergywithouttransferring
matterandthesevibrationsarerepeatedperiodically.
Is a wave where the direction of vibrations of particles is perpendicular to the Anytypesofwavesexceptsoundwave!!
propagationofwave
Forexample:electromagneticwave,light,waterwave
#producecrestsandtroughs
Longitudinalwave
Isawavewherethedirectionofvibrationsofparticlesisparalleltothepropagation Soundwaveonly
ofwave
#Produceaseriesofcompressionsandrarefactions
Dampedoscillations
Isanoscillationwhereitsamplitudedecreaseswithtimebutthefrequencyremains
constantandthisvibrationwillcometoastop.
Resonance
Is the vibration where is forced frequency is equal to the natural frequency of the
object.
Asoundwhereitdependsonitsamplitude
Asoundwhereitdependsonitsfrequency
Refraction,diffraction,reflectionandinterference
Refractionpassesthrough/seethroughthemedium
frequencyconstant
decreaseinspeed,wavelength(deeptoshalloworlessdensetodenser)
Loudnessofsound
Pitchofsound
Phenomenonofwave
Refraction
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Example:Bridgefallsbywindblow
PHYSICSTERM
Diffraction
DEFINITION
Diffractionpassesthroughthegaporhole
frequency,wavelengthandspeedconstant
changeinshapeonlydependsonsizeofgap
EXAMPLE/QUESTION
Wavelengthissmallerthansizeofgap
diffractioneffectislessobvious
strongerenergyofwaveenteringthegap
Wavelengthisbiggerthansizeofgap
diffractioneffectismoreobvious
lesserenergyofwaveenteringthegap
Reflection
Interference
Diffractiongoandreboundedbyshinnysurfaceorreflector
frequency,wavelengthandspeedconstant
changeindirectionofmovingonly
Interferenceresultantofallwaves
AntinodeAlinejoiningalltheconstructivepoints
NodeAlinejoiningallthedestructivepoints
Wavelength,=
ax
D
a=sizeofgap/distancebetweensource
D=distancebetweenthescreenwithgaps
Monochromaticlight
Isonewavelengthoronecolouroflight
Coherent
Samefrequencyandsamephase
Electromagneticspectrum
andfrequencies.
andkillscancercells
Xray(Scanning),ultraviolet(detectforgenote),
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PHYSICSTERM
Charge
DEFINITION
FORM5CHAPTER2ELECTRICITY
Isthetotalamountofelectronsflow
EXAMPLE/QUESTION
Charge,Q=nen=numberofelectrons,e=1.67x1019C
Charge,Q=ItI=current,t=timeinseconds
UnitforchargeiscoulomborC
I=Q/t
UnitforcurrentisampereorA
##Lowcurrentifhighresistance!!
Higherresistanceproducemoreheat
morevoltage
lowercurrent
Unitforresistanceisohmor
Current
Istherateofelectronflow
Resistance
Istheoppositionofcurrentflow.
Note:Resistance,R=L/A
Higherresistancehigherresistivity,longerwirelengthbutthinnerwire
Coppergoodconductorofelectricity
Aluminiumgoodaselectriccableascheapandlowrustingrate
Tungstengoodasfilamentbulbasitcanionizeeasilytogiveoutray
Copperveryhighresistanceandthusitisgoodasheatingelement
Is the work done when one coulomb of charge passes from one point to another Voltage,V=IR
point
##Morevoltageifhighresistance!!
##Highvoltagedoesnotmeanhighcurrent!!
##Highvoltagecansayhighresistance!!
States that the potential difference across an ohmic conductor is directly OhmslawmeansV=IR
proportional to its current flow such that the temperature and other physical
quantityarekeptconstant!
##AnytypesofwiresobeyOhmslaw
##Bulbandheatingmaterialsdoesnotobeyohmslaw
Voltage
Potentialdifference
Ohmslaw
Electricalenergy
Totalworkdonetomoveonecoulombofchargeinonesecond
Electricpower
Istherateofelectricenergy
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Electricalenergy,E=VIttinsecond
InunitJouleorJ
Power,P=IV=V2/R=I2/R
InunitWattorW
PHYSICSTERM
240V,300W
Fuse
Electromagnet
Righthandgriprule
Flemingslefthandrule
Flemingsrighthandrule
Lenzslaw
Inducedcurrent
Thermionicemission
Cathoderayoscilloscope
DEFINITION
EXAMPLE/QUESTION
Means the device is supplied with potential difference of 240 V and releases the 240V,300WV=240
powerat300Joulesinonesecond
P=300
FromP=IV
300=I(240)
I=1.25A
Isanelectriccomponentinstalledinelectricsockettosurgethepowerdownonce Fusemustcangethoteasily
canbeburnteasily
lowmeltingpoint
highresistanceandproducesheatfast
FORM5CHAPTER3ELECTROMAGNETISM
Istheflowofelectricaroundthecoilofwirewhichproducemagnet
Statethatforacurrentcarryingconductor,thethumbwillpointtothedirectionof
currentflowwhereastherestoffingerwillpointtothemagneticdirection
Statethatwhenthethumb,forefingerandmiddlefingerareextendedattheright Forelectricmotorwhichcarriescurrent
angletoeachothersforsystemwhichcarriescurrent,thenthethumbwillshowthe Thumbdirectionofforce
magnetic force, the forefinger will point to the direction of magnetic field and the Forefingerdirectionofmagnet(fromnorthtosouth)
middlefingerwillpointtothecurrentdirection.
Middlefingercurrentdirection
Statethatwhenthethumb,forefingerandmiddlefingerareextendedattheright Forelectricdynamowhichgenerateelectriccurrent
angle to each others for system generating current, then the thumb will show the Thumbdirectionofforce
magnetic force, the forefinger will point to the direction of magnetic field and the Forefingerdirectionofmagnet(fromnorthtosouth)
middlefingerwillpointtothecurrentdirection.
Middlefingercurrentdirection
States that the direction of induced current is always opposing to the direction determinethedirectionofinducedcurrent
whichproducesthecurrent.
determinethemagneticpole(northorsouth)
Statesthatthemagnitudeofinducedcurrentisalwaysdirectlyproportionaltothe determinethemagnitudeofinducedcurrent
rateofmagneticfluxbeingcutbymovingmagnetinsolenoid
Is the produce of current by movement of magnet bar in solenoid and there is no
physicalcontactbetweenthem
FORM5CHAPTER4ELECTRONICS
Istheprocessofreleasingelectronsfromtheheatedcathode
Moreelectronscanbereleasedif
cathodeisheatedathightemperature
moresurfaceareaisexposedtoheat
typeofmaterialwhichcanproducemoreelectrons
Is a device used to display waveform, measure short time intervals or to measure
thepotentialdifference
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PHYSICSTERM
Rectification
Semiconductor
Logicgate
Spontaneous
Random
Halflife
Whatisalpha
Whatisbeta
Whatisgamma
DEFINITION
EXAMPLE/QUESTION
Istheprocessofchangingalternatingcurrenttodirectcurrent
Usediode
Isamaterialwhichhastheconductivitybetweentheconductorandinsulatorandits ptypesemiconductorsilicondopedwithtrivalent
conductivitycanbeimprovedbyincreasingitstemperature.
majoritychargecarrierofhole
ntypesemiconductorsilicondopedwithpentavalent
majoritychargecarrierwhichis
calledelectron
Isgatewhichhasoneormorethanoneinputsbutwithonlyoneoutputandlogic
gatesarereferredtoswitch
Isunstablesubstancewhichhassameprotonnumberbutdifferentnucleonnumber
Is the spontaneous and random disintegration of unstable substance to become
Meanshappenautomaticallybyitselfwithouttriggeredbyanyexternalsourcelike
temperatureorpressure
whenitwillhappen
Alphaisheliumparticle
Range0fewcm(canbestoppedbypaper)
highionizingpower(canchangethestructureof)
positiveheavychargewithsmalldeflectiontoward
negativeplate
deflectupfromthemagneticfield
lowpenetratingpower
movesstraightin
Betaisafastmovingelectronbeam
Range0fewm(canbestoppedbyaluminiumfoil)
mediumionizingpowerandpenetratingpower
negativelightchargewithbigdeflectiontoward
positiveplate
deflectdownfromthemagneticfield
Gammaisanenergeticelectromagneticray
Range0fewhundredm
lowionizingpower
veryhighpenetratingpower(killthe.)
nochargeandnodeflectionneitherinelectricfield
normagneticfield
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10
PHYSICSTERM
Nuclearfusion
Nuclearfission
Chainreaction
DEFINITION
EXAMPLE/QUESTION
is the combining of two lighter nuclei to form a heavier nucleus, releasing a vast
amountofenergyduringtheprocess
isthesplittingofaheavynucleusintotwolighternucleiafterthenucleusofanatom 1 n 235U 91Kr 142Ba 3 1 n Energy
0
92
36
56
0
is bombarded with a neutron with the release of a large amount of energy during
theprocess.
isaselfsustainingreactioninwhichtheproductsofareactioncaninitiateanother Theprocessisexpandingandnonstopasthenumberof
similar reaction. For instance, as uranium atoms continue to split, a significant neutronskeepsonmultiplyingwithtime
amountofenergyisreleasedduringeachreaction.Theheatreleasedisharnessed
andusedtogenerateelectricalenergy.
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11
PART2:MASTERINGTHEPROBLEMSOLVING[EXPLAININGTHEPROCESS]
1
Diagram1showsthephenomenonofseabreeze.
1st:Betaisusedforpaperdetectionasithasmediumpenetratingpoweranditspenetration
powervarieswiththethicknessofpaper.
2nd:Betaislocatedunderneaththepaper.Theratemeter(GeigerMullertube)whichisconnected
tothesignalgeneratorislocatedabovethepaper.
penetratedthroughthepaper.
betacanpenetratethroughit.So,thecompressorwillbetriggeredtostopcompressingthe
paper.
betacannotpenetratethroughit.So,thecompressorwillbetriggeredtocompressthepaper
Diagram1
Using the concept of specific heat capacity, explain how the phenomenon of sea
breezeoccurs.
Seabreeze
1st:Seabreezehappensduringdaytime.
2nd:Thelandhaslowerspecificheatcapacity
thanthesea.
3rd:Thelandisfastergettinghotthanthe
seawhenexposedtosunlight.
4th:Theairmoleculesatlandarefaster
heatedup.
5th:Thehotairmoleculesatlandwillriseup
andreplacedwithcoolairfromthesea.
6th:Movementofcoolairmoleculesfromthe
seatothelandiscalledtheseabreeze.
Landbreeze[Extranote]
1st:Landbreezehappensduringnighttime.
2nd:Thelandhaslowerspecificheatcapacity
thanthesea.
3rd:Theseaisslowergettingcoldthanthe
seaatnightime.
4th:Theairmoleculesatseaareslowerbeing
cooleddown.
5th:Thehotairmoleculesatseawillriseup
andreplacedwithcoolairfromtheland.
6th:Movementofcoolairmoleculesfromthe
landtotheseaiscalledthelandbreeze.
Diagram2showsarelayusedinanelectricalcircuit.
Diagram2
Explaintheworkingprincipleofrelayswitch.
Diagram4showsaBunsenburner.
Diagram4
ExplainhowtheBunsenburnercanproduceasmallblueflameorbigyellowreddish
flame.
Smallblueflame
Whentheairholeisopened,thereisahigh
airspeedflowbetweentheinnerofBunsen
burnerwiththesurroundingair.
AccordingtoBernoullisprinciple,highair
speedwillresultinlowpressureinsidethe
Bunsenburner.
Thislowairpressurecannotpushupthe
flamehighandcausesasmallflame
Theflameisbluebecausethereisa
completecombustionastheairholeis
opened
1st:Theworkingprincipleofrelayswitchisbasedontheelectromagnetconcept.
2nd:Therelayswitchhasthesoftironcoreinsidewiththecoilofwire.
3rd:Whenasmallcurrentflowsintotherelayswitch,thesystemismagnetizedandproduces
magneticforce.
4th:Themagneticforcewillpulltheswitchandcompletethecircuit.
5th:Whenthecurrenttotherelayswitchiscutoff,thesystemisdemagnetizedandtheswitchis
released.
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Explain the arrangement of the paper thickness detector apparatus and state how
Explaintheprocessofdetectionofpaperthicknessinfactorywithuseofonenamed
12
Bigyellowreddishflame
Whentheairholeisclosed,thereisnoair
speedflowbetweentheinnerofBunsen
burnerwiththesurroundingair.
AccordingtoBernoullisprinciple,lowair
speedwillresultinhighpressureinsidethe
Bunsenburner.
Thishighairpressurecanpushuptheflame
highandresultinbigflame
Theflameisyellowreddishbecausethe
combustionisnotcompleteastheairholeis
closed.
Youaregiventwocoils,P andQ,with100turnsand500turnsrespectively,asolid
coreanda240Va.c.powersupply.Usingallthematerialssuppliedandwiththeaid
oflabeleddiagram,showshowyoucanbuildasimplestepdowntransformer.
Diagram8.1andDiagram8.2showsanexperimenttostudytherelationshipbetween
thepressureandvolumeofairtrappedinanairtightcontainer.Thepistonsforboth
diagramsarepusheddownslowly.
AsimplestepdowntransformercanbeconstructedusingasoftUshapeironcorewithnumber
ofinputcoilsismorethantheoutputcorewhichisusingana.c.inputvoltageasshownbelow:
Remember:Transformermustuseinputa.cvoltage
Diagram8.1Diagram8.2
Theexperimentaboveisusuallyapplicableifagasexpandsorcompressedslowly.
Whyisthisso?
Explain how you would go to escape from being chased by a bull based on one
concept.
1st:TheBoyleslawisapplicableifthetemperatureofthegasisconstant.
2nd:Thegasmustbeexpandedorcompressedslowlyastoreducethecollisionbetweenthe
moleculesofgas.
3rd:Thecollisionofmoleculesincreasethefrictionofmoleculesbetweenthemandthisfriction
willproduceheat.
1st:Iwillperformmyruninzigzagdirectionswithnodefinitedirectionofrun.
2nd:Thebullhasbiggermassifcomparedtome.
4th:Duetoinertia,thebullishardertochangeitsdirectionandwilllosecontrolandfall
Diagram7.1and7.2 showtwoidenticalblockA andblockB hangingbystringX and
string Z. In Diagram 7.1, the string W is given by an increasing gradually of pulling
force.InDiagram7.2,thestringYisgivenbyasharppullandfast.
1st:reducespowerlossduringtransmission
2nd:thesupplyofelectricityismorestableandreliable
3rd:electricitycanbedistributedtodifferentusersaccordingtothevoltagerequirement
4th:maintenanceandrepairworkofpowerstations,cableandpylonscanbedoneatanytime
Diagram7.1:PulledgentlyDiagram7.2:Pulledveryfast
(a) InDiagram7.1,explainwhatwillhappenifthestringW isgivenbyanincreasing
10
(b) InDiagram7.2,explainwhatwillhappenifthestringWisgivenbyasharppull
andfast.
IfthestringWisgivenbyasharppullandfast,thestringYitselfwillcrack.
Reason:
IfthestringYisgivenbyasharppullandfast,thestringYwillpossessaninertiawhichwill
tendtoresistthesuddenpullexertingontoit.Asaresult,stringYwillcrack.
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Explainwhyaspringiselastic?
1st:Undernormalcircumstances,thespringhasitsattractiveandrepulsiveforcesbetween
neighbouringatomswhicharebalancingoneanother.
2nd:Whenthespringisstretched,theattractiveforcesbetweenneighbouringatomsactto
againsttheforcethattriestoseparatethem.
3rd:Whenthespringiscompressed,therepulsiveforcesbetweenneighbouringatomsactto
resisttheforcethattriestobringthemtogether.
4th:Itisthecombinedactionoftheattractiveandrepulsiveforcesthatenablethespringto
resumetoitsoriginalshapeandsizewhentheexternalforcethatactsonitisremoved.
Reason:
ThepullingforcewillbetransferredfromstringWtothestringX.
StringXhastosupportthepullingforceandalsohastosupporttheweightofblockAat
thesametime.So,stringXwillcrack.
Largegeneratorsareusedtogeneratetheelectricitythatwillbeusedinthecountry.
ElectricityisdistributedthroughoutthecountrybytheNationalGridNetwork.
totheconsumers.
13
11 A small amount of impurities are normally added into the pure crystal of
semiconductorastoimprovetheconductivityofsemiconductor.
(a) Withaidofdiagram,explainhowyougotoproduceaptypesemiconductor.
12
1st:Asemiconductorlikessiliconhasfourvalenceelectrons.
2nd:Toproduceaptypesemiconductor,itmustbedopedwithelementwhichhasthree
valenceelectrons.
3rd:Elementwithindiumatomisdopedintothesiliconandproducesanemptyspacein
indiumasshownbelow,
Diagram12 showsthepatternofseawaveswhenapproachingthebeach.
Diagram12
ExplainintermsofthewavephenomenainDiagram12,whythewaterwavesfollow
theshapeofthebeachasitapproachestheshore.
1st:Whenthewavesrefractfromdeepseatoshallowersea,boththewavelengthandenergy
decrease.
2nd:Therefore,itbecomesweakerandfollowstheshapeofthebeach.
13
4th:Thisemptyspaceiscalledtheholeandbecomethemajoritychargecarriertothe
semiconductor.
5th:Asaresult,thesemiconductorbecomesptypewithmajoritychargecarrierwhichis
calledthehole.
Diagramshowsasoundwaveproducedbyvibrationofatuningfork.Thesoundwave
travelsinair.
WiththehelpofDiagram13,explainhowthesoundwaveisproduced.
1st:Whenatuningforkvibrates,airmoleculeswillvibrate.
2nd:Whenthetuningforkmovesforwards,theairiscompressed.
3rd:Whenthetuningforkmovesbackwards,theairlayersarepulledapartand
causetherarefaction.
4th:Therefore,aseriesofcompressionandrarefactionswillproducesound.
5th:Thesoundenergyispropagatedthroughtheairarounditintheformofwaves.
11 (b) Withaidofdiagram,explainhowyougotoproducentypesemiconductor.
1st:Asemiconductorlikessiliconhasfourvalenceelectrons.
2nd:Toproduceantypesemiconductor,itmustbedopedwithelementwhichhasfive
valenceelectrons.
3rd:Elementwitharsenicatomisdopedintothesiliconandproducesanextraelectron
aroundthearsenicasshownbelow,
14
Diagramshowsatransformer.Atransformerisoperatedbasedontheprincipleof
electromagneticinduction.
Explaintheworkingprincipleoftransformer.
1st:Theworkingprincipleoftransformerisbasedonelectromagneticinductionwiththeinput
voltagemustbealternatingcurrent.
2nd:Whenanalternatinginputcurrentflowsinprimarycoil,itinducesamagneticfluxaround
thecoil.
3rd:Thisinducedmagneticfluxwillbeinducedtothesecondarycoil.
4th:Themagnitudeofe.m.f.inducedtothesecondarycoilsdependsonthenumberofsecondary
coilsbesidesthecoreislaminatedorsoftornot.
5th:Thisinducede.m.fwillproduceaninducedvoltageandalsoinducedcurrenttothebulbto
lighton.
4th:Thisextraelectronwillmovefreelyandbecomethemajoritychargecarriertothe
semiconductor.
5th:Asaresult,thesemiconductorbecomesntypewithmajoritychargecarrierwhichis
calledtheelectron.
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14
1
Diagram1.1showsanverniercallipers.
10 11 12
Diagram 2.1 shows a student taking the reading of thermometer at three
differentpositions,P,QandR.
13 14 15
0 1 2 3 4 5 6 7 8 9 10
Mainscale
Vernierscale
Diagram1.1
Namethephysicalquantitybeingmeasuredbyverniercalipers.
(a)
(b)
(i)
NamethepartlabeledZ.
(ii)
StatethefunctionofpartZ.
(c)
Whatisthesensitivityoftheverniercalipers?
(d)
(a)
(b)
Temperature
Diameter
Tail
Vectorquantity
Whatisthesmallestscaledivisionofthescaleofthethermometerin
Diagram2.1?
(c)
(d)
scaleofthermometer?
(e)
What is the reading of the thermometer based on your position in
2(d)?
(f)
Explainwhythemercuryisusedinthermometer?
0.01cm
0.1C
Nameonemeasuringinstrumentwhichismoresensitivethanvernier
calipers.
Micrometerscrewgauge
(e)
Whatisthetypeofphysicalquantityyounamein2(a)?Tickthecorrect
Scalarquantity
Tomeasurethedepth
Diagram2.1
Namethephysicalquantitybeingmeasuredbythermometer.
AtpositionQ
closed.
28.7C
Itissensitivetotheheat.
Itdoesnotsticktothewallofcapillarytubeinthermometer.
Itissensitivetowiderangeoftemperature.
Itisagoodheatconductor
Diagram1.2
BasedonDiagram1.2:
(i) Namethetypeoferroroccurred.
(ii)
Nameoneprincipleinvolvedforthemeasuringofthermometer.
(h)
Makethecapillarytubenarrowerwiththinnerstem.
Statethevalueoftheerror.
(i)
Nameatypeoferrorduetothewrongpositioningofeyesduringscale
Parallaxerror
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(g)
Thermalequilibriumconcept
Zeroerror
15
Diagram3.1showsanimageisformedbyacurvedmirror.
(g)
Explain how to use a concave mirror to heat up water in a container
usingsolarenergy.
1st:concavemirrorcanreflecttheparallelsunlight.
2nd:Thereflectedraysareconvergingtofocalpoint
3rd:Thefocalpointisplacedwiththewatertobeheated
(a)
(b)
Diagram3.1
Namethelightphenomenoninvolved.
(h)
Diagram 3.3 shows a microscope. You are given two convex lenses P
andQ,withfocallengthsof20cmand5cmrespectively.Bothofthe
lensesareusedtobuildamicroscope.
Reflection
NamethetypeofmirrorasshowninDiagram3.1.
Convexmirror
(c)
(i)
CompletetheraydiagraminDiagram3.2toshowtheformation
ofimage.
Diagram3.3
Whatismeantbyfocallength?
FocallengthisthedistancebetweenthecentreoflenswiththeprincipalF.
Diagram3.2
(ii)
(d)
Stateoneuseofthismirrorindailylife
Stateonecharacteristicoftheimageformed.
Distancebetweentheobjectiveand
eyepiecelensmustbebiggerthanthe
sumoffocallengthsofbothlens
Storethemicroscopeatcoolanddryplace
Asmirrorinsidethesupermarkettoviewunwantedactivitiesofshoplifter
(e)
(f)
Haswiderviewofvision
Whathappentothesizeofimageiftheobjectisplacednearertothe
mirror?
Installoneconcavemirrorunderneaththe
slaid
Sizeofimageincreases
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Using an appropriate physics concept, suggest and explain suitable
modifications to enable the microscope to form brighter and clear
image.Yourmodificationscanbeemphasizedonthefollowingaspects:
Theselectionoflensasobjectivelensandasaneyepiece
Thediameterofthelens
Thedistancebetweentheobjectivelensandeyepiece
Conditionoftheplacetostorethemicroscope
Suggestion
Shorterfocallengthasobjectivelens
Longerfocallengthaseyepiecelens
Diameteroflensmustbebig
Upright//Diminished//Virtual
16
Reason
Producebigimagemagnification
Morerefractionoflightsandthusthe
imageisbrightandclearq
Preventtheactivitiesoffungusonthe
lenses
Thereflectedrayisconvergingtothe
slaid
Diagram 5.4 shows an astronomical telescope to be used to view distant
objects.
(a)
Whatismeantbyfocallength?
(b)
UsingDiagram6.1andDiagram6.2,compare;
(i) Thefocallengthofthelens.
(ii)
(iii) Theheightofimage,h1andh2.
(iv) State the relationship between the object distance and the
heightofimage.
FocallengthisthedistancebetweenthecentreoflenswithitsprincipalF.
Boththelenseshavethesamefocallength.
Diagram5.4
Table5.4showsthecharacteristicsoffourdifferenttelescopes.
Telescope Typeof
Focallengthof
Poweroflens Diameterof
lens
objectivelens
lens
S
Convex
40
10
5.0
T
Concave
10
40
5.0
U
Convex
10
40
2.5
V
Concave
40
10
2.5
Explainthesuitabilityofeachcharacteristicofthetelescopeanddetermine
the most suitable telescope to be used to observe very far object. Give
reasonforyourchoice.
u1isshorterthanu2.
h1islongerthanh2.
Shortertheobjectdistance,longertheheightofimage.
(v)
(c)
Diagram6.3showsanobject,Oplacedatthefrontofaconcavelensof
focallength2cm.Thelightraysoftheobjectpassingthroughthelens
usingthelightphenomenonin6(b)(ii).
Reason
Thelightisconvergingtoonepoint
Toviewdistantobject
Toproducebigmagnificationofimage
More refraction of lights and thus the
imageisbrighterandclearer
So,thetelescopeSischosenbecauseitusesconvexlens,focal lengthoftheobjectivelensis
long,powerofthelensisbigandthediameterofthelensisbig.
Diagram6.1andDiagram6.2showtwoidenticalobjectslocatedatdifferent
positionsinfrontofidenticalconvexlens.Realimageswithdifferentheight
areproduced.
(i)
Diagram6.3
Sketchraydiagramoftheobjecttoshowanimageisformed.
(ii)
Statethreecharacteristicsoftheimageformed.
(iii) Stateoneuseofconcavelens.
Diagram6.1
Name the light phenomenon that occurs in Diagram 6.1 and
Diagram6.2.
Refraction
Characteristics
Useconvexlens
Focallengthofobjectivelensmustbebig
Poweroflensisbig
Diameteroflensmustbebig
Theobjectdistance,u1andu2.
Upright//Diminished//Virtual
Tomakespectaclelens
Diagram6.2
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17
Diagram 7.1 shows a method used to detect leakage of pipes lay
underground. A little radioisotope substance is dissolved in the water that
flowsinthepipes.AGeigerMullertubewhichisconnectedtotheratemeter
is then moved over the pipes according to the layout plan of the
undergroundpipes.
(ii)
Alphaispositivechargeandthusdeflectedtonegativeplate.
Betaisnegativechargeandthusdeflectedtopositiveplate.
(iii) Calculatethenumberofparticleandparticlethatemittedin
theThorium234decaysafterwritingthedecayequation.
234
90
4
0
Th 226
88 Ra +2( 2 He )+2( 1 e )+Energy
So,releasetwoalphasandtwobetas.
Diagram7.1
LocationofGeigerMullerTube A
B
C
D
E
F
290 295 284 372 290 216
(countsperminute)
Table7.1
(b)
(c)
Based on Table 7.1, state the location on the pipe where the leakage
Diagram 7.2 shows a nuclide Thorium234,
234
90
Th is placed in a
88 Ra
byemittingparticleandparticle.
Thoriumnuclide
Bekas
(i)
(v)
(d)
Stateof
power
matter
P
5minutes
Low
Gamma
Liquid
Q
8days
High
Alpha
Solid
R
6hours
Low
Gamma
Liquid
S
5years
Low
Beta
Solid
T
7hours
High
Alpha
Liquid
Table7.1
Asamedicalofficer,youarerequired todeterminethemostsuitable
Definethemeaningofhalflife.
Reason
Does not give long term effect to the
patientasitdecaysfast
Theionizingpowermustbelow
Does not change the structure of
substantialcells
Has high penetrating power which can
killthecancercells
Stateofmatterisliquid
Easy to put into the area of tumor by
injection
So, the radioisotope P is chosen as its halflife is short, has low ionizing power, use
Diagram7.2
InDiagram7.2,drawthepathofparticleandparticle.
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Characteristics
Useshorthalflife
Nuklidathorium
Container
Thorium234 has halflife of 20 days and initial mass of 48 g.
CalculatethemassofundecayedThorium234after60days.
LocationDshowstheleakagetakesplace.
comingoutfromtheleakagetobedetectedbyratemeter.
(iv)
48g24g12g6g
202020
So,massundecayedis20g.
Radioisotope is an unstable nucleus which has same proton number but different
nucleonnumber.
18
Neptunium239( 239
93 Np )has93protonsanddecaystonuclideXwithproton
numberof94.
ThemassofNeptunium=239.04251am.u
massofnuclideX=239.02312a.m.u
Massofelectron=0.00054a.m.u,1a.m.u=1.67x1027kg,
Speedoflight,c=3x108ms1
(a) Whatisthemeaningofprotonnumber?
(b)
Protonnumberisthetotalnumberofprotoncontainedinnucleus.
Betaparticle
(c)
WritethedecayequationforthedecayofNeptunium239.
239
93 Np
0
239
94 X + 1 e +Energy
(d)
(i)
Statethetypeofnuclearprocessfor8(c).
(ii)
Calculatethemassdefect,inkg,inthisnuclearprocess.
Nuclearfission
Characteristics
Typeofreactionisfission
Halflifeofnuclearfuelmustbelong
Reason
Reactioncanoccuratlowtemperature
Canbeusedforlongertimewiththeleast
ofreplacement
Specificheatcapacitymustbelow
Heatupfaster
Materialofshieldmustbeconcrete
So,thedesignRischosenbecauseitstypeofreactionisfission,halflifeofthefuelis
long,specificheatcapacityofgasislowandmaterialoftheshieldisconcrete.
Totalmassbefore=239.04251a.m.u
Totalmassafter=239.02312a.m.u+0.00054a.m.u
=239.02366a.m.u
Totalmassdefect=239.04251a.m.u239.02366a.m.u
=0.01885a.m.u
=0.01885x1.67x1027kg
=3.14795x1029kg
(iii) Calculatethetotalenergyreleasedinthisprocess.
FromEinsteinstheory,E=mc2
So,energy=(3.14795x1029)(3x108)2J
=2.83x1012J
(e)
Diagram 8.1 shows the schematic diagram of a nuclear reactor at a
nuclearpowerstation.
Diagram8.1
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Table8.1showsthefourdesignsP,Q,RandSofnuclearreactorwith
differentspecifications.
Design Typeof
Halflifeof
Specificheat
Materialof
reaction nuclearfuel capacityofgas shield
P
Fusion
Long
Low
Brick
Q
Fission
Short
High
Concrete
R
Fission
Long
Low
Concrete
S
Fusion
Short
High
Brick
Table8.1
You are required to determine the most suitable design of nuclear
reactorsothatthenuclearenergycanbeusedefficientlyandsafelyin
the generation of electricity. Determine the most suitable design and
justifyyourchoice.
(f)
Ionising
ray
matter
power
P
7hours Alpha
Solid
High
Q
7years Gamma
Solid
Low
R
10days Gamma Liquid
High
S
8years
Beta
Solid
Low
Table8.2
Asafactoryengineer,youarerequiredtodeterminethemostsuitable
radioisotope that can be used to detect the volume of guava juice in
Characteristics
Reason
Uselonghalflife
Canbeusedforlongertimewithlessrefillmentcost
Usegammaray
Canpenetratethetintosee
Physicalstateissolid
Easytohandle
Ionizingpowerislow
Doesnotchangethetasteofjuiceinside
haslowionizingpower.
19
Diagram9.1andDiagram9.2showmovementsofidenticalbarmagnetinto
thesolenoid withthesameforceastoproduce current.Bothsolenoidsare
Coilofinsulatedcopperwire
Gegelungdawaikuprumbertebat
Magnet
Magnet
Softironcore
Terasbesilembut
Diagram9.1Diagram9.2
(a) Underline the correct answer in the bracket to complete the sentence
below.
The method of producing current without electrical supply is called
(electromagnet,electromagneticinduction).
(b) OnDiagram9.1andDiagram9.2:
(i) StatethepolarityofregionP.
(ii) Namethelawusedtodeterminethepolarityin9(b)(i).
(c)
BasedonDiagram9.1andDiagram9.2,compare:
(i) Thenumberofturnsofcoils
Diagram9.3
Explainhowthebicycledynamoworkstoproducealternatingcurrent
tolightupthelamp.
1st:Thecoilrotateswithinthemagneticfield.
2nd:Magneticfieldlinescutbythecoil.
3rd:Cutofmagneticfieldlinescausestheinducedcurrentflowinthecircuit
(g)
Diagram9.4showsamovingcoilammeterwhichislesssensitive.
Northpole
Lenzslaw
Diagram9.4
Explainhowyouwoulddesignamovingcoilammeterthatcanfunction
better.Inyourexplanation,emphasizethefollowingaspects:
NumberofturnsofcoilMaterialofcore
ShapeofthemagnetStiffnessofhairspring
Typeoftheammeterscale
NumberofturnsofcoilsinDiagram9.1ismorethaninDiagram9.2
(ii) Deflectionofthepointerofthegalvanometer
DeflectionofpointerofgalvanometerishigherinDiagram9.1thaninDiagram
9.2
(d)
Statetherelationshipbetweenthenumberofturnsofcoilsand
(i) deflectionofthepointerofthegalvanometer
(ii) magnitudeofinducedcurrent
(e)
State what will happen to the deflection of galvanometer if a soft
magnetisused?
(f)
Morenumberofturnsofcoils,moredeflectionofthepointerofgalvanometer
Suggestion
Morenumberofturnsofcoils
Softironcorematerial
Curvemagnet
Lowstiffnessofhairspring
Stripmirrorunderthepointer
Morenumberofturnsofcoils,highermagnitudeofinducedcurrent
Deflectionofgalvanometerwillincrease
Diagram9.3showsacrosssectionofabicycledynamowhichhastwo
magnets with difference pole, a coil of insulated copper wire. The
outputofthedynamoisconnectedtothebicyclelamp.
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20
Reason
Increasemagneticfieldstrength
Canbemagnetisedordemagnetizedeasily
Thepointeriseasiertodeflect
Toavoidparallaxerror
Tocorrectthezeroerror
10
Diagram10.1showsthreetransformersX,YandZ.Eachtransformerhasits
own specific number of turns of primary and secondary coils respectively.
Each transformer is connected to 240 V alternating current suppliers. The
output for each transformer is connected to electric filament bulbs in
differentarrangement.
(d)
A student wants to build a simple transformer. Table 10.1 shows the
characteristicoffourcores.
Core
Shapeofthe Materialof
Typeof
Thicknessof
core
core
core
wire
A
Softiron
Laminated
Thick
B
Steel
Single
Thin
C
D
TransformerXTransformerYTransformerZ
Diagram10.1
(a) Whatistransformer?
(b)
(i)
StatethetypeoftransformerusedinDiagram10.1.
(ii)
Stateonereasonwhythea.c.voltageissupplied.
Sothatthevoltagecanbealternatedaccordingtothechangingmagneticflux
beinginducedtosecondarycoilintransformer
(iii) Statethefunctionofdiodeinthecircuit.
(b)
Statetheprincipleappliedfortheworkingprincipleoftransformer.
(c)
Thin
Thick
Characteristics
Reason
ShapeofcoreisUshape
Centralizethemagnetfromleaking
Usesoftironcorematerial
Canbemagnetizedordemagnetizedeasily
Thecoreislaminated
Toreduceeddycurrent
Usethickwire
Lowresistance
thickwire.
Stepdowntransformer
Single
Laminated
Table10.1
Explain the suitability of the characteristics given so that it can produce the
mostefficienttransformer.Choosethebestcoreandjustifyyourchoice.
Steel
Softiron
Tochangethea.cvoltagetod.cvoltage
(e)
Diagram10.2 showsacrosssectionofamovingcoilmicrophone.
Electromagneticinduction
Whentheswitchison,0.25Acurrentflowsthroughtheprimarycoilin
eachtransformer.Allthebulbslightsupnormally.
(i) Calculatetheinputpower.
Inputpower,P=IV=0.25x240
=60W
(ii)
All the transformers in Diagram 10.1 have the same output
voltage.Calculateitsoutputvoltage.
Diagram10.2
Using an appropriate concept in physics, suggest and explain suitable
modifications or ways to enable the microphone to detect sound effectively
andgeneratebiggercurrentbasedonthefollowingaspect:
(i)thicknessofdiaphragm(ii)strengthofthematerialfordiaphragm
(iii)numberofturnsofcoil(iv)diameterofthewireofcoil
(v)strengthofmagnet
Byratio:1200turns240V
800turns?
So,1200/800=240/?
?=160V
(iii) CalculatetheoutputpowerforallthetransformersX,YandZ.
OutputpowerfortransformerX=12W+12W+12W=36W
OutputpowerfortransformerY=24W+12W+12W=48W
OutputpowerfortransformerZ=18W+18W+18W=54W
Characteristics
Usethickerdiaphragm
Highstrengthofdiaphragm
Usemoreturns
Biggerdiameterofwirecoil
Higherstrengthofmagnet
(iv) Between transformer X, Y and Z, which one has the highest
efficiency?Why?
Transformer Z. Because its output power is closed to input power with its
efficiencyof90%
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21
Reason
Canwithstandhighpressureofsoundvibrations
Longlasting
moremagneticfluxcut
Reduceresistance
Producestrongermagneticflux
11
Diagram 11.1 shows an apparatus used to investigate one physics concept.
When an air is blown from region P, the water level in the arms of tube
changesasshown.
(a)
(b)
Diagram11.1
Namethephysicsprincipleinvolved.
(f)
Table 11.1 shows four Bunsen burners, P, Q, R and S, with different
specifications.
Bunsen StructureofBunsenburner Meltingpoint Density of
burner
ofmaterial
material
P
High
High
High
Low
High
Low
Low
Low
Bernoullisprinciple
(i)
ComparetheairspeedatregionPandregionQ.
AirspeedatregionPislowerthanatregionQ.
(ii)
ComparetheairpressureatregionPandregionQ.
AirpressureatregionPishigherthanatregionQ.
regionQ.
AccordingtoBernoullisprinciple,regionwhichhashighairspeedwillhaslow
pressureandviceversa.
AtregionP,theairspeedislowbutwithhighpressure.
AtregionQ,theairspeedishighbutwithlowpressure.
(c)
CalculatethedifferenceinwaterpressurebetweenregionPandregion
Q.Giventhedensityofwateris1000kgm3.
Table11.1
You are required to determine the most suitable Bunsen burner that
canproducebiggerblueflameandportable.
Study the specifications of all the four Bunsen burners from the
followingaspects:
(a)Sizeofgasnozzle(b)Sizeoforifice
(c)Meltingpointofthematerial(d)Densityofthematerial
Explainthesuitabilityoftheaspects.Justifyyourchoice.
Differenceinwaterpressure=hg
=(0.05)(1000)(10)Pa
=500Pa
(d)
Suggest three ways by which the difference in water pressure can be
Suggestion1:increasethespeedofairflow
Reason:differenceinairpressurebetweenPandQwillbebigger
Suggestion2:reducethediameteroftubeQ.
Reason:higherspeedproduceatQresultinlowerpressureatQ
Suggestion3:reducethediameterofarms
Reason:thearmswillbemoresensitivetosmallchangeinpressure
(e)
Characteristics
Reason
Smallgasnozzle
Producehighairspeed
Sizeoforificeisbig
MoreairflowintotheBunsenburner
Highmeltingpoint
Canwithstandhightemperaturewithoutmelt
Lowdensity
Lighterandportable
So,BunsenburnerRischosenbecauseitssmallgasnozzle,smallorifice,hashigh
meltingpointandlowdensitymaterial.
Whatwillhappentothewaterlevelifthenonuniformhorizontaltube
isreplacedwithuniformhorizontaltube?
Thedifferenceinwaterlevelwillbereversed.
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12
Diagram12.1showsanordinarybicycle.
(c)
Diagram12.4
Theobjectiveistotraveltherocketasfaraspossible.Usingtheappropriate
physicsconcepts,suggestandexplainthefollowingaspects:
(i)Theshapeoftherocket(ii)Volumeofwatertobefilledtobottle
(iii)Thestabilityofthemotion(iv)Densityofmaterialtotherocket
(v)Angleoflaunching
Diagram12.1
You are required to give some suggestions to enable the cyclist to ride the
bicycle safely at higher speed. Your explanations are based on following
characteristics:
(i)
Massofbicycle
(ii)
Typeofmaterialforthebodyofbicycle
(iii)
Widthoftyres
(iv)
Typeofbrakingdistance
(v)
Theattireoraccessoriesforthecyclist
Suggestion
Themassofbicyclemustbesmall
Materialforbodymustbelowdensity
Widthoftyresmustbebig
Usehydraulicdiscbrake
Weartightattire
Wearglove
Wearhelmet
(b)
Suggestion
Theshapeofrocketisaerodynamics
1/3ofthebottleisfilledwithwater
Densityofthematerialislow
Launchat45fromhorizontal
Reason
Smallmassproducehigheracceleration
Lighterforcyclisttoride
Stablewhileriding
Canstopthebicycleinshorterdistanceand
shorterbrakingtime
Reduceairfriction
Bettergriponhandles
Diagram 12.2 and Diagram 12.3 shows the situation of the canopy of
thelorrybeforelorrymovesandwhenthelorrymovesatahighspeed.
Diagram12.2Diagram12.3
Explainwhythecanopyofthelorryliftsupwhenthelorrymovesata
highspeed.Nametheprincipleinvolved.
(d)
Reason
Reducesairfrictionwhileflying
Gainmomentumtolaunch
Stablewhileflyingandnotwobble
Nottooheavytofly
Getmaximumprojectilewithmaximum
distancetraveled.
Diagram12.5 showsfourracingcars,P,Q,R andS,withdifferentspecifications.
Car
Shape
Ridgeson
Engine
Materialfor
tyre
power
thecarbody
P
Yes
518kW
Lightand
elastic
Aerodynamics
Q
745kW
Heavyandstiff
None
Aerodynamics
R
Yes
518kW
Heavyand
elastic
Invertedaerofoil
S
None
745kW
Lightandstiff
Invertedaerofoil
Diagram12.5
Youarerequiredtoinvestigatethespecificationsgivensothatthecarcanrun
veryfast.Determinethemostsuitablecarandjustifyyourchoice.
Characteristics
Reason
Theshapeisinvertedaerofoil Producedownwardforce
Noridgeontyre
Canmovefast
Enginepowerisbig
Producebigacceleration
Materialislightandstiff
Nottooheavytomovebutyetcanwithstandforce
So,carSischosenbecauseitisinvertedaerofoil,noridgeontyres,enginepowerisbig
andmaterialofbodyislightandstiff.
1st:Beforelorrymoves,theairspeedoutsideandinsidethecanopyissame.
2nd: When the lorry moves, the air speed outside the canopy is high causing low
pressureoutside.Insidethecanopy,theairspeedislowbutwithhighpressure.
3rd:HighpressureinsidethecanopypushesupthecanopyasshoninDiagram12.3
4th:PhysicsprincipleinvolvedistheBernoullisprinciple.
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23
13
Diagram 13.1 shows a submarine floating in sea water due to the effect of
buoyantforce.
(a)
(b)
(ii)
(iii) Deducetherelationshipbetweentheweightoftheshipandthe
weightofthewaterdisplaced.
(iv) A ship that travels round the world will has Plimsoll symbol as
showninDiagram13.4.
Explainhowasubmarineisabletosubmergeintodeepseawater.
Explainwhytheboatisabletofloat?
Weightoftheshipisequaltotheweightofwaterdisplaced
1st:Tosubmerge,thelowervalveoftheballasttankisopenedtoletinthewater.
2nd:Theuppervalveisopentoletoutthetrappedairinsidetheballasttank
3rd:Whentheweightofsubmarineishigherthanbuoyantforce,thesubmarinestarts
tosubmerge.
4th:Theloweranduppervalvesareclosedwhenthesubmarinehasreachedthedepth
required.
(c)
(i)
Theboatdisplacesthewaterandthusgainsthesamebuoyantforcetofloat.
Buoyantforceisanupthrusttotheobjectfromwatercausingtheobjecttofloat.
SeaRiver
Diagram13.3
Nametheprincipleappliedforthefloatingoftheboat.
Archimedesprinciple
Diagram13.1
Whatisthemeaningofbuoyantforce?
Diagram13.2showstheairballoonwhichisusedasaweatherballoon
atmosphere.
Diagram13.4
StatethecommonfunctionofthePlimsollline.
To guide navigator the maximum weight load limits that can still be safely
(i)
Diagram13.2
StatetheArchimedesprinciple.
Archimedes principle states that the when the object is immersed partially or
whollyintofluid,theweightofwaterdisplacedisequaltotheweightofobject
beingimmersed.
(ii)
Explainwhyaweatherballoonthatisrisingupintheairwillstop
atcertainaltitude.
1st:Densityofairdecreasesasthealtitudeincreases
2nd:Buoyantforcebecomesmaller
3rd:Atcertainheight,theweightofairdisplacedisequaltotheweightof
balloon.
4th:Therefore,nonetforcetopushtheballoonup.
(d)
(e)
Youarerequiredtogivesomesuggestionsonhowtodesigntheboatin
Diagram 13.3 as to increase the floating force and safer. Explain the
suggestionsbasedonthefollowingaspects:
Materialused
Shapeofboat
Densityofboat
Safetyfeature
Suggestion
Materialusedmustbewithlowrusting
rate(fibrecomposite)
Shapeofboatisstreamline
Densityofboatislight
Haslifejacketandtyre
Installperiscope
Diagram 13.3 shows two boats of the same weight floating on the
surfaceofwaterintheseaandintheriver.
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24
Reason
Longlastingwithoutrust
Reducewaterfrictionwhileflying
Forpassengertofloatinemergencycase
Toviewhiddenobjectwhichmaycollide
theboat
14
(a)
Table 14.1 shows four hot air balloons P, Q, R and S with different
features.
Balloon Sizeand
Numberof
Typeof
Temperatureof
volume
burners
balloon
airinside
fabric
P
Smalland
1
Synthetic
100C
3
800m
nylon
Q
Largeand
2
Synthetic
120C
3
2500m
nylon
R
Largeand
1
Canvas
60C
2500m3
S
Smalland
2
Canvas
70C
800m3
Table14.1
You are required to investigate the hot air balloon which is able to
carry three or four people to a higher altitude in a shorter time.
Determinethemostsuitableballoonandjustifyyourchoice.
Reason
Candisplacemorewaterandthusgains
morebuoyantforce
Usemoreairtank
Cantrapmoreairtogainmorebuoyant
force
Higherpressurethatcanbetolerated
Cansubmergedeeperwithoutcrackdue
waterpressure
Shapeofsubmarineisstreamline
Reducewaterfrictionwhilemoving
So,thesubmarineQischosenbecauseithashighvolumeofballasttankwithmoreair
tanks,cantoleratehigherpressureandisinstreamlineshape.
Reason
Candisplacemoreairandgainshigherbuoyant
force
Usemoreburners
Fasterheatingtheairinsidetheballoon
Usesyntheticnylon
Resistancetoheatwithoutmelt
Hightemperatureinsideballoon
Hotairislighter
So,theballoonQischosenbecauseitisbigsizewithhighvolume,usemoreburners,
(b)
withaheavybox.Thevolumeoftheimmersedportionoftheboatis
5.0m3.
(i) Calculatethebuoyantforceexertedtotheboat.
[Densityofseawateris1020kgm3]
Fromformula,buoyantforce,F=Vg
=1020x5x10
=51000N
(ii)
Calculate the maximum weight of the box so that the boat will
notsinkcompletely.
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You are asked to investigate the characteristics of four submarines
showninTable14.2.
Volumeof Number
Maximum
Shapeofsubmarine
ballasttank ofairtank pressuretobe
tolerated
3000litre
15
4.5atm
P
2500litre
30
6.0atm
Q
350litre
3
6.1atm
R
400litre
1
2.0atm
S
Table14.2
Explain the suitability of each characteristic of the submarines and
determinethesubmarinewhichcantravelfaster,staylongerindeeper
seawaterandabletocarrymorecrew.Givereasonsforyourchoice.
Characteristics
Volumeofballasttankishigh
Characteristics
Bigsizeandhighvolume
(c)
25
15
Diagram15.1
Diagram15.2
Diagram15.3
Diagram15.1showsoneendofaspringisfixedtoawoodenblock.
Diagram15.2showsthespringiscompressedbyasteelballofmass0.52kg
usingaforceF.
Diagram15.3showsthesteelballmovesaftertheforce,Fisremoved.
[Thespringconstant=50Nm1]
(a)
Whatismeantbyelasticity?
(ii)
Theelasticityofaspringcanbeexplainedbyonelaw.Statethat
law.
Hookeslaw
(d)
Whenthespringiscompressed,itslengthdecreasesandreturnsback
to its original length after compressive force is removed due to
elasticity property of a material. Based on the forces between atoms,
explainwhythespringiselastic.
(e)
Diagram15.4showsatrampoline.Itusestheelasticpropertyofa
materialtorebounceapersonupwards.
RefertoabovenotePage13..
Whatismeantbyforce?
(b)
(i)
Nametheformofenergystoredincompressedspringasshown
inDiagram15.2.
(ii)
CalculatethevalueofF.
Diagram15.4
Youarerequiredtogivesomesuggestionstoimprovethedesignofthe
trampolinesothatitcanbeusedbythechildrensafelyandcanjump
higher.Explainthesuggestionsbasedonthefollowingaspects:
(i)thenumberofspringused
(ii)springconstant
(iii)thematerialusedforframe
(iv)thematerialusedforfabric
(v)extrafittingordesignofthetrampolinetoensuresafety
Elasticpotentialenergy
FromF=kx
=50x(0.200.15)
=2.5N
(iii) Statetheconversionofenergywhentheballisreleased.
(iv) Calculatethespeedofball,v.
Elasticpotentialenergyischangedtokineticenergy
Workdonetocompressspring=kx2
=(50)(0.200.15)2
=0.0625J
Fromkineticenergy:mv2=0.0625
(0.52)v2=0.0625
v=0.49ms1
(v)
Suggestion
Usemorespringinparallel
Lowspringconstant
Usenylonforfabric
Hasnetaroundtheedgeoftrampoline
Statetheprincipleyouusedtofind(iv).
Principleofconservationofenergy
(i)
Elasticity is the ability of an object to resume to its original state once the
appliedforceisremovedwithelasticlimitisnotexceeded.
Forceistheproductofmassandacceleration
(c)
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26
Reason
Increasetheelasticityofspring
Canextendorcompressmore
Strongmaterialwhichdoesnotbreak
Doesnottear
Topreventthechildrenfromslippingout
tothefloor
16
AtypicaltorchlightwithbatteriesisshowninDiagram16.1.
(a)
(i)
(ii)
DrawacircuitdiagramforthetorchlightinDiagram16.1.
Reason
To reduce the resistance of the power socket
extensionastheresistanceincreaseswithlength
Once one of the fuse is burnt due overloaded of
that socket, the rest of the sockets can still
function
There is a power surge To shut down the whole current to the power
protection
socketifthereisaelectriccurrentleakage
Must have head plug To flow the unused current to the earth so that
earthing
theuserwillnotbeelectrocuted
So,thetypeSischosenbecauseithasnoextensioncordlength,havefusefor
Directcurrent
(b)
Characteristics
No extension cord
lengthisneed
Must have fuse for
everysocket
Diagram16.1
Namethetypeofcurrentusedinthetorchlight.
(c)
Diagram16.2showsatwodoorrefrigeratorforhouseholduse.
Energy efficiency and safety are important considerations in the
purchaseanduseofelectricalproductsandappliances.
(i) Afuseinapowerplugislabeled8A.
Whatdoesthelabel8Amean?
(ii)
Table16.1showsfourtypesofpowersocketextensionsP,Q,R
andSavailableinasupermarket.
Type
Type
Extension
cord
length
5m
None
5m
None
Number
offuse
Four
sockets
onefuse
Four
sockets
onefuse
Each
socket
onefuse
Each
socket
onefuse
Power
Surge
Protection
Available
Not
Available
Not
available
Available
Earthing
system
Diagram16.2
Using the knowledge about heat flows, explain the modification
needed to produce a refrigerator which is constantly cold, energy
savingandlastingforthepurposeofkeepingthefreshnessofthefood
stored in it. Your modification should be based on the following
characteristics:Materialusedtomakethedesk,Typeoflampusedin
refrigerator, Power of the refrigerator, Air circulation in the
refrigerator,Specificheatcapacityofthecoverofrefrigerator
Not
available
Available
Not
available
Available
Characteristics
UseLEDlamporlampoflowpower
andjustlightwhenthedoorisopen
Usehighpowerofrefrigerator
Reason
Doesnotrustandlonglasting
Does not heat up the refrigerator inside due
tothelampofbulb
Therefrigeratorwillbemorecoldtokeepthe
freshnessoffoodstored
Thefreezermustbeinstalledatthe The cool air has higher density will move
highestpartinsidetherefrigerator downtocoolthelowerpartoffoodandthus
causes the air circulation. Hot air at lower
part will move up and cooled down by the
freezer.
The specific heat capacity of the Slower conducting the heat outside into the
coverofrefrigeratormustbehigh.
refrigerator.
Table16.1
Using physics concepts, explain the suitability of the power
socketextensionsforeachaspectwhichcanbeusedsafelyand
efficiently for normal home use. Determine the most suitable
socketextensionandjustifyyourchoice.
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17
Diagram17.1showsahairdryerlabelled240V,500Wconnectedtoathree
pinplug.Diagram17.2showsthefuseinthethreepinplug.
(b)
(i)
Whatisthemeaningofthelabel0.5Aonthefuse?
(ii)
Table17.1showsthespecificationofafewmetalstobeusedas
afusewire.
Diameter Resistivity
Metal Melting Specificheat
point/C capacity/Jkg1C1
W
1100
900
Big
Low
X
600
900
Small
High
Y
1100
240
Big
Low
Z
700
240
Small
High
Table17.1
Explain the suitability of each characteristic of the four metals
and determine the most suitable metal to be used as the fuse
wire.Givereasonsforyourchoice.
Reason
Easymeltwhenhot
Fastergettinghot
Produceshigherresistancewhichcan
convertcurrenttoheat
Theresistivitymustbehigh
Produceshighresistance
So,themetalZischosenbecauseitsmeltingpointislow,specificheatcapacity
islow,smalldiameteranditsresistivityislow.
(d)
Fusetakessometimetomeltorblow.Afastblowingfuseisrequired
toprotectsemiconductorequipmentswhichcannotstandhighcurrent
surge for too long. When a fuse blows, sparking may occur and
produceshightemperature.
Reason
Hashigherresistancewhichcangethotfaster
Easilygetbrokenwhenhot
Enoughtobreakdownthe240V,2000Wsemiconductor
devicewhichneedsonlyacurrentof8.33A
Themeltingpointmust Fastergettingmeltandshortopenthecircuitwhenthe
below
circuitoverheated
ofglass,fuseratingis10Aandthemeltingpointislow.
##Remember:Thefuseitselfmustbecangethotfasterwithhighresistanceandeasily
getmeltsothatthecircuitisshortopenandthusprotecttheelectricalcomponentfrom
gettingburnt.
Giventhehairdryerlabelled240V,500W
VoltagePower
FromPower,P=IV
500=I(240)
I=2.08A
(iii) Calculatetheenergyusedbythehairdryerwhenitisswitched
onfor10minutes.
Characteristics
Thethicknessoffuse
wireisthin
Thecartridgetyemust
beglass
Thefuseratingis10A
ThehairdryerinDiagram17.1isswitchedon.
(i) Calculatethecurrentflowingthroughthehairdryer.
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Table 17.2 shows the specifications of five fuses that can be used to
protectasemiconductordevice.
Fuse Thicknessoffuse
Cartridge
Rating Melting
wire
tye
point
P
Medium
Rubber
10A
Medium
Q
Thin
Glass
10A
Low
R
Thin
Ceramic
13A
Low
S
Thick
Plastic
10A
High
Table17.2
Determine the most suitable fuse to protect a 240V, 2000 W
semiconductormaterialdevice.Studythespecificationsofallfivefuses
givenandjustifyyourchoice.
Characteristics
Themeltingpointmustbelow
Specificheatcapacitymustbelow
Thediameterofmetalmustbelow
(c)
Statewhetherthe0.5Afuseissuitabletobeusedintheplug.
Power=Energy/Time
Energy=PowerTime
=500(1060)J
=300000J@300kJ
Hashighmeltingpoint,hashighresistance,haslowspecificheatcapacity
(ii)
Not suitable. The hair dryer needs 2.08 A to function. This current of 2.08 A
whichpassestothefuseof0.5Awillburnthefusedirectlybeforeworkingup
thehairdryer.
Diagram17.1Diagram17.2
(a) State three properties of the material of the heating element in the
hairdryer.
28
18
19
Diagram 18.1 and Diagram 18.2 shows an experiment to study the
relationship betweenthepressureandvolumeofairtrappedinanairtight
container.Thepistonsforbothdiagramsarepusheddownslowly.
Diagram18.1Diagram18.2
(a) StatethephysicalquantitybeingmeasuredbyBourdongauge.
(b)
BasedonDiagram18.1andDiagram18.2;
(i) Comparethevolumeofthegasintheairtightcontainer
(ii)
(iii) Temperatureofthegasintheairtightcontainer
(i)
(ii)
(ii)
StatethesensitivityoftheBourdonGauge.
(b)
(i)
State the correct position of the eye while taking reading from
theBourdonGauge?
(ii)
shownbytheBourdongauge?
(c)
The round bottom flask is then heated. Would the reading of the
Bourdongaugeincreaseordecrease?
(d)
(i)
Stateonelawthatyouusedin(c).
(ii)
Definethelawyoustatein(d)(i).
Increases
Statethegaslawinvolved.
Pressurelaw
Boyleslaw
(i)
165Pa
The temperature of the gas in the airtight container for both diagram are
same.
(c)
(a)
Perpendiculartothescale
5Pascal[#smallestscale]
Volume of gas in the airtight container in Diagram 18.1 is more than in
Diagram18.2
(iii) Definethegaslawyounamein(c)(ii).
Pressurelawstatesthatforafixedmassofgas,thepressureofgasisdirectly
proportional to its absolute temperature such that the volume of gas is kept
constant.
Boyleslawstatesthatforafixedmassofgas,thepressureofgasisinversely
proportionaltoitsvolumeatconstanttemperature.
(d)
(e)
An experiment is carried out to investigate the relationship between
the pressure, P and the temperature, of a fixed mass of a gas as
showningraphbelow.
Asyringecontains50m3ofairatapressureof100kPa.Thispistonis
pulled outwards slowly so that the air expands. What would be the
volumeoftheairwhentheairpressuredropsto80kPa?
(i)
FromBoyleslaw:P1V1=P2V2
(100)(50)=80(V2)
V2=62.5m3
What is the value of temperature, To when the pressure of the
gasiszero?
(ii)
WhatisthenamegiventoTo?
The experiment above is usually applicable if a gas expands or
compressedslowly.Whyisthisso?
TheBoyleslawisapplicableifthetemperatureofthegasisconstant.Thegasmustbe
expandedorcompressedslowlyastoreducethecollisionbetweenthemoleculesofgas.
The collision of molecules increase the friction of molecules between them and this
frictionwillproduceheat.Iftheheatproduced,thenthetemperatureofthegasisnot
(e)
273C
Diagram19.1
Whatismeantbysensitivity?
Gaspressure
Diagram19.1showsaBourdongaugewhichisusedtomeasuregaspressure
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Absolutezerotemperature
29
20
Diagram20.1showsacarandalorrystoppingataredtrafficlight.Whenthe
thelorry.
P
Q
M
N
S
Diagram20.1Diagram20.2
(a) Whatismeantbymass?
Massisthequantityofmattercontainedbyanobject.
(b)
BasedonDiagram20.1andDiagram20.2,comparethemassesofthe
and the way it can start moving from rest to deduce a concept in
physicswithregardtothemotionofobjects.
(c)
(d)
Definethephysicsconceptyounamein(b).
forwardswhenacarheisdrivingcomestoasuddenhalt.
(g)
Whenthecarcomestoasuddenhalt,theinertiaisverybigactingontothedriver.The
inertiawillcausethedrivertocontinuetomoveforwardalthoughthecarhasstopped.
Asaresult,thedriverwillbesurgedforward.
(e)
(f)
Reason
Topreventthebackhoefromsinkingintosoft
ground
Liquidhasverylowcompressibility
Diagram 20.4 shows a transformer connected between a 240 V a.c.
powersupplyandtwolightbulbs.Thebulbsareatnormalbrightness
You are asked to investigate the characteristics of each backhoes in
Table20andchooseabackhoethatcandoheavyworks,anexampleof
a backhoe is shown in Diagram 20.3. Explain the suitability of the
characteristics each backhoes. Determine the most suitable backhoe.
Givereasonforyourchoice
(1)
Diagram20.4
StatethetypeoftransformerinDiagram20.4.
(2)
Whatistheoutputvoltageofthetransformer?
6V
(3)
Calculatetheefficiencyofthetransformer?
Inputpower,P=IV=2400.25=60W
Outputpower,P=12W+24W=36W
Efficiency=(36/60)100%=60%
(4)
Explainwhythetransformermustusea.c.inputvoltage?
Sothatthevoltagecanbechangedeasily
Diagram20.3
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Stepdowntransformer
Centre
of
gravity
Large
Low
Large
Low
Small
High
Medium
Low
Medium
High
Explainhowyouwouldgotoescapefrombeingchasedbyabullbased
ononeconcept.
1st:Iwillperformmyruninzigzagdirectionswithnodefinitedirectionofrun.
2nd:Thebullhasbiggermassifcomparedtome.Thismeansthatthebullwillhas
3rd:Duetoinertia,thebullishardertochangeitsdirectionandwilllosecontrolandfall
Base
area
Fluidusedinthehydraulicsystem
isliquid
Themassmustbebig
Thebackhoewillbemorestable
Thebaseareamustbebig
Ensurethatthebackhoewillnotcollide
Thecentreofgravitymustbelow Stablefromfalling
So,thebackhoePischosenbecauseitstyreisbig,fluidusedinhydraulicsystemis
liquid,haslargemass,thebaseareaislargeandhaslowcentreofgravity.
Inertiaisthetendencyofanobjecttoresistthesuddenchangeexertingontheobject.
Fluidusedin Mass
hydraulic
system
Large
Liquid
Large
Large
Liquid
Small
Large
Gas
Large
Medium
Liquid
Large
Large
Liquid
Large
Characteristics
Thesizeoftyremustbebig
The mass of the lorry is more than the car causing the lorry has less ability to speed
physicsconceptinvolvedistheinertia.
Typeof
Sizeof
Backhoe tyre
30
21
Diagram21.1showsabrightspot,M,formedonthescreenonthecathode
rayoscilloscope,CRO,whenitisswitchedon.
(a)
(b)
CalculatethevalueofthevoltageshownbyN.
Valueofvoltage=5Vperdivision2divisions
=10V
Diagram21.1
Whatisthemeaningofcathoderay?
(f)
TheCROinDiagram21.2isconnectedtoalternatingcurrentsupply,a.c
andthetimebaseissetoff.
On Diagram 21.3 below, sketch the output waveform that will be
displayedonthescreen.
Diagram21.3
Cathoderayisafastmovingelectronbeam
Whatisthemeaningofthermionicemission?
Thermionic emission is the process of releasing electrons from a heated cathode
surface.
(c)
Nametheparticlethatproducesthebrightspot,M,whenithitsthe
fluorescentscreenoftheCRO.
Electron
(d)
(e)
(g)
StateonecommonfunctionofCRO.
To display waveform //To measure short time interval // To measure the potential
difference
Explainhowtoproduceabrightspot,MonthescreenofCRO?
Thed.cvoltageissuppliedtotheCROwiththetimebaseissetoff.
isconnectedtotheYinputoftheCRO.
(h)
Diagram 21.4 and Diagram 21.5 show two circuits which consist of
identicalammeters,drycellsandsemiconductordiodes.
Diagram21.4Diagram21.5
(1) Nameanexampleofpuresemiconductormaterial.
(2)
Silicon
thecurrentflow
Diagram21.2
TheYgainoftheCROissetat5V/divisionwiththetimebaseisset
off.
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(3)
Toallowtheflowofcurrentinonedirectiononly
31
22
Diagram22.1showsthepatternofseawaveswhenapproachingthebeach.
ItisobservedthatthefourpendulumsB,C,DandEwilloscillatewith
differentamplitudesbutwiththesamefrequency
(i) Whatisthemeaningofamplitude?
(ii)
(iv) Namethephenomenonstatedin22(d)(iii).
(e)
Ultrasonicechoesarewiselyusedinmedicinetoseetheinternalorgansof
insidethebody.Diagram22.3showstheuseofultrasoundscanneracrossthe
motherswombtoseetheunbornbabies.
Amplitudeisthemaximumdisplacementofwavefromtheequilibriumposition
(a)
Diagram22.1
NamethewavephenomenonshowninDiagram22.1.
PendulumC
HasthesamelengthwiththependulumA
Refraction
(b)
Explainintermsofthewavephenomenain22(a),whythewaterwaves
followtheshapeofthebeachasitapproachestheshore.
Resonance
Whenthewavesrefractfromdeepseatoshallowersea,boththewavelengthandthe
energydecrease.Therefore,itbecomesweakerandfollowstheshapeofthebeach.
(c)
Diagram22.1showstheseashoreofafishingvillage.Duringtherainy
season, waves are big. One year the waves eroded the seashore,
causedthejettytocollapseanddamagedthefishermensboats.
To prevent similar damage in the future, the fishermen suggest
building retaining walls and relocating the jetty. You should use your
knowledgeofreflection,refractionanddiffractionofwavestoexplain
thesesuggestion,toincludetheaspects:
i.thedesignandstructureoftheretainingwall
ii.thelocationofthenewjetty
iii.thesizeorenergyofthewaves.
Suggestion
Reason
Designaninclinedconcretebarrier
Jettyisbuildatbay
Builtdiffractionbarrier(slitvery
small)forshippassage
Barrierisbuiltfromastrong
material/concrete
Designthehigherbarrier
Ultrasonic waves
Gelombang ultrasonik
(d)
Diagram22.3
Table22showsthecharacteristicsoftheultrasoundscannerW,X,YandZ.
Waveisrefractedwithdecreasingwavelength
Waveatbayiscalmer
Lesswaveenergypassingthrough
Typeofwave
W
X
Y
Z
Mechanical
Mechanical
Electromagnet
Electromagnet
Frequencies
range(Hz)
<20000
>20000
<20000
>20000
Penetrating
power
High
Low
Low
Low
Ionizing
power
Low
Low
High
High
Characteristics
Reason
Usemechanicalwave
Ultrasonicneedsmediumtotravel
Usehighfrequency
Theimagescannedisclear
Uselowpenetratingpower
Doesnothurtthefoetus
Uselowionizingpower
Doesnotchangethestructureofcellsoffoetus
So, scanner X is chosen because it use mechanical wave, use high frequency, low
penetratingpowerandlowionizingpower
Diagram22.2showsaBartonspendulumwhichconsistsoffivesimple
pendulums hanging on a horizontal string. When A is pulled and
released,itwillcausetheotherfourpendulumstooscillate.
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Scanner
Table22
Explain the suitability of each characteristic in Table 22 that can be used as
ultrasound scanner to scan the image of foetus safely. Determine the most
suitableultrasoundscannertobeusedandhence,justifyyourchoice.
Noteasilycorrosive/broken//canpreventa
stronghardwave
Watercannotoverflow
Diagram22.2
Ultrasound scanner
Pengimbas ultrasonik
Foetus
Whichpendulumoscillateswiththemaximumamplitude?
32
23
Diagram 23.1 and Diagram 23.2 shows water waves passing through the
entranceoftwodifferentharbours.
(i)
Namethephenomenoninvolved.
(ii)
What will happen to the frequency, wavelength and speed of
waveafterpassingthroughthegap?
Refraction
Frequency:unchanged
Wavelength:decreases
Speedofwave:decreases
Diagram23.1Diagram23.2
(a) Namethetypeofwaveofwaterwave.
(b)
(i)
Namethephenomenoninvolvedinbothdiagrams.
(ii)
What will happen to the frequency, wavelength and speed of
waveafterpassingthroughthegap?
(iii) CompletethewavepatterninDiagram23.3.
Transversewave
Diffraction
Frequency:unchanged
Wavelength:unchanged
Speedofwave:unchanged
(c)
(f)
BetweenDiagram23.1andDiagram23.2,whichoneshowstheobvious
diffractioneffect?Explainwhy?
Deep area
Shallow area
Kawasan dalam
Kawasan cetek
of water appearing on the road ahead, but the pool of water
disappearsasthecarapproachesit.
Diagram23.1showsobviousdiffractioneffect.
Becausethesizeofgapislessthanthewavelengthofthewave
(d)
BetweenDiagram23.1andDiagram23.2,whichoneshowsthebigger
energywaveenteringthegap?
Diagram23.2
(e)
(i)
Diagram23.4
Namethisnaturalphenomenonasobservedbythedriver.
Mirage
Diagram23.3showsthewavesenteringtwodifferentmediums.
(ii)
Statethephysicsconceptthatisinvolvedinthisphenomenon.
Totalinternalreflection
(iii) When light rays propagates from a denser medium to a less
dense medium, state what happen to the direction of the
refractedrays.
(iv) Stateoneapplicationofthisphenomenon.
Refractedawayfromnormal
Opticalfibre
Diagram23.3
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33
PAPER3EXPERIMENTS
DEPTHWITHPRESSURE
Diagram 1.1 shows a scuba diving in a sea notices that the water pressure
acted on his eardrums is greater when he dives at greater depth.
(a)
(b)
(c)
Thewaterpressureisinfluencedbythedepthofwater
Whenthedepthofwaterincreases,itswaterpressureincreasesalso
Tofindtherelationshipbetweenthedepthwiththewaterpressure
(i)
ManipulatedVariable:depthofwater
(ii)
RespondingVariable:waterpressure
ConstantVariable:densityofwater
Apparatus: Metre rule, manometer, water, rubber tube, measuring cylinder,
(iii)
thistlefunnel,rubbersheet
Materials:water
(iv)
Diagram 1.1
Based on your knowledge of the pressure and observation above:
[1 mark]
(a) State one suitable inference.
(b)
State one suitable hypothesis.
(c)
With the use of apparatus such as thistle funnel, a manometer and other
apparatus, describe an experiment framework to investigate the
hypothesis stated in 1(b).
In your description, state clearly the following:
[1 mark]
(i)
(ii)
(iii)
(iv)
(v)
The procedure of the experiment which include the method of
controlling the manipulated variable and the method of measuring
the responding variable
(vi)
(vii)
The way you would analyse the data
(v)
(vi)
(vii)
OperationalDefinitions:
Thedepthofwaterismeasuredusingmetrerule
Thewaterpressureismeasuredfromdifferenceheightbetweenthecolumn
ofwaterinmanometermeasuredusingmetrerule
The experiment is started by lowering the thistle funnel into the water to
manometerisrecorded.
Theprocedureisrepeatedwiththedepthsof4cm,6cm,8cm,10cmand12
metrerule.
Depthx/cm Differenceinheightofcolumn,h/cm
2
10
12
Agraphofdifferenceinheightofwatercolumnagainstthedepthisplotted.
[10 marks]
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34
DENSITYOFLIQUIDWITHPRESSURE
Diagram 2.1 shows a scuba diving in a sea notices that the water pressure
acted on his eardrums is greater compared to fresh water at same depth.
(a)
(b)
(c)
Thewaterpressureisinfluencedbythedensityofwater
Whenthedensityofwaterincreases,itswaterpressureincreasesalso
Tofindtherelationshipbetweenthedensityofliquidwiththewaterpressure
(i)
ManipulatedVariable:densityofwater
(ii)
RespondingVariable:waterpressure
ConstantVariable:depthofthistlefunnelimmersed
Apparatus: Metre rule, manometer, water, rubber tube, measuring cylinder,
(iii)
thistlefunnel,rubbersheet
Materials:water,salts
(iv)
Diagram 2.1
Based on observation above:
(a) State one suitable inference.
(b)
State one suitable hypothesis.
(c)
With the use of apparatus such as thistle funnel, salts, a manometer and
other apparatus, describe an experiment framework to investigate the
hypothesis stated in 2(b).
In your description, state clearly the following:
[1 mark]
[1 mark]
(i)
(ii)
(iii)
(iv)
(v)
The procedure of the experiment which include the method of
controlling the manipulated variable and the method of measuring
the responding variable
(vi)
(vii)
The way you would analyse the data
(v)
(vi)
(vii)
[10 marks]
OperationalDefinitions:
Thedepthofthistlefunnelisfixedat10cm.
Thewaterpressureismeasuredfromdifferenceheightbetweenthecolumn
ofwaterinmanometermeasuredusingmetrerule
Densityofwaterisdeterminedfromthemassofsaltsdissolved
Theexperimentisstartedbydissolving200gofsaltintothewaterwithfixed
volumeofV.Thethistlefunnelimmersedintothewatertodepth10cm.The
reading of difference in height of water column, h, of the manometer is
recorded.
Theprocedureisrepeatedwiththeamountofsaltdissolvedof400g,600g,
800g and 1000g and the respective reading of the manometer are read
respectivelyfrommetrerule.
Amountof
Differenceinheightofcolumn,h/cm
salt,m/g
200
400
600
800
1000
Agraphofdifferenceinheightofwatercolumnagainstthemassofsaltis
plotted.
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35
DEPTH WITH READING OF SPRING BALANCE (BUOYANT FORCE)
Diagram 3.1 shows a boy lifted up a rock in the sea water.
Diagram 3.2 shows the boy lifted up the rock at the surface of the sea water.
He feels much heavier than before.
Diagram 10.1
Diagram 10.2
Based on your knowledge of the buoyant force and observation above:
[1 mark]
(a) State one suitable inference.
(b)
State one suitable hypothesis.
(c)
With the use of apparatus such as tall beaker, spring balance, a metal
rod and other apparatus, describe an experiment framework to
investigate the hypothesis stated in 3(b).
In your description, state clearly the following:
(a)
(b)
(c)
Theimmersedistanceaffectsthebuoyantforce
Whentheimmersedistanceincreases,itsbuoyantforceincreasesalso
Tofindtherelationshipbetweentheimmersedistancewiththebuoyantforce
(i)
ManipulatedVariable:immersedistance
(ii)
ConstantVariable:densityofwater
(iii)
Materials:string,tapwater
(iv)
[1 mark]
(i)
(ii)
(iii)
(iv)
(v)
The procedure of the experiment which include the method of
controlling the manipulated variable and the method of measuring
the responding variable
(vi)
(vii)
The way you would analyse the data
(v)
(vi)
(vii)
Operationaldefinition
Theimmersedistanceismeasuredusingmetrerule
d=2cmandthebuoyantforceactingtothewoodenblockcanbecountedby
balance,Farerecorded.
Immersedepth,d/cm
Buoyantforce,F/N
2
10
12
Agraphofbuoyantforceagainsttheimmersedistanceisplotted.
[10 marks]
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36
TRANSFORMER (NUMBER OF TURNS OF SECONDARY COILS WITH INDUCED
CURRENT/VOLTAGE)
Diagram 4.1 shows a substation in a residential area in Shah Alam. The number of
turns of primary coil of the transformer is 200 with the voltage of 450V while the
numberofturnsofsecondarycoilofthetransformeris107withthevoltageof240V.
(a)
(b)
(c)
Themagnitudeofinducedcurrentdependsonthenumberofturnsofsecondarycoils
When the number of turns of secondary coils increases, the magnitude of induced
currentincreasesalso
To investigate the relationship between number of turns of secondary coils
(i)
withthemagnitudeofinducedcurrent
MV:numberofturnsofthesecondarycoil
(ii)
RV:Magnitudeofinducedcurrentorpotentialdifference
CV:numberofturnsofprimarycoils/strengthofmagnetused
Apparatus:softiron,ammeters/voltmeter,Cshapemagnetbars,a.cpower
(iii)
supply,bulb,connectingwires
(iv)
Diagram4.1
Basedontheaboveobservation;
(a) State one suitable inference.
(b)
(c)
With the use of apparatus such as ammeter, voltmeter, constantan wire,
metre rule and other apparatus, describe an experiment framework to
investigate the hypothesis stated in 4(b).
In your description, state clearly the following:
[1 mark]
[1 mark]
(i)
(ii)
(iii)
(iv)
(v)
The procedure of the experiment which include the method of
controlling the manipulated variable and the method of measuring
the responding variable
(vi)
(vii)
The way you would analyse the data
[10 marks]
OperationalDefinitions:
Theinducedcurrentismeasuredusingammeter
Setuptheapparatusasshown,witha240Vaccurrentsupplywith50turns
ontheprimarycoil.
Setthesecondarycoilsothatthenumberofturnsn=20
Switchonthepowersupply,measurethecurrent,I(withtheammeter)that
passesthroughthesecondarycoil.
Repeatstep2and3forn=40,60,80and100turns.
Numberofsecondarycoils,n Inducedcurrent,I/A
20
40
60
80
100
Graph of induced current, I/A against the number of secondary coils, n is
plotted.
(v)
(vi)
(vii)
-SWING OF PENDULUM BOB
-MASS/VOLUME OF WATER WITH BOILING TIME
-ANGLE OF INCIDENT WITH ANGLE OF REFRACTION
-DISTANCE BETWEEN TWO SPEAKERS WITH DISTANCE BETWEEN
TWO LOUD SOUNDS
-NUMBER OF TURNS WITH INDUCED CURRENT (LENZS LAW)
-SPEED/HEIGHT OF MAGNET BAR DROPPED WITH INDUCED CURRENT
(LENZS LAW) | 22,576 | 75,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-35 | latest | en | 0.427824 |
https://www.unitsconverters.com/en/Secondsfrom0to100mph-To-Milepersquaresecond/Unittounit-3450-3431 | 1,653,805,856,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00508.warc.gz | 1,212,262,394 | 41,126 | Formula Used
1 Meter per Square Second = 44.705 Seconds from 0 to 100 mph
1 Meter per Square Second = 0.000621371192237334 Mile per Square Second
1 Seconds from 0 to 100 mph = 1.3899366787548E-05 Mile per Square Second
Seconds from 0 to 100 mphs to Mile per Square Seconds Conversion
s stands for seconds from 0 to 100 mphs and mi/s² stands for mile per square seconds. The formula used in seconds from 0 to 100 mphs to mile per square seconds conversion is 1 Seconds from 0 to 100 mph = 1.3899366787548E-05 Mile per Square Second. In other words, 1 seconds from 0 to 100 mph is 71946 times smaller than a mile per square second. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
Convert Seconds from 0 to 100 mph to Mile per Square Second
How to convert seconds from 0 to 100 mph to mile per square second? In the acceleration measurement, first choose seconds from 0 to 100 mph from the left dropdown and mile per square second from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from mile per square second to seconds from 0 to 100 mph? You can check our mile per square second to seconds from 0 to 100 mph converter.
How to convert Seconds from 0 to 100 mph to Mile per Square Second?
The formula to convert Seconds from 0 to 100 mph to Mile per Square Second is 1 Seconds from 0 to 100 mph = 1.3899366787548E-05 Mile per Square Second. Seconds from 0 to 100 mph is 71947.6221310886 times Smaller than Mile per Square Second. Enter the value of Seconds from 0 to 100 mph and hit Convert to get value in Mile per Square Second. Check our Seconds from 0 to 100 mph to Mile per Square Second converter. Need a reverse calculation from Mile per Square Second to Seconds from 0 to 100 mph? You can check our Mile per Square Second to Seconds from 0 to 100 mph Converter.
How many Meter per Square Second is 1 Seconds from 0 to 100 mph?
1 Seconds from 0 to 100 mph is equal to 1.3899E-05 Meter per Square Second. 1 Seconds from 0 to 100 mph is 71947.6221310886 times Smaller than 1 Meter per Square Second.
How many Kilometer per Square Second is 1 Seconds from 0 to 100 mph?
1 Seconds from 0 to 100 mph is equal to 1.3899E-05 Kilometer per Square Second. 1 Seconds from 0 to 100 mph is 71947.6221310886 times Smaller than 1 Kilometer per Square Second.
How many Micrometer per Square Second is 1 Seconds from 0 to 100 mph?
1 Seconds from 0 to 100 mph is equal to 1.3899E-05 Micrometer per Square Second. 1 Seconds from 0 to 100 mph is 71947.6221310886 times Smaller than 1 Micrometer per Square Second.
How many Mile per Square Second is 1 Seconds from 0 to 100 mph?
1 Seconds from 0 to 100 mph is equal to 1.3899E-05 Mile per Square Second. 1 Seconds from 0 to 100 mph is 71947.6221310886 times Smaller than 1 Mile per Square Second.
Seconds from 0 to 100 mphs to Mile per Square Seconds Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like acceleration finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like s to mi/s² through multiplicative conversion factors. When you are converting acceleration, you need a Seconds from 0 to 100 mphs to Mile per Square Seconds converter that is elaborate and still easy to use. Converting Seconds from 0 to 100 mph to Mile per Square Second is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Seconds from 0 to 100 mph to Mile per Square Second conversion along with a table representing the entire conversion.
Let Others Know | 1,016 | 4,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-21 | latest | en | 0.859018 |
https://coolconversion.com/geometry/circle/area/What-is-the-area-of-a-circle-with-_circumference-_2.1_yard_%3F | 1,534,556,184,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213247.0/warc/CC-MAIN-20180818001437-20180818021437-00244.warc.gz | 661,381,815 | 23,333 | Please get in touch with us if you:
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4. Anything else ...
# What is the area of a circle with circumference 2.1 meters?
Here is the answer to questions like: how to find the area of a circle with circumference 2.1 meters?
Use the this circle area calculator below to find the area of a circle given its circumference, or other parameters. To calculate the area, you just need to enter a positive numeric value in one of the 3 fields of the calculator. You can also see at the bottom of the calculator, the step-by-step solution.
### Inputs:
Radius: or Diameter: or Circumference: Unit: centimeter foot inch kilometer meter mile yard
### Result:
The area of a circle with circumference 2.1 is 0.3509
### Area of a cicle in terms of radius:
Area = π . r2 = 3.14 × 0.332 = 0.35 square meters.
### In terms of diameter:
Area = π . (d/2)2 = 3.14 × (0.67/2)2 = 3.14 × (0.33)2 = 0.35 square meters.
### In terms of circumference:
Area = C2/(4π) = 2.12/ = 4.41/(4 × 3.14) = 4.41/(12.56) = 0.35 square meters.
Note: for simplicity, some results may be rounded to nearest hundredth and π was rounded to 3.14. See formula details below in this page.
### A circle of radius = 0.3342 or diameter = 0.6685 or circunference = 2.1 meters has an area of 0.3509 square meter which is equal to:
• 3.509E-7 square kilometer (km²)
• 3509 square centimeter (cm²)
• 8.671E-5 acre (ac)
• 3.509E-5 hectare (ha)
• 1.355E-7 square mile (mi²)
• 0.4197 square yard (yd²)
• 3.777 square foot (ft²)
• 543.9 square inch (in²)
• ## Formula for area of a circle
Here a three ways to find the area of a circle (formulas):
### Circle area formula in terms of circumference
See below some definitions related to the formulas:
### Circumference
Circumference is the linear distance around the circle edge.
The radius of a circle is any of the line segments from its center to its perimeter. The radius is half the diameter or r = d/2.
### Diameter
The diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints lie on the circle. The diameter is twice the radius or d = 2r.
### The Greek letter π
π represents the number Pi which is defined as the ratio of the circumference of a circle to its diameter or π = C / d. For simplicity, you can use Pi = 3.14 or Pi = 3.1415. Pi is an irrational number. The first 100 digits of Pi are: 3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 ...
### Note:
If you input the radius in centimeters, you will get the answer in square centimeters (cm²), if in inches, will get the answer in square inches (in²) and so on ...
Circumference is often misspeled as circunference.
You can use our area unit converter to see the results in other units.
### Disclaimer
While every effort is made to ensure the accuracy of the information provided on this website, we offer no warranties in relation to these informations. | 868 | 3,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-34 | latest | en | 0.808481 |
https://www.trekbbs.com/threads/u-s-s-ariel-federation-shuttlecarrier.144181/page-2 | 1,500,984,464,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425144.89/warc/CC-MAIN-20170725102459-20170725122459-00321.warc.gz | 864,620,643 | 30,847 | # U.S.S. Ariel - Federation Shuttlecarrier
Discussion in 'Fan Art' started by Cary L. Brown, Jun 27, 2011.
Joined:
Oct 14, 2005
Location:
Austin, Texas
How to make a B/C deck superstructure
Another poster here, "Circusdog," has decided to start working on his own version of the classic 1701, and like me, he's using a CAD package (as opposed to a surface-modeler like Maya or Lightwave or the like).
He was having some trouble figuring out the best way to make the unusual shape of the B/C deck superstructure, and I've offered to post my step-by-step here to help him figure it out.
First... here's the topside of the Ariel's primary hull. I originally created it solely as a set of revolved shapes, and wasn't really planning to do this detail yet (since my focus remains on the secondary hull's complex topside). But I'll have to do it sooner or later, so I've gone ahead and done it now.
The first thing to do is to turn the revolved section into a smaller region, as seen below. It's important to make sure that you have the angle of the absent area as a changeable parameter, so you can simply adjust the angle of that cut-out area (and the tangent angle of the new section) on-the-fly as you tweak the shape.
Now, with the TOS Enterprise, the cylindrical "core" is perfectly adequate, but for TMP-era ships, you can't just sweep around a cylinder and get the shape right, so I added an additional volume to provide the "core" around which I'll be sweeping my shape. This should have a surface which is normal to the shape you want to achieve. I usually use conics to create this sort of shape... this path is defined by two symmetrical conic section, normal to the cut faces.
Now, you need to create the structure you'll be sweeping... a path, and at at least three sketches for cross-sections, as seen below. If you're not happy with the shape you're able to create using three sections, you can always add more (ensuring that they're symmetrical, obviously!). Each section needs to have the same number of drawing entities, and should have its origin at the same point, facing the same direction. I set my section origins in the innermost corner, facing upwards. If you don't get the origin right, the shapes will "twist" as the shape extrudes along the path, and you'll get a MESS.
(Note that you want the new, swept surfaces to be tangent to the existing surfaces. You need to have a surface adjacent to each of those sketch curve edges, so you can establish the tangency of the swept shape against those surfaces.)
I'll show the result of that sweep in no-hidden-line wireframe mode. Now, in this case, I wanted my upper "lip" and lower "lip" for the inset region to be unrelated, so I left the lower lip out of the first sweep. (This is not necessary for the TOS ship, obviously!)
I created a sweep profile for the lower lip... this is a 2D sketch I created, projected onto the surfaces of the inset region. I created three profiles, again, just as before.
Here's the result. Note that I've added a few rounds to help make it look more like a real object... the rounds were not part of the original sweeps.
At this point, you start comparing your shape to the sketches (which you should have been doing all along, obviously, but now you make it into the "final" shape). You tweak the various parameters in your shapes... the "cut-out angle" for example... and adjust things to get as close as you can to the drawn images.
The following two images allow you to compare my final result to the "trace sketch reference" I'm using. The lateral view is nearly perfect... and I'm not inclined to make any further tweaks here.
The top-down view is very close... but not totally exact. As I said before, I could add more sections to match the top-down view more closely, but I don't WANT to do that... I actually like the look I've got here better.
And that's all there is to it. Hope that helps... or at least is of some interest to the rest of you!
2. ### Science OfficerLieutenant CommanderRed Shirt
Joined:
Oct 28, 2009
Location:
United Kingdom
Hi Cary,
Thanks for the tutorial - I never considered doing something like that to create the B/C decks. In C4D I think there are various NURBs options that can achieve the same thing.
A question on the your B/C/Bridge decks. Are the drawings meant to represent the exact structure used on the TMP NCC-1701 or a variation?
I ask because that lower lip (below the windows) looks very low. Looking at the side view drawing, the lip on the LHS of the structure is inconsistent with the line drawn across (as on the LHS it's twice as high).
Cheers,
S.O.
Joined:
Oct 14, 2005
Location:
Austin, Texas
Hi S.O.
No, it's not a 1:1 match to the Enterprise's superstructure. In fact, if you review the earlier entry where I lay out decklines, you'll see that this entire section (what on the Enterprise is B and C decks) ONLY includes B deck on this ship.
I didn't create the drawings, of course... also as mentioned earlier, this is Aridas' design, and I'm trying to replicate it as closely as possible, so I'm trying to match his drawings.
FYI, many ships designed in the post-TWOK era were similar to, but not identical to, the 1701. A great example of this is the Federation class Dreadnought (uprated to TMP-era specs). Most people look at this and see the Enterprise with only a third nacelle added and different engine attachment, but this is not true.
Major differences in the TMP-era dreadnought from the TMP-era Enterprise include:
a) A dramatically larger B/C deck structure... nearly twice the volume of the one on the 1701.
b) A different bridge module, with an extended aft section incorporating dual docking ports.
c) A much larger secondary hull... fully 1/3 longer.
d) An extended "lip" at the landing bay doors.
e) two aft-firing photon torpedo launchers (inline with the forward ones)
That's in addition to the top-mounted nacelle and the different lower nacelle mounting.
This is what was done during that time period. Similar shapes and functions... similar STYLE, really... but SIMILAR, not IDENTICAL.
In the case of the Ariel, the protruding stuff on top is actually smaller than on the Enterprise. But the saucer is quite a bit larger in diameter, and has a much different shape (with no undercut). It's only 8 decks thick, though... I'm anxious to do a volume check on it, and to compare this to the TMP 1701 saucer's volume.
In my personal worldview, the inset ring around both the TMP 1701's B/C deck superstructure and this ship's B-deck superstructure is, in fact, the main subspace antenna. A similar device was inside of the TOS ship's B/C deck superstructure, as well... which is the sole reason for this structure's rather unique shape.
Obviously, by the time the Excelsior launched, a different antenna configuration had been introduced which didn't require such a massive, exposed slab of antenna material.
Joined:
Sep 19, 2007
Location:
Boyertown, PA as of July 2011
awesome 3D work so far
Joined:
Oct 14, 2005
Location:
Austin, Texas
Docking Ports - Dimensions?
I need some help here. I know that it's been well-established, the exact size of the TMP-era starship docking ports. Unfortunately, I can't seem to find this information anywhere
(Google used to be useful, but now you only get results based on advertising revenue to Google... so no matter how I phrase my query I get the same damned list of WRONG answers every time!)
I'm most specifically looking for the actual enclosed diameter. I could derive it, or take my best guess, but I KNOW that people have figured this out previously, so why re-invent the wheel?
Also... if anyone has a model they've made, using TMP warp nacelles as separate components in their mix, and is willing to share them, I'd be very grateful. OBJ, DXF, Maya, Lightwave... anything like that should be useable, as long as it's NURBS based (not polygonal, or "sub-div", based... those won't work for me).
I'll give credit where due, no worries about that.
Last edited: Jul 2, 2011
6. ### Bernard GuignardFleet CaptainFleet Captain
Joined:
Jan 25, 2005
Location:
Ontario
Re: Docking Ports - Dimensions?
Hello Cary
Here are some dimensions that I took off an Autocad Bridge blueprint that was sent to me it looks pretty good to me
Joined:
Oct 14, 2005
Location:
Austin, Texas
The past several days have been spent experimenting with the secondary hull shape, with some input from Aridas. In the end, I ended up keeping pretty much the shape I'd come up with originally, but it was worthwhile effort in any case.
The one significant change in the secondary hull, noticeable to anyone but me anyway, is the addition of the deflector housing. I'll end up "re-ordering" the model to put this after the hull solidification once I get the inner shape worked out... but for now, I'm using it for reference to ensure that the hull surface entirely encompasses this, complete with a 1m thick anti-radiation wall around the entire compartment. (The dish itself, and the various supporting hardware for it, are not yet present, of course... but this will end up looking very much like the TMP Enterprise's dish, just less exposed.)
In addition, doing my B-deck structure got me onto primary hull detailing. Eventually, I'm going to split the model into two, but for now (until the neck structure is final), it's all in one file. Eventually, I plan to have four components... a saucer, a secondary hull, and a pair of nacelles. I'm going to get those patterned up with an SLA machine, so I can have a physical copy of this on my shelf!
On the primary hull, I got the shapes final on the upper decks (A deck and B deck), as you can see here. There are some fine details to be added, but the overall form you see here is "final."
On the underside, I got pretty far along on the lower scanner platform and sensor dome. I think I have a little bit more shape-tweaking to do, and I need to add the surface vents to each of the four legs of this shape, and then create the "scanner windows" and other details inside of each of the four scanner ports. And, obviously I think, the dome needs some detailing added as well!
This is essentially the same technology, even likely the exact identical hardware, as on the Enterprise, but the housing is not shaped identically at this point (again, matching up the shape from the port view on the drawing I provided in my first post in this thread).
Finally, the standard orthos, for comparison to the drawing views:
And finally, one somewhat nicer image...
Last edited: Jul 2, 2011
Joined:
Mar 22, 2010
Thanks ever so much for doing this. I love all the other Planet of the Titans entries but have wanted to see a 3d model of this for a very long time. I like how you made her a through deck design. That secondary hull is quite fluid.
Joined:
Oct 14, 2005
Location:
Austin, Texas
I'm afraid you've gotten a misconception... remember, this is not finished... it's a "work in process."
The Ariel is NOT a "through-deck carrier." You can see the bay outlines from the front, because I want to be able to see that surface (with those openings) as I create the remaining volume of the secondary hull.
There WILL be some "from the front" landing capabilities, however. I have discussed this with Aridas, and if you look at the top-down view, you'll see four squares with four smaller squares visible in them. Those are lift platforms. You can best visualize these by thinking about the landing platforms seen on "Space 1999" for the Eagles. But they are on the "forward side" of the secondary hull... so in THAT sense you will "sort of" get what you were looking for.
I guess I really do need to get my secondary hull shape finished, don't I?
I've already worked out, in my mind, the internal layout I plant for this ship to have. Aridas has given me some feedback on what he was envisioning, and I'm pretty comfortable with how I'm going to make it work, and only have to implement that (but I can't do that until I finish the solid form of the secondary hull).
Maybe I'll spend a few hours tonight and try to make that work, huh?
10. ### beamMeCommodore
Joined:
Mar 17, 2011
Location:
Europa
Good lord, that's ugly.
The ship - not the modelling.
Joined:
Oct 14, 2005
Location:
Austin, Texas
Hey, to each their own... I'd wager that there are ships you think are good-looking that I'd feel the same way over. I've always liked this one, personally...
Joined:
Oct 14, 2005
Location:
Austin, Texas
I've been struggling to get the "wing back" surface to work. This is one of the areas where Pro/E, frankly, sucks... surface-based modeling. I've gotten the whole thing set up, and I can create the curved surface I want, but the moment I try to assign tangency conditions, the thing flips out. There doesn't seem to be rhyme or reason behind it, though I'm sure I'll figure it out (Pro/E is full of annoying "long-term bugs" that I've had to learn ways around over the years!)
So... at the moment, my surfaces are NOT tangent to the adjacent surfaces, as I intend them to be. But you can at least get the general idea.
Also, as I always knew would be the case, once I started doing this, my neck (but not the surfaces I created to make it up) failed. So, I've got my surfaces visible, but the nice, clean structure I've had before isn't there. Mainly, you can see the neck structure extending well into regions where it doesn't "really" exist, at the moment at least. I'll fix this once I've got the secondary hull top surface fixed.
Here's a perspective view showing the new surfaces. Again, these are SUPPOSED to be tangent to the top, aft surface of the secondary hull, but they're not. Also, I've got some odd inflection event occurring out near the nacelle pylons. I'll be fixing all of that, of course. Or... maybe I'll just take this out of Pro/E to create the surface, and then import that surface back in? I'm just really annoyed at how "uncooperative" this one feature is, especially compared to how nicely this sort of thing works in other packages I've used. (I guess nothing can be perfect!)
Next, here's the view from ahead... you can see the relationship between the neck and the body fairly well from this view. Note that there will not be a sharp intersection... instead, I'll be adding a fillet to blend the neck into the body, with the upper extent of that fillet being where you can see the blue line.
Finally, the port-side elevation. This shows, better than the other views, how nicely I've managed to match the "original design intent" so far. I know what I have to do (and I know how to do it in Lightwave, Maya, or yes, even AutoDesk INVENTOR)... but Pro/E is really fighting back with me on this.
Once I can get the surface to work as I want it to, I'll solidify it, then restore the neck, blend the two... and then I'll be able to start dealing with detailing.
I've decided that the nacelle is something I'm going to get creative with... since nobody has been able to provide me with a useable LN-64 nacelle set I can modify, I've decided to create "family table" for my nacelle.
There will, literally, only be one "part" I'll be creating, but I can create additional features which can be turned on or off (top-side or bottom-side control reactor rib, interior grills, intercooler, etc), and I can assign parameters that I can feed in to adjust different elements (in this case, the starting point and ending point for the location of the control reactor rib).
I can then just type entries into the "family table" (basically a spreadsheet" to create new "instances" that I can use as components in other assemblies.
This means I'll do one LN-64 nacelle... and I'll be able to use that part for a left-hand, right-hand, middle, or "solo" nacelle... mounted from the top or from the bottom... and mounted with any configuration of pylon we've ever seen.
Which, I think, will lead me inevitably into making more TMP-era ships. It's mainly been the nacelle that I've been avoiding, re: ships from this era. So... now, I think I can just do it one time, and never have to worry about it again.
13. ### MLJamesLieutenant CommanderRed Shirt
Joined:
Sep 26, 2010
Location:
Riverbank, CA
According to the original blueprints for Ariel, she doesn't use a standard LN-64 nacelle. Her nacelles are larger in all three dimensions, but, unfortunately, the percentage increase is different for length, beam, and draft. You won't be able to simply apply one factor to a basic LN-64 if you want her to be completely accurate.
Joined:
Oct 14, 2005
Location:
Austin, Texas
FYI... I'm leaving the original text of my post, just for context... but adding to the end.
Well, that's quite an interesting claim you make, MLJames.
First off, there are no "official blueprints." The closest thing to that is the two-view sheet (and descriptive page) I provided at the outset of this thread. Do you have other blueprints which you believe are "more official" than those two sheets?
Second... the original designer of the Ariel, Aridas Sofia, is working with me as I do this. If HE chose to tell me that the nacelles I'm using are wrong, that would be his prerogative. But nobody else can say anything of the sort.
Third, the text on the "descriptive sheet" I copied at the top of this thread reads as follows:
Now... that was written by Aridas Sofia, who (in case you've forgotten) is the guy who designed the Ariel.
The LN-64 (including mods 1, 2, and 3) and the LN-65 are externally identical engines. The LN-68 is subtly different, mainly having a sloped top-front area. These various nacelle concepts have been around in "fannon" works for a long time. I'm not inventing them myself, in other words.
SO... given what I've just presented, perhaps you can support your own claim with something more than just the claim itself?
In particular, I'd love to see the "official blueprints" for this ship, as you mentioned, and know who published them, who drew them, etc, etc.
********
EDIT: Okay, MLJames didn't give me enough information, but it turns out that he was actually correct...
Here is an LN-64 (under construction) in the location, along with the backdrop of the engine Aridas put into the design... and even though other sources I've read describe the LN-65 as being in the same housing as the LN-64, clearly this isn't the same unit.
I wish MLJames had been more clear, but it turns out he was right. Apology issued...
Last edited: Jul 4, 2011
15. ### MLJamesLieutenant CommanderRed Shirt
Joined:
Sep 26, 2010
Location:
Riverbank, CA
No biggie. I honestly didn't think I needed to go into much more detail, because the info I was citing was right off of aridas' pages from the development chart. That was the "original blueprints" I was talking about, and it was one of the images in your first post.
16. ### Bernard GuignardFleet CaptainFleet Captain
Joined:
Jan 25, 2005
Location:
Ontario
While it looks to be a little longer on the elevation view
how does the warp engine look on the dorsal view looks
like All you would have to do is Stretch out an LN-64 in
the elevation view. Cary I've collected a wack load of
warp engine drawings some from Gary Kerr would you like
me to send you those for reference? if so PM me with your
e-mail address. Great work by the way on the nacelle
I know that even drawing them in 2d is a pain
Joined:
Oct 14, 2005
Location:
Austin, Texas
Thanks, Bernard... I'll PM you with that info momentarily.
Basically, I made a copy of my LN-64 and edited the features making it up, and got an LN-65.
My first step in the nacelle creation was simply creating a "box" of datum planes representing the dimensions that Aridas listed on his drawing. I then took his two provided views and scaled the images of the nacelles in those views to exactly fit in the boxes (granted, with lines having a fairly large thickness at this scale, I just "eyeballed" things so that my real planes, or real lines for that matter, fall at about the midpoint of any of the drawn lines).
I also used the LN-64 front view, rear view, and bottom view... tweaked and stretched to best match the shape of the "box" in their particular view. They turned out pretty close, in most respects, so I was easily able to interpolate the LN-65 from the LN-64 in those views.
One thing I noticed is that the nacelles in the drawing seem to be tilted slightly nose-downwards. I'm not going to do that, however... not unless Aridas intended them to "nose down" like that. It's only by about two and a half degrees, but it's still noticeable.
One thing you'll likely notice is that even the new "LN-65" nacelle doesn't perfectly match the drawing. This is because I used the written dimensions to determine size, and the "on-print" size does not match the value given in the table.
Similarly, you may have noticed that my primary hull diameter looks "small" in the top-down view (and has its axis forward of where it is in the print by a small amount). Again, in this case, I kept the overall size of the ship as my "stake in the ground," and then treated each sub-component as being built, again, in accordance with the overall sizes listed. The amount each is off is Small , but noticeable when compared to the drawing.
For reference, here's the LN-64:
And here is the LN-65:
Now, when you look at these images, remember that I have only the "untrimmed" surface for the neck... so please disregard the excess "neck material" extending below the top of the secondary hull.
Here are the top view and the port and front elevation views.
Finally, while I'm posting, I thought I'd just toss up the current state of my nacelles... the LN-65 is shown, but the LN-64 is in the same state.
Again, I'm going to model the basic nacelle, then create some additional features ("inside grill," "outside light panel," three different intercooler fins, and two different, and parametrically-modifiable, control reactor "ribs" (top and bottom)). I'll then be able to make any particular installation of these to any ship I want, in the future, and will only have to tweak one part in order to get port, starboard, solo, or center-mounted nacelles... using a trick called a "family table" which is common in CAD.
Joined:
Mar 22, 2010
Good job on the nacelles. I see what you meant earlier, in that you are keeping the flowing form of the secondary hull and not building a spolier-like terrace in the back for the shuttles.
You know, there is nothing that says you have to have an Excelsior styled recessed sensor/nav-deflector dish. A tesselated skin-tight phased array of plating that merges into the aztec might work. With Shuttle standing down, I just noticed how the secondary hull evokes the STS orbiter after a fashion.
I mentioned the similarity between ST 6 and Atlantis last flight here
http://federationreference.prophpbb.com/post11560.html#p11560
Nice TOS nacelles
http://img221.imageshack.us/img221/2273/nacellewip.jpg
Aridas also had the idea for ACE nacelles. I wonder if that might be easier to do:
http://federationreference.prophpbb.com/topic6-220.html#p11575
Joined:
Oct 14, 2005
Location:
Austin, Texas
A quick update...
I've been focusing on the warp nacelles for several days now (or rather, what little time I've spent on this has been on the nacelles).
The "base" nacelles are ALMOST done now. I've done everything except for the coppery shapes near the front.
After that, I'll create separate features for the three different types of intercoolers (port, starboard, symmetrical), the inboard grills (port and starboard) and external "light panel" (port and starboard), and the upper and lower "control reactor ribs." I'll set those up as variable entries, so I can use this as all of the nacelles, anywhere this type is used.
Again, this is the very slightly larger LN-65. I haven't done all this on the 64 yet, but I do plan to.
Joined:
Oct 14, 2005
Location:
Austin, Texas
Saucer Internal Layout - Starting off
Okay, even though I don't plan on doing a total, complete interior for this ship, I did want to figure out generally how it would be laid out internally. So, I've cut spaces into the interior of the saucer to help me with this.
You can see that there are eight total decks in this saucer, including the relatively small bridge at the top and the sensor/scanner operations center at the bottom. There's also a lot of "wasted space" (from a habitation standpoint, not from a "storage and equipment" standpoint, realize!) due to the slope of the hull surfaces. I've chosen to cut off the habitable spaces based upon a minimum 2m (~6ft) headroom limit. Of course, those sub-height areas will be of limited use... lower-level ones will be store-rooms, mainly, and upper-level ones can be cabins.
I put a 6' tall figure in the bridge cavity. It matches up pretty well with the TMP bridge set, I think... but it's worth noting that the latter-film bridges would not fit into this space.
The shape of the saucer precludes two full decks, with proper interstitial spaces, in the outer rim, unless the decks in the saucer are much shorter than is normal (say, 7.5' decks?). Right now, you've got 9.5' ceiling height, which is pretty much what we've seen in every series and every film.
So, what I've done is provided for a "high-bay" out there. I'm thinking that the "high bay" will only exist adjacent to the window clusters, and will represent recreation/lounge facilities. A catwalk along the outer wall will allow those windows to make sense, while there's no other way I can make them make any sense (unless I just treat them as sensors and ignore the idea of windows completely). | 6,107 | 25,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-30 | longest | en | 0.970134 |
https://nugae.wordpress.com/og-3/ | 1,501,181,012,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549429417.40/warc/CC-MAIN-20170727182552-20170727202552-00660.warc.gz | 661,691,492 | 30,231 | # The inadequacy of the circular vernier
Many people have thought of making clocks by taking a linear vernier and bending it into a circle. This is not an adequate solution, and here are the reasons why.
A circular vernier has a static disc and a rotor that look like this:
This is what happens when the rotor rotates. The first picture is the rotor on its own; the second picture is the rotor on top of the static disc.
The resultant pattern does indeed rotate 12 times as fast as the rotor, just as the minute hand of a clock or watch rotates 12 times as fast as the hour hand.
But the pattern is far from being a simple hand pointing in a specific direction. It is impossible to read it with any accuracy. For example, here are two snapshots, one taken at 20 past the hour and one at 25 past the hour:
Even accepting that the “hands” in this picture are white on a black background, it is impossible to work out where they are pointing.
The vernier principle is necessary for the construction of clock displays with a single moving part, but it is not sufficient. The resulting displays do not work.
Optical gearing adds advanced mathematics, optics, and accurate engineering to produce workable clock displays with only one moving part.
The Principles of Optical Gearing | 271 | 1,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-30 | longest | en | 0.94259 |
https://www.includehelp.com/scala/convert-hex-string-to-long.aspx | 1,656,692,824,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00679.warc.gz | 882,906,488 | 37,580 | # How to convert hex string to long in Scala?
Scala | Convert Hex String to Long: Here, we are going to learn how to convert hex string to long in Scala?
Submitted by Shivang Yadav, on May 28, 2020
## Hex String
Hex String also is known as the hexadecimal string is a string of hexadecimal digits i.e. base 16 numbers.
Example:
``` string = "49AD1"
```
## Long Integers
Long integers are storages for large integer values. It can store 64-bit signed integer value. The maximum value it can store is 9223372036854775807.
## Convert hex string to long in Scala
We can convert a hex string to long in Scala by following these steps,
• Step1: We will convert the hex string to integer values using the parseInt() method.
• Step2: Then, we will convert this integer value to long value using the toLong method.
Program to convert hex string to long in Scala
```object MyObject {
def main(args: Array[String]) {
val hexString : String = "42e576f7"
println("The Hex String is " + hexString)
val intVal : Int = Integer.parseInt(hexString, 16)
val longInt = intVal.toLong
println("HexString to Long value : " + longInt)
}
}
```
Output:
```The Hex String is 42e576f7
HexString to Long value : 1122334455
```
Explanation:
In the above code, we have created a hexadecimal string named hexString. Then we have converted hexString to integer value using the parseInt() method of the Integer class and store it to intVal. This intVal is converted to a Long Integer value using the toLong method and prints the value using println method.
Preparation
What's New
Top Interview Coding Problems/Challenges! | 392 | 1,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-27 | longest | en | 0.65893 |
https://numbersworksheet.com/adding-unlike-rational-numbers-worksheet/ | 1,716,881,154,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00714.warc.gz | 361,021,745 | 14,473 | # Adding Unlike Rational Numbers Worksheet
A Reasonable Numbers Worksheet will help your child become more knowledgeable about the principles right behind this rate of integers. In this worksheet, college students can remedy 12 various troubles related to rational expressions. They are going to discover ways to grow a couple of numbers, group them in pairs, and figure out their products. They will also practice simplifying realistic expression. Once they have perfected these methods, this worksheet will be a important instrument for advancing their scientific studies. Adding Unlike Rational Numbers Worksheet.
## Realistic Numbers really are a ratio of integers
There are two forms of phone numbers: rational and irrational. Realistic phone numbers are understood to be entire phone numbers, while irrational phone numbers do not recurring, and also have an endless number of digits. Irrational figures are no-zero, low-terminating decimals, and square origins which are not excellent squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To establish a rational number, you must know just what a realistic quantity is. An integer can be a complete quantity, and a rational quantity is actually a proportion of two integers. The percentage of two integers may be the variety ahead divided by the amount at the base. If two integers are two and five, this would be an integer, for example. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be made in to a portion
A reasonable variety includes a numerator and denominator which are not no. Because of this they can be conveyed being a small fraction. Together with their integer numerators and denominators, realistic numbers can furthermore have a unfavorable importance. The negative importance must be placed to the left of and its total importance is its length from no. To make simpler this instance, we shall state that .0333333 is a portion that can be published like a 1/3.
Together with adverse integers, a realistic number can even be produced in to a small percentage. For instance, /18,572 can be a realistic variety, whilst -1/ is not. Any fraction comprised of integers is realistic, given that the denominator fails to consist of a and will be created for an integer. Also, a decimal that ends in a point can be another realistic variety.
## They are sensation
Despite their label, logical numbers don’t make much sense. In mathematics, they are solitary entities by using a special size in the amount line. Which means that once we count up anything, we can easily purchase the dimensions by its rate to its original volume. This holds accurate even though there are endless rational phone numbers involving two certain phone numbers. In other words, numbers should make sense only if they are ordered. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To get the duration of a pearl, for example, we could count up its breadth. Just one pearl weighs twenty kgs, which is a realistic quantity. Moreover, a pound’s bodyweight is equal to 15 kilos. Hence, we should certainly split a lb by 15, without the need of be worried about the length of an individual pearl.
## They may be indicated as being a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal amount could be composed as being a numerous of two integers, so 4 times 5 is equal to 8-10. An identical issue involves the repetitive fraction 2/1, and either side ought to be separated by 99 to find the right respond to. But how can you have the conversion? Here are several examples.
A logical variety may also be designed in many forms, which include fractions and a decimal. A great way to stand for a logical variety within a decimal is to break down it into its fractional comparable. There are three ways to split a rational variety, and all these ways brings its decimal equal. One of these brilliant approaches is usually to break down it into its fractional equal, and that’s what’s referred to as a terminating decimal. | 877 | 4,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-22 | latest | en | 0.95402 |
https://crm2.univ-lorraine.fr/mathcryst/uberlandia2012.php | 1,632,374,894,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.10/warc/CC-MAIN-20210923044248-20210923074248-00246.warc.gz | 235,460,404 | 8,325 | # Scientific Program
#### FUNDAMENTALS
• Introduction to matrix algebra.
• Basics of group theory. Space groups and space group types.
#### BASIC CONCEPTS
• Affine spaces vs. vector spaces
• Mappings: affine and Euclidean; isometries
• Sets; homomorphisms, isomorphisms, automorphisms
• Symmetry operations as special case of isometries
• Introduction to group theory: abstract groups, subgroups, cosets
#### CRYSTALLOGRAPHY IN DIRECT SPACE
• Periodic structure of the crystalline matter
• Crystal lattice vs. crystal pattern and crystal structure
• Crystallographic calculations in direct space
• Symmetry directions in a lattice
• Unit cells: primitive cells, multiple cells, conventional cells in 2D and 3D
• Crystal families
• Symmetry groups and types of symmetry in direct space
• morphological symmetry
• symmetry of physical properties
• symmetry of lattices
• symmetry of the unit cell content
• symmetry of crystallographic patterns
• Hermann-Mauguin symbols for point groups
• Lattice systems
• Crystal systems
• Stereographic projection and the morphology of crystals
• Types of crystallographic point groups through the stereographic projection
• Symmetry operations with a glide component
• Space groups and types of space groups
• Exercises on the space group diagrams from Volume A of the International Tables for Crystallography
• Online tools for crystallography
#### PHYSICS OF DIFFRACTION AND CRYSTALLOGRAPHY IN RECIPROCAL SPACE
• Interaction of X-rays with crystalline matter
• Fourier transforms and convolutions
• Crystallographic calculations in reciprocal space
• Diffraction symmetry: Laue classes, Friedel's law, resonant scattering
• Integral, zonal and serial reflection conditions and their use to determine the space-group type
• Structure solution and refinement: introductory strategies. Introduction to JANA software. Examples.
#### CRYSTALLOGRAPHY ONLINE
• Introduction to the Bilbao Crystallographic Server
• Space Groups Retrieval Tools
• Group - Subgroup Relations of Space Groups
• Structure utilities and computer tools: different structure descriptions and structural relationships
• Tools for the study of structural phase transitions on the Bilbao Crystallographic Server
#### TWINNED CRYSTALS
• Reticular classification of twins and its extensions
• Methods of calculations of twin elements and derivation of twin laws.
• The Coincidence-Site Lattice theory: application to the derivation of twin lattices and their classification in terms of twin index and obliquity.
• Translational and point symmetry of twins: twin point groups and their chromatic features.
• Diffraction patterns of twinned crystals.
Details of the program
Preparatory SessionInternational School on Fundamental Crystallography (ISFC)Workshop on twinned crystalsISFC (cont.)
Day/TimeSunday
(Nov. 25)
Monday
(Nov. 26)
Tuesday
(Nov 27)
Wednesday
(Nov 28)
Thursday
(Nov 29)
Friday
(Nov 30)
Saturday
(Dec 01)
Sunday
(Dec 02)
Monday
(Dec 3)
8:30 - 9:00Registration Registration/
Opening
Tutorial on point groups Nespolo/Aroyo Group-subgroup relations Aroyo Physics of Diffraction (I) Fourier transforms. Space group in reciprocal space. Estevez Search for pseudosymmetries Aroyo Twinned crystals - Introduction and definitions Nespolo Free Tutorial on Jana2006 (III) Twinned crystals - basic examples Dusek
9:00 - 10:00 Introduction to abstract groups Nespolo
10:00 - 10:30Coffee break Coffee break Coffee break Coffee break Coffee break Coffee break Coffee break Coffee break
10:30 - 12:30Matrix Algebra Estevez Crystallographic symmetry in E2 Nespolo Crystallographic symmetry in E3 - part II: space groups Nespolo Bilbao Crystallographic Server for subgroups Aroyo Data processing and reduction Suescun Symmetry-mode analysis with Amplimodes Aroyo Twinned crystals - Calculations of twin parameters Nespolo Tutorial on Jana2006 (VI) Twinned crystals - advanced examples Dusek
12:00 - 14:00Lunch Lunch Lunch Lunch Lunch Lunch Lunch Lunch
14:00 - 16:00Fourier Transform Estevez/Serrano Crystallographic symmetry in E3 - part 1: point groups Nespolo Exercises on space groups construction and use of Volume A of the International Tables Nespolo Structure Utilities Aroyo Physics of Diffraction (II). Systematic absences, reflection conditions and space group determination Granado Tutorial on Jana2006 (I) Introduction to Jana2006. Basic single crystal example Dusek Twinned crystals - polychromatic symmetry Nespolo Tutorial on Jana2006 (V) Simple modulated structure application of rigid body for powders Dusek
16:00 - 16:30Coffee break Coffee break Coffee break Coffee break Coffee break Coffee break Coffee break Coffee break
16:30 - 18:30Diffraction Physics Estevez/Serrano Tutorial on point groups Nespolo/Aroyo Bilbao Crystallographc Server: retrieval of space group information Aroyo Structure Utilities Aroyo Strategies for structure solution and refinement Suescun/Granado Tutorial on Jana2006 (II). Basic powder example. Disorder Dusek Twinned crystals - Tutorial on the software GEMINOGRAPHY Nespolo Closing
Welcome reception
## Language
The official language of the Schools was English. No simultaneous interpretation was provided. However, local tutors helped the participants during the exercises.
### Venue
The School was held at the Institute of Physics of the Federal University of Uberlândia, in the southeast of Brazil. PDF of the Campus Map available for download.
### International Program Committee
• Prof. Massimo Nespolo, Université de Lorraine, France, Chair.
• Prof. Mois Ilia Aroyo, Universidad del País Vasco, Bilbao, Spain.
• Prof. Ernesto Estevez Rams, Universidad de la Habana, Cuba.
• Prof. Raimundo Lora Serrano, Universidade Federal de Uberlândia, Brazil.
• Prof. Michal Dusek, Academy of Sciences. Czech Republic.
• Prof. Leopoldo Suescun, Universidad de la República, Uruguay.
• Prof. Rômulo Simões Angélica, Universidade Federal do Pará, Brazil.
### Lecturers
• Prof. Massimo Nespolo, Université de Lorraine, France.
• Prof. Mois Ilia Aroyo, Universidad del País Vasco, Bilbao, Spain.
• Prof. Ernesto Estevez Rams, Universidad de la Habana, Cuba.
• Prof. Michal Dusek, Academy of Sciences. Czech Republic.
• Prof. Leopoldo Suescun, Universidad de la República, Uruguay.
• Prof. Raimundo Lora Serrano, Universidade Federal de Uberlândia, Brazil.
### Local Organizing Committee
• Prof. Raimundo Lora Serrano, Chair
• Prof. Omar de Oliveira Diniz Neto
• Prof. Augusto Miguel Alcalde Milla
• Prof. Silvana Guilardi
• Prof. Eduardo de Faria Franca
• Prof. José de los Santos Guerra.
• Vinicius Soares O. de Sousa (Secretary)
### Photos
A selection of ptohos is available at the IUCr gallery website (three pages, clic on the "Next" link in the bottom-right corner). | 1,632 | 6,733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-39 | latest | en | 0.671748 |
http://www.buenastareas.com/ensayos/Vyaderm/3472564.html | 1,511,215,248,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00276.warc.gz | 354,934,117 | 12,769 | Solo disponible en BuenasTareas
• Páginas : 11 (2666 palabras )
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• Publicado : 12 de febrero de 2012
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[pic]
EVA – A CATHARTIC CHANGE?!
Prof. Zhaoyang Gu
Course
45-701
By:
Neha Arya
Marc Brands
Anil Konjalwar
EVA: North American Dermatology Division
We will first calculate the 1999 actual EVA retroactively and if our figure matches Vyaderm’s then we will use that method to calculate EVA for 2000 and 2001.
1999 EVA Calculation:|In ‘000s |1999 |
|Operating earnings |20,000 |
|add R& D expense |20,000 |
|minus R & D amort. |14,972.8 |
|NOPBT |27530.86|
|less taxes |7875 |
|NOPAT |19655.86 |
|In ‘000s |1999 |
|NOA |110000 |
|Capital|152141.46 |
EVA = NOPAT – [Capital * Cost of Capital]
EVA = 19655.86 – [152141.46 * 0.11]
EVA = 2920.29 ~ 2920
Since this is the figure calculated by Vyaderm, we have arrived at the figures presented below using the same technique.
The figures requested by Maurice Vedrine:
• 2000 EVA for the North American Division – \$31,361,000
• 2000 EVA bonuspayout - \$252,000
• 2001 EVA for the North American Division – \$(6,587,000)
• 2001 EVA bonus payout - \$0
The numbers used for the calculation of division EVA are presented in Exhibit 1. These were used to arrive at the report given in Exhibit 2. The methodology and assumptions for 2001 are outlined in Exhibit 3.
As is evident from the ending bank balance of the manager for 2001, notonly will she not get any bonus for the year 2001 but will have to work off the negative balance for the year 2002 and onwards.
However, the lump sum she gets in 2000 is very large compared to her usual annual bonus. In fact it is more than 4 times her bonus in 1999. Since she knows her bonus may not materialize for many years (it depends on factors that are not entirely in her control), what’sstopping her from taking the bonus this year and quitting the next?
The current system is obviously flawed but not irredeemably so. One suggestion is having a cap and a threshold for bonus payouts of exceptionally good and bad years respectively. This does not mean that the manager’s earned bonus will be capped for good years, it will just stay in his bonus bank. Similarly for really bad yearssome threshold payout will still be made and in case the bank balance is zero or negative it will be treated as a loan from the company which has to be paid back interest free, the manager will have the option of refusing this loan.
The next sections deal with a thorough analysis of EVA at Vyaderm and our recommendations based on the above anomaly in the system.
The Merits of EVA: AnEvaluation
To evaluate the claims made in the Executive presentation, we have looked at all six claims individually and tried to analyze each in the context of Vyaderm. The first three points are related to the traditional system.
1. Caps limit incentive for exceptional performance
Agree
o Once a salesperson/manager achieves their cap, there is no incentive to go beyondthe cap. Though there is no explicit reference to how the caps were set in Vyaderm, but the case does mention that the old compensation system included an annual bonus. 50% of the bonus was based on objective operative results. The other 50% was based on subjective evaluation of the manager’s personal contribution. Furthermore, an inherent flaw in the system was managers spending more time... | 862 | 3,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-47 | latest | en | 0.866819 |
http://mathhelpforum.com/advanced-algebra/83922-spanning-sets.html | 1,481,066,193,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542002.53/warc/CC-MAIN-20161202170902-00421-ip-10-31-129-80.ec2.internal.warc.gz | 151,085,733 | 9,873 | 1. Spanning sets?
I'm confused about spanning sets...I can't find any information about them from my book and I must have forgotten going over it in class.
Here's the question:
"W is the subspace of R^4 consisting of all vectors of the form [a,b,a-b,a+c] where a, b, c are in R. Find a spanning set for W."
A spanning set is a set of vectors whose linear combination is the subspace W (is this correct?)
So would this be a spanning set?
{a[1,0,1,1], b[0,1,-1,0], c[0,0,0,1]}
2. Originally Posted by paulrb
I'm confused about spanning sets...I can't find any information about them from my book and I must have forgotten going over it in class.
Here's the question:
"W is the subspace of R^4 consisting of all vectors of the form [a,b,a-b,a+c] where a, b, c are in R. Find a spanning set for W."
A spanning set is a set of vectors whose linear combination is the subspace W (is this correct?)
So would this be a spanning set?
{a[1,0,1,1], b[0,1,-1,0], c[0,0,0,1]}
you're almost correct! the set of three vectors $\begin{bmatrix} 1 \\ 0 \\ 1 \\ 1 \end{bmatrix}, \ \begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix}, \ \begin{bmatrix}0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$ is a spanning set for $W.$ (it's clear that these three vectors are linearly independent. so $\dim W=3.$) | 395 | 1,273 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2016-50 | longest | en | 0.907544 |
https://ru.scribd.com/document/405741785/JEE-Main-2019-Mathematics-April-Attempt-Shift-1-10th-April-2019 | 1,566,129,127,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313803.9/warc/CC-MAIN-20190818104019-20190818130019-00474.warc.gz | 636,882,112 | 69,778 | You are on page 1of 19
PAPER-1 (B.E./B. TECH.
JEE (Main) 2019
COMPUTER BASED TEST (CBT)
Questions & Solutions
Date: 10 April, 2019 (SHIFT-1) | TIME: 09.30 A.M. to 12.30 P.M.
Duration: 3 Hours | Max. Marks: 360
SUBJECT: MATHEMATICS
_______________________________________________________________________________________________________________
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| JEE(MAIN) 2019 | DATE : 10-04-2019 (SHIFT-1) | PAPER-1 | OFFICIAL PAPER | MATHEMATICS
PART : MATHEMATICS
Straight Objective Type (lh/ks oLrqfu"B izkj)
This section contains 30 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 30 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. The region represented by |x – y| 2 and |x + y| 2 is bounded by a :
(1) rhombus of area 8 2 sq. units (2) square of side length 2 2 units
(3) square of area 16sq. units (4) rhombus of side length 2 units
|x – y| 2 rFkk |x + y| 2 }kjk iznf'kZr
{ks=k ftlds }kjk izfrc) (bounded) gS] og gS :
(1) ,d leprqHkaqZt ftldk {ks=kQy 8 2 oxZ bdkbZ gSA
(2) ,d oxZ ftldh Hkqtk dh yEckbZ 2 2 oxZ bdkbZ gSA
(3) ,d oxZ ftldk {ks=kQy 16 oxZ bdkbZ gSA
(4) ,d leprqHkaqZt ftldh Hkqtk dh yEckbZ 2 oxZ bdkbZ gSA
Ans. (2)
Sol.
y
x–y=–2
B (0,2)
x–y=2
C A
x
(0,0)
(2,0)
x+y=2
(–2,0)
D (0,–2)
x+y=–2
ABCDisa square
1
Area 4 2 2 8
2
side = 2 2
3 13 5 (13 23 ) 7 13 23 33
2. The sum + + + ……… upto 10th term, is :
12 12 22 12 22 32
3 13 5 (13 23 ) 7 13 23 33
+ + + ……… ds izFke nl inksa dk ;ksxQy gS %
12 12 22 12 22 32
Ans. (1)
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| JEE(MAIN) 2019 | DATE : 10-04-2019 (SHIFT-1) | PAPER-1 | OFFICIAL PAPER | MATHEMATICS
Sol. General term of given series (T n)
(2n 1)(13 23 33.....n3 )
Tn =
(12 22 ......n2 )
2
(n)(n 1)
(2n 1)
Tn = 2
n(n 1)( 2n 1)
6
3 3 2
Tn = n(n + 1) Tn = [n + n]
2 2
10
3 10(11)( 21) 10.11
T
3
Sum of series = = = [440] = 660
2 2
n
n 1
6 2
(n 1)1/ 3 (n 2)1/ 3 (2n)1/ 3
3. lim .... is equal to :
n 4/3
n4 / 3 n4 / 3
n
(n 1)1/ 3 (n 2)1/ 3 (2n)1/ 3
lim .... cjkcj gS %
n 4/3
n4 / 3 n4 / 3
n
3 3 3 4 4 4
(1) (2)4/3 – (2) (2)4/3 – (3) (2)4/3 (4) (2)3/4
4 4 4 3 3 3
Ans. (1)
1
1/ 3
2
1/ 3
n
1/ 3
1
n
1/ 3
r 1
Sol. lim1 1 ..... 1 = lim
n n n n
1
n n r 1 n n
1
. 1 x
1 1 3
(1 x)
4/3
= 1/ 3
dx = = (2 4 / 3 – 1)
4/3 0 4
0
x sin cos x sin 2 cos 2
4. If 1 = sin x 1 and sin x 1 , x 0 ; then for all 0, :
2
cos 1 x cos 2 1 x
x sin cos x sin 2 cos 2
;fn 1 = sin x 1 rFkk sin x 1 , x 0 ; rks lHkh 0, ds fy, :
2
cos 1 x cos 2 1 x
(1) 1 – 2 = x (cos2– cos4) (2) 1 + 2 = – 2x3
(3) 1 + 2 = – 2(x3 + x – 1) (4) 1 – 2 = – 2x3
Ans. (2)
Sol. 1 = x(–x2 – 1) – sin(–xsin – cos) + cos(–sin + xcos)
= –x3 – x + xsin2+ sincos – sincos + xcos2
= –x3
Similarly 2 = –x3
1 + 2 = –2x3
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| JEE(MAIN) 2019 | DATE : 10-04-2019 (SHIFT-1) | PAPER-1 | OFFICIAL PAPER | MATHEMATICS
dx x 1 f ( x)
5. If (x 2x 10)
2 2
= A tan–1
3
2
x 2x 10
+C
Where C is a constant of integration, then
1 1
(1) A = and f(x) = 9 (x – 1) (2) A = and f(x) = 9(x –1)2
27 54
1 1
(3) A = and f(x) = 3(x – 1) (4) A = and f(x) = 3(x – 1)
81 54
dx x 1 f ( x)
;fn
( x 2x 10)
2 2
= A tan–1
2
3 x 2x 10
+C
tgk¡ C ,d lekdyu vpj gS] rks %
1 1
(1) A = rFkk f(x) = 9 (x – 1) (2) A = rFkk f(x) = 9(x –1)2
27 54
1 1
(3) A = rFkk f(x) = 3(x – 1) (4) A = rFkk f(x) = 3(x – 1)
81 54
Ans. (4)
dy dx
Sol. ( x 2x 10)
2 2
(( x 1)2 9)2
Put x – 1 = 3tan
dx = 3sec2 d
3 sec 2 d 1 d 1 1 1 cos 2 1 1
(9 sec 2
)2
27 sec 2
=
27
cos2 d =
27 2
d =
54 2 sin 2 C
1 1 x 1 3( x 1)
= tan 2 C
54 3 x 2x 10
x y 1 z 1 3
6. If the length of the perpendicular from the point (, 0, ) ( 0) to the line, = = is ,
1 0 1 2
then is equal
x y 1 z 1 3
;fn fcUnq (, 0, ) ( 0) ls js[kk = = ij [khaps x, yac dh yackbZ gS] rks cjkcj gS %
1 0 1 2
(1) 1 (2) – 1 (3) 2 (4) –2
Ans. (2)
P(, 0, )
Sol.
dr's (–,–1, + + 1)
M (1,0, –1)
(,1, – –1)
PM is perpendicular to given line
1
– + 0 – – – 1 = 0 = –
2
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| JEE(MAIN) 2019 | DATE : 10-04-2019 (SHIFT-1) | PAPER-1 | OFFICIAL PAPER | MATHEMATICS
1 3
M – ,1, –
2 2
3
PM =
2
2 2
1 3 3 1 9 3
1 = 22 + + + 1 +2 + 3 + = 22 + 4 + 2 = 0
2 2 2 4 4 2
( + 1)2 = 0 = – 1
(1 i)2 2
7. If a > 0 and z = , has magnitude , then z is equal to
ai 5
(1 i)2 2
;fn a > 0 rFkk z = , dk ifjek.k (magnitude) , gS] rks z cjkcj gS %
ai 5
1 3 1 3 1 3 3 1
(1) – – i (2) + i (3) – i (4) – – i
5 5 5 5 5 5 5 5
Ans. (1)
( 2 )2 2
Sol. |z| = a2 + 1 = 10 a = 3
a 1
2 5
(1 i)2
Hence, z =
3i
(1 i)2 ( 2i)( 3 i) 1 3i
z = = =
3i 10 5
x4 1 x3 k 3
8. If lim = lim 2 , then k is :
x 1 x 1 x k x k 2
x4 1 x3 k 3
;fn lim = lim 2 , rks K cjkcj gS %
x 1 x 1 x k x k 2
4 3 3 8
(1) (2) (3) (4)
3 2 8 3
Ans. (4)
3k 2 8
Sol. 4= , k
2k 3
9. Let f(x) = ex – x and g(x) = x2 – x, x R. Then the set of all x R, where the function h(x) = (fog) (x)
is increasing, is :
ekuk f(x) = ex – x rFkk g(x) = x2 – x, x R. rks lHkh x R, ftuds fy, Qyu h(x) = (fog) (x) o/kZeku gS] dk
leqPPk; gS :
1 1
(1) 0 [1, ) (2) , 0 [1, )
2 2
1 1
(3) – 1, , (4) [0, )
2 2
Ans. (1)
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2
x
Sol. h(x) = fog(x) = f(g(x)) = ex x2 x
2
x 2
x
h'(x) = ex (2x 1) 2x 1 0 = (2x – 1)( ex 1) 0
1
Case-I : x & x2 – x 0 x 1
2
or
1
Case-II : x and x2 – x 0
2
1 1
0x So x 0 [1, )
2 2
10. If Q (0, – 1, –3) is the image of the point P in the plane 3x – y + 4z = 2 and R is the point (3, –1, –2),
then the area (in sq. units) of PQR is :
;fn fcUnq P dk lery 3x – y + 4z = 2 esa izfrfcEc Q (0, – 1, –3) gS rFkk R(3, –1, –2), ,d vU; fcUnq gS] rks
PQR dk {ks=kQy ¼oxZ bdkb;ksa esa½
91 65 91
(1) (2) (3) (4) 2 13
4 2 2
Ans. (3)
Sol.
P(0,–1,–3)
(–3,–1,–2) R M
3x – y + 4z – 2 = 0
1 – 12 – 2 13
PM = =
9 1 16 2
13 7 1 13 7 91
PR = 9 1 = 10 RM = 10 – = PQR = 2 × × × =
2 2 2 2 2 2
11. If a1 , a2 , a3, ………., an are in A.P. and a1 + a4 + a7 + ………+ a16 = 114, then a1 + a6 + a11 + a16 is
equal to :
;fn a1 , a2 , a3, ………., an ,d lekUrj Js<+h esa gSa rFkk a1 + a4 + a7 + ………+ a16 = 114 gS] rks a1 + a6 + a11
+ a16 cjkcj gS %
(1) 64 (2) 38 (3) 76 (4) 98
Ans. (3)
Sol. a1, a2,…….a16 are in A.P.
a1 + a4 + a7 + a10 + a13 + a16 = 114
a1 + a16 = a4 + a13 = a7 + a16 = a5 + a12 = a6 + a11
3(a6 + a11) = 114
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a6 + a11 = 38
(a1 + a6 + a11 + a16) = 2(a6 + a11) = 2 × 38 = 76
sin(p 1)x sin x
, x0
x
12. If f(x) = q , x0
xx x
2
, x0
x3 / 2
is continuous at x = 0, then the ordered pair (p, q) is equal to :
sin(p 1)x sin x
, x0
x
;fn f(x) = q , x0
xx x2
, x0
x3 / 2
x = 0 ij larr gS] rks Øfer ;qXe (p, q) cjkcj gS %
3 1 5 1 3 1 1 3
(1) , (2) , (3) , (4) ,
2 2 2 2 2 2 2 2
Ans. (1)
Sol. f(0–) = f(0) = f(0+)
lim = q = lim
h0 h h0 h h
(p 1) sin(p 1)h sinh h 1 1 h 1 1
lim = p + 1 + 1 = q = lim = lim
h 0 (p 1)h h h0 h h0 h( h 1 1)
1 3 1
p+2=q= p = ,q
2 2 2
x2 y2 9
13. If the line x – 2y = 12 is tangent to ellipse + = 1 at the point 3, , then the length of the latus
a2 b2 2
rectum of the ellipse is :
x2 y2 9
;fn js[kk] x – 2y = 12 nh?kZo`Ùk 2
+ 2
= 1 dks fcUnq 3, ,ij Li'kZ djrh gS] rks blds ukfHkyEc dh yEckbZ gS :
a b 2
(1) 9 (2) 8 3 (3) 12 2 (4) 5
Ans. (1)
9 x2 y2
Sol. 3, – lies on 2 2 1
2 a b
9 81
2
1
a 4b2
9
Equation of tangent at 3, –
2
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9
y. –
2 1
x.3 2
2
a b
x y
compare with 1
12 – 6
a2 2b 2
12 and 6
3 9
a = 6 and b = 3 3
2b 2 2 27
L.R. = 9
a 6
14. All the pairs (x, y) that satisfy the inequality
1
2 sin x 2 sinx 5 .
2
1 also
sin2 y
4
satisfy the equation :
og lHkh ;qXe (x, y) tks vlfedk
1
2 sin x 2 sinx 5 .
2
1 dks larq"V djrs gS] fuEu esa ls fdl lehdj.k dks Hkh larq"V djrs gSa \
sin2 y
4
(1) sin x = |siny| (2) 2|sin x| = 3 siny
(3) sin x = 2 sin y (4) 2 sin x = 2 sin y
Ans. (1)
(sin x –1)2 4 2
Sol. 2 4sin y
2sin2y (sin x – 1)2 4
2sin2y [0, 2]
Hence 2sin2y = (sin x – 1)2 4 , for |siny| = 1 and sinx = 1
|siny| = sinx
15. If and are the roots of the quadratic equation, x2 + xsin – 2sin = 0, 0 , , then
2
12 12
–12 12 .( )24is equal to :
;fn f}?kkrh lehdj.k] x2 + xsin – 2sin = 0, 0 , , ds ewy rFkk gSa] rks
2
12 12
–12
12 .( )24
cjkcj gS %
212 212 212 26
(1) (2) (3) (4)
(sin 8)6 (sin 8)12 (sin 4)12 (sin 8)12
Ans. (2)
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Sol. x2 + xsin – 2sin = 0, has roots and
12 12 12 .12 ( )12
( 12 12 )( )24 ( )24 ( )24
( 2 sin )12 212
= =
sin 2 8 sin
24
(8 sin )12
16. Let f : R R be differentiable at c R and f(c) = 0. If g(x) = |f(x)|, then at x = c, g is :
(1) not differentiable (2) differentiable if f(c) = 0
(3) not differentiable if f (c) = 0 (4) differentiable if f(c) 0
ekuk f : R R, c R ij vodyuh; gS rFkk f(c) = 0 gS ;fn g(x) = |f(x)|, rks x = c, ij g :
(1) vodyuh; ugha gS (2) vodyuh; gS ;fn f(c) = 0
(3) vodyuh; ugha gS ;fn f (c) = 0 (4) vodyuh; gS ;fn f(c) 0
Ans. (2)
Sol. g(x) = |f(x)|
f ( x ) – f (c ) f (x)
g'(c+) = lim = lim = ± f'(c)
x c x–c x c x–c
f ( x ) – f (c )
g'(c–) = lim–
x c x–c
f (x)
= lim–
x c x–c
= ± f'(c)
for g (x) to be differentiable at x = c.
f'(c) must be 0. Else it is non-differentiable.
17. Let A(3, 0, –1), B(2, 10, 6) and C(1,2,1) be the vertices of a triangle and M be the midpoint of AC. If G
divides BM in the ratio, 2 : 1, then cos( GOA) (O being the origin) is equal to :
ekuk ,d f=kHkqt ds 'kh"kZ fcUnq A(3, 0, –1), B(2, 10, 6) rFkk C(1,2,1) gSa rFkk AC dk e/;fcUnq M gSA ;fn G, BM
dks 2 : 1 ds vuqikr esa foHkkftr djrk gS] rks cos( GOA) (O ewyfcUnq gSSa) cjkcj gS %
1 1 1 1
(1) (2) (3) (4)
6 10 2 15 15 30
Ans. (3)
Sol. A (3,0–1,), B (2,10,6) and C (1,2,1)
C (1,2,1)
m (2, 4,2)
m
B(2,10,6)
A(3, 0,–1)
0(0, 0,0)
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G is centroid of from given information .
OA . OG = OA OG cos 2î 4 ĵ 2k̂ . 3î – k̂ = 2 6. 10 cos
6 – 2 = 2 3. 5.2 cos
1
cos =
15
18. If the circles x2 + y2 + 5Kx + 2y + K = 0 and 2(x2 + y2) + 2Kx + 3y – 1 = 0, (K R), intersect at the points
P and Q, then the line 4x + 5y – K = 0 passes through P and Q, for :
(1) exactly one value of K (2) exactly two values of K
(3) no value of K (4) infinitely many values of K
;fn o`Ùkksa x2 + y2 + 5Kx + 2y + K = 0 rFkk 2(x2 + y2) + 2Kx + 3y – 1 = 0, (K R), ds izfrPNsnu fcUnq P rFkk Q
gSa] rks js[kk 4x + 5y – K = 0 ds fcUnqvksa P rFkk Q ls gksdj tkus ds fy, %
(1) K dk ek–=–k ,d eku gSA (2) K ds ek–=k nks eku gSA
(3) K dk dksbZ eku ugha gSA (4) K ds vuUr eku gSa
Ans. (3)
Sol. equation of common chord is S1 – S2 = 0
3y 1
4kx + k 0
2 2
Which is identical to
4x + 5y – k = 0 (given)
4k 3/2 K 1/ 2
Hence = =
4 5 –k
1
k
3 3 2 k 5
k= and
10 10 k 13
There is no value of k which satisfy simultaneously
19. If the coefficients of x2 and x3 are both zero, in the expansion of the expression (1 + ax + bx 2) (1 – 3x)15
in powers of x, then the ordered pair (a, b) is equal to :
;fn x dh ?kkrksa (powers) esa] O;atd (1 + ax + bx2) (1 – 3x)15 ds izlkj esa x2 rFkk x3 nksuksa ds xq.kkad 'kwU; ds cjkcj
gSa] rks Øfer ;qXe (a, b) cjkcj gS %
(1) (28, 315) (2) (28, 861) (3) (–21, 714) (4) (–54, 315)
Ans. (1)
Sol. Coefficient of x2 in (1 + ax + bx2)(1 – 3x)15
= 15C2(–3)2 + a. 15C1(–3) + b.15C0 = 0
15.14
.9 – 3.a.15 + b = 0
2
15 × 63 – 45a + b = 0 ……..(1)
Coefficient of x3 in (1 + ax + bx2)(1 – 3x)15
= 15C3(–3)3 + a. 15C2(–3)2 + b.15C1 (–3) = 0
15.14.13 2 15.14
.3 – a.3. + 15.b = 0
32 2
7.13.3 – 21a + b = 0 ……(2)
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by using (1) – (2)
672 – 24a = 0 a = 28
Hence b = 315
20. If a directrix of a hyperbola centred at the origin and passing through the point (4, –2 3 ) is 5x = 4 5
and its eccentricity is e, then :
,d vfrijoy; dk dsUnz ewyfcUnq ij gS rFkk ;g fcUnq (4, –2 3 ) ls gksdj tkrk gSA ;fn bldh ,d fu;rk
(directrix) 5x = 4 5 gS rFkk bldh mRdsUnzrk e gS] rks %
(1) 4e4 – 12e2 – 27 = 0 (2) 4e4 + 8e2 – 35 = 0
(3) 4e4 – 24e2 + 27 = 0 (4) 4e4 – 24e2 + 35 = 0
Ans. (4)
x2 y2 16 12
Sol. – 1 2
– 1 ….(1)
a 2
b 2
a b2
4 a
16e2 = 5a2 ….(2)
5 e
a2
(1) 16 – 12 = a2
b2
12 16e 2
16 – =
e2 – 1 5
80(e2 – 1) – 60 = 16e2 (e2 – 1)
16e4 – 16e2 = 80e2 – 140
16e4 – 96e2 + 140 = 0
4e4 – 24e2 + 35 = 0
21. If for some x R, the frequency distribution of the marks obtained by 20 students in a test is :
Marks 2 3 5 7
Frequency (x + 1)2 2x – 5 x2 – 3x x
then the mean of the marks is :
;fn fdlh x R, ds fy,] 20 fo|kfFkZ;ksa }kjk ,d ijh{kk esa izkIr vadks dk ckjackjrk caVu gS :
vad 2 3 5 7
ckjackjrk (x + 1)2 2x – 5 x2 – 3x x
rks vadks dk ek/; gS :
(1) 2. 5 (2) 2.8 (3) 3.0 (4) 3.2
Ans. (2)
Sol. Number of students = 20
(x + 1)2 + (2x – 5) + x2 – 3x + x = 20
2x2 + 2x – 24 = 0 x2 + x – 12 = 0
x = –4, 3 , but x can not be negative
x=3
2 16 3 1 5 0 7 3 56
Avg. marks = = 2.8
20 20
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2
22. The value of [sin 2x(1 cos3x)] dx, where [t] denotes the greatest integer function, is :
0
2
[sin 2x(1 cos3x)] dx, dk eku gS] tgk¡ [t] ekgÙke iw.kk±d Qyu dks O;Dr djrk gS&
0
(1) 2 (2) – 2 (3) – (4)
Ans. (3)
2
Sol. = [sin 2x(1 cos3x)] dx
0
Apply a + b – x
2
= [ sin 2x(1 cos3x)] dx
0
2
2 = [sin 2x(1 cos3x)] [ sin 2x(1 cos3x)] dx
0
2 = –2
= –
23. ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC.
If the angles of elevation of the top of the tower at A and B are cot –1 (3 2 ) and cosec–1 respectively,
then the height of the tower (in metres)is :
ABC ,d f–=kHkqtkdkj ikdZ gS ftlesa AB = AC = 100 ehVj gSA BC ds e/; fcanq ij ,d lh/kh ehukj [kM+h gSA ;fn
ehukj ds f'k[kj ds fcanqvksa A rFkk B ij mUu;u dks.k Øe'k% cot–1 (3 2 ) rFkk cosec–1(2 2 ) gSa] rks ehukj dh
Å¡pkbZ ¼ehVjksa esa½ gS %
100
(1) (2) 25 (3) 10 5 (4) 20
3 3
Ans. (4)
P
A
Sol. 100 m 100 m h
1002 – x 2
B C A
x M x 1002 – x 2 M
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x 2 h2
h
B
x M
cot = 3 2 ……..(i)
equation ….(i)
1002 – x 2
= 3 2
h
1002 – x2 = 18h2 ……..(iii)
equation (ii)
x 2 h2
= 2 2 x2 + h2 = 8h2 x2 = 7h2 ……(iv)
h
equation (iii) and (iv)
100 100
1002 – 7h2 = 18h2 h2 = = 400
25
h = 20
24. Which one of the following Boolean expressions is a tautology ?
cwys ds fuEu O;atdks esa ls dkSulk ,d] ,d iqu:fDr gS ?
(1) (p q) (p q) (2) (p q) (p q) (3) (p q) (p q) (4) (p q) (pq)
Ans. (4)
Sol. (1) (p q) (p ~q) = p (q~q) = p f = p
(2) (p q) (p q)
p q ~p ~q p q p ~q (p q) (p ~q)
T T F F T F F
T F F T T T T
F T T F T T T
F F T T F T F
(3) (p q) (p q) = p (q ~q) = p t = p
(4) (p q) (pq) = p (q ~ q) = p t = t
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25. Assume that each born child is equally likely to be a boy or a girl. If two families have two chidren each,
then the conditional probability that all children are girls given that at least two are girls is :
ekuk izR;sd tUe ysus okys cPps dk yM+dk vFkok yM+dh gksuk lelaHkkO; gSA ekuk nks ifjokjksa esa izR;sd esa nks cPps
gSaA ;fn ;g fn;k x;k gS fd de ls de nks cPps yM+fd;ka gS] rks lHkh cPpksa ds yM+dh gksus dh lizfrca/k izkf;drk gS%
1 1 1 1
(1) (2) (3) (4)
11 10 17 12
Ans. (1)
Sol. A B
G G G G 1
G G G B 4C1
G G B B 4C2
1 1
Required probability =
1 C1 C2
4 4 11
26. The line x = y touches a circle at the point (1, 1). If the circle also passes through the point (1, –3), then
its radius is :
;fn js[kk x = y ,d o`Ùk dh fcUnq (1, 1). ij Li'kZ djrh gSA ;fn o`Ùk fCkUnq (1, –3) ls xqtjrk gS] rks bldh f=kT;k gS :
(1) 3 2 (2) 3 (3) 2 2 (4) 2
Ans. (3)
Sol. P(1,1)
(1, –3)
y=x
Family of circle touching a given line at a given point
(x – 1)2 + (y – 1)2 + (x – y) = 0
passes through (1, – 3)
so 0 + 16 + (1 + 3) = 0 = – 4
so required circle
x2 + 1 – 2x + y2 + 1 – 2y – 4x + 4y = 0
x2 + y2 – 6x + 2y + 2 = 0
r= 9 1– 2 = 2 2
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dy
27. If y = y(x) is the solution of the differential equation = (tan x – y)sec2 x, x , , such that
dx 2 2
y(0) = 0, then y is equal to :
4
dy
;fn y = y(x) vody lehdj.k = (tan x – y)sec2 x ,x , , tcfd y(0) = 0, dk gy gS] rks
dx 2 2
y cjkcj gS :
4
1 1 1
(1) e – 2 (2) –2 (3) 2 + (4) –e
e e 2
Ans. (1)
dy
Sol. + y(sec2x) = tanx sec2x
dx
I.F. = e
sec2 xdx
e tan x
solution is
e
tan x
y(etanx) = tan x sec 2 x dx
Put tanx = t
t.e dt = t.e – e + c = e (t – 1) + c
t
y(etanx) = t t t
yetanx = etanx(tanx – 1) + c
0 = –1 + c c=1
Hence y = e – 2
4
28. The number of 6 digit numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible
by 11 and no digit is repeated, is :
vadks (digit) 0, 1, 2, 5, 7 rFkk 9 ds iz;ksx ls N% vadks okyh ,slh la[;kvksa tks 11 ls HkkT; gksa rFkk ftuesa dksbZ Hkh
vad nksckjk u vk,] dh la[;k gS % :
(1) 36 (2) 72 (3) 48 (4) 60
Ans. (4)
Sol. a b c d e f
9 + 7 + 5 + 2 + 1 + 0 = 24
|a + c + e – (b + d + f)| = 0 or a multiple of 11
and (a + c + e) – (b + d + f) can be
21 – 3 = 18 (minimum b + d + f)
20 – 4 = 16
17 – 5 =14
18 – 6 = 12
17 – 7 = 10
16 – 8= 8
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15 – 9 = 6
14 –10 = 4
13 –11 = 2
12 – 12 = 0
a + c + e = 12
and b + d + f = 12
so there is only one possible way
{a, c, e} and {b, d, f} = {7, 5, 0}
or {a, c, e} and {b, d, f} = {7, 5, 0}
= {9, 2,1}, {7, 5, 0}
number of ways = 2.3!.3! – 2! 3! = 72 – 12 = 60
29. Let f(x) = x2 , x R for any A R, definer g(A) = {x R : f(x) A}. If S = [0, 4], then which one of the
following statements is not true ?
ekuk f(x) = x2 , x R fdlh Hkh A R, ds fy, g(A) = {x R : f(x) A} gSA ;fn S = [0, 4] gS] rks fuEu esa ls
dkSu lk ,d dFku lgh ugha gS \
(1) f(g(S)) = S (2) f(g(S)) f(S) (3) g(f(S)) S (4) g(f(S)) = g(S)
Ans. (4)
Sol. f(s) = s2 0 f(s) 16 ……..(i)
g(s) = {x : x R, x2 S}
= {x : x2 [0, 4]}
–2 g(s) 2 …….(ii)
from (i) 0 f(s) 16
g(f(s)) = {x : f(x) f(s)}
= {x : x2 [0, 16]}
= {x : –4 x 4]}
–4 g(f(s)) 4 …….(iii)
from (ii) –2 g(s) 2 0 (g(s))2 4
f(g(s)) = g(s)2
0 f(g(s)) 4 …..(iv)
from (iv) and (i), (2) is true
from (iv) and S [0, 4], (1) is true
from (iii) and (ii), (4) is False
from (iii) and S [0, 4], (3) is true
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30. If the system of linear equations
x+y+z=5
x + 2y + 2z = 6
x + 3y + z = , (, R), has infinitely many solutions, then the value of + is :
;fn jSf[kd lehdj.k fudk;
x+y+z=5
x + 2y + 2z = 6
x + 3y + z = , (, R), ds vuUr gy gS rks + dk eku gS :
(1) 9 (2) 7 (3) 12 (4) 10
Ans. (4)
1 1 1 1 1 1
Sol. D = 1 2 2 = – 1 0 0 = 1.( – 3)
1 3 1 3
5 1 1 5 1 1
D1 = 6 2 2 = – 4 0 0 = 4.( – 3)
µ 3 µ 3
1 5 1 1 5 1
D2 = 1 6 2 = 0 1 1 = –2–µ+6=–µ+4
1 µ 0 µ–6 –2
1 1 5 1 1 5
D3 = 1 2 6 = 0 1 1 = µ–6–1=µ–7
1 3 µ 0 1 µ–6
for infinitely many solutions
D = 0, D1 = 0, D2 = 0, D3 = 0
= 3, = 3, – µ = – 4, µ = 7
=3&µ=7
x+y+z=5 ………..(i)
x + 2y + 2z = 6 ………..(ii)
x + 3y + 3z = 7 ………..(iii)
from (i) and (ii) y + z = 1 x = 4 which satisfy (iii) equation hence there are infinite number of solutions
+ µ = 10
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A recent survey of all auto accident victims in Dole County
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A recent survey of all auto accident victims in Dole County [#permalink]
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15 Aug 2003, 06:13
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1. A recent survey of all auto accident victims in Dole County found that, of the severely injured drivers and front-seat passengers, 80 percent were not wearing seat belts at the time of their accidents. This indicates that, by wearing seat belts, drivers and front-seat passengers can greatly reduce their risk of being severely injured if they are in an auto accident.
The conclusion above is not properly drawn unless which of the following is true?
The answer: Of all the drivers and front-seat passengers in the survey, more than 20 percent were wearing seat belts at the time of their accidents.
Why is it an assumption in order for the conclusion to be valid?
2. Which of the following best completes the passage below?
Sales campaigns aimed at the faltering personal computer market have strongly emphasized ease of use, called user-friendliness. This emphasis is oddly premature and irrelevant in the eyes of most potential buyers, who are trying to address the logically prior issue of whether----
C. currently available models are user-friendly enough to suit them
E. they have enough sensible uses for a personal computer to justify the expense of buying one
The answer is E. Why is E better than C?
Thank you
If you have any questions
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15 Aug 2003, 08:12
the first) you have three groups of people, not two
(1) all the drivers and front-seat passengers in the survey (X)
(2) the severely injured drivers and front-seat passengers (Y)
(3) the severely injured not wearing belts (Z=0.8Y)
Z can be a tiny share of X. So wearing belts just slightly affect your chances. You have to prove that Z is significant share of X. I think this is a rationale. But the question is difficult.
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15 Aug 2003, 18:40
Joviest, pls give us all options w/o answers in the next question on the forum. It makes it more interesting for us to solve!
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15 Aug 2003, 20:55
Joviest wrote:
1. A recent survey of all auto accident victims in Dole County found that, of the severely injured drivers and front-seat passengers, 80 percent were not wearing seat belts at the time of their accidents. This indicates that, by wearing seat belts, drivers and front-seat passengers can greatly reduce their risk of being severely injured if they are in an auto accident.
The conclusion above is not properly drawn unless which of the following is true?
The answer: Of all the drivers and front-seat passengers in the survey, more than 20 percent were wearing seat belts at the time of their accidents.
Why is it an assumption in order for the conclusion to be valid?
Let's assume that only 20% of people where seat belts. Since seat belts do not CAUSE accidents, we can assume that people that get into accident will has the same ratio of seat belt wearers to non seat belt wearers. Hence, it WOULD BE EXPECTED that 80% of the accidents would involve those not wearing seat belts. However, if much more that 20% wear seat belts, then if the number of people in accidents who do not have belts on is 80%, that would be disproportionate to the general pool of people, hence strengtheni9ng the conclusion.
As a corollary, suppose only 1% of the population wore seat beats. Then the 20% of those found wearing seat belts in an accident would be disproportionately LARGE and one could reasonably conclude that WEARING SEAT BELTS actually causes the injuries!!!
Hence, the simple fact that 80% of accident victims did not wear seat belts is not enough by itself to conclude that seat belts work, however, whether or not this proportion is disproportionate to the general population is.
A is the only one that seems to show this disproportionality.
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Last edited by AkamaiBrah on 19 Aug 2003, 00:26, edited 1 time in total.
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17 Aug 2003, 04:05
I repost those two questions as follows:
1. A recent survey of all auto accident victims in Dole County found that, of the severely injured drivers and front-seat passengers, 80 percent were not wearing seat belts at the time of their accidents. This indicates that, by wearing seat belts, drivers and front-seat passengers can greatly reduce their risk of being severely injured if they are in an auto accident.
The conclusion above is not properly drawn unless which of the following is true?
(A) Of all the drivers and front-seat passengers in the survey, more than 20 percent were wearing seat belts at the time of their accidents.
(B) Considerably more than 20 percent of drivers and front-seat passengers in Dole County always wear seat belts when traveling by car.
(C) More drivers and front-seat passengers in the survey than rear-seat passengers were very severely injured.
(D) More than half of the drivers and front-seat passengers in the survey were not wearing seat belts at the time of their accidents.
(E) Most of the auto accidents reported to police in Dole County do not involve any serious injury.
2. Which of the following best completes the passage below?
Sales campaigns aimed at the faltering personal computer market have strongly emphasized ease of use, called user-friendliness. This emphasis is oddly premature and irrelevant in the eyes of most potential buyers, who are trying to address the logically prior issue of whether----
(A) user-friendliness also implies that owners can service their own computers
(B) personal computers cost more the more user-friendly they are
(C) currently available models are user-friendly enough to suit them
(D) the people promoting personal computers use them in their own homes
(E) they have enough sensible uses for a personal computer to justify the expense of buying one
Thanks
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17 Aug 2003, 05:47
Sales campaigns aimed at the faltering personal computer market have strongly emphasized ease of use, called user-friendliness. This emphasis is oddly premature and irrelevant in the eyes of most potential buyers, who are trying to address the logically prior issue of whether----
E is better than C because the first issue that comes to mind while making a purchase is to try & estimate the uses of the product and if it is really going to be used. For example, if I were to buy a music system, I would first decide if I really would use it, as only then will I want to buy it. Add ons like surround speakers etc would be deciding factors only if Im convinced I have a use for the product.
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19 Aug 2003, 00:30
Joviest wrote:
I repost those two questions as follows:
2. Which of the following best completes the passage below?
Sales campaigns aimed at the faltering personal computer market have strongly emphasized ease of use, called user-friendliness. This emphasis is oddly premature and irrelevant in the eyes of most potential buyers, who are trying to address the logically prior issue of whether----
(A) user-friendliness also implies that owners can service their own computers
(B) personal computers cost more the more user-friendly they are
(C) currently available models are user-friendly enough to suit them
(D) the people promoting personal computers use them in their own homes
(E) they have enough sensible uses for a personal computer to justify the expense of buying one
Thanks
The wording of the question implies that there is a possibility that the whole idea of "user-friendliness" is "premature" hence irrelevant. If a user has not yet determined whether there are any "uses" for a computer, the question of whether or not it is "user-friendly" is moot. Hence, E is the answer.
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Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
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11 Jun 2010, 07:14
its A
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13 Jul 2010, 21:54
takers any?
Re: Two CR Questions [#permalink] 13 Jul 2010, 21:54
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Answer to Question #23427 in Inorganic Chemistry for Mallori
Question #23427
How many electrons must pass through the circuit for 767 coulombs of charge
1
2013-02-01T08:00:06-0500
The magnitude of the electrical charge of one mole of elementary charges(approximately 6.022 × 1023 , or Avogadro's number ) is known as a Faraday unit of charge (closely related to the Faraday constant ).
One Faraday is equal to 96485.3399 coulombs. In terms of Avogadro's number NA, one coulomb is equal to approximately 1.036 × NA × 10−5 elementary charges.
760 coulomb * 1.036 × NA × 10−5 = 7.8736 × NA × 10−3
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https://www.si.com/vault/1996/07/01/208599/emmitt-unplugged-high-voltage-running-back-emmitt-smith-of-the-dallas-cowboys-may-look-as-if-hes-relaxing-but-dont-be-fooled | 1,542,528,437,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744320.70/warc/CC-MAIN-20181118073231-20181118095231-00141.warc.gz | 973,179,592 | 65,217 | # Emmitt Unplugged
High-voltage running back Emmitt Smith of the Dallas Cowboys may look as if he's relaxing, but don't be fooled.
High-voltage running back Emmitt Smith of the Dallas Cowboys may look as if he's relaxing, but don't be fooled.
June 30, 1996
Emmitt Smith was home alone one night in April, minding his own business, when the old picture started crowding his head again. It was Walter Payton, the former Chicago Bears great who rushed for more yards than any back in NFL history. Payton was wearing his familiar number 34. He had that big C on the side of his helmet, and he was chomping on that funny-looking mouthpiece. And he was running, running because that's how Emmitt sees him whenever the picture comes back: Payton slipping past the defense, Payton finding open field, Payton scoring a touchdown.
This night a torrent of numbers came trailing after the picture, and Emmitt felt compelled to write them down. He got out a pen and a piece of paper and made a note that Payton had rushed for almost 17,000 yards in his career. Next to that figure Emmitt scribbled the number 9,000, which is approximately how many yards he has gained since joining the Dallas Cowboys in 1990.
On the page the difference seemed incredible. Payton's figure looked huge, almost epic, while Emmitt's seemed small and insignificant. After considering the disparity for a while, Emmitt did some simple arithmetic. If over the next five years you gain 1,500 yards a season, he said to himself, that'll give you another 7,500.
He was writing furiously now, his face crimped with intensity. It was history, after all, that he was trying to draw a bead on: a place where no runner had ever gone before. Add the 7,500 to the 9,000 you've already gained, he continued, and your total is 16,500.
It was still a few hundred yards short of Payton's mark. But there was something else to consider. In five years Emmitt would be 31. Payton finished his playing career at 33.
"I can be there," Emmitt said later, when he found himself fixating on the great Walter Payton again. "But I've got to hit one helluva pace. I've got to get ahead of the curve...and I can do that. There's time, there's time."
What makes Emmitt run? Why is he in such a hurry? What, exactly, does he expect to find when he gets where he's going?
Today he'll find a new \$2 million manse in Addison, Texas, just north of Dallas, because that's where he's headed now, wheeling through traffic with an intensity of purpose that would shame even the Indy 500 boys. Emmitt moved into the house over the Christmas holidays, and he really couldn't be happier, despite the fact that the curtains and bedroom suite haven't come yet and the lawn's a little spotty. Emmitt spared no expense in building the place, for this was to be the home of a man who in only six pro seasons had already won three Super Bowls and four rushing titles and who last year set the NFL record for most touchdowns in a season, 25. To reflect his great good fortune, Emmitt wanted something altogether spectacular. And so now he has it: a home suited for Jay Gatsby, or at least for the best football player in the game today.
Until he made the move, Emmitt occupied fairly modest accommodations for a man of his means, in keeping with his reputation for being someone who likes to stay close to his money. He rented apartments, one after another, never paying more than \$800 a month for two bedrooms. But when the annual value of his endorsement deals came to equal that of his \$3.4 million salary with the Cowboys, Emmitt threw some money at an architect and said, "O.K., build it," and after 13 torturous months it was ready.
To celebrate, Emmitt's family came out from Pensacola, Fla., over the holidays, and some 30 guests stayed at the house. Emmitt's mother, Mary Smith, always the paragon of modesty, took her shoes off before climbing the staircase, not wanting to scratch the mahogany. "Emmitt, you should cover this with some carpet," she told him. And Emmitt answered, "Mama, you don't cover wood that pretty with carpet." What was amazing was how easily the house held everyone. On Christmas morning the family held hands and sang songs, and everybody was just so happy that Emmitt was happy. He remembered how, when he was a boy, he mowed lawns for money and sold pecans and aluminum cans and squirreled away his savings in a beat-up shoebox and dreamed a house like this one, dreamed it from top to bottom.
Today traffic is backed up, and every time Emmitt sees an opening, somebody fills it. Only a couple of miles from his house he comes to a tollbooth and hardly stops, throwing two quarters at the toll machine—one for himself, one for the reporter following in the car behind him. Emmitt puts an arm out the window and waves for the reporter to drive on, but the guy doesn't understand. He must think Emmitt's pointing at the sky or something, because he stops and feeds the machine a third quarter, and by then Emmitt is long gone, weaving through traffic, blowing past everybody.
Emmitt drives like he runs the football, which is to say hard and without pretense, and in minutes he has lost the reporter, and the reporter has lost him, and all Emmitt can do is pull over and wait in the parking lot of a shopping mall, and everybody knows how he hates to wait, how waiting to him is the slowest form of death.
Back in 1990, his rookie year, Emmitt had to be the first one off the ground after he was tackled. He was that impatient. He would push everybody off and race back to the huddle to see if Dallas quarterback Troy Aikman would let him run the ball again. The defense couldn't understand why Emmitt was in such a hurry, unless he had an attitude problem. But it wasn't that, and it wasn't about being young, either. "I just had this energy," Emmitt says. "I had this...enthusiasm." Then one day after a game, his father, Emmitt Smith Jr., pulled him aside and gave him a look. Emmitt's dad doesn't talk much, and sometimes when he gives you a look, it's like he's giving you a smack in the chops. He wondered why Emmitt couldn't be patient and wait on the ground like a normal person. "Son," he said, "you know how much energy you waste trying to be the first one up like that?"
Emmitt thought about it for a minute. "You're right," he finally said. And the next week, when he was tackled, he lay there and looked at the world and admired it and felt altogether grateful for his place in it.
"You hurt, Emmitt?" came a voice. It was a guy on the other team. Everybody had gotten off Emmitt by then, and he was still lying there, just as his father had told him to.
"Hurt?" Emmitt said. "No, I ain't hurt." And then a remarkable thing happened. Before Emmitt could rise on his own power, the guy was offering him a hand. He was helping Emmitt up, expending some of his own energy, conserving Emmitt's.
There was a powerful lesson in that, if you had time to look. But who has time anymore, least of all Emmitt, the busiest man in all of football? And now in the parking lot of the shopping mall, after having blown about 15 minutes of his valuable time, Emmitt is starting to get jumpy. He wanted to give the reporter a tour of his house—let him meet the dogs, show him the minibars and the porcelain doll collection, walk him through the master bedroom suite and bath. Emmitt drives to where his car can be seen from the street, and at long last the reporter arrives. "Why'd you take off like that?" the guy asks, sticking his head out his car window.
But Emmitt doesn't answer. He shifts into drive and gives his Lexus a long, tall drink of unleaded.
The truth is, Emmitt's been on the run since the day he was born, his body clock ticking faster than everybody else's. The pace hasn't hurt him. He's entering the last year of his contract with the Cowboys, and after the 1996 season he's likely to sign a new deal with Dallas making him the highest-paid player in football. "That's all up to Jerry," Emmitt says, referring to the team's owner, Jerry Jones.
The Cowboys aren't all there is to Emmitt, though. He's also a businessman who at last count presided over four enterprises: Emmitt, Inc. (a collectibles and trading-card company), Emmitt Smith Communications Corporation (which sells and leases cellular phones and pagers), Emmitt Zone, Inc. (which licenses his name to outfits such as Reebok, Starter and Scoreboard) and Emmitt Smith Charities (which handles his humanitarian endeavors). With all that—and football to boot—there's hardly a day on Emmitt's calendar that isn't filled a year in advance.
"If you were to ask me what Emmitt will be doing at two in the afternoon on November the 14th of this year, I could look at his calendar and tell you," says Horace Irwin Jr., executive vice president of Emmitt Smith Communications. "He really doesn't have a life of his own. People call and say, Can Emmitt do this? or Can Emmitt do that? and we look at his schedule, and the date's already taken. We feel bad about it, having to tell them no all the time, but Emmitt's a busy man."
Despite his commitments Emmitt will surprise you on occasion. Take, for instance, the grand opening last October of his Dallas cellular phone business. Emmitt arrived at the Preston Road address at seven in the morning, before most of his employees, his eyes bright with expectation. The phones started to ring, and Emmitt answered one. "Emmitt Smith Communications," he said. "This is Emmitt."
His name inspired silence on the other end. It couldn't be Emmitt, not the one who plays for the Cowboys.
Later that day Emmitt returned to the store and asked his staff how things were going. Everybody said, "Great." Sales were brisk, customers seemed satisfied. Emmitt was wearing a serious expression on his face. He asked for the folders of 10 customers. The folders held invoices with the names and phone numbers of people who had bought things at the store that day.
Emmitt went into an office and closed the door. He opened a file and found a name and called the person and said, "Hello, this is Emmitt Smith of Emmitt Smith Communications. You came by our business today, and I just wanted to know how the service was. Did everything go all right?"
Again, some of those he called didn't believe it was really Emmitt. They laughed and stammered and succumbed to mild fits of hysteria. Didn't he have more important things to do? With the football season in full swing, could he possibly have time to phone a complete stranger and ask how his new cellular service was working out?
"If we can be of any help," Emmitt said, "please don't hesitate to call or come by."
"There are so many qualities that Emmitt possesses, and the most impressive one is that he's a good person," says Joe Brodsky, the Cowboys' running backs coach. "Emmitt was brought up properly. He knows where he's going at all times and how to get there. He's what I call a complete grinder, meaning he's totally focused and will do anything it takes to win. I don't think there's a selfish bone in Emmitt's body. He'd give it all up—everything—for the team."
Back when Emmitt was a student at Escambia High in Pensacola, his football coach used to say, "It's a dream until you write it down. Then it's a goal." And Emmitt never forgot that. He all but branded the words on his heart, and he was always writing things down. When Emmitt was a rookie, before he had even played his first game, one of his Dallas teammates, safety James Washington (now with the Washington Redskins), stopped by to visit him at his apartment and found Emmitt sitting and writing things like "Rookie of the Year" and "leading rusher" on a piece of paper.
"What are you doing?" Washington asked.
Emmitt held up the list. "This is what I want to accomplish this season."
Emmitt is no different when it comes to business. Write it down and it isn't a dream—it's a goal. "One day not long ago," says Everett Brooks, marketing manager for Emmitt Smith Communications, "Emmitt came in and sat down and started putting together an organization chart detailing exactly where he wants to be in six months. He wrote it all out, in detail: I want this person here, and I want this person to be doing this. He was projecting future job assignments for people. He knows what he wants. And nothing will stop him."
One of Emmitt's goals is to be as successful off the field as he is on it. That requires dedication. That requires go go go. And, sometimes, that requires pulling people aside and asking the hard questions.
"Tell me something, Everett," Emmitt said to Brooks one day recently. "Why do you have so much inventory in here?" They were standing in the stockroom.
"Well," Brooks answered, looking around, "we're having a promotion this coming Friday."
"O.K.," Emmitt said. "Fine. Keep up the good work."
If God is in the details, as they say, then Emmitt will learn the details. He will learn all he needs to know to stay ahead of the game. To achieve this end he has surrounded himself with good people. In fact, his lawyer, Mike Ferguson, is a retired brigadier general of the U.S. Army. Most players are content to hire a lawyer bright enough to have passed the bar exam, but not Emmitt. Emmitt's lawyer, you don't know whether to shake his hand or hit the dirt and give him 10 push-ups. And what does Emmitt call the man?
"Mike," Ferguson says. "He did ask me once, early on, if he should call me General, but, no, I said Mike was fine."
Somebody comes to Emmitt with a business proposition, Ferguson checks it out. Restaurants, nightclubs, shopping centers, you name it: Emmitt has been asked to invest in them all. One group even wanted him to put some money into a satellite. At the end of each year Ferguson convenes a summit conference of Emmitt's business associates. "His corporate entities are represented," Ferguson says. "His financial advisers, his accountants, his corporate general managers. Emmitt is at every meeting, making decisions."
A few years ago Emmitt was earning only about \$500,000 a year in endorsement money. Other celebrity athletes were making fortunes; Michael Jordan led the pack with roughly \$40 million a year. Emmitt decided he could be doing better. He gave his agent six months to come up with a plan to market him better, and when the agent failed to satisfy him, Emmitt hired somebody else. "I want to be the Michael Jordan of football," he told Werner Scott of the Advantage Marketing Group, a Dallas firm that now oversees Emmitt's commercial ventures.
After Emmitt was named MVP of Super Bowl XXVIII in January 1994, Scott organized a summit of his own at Walt Disney World in Orlando. He called it the Team Emmitt Summitt, and he brought together marketing reps from companies that either had a relationship with Emmitt or wanted to develop one. The group spent a couple of days trying to decide how best to sell Emmitt to the world, and Emmitt was there for every meeting, taking notes, pitching ideas of his own.
"I don't want to overexploit myself," he says now. "But to sit here and be in the position I'm in and not really explore my opportunities and maximize them, I'd be wasting my talent. I'd be dumb." Since 1995 he has earned as much from endorsements as he has from the Cowboys. "Maybe in the near future," he says, "we can double or triple that. Maybe next year."
If he can wait that long.
The pace, the drive, the hunger—they're nothing new to Emmitt. He hustled even as a kid, always a step ahead of others his age, always looking for records to topple.
Nine months old, for instance, and he had already figured out how to climb out of his crib. His family would be eating in the kitchen or sitting in the den, and Emmitt would appear with a look of hot, eager triumph on his face. Try, the look seemed to say. Try and catch me. They'd run at him with arms outstretched, and he'd scamper just past their grasp, an escape artist even then.
A month or two later, not even a year old, Emmitt started working on his game. His mother had noticed that he quieted down whenever there was football on TV, so she took to placing him in a swing in front of the set every time a game was on. Emmitt would sit there transfixed, his dark, wobbling eyes following the action to all points on the screen. He had the keenest concentration, as if he were memorizing plays, sets, formations. "Football," Emmitt says. "I remember it before anything else. Sitting there watching, wanting to play. It's my earliest memory. Before anything else, there was football."
At two he was walking the edges of curbs and the tops of fences, his arms spread out like the wings of an airplane, and not even a big, fast wind could knock him down. On the playground you couldn't tackle him, either. Emmitt was like one of those inflatable punching dummies: No matter how hard you hit the thing, it never fell over. At eight his legs were thick and knotted with muscle, his shoulders nicely rounded. He played mini-mite football, but there was nothing mini or mite about him.
His father drove a Pensacola city bus, and his mother was a document clerk for a bank. The Smiths were low-income, so for a time they lived in a housing project on busy Cervantes Street. Even though they didn't have much money, Emmitt knew what made the world go around. His grandfather would take him to the bank and lecture him on the importance of hard work, sacrifice, hustle. Emmitt loved going to the bank. He wanted things, and he was willing to work to get them. When he was in high school and his mother couldn't afford to buy him the designer clothes he favored, he took on menial jobs and scrimped and saved and bought the clothes himself.
"That was a valuable lesson," he would say many years later. "It gave me a sense of independence. It showed me that if I earn my own money, then I've got the right to spend it on whatever I want, and nobody can say anything about it."
True to form, Emmitt even beat the rest of the field to puberty. He had whiskers at 12. They came in dense, furry bunches, while the whiskers of all the other kids could be counted one-two-three. "Eighth grade, and he looked like a grown man," says Johnny Nichols, one of Emmitt's oldest and best friends.
At 13 Emmitt was practicing the moves he uses today. You gave him the ball and he could see things: the play developing as if in slow motion, the hole that everybody else saw as only a crease, the very intent of the defense.
Jimmy Nichols used to go to middle school track meets to see Emmitt run. Jimmy is Johnny's father, and back in those days he was the offensive coordinator at Escambia High. "The guy could really go," Jimmy says of Emmitt. "Rules kept us from talking to him, but I kept hearing that he was coming to Escambia, and you can just about imagine how excited that made me. His speed was impressive, and most people don't think he has great speed. I will say this about that: Nobody ever caught Emmitt from behind. It's deceptive speed. He does what he has to do. Not only could he run as an eighth-grader, but the way the kid handled and presented himself set him apart. This was a young man as a 13-year-old."
Dwight Thomas, then Escambia's head coach, had the same first impression. He once told a reporter from USA Today that, after meeting Emmitt, he dug up the kid's birth certificate just to make sure he wasn't as old as he seemed.
Fourteen was when Emmitt first ventured into a weight room and settled under a barbell. He liked it: the warm iron bar, the clattering plates. In a month he was squatting 400 pounds and clean-and-jerking 275. Coaches and players watched in heart-hammering wonder, their mouths forming circles. "Fourteen," they whispered.
At the start of two-a-days, Thomas and Nichols gave Emmitt a locker in the varsity dressing room. They also gave him two weeks to prove himself deserving of the honor, but Emmitt took care of that at the first few practices. "Since nobody could exactly tackle the guy...," recalls Nichols.
The year before, Escambia had won only one of 10 games, but during Emmitt's freshman season, in 1983, the Gators went 7-3 and barely missed the playoffs. Emmitt carried the team, just as he would for the next few years. Back in those days he was barely 5'9", but he could dunk a basketball. His thighs were almost as big around then as they are now: 27 inches. Emmitt went to stores and ogled the yawning racks of jeans, not one pair of which he could pull up past his knees. "Mama," he complained, "why can everybody wear jeans but me?"
Sixteen, and the college recruiters started to gather along the sidelines to watch him practice. They came by the dozens, all manner of recruiters from all kinds of schools: hustlers and dreamers, losers and champions and also-rans. It was a rare autumn Friday night that Emmitt didn't rush for more than 100 yards, but that wasn't the only reason people went to see him. A sort of folklore had grown up around him, and people wanted to be able to say they had seen him play way back when.
He was so much fun to watch, the legend goes, that a coach for rival Gulf Breeze High complained one night when Nichols pulled Emmitt out of the game. It was Escambia's policy to bench its starters as soon as the game got out of reach, and in this one the outcome was settled early in the third quarter. "It was something like 45-0," Nichols recalls. "And after the game the coach for the other team is mad as hell. He comes over and says, 'Jimmy, what are you doing taking Emmitt out?' I said, 'Look here, I have no intention of embarrassing you and your kids.' He said, 'Come on! I like watching the kid run!'"
In four years at Escambia, Emmitt gained slightly more than 8,800 yards rushing, the fourth-best career total ever for a schoolboy running back. Parade magazine named him the national high school Player of the Year in 1986, and USA Today, in doing the same, put him on its lead sports page. FLORIDA PREP BECOMING A LEGEND, FAST, said the headline.
"We do three things here on offense," Thomas said. "We hand the ball to Emmitt, we pitch the ball to Emmitt, and we throw the ball to Emmitt."
As the Gatorade Player of the Year, too, Emmitt won an all-expenses-paid trip for two to Super Bowl XXI in Pasadena. He took Johnny Nichols along, and they stayed in a luxury hotel and really lived it up for a few days. More than 120 million people watched the game on TV, but there Emmitt was, in person, on the floor of the Rose Bowl. As thrilled as he was by the experience, he was able to keep it in perspective. Super Bowls were his destiny. "One day," he said, turning to Nichols with a look of absolute conviction, "I'm going to play in a game like this."
But first he had college, and that was Florida. Emmitt broke a personal streak when he failed to start in his first game as a freshman. When Gators coaches recruited him, they had promised that he would start right away, and yet in the opener, against Miami, Emmitt found himself watching from the sideline until the fourth quarter. Didn't they know he didn't like to wait? Couldn't they see what a hurry he was in?
After the game coach Galen Hall tried his best to calm Emmitt down. Didn't want to put too much pressure on you too soon, was basically what he said. Didn't want you to go losing your confidence. Emmitt didn't start the second game, against Tulsa, either, but when he got the call in the first quarter, he showed them, he showed everybody. Emmitt gained a total of 109 yards and scored a couple of touchdowns, one of them on a 66-yard run. Hall let him start the third game, against Alabama, and Emmitt responded with 224 yards and two touchdowns and led the Gators to a 23-14 upset. No Florida player had ever gained as many yards on the ground in one game, which to Emmitt sounded more like it: He was back on the fast track, waiting for no one.
In Game 7, in mid-October, Emmitt passed the 1,000-yard mark, becoming the first runner in college football history to get there that fast. "He definitely was the big man around Gainesville," says Johnny Nichols, then Emmitt's roommate, "but it never went to his head. He was the same Emmitt we all knew before. Humble."
Emmitt decided to go pro after his junior year. The football program at Florida was going through tough times. Hall resigned during the 1989 season, and NCAA probation was looming. And there was one more thing that helped Emmitt make up his mind: For the first time the NFL would allow teams to draft underclassmen, and Emmitt was a lock to go high.
The day of the draft he gathered with family and friends at a beach house in Pensacola. Whenever the phone rang, Mary answered, then passed the receiver to Emmitt, who retreated to a bedroom and talked for a minute before coming back out. Each call carried the weight of enormous possibility, and it racked the relatives' nerves just to hear the phone ring. Sixteen players were chosen before Cowboys coach Jimmy Johnson called. Mary greeted him with a polite but anxious hello and handed him over to Emmitt. After talking in the bedroom, Emmitt emerged with a grave expression on his face.
"Well, what did he say?" somebody asked.
Emmitt was quiet for a second. Then he said, "He wanted to know if I wanted to wear a star on my helmet."
Everybody screamed and hollered, and Emmitt went out by himself to look at the Gulf of Mexico and ponder the future.
The gate, which rolls from side to side on a track, is a little slow in opening but plenty tall enough to keep undesirables out. Last season, after the Cowboys beat the Green Bay Packers for the NFC championship, Emmitt came home to find a gift from a neighbor named Jim sitting in front of the gate. It was a magnum of champagne in a silver-plated bucket. What a good man Jim was, to do that. Later, when Emmitt saw Jim, he asked if Jim wanted the bucket back. The thing looked expensive, and Emmitt wanted to make sure...well, did he want it? Jim said no way, and Emmitt said, "Then you'll have to come over when I open the champagne. We'll celebrate together."
Life was good and getting better all the time. Emmitt wasn't even 30 yet, and already he owned a champagne bucket. You should check the record books on that one, too. It's usually your older crowd that owns champagne buckets.
Emmitt lives by himself in the house, all 13,000 square feet of it. There's a woman he dates, but there's much to do yet before he even thinks about getting married. For now it's just Emmitt and his two Akitas. When they see him driving up today they scramble to the front of their kennel and snort around and generally do their best to let him know how much they like him. Emmitt named them after a couple of characters in a movie nobody saw.
"That one's Tango, and that one's Cash," he says, but he doesn't point, so it's hard to tell who is who. The dogs don't seem to know either.
Emmitt walks through the garage, where he thinks he might build a model-railroad town one of these days. It would be more than a town, actually—more like a little version of the world in this garage, with about 14 stops. One stop might be a place in the mountains, another a resort by the ocean. Emmitt has even considered including the Dallas location where President Kennedy was shot. He would put in the Texas School Book Depository and the grassy knoll and the motorcade. It's all down the road a bit, but he's determined to build the thing. "Maybe when I'm married and have kids," he says.
The house is extremely clean and beautiful. Emmitt could stick a sign in the yard that says LIFE AT THE TOP and sell tickets to sightseers and pay off the mortgage right there. He goes to the den, where pictures of his family stand in large frames. There are his three brothers, his two sisters. There's Mary. There's his dad. And there's Emmitt with Roy Jones Jr., the super middleweight champion. Jones is from Pensacola, too, and the two of them have been friends for years. People back home are always telling Emmitt that Roy did this and Roy did that for Pensacola and that Roy still lives in Pensacola. Emmitt says, "But my work is in Dallas. I have to live there. I can't play the Super Bowl in Pensacola, can I?" But that isn't explanation enough. They still take their swipes.
"I honestly believe people in Pensacola will not be happy until I fall flat on my face," Emmitt says, in a rare dark moment. He hardly ever gets back to his hometown. Four or five days a year, maybe. Once when he was there he went out to a club, and one of his old pals from high school joined him, and after a while the old friend turned to him and said, "Emmitt, can I have your autograph?" The guy had one of those looks, too, as if he had just glimpsed heaven and wanted to snatch a puff of cloud. And the look, coming from a friend, irritated Emmitt no end. "Man, what do you need my autograph for?" he said. "You know me better than anybody in here."
"Shoot, you're a superstar," Emmitt's buddy said.
"I'm Emmitt. We played ball together. You've been in my house."
"Yeah," the guy said, "but you've been on Arsenio Hall." They talked on for a while, Emmitt growing more depressed by the second. When he got back to his parents' house that night, he slept on the couch in the living room. It was where he'd slept sometimes when he was a kid. "You're stargazing," Emmitt had told his friend, and it all made Emmitt wonder, Is it possible to get too big too fast? If you run way out ahead of everyone else, where does that leave you? All alone in a nightclub crowded with people, one of them shoving a scrap of paper at you and saying, "Sign here, Mr. Superstar"?
But, hey, Emmitt has moved to another room of his new house now. And you should see it. "This is my kitchen," he says. "This cost pretty good. Come to think of it, somewhere between \$75,000 and \$100,000. Those cabinets, that's mahogany with a high gloss on top. Check out those stools. Go ahead, pick one up. Heavy as hell, ain't it? That refrigerator door? That's mahogany, too. See the finish?" He gives a nod. "Right. High gloss."
He moves from room to room and eventually makes his way upstairs. "This is my little boy's room," he says.
"Well, when I have a little boy...."
In the game room he pauses before an enormous fish tank and bends over, staring through the glass. "Some of these things in here," he says, "they look dead, but they're not." He points to a fluttering wisp of something on a rock. "Now that—yes, that is a skeleton. If it looks dead, it's because it is."
In spite of everything that a new house demands and all that this one has to offer, Emmitt still gets restless on occasion. How long can you look at fish? How many movies can you watch? It was on one such night that he rounded up the pen and the piece of paper and compared his numbers with Payton's. "It's a dream until you write it down," his coach told him half a lifetime ago. "Then it's a goal." Somewhere in the heat of all that scribbling, lost in the numbers, was the place to which Emmitt was running.
To end the tour today, he goes through a glass door and stands outside on the balcony. It's getting late, and the shadows are long, and the view is all manner of lovely. You can see Emmitt's barbecue and smoker down to the right. Straight ahead, occupying a big chunk of the backyard, is Emmitt's swimming pool. It's a little cold to go for a dip, but the pool is in perfect shape, a dream. And, wait, what's that down there on the bottom?
Emmitt leans over the rail to get a better look. "Oh," he says. "That's a friend of mine." The image is made of blue and white tiles, and though it lacks the precision of a snapshot, it's so familiar that even 20,000 gallons of water can't distort it.
Bursting through a Cowboys star, kicking a knee high, is Emmitt, the ball tucked against his body for safekeeping. He's running, of course, running for the goal line, running for the victory, running as if to forever.
HOLE YARDS PAR R1 R2 R3 R4
OUT
HOLE YARDS PAR R1 R2 R3 R4
IN
Eagle (-2)
Birdie (-1)
Bogey (+1)
Double Bogey (+2) | 7,501 | 32,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-47 | latest | en | 0.99219 |
http://www.haskell.org/haskellwiki/index.php?title=Sinc_function&oldid=2541 | 1,406,374,009,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997901076.42/warc/CC-MAIN-20140722025821-00109-ip-10-33-131-23.ec2.internal.warc.gz | 764,093,352 | 5,064 | # Sinc function
## Sinc function
The sinc function sin(x) / x is a useful function that is a little tricky to use because it becomes 0/0 as x tends to 0. Here is an implementation taken from the Boost library.
```epsilon :: RealFloat a => a
epsilon = encodeFloat 1 (fromIntegral \$ 1-floatDigits epsilon)
{- Boosted from Boost http://www.boost.org/boost/math/special_functions/sinc.hpp -}
sinc :: (RealFloat a) => a -> a
sinc x | (abs x) >= taylor_n_bound = (sin x)/x
| otherwise = 1 - (x^2/6) + (x^4/120)
where
taylor_n_bound = sqrt \$ sqrt epsilon
``` | 170 | 557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2014-23 | longest | en | 0.658591 |
http://www.beatthegmat.com/og-17-t293465.html | 1,508,456,381,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823482.25/warc/CC-MAIN-20171019231858-20171020011858-00652.warc.gz | 401,702,253 | 33,923 | • 5 Day FREE Trial
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## OG-17
tagged by: Brent@GMATPrepNow
This topic has 3 expert replies and 1 member reply
Joy Shaha Senior | Next Rank: 100 Posts
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#### OG-17
Sat Nov 19, 2016 1:13 pm
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
A technician makes a round-trip to and from a certain service center by the
same route. If the technician completes the drive to the center and then
completes 10 percent of the drive from the center, what percent of the
round-trip has the technician completed?
A) 5% B) 10% C) 25% D) 40% E) 55%
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melguy Master | Next Rank: 500 Posts
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Sat Nov 19, 2016 6:52 pm
Let the drive from A to B is 100 Kms
A -> B = 100
B -> A = 100
(Total = 200)
Driver has completed A -> B (100 Kms) + 10% of B -> A (10 Kms)
100+ 10 = 110
How percent has he covered out of the total distance (200 kms)
110
----- X 100
200
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Brent@GMATPrepNow GMAT Instructor
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Sun Nov 20, 2016 5:14 am
Joy Shaha wrote:
A technician makes a round-trip to and from a certain service center by the
same route. If the technician completes the drive to the center and then
completes 10 percent of the drive from the center, what percent of the
round-trip has the technician completed?
A) 5%
B) 10%
C) 25%
D) 40%
E) 55%
We can use the answer choices to our advantage (ALWAYS check the answer choices before performing any calculations - see the General GMAT Math Strategies video below)
The TOTAL distance for the round trip = (distance from home to service center) + (distance from service center to home)
Once the technician drove the distance from home to service center, he/she has completed HALF (50%) of the round trip.
Once the technician drives a portion of distance from service center to home, he/she has completed more than HALF of the round trip.
So, the correct answer is GREATER THAN 50%
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Fri Nov 25, 2016 3:10 pm
I'd hop on the answers here. Since the technician has already traveled at least half of the total trip, he must have gone > 50% of the way.
With that in mind, only E works!
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Sun Nov 27, 2016 5:50 pm
Joy Shaha wrote:
A technician makes a round-trip to and from a certain service center by the
same route. If the technician completes the drive to the center and then
completes 10 percent of the drive from the center, what percent of the
round-trip has the technician completed?
A) 5% B) 10% C) 25% D) 40% E) 55%
We can let the distance to the center = d, so the roundtrip distance = 2d.
Since the technician has driven to the center and completed 10 percent of the drive home, he has gone d + 0.1d = 1.1d. Thus, the percent completed is:
(1.1d)/(2d) x 100
11/20 x 100 = 11 x 5 = 55 percent.
Alternate solution:
As Brent has already mentioned, after the technician has driven to the center, he or she has completed half, or 50% of the round trip. We know that the technician will drive a bit more, and without doing the math, we see that the only logical answer choice is 55%.
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See More Top Beat The GMAT Experts | 1,681 | 6,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-43 | longest | en | 0.882088 |
https://discuss.leetcode.com/topic/17745/8-ms-non-bfs-c-solution | 1,516,299,297,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887535.40/warc/CC-MAIN-20180118171050-20180118191050-00109.warc.gz | 666,788,713 | 10,435 | 8 ms non-BFS C++ solution
• Just want to share my O(N^2) C++ solution. The idea is to find all possible islands and remove the duplicates. cnt marks the island number each element belongs to. mapping maps the island number to the smallest connected island number.
``````class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
if (grid.size() == 0) return 0;
vector<vector<int>> cnt(grid.size(), vector<int>(grid[0].size(), 0));
int max = 0;
int dup = 0;
map<int, int> mapping;
for (int i = 0; i < grid.size(); i++) {
for (int j = 0; j < grid[i].size(); j++) {
if (grid[i][j] == '1') {
int a = (i == 0) ? 0: cnt[i-1][j];
int b = (j == 0) ? 0: cnt[i][j-1];
if (a == 0 && b == 0) {
max++;
cnt[i][j] = max;
mapping[max] = max;
} else if (a == 0) { // b != 0
cnt[i][j] = b;
} else if (b == 0) { // a != 0
cnt[i][j] = a;
} else if (a == b) {
cnt[i][j] = a;
} else { // both not zero and not equal
if (mapping[a] != mapping[b]) {
dup++;
if (mapping[a] > mapping[b]) {
remap(mapping, a, mapping[b]);
} else {
remap(mapping, b, mapping[a]);
}
}
cnt[i][j] = mapping[a];
}
}
}
}
return max - dup;
}
void remap(map<int,int>& mapping, int key, int val) {
mapping[mapping[key]] = val; // important
mapping[key] = val;
} };``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 444 | 1,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-05 | latest | en | 0.444135 |
http://www.talkstats.com/threads/advice-on-methodology-and-analysis-regression.71567/ | 1,550,890,732,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249434065.81/warc/CC-MAIN-20190223021219-20190223043219-00287.warc.gz | 444,012,388 | 9,120 | # Advice on methodology and analysis, regression.
#### Action
##### New Member
Dear all,
For my Master’s thesis, I have some difficulties with the statistical part of my research.
I want to analyze the relationship between two variables: Investment in Innovation (R&D) --> Concentration Ratio
1. Research & Development (R&D) Investments in an industry
2. Concentration Ratio of the industry (Herfindahl index or Concentration Ratio by 4/50 largest firms)
Now, my supervisor is very busy yet told me I should use:
Regression Formula = H(it)-H(it-1)= a+b*(R&D/VA)+ c*(GFCF/VA)
DV = Change in Concentration ratio (Herfindahl index or Concentration Ratio index for 4/tm50)
- i = Industry (in NAICS industry code)
- t = Time (1997-2012, with a 5 year interval; 1997->2002->2007->2012)
- H = Herfindahl index for each year. (a measure of concentration)
- RD= (R&D) Research and development investment
- VA= Value Added
- GFCF= Gross Fixed Capital Formation
My questions:
1. Can you guys give me any advice? And what would my hypotheses be?
2. How do I include a dummy variable into this? Any suggestions on variables to use? I am considering on using: R&D intensive industry or not (1=yes, 0=no).
3. How would I go on about this, in Spss and such. Basically, I am asking for advice on the analysis.
I hope I was clear enough, please ask for further details. Thank you! | 353 | 1,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-09 | longest | en | 0.886186 |
https://cryptobygo.netlify.app/salt7507zaw/what-is-the-price-weighted-index-return-85 | 1,726,602,692,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00498.warc.gz | 163,592,978 | 10,352 | ## What is the price- weighted index return
14 Jun 2018 In a price-weighted index, the index is influenced by the individual share price of each listed stock. It differs from a market weighted index in the In a price-weighted index, a stock that increases from \$110 to \$120 will have a greater effect on the index than a stock that increases from \$10 to \$20, even though the percentage move is greater A price-weighted index gives influence to each of the companies in the index based on its share price, not its total market value. For example, if Company A's stock trades at \$90 per share and Company's B's stock trades at \$30 per share, Company A's stock is weighted three times as heavily as Company B's.
A capitalization-weighted (or "cap-weighted") index, also called a market-value- weighted index is a stock market index whose components are weighted 18 May 2018 This allows the construction of indexes that will track the average stock price performance of a specific sector or market. One of the most popular Divide the gain or loss by the initial value to figure the rate of return for the index. Continuing the example, divide \$10 by \$100 to get a return rate of 0.1. Multiply the 6 Jun 2019 For example, let's assume that the following companies are in the XYZ price- weighted index: A price-weighted index is simply the sum of the A price-weighted index is a type of stock market index in which each component of the index is weighted according to its current share price. In price-weighted 15 Jan 2020 The Dow Jones Industrial Average (DJIA) is a type of price-weighted index that currently measures the stock performance of 30 large 3 Jul 2019 A price-weighted index is a stock market index in which the index return is skewed towards the company with highest stock price i.e. Google.
## Calculation: Price-weighted index = [(\$16 + \$61 + \$38)/3] - [(\$21 + \$49 + \$37)/3]/[(\$21 + \$49 + \$37)/3] = 7.48 percent Example: A price-weighted index consists of stocks A, B, and C which are priced at \$38, \$21, and \$26 a share, respectively.
15 Jan 2020 The Dow Jones Industrial Average (DJIA) is a type of price-weighted index that currently measures the stock performance of 30 large 3 Jul 2019 A price-weighted index is a stock market index in which the index return is skewed towards the company with highest stock price i.e. Google. An index of a group of securities computed by calculating a weighted average of the returns on each security in the index, where the weights are proportional to Each index contains index returns with and without dividends, index weights and counts. The Equal-Weighted Index is an Equal-Weighted Portfolio built each The rate of return would be: (70 - 62.5) / 62.5 = 12%. Stock split. Price- weighting is simple, but a price-weighted index has a downward bias. High- priced
### A value-weighted index assigns a weight to each company in the index based on its value or market capitalization. Follow the example and you will learn how a value weighted index number is calculated.
24 Nov 2019 The components of a market value-weighted index are weighted in to achieve returns above the common market value-weighted index… 4 Jan 2019 The price-weighted index Price-weighted indexes aren't particularly c. Why does is become harder to maintain the same level of return in the An equally weighted index weights each stock equally regardless of its market return over long periods after expenses vs. market cap weighted indexes. the weights applied to the sample securities (that is, price-weighted, value- weighted, or The S&P 500 is a value-weighted index; that is, each stock's return. This paper issues a warning that compounding daily returns of the Center for Research in Security Prices. (CRSP) equal-weighted index can lead to surprisingly 3 Jan 2020 The allure of equally weighted indexes is that the investor is getting a broader representation of the index constituents, or increased diversification An equally weighted index holds the same dollar amount of each security, making it easy for you to track performance. Businessman with newspaper. Equally
### Divide the gain or loss by the initial value to figure the rate of return for the index. Continuing the example, divide \$10 by \$100 to get a return rate of 0.1. Multiply the
In a price-weighted index, a stock that increases from \$110 to \$120 will have a greater effect on the index than a stock that increases from \$10 to \$20, even though the percentage move is greater A price-weighted index gives influence to each of the companies in the index based on its share price, not its total market value. For example, if Company A's stock trades at \$90 per share and Company's B's stock trades at \$30 per share, Company A's stock is weighted three times as heavily as Company B's. Price-weighted indices display the average value of a stock without regard to the number of shares purchased or the magnitude of the stock's price. Changes in a price-weighed index allow you to track increases or decreases in the index. And from this information you may derive its rate of return. This rate may be calculated as a period return or annualized to compare with the annual rates of return of other investments. A price-weighted index is a type of stock market index in which each component of the index is weighted according to its current share price. In price-weighted indices, companies with a high share price have a greater weight than those with a low share price. What is Price-Weighted Index? A price-weighted index is a stock market Index in which companies’ stocks are weighted according to their share price. A price-weighted index is mostly influenced by stock which has a higher price and such stock receives greater weight in the index regardless of companies issuing size or number of outstanding Shares. A price-weighted index is an index in which the member companies are weighted in proportion to their price per share, rather than by number of shares outstanding, market capitalization or other factors. The Dow Jones Industrial Average (DJIA) is a price-weighted index.
## The rate of return would be: (70 - 62.5) / 62.5 = 12%. Stock split. Price- weighting is simple, but a price-weighted index has a downward bias. High- priced
An equally weighted index weights each stock equally regardless of its market return over long periods after expenses vs. market cap weighted indexes. the weights applied to the sample securities (that is, price-weighted, value- weighted, or The S&P 500 is a value-weighted index; that is, each stock's return.
A capitalization-weighted index is a type of market index with individual components, or securities, weighted according to their total market capitalization. Market capitalization uses the total market value of a firm's outstanding shares. The calculation multiples outstand shares by the current price of a single share. A price-weighted average is a simple mathematical average of several stock prices, and is often used to construct a price-weighted index. Perhaps the most well-known stock index in the U.S., the Dow Jones Industrial Average is a price-weighted index. With a price-weighted index, the index trading price is based on the trading prices of the individual securities (stocks) that comprise the index basket (known as components). In other words, the stocks with the higher prices will have more impact on the movement of the index than stocks with lower prices, since their price is "weighted" higher. Example: A price-weighted index consists of stocks A, B, and C which are priced at \$38, \$21, and \$26 a share, respectively. The current index divisor is 2.7. What will the new index divisor be if stock B undergoes a 3-for-1 stock split Price-weighted indices display the average value of a stock without regard to the number of shares purchased or the magnitude of the stock's price. Changes in a price-weighed index allow you to track increases or decreases in the index. And from this information you may derive its rate of return. A price-weighted index is a type of stock market index in which each component of the index is weighted according to its current share price. In price-weighted indices, companies with a high share price have a greater weight than those with a low share price. | 1,810 | 8,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-38 | latest | en | 0.912663 |
https://thirdspacelearning.com/secondary-resources/gcse-maths-worksheet-area-of-compound-shapes/ | 1,716,311,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058504.35/warc/CC-MAIN-20240521153045-20240521183045-00331.warc.gz | 499,531,657 | 37,506 | Maths Resources GCSE Worksheets
# Area Of Compound Shapes Worksheet
• Section 1 of the area of compound shapes worksheet contains 15 skills-based area of compound shapes questions, in 3 groups to support differentiation
• Section 2 contains 4 applied area of compound shapes questions with a mix of worded problems and deeper problem solving questions
• Section 3 contains 4 foundation and higher level GCSE exam style area of compound shapes questions
• Answers and a mark scheme for all questions are provided
• Questions follow variation theory with plenty of opportunities for students to work independently at their own level
• All questions created by fully qualified expert secondary maths teachers
• Suitable for GCSE maths revision for AQA, OCR and Edexcel exam board
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### Area of compound shapes at a glance
A compound shape is made from two or more simple shapes. Area is the amount of space covered by a 2D shape and is measured in square units – metric units of area include mm², cm² and m². To find the area of compound shapes, we need to break the shape into the component parts and find the area of each part. Sometimes the terminology ‘area of composite shapes’ is used which means exactly the same thing.
Rectilinear shapes are shapes with straight sides and right angles – i.e. made from rectangles. To find the area of a rectilinear shape, we need to find the area of each rectangle that makes up the shape, then find the sum of them in order to find the total area.
Other compound shapes can be made using simple 2D shapes – for example, a triangle placed on top of a parallelogram. Use the standard area formulae to calculate the area of each simple shape, then sum them to find the total area. In some problems, students may be required instead to use subtraction to find the area left when one shape is removed from another. This is particularly common in exam style questions and word problems – for example, finding the area of grass in a rectangular garden with a circular pond.
In addition to whole-number lengths and areas, it is important that students are confident working with decimals, particularly rounding an answer to a given number of decimal places.
Looking forward, students can then progress to additional area worksheets and other geometry worksheets, for example an angles in polygons worksheet or area and circumference of a circle worksheet.
For more teaching and learning support on Geometry our GCSE maths lessons provide step by step support for all GCSE maths concepts.
## Do you have students who need additional support to achieve their target GCSE maths grade?
There will be students in your class who require individual attention to help them succeed in their maths GCSEs. In a class of 30, it’s not always easy to provide.
Help your students feel confident with exam-style questions and the strategies they’ll need to answer them correctly with personalised online one to one tutoring from Third Space Learning
Lessons are selected to provide support where each student needs it most, and specially-trained GCSE maths tutors adapt the pitch and pace of each lesson. This ensures a personalised revision programme that raises grades and boosts confidence.
GCSE Revision Programme | 697 | 3,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-22 | longest | en | 0.925 |
https://www.instasolv.com/question/5-let-k-be-a-real-number-such-that-k-0-if-a-and-b-are-non-zero-complex-nu-opppnr | 1,610,942,239,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00296.warc.gz | 851,148,388 | 11,054 | 5. Let k be a real number such that...
Question
# 5. Let k be a real number such that k 0. If a and B are non zero complex numbers satisfying a + B = - 2k and Of + BP = 4k? - 2k, then a quadratic equation having a+ and C+B as its roots is equal to (B)x - 46x + 4k=0 (A) 4x2 - 4kx + k=0 (C) 4kx² - 4x + k = 0 (D) 4kx² - 4x + 1 = 0
JEE/Engineering Exams
Maths
Solution
98
4.0 (1 ratings)
The Oudrahic equahon is ( x^{2}-(alpha+beta) x+alpha_{1} beta ) ( =x^{2}-left(frac{20+beta}{alpha}+frac{alpha+beta}{beta}right) x+frac{(alpha+beta)^{2}}{alpha beta} ) ( Rightarrow x^{2}-left(frac{8^{2}+alpha beta+alpha^{2}+alpha beta}{alpha beta}right) x+frac{(alpha+B)^{2}}{alpha beta} ) | 281 | 676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-04 | latest | en | 0.6383 |
http://openstudy.com/updates/50d320d8e4b069abbb719685 | 1,444,421,858,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737935292.75/warc/CC-MAIN-20151001221855-00136-ip-10-137-6-227.ec2.internal.warc.gz | 234,442,502 | 11,110 | 1. jason10
hmmm same here then1
2. surbhimahajan93
for q6 what is the answer for vmin if v0=10V,T=1ms,rw=2ohm,l=1mh,RL=3OHM
3. chiragsala
Q 6 answers got: a) 0.5 b) 0.5 c) 4.38 d)0.67
4. surbhimahajan93
mine are 0.2 0.2 6 but the last one is wrong
5. chiragsala
@surbhimahajan93 it can be 54.15m or 0.054
6. surbhimahajan93
its my final check.gimme the formula.
7. chiragsala
simulate it dude.
8. chiragsala
how you did 3rd part.... once tell for these values... Vo=5V, Rw=1ohms, Rl=1ohms, L=1mH
9. chiragsala
no thats wrong.
10. surbhimahajan93
for ur values 2.5
11. surbhimahajan93
2.5 is incorrect??
12. surbhimahajan93
i did mine with that formulae and got 6.got a green check
13. vim0409
Anyoone to finish Q4 part 2 and 3
14. dancios
could somebody can give me values for q6 ? i have something wrong 2 checks already :) Vo=10V, T=0.5ms, Rw= 1 ohms, Lw=0.5mH, Rl=4 ohms.
15. brunofornari
V0=5V, T=1ms, Rw=1Ω, Lw=1mH and RL=1Ω. | 405 | 959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2015-40 | longest | en | 0.604118 |
http://www.stata.com/statalist/archive/2004-05/msg00839.html | 1,495,606,346,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607802.75/warc/CC-MAIN-20170524055048-20170524075048-00182.warc.gz | 671,970,758 | 2,783 | # st: multiple )))brackets, is there a more efficient way?
From "Annelies Vos" To statalist@hsphsun2.harvard.edu Subject st: multiple )))brackets, is there a more efficient way? Date Tue, 25 May 2004 08:35:06 +0200 (MEST)
```Dear all,
in the FAQs I found the following very useful recommendation:
. generate byte a = 1 if y <= 20
. replace a = 2 if y > 20 & y <= 30
. replace a = 3 if y > 30 & y <= 40
. replace a = 4 if y > 40 & y <.
do the following:
. #delim ;
. generate byte a =
cond(y<=20, 1,
cond(y<=30, 2,
cond(y<=40, 3,
cond(y<., 4,
. ))));
However, the variable I want to use it for (nationality) has many
values (every country in the world), which should be recoded into
countrygroups. I don't really like the idea of having to count the
number of "opening brackets": "(" , to know with how many "closing
brackets": ")" I should end. Is there any easier solution for this?
to explain a piece of my syntax:
> #delim;
> generate byte origin =
> cond(natio==3, 7,
> cond(natio==8, 10,
> cond(natio==12, 10,
> cond(natio==14, 8,
> cond(natio==28, -9,
> cond(natio==54, 6,
> cond(natio==69, 10,
> cond(natio==82, 8,
> cond(natio==139, 10,
> cond(natio==141, 10,
...etcetera
...which I would like to end on another way than:
> . ))))))))))
Thanks for any suggestions,
Annelies Vos
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 484 | 1,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-22 | latest | en | 0.758906 |
https://brainmass.com/physics/classical-mechanics/binary-decimal-system-532536 | 1,660,227,797,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571472.69/warc/CC-MAIN-20220811133823-20220811163823-00212.warc.gz | 172,720,217 | 74,606 | Explore BrainMass
# Binary and decimal system
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1. Convert the following decimal numbers to binary
a. 81_10
b. 248_10
2. Convert the following octal numbers to binary
a. 47_8
b. 214_8
3. Convert the following hexadecimal numbers tobinary
a. BA3_16
b. F3B4_16
4. Convert the following binary numbers to decimal numbers
a. 1 1 1 0 1 1 0 1 1 0 1 1 0 1
b. 1 0 0 0 1 1 0 1 1 0 1 0 1 1
5. Convert the following hexadecimal numbers to octal numbers
a. 3CF2_16
b. B7F23_16
https://brainmass.com/physics/classical-mechanics/binary-decimal-system-532536
## SOLUTION This solution is FREE courtesy of BrainMass!
1.
a. 81_10 = 64 + 16 + 1 = 1010001_2.
b. 248_10 = 128 + 64 + 32 + 16 + 8 = 111111000_2.
2.
a. 47_8 = 100111_2.
b. 214_8 = 10001100_2.
3.
a. BA3_16 = 101110100011_2.
b. F3B4_16 = 1111001110110100_2.
4.
a. 11101101101101_2 = 1 + 4 + 8 + 32 + 64 + 256 + 512 + 2048 + 4096 + 8192 = 15197_10.
b. 10001101101011_2 = 1 + 2 + 8 + 32 + 64 + 256 + 512 + 8192 = 9067_10.
5.
a. 3CF2_16 = 11110011110010_2 = 36362_8.
b. B7F23_16 = 10110111111100100011_2 = 2677443_8.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here! | 552 | 1,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-33 | longest | en | 0.585522 |
https://getrevising.co.uk/revision-tests/maths_unit_a2d_and_3d_pythagoras | 1,526,969,180,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864626.37/warc/CC-MAIN-20180522053839-20180522073839-00481.warc.gz | 563,636,483 | 13,302 | # Maths unit A:2D and 3D Pythagoras
HideShow resource information
• Created by: rvsn
• Created on: 16-11-14 22:51
What triangles does this theorem work on?
Right-angled triangles
1 of 5
What is the theorem?
a^2+b^2=c^2
2 of 5
What does c represent?
The hypotenuse (the longest side)
3 of 5
How do you do this in cubes/cuboids?
Use multiple right-angled triangles until you reach the side you need to find
4 of 5
How do you do this in square based pyramids?
You will do the same thing as a cube/cuboid except you have to find the midpoint of the side in the middle of the base at some point
5 of 5
## Other cards in this set
### Card 2
#### Front
What is the theorem?
a^2+b^2=c^2
### Card 3
#### Front
What does c represent?
### Card 4
#### Front
How do you do this in cubes/cuboids?
### Card 5
#### Front
How do you do this in square based pyramids? | 261 | 864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-22 | latest | en | 0.898797 |
https://www.esaral.com/q/a-pole-stands-vertically-inside-a-triangular-park-abc-66679 | 1,723,192,367,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640762343.50/warc/CC-MAIN-20240809075530-20240809105530-00085.warc.gz | 599,481,855 | 11,587 | # A pole stands vertically inside a triangular park ABC.
Question:
A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of
the pole from each corner of the park be $\frac{\pi}{3}$.
If the radius of the circumcircle ot $\triangle \mathrm{ABC}$ is 2, then the height of the pole is equal to :
1. $\frac{2 \sqrt{3}}{3}$
2. $2 \sqrt{3}$
3. $\sqrt{3}$
4. $\frac{1}{\sqrt{3}}$
Correct Option: , 2
Solution:
Let $P D=h, R=2$
As angle of elevation of top of pole from
$\mathrm{A}, \mathrm{B}, \mathrm{C}$ are equal So
D must be circumcentre
of $\triangle \mathrm{ABC}$
$\tan \left(\frac{\pi}{3}\right)=\frac{\mathrm{PD}}{\mathrm{R}}=\frac{\mathrm{h}}{\mathrm{R}}$
$h=R \tan \left(\frac{\pi}{3}\right)=2 \sqrt{3}$ | 248 | 761 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-33 | latest | en | 0.528085 |
http://en.wikibooks.org/wiki/Haskell/Strictness | 1,427,928,116,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131309963.95/warc/CC-MAIN-20150323172149-00067-ip-10-168-14-71.ec2.internal.warc.gz | 96,059,335 | 10,592 | Strictness (Solutions)
## §Difference between strict and lazy evaluation
Strict evaluation, or eager evaluation, is an evaluation strategy where expressions are evaluated as soon as they are bound to a variable. For example, with strict evaluation, when `x = 3 * 7` is read, 3 * 7 is immediately computed and 21 is bound to x. Conversely, with lazy evaluation values are only computed when they are needed. In the example `x = 3 * 7`, 3 * 7 isn't evaluated until it's needed, like if you needed to output the value of x.
## §Why laziness can be problematic
Lazy evaluation often involves objects called thunks. A thunk is a placeholder object, specifying not the data itself, but rather how to compute that data. An entity can be replaced with a thunk to compute that entity. When an entity is copied, whether or not it is a thunk doesn't matter - it's copied as is (on most implementations, a pointer to the data is created). When an entity is evaluated, it is first checked if it is thunk; if it's a thunk, then it is executed, otherwise the actual data is returned. It is by the magic of thunks that laziness can be implemented.
Generally, in the implementation the thunk is really just a pointer to a piece of (usually static) code, plus another pointer to the data the code should work on. If the entity computed by the thunk is larger than the pointer to the code and the associated data, then a thunk wins out in memory usage. But if the entity computed by the thunk is smaller, the thunk ends up using more memory.
As an example, consider an infinite length list generated using the expression `iterate (+ 1) 0`. The size of the list is infinite, but the code is just an add instruction, and the two pieces of data, 1 and 0, are just two Integers. In this case, the thunk representing that list takes much less memory than the actual list, which would take infinite memory.
However, as another example consider the number generated using the expression `4 * 13 + 2`. The value of that number is 54, but in thunk form it is a multiply, an add, and three numbers. In such a case, the thunk loses in terms of memory.
Often, the second case above will consume so much memory that it will consume the entire heap and force the garbage collector. This can slow down the execution of the program significantly. And that, in fact, is the main reason why laziness can be problematic.
Additionally, if the resulting value is used, no computation is saved; instead, a slight overhead (of a constant factor) for building the thunk is paid. However, this overhead is not something the programmer should deal with most of the times; more important factors must be considered and may give a much bigger improvements; additionally, optimizing Haskell compilers like GHC can perform 'strictness analysis' and remove that slight overhead.
## §References
Strictness Solutions to exercises Haskell Performance edit this chapter Haskell edit book structure | 638 | 2,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2015-14 | latest | en | 0.941168 |
https://www.ademcetinkaya.com/2023/04/lvwr-livewire-group-inc-common-stock.html | 1,686,323,789,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656737.96/warc/CC-MAIN-20230609132648-20230609162648-00768.warc.gz | 711,415,098 | 60,289 | Outlook: LiveWire Group Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Hold
Time series to forecast n: 10 Apr 2023 for (n+6 month)
Methodology : Inductive Learning (ML)
## Abstract
LiveWire Group Inc. Common Stock prediction model is evaluated with Inductive Learning (ML) and Chi-Square1,2,3,4 and it is concluded that the LVWR stock is predictable in the short/long term. According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Hold
## Key Points
1. Is it better to buy and sell or hold?
3. What are the most successful trading algorithms?
## LVWR Target Price Prediction Modeling Methodology
We consider LiveWire Group Inc. Common Stock Decision Process with Inductive Learning (ML) where A is the set of discrete actions of LVWR stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Chi-Square)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Inductive Learning (ML)) X S(n):→ (n+6 month) $∑ i = 1 n s i$
n:Time series to forecast
p:Price signals of LVWR stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## LVWR Stock Forecast (Buy or Sell) for (n+6 month)
Sample Set: Neural Network
Stock/Index: LVWR LiveWire Group Inc. Common Stock
Time series to forecast n: 10 Apr 2023 for (n+6 month)
According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Hold
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for LiveWire Group Inc. Common Stock
1. The fair value of a financial instrument at initial recognition is normally the transaction price (ie the fair value of the consideration given or received, see also paragraph B5.1.2A and IFRS 13). However, if part of the consideration given or received is for something other than the financial instrument, an entity shall measure the fair value of the financial instrument. For example, the fair value of a long-term loan or receivable that carries no interest can be measured as the present value of all future cash receipts discounted using the prevailing market rate(s) of interest for a similar instrument (similar as to currency, term, type of interest rate and other factors) with a similar credit rating. Any additional amount lent is an expense or a reduction of income unless it qualifies for recognition as some other type of asset.
2. In accordance with paragraph 4.1.3(a), principal is the fair value of the financial asset at initial recognition. However that principal amount may change over the life of the financial asset (for example, if there are repayments of principal).
3. If a financial instrument that was previously recognised as a financial asset is measured at fair value through profit or loss and its fair value decreases below zero, it is a financial liability measured in accordance with paragraph 4.2.1. However, hybrid contracts with hosts that are assets within the scope of this Standard are always measured in accordance with paragraph 4.3.2.
4. Rebalancing refers to the adjustments made to the designated quantities of the hedged item or the hedging instrument of an already existing hedging relationship for the purpose of maintaining a hedge ratio that complies with the hedge effectiveness requirements. Changes to designated quantities of a hedged item or of a hedging instrument for a different purpose do not constitute rebalancing for the purpose of this Standard
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
LiveWire Group Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. LiveWire Group Inc. Common Stock prediction model is evaluated with Inductive Learning (ML) and Chi-Square1,2,3,4 and it is concluded that the LVWR stock is predictable in the short/long term. According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Hold
### LVWR LiveWire Group Inc. Common Stock Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementBa2Caa2
Balance SheetBaa2Ba2
Leverage RatiosBaa2C
Cash FlowCCaa2
Rates of Return and ProfitabilityCCaa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 86 out of 100 with 568 signals.
## References
1. Athey S. 2019. The impact of machine learning on economics. In The Economics of Artificial Intelligence: An Agenda, ed. AK Agrawal, J Gans, A Goldfarb. Chicago: Univ. Chicago Press. In press
2. Çetinkaya, A., Zhang, Y.Z., Hao, Y.M. and Ma, X.Y., Trading Signals (WTS Stock Forecast). AC Investment Research Journal, 101(3).
3. J. G. Schneider, W. Wong, A. W. Moore, and M. A. Riedmiller. Distributed value functions. In Proceedings of the Sixteenth International Conference on Machine Learning (ICML 1999), Bled, Slovenia, June 27 - 30, 1999, pages 371–378, 1999.
4. Li L, Chen S, Kleban J, Gupta A. 2014. Counterfactual estimation and optimization of click metrics for search engines: a case study. In Proceedings of the 24th International Conference on the World Wide Web, pp. 929–34. New York: ACM
5. Candès E, Tao T. 2007. The Dantzig selector: statistical estimation when p is much larger than n. Ann. Stat. 35:2313–51
6. Imbens GW, Lemieux T. 2008. Regression discontinuity designs: a guide to practice. J. Econom. 142:615–35
7. C. Claus and C. Boutilier. The dynamics of reinforcement learning in cooperative multiagent systems. In Proceedings of the Fifteenth National Conference on Artificial Intelligence and Tenth Innovative Applications of Artificial Intelligence Conference, AAAI 98, IAAI 98, July 26-30, 1998, Madison, Wisconsin, USA., pages 746–752, 1998.
Frequently Asked QuestionsQ: What is the prediction methodology for LVWR stock?
A: LVWR stock prediction methodology: We evaluate the prediction models Inductive Learning (ML) and Chi-Square
Q: Is LVWR stock a buy or sell?
A: The dominant strategy among neural network is to Hold LVWR Stock.
Q: Is LiveWire Group Inc. Common Stock stock a good investment?
A: The consensus rating for LiveWire Group Inc. Common Stock is Hold and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of LVWR stock?
A: The consensus rating for LVWR is Hold.
Q: What is the prediction period for LVWR stock?
A: The prediction period for LVWR is (n+6 month) | 1,883 | 7,773 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-23 | latest | en | 0.836006 |
http://mathhelpforum.com/trigonometry/42455-can-anyone-help-print.html | 1,529,323,463,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859766.6/warc/CC-MAIN-20180618105733-20180618125733-00612.warc.gz | 199,984,806 | 3,134 | # can anyone help
• Jun 25th 2008, 05:11 PM
Mr. red
can anyone help
Many calendars number the days of the year. January 1st gets number one, january 2nd gets number 2, january 31st gets 31, february 1st gets 32, december 31st gets 365 in a non-leap year, etc. mary ann noticed that on the northern hemisphere island where she lived there was 14 hours, 42 minutes of sunlight on the longest day of the year and 9 hours, 18 minutes on the shortest. created a sin function or cosine function that would allow someone to input the day number and get the hours of sunlight. explain your reasoning for choosing the parameters that you have. (Headbang)
• Jun 26th 2008, 01:31 AM
thelostchild
NB this is my way of approaching the problem, it may not necessarily be correct so dont take my answer to be perfect :)
ok we need a sine function the most general form of this possible is
$\displaystyle y=A \sin(\omega x + \alpha) + B$
first of all it would need a period of 365 days (assuming we havn't got a leap year) therefore as
$\displaystyle \omega = \frac{2 \pi}{T}=\frac{2 \pi}{365}$
looking it up on wikipedia the longest day of the year is on 21st June, using your numbering system this would be day 172
also the shortest day is on 21st December, so that would be day 355.
we know the function must be a maximum on day 172, so therefore
$\displaystyle \sin(\frac{2 \pi}{365} \times 172 + \alpha)=1$
solving this for alpha gives us $\displaystyle \alpha = -\frac{323}{730} \pi$
ok were half way there!
the longest day of the year is 14 hours 42 mins = 14.7 hours
therefore when the function is a maximum y=14.7
sine has is max value of 1 so
$\displaystyle 14.7=A+B$
the shortest day of the year is 9 hours 18 mins=9.3 hours
sine has a min value of -1 so
$\displaystyle 9.3=-A+B$
solving the two equations gives us A=2.7 and B=12
so the final equation is
$\displaystyle y=2.7 \sin (\frac{2 \pi}{365}x-\frac{323 \pi}{730})+12$ | 567 | 1,937 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-26 | latest | en | 0.89416 |
https://www.etcnbusiness.com/ebooks/3-restricted-connectivity-of-graphs-with-given-girth | 1,555,841,946,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530527.11/warc/CC-MAIN-20190421100217-20190421122217-00140.warc.gz | 673,454,340 | 9,673 | By Guo L.-T.
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Additional resources for 3-restricted connectivity of graphs with given girth
Sample text
1(iii) and by induction, we may assume that D has no cut vertices and that n ~ 3. H every directed cycle of D has length two, then since D is minimally directed and by n ~ 3, at least one of the two vertices in a directed cycle of length two is a cut vertex of D. Therefore, D must have a directed cycle C of length m ~ 3. We may assume that D =F C, and son- m ~ 1. Thus D/C has n- m + 1 ~ 2 vertices, and so by induction, D I a has at least two cyclic vertices, 11t and 112 (say). H both 11t, V2 E v (D) - V( C)' then they are both cyclic vertices of D.
Then f(n,e) = k -1. 12) Proof Let A= (ai;) E ~·(n,e). 2, it suffices to show that p(A) ~ k-1, and that p(A) = k -1 if and only if A is similar to the matrix in the theorem. Since A E ~*(n,e), we may assume that A= ( z ~t), where r < k -1 and where all the entries in the first column of A 1 are 1's. Since ~ is symmetric, there is an orthonormal matrix U such that UT ~U is a diagonalized matrix. Let V be the direct sum of U and In-ro and let R be the r by r matrix whose (1, 1) entry is an rand whose other entries are zero.
Hence each A; has equal row sums, so p(A) is the maximum row sum of A. j1 + 8e is an integer, and e =(k 2 ). Then p(A) = k -1, and it follows that there is one nonzero block At=~· This completes the proof. O = Let M (m;J) EM,. and let r; and 8; be the ith row sum and the ith column sum of M, respectively, for each i 1,2, · · · , n. n. 5 (Gregory, Shen and Liu [100]) Let M (m;J) E M,. 14) My=pxo Let x = (x 1 . )T and y = (y1 ° • oy; o. 21, we = E 1s;s,. 14) and Cauchy-Schwarz inequality, Matrices and Graphs for each i 31 = 1,2, · · · ,n, p2x~ (E m;m)2 = ( E = i E ~ m;m)2 J:mwFO E yJ m~; j:m;s¢0 j:m;s¢0 = r1(1- :E yJ). | 933 | 3,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-18 | latest | en | 0.957287 |
http://chemistry.about.com/od/workedchemistryproblems/a/Electrochemical-Cell-Emf-Example-Problem.htm | 1,408,651,223,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500821289.49/warc/CC-MAIN-20140820021341-00183-ip-10-180-136-8.ec2.internal.warc.gz | 38,361,520 | 14,314 | • Share
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# Electrochemical Cell EMF Example Problem
## How to Calculate Cell EMF for an Electrochemical Cell
EMF is the net voltage of half-reactions in an electrochemical cell.
Maria Toutoudaki, Getty Images
The cell electromotive force, or cell EMF is the net voltage between the oxidation and reduction half-reactions taking place between two redox half-reactions. Cell EMF is used to determine whether or not the cell is galvanic. This example problem shows how to calculate the cell EMF using standard reduction potentials.
The Table of Standard Reduction Potentials is needed for this example.
Problem:
Consider the redox reaction:
Mg(s) + 2 H+(aq) → Mg2+(aq) + H2(g)
a) Calculate the cell EMF for the reaction.
b) Identify if the reaction is galvanic.
Solution:
Step 1: Break the redox reaction into reduction and oxidation half-reactions.
Hydrogen ions, H+ gain electrons when forming hydrogen gas, H2. The hydrogen atoms are reduced by the half-reaction:
2 H+ + 2 e- → H2
Magnesium loses two electrons and is oxidized by the half-reaction:
Mg → Mg2+ + 2 e-
Step 2: Find the standard reduction potentials for the half-reactions.
Reduction: E0 = 0.0000 V
The table shows reduction half-reactions and standard reduction potentials. To find E0 for an oxidation reaction, reverse the reaction.
Reversed reaction:
Mg2+ + 2 e- → Mg
This reaction has a E0 = -2.372 V.
E0Oxidation = - E0Reduction
E0Oxidation = - (-2.372 V) = + 2.372 V
Step 3: Add the two E0 together to find the total cell EMF, E0cell
E0cell = E0reduction + E0oxidation
E0cell = 0.0000 V + 2.372 V = +2.372 V
Step 4: Determine if reaction is galvanic. Redox reactions with a positive E0cell value are galvanic.
This reaction's E0cell is positive and therefore galvanic.
The cell EMF of the reaction is +2.372 Volts and is galvanic.
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# Stereo Map Disparity Values
I successfully generated my depth map and printed out some x,y coordinates and its corresponding disparity. Through disparity, we could integrate distance measurement. The problem is, how can I have static disparity values? when I output them, its disparity value fluctuates? Is this normal or i would have some adjustments?
Btw, i am using 2rgb cameras in a live feed
So i click a specific point in the map, (the red circles are the disparity values in an x,y coordinate) As you can see the 4th & 5th circle show different disparity values. It fluctuated from the first 3
str(disp[y,x]), str(filteredImg[y,x] y x coordinate y x in the disparity map
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Not that much information are provided, so I try to be exhaustive as much as I can.
Starting from the beginning, you get the disparity map from the left and right images. This disparity map is static, that means that if you run the disparity map computation over and over with the same L and R images and the same algorithm you'll have always the same disparity map (if you don't change the left and right images, of course!).
What you mean for output, you display the disparity map or in terms of values?
In the first case the disparity map must look always the same if you use the same images and the same algorithm each execution. In that case, if you encounter changes, I'll probably check if there is some randomized things inside your algorithm.
If you mean printing value maybe you're just getting the value in the bad way. Check the matrix type and verify that you access it with the correct data type or you'll get wrong values. For instance, a CV_64F can't be printed as a CV_8U matrix.
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## Comments
Thankyou for the effort & sorry for giving less information. So i am not using still images, my input are from live video feed from two logitech cameras.
In my observation my rectification was complete okay. No distortion at all so i am posting the code where the disparity map was generated. Note: dispL & dispR are the output from stereo.compute and grayL & grayR are the output from cv2.remap
# Using the WLS filter
filteredImg= wls_filter.filter(dispL,grayL,None,dispR)
filteredImg = cv2.normalize(src=filteredImg, dst=filteredImg, beta=1, alpha=255,
norm_type=cv2.NORM_MINMAX, dtype=cv2.CV_32F);
filteredImg = np.uint8(filteredImg)
cv2.imshow('Disparity',filteredImg)
Moreover...
( 2019-02-19 10:50:21 -0500 )edit
Through a def function using mouse click button i can output the x,y coordinate & its disparity values from filteredImg.
def coords_mouse_disp(event,x,y,flags,param): if event == cv2.EVENT_LBUTTONDBLCLK: print (str(x),str(y),str(disp[y,x]),str(filteredImg[y,x]))
Then from python shell, filteredImg[y,x] or the disparity values in a given point. Is not static. (please see attached image above, ill edit my post) thankyou @HYPEREGO
( 2019-02-19 10:56:15 -0500 )edit
The algorithms for stereo match and disparity are quite sensitive to frame-by-frame or camera-to-camera illumination changes and imager noise, especially if there are over/underexposure areas. This may cause unstable ripples, waves, and artifacts along edges. Controlling illumination and using L+R images collected under identical illumination (at same time) may help. Pulising illumination, unsynchronized L+R exposure start, glare/reflection/transparency, rolling shutters can contribute to this.
Also, self-similar pattern matching (especially when the window size is too small) can result in all-out confusion (ex: stereo match disparity of a checkerboard or dot pattern, or low contrast areas of the image - a random projector pattern can help here if contrast is adequate).
( 2019-02-19 14:04:17 -0500 )edit
Oh so the light exposure in the environment or the L+R images from my calibration can be the factor? Can it be adjusted through WLS filter or StereoSGBM parameters?
( 2019-02-19 18:35:16 -0500 )edit
2
Since you're using live stream video, it is ok that in a certain point the disparity value change, due to what opalmirror have underlined; algorithm can have some uncertainty, especially SGBM and similar.
Let's try this: run the algorithm using two static images (try Tsukuba on Middlebury dataset, for example) and see if there are changes on values, so you can validate your result and eventually think about how to fix it; since there are only a small difference between the disparity values, maybe just a post-processing will work.
( 2019-02-20 04:33:22 -0500 )edit
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Asked: 2019-02-19 07:28:55 -0500
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1. Jul 19, 2009
bobo105
I want to know the relationship between the resistive forces of a gyroscopic exercise tool and the revolutions per minute of the rotor.
2. Jul 19, 2009
rcgldr
It generates from 32lbs to 35 lbs of torque when speeds of 9,000 - 13,000 RPMs are reached. I assume this means 32 ft-lbs to 35 ft lbs of torque, however the torque is zero unless you try to induce a torque, so I assume this is peak torque.
http://www.basegear.com/powerballgyro.html [Broken]
these powerballs are made primarily of metal and have twice the weight of "regular" Powerball models and are more challenging to operate since up to 250Nm of torque is produced (250Nm = 184.4 ft lb).
http://en.wikipedia.org/wiki/Powerball_(exercise_tool)
The interface between the narrow track and the ends of the gyro axis would wear out (loss of friction), resulting in an unusable powerball after a relatively short period of time. I don't know if this issue was ever fixed.
Last edited by a moderator: May 4, 2017
3. Jul 20, 2009
bobo105
Thanks, but is there a formula that links the revs/min of the rotor of the gyroscope to the torque?
$$T = I \alpha$$
$$T = \frac{mr^2}{2} \times \frac{\omega}{t}$$
$$\omega = \frac{2 \pi R}{60}$$
$$T = \frac{mr^2 \pi R}{60t}$$
where R = revs/min, r = radius, m = mass, ω = angular velocity, t = time
So far I have found that a torque of 0.25Nm is required to rotate the gyroscope to 17000 revs/min in 1 second (with a mass of 0.45kg and a radius of 0.025m). However, I want to know, for example, how much torque is required to rotate the gyroscope at a constant 15000 revs/min.
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FINC 600 Week 3 Practice Quiz
\$12.99
FINC 600 Week 3 Practice Quiz Question 1 Which of the following portfolios have the least risk? Question 2 If the average annual rate of return for common stocks is 11.7%, and for treasury bills it is 4.0%, what is the market risk premium? Question 3 Spill Oil Company’s stocks had -8%, 11% and 24% rates of return during the last three years respectively; calculate the average rate of return for the stock. Question 4 Given the following data: risk-free rate = 4%, average risk premium = 7.7%. Calculate the required rate of return: Question 5 The unique risk is also called the: Question 6 Market risk is also called: I) systematic risk, II) undiversifiable risk, III) firm specific risk. Question 7 Stock A has an expected return of 10% per year and stock B has an expected return of 20%. If 40% of the funds are invested in stock A, and the rest in stock B, what is the expected return on the portfolio of stock A and stock B? Question 8 If the correlation coefficient between stock C and stock D is +1.0% and the standard deviation of return for stock C is 15% and that for stock D is 30%, calculate the covariance between stock C and stock D. Question 9 The beta of market portfolio is: Question 10 The distribution of returns, measured over a short interval of time, like daily returns, can be approximated by: Question 11 Florida Company (FC) and Minnesota Company (MC) are both service companies. Their historical return for the past three years are: FC: – 5%,15%, 20%; MC: 8%, 8%, 20%. If FC and MC are combined in a portfolio with 50% of the funds invested in each, calculate the expected return on the portfolio. Question 12 Suppose you invest equal amounts in a portfolio with an expected return of 16% and a standard deviation of returns of 20% and a risk-free asset with an interest rate of 4%; calculate the expected return on the resulting portfolio: Question 13 Suppose you invest equal amounts in a portfolio with an expected return of 16% and a standard deviation of returns of 20% and a risk-free asset with an interest rate of 4%; calculate the standard deviation of the returns on the resulting portfolio: Question 14 The correlation measures the: Question 15 The security market line (SML) is the graph of:
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## Presentation on theme: "Chapter 8 Risk and Return. Topics Covered Markowitz Portfolio Theory Risk and Return Relationship Testing the CAPM CAPM Alternatives."— Presentation transcript:
Chapter 8 Risk and Return
Topics Covered Markowitz Portfolio Theory Risk and Return Relationship Testing the CAPM CAPM Alternatives
Markowitz Portfolio Theory Combining stocks into portfolios can reduce standard deviation, below the level obtained from a simple weighted average calculation. Correlation coefficients make this possible. efficient portfolios The various weighted combinations of stocks that create this standard deviations constitute the set of efficient portfolios.
Markowitz Portfolio Theory Price changes vs. Normal distribution Microsoft - Daily % change 1990-2001 Proportion of Days Daily % Change
Markowitz Portfolio Theory Standard Deviation VS. Expected Return Investment A % probability % return
Markowitz Portfolio Theory Standard Deviation VS. Expected Return Investment B % probability % return
Markowitz Portfolio Theory Standard Deviation VS. Expected Return Investment C % probability % return
Markowitz Portfolio Theory Standard Deviation VS. Expected Return Investment D % probability % return
Markowitz Portfolio Theory Coca Cola Reebok Standard Deviation Expected Return (%) 35% in Reebok Expected Returns and Standard Deviations vary given different weighted combinations of the stocks
Efficient Frontier Standard Deviation Expected Return (%) Each half egg shell represents the possible weighted combinations for two stocks. The composite of all stock sets constitutes the efficient frontier
Efficient Frontier Standard Deviation Expected Return (%) Lending or Borrowing at the risk free rate ( r f ) allows us to exist outside the efficient frontier. rfrf Lending Borrowing T S
Efficient Frontier Example Correlation Coefficient =.4 Stocks % of PortfolioAvg Return ABC Corp2860% 15% Big Corp42 40% 21% Standard Deviation = weighted avg = 33.6 Standard Deviation = Portfolio = 28.1 Return = weighted avg = Portfolio = 17.4%
Efficient Frontier Example Correlation Coefficient =.4 Stocks % of PortfolioAvg Return ABC Corp2860% 15% Big Corp42 40% 21% Standard Deviation = weighted avg = 33.6 Standard Deviation = Portfolio = 28.1 Return = weighted avg = Portfolio = 17.4% Let’s Add stock New Corp to the portfolio
Efficient Frontier Example Correlation Coefficient =.3 Stocks % of PortfolioAvg Return Portfolio28.150% 17.4% New Corp30 50% 19% NEW Standard Deviation = weighted avg = 31.80 NEW Standard Deviation = Portfolio = 23.43 NEW Return = weighted avg = Portfolio = 18.20%
Efficient Frontier Example Correlation Coefficient =.3 Stocks % of PortfolioAvg Return Portfolio28.150% 17.4% New Corp30 50% 19% NEW Standard Deviation = weighted avg = 31.80 NEW Standard Deviation = Portfolio = 23.43 NEW Return = weighted avg = Portfolio = 18.20% NOTE: Higher return & Lower risk How did we do that? DIVERSIFICATION
Efficient Frontier A B Return Risk (measured as )
Efficient Frontier A B Return Risk AB
Efficient Frontier A B N Return Risk AB
Efficient Frontier A B N Return Risk AB ABN
Efficient Frontier A B N Return Risk AB Goal is to move up and left. WHY? ABN
Efficient Frontier Return Risk Low Risk High Return High Risk High Return Low Risk Low Return High Risk Low Return
Efficient Frontier Return Risk Low Risk High Return High Risk High Return Low Risk Low Return High Risk Low Return
Efficient Frontier Return Risk A B N AB ABN
Security Market Line Return Risk. rfrf Risk Free Return = Efficient Portfolio Market Return = r m
Security Market Line Return. rfrf Risk Free Return = Efficient Portfolio Market Return = r m BETA1.0
Security Market Line Return. rfrf Risk Free Return = BETA Security Market Line (SML)
Security Market Line Return BETA rfrf 1.0 SML SML Equation = r f + B ( r m - r f )
Capital Asset Pricing Model R = r f + B ( r m - r f ) CAPM
Testing the CAPM Avg Risk Premium 1931-65 Portfolio Beta 1.0 SML 30 20 10 0 Investors Market Portfolio Beta vs. Average Risk Premium
Testing the CAPM Avg Risk Premium 1966-91 Portfolio Beta 1.0 SML 30 20 10 0 Investors Market Portfolio Beta vs. Average Risk Premium
Testing the CAPM High-minus low book-to- market Return vs. Book-to-Market Dollars Low minus big http://mba.tuck.dartmouth.edu/pages/faculty/ken.french/data_library.html
Consumption Betas vs Market Betas Stocks (and other risky assets) Wealth = market portfolio Market risk makes wealth uncertain. Stocks (and other risky assets) Consumption Wealth Wealth is uncertain Consumption is uncertain Standard CAPM Consumption CAPM
Arbitrage Pricing Theory Alternative to CAPM Expected Risk Premium = r - r f = B factor1 (r factor1 - r f ) + B f2 (r f2 - r f ) + … Return= a + b factor1 (r factor1 ) + b f2 (r f2 ) + …
Arbitrage Pricing Theory Estimated risk premiums for taking on risk factors (1978-1990)
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# AC CIRCUIT DIAGRAM FOR MOUSE TRAPPER
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A measure of the degree of linear association between a pair of random variables.
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### Is a more robust Covariance estimation possible?
I'm working on a mean-variance optimization problem, but instead of financial securities I'm choosing a 'portfolio' of N athletes. It is a 1-period optimization problem over one generic statistic ... | 1,880 | 8,687 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-05 | latest | en | 0.925363 |
https://socratic.org/questions/how-do-you-find-area-of-sector-if-angle-is-0-4-rad-and-adjacent-angle-radius-is- | 1,685,744,027,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648858.14/warc/CC-MAIN-20230602204755-20230602234755-00403.warc.gz | 590,048,120 | 5,676 | # How do you find area of sector if angle is 0.4 rad and adjacent angle/radius is 10?
Sep 7, 2017
$20 {\text{ units}}^{2}$
#### Explanation:
Area of sector is
${A}_{s} = \frac{1}{2} {r}^{2} \theta$
$r = \text{the radius}$
$\theta = \text{the angle in radians}$
${A}_{s} = \frac{1}{2} \times {10}^{2} \times 0.4$
${A}_{s} = \frac{1}{2} \times 100 \times 0.4 = \frac{1}{2} \times 40 = 20 {\text{units}}^{2}$ | 170 | 413 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-23 | latest | en | 0.595524 |
http://sportstatist.com/understanding-and-managing-your-risks-of-ruin/ | 1,555,766,638,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529813.23/warc/CC-MAIN-20190420120902-20190420142902-00193.warc.gz | 167,571,435 | 25,497 | # Understanding and managing your risks of ruin
The risk of ruin concept is well-known in finance and investing but how can it be applied to sports betting? Why should even profitable bettors be aware of the concept? Read on to find out.
In his book Skin in the Game, Nassim Nicholas Taleb presents the following thought experiment, which I have adapted very slightly.
100 persons go to a Casino, to gamble a certain set amount each. Some may lose, some may win, and we can infer at the end of the day what the “edge” is, that is, calculate the returns simply by counting the money left with the people who return. We can thus figure out if the casino is properly pricing the odds. Now assume that gambler number 28 goes bust. Will gambler number 29 be affected? No.
You can safely calculate, from your sample, that about 1% of the gamblers will go bust. And if you keep playing and playing, you will be expected to have about the same ratio, 1% of gamblers over that time window.
Now instead suppose only you go to the Casino a hundred days in a row, starting with a set amount. On day 28 you go bust. Will there be day 29? No. No matter how lucky you are, you can safely calculate that you have a 100% probability of eventually going bust.
The probabilities of success from the collection of people do not apply you. Let us call the first set ensemble probability and the second one time probability (since one is concerned with a collection of people and the other with a single person through time). Now, when you read blogs by betting analysts or tipsters based on the long term expected returns, beware. Even if their forecasts were true, no person can achieve the expected returns of the market unless he has infinite pockets. They are conflating ensemble probability and time probability. If a bettor has to eventually reduce his staking because of losses, his returns will be divorced from those theoretically expected, period.
Of course it’s a statement of the obvious that casino gamblers who play games designed purely around statistical algorithms (roulette and craps for example) will inevitably go bust in the end. The same will be true for unskilled bettors who will eventually be ground down by the bookmaker’s margin.
Betting, however, is different to the casino because there is the theoretical possibility for players to show profitable expectation, provided they are better able to determine ‘true’ outcome probabilities than those who lay the odds.
Nevertheless, Taleb’s thought experiment is useful in this respect: it reminds even advantage players to think about the possibility of ruin. You might be a superb forecaster, but if an unlucky sequence of play wipes out your bankroll, that’s it. There will be no next day.
In the remainder of this article, therefore, I want to spend a little time examining the risks of ruin faced by advantage bettors.
Betting is a combination of luck and skill. How do we know when a bettor shows any of the latter? Last year I discussed the use of a statistical test – the Student’s t-test – that might help us answer this question. By its nature this test can’t actually tell us directly whether a bettor is skilled at forecasting and achieving a long term profitable expectation.
All it can do is calculate the likelihood of a set of profits and losses occurring if nothing other than chance is operating. When that likelihood is small, however, statisticians will happily make a leap of faith and assume that something other than chance is probably at work. Typical benchmark figures applied in this context include 5% and 1%. In other words, when there is a less than 1% probability that a betting record will have arisen by chance we might argue that it’s likely the bettor responsible for it is exhibiting some skill.
There are numerous problems with making such a bold prediction, not least the issue of survivorship bias: we often only get to see the best records, not all the other mediocre or losing ones. If there are 100 bettors in a population and the best one has a 1-in-100 profits history, what is that really telling us? Unfortunately, we rarely get to see the whole population.
Nevertheless, for this article I will assume that a 1-in-100 record provides some evidence of an advantage bettor. Whether it really does is perhaps not so relevant to what follows. Since lucky bettors will ultimately regress to the mean, the following data might be considered best case scenarios.
### The probability of ruin
What is the probability a bettor will face ruin during a series of bets? This depends on a number of variables, including how skilled (or lucky) they are, how long they bet for, the odds they choose to bet at and the stakes they choose to place. It is self-evident that the more superior (less unlucky?) the bettor, the less likely they are to lose all their bankroll.
We should also be familiar with the fact that the longer the odds we bet, the greater the variance in outcomes. Greater variance means a wider range of possible profits or losses, greater risks and hence a higher probability of ruin, all other things being equal. And the larger our stakes are as a proportion of our bankroll, unsurprisingly the greater the likelihood that an unlucky streak will wipe us out.
Bettors who bet longer odds will also typically show larger percentage returns than those betting at shorter odds because of the greater variances. To put it another way, an equivalent return from longer odds has a greater contribution from luck. It is for this reason that you will find racing tipsters (who typically bet longer odds) with bigger published returns than sports betting tipsters (who typically bet shorter odds).
Of course, for the same reason you will find racing tipsters at the bottom of tipster league tables with larger losses (because of the influence of bad luck). The following table shows the profit expectations for profitable 1-in-100 bettors placing 1,000 wagers for different odds, calculated using my t-test calculator.
Odds Expected Return on Investment 1.25 103.48% 1.5 105.06% 2 107.35% 3 110.67% 5 115.53% 10 124.31%
From a 10,000-run Monte Carlo simulation the following chart illustrates how the risk of ruin varies for different 1-in-100 bettors placing up to 1,000 wagers with varying betting odds and stake sizes. It is assumed the bettor starts with a bankroll of 100 and stake sizes are fixed.
Unsurprisingly, larger stakes increase the risk of ruin dramatically, especially when betting at longer odds. Let’s say you have a preference for betting on racing longshots at prices of ten. Despite holding a theoretical profit expectation of over 24%, you’ll never even make it to 1,000 wagers on more than 60% of occasions if you bet ten units stakes from a starting bankroll of 100. Naturally, the majority of longshot bettors will choose to appropriately scale down the size of their stakes, but that necessarily implies they will make less absolute profit.
### The impact of managing ruin-risk on absolute profits.
Let’s suppose we consider a 1% risk of ruin to be the maximum acceptable. For a bettor betting odds of ten, their stakes can be no more than one unit for a 100 starting bankroll. By contrast, for the bettor betting much shorter odds of 1.25, they can risk stakes of six units. Consequently, despite the longshot bettor having a much higher profit expectation, they will actually end up with about the same absolute profit as the bettor who bets on strong favourites.
Another example: suppose the acceptable ruin-risk is about 20%. A bettor betting odds of five should stake about five units. In contrast, the bettor placing bets at odds of 1.5 can stake about 15 units. The former has a profit expectation about three times the latter. Hence, again they’ll both end up with about the same absolute profit expectation.
There is a clear and on obvious conclusion here: for bettors with equivalent skill levels, accepting equivalent risks of ruin implies that it makes very little different what odds you bet at. Targeting longer odds because they offer more percentage profit expectation must be counter balanced by managing an increased risk of ruin by reducing stake sizes.
### The influence of ruin on profit expectancy
Taleb followed his thought experiment on ensemble versus time probability by quoting Warren Buffet:
“In order to succeed, you must first survive.”
He followed it with his own interpretation:
“The presence of ruin does not allow cost-benefit analyses.”
Buffet and Taleb are right. Ruin changes the expectancy calculation. When we attempt to estimate expected profitability, we do so by ignoring all those occasions where our betting sequence is cut short. The estimated profit expectancies in the table above all assume 1,000 wagers. Evidently, when the probability of ruin is non-zero not all sequences will last that long.
Consider a bettor placing wagers at odds of two at stakes of 10 units. They face a 23% risk of ruin within 1,000 wagers. In my 10,000 betting history simulations, 95% of those sequences that experienced ruin theoretically went on to show a profit with an average return on investment of 105.74%. But in reality, those profits would never have happened, because there is no next day for the ruined bettor.
If we want to keep real expectations as close as possible to their theoretical counterparts, we must minimise the risk of ruin to very small levels. Intuitively, most serious bettors understand this, but the analysis here will hopefully quantify these ideas.
What is very clear is that for even skilled advantage players, to hold the probability of ruin to levels significantly below 1% we cannot entertain stake sizes much beyond 1% of an initial bankroll, and smaller still if betting at longer odds. And when we actively think about our risk of ruin it becomes clear that the type of odds we might choose to wager makes very little difference at all.
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Source: pinnacle.com | 2,127 | 10,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-18 | longest | en | 0.958259 |
http://www.coursehero.com/file/7098221/ECE-201-Exam-2-Solving-Techniques/ | 1,411,011,583,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657125488.38/warc/CC-MAIN-20140914011205-00072-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 444,565,766 | 16,659 | This preview has intentionally blurred parts. Sign up to view the full document
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Thev & Nort Solving Techniques For any Linear Resistive Network there exists: Thevenin Equiv. Network Norton Equivalent Network V OC- open-circuit voltage appearing across the terminals of the network, and R TH- Thevenin equivalent resistance when all independent sources are deactivated. I SC current through load when replacing load with a short circuit Equations Solve for R TH 1. Deactivate all ind. srcs (leave dep. srcs) 2. Remove the load R L in the circuit and look at resistance seen through nodes A & B *In circuits w. dependent sources, may need to apply a voltage or current source between A & B, then apply the definition R TH = V OC / I SC Finding V TH 1. Remove load R L in original circuit & relate voltage V AB to rest of network. a. Sources remain unchanged. b. Use methods such as superposition, mesh, nodal, V-Division, etc Finding I SC 1. Replace load R L in original circuit with a short circuit. a. Sources remain unchanged. Coefficient Approach V-I Char. Of Thev: V AB = R TH I A + V OC V-I Char. Of Nort: I A = (1/R TH )V AB- I SC 1. Obtain an equation in one of the above forms allows us to match the coefficients of the above equations to determine R TH , V OC , I A , or G TH . Measured Data 1. Substitute the given data into V-I Char. Eqs: Thev: V AB = R TH I A + V OC or Nort: I A = (1/R TH )V AB- I SC 2. Put into matrix form 3. Solve for remaining variables. General Solving Procedures 1. Have ways to find V OC , I SC , R TH directly using above techniques 2. After solving for 2 of the 3, below are listed ways to solve for 3 rd Thev and Nort equiv w/ Dep. Srcs & w/out Ind. Srcs *Extra condition for constructing Thev and Nort equivalents for active networks all controlling voltages or currents must be within the 2-terminal network whose Thev/Nort equiv are being sought. 1. Since there are no ind. internal srcs, the Thev equiv. consists of a single resistance, R TH (V OC = I SC = 0) 2. Write an equation(s) to relate the terms in the 2-terminal network via previous solving techniques (KCL, KVL, Mesh, Nodal, etc) 3. If one equation, match coeficients with the V-I char. equations of a Thev. Or Nort. Equiv. network. 4. If set of equations, solve using matrix operations. Maximum Power Transfer Theorems, Requirements, Conditions, Assumptions, and Definitions 1. Always put circuit in Thevenin or Norton Equivalent form 2. Fixed R TH Equations Always True Under M.P.T.... View Full Document
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Sign up now to access the rest of the document | 672 | 2,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2014-41 | latest | en | 0.848079 |
https://mm-to-inches.net/centimetres-to-inches/ | 1,675,783,771,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500619.96/warc/CC-MAIN-20230207134453-20230207164453-00784.warc.gz | 410,355,776 | 15,572 | # Centimetres to Inches
We measure everything we use. Like the grocery, clothes, land, medicine we purchase. But, here, we discuss measuring length or height with the help of two units: Centimetres to Inches. What do we measure? we measure the weight, size, length, distance. Some measurements are smaller while others are bigger. Centimetres and Inches are the two crucial units we always use to measure the height or length of various objects.
Inches
## Important points for Converting Centimetres to Inches
1. we use various tools to measure the length of objects like measuring tapes, rulers etc.
2. The unit Centimetre is a bigger unit than Inches.
3. The unit Inches is a smaller unit than Centimetre.
## What is Centimetre?
A Centimetre is one of the units of length. A centimetre is used to measure the length of the objects. The Centimetre is the base unit of the length, and the abbreviation used to write Centimetre is ‘cm’. One Centimetre is equal to 0.393701 inches, and one Metre is equal to a hundred Centimetre. The unit Centimetre is smaller than Metre but a bigger unit if we compare it to inches. It is being asked How long is 27 CM in Inches? So, the answer to this question is 0.393701 inches.
## What is Inch?
An Inch is one of the smaller units of length. An Inch is used to measure the length of the various objects. The Inch is the important unit of length in the United Customary Units and in the British Imperial System of Units. And the abbreviation used to write Centimetre is ‘cm’. One Centimetre is equal to 0.393701 inches, and one Metre is equivalent to a hundred Centimetre. The unit Centimetre is smaller than Metre but a bigger unit if we compare it to inches. It is being asked How to convert Hexadecimal to Decimal and How long is 27 CM in Inches? So, the answer to this question is 0.393701 inches.
#### Common Conversion of Centimetres to Inches
Centimetres Exact size to Inches
1 Cm 0.393701 in
2 Cm 0.787402 in
3 Cm 1.181103 in
4 Cm 1.574804 in
5 Cm 1.9685050 in
6 Cm 2.362206 in
7 Cm 2.755907 in
8 Cm 3.149608 in
9 Cm 3.5433090 in
10 Cm 3.937010 in
11 Cm 4.330711 in
12 Cm 4.724412 in
13 Cm 5.118113 in
14 Cm 5.511814 in
15 Cm 5.905515 in
16 Cm 6.299216 in
17 Cm 6.6929170 in
18 Cm 7.0866180 in
19 Cm 7.480319 in
20 Cm 7.874020 in
21 Cm 8.267721 in
22 Cm 8.661422 in
23 Cm 9.055123 in
24 Cm 9.448824in
25 Cm 9.842525 in
Centimetres Exact size to Inches
26 cm 10.236226 in
27 cm 10.629927 in
28 cm 11.023628 in
29 cm 11.417329 in
30 cm 11.81103 in
31 cm 12.204731 in
32 cm 12.598432 in
33 cm 12.992133 in
34 cm 13.385834 in
35 cm 13.779535 in
36 cm 14.173236 in
37 cm 14.5669370 in
38 cm 14.9606380 in
39 cm 15.354339 in
40 cm 15.748040 in
41 cm 16.141741 in
42 cm 16.535442 in
43 cm 16.929143 in
44 cm 17.322844 in
45 cm 17.716545 in
46 cm 18.110246 in
47 cm 18.503947 in
48 cm 18.897648 in
49 cm 19.291349 in
50 cm 19.68505 in
Centimetres Exact size to Inches
51 cm 20.078751 in
52 cm 20.472452 in
53 cm 20.866153 in
54 cm 21.259854 in
55 cm 21.653555 in
56 cm 22.047256 in
57 cm 22.440957 in
58 cm 22.834658 in
59 cm 23.228359 in
60 cm 23.62206 in
61 cm 24.015761 in
62 cm 24.409462 in
63 cm 24.803163 in
64 cm 25.196864 in
65 cm 25.590565 in
66 cm 25.984266 in
67 cm 26.377967 in
68 cm 26.7716680 in
69 cm 27.1653690 in
70 cm 27.559070 in
71 cm 27.9527710 in
72 cm 28.3464720 in
73 cm 28.7401730 in
74 cm 29.1338740 in
75 cm 29.5275750 in
Centimetres Exact size to Inches
76 cm 29.9212760 in
77 cm 30.3149770 in
78 cm 30.7086780 in
79 cm 31.1023790 in
80 cm 31.496080 in
81 cm 31.8897810 in
82 cm 32.283482 in
83 cm 32.677183 in
84 cm 33.070884 in
85 cm 33.464585 in
86 cm 33.858286 in
87 cm 34.251987 in
88 cm 34.645688 in
89 cm 35.039389 in
90 cm 35.43309 in
91 cm 35.826791 in
92 cm 36.220492 in
93 cm 36.614193 in
94 cm 37.007894 in
95 cm 37.401595 in
96 cm 37.795296 in
97 cm 38.188997 in
98 cm 38.582698 in
99 cm 38.976399 in
100 cm 39.3701 in
## Convert Centimetres to Inch
Converting the two units, from Centimetres to Inch, is very easy. We need to follow a few easy steps. We all know that one Centimetre is equal to 0.393701 inches. Every time it is asked, How tall is 27 CM in Inches? or How long is 27 CM in Inches? The answer is 10.6299 inches. But we will do more examples to learn much better. So, let’s begin with the learning process.
1. Convert 40 cm to the Inch?
= 40 X 0.393701 = 15.748031496
= 15.74804 inches
2. Convert 105 cm to the Inch?
= 105 X 0.393701 = 41.338605
= 41.338 inches
3. Convert 20.5 cm to the Inch?
= 20.5 X 0.393701 = 8.0708
= 8.0708 inches
What is 27cm to the Inch?
10.62992126 inches
How many inches is 27 cm?
10.62992126 inches
How to Convert 27cm in inches?
= 27 X 0.393701 = 10.6299
10.6299 inches
27 centimetres to inches
10.62992126 inches
What size is 27 CM to inches?
10.62992126 inches
How tall is 27 CM in Inches?
10.62992126 inches
How big is 27 CM?
10.62992126 inches
## Conclusion
I hope this calculator for converting unit Centimetres to Inches helps get the solution’s easy result. Keep practising and keep learning. “Happy Learning To You”. | 1,799 | 5,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-06 | latest | en | 0.905103 |
https://math.stackexchange.com/questions/3198683/sum-of-product-of-stirling-numbers | 1,560,682,124,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00181.warc.gz | 517,983,610 | 32,529 | # Sum of product of Stirling numbers
We have for $$n>0$$, $$k>0$$ $$\sum\limits_{j=1}^{\min(n,k)}(j!)^2{n\brace j}{k+1\brace j+1}=\sum\limits_{j=0}^{\min(n,k-1)}j!(j+1)!{n+1\brace j+1}{k\brace j+1}$$ How can we prove it?
• Do you have any context? – Phicar Apr 23 at 20:18
• For the second sum "Masanobu Kaneko. Multiple zeta values, poly-Bernoulli numbers, and related zeta functions. Nagoya Math Journal, 153:189-201, 1999". First one is mine and based on previous question. – user514787 Apr 23 at 20:21
Using the recurrence relation for the Stirling numbers of the second kind $$\begin{eqnarray*} {k+1\brace j+1}=(j+1){k\brace j+1}+{k\brace j}. \end{eqnarray*}$$ The sum becomes $$\begin{eqnarray*} S&=&\sum\limits_{j=1}^{\min(n,k)}(j!)^2{n\brace j}{k+1\brace j+1} \\ &=& \sum\limits_{j=1}^{\min(n,k)}(j!)((j+1)!){n\brace j}{k\brace j+1} +\sum\limits_{j=1}^{\min(n,k)}(j!)^2{n\brace j}{k\brace j} \end{eqnarray*}$$ shift the second summation variable by $$1$$ $$\begin{eqnarray*} S&=& \sum\limits_{j=0}^{\min(n,k-1)}(j!)((j+1)!){k\brace j+1} \left({n\brace j}+(j+1){n\brace j+1}\right)\\ \end{eqnarray*}$$ Now use the recurrence formula again $$\begin{eqnarray*} {n+1\brace j+1}={n\brace j}+(j+1){n\brace j+1} \end{eqnarray*}$$ & we are done. | 530 | 1,248 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-26 | latest | en | 0.465431 |
http://www.solutioninn.com/a-city-in-ohio-is-considering-replacing-its-fleet-of | 1,508,300,903,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822739.16/warc/CC-MAIN-20171018032625-20171018052625-00266.warc.gz | 580,939,507 | 8,526 | A city in Ohio is considering replacing its fleet of
A city in Ohio is considering replacing its fleet of gasoline-powered automobiles with electric cars. The manufacturer of the electric cars claims that this municipality will experience significant cost savings over the life of the fleet if it chooses to pursue the conversion. If the manufacturer is correct, the city will save about \$1.5 million dollars. If the new technology employed within the electric cars is faulty, as some critics suggest, the conversion to electric cars will cost the city \$675,000. A third possibility is that less serious problems will arise and the city will break even with the conversion. A consultant hired by the city estimates that the probabilities of these three outcomes are 0.30, 0.30, and 0.40, respectively. The city has an opportunity to implement a pilot program that would indicate the potential cost or savings resulting from a switch to electric cars. The pilot program involves renting a small number of electric cars for three months and running them under typical conditions. This program would cost the city \$75,000.
The city’s consultant believes that the results of the pilot program would be significant but not conclusive; she submits the values in the file P06_72.xlsx, a compilation of probabilities based on the experience of other cities, to support her contention. For example, the first row of her table indicates that given that a conversion to electric cars actually results in a savings of \$1.5 million, the conditional probabilities that the pilot program will indicate that the city saves money, loses money, and breaks even are 0.6, 0.1, and 0.3, respectively. What actions should the city take to maximize its expected savings? When should it run the pilot program, if ever?
Membership | 361 | 1,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-43 | latest | en | 0.949489 |
https://www.scribd.com/document/29904023/Guided-Training-Exercise-Sample | 1,575,924,863,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540523790.58/warc/CC-MAIN-20191209201914-20191209225914-00309.warc.gz | 851,023,800 | 70,026 | You are on page 1of 23
# SAMCEF Guided Training Exercise 1
Objective:
This is a guided exercise to use SAMCEF Bacon command language to analyse the stresses
created by a force acting on a structure. The aim of this exercise is for the user to familiarise
themselves with the operation of SAMCEF.
## Document Reference : SAMCEF_GuidedTraining_Exec_01_2010
Author : Tony Fong
Date : 26th March 2010
Introduction
1 Introduction
This is a guided exercise to use SAMCEF Bacon command language to analyse the stresses created by
a force acting on a structure. The aim of this exercise is for the user to familiarise themselves with
the operation of SAMCEF.
2 Description
The structure is a 0.3m x 0.1m x 0.01m steel plate built-in to a wall with a point force acting on the
outer edge of the block as shown in the diagram below. A plane stress hypothesis is adopted, and a
linear analysis is performed.
## 3 Table of utilised commands
Pre-processing: .DEL .POIN .DROI .CONTOU .DOMAIN .GENE .HYP .MAT .SEL .AEL .PHP .CLM
Analyse: .FIN
Post-processing: .DOC .DES
Contents
1 Introduction .................................................................................................................................... 2
2 Description ...................................................................................................................................... 2
3 Table of utilised commands ............................................................................................................ 2
4 Procedure........................................................................................................................................ 4
4.1 Starting SAMCEF ..................................................................................................................... 4
4.2 Pre-processing......................................................................................................................... 6
4.2.1 Creating the model ......................................................................................................... 6
4.2.2 Pre-mesh ....................................................................................................................... 10
4.2.3 Meshing......................................................................................................................... 12
4.2.4 Hypothesis and Material ............................................................................................... 13
4.2.5 Model structure property ............................................................................................. 14
4.2.6 Apply Boundary conditions and load ............................................................................ 17
4.3 Analyse .................................................................................................................................. 19
4.3.1 Execute Solver ............................................................................................................... 19
4.4 Post-Processing ..................................................................................................................... 20
4.4.2 Display result ................................................................................................................. 20
4.5 Quitting SAMCEF ................................................................................................................... 21
5 Command lines ............................................................................................................................. 22
6 Summary of the commands used ................................................................................................. 23
Starting SAMCEF
4 Procedure
4.1 Starting SAMCEF
1. Double click on the SAMCEFTM icon on the desktop to start the software. It may also be
started using the Start button at the bottom left of your Windows desktop.
## Use the following sequence of clicks:
Start -> All Programs -> Samtech -> Samcef v14.0 -> Samcef v14.0
The SAMCEF start menu and the command window will appear on your display.
2. Create a file name for the project, it is good practice to assign a file name that is meaningful and
associated with the project work. In this case, exo1 means exercise 1.
4
Starting SAMCEF
## 3. In order to create and run this exercise, the applicable
modules need to be selected for the analysis. In every
case, “bacon” should be selected for Pre-processing
(setting up the model), and “baconpost” for Post-
processing (reviewing results). The solver module
which is selected will depend on the type of analysis
being carried out, in this case a linear analysis, so the
linear solver “asef” is selected.
## Select “Add Module” tab on the top of the menu
and select “bacon”
and select “asef”
## Select “Add Module” tab on the top of the menu
and select “baconpost”
*Note: The selection of modules must be performed as instructed in the above sequence.
## 4. Once the modules are selected, the SAMCEF start
menu will appear with the selected modules listed in
the left hand column as shown in the diagram.
## To proceed to the pre-processing stage
Click RUN
5
Pre-processing – Creating the model
4.2 Pre-processing
4.2.1 Creating the model
4.2.1.1 Create reference points
1. Create 4 reference points for the outer edges of the model.
## In the command window below, type the following commands
.DEL.*
Press <Enter>
.POIN
Press <Enter>
I 1 X 0 Y 0
Press <Enter>
I 2 X 300 Y 0
Press <Enter>
I 3 X 300 Y 100
Press <Enter>
I 4 X 0 Y 100
Press <Enter>
*Note: There must be a space in between the parameters, for example - I_1_X_0_Y_0
In order to save the environment and the cost of the printing for this exercise, the command window
colour has been altered to white background and black text.
*Command information:
.DEL.* - Delete all the previous commands for a fresh session
.POIN – Generation of points (2D)
I 1 X 0 Y 0 – Point ID number 1 location X = 0 and Y = 0
*Click on the command above for additional information.
## Type the following command in the command window
VI
Press <Enter>
6
Pre-processing – Creating the model
The SAMCEF Graphic Window will appear to show the created points.
Explanation:
## 4.2.1.2 Connect points with lines
1. Connect the reference points with lines to create the plate.
## In the command window, type the following commands
.DROI
Press <Enter>
I 1 POINTS 1 2 3 4 1
7
Pre-processing – Creating the model
Press <Enter>
*Command information:
.DROI - generation of straight line segments (2D)
I 1 POINTS 1 2 3 4 1 – Line ID number 1 connecting point 1 to 2, and line ID number 2 connecting
point 2 to 3, and so on to create a rectangular box section.
## Type the following command in the command window
VI
Press <Enter>
The SAMCEF Graphic Window will appear to show the created model.
8
Pre-processing – Creating the model
Explanation:
9
Pre-processing – Pre-mesh
4.2.2 Pre-mesh
4.2.2.1 Create contour and domain
1. These steps consist of creating a closed contour (outline of the mesh) and a domain (part of the
surface delimited by the outline to mesh). In this case, an auto contour is used to create a
contour by automatically search for a closed contour lines. Similarly, an auto domain is used to
create a domain by automatically search for a closed contour.
## In the command window, type the following commands
.CONTOU AUTO
Press <Enter>
.DOMAIN AUTO
Press <Enter>
*Note:
“Outline: 1 Lines: 1. 2. 3. 4.” Indicates a closed contour is created with Line 1, 2, 3 and 4.
“Domain: 1. Outlines(s) : 1.” Indicates a domain is created on contour 1.
*Command information:
.CONTOU AUTO – Generation of closed contour
.DOMAIN AUTO – Generation of domain
## Type the following command in the command window
VI
Press <Enter>
*Note: The program can run without visualising the created part, it is just an illustration for the
user to understand what is being created.
10
Pre-processing – Pre-mesh
The SAMCEF Graphic Window will appear to show the created contour and domain indicating C1
(contour 1) and D1 (domain 1).
Explanation:
11
Pre-processing – Meshing
4.2.3 Meshing
4.2.3.1 Generate Mesh
1. The following steps are to create a mesh for the model, a free/auto mesh generator is used for
this exercise. The automatic mesh utility selects the most appropriate mesh generator
(transfinite, triangulation, etc.) for the model.
## Type the following command in the command window
.GENE MAILLE
Press <Enter>
*Command information:
.GENE – Automatic mesh generation
MAILLE – Free mesh generation
## 2. To visualise the created mesh
*Note: The program can run without visualising the created part, it is just a pure illustration for
the user to understand what is being created.
## Type the following command in the command window
VI
Press <Enter>
The SAMCEF Graphic Window will appear to show the created mesh for the model
12
Pre-processing – Hypothesis and material
## 4.2.4 Hypothesis and Material
4.2.4.1 Create Hypothesis
1. A plane stress Shell hypothesis is adopted as stated in the exercise description. This hypothesis
is selected due to the relative thickness of the plate.
## Type the following command in the command window
.HYP MEMBRANE BIDIM
Press <Enter>
*Command information:
.HYP – Modelling hypothesis
MEMBRANE BIDIM - 2D plane stress hypothesis for the selected elements that have degrees of
freedom along two structural directions.
## 4.2.4.2 Create Material
1. This section is to create a material for the model. In this case, it is Steel with:
Young Modulus = 2.1E5 N/mm2 and Poisson ratio = 0.3
## Type the following command in the command window
.MAT NOM “STEEL” YT 2.1E5 NT 0.3
Press <Enter>
*Command information:
.MAT – Material property
NOM – Material name, in this case “STEEL”
YT – Young Modulus, in this case 2.1E5 N/mm2
NT – Poisson ratio. In this case 0.3
13
Pre-processing – Model Structure properties
## 4.2.5 Model structure property
4.2.5.1 Create group
1. Nodes and elements are grouped for the convenience of assigning properties to the model as a
whole.
## Type the following command in the command window
.SEL GROUP “ALL” MAILLE TOUT
Press <Enter>
*Command information:
.SEL - Selecting groups of nodes, cells, interfaces and faces
GROUP – Group “name”, in this case ALL
MAILLE TOUT – all meshed elements
## Type the following command in the command window
VI GROUP “ALL”
Press <Enter>
*Command information:
VI GROUP “ALL” - Visualisation of groups. In order to allow several groups to be simultaneously
visualised, the program assigns a palette colour to each group.
The SAMCEF Graphic Window will appear to show the selected group.
14
Pre-processing – Model Structure properties
## 4.2.5.2 Assign material properties
1. Assign the Steel material created in the previous command to the model
## Type the following command in the command window
.AEL GROUP “ALL” MAT “STEEL”
Press <Enter>
*Command information:
.AEL – Element or cell attributes
GROUP “ALL” MAT “STEEL” – Assign STEEL material to a group named “ALL”
## 4.2.5.3 Define model thickness
1. Define the thickness of the Shell elements on the meshed group
## Type the following command in the command window
.PHP GROUP “ALL” THICK VAL 10
Press <Enter>
*Command information:
.PHP – Define physical property
GROUP “ALL” THICK VAL 10 – Assign a thickness of 0.01m to the group named “ALL”
## Type the following command in the command window
VI THICK
Press <enter>
15
Pre-processing – Model Structure properties
The SAMCEF Graphic Window will appear to show the panel thickenss
16
Pre-processing – Apply boundary conditions and load
## 4.2.6 Apply Boundary conditions and load
1. Apply a fixed boundary condition for line 4 that represents the built-in condition
## Type the following command in the command window
.CLM FIX LIGNE 4
Press <Enter>
*Command information:
.CLM - Defining mechanical boundary conditions
FIX LIGNE 4 – To impose fixation on line 4
## Type the following command in the command window
VI FIX
Press <Enter>
The SAMCEF Graphic Window will appear to show the applied boundary conditions
17
Pre-processing – Apply boundary conditions and load
## Type the following command in the command window
CHA POINT 2 NC 1 V 0 -100 0
Press <Enter>
*Command information:
POINT 2 – At point ID 2
NC 1 – Load case ID number 1
V 0 -100 0 – At the of axis with a load of X = 0, Y = - 100N, Z = 0 , negative value is the force acting in
the opposite direction along the axis
VI CHA NC 1
Press <Enter>
## The SAMCEF Graphic Window will appear to show the load
18
Analyse – Execute Solver
4.3 Analyse
4.3.1 Execute Solver
1. To execute solver
## Type the following command in the command window
.FIN 1
Press <Enter>
*Command information:
.FIN 1 - This command generate a SAMCEF file and a database file. These form the data files for the
analysis modules.
19
Post-processing – Load & Display results
4.4 Post-Processing
## Type the following command in the command window
.DOC DB “exo1”
Press <Enter>
*Command information:
.DOC - This command retrieve data from different SAMCEF types of files
DB “exo1” – Read data from the database names “exo1”
## 4.4.2 Display result
1. To display result of the processed data
## Type the following commands in the command window
.DES CODE 163 REFE 4 1 ; MODU DEPLA ; VI
Press <Enter>
*Command information:
.DES – Display of result
CODE 163 – Displacement vector
REFE 4 1 – Reference to load case (inferred by command choice 4) ID number 1
MODU DEPLA - Computes the displacement magnitude
VI – Visualise
20
Post-processing – Load & Display results
The SAMCEF Graphic Window will appear to show the results of the simulation.
The calculated response of the metal plate with the load is as shown in the above diagram, and this
is the end of the exercise. To quit the program, follow the Quitting SAMCEF procedure in the next
section.
## 4.5 Quitting SAMCEF
1. To quit SAMCEF
## Type the following command in the command window
.STOP
Press <Enter>
21
Command lines
5 Command lines
.DEL.*
.POIN
I 1 X 0 Y 0
I 2 X 300 Y 0
I 3 X 300 Y 100
I 4 X 0 Y 100
VI
.DROI
I 1 POINTS 1 2 3 4 1
VI
.CONTOU AUTO
.DOMAIN AUTO
VI
.GENE MAILLE
VI
.HYP MEMBRANE BIDIM
.MAT NOM "STEEL" YT 2.1E5 NT 0.3
.SEL GROUP "ALL" MAILLE TOUT
VI GROUP "ALL"
.AEL GROUP "ALL" MAT "STEEL"
.PHP GROUP "ALL" THICK VAL 10
VI THICK
.CLM FIX LIGNE 4
VI FIX
CHA POINT 2 NC 1 V 0 -100 0
VI CHA NC 1
.FIN 1
.DOC DB “EXO1”
.DES CODE 163 REFE 4 1 ; MODU DEPLA ; VI
22
Summary of the commands used
## 6 Summary of the commands used
Command Descriptions
Pre-processing
.DEL Clear all previous commands
.POIN Generate points (2D)
.DROI Generate straight line (2D)
.CONTOU Generate contour
.DOMAIN Generate domain
.GENE Generate mesh
.HYP Modelling hypothesis
.MAT Material property
.SEL Selecting groups of nodes, cells, interfaces and faces
.AEL Element attribute
.PHP Model physical property
.CLM Defining mechanical boundary conditions and load
Analyse
.FIN Execute solver
Post-processing
.DOC Retrieve data
.DES Display of result
23 | 3,619 | 15,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-51 | latest | en | 0.618612 |
https://www.dai.ed.ac.uk/groups/ssp/bookpages/quickprolog/node23.html | 1,679,515,478,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00720.warc.gz | 810,457,960 | 2,566 | Next: Useful Tip 4 Up: Efficiency Considerations Previous: Useful Tip 2
## Useful Tip 3
Take advantage of tail recursion optimisation. This is a technique built into many Prolog systems which allows the space allocated to the head of a clause to be used for the last subgoal, provided that there is no chance of further results on backtracking by the time the last subgoal gets called. Consider, for example, the `append/3` predicate:
```append([H|T], L1, [H|L2]):-
append(T, L1, L2).
append([], L, L).
```
For this predicate, if we are trying to satisfy a goal such as:
```| ?- append([a,b], [c,d], X).
```
and have managed to match on the head of the first clause and are at the point of calling the last (and only) subgoal of that clause then we know that there would be no other ways of finding a solution (since there are no other subgoals within the clause and the 2nd clause wouldn't match to the goal). If the Prolog interpreter is smart enough to spot this, it can get rid of any choice points for the current goal (`append([a,b], [c,d], X)`) and replace it in memory with the outstanding subgoal (`append([b], [c,d], X)`). This means that a solution to the `append/3` goal can be found using constant memory space, rather than requiring space which increases linearly with the depth of recursion. Of course, this all depends on the interpreter being able to recognise when tail recursion optimisation can be applied - a task which isn't always easy. The `append/3` example, above, is comparatively simple but even it isn't straightforward. Suppose that the goal had been:
```| ?- append(X, Y, [a,b]).
```
Then tail recursion optimisation wouldn't apply because the program isn't deterministic at the point where the subgoal of the 1st clause gets called. Some Prolog systems have fancy indexing systems which allow them to detect determinism quite effectively. However, there is one way in which you can make it clear to the Prolog interpreter that no choice points remain at the point where the last subgoal of a clause is about to be called....add a cut. For example, suppose we have the following program which is designed to gobble up space on the machine:
```gobble_space:-
gobble,
gobble,
gobble,
gobble,
gobble,
gobble,
gobble,
gobble,
statistics(local_stack, L), write(L), nl,
gobble_space.
gobble_space.
gobble.
gobble:-
gobble_space.
```
The behaviour of this predicate is as follows:
```| ?- gobble_space.
[380,16000]
[652,15728]
......SOME TIME LATER.....
[4190812,2468]
[4191084,2196]
{ERROR: Memory allocation failed}
{ Execution aborted }
```
i.e. It eats up a lot of space and eventually exceeds the limits of the system. Now let's try the same program but with a cut in the first clause of `gobble_space/0` to permit tail recursion optimisation.
```gobble_space:-
gobble,
gobble,
gobble,
gobble,
gobble,
gobble,
gobble,
gobble,
statistics(local_stack, L), write(L), nl, !,
gobble_space.
gobble_space.
```
The behaviour of this predicate is:
```| ?- gobble_space.
[380,16000]
[380,16000]
......SOME TIME LATER.....
[380,16000]
[380,16000]
```
i.e. It doesn't eat up space and will continue to run indefinitely (or at least for a very long time).
Next: Useful Tip 4 Up: Efficiency Considerations Previous: Useful Tip 2
Dave Stuart Robertson
Tue Jul 7 10:44:26 BST 1998 | 852 | 3,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-14 | latest | en | 0.927787 |
https://www.airmilescalculator.com/distance/tks-to-nkm/ | 1,603,741,378,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891624.95/warc/CC-MAIN-20201026175019-20201026205019-00677.warc.gz | 602,286,243 | 16,100 | # Distance between Tokushima (TKS) and Nagoya (NKM)
Flight distance from Tokushima to Nagoya (Tokushima Airport – Nagoya Airfield) is 153 miles / 246 kilometers / 133 nautical miles. Estimated flight time is 47 minutes.
Driving distance from Tokushima (TKS) to Nagoya (NKM) is 191 miles / 307 kilometers and travel time by car is about 3 hours 42 minutes.
## Map of flight path and driving directions from Tokushima to Nagoya.
Shortest flight path between Tokushima Airport (TKS) and Nagoya Airfield (NKM).
## How far is Nagoya from Tokushima?
There are several ways to calculate distances between Tokushima and Nagoya. Here are two common methods:
Vincenty's formula (applied above)
• 152.917 miles
• 246.097 kilometers
• 132.882 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 152.759 miles
• 245.842 kilometers
• 132.744 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Tokushima Airport
City: Tokushima
Country: Japan
IATA Code: TKS
ICAO Code: RJOS
Coordinates: 34°7′58″N, 134°36′25″E
B Nagoya Airfield
City: Nagoya
Country: Japan
IATA Code: NKM
ICAO Code: RJNA
Coordinates: 35°15′18″N, 136°55′26″E
## Time difference and current local times
There is no time difference between Tokushima and Nagoya.
JST
JST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 48 kg (105 pounds).
## Frequent Flyer Miles Calculator
Tokushima (TKS) → Nagoya (NKM).
Distance:
153
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
153
Round trip? | 474 | 1,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-45 | latest | en | 0.787353 |
https://www.vedantu.com/maths/properties-of-triangles-formulas | 1,696,274,551,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511002.91/warc/CC-MAIN-20231002164819-20231002194819-00695.warc.gz | 1,133,406,285 | 38,737 | Courses
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# Properties of Triangle Formula
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Join Vedantu’s FREE Mastercalss
## Introduction:
Any three-sided polygon having three edges and three vertices is referred to as a triangle in geometry.
The fact that a triangle's interior angles add up to $180^\circ$ is its most crucial characteristic. Certain fundamental ideas, including the Pythagorean Theorem and trigonometry, depend on the characteristics of triangles.
## What are the Properties of a Triangle?
We must understand the many sorts of triangles in order to learn about the properties of triangles. Although all triangles have some characteristics, some of these characteristics depend on the sides and angles of the triangle.
## Angle Sum Property:
The angle sum property states that the sum of a triangle's three interior angles is always $180^\circ$.
Angle Sum Property
In the given triangle, $\angle A + \angle B + \angle C = 180^\circ$
## Triangle Inequality Property
The length of a triangle's two sides added together is longer than its third side, according to the triangle inequality theorem.
Inequality Property
## Pythagoras Theorem:
The hypotenuse square of a right-angled triangle is equal to the sum of the squares of the other two sides, according to the Pythagoras theorem. Mathematically, it is expressed as $Hypotenus{e^2} = Bas{e^2} + Altitud{e^2}$. See the altitude, base, and hypotenuse in the illustration below.
Side Opposite to the Greater Angle is the Longest Side:
Look at the triangle below to better grasp this triangle's property that the longest side is the one that is opposite the largest angle. B is the largest angle in this triangle. The side AC is hence the longest side.
## Exterior Angle Property:
The exterior angle of a triangle is always equal to the sum of the interior opposite angles, according to the outside angle theorem. Exterior angle (e) of the triangle presented is equal to $\angle a + \angle b$
Exterior Angle Property
It should be remembered that a triangle can expand its three external angles, and the sum of all these exterior angles is $360^\circ$.
## Important Notes for Triangle:
• What is the formula of triangle area? The fundamental formula for determining a triangle's area is ${\rm{Area of triangle = }}\dfrac{1}{2} \times Base \times Height$.
• A triangle's perimeter is the sum of the lengths of its three sides.
## Conclusion:
To learn about the properties of triangles, we must first understand the many types of triangles. Although all triangles have some characteristics, some of these features are dependent on the triangle's sides and angles.
## Solved Example:
Example 1: A triangle has two angles that are $75^\circ$ and $60^\circ$ in length. Determine the third angle's measurement.
Solution: Two angles in a triangle have measurements of an$75^\circ$and $60^\circ$.
Sum of two angles is $135^\circ = 75^\circ + 60^\circ$.
Using a triangle's characteristics, we can determine that the total of its three angles equals 180°.
So the third angle will be $180^\circ -135^\circ =45^\circ$.
Example 2: Tim is trying to build a triangle with sides that are 5 cm, 4 cm, and 9 cm long. Can he accomplish it?
Solution: The sides are 5 cm, 4 cm, and 9 cm long.
5 cm + 4 cm= 9 cm
In this case, the third side is equal to the total of the two smaller sides. However, according to the triangle inequality theorem, any two sides should add up to more than the third side.
Tim won't be able to build a triangle with sides of 5 cm, 4 cm, and 9 cm, according to the triangle's characteristics.
Example 3: The sides of a triangle are given as 3 cm, 4 cm, and 5 cm. Calculate the perimeter of the triangle.
Solution: Sides of the triangle are: x = 3 cm, y = 4 cm and z = 5 cm
The perimeter of the triangle is given by
$\begin{array}{l}P = x + y + z\\P = 3 + 4 + 5\\P = 12cm\end{array}$
Therefore, the perimeter of the given triangle is 12 cm.
Last updated date: 28th Sep 2023
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## FAQs on Properties of Triangle Formula
1. In maths, how many different kinds of triangles are there?
In general, there are six different kinds of triangles. They are right-angled triangles, acute triangles, obtuse triangles, scalene triangles, isosceles triangles, and equilateral triangles.
2. What is a Right-Angle Triangle?
A triangle that has one of the interior angles as $90^\circ$ is a right-angled triangle.
3. What is the Area of a Triangle?
The area of a triangle is equal to half of the product of its base and height. It is the space enclosed by the sides of the triangle and is expressed with the formula, ${\rm{Area of triangle = }}\dfrac{1}{2} \times Base \times Height$ . The area of a triangle is expressed in square units.
4. Describe the Angle Sum Property of a Triangle?
According to the Angle sum property of a triangle, the sum of the interior angles of a triangle is always$180^\circ$ . For example, if the 3 interior angles of a triangle are given as$\angle a,\angle b{\rm{ and }}\angle c$ , then this property can be expressed as, $\angle a + \angle b + \angle c = 180^\circ$
5. What are the Properties of a Right-angled Triangle?
The properties of a right-angled triangle are given as follows:
The largest angle is always $90^\circ$ which means it cannot have any obtuse angle.
The largest side is called the hypotenuse which is always the side opposite to the right angle. | 1,320 | 5,457 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2023-40 | longest | en | 0.889332 |
https://prerelease.barnesandnoble.com/w/gmat-official-guide-2018-quantitative-review-gmac-graduate-management-admission-council/1126406197?ean=9781119387497 | 1,563,319,024,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524972.66/warc/CC-MAIN-20190716221441-20190717003441-00182.warc.gz | 525,314,051 | 37,770 | # GMAT Official Guide 2018 Quantitative Review: Book + Online
## Paperback
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## Overview
A supplement to the Official Guide with 300 additional quantitative questions
The GMAT Official Guide Quantitative Review provides targeted preparation for the mathematical portion of the GMAT exam. Designed by the Graduate Management Admission Council, this guide contains 300 real GMAT questions from past exams including 45 never-before-seen questions, plus the following features:
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Don’t waste time practicing on fake GMAT questions. Optimize your study time with the GMAT Official Guide 2018 Quantitative Review using real questions from actual past exams.
## Product Details
ISBN-13: 9781119387497 Wiley 06/19/2017 240 1,171,597 8.50(w) x 10.80(h) x 0.60(d)
1.0 What is the GMAT® Exam? 2
1.0 What is the GMAT® Exam? 3
1.1 Why Take the GMAT® Exam? 3
1.2 GMAT ® Exam Format 4
1.3 What Is the Content of the Exam Like? 5
1.4 Integrated Reasoning Section 6
1.5 Quantitative Section 6
1.6 Verbal Section 7
1.7 Analytical Writing Assessment 7
1.8 What Computer Skills Will I Need? 7
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2.0 How to Prepare 10
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2.1 How Can I Best Prepare to Take the Test? 11
2.2 What About Practice Tests? 11
2.3 Where Can I Get Additional Practice? 12
2.4 General Test-Taking Suggestions 12
3.0 Math Review 14
3.0 Math Review 15
3.1 Arithmetic 16
3.2 Algebra 27
3.3 Geometry 35
3.4 Word Problems 47
4.0 Problem Solving 56
4.0 Problem Solving 57
4.1 Test-Taking Strategies 58
4.2 The Directions 58
4.3 Sample Questions 60
5.0 Data Sufficiency 148
5.0 Data Sufficiency 149
5.1 Test-Taking Strategies 150
5.2 The Directions 152
5.3 Sample Questions 154 | 633 | 2,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-30 | latest | en | 0.803298 |
http://www.fearofphysics.com/Probs/mech086.html | 1,534,479,678,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211664.49/warc/CC-MAIN-20180817025907-20180817045907-00333.warc.gz | 500,350,661 | 3,540 | [Homework Help Home]
A boy runs directly toward the right as shown, and jumps onto a merry-go-round (MGR) which is initially at rest.
He lands at the position of the blue dot, which is a distance d= m from the center of the MGR.
The boy has a mass m= kg and runs with a speed of v0= m/s. The MGR has a mass of M= kg and a radius R= m.
What angular speed does the MGR + boy have after the boy lands?
[Homework Help Home] | 117 | 424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-34 | longest | en | 0.961055 |
https://nrich.maths.org/public/topic.php?code=-39&cl=2&cldcmpid=994 | 1,606,679,859,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141202590.44/warc/CC-MAIN-20201129184455-20201129214455-00028.warc.gz | 427,569,757 | 10,012 | # Resources tagged with: Combinations
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Broad Topics > Decision Mathematics and Combinatorics > Combinations
### A Rod and a Pole
##### Age 7 to 11 Challenge Level:
A lady has a steel rod and a wooden pole and she knows the length of each. How can she measure out an 8 unit piece of pole?
##### Age 7 to 11 Challenge Level:
Three dice are placed in a row. Find a way to turn each one so that the three numbers on top of the dice total the same as the three numbers on the front of the dice. Can you find all the ways to do. . . .
### Penta Post
##### Age 7 to 11 Challenge Level:
Here are the prices for 1st and 2nd class mail within the UK. You have an unlimited number of each of these stamps. Which stamps would you need to post a parcel weighing 825g?
### Zargon Glasses
##### Age 7 to 11 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Super Value Shapes
##### Age 7 to 11 Challenge Level:
If each of these three shapes has a value, can you find the totals of the combinations? Perhaps you can use the shapes to make the given totals?
### Rod Measures
##### Age 7 to 11 Challenge Level:
Using 3 rods of integer lengths, none longer than 10 units and not using any rod more than once, you can measure all the lengths in whole units from 1 to 10 units. How many ways can you do this?
### Prison Cells
##### Age 7 to 11 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Dice and Spinner Numbers
##### Age 7 to 11 Challenge Level:
If you had any number of ordinary dice, what are the possible ways of making their totals 6? What would the product of the dice be each time?
### Six Is the Sum
##### Age 7 to 11 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### The Puzzling Sweet Shop
##### Age 5 to 11 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Two Egg Timers
##### Age 7 to 11 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
### Hubble, Bubble
##### Age 7 to 11 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
### Magic Triangle
##### Age 7 to 11 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### Sending Cards
##### Age 7 to 11 Challenge Level:
This challenge asks you to investigate the total number of cards that would be sent if four children send one to all three others. How many would be sent if there were five children? Six?
### Throw a 100
##### Age 7 to 11 Challenge Level:
Can you score 100 by throwing rings on this board? Is there more than way to do it?
### On Target
##### Age 7 to 11 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
### Five Coins
##### Age 5 to 11 Challenge Level:
Ben has five coins in his pocket. How much money might he have?
### Polo Square
##### Age 7 to 11 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Team Scream
##### Age 7 to 11 Challenge Level:
Seven friends went to a fun fair with lots of scary rides. They decided to pair up for rides until each friend had ridden once with each of the others. What was the total number rides?
### Octa Space
##### Age 7 to 11 Challenge Level:
In the planet system of Octa the planets are arranged in the shape of an octahedron. How many different routes could be taken to get from Planet A to Planet Zargon?
### Elf Suits
##### Age 7 to 11 Challenge Level:
If these elves wear a different outfit every day for as many days as possible, how many days can their fun last?
### Sealed Solution
##### Age 7 to 11 Challenge Level:
Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?
### Calendar Cubes
##### Age 7 to 11 Challenge Level:
Make a pair of cubes that can be moved to show all the days of the month from the 1st to the 31st.
### Cat Food
##### Age 7 to 11 Challenge Level:
Sam sets up displays of cat food in his shop in triangular stacks. If Felix buys some, then how can Sam arrange the remaining cans in triangular stacks?
### It Figures
##### Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### The Money Maze
##### Age 7 to 11 Challenge Level:
Go through the maze, collecting and losing your money as you go. Which route gives you the highest return? And the lowest?
### Finding Fifteen
##### Age 7 to 11 Challenge Level:
Tim had nine cards each with a different number from 1 to 9 on it. How could he have put them into three piles so that the total in each pile was 15?
### Home City
##### Age 7 to 11 Challenge Level:
Use the clues to work out which cities Mohamed, Sheng, Tanya and Bharat live in.
### Bean Bags for Bernard's Bag
##### Age 7 to 11 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Halloween Investigation
##### Age 7 to 11 Challenge Level:
Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
### Making Cuboids
##### Age 7 to 11 Challenge Level:
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
### Sweets in a Box
##### Age 7 to 11 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### More Plant Spaces
##### Age 7 to 14 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
##### Age 7 to 11 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
### Chocs, Mints, Jellies
##### Age 7 to 11 Challenge Level:
In a bowl there are 4 Chocolates, 3 Jellies and 5 Mints. Find a way to share the sweets between the three children so they each get the kind they like. Is there more than one way to do it?
### Wag Worms
##### Age 7 to 11 Challenge Level:
When intergalactic Wag Worms are born they look just like a cube. Each year they grow another cube in any direction. Find all the shapes that five-year-old Wag Worms can be.
### Penta Primes
##### Age 7 to 11 Challenge Level:
Using all ten cards from 0 to 9, rearrange them to make five prime numbers. Can you find any other ways of doing it?
### The Twelve Pointed Star Game
##### Age 7 to 11 Challenge Level:
Have a go at this game which involves throwing two dice and adding their totals. Where should you place your counters to be more likely to win?
### More and More Buckets
##### Age 7 to 11 Challenge Level:
In this challenge, buckets come in five different sizes. If you choose some buckets, can you investigate the different ways in which they can be filled?
### Button-up Some More
##### Age 7 to 11 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### More Children and Plants
##### Age 7 to 14 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### Delia's Routes
##### Age 7 to 11 Challenge Level:
A little mouse called Delia lives in a hole in the bottom of a tree.....How many days will it be before Delia has to take the same route again?
### Calcunos
##### Age 7 to 11 Challenge Level:
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
### Ice Cream
##### Age 7 to 11 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
### Plates of Biscuits
##### Age 7 to 11 Challenge Level:
Can you rearrange the biscuits on the plates so that the three biscuits on each plate are all different and there is no plate with two biscuits the same as two biscuits on another plate?
### Street Party
##### Age 7 to 11 Challenge Level:
The challenge here is to find as many routes as you can for a fence to go so that this town is divided up into two halves, each with 8 blocks.
### Newspapers
##### Age 7 to 11 Challenge Level:
When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different?
### Three Way Mix Up
##### Age 5 to 11 Challenge Level:
Jack has nine tiles. He put them together to make a square so that two tiles of the same colour were not beside each other. Can you find another way to do it?
### Three by Three
##### Age 5 to 11 Challenge Level:
Arrange 3 red, 3 blue and 3 yellow counters into a three-by-three square grid, so that there is only one of each colour in every row and every column
### Train Carriages
##### Age 5 to 11 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done? | 2,482 | 10,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-50 | latest | en | 0.899506 |
https://numberworld.info/26496 | 1,679,729,123,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00129.warc.gz | 482,648,405 | 4,060 | # Number 26496
### Properties of number 26496
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 * 23
Divisors:
1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 23, 24, 32, 36, 46, 48, 64, 69, 72, 92, 96, 128, 138, 144, 184, 192, 207, 276, 288, 368, 384, 414, 552, 576, 736, 828, 1104, 1152, 1472, 1656, 2208, 2944, 3312, 4416, 6624, 8832, 13248, 26496
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 32:
ps0
sin(26496)
-0.1912547901073
cos(26496)
0.98154042466982
tan(26496)
-0.1948516691726
ln(26496)
10.184749057185
lg(26496)
4.4231803151048
sqrt(26496)
162.77591959501
Square(26496)
### Number Look Up
Look Up
26496 which is pronounced (twenty-six thousand four hundred ninety-six) is a great number. The cross sum of 26496 is 27. If you factorisate 26496 you will get these result 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 * 23. 26496 has 48 divisors ( 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 23, 24, 32, 36, 46, 48, 64, 69, 72, 92, 96, 128, 138, 144, 184, 192, 207, 276, 288, 368, 384, 414, 552, 576, 736, 828, 1104, 1152, 1472, 1656, 2208, 2944, 3312, 4416, 6624, 8832, 13248, 26496 ) whith a sum of 79560. The figure 26496 is not a prime number. The number 26496 is not a fibonacci number. The figure 26496 is not a Bell Number. The number 26496 is not a Catalan Number. The convertion of 26496 to base 2 (Binary) is 110011110000000. The convertion of 26496 to base 3 (Ternary) is 1100100100. The convertion of 26496 to base 4 (Quaternary) is 12132000. The convertion of 26496 to base 5 (Quintal) is 1321441. The convertion of 26496 to base 8 (Octal) is 63600. The convertion of 26496 to base 16 (Hexadecimal) is 6780. The convertion of 26496 to base 32 is ps0. The sine of 26496 is -0.1912547901073. The cosine of 26496 is 0.98154042466982. The tangent of the figure 26496 is -0.1948516691726. The square root of 26496 is 162.77591959501.
If you square 26496 you will get the following result 702038016. The natural logarithm of 26496 is 10.184749057185 and the decimal logarithm is 4.4231803151048. that 26496 is very unique number! | 935 | 2,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-14 | latest | en | 0.715443 |
https://www.physicsforums.com/threads/finding-the-fourier-series-of-a-step-function.998038/ | 1,701,509,926,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100381.14/warc/CC-MAIN-20231202073445-20231202103445-00276.warc.gz | 1,050,568,278 | 16,884 | # Finding the Fourier Series of a step function
• Tony Hau
In summary, the conversation discusses the answer in a textbook that presents a formula for a trigonometric series. The formula involves two series, but the individual discussing the content is unsure about where the constant term of ##\frac{1}{4}## comes from. They question whether it should be ##\frac{1}{2}## instead. The expert summarizer explains that the constant term is the average value of ##f(x)## over the interval of interest, and since the function has a value of 1 on a fourth of the interval, the average value is ##\frac{1}{4}##. This is not mentioned in the book, but it is a useful property to remember when double-checking calculations.
#### Tony Hau
Homework Statement
Given : ## f(x) = \begin{cases}
0, & -\pi \lt x \lt 0 \\
1, & 0 \lt x \lt \frac{\pi}{2} \\
0, & \frac{\pi}{2} \lt x \lt \pi
\end{cases} ##,
Find the Fourier Series of ##f(x)##.
Relevant Equations
##a_n =\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx)dx##
##b_n =\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx)dx##
The answer in the textbook writes: $$f(x) = \frac{1}{4} +\frac{1}{\pi}(\frac{\cos(x)}{1}-\frac{\cos(3x)}{3}+\frac{\cos(5x)}{5} \dots) + \frac{1}{\pi}(\frac{\sin(x)}{1}-\frac{2\sin(2x)}{2}+\frac{\sin(3x)}{3} + \frac{\sin(5x)}{5}\dots)$$
I am ok with the two trigonometric series in the answer. However, I don't understand where that ##\frac{1}{4}## comes from.
Since the formula for ##a_0## is ##a_0 =\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}} dx##, which gives ##\frac{1}{2}##, isn't that constant equal to ##\frac{1}{2}## instead of ##\frac{1}{4}##.
Tony Hau said:
Since the formula for ##a_0## is ##a_0 =\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}} dx##, which gives ##\frac{1}{2}##, isn't that constant equal to ##\frac{1}{2}## instead of ##\frac{1}{4}##.
The first term is the series is ##a_0/2##, not ##a_0##.
Delta2 and Tony Hau
etotheipi said:
The first term is the series is ##a_0/2##, not ##a_0##.
Thanks. No joke but this thing has bothered me for the whole afternoon... :)
Delta2 and etotheipi
Tony Hau said:
Thanks. No joke but this thing has bothered me for the whole afternoon... :)
That's annoying... but at least you will never forget it again!
DrClaude, Delta2 and Tony Hau
Note that the constant term of the Fourier series is simply the average value of ##f(x)## over the interval of interest (which is ##[-\pi,\pi]## in this case). Since your ##f(x)## has value ##1## on one fourth of this interval (namely on ##[0,\pi/2]##) and value 0 elsewhere, it's clear that the average value of ##f(x)## is ##1/4##. This is a useful property to keep in mind when double-checking your calculation.
Tony Hau, etotheipi, DrClaude and 1 other person
jbunniii said:
Note that the constant term of the Fourier series is simply the average value of ##f(x)## over the interval of interest (which is ##[-\pi,\pi]## in this case). Since your ##f(x)## has value ##1## on one fourth of this interval (namely on ##[0,\pi/2]##) and value 0 elsewhere, it's clear that the average value of ##f(x)## is ##1/4##. This is a useful property to keep in mind when double-checking your calculation.
Thanks. This is a nice interpretation that is not mentioned in the book. | 1,028 | 3,216 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2023-50 | latest | en | 0.727254 |
http://ccom.ucsd.edu/~jdilts/teaching/10A-Winter17/hw.php | 1,542,378,521,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743046.43/warc/CC-MAIN-20181116132301-20181116154301-00110.warc.gz | 56,924,865 | 2,189 | ### Homework 0
The following exercises are due Friday, Jan 13:
1. Turn in a pdf to Gradescope.com containing your name and the message "Hello World!"
### Homework 1
This assignment is due Friday, Jan 20:
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### Homework 9
This assignment is due Friday, Mar 17:
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### Homework 10
This assignment is due Monday, Mar 20:
Solutions are here.
Suggested textbook problems (Don't turn these in. Instruction on finding the WileyPlus course here.):
Section 1.1: 1, 3, 7, 11, 21, 43, 61
Section 1.2: 5, 15, 17, 39, 51, 57
Section 1.3: 1, 9, 19, 23, 29, 47, 51, 73, 78
Section 1.4: Solve as many of 1-30 as you need to practice! 45, 56, 63
Section 1.5: 1, 11, 23, 27, 43, 45, 58, 68, 69
Section 1.6: 3, 5, 7, 9, 15, 28, 37, 51
Section 1.7: 1, 9, 15, 25, 37, 39, 45
Section 1.8: 2, 13, 15, 65, 75, 89, 93
Section 2.1: 1, 9, 11, 13, 19, 25, 27, 29, 33, 34, 35
Section 2.2: 3, 11, 12, 14, 21, 23, 35, 37, 53, 57
Section 2.3: 3, 11, 28, 41, 49, 51, 59
Section 2.4: 1, 3, 6, 13, 21, 29, 42
Section 2.5: 1, 3, 5, 15, 19, 21, 34, 36
Section 2.6: 1, 13, 16, 18,
Section 3.1: 3, 5, 7, 11, 19, 23, 33, 42, 68, 89, 91
Section 3.2: As many of 1-24 as you need, 31, 39, 47, 49, 50, 51, 52, 53
Section 3.3: 1, As many of 3-30 as you need, 31, 40, 41, 56, 59, 71
Section 3.4: As many of 1-56 as you need, 57, 65, 69, 74, 77, 91
Section 3.5: As many of 2-47 as you need, 49, 54, 56, 73
Section 3.6: As many of 1-41 as you need, 43, 49, 66, 67
Section 3.9: 1, 3, 8, 11, 20, 22, 47
Section 3.7: As many of 1-30 as you need, 35, 39
Section 4.1: 1, 4-8, 16-19, 27, 35, 37, 55-57
Section 4.2: 1, 4-25 if you need them, 37, 41, 53, 54, 62
Section 4.3: 1, 4, 8, 10, 17, 32, 34, 44
Section 4.7: As many of 1-15 as you need. 16, 20, 27, 32, 37, 39, 45, 46, 47, 67 | 1,068 | 2,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2018-47 | latest | en | 0.953721 |
https://ellejae.com/archive/2caa12-second-order-differential-equation-with-variable-coefficients | 1,718,611,392,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861698.15/warc/CC-MAIN-20240617060013-20240617090013-00835.warc.gz | 202,870,496 | 16,152 | Physics Behind Seatbelts, Brown Middle School Dress Code, Online Clothing Wholesalers, Westlake High School Football Ohio, Mass Effect 1 Walkthrough, Barcelona Sants Station Map, St Brendan Catholic School Calendar, Black Vulture South Carolina, Southern Baked Mac And Cheese With Bread Crumbs, Sahuagin Baron Dnd 5e Stats, Mens Rugby Clothing, " /> Physics Behind Seatbelts, Brown Middle School Dress Code, Online Clothing Wholesalers, Westlake High School Football Ohio, Mass Effect 1 Walkthrough, Barcelona Sants Station Map, St Brendan Catholic School Calendar, Black Vulture South Carolina, Southern Baked Mac And Cheese With Bread Crumbs, Sahuagin Baron Dnd 5e Stats, Mens Rugby Clothing, " />
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1\\ the sum of all diagonal elements: ${\text{tr}\left( {A\left( \tau \right)} \right) }={ {a_{11}}\left( \tau \right) + {a_{22}}\left( \tau \right) + \cdots }+{ {a_{nn}}\left( \tau \right).}$. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. This website uses cookies to improve your experience. Finding particular solution to a linear first order ODE, Second order linear ODE with polynomial coefficients, Mentor added his name as the author and changed the series of authors into alphabetical order, effectively putting my name at the last. }\], After integration we obtain the vector $$\mathbf{C}\left( t \right).$$. 0&{{e^x}}&y^{\prime\prime} \]. \], As a result, we get the following expression for $$z:$$, $= {{x^2}\left( {{C_1}x + {C_2}} \right) } {{x_{n1}}\left( t \right)}&{{x_{n2}}\left( t \right)}& \vdots &{{x_{nn}}\left( t \right)} {\frac{{{y_2}}}{{{y_1}}} = {C_1}x + {C_2},\;\;}\Rightarrow = {z’ + z’ + xz^{\prime\prime} } {{y^{\prime\prime} = 2z + 2xz’ + 2xz’ }+{ \left( {{x^2} + 1} \right)z^{\prime\prime} }} = {{C_1}{x^3} + {C_2}{x^2},} \end{array}} \right| = 0.} After substituting the original equation becomes: \[\require{cancel} How to solve this puzzle of Martin Gardner? = {\int\limits_{{x_0}}^x {\left( {\frac{{ – 4t}}{{{t^2}}}} \right)dt} } You also have the option to opt-out of these cookies. It is mandatory to procure user consent prior to running these cookies on your website. Sync all your devices and never lose your place. Based on the structure of the equation, we can try to find a particular solution in the form of a quadratic function: \[{{y’_1} = 2Ax + B,\;\;}\kern-0.3pt{{y^{\prime\prime}_1} = 2A. 3, 1959, pp. A set of two linearly independent particular solutions of a linear homogeneous second order differential equation forms its fundamental system of solutions. {{x^3}z^{\prime\prime} = 0.} These cookies will be stored in your browser only with your consent. {\Phi \left( t \right)\mathbf{C’}\left( t \right) = \mathbf{f}\left( t \right).} {{x\left( {{e^x}y^{\prime\prime} – {e^x}y’} \right) }-{ 1 \cdot \left( {{e^x}y^{\prime\prime} – {e^x}y} \right) = 0,\;\;}}\Rightarrow {{x_n}\left( t \right)} { \frac{1}{2}\arctan x + {C_2}} \right] } = {x\left( {{C_1}x + {C_2}} \right) } = { – 4\ln x + 4\ln {x_0} }$, Given that the matrix $$\Phi \left( t \right)$$ is nonsingular, we multiply this equation on the left by $${\Phi^{ – 1}}\left( t \right):$$, ${{{\Phi^{ – 1}}\left( t \right)\Phi \left( t \right)\mathbf{C’}\left( t \right) }={ {\Phi^{ – 1}}\left( t \right)\mathbf{f}\left( t \right),\;\;}}\Rightarrow {\mathbf{C’}\left( t \right) = {\Phi^{ – 1}}\left( t \right)\mathbf{f}\left( t \right). {{\cancel{\Phi’\left( t \right)\mathbf{C}\left( t \right)} + \Phi \left( t \right)\mathbf{C’}\left( t \right) }} An order linear ordinary differential equation with variable coefficients has the general form of Most ordinary differential equations with variable coefficients are not possible to solve by hand. {y = \left( {{x^2} + 1} \right)z } Why didn't Crawling Barrens grow larger when mutated with my Gemrazer? Asymptotic solutions of nonlinear second order differential equations with variable coefficients Asimptoticheskie resheniia nelineinykh differentsial'nvkh uravnenii utorogo poriadka s peremennymi koeff itsientaihi: PMM vol. Divide both sides of the equation by $$y_1^2 = {x^2}$$ (assuming that $$x \ne 0$$). {{\left( {\frac{{{y_2}}}{{{y_1}}}} \right)^\prime } = \frac{{{C_1}}}{{x \cdot {x^2}}} = \frac{{{C_1}}}{{{x^3}}},\;\;}\Rightarrow {{a_{21}}\left( t \right)}&{{a_{22}}\left( t \right)}& \vdots &{{a_{2n}}\left( t \right)}\\ Making statements based on opinion; back them up with references or personal experience. MathJax reference. linear homogeneous second order equation with variable coefficients. Thanks in advance. y′′ +a1(x)y′ +a2(x)y = f (x), where a1(x), a2(x) and f (x) are continuous functions on the interval [a,b]. \end{array}} \right|} So a base of the associate homogeneous equaction it's c_1\cdot sec(x)+c_2\cdot xsec(x) (sec(x)=\frac{1}{\cos x}). Get Differential Equations now with O’Reilly online learning. Any system of $$n$$ linearly independent solutions $${\mathbf{x}_1}\left( t \right),$$ $${\mathbf{x}_2}\left( t \right), \ldots ,$$ $${\mathbf{x}_n}\left( t \right)$$ is called a fundamental system of solutions. The vector functions $${\mathbf{x}_1}\left( t \right),{\mathbf{x}_2}\left( t \right), \ldots ,{\mathbf{x}_n}\left( t \right)$$ are linearly dependenton the interval $$\left[ {a,b} \right],$$ if there are numbers $${c_1},{c_2}, \ldots ,{c_n},$$ not all zero, such that the following identity holds: \[ {{c_1}{\mathbf{x}_1}\left( t \right) + {c_2}{\mathbf{x}_2}\left( t \right) + \cdots }+{ {c_n}{\mathbf{x}_n}\left( t \right) \equiv 0,\;\;}\kern-0.3pt {\forall t \in \left[ {a,b} \right].} {W\left( x \right) = {W_{{y_1},{y_2}}}\left( x \right) } {{y’_2}{y_1} – {y_2}{y’_1} = {C_1}{x^4}} \right|:y_1^2,\;\;}\Rightarrow = { – \ln \frac{{{x^4}}}{{x_0^4}}.} {{y’_1}\left( x \right)}&{{y’_2}\left( x \right)} {{y_1}\left( t \right)} The associated homogeneous equation is written as. {{y’_2} = {\left( {{e^x}} \right)^\prime } = {e^x},\;\;}\kern-0.3pt {{y^{\prime\prime}_2} = {\left( {{e^x}} \right)^\prime } = {e^x}.} \end{array}} \right| } {{y_2} = {y_1}\left( {{C_1}x + {C_2}} \right) } {{x_2}\left( t \right)}\\ {W\left( x \right) = {W_{{y_1},{y_2}}}\left( x \right) } where $$\mathbf{C}$$ is an $$n$$-dimensional vector consisting of arbitrary numbers. In Monopoly, if your Community Chest card reads "Go back to ...." , do you move forward or backward? = {{C_1}x }+{ {C_1}\left( {{x^2} + 1} \right)\arctan x }+{ {C_2}\left( {{x^2} + 1} \right) } The relevant examples are given below. . Did Star Trek ever tackle slavery as a theme in one of its episodes? \cdots & \cdots & \cdots & \cdots \\ For the case of two functions, the linear independence criterion can be written in a simpler form: The functions $${y_1}\left( x \right),$$ $${y_2}\left( x \right)$$ are linearly independent on the interval $$\left[ {a,b} \right],$$ if their quotient in this segment is not identically equal to a constant: \[\frac{{{y_1}\left( x \right)}}{{{y_2}\left( x \right)}} \ne \text{const.}$. = {\frac{{{C_1}x}}{{{x^2} + 1}} }+{ {C_1}\arctan x }+{ {C_2},} = {{C_1}\exp \left( { – \int {\frac{{{a_1}\left( x \right)}}{{{a_0}\left( x \right)}}dx} } \right).} Click or tap a problem to see the solution. (\cos x)y''-2(\sin x)y'-(\cos x)y=e^x }\], Substituting the given functions and their derivatives, we obtain, $These cookies will be stored in your browser only with your consent. \[ \cdots & \cdots & \cdots & \cdots \\ Hence, find the general soluion of he differential equation. {{x_1}\left( t \right)}\\ If a particular solution $${y_1}\left( x \right) \ne 0$$ of the homogeneous linear second order equation is known, the original equation can be converted to a linear first order equation using the substitution $$y = {y_1}\left( x \right)z\left( x \right)$$ and the subsequent replacement $$z’\left( x \right) = u.$$. If $${y_1}\left( x \right),{y_2}\left( x \right)$$ is a fundamental system of solutions, then the general solution of the second order equation is represented as, \[{y\left( x \right) }={ {C_1}{y_1}\left( x \right) + {C_2}{y_2}\left( x \right),}$. | 2,762 | 8,091 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-26 | latest | en | 0.760022 |
https://www.paynut.org/what-is-the-abbreviation-for-feet-and-inches/ | 1,656,970,192,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104496688.78/warc/CC-MAIN-20220704202455-20220704232455-00770.warc.gz | 983,625,406 | 12,977 | What is the abbreviation for feet and inches?
What is the abbreviation for feet and inches?
The IEEE usual image for a foot is “ft”. In some instances, the foot is denoted via a major, frequently approximated by an apostrophe, and the inch by means of a double high; for example, 2 feet 4 inches is every now and then denoted as 2′ 4″.
Does the abbreviation for inches have a duration?
Actually, the abbreviation for inches is a double high; even in the case you utilize the citation mark as abbreviation, the duration is placed after the abbreviation. I’m 6′2″.
How do you write an inch and a part?
in informal writing – the image or abbreviated shape is not unusual, and other people ceaselessly can’t be troubled to put in ½ . In speech, I would be expecting people to mention ‘half an inch’ (although written 0.5″) most of the time, but when they are saying it in decimals, then it might be ‘(0/nought/oh) level five inches’.
Is Five cm equal to two inches?
* The inches fraction outcome is rounded to the nearest 1/Sixty four fraction….Centimeters to inches conversion table.
Centimeters (cm) Inches (“) (decimal) Inches (“) (fraction)
2 cm 0.7874 in 25/32 in
3 cm 1.1811 in 1 3/16 in
4 cm 1.5748 in 1 37/Sixty four in
Five cm 1.9685 in 1 31/32 in
What dimension is 7 cm in inches?
Centimeters to Inches table
Centimeters Inches
5 cm 1.97 in
6 cm 2.36 in
7 cm 2.76 in
8 cm 3.15 in
How many inches are there in 6 35 centimeters?
2.5 inches
What measurement is 20×30 cm in inches?
Standard Metric ‘CM’ image frame sizes
Centimetres (cm) Millimetres (mm) Inches
20 x 20 cm 200 x 200 mm 7.87″ x 7.87″
20 x 25 cm 2 hundred x 250 mm 7.87″ x 9.84″
24 x 30 cm 240 x three hundred mm 9.45″ x 11.81″
25 x 30 cm 250 x 300 mm 9.84″ x 11.81″
How many sq. inches are in 1 inch?
Square inch to Inch Calculator
1 in2 = 1 inch 1 inch =
8 in2 = 2.8284 inch 8 inch =
Nine in2 = Three inch 9 inch =
10 in2 = 3.1623 inch 10 inch =
Eleven in2 = 3.3166 inch Eleven inch =
How do I convert inches to square inches?
Multiply duration × width. For example, let’s say that, for an oblong area, you measure a period of 4 inches and a width of 3 inches. In this case, the space inside your rectangle is 4 × 3 = 12 square inches. | 680 | 2,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-27 | latest | en | 0.885843 |
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# News from Mathnasium of Leaside
### Questions of the Week March 20th to 24th
Mar 23, 2017
Lower Elementary:
Question: A baby T-Rex weighs 3 pounds when it hatches. After a week, it weighs 6 pounds. At two weeks, it weighs 12 pounds. At three weeks, it weighs 24 pounds. If the pattern continues, how much will the baby T-Rex weigh when it is 5 weeks old?
Upper Elementary:
Question: A paleontologist found five triceratops nests. The first nest had 12 eggs. The second nest had 14 eggs. The third nest had 12 eggs. The fourth nest had 16 eggs. The fifth nest had 21 eggs. What is the average number of eggs in a triceratops nest?
Middle School:
Question: A pterodactyl’s wingspan is 30 feet. Its height is 6 feet. If a twelve-year-old kid is 5 feet tall, then how wide would their “wingspan” (the length from fingertip to fingertip with arms spread out) be if it were proportional to that of a pterodactyl?
Algebra and Up:
Question: The Pachycephalosaurus was a dome-headed dinosaur that fought by running into its competitors head-first like a bighorn sheep. If two pachycephalosaurs run toward each other, each traveling 30 miles per hour, from a quarter mile apart, then how long will it take for them to collide? | 357 | 1,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-23 | latest | en | 0.955055 |
https://community.powerbi.com/t5/Desktop/Warning-when-Market-Share-falls-below/m-p/480310 | 1,638,949,173,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363445.41/warc/CC-MAIN-20211208053135-20211208083135-00497.warc.gz | 257,203,299 | 100,149 | cancel
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Helper I
## Warning when Market Share falls below
Hi All
I have a matrix where it displays my clients and how much they have purchased with us as well as with competitors.
The challenge that I have at the moment is that we have more than 1000 clients on this matrix and I would like to easily identify those clients who are falling behind in the Market Share.
i.e. It could be a filter where it would ONLY display clients who have less than 50% market share with us the ABC CO.
This way I would not require to scroll down the matrix to identify these clients.
Have any of you come across something similar?
I would appreciate any assistance on this
Regards
Rodrigo
1 ACCEPTED SOLUTION
Solution Sage
Rather than use % Row total you need to write a few measures.
```Revenue = SUM(Dashboard[Base Amount])
Share = CALCULATE([Revenue],Dashboard[Bought_from_us_or_Competitor]="ABC CO") / [Revenue]```
With these measures and using [Share] in your visual you can sort by Market Share and put the lowest on top. See link
https://1drv.ms/u/s!AuCIkLeqFmlhhJh6m05vmpRbIHQVcw
5 REPLIES 5
Solution Sage
Rather than use % Row total you need to write a few measures.
```Revenue = SUM(Dashboard[Base Amount])
Share = CALCULATE([Revenue],Dashboard[Bought_from_us_or_Competitor]="ABC CO") / [Revenue]```
With these measures and using [Share] in your visual you can sort by Market Share and put the lowest on top. See link
https://1drv.ms/u/s!AuCIkLeqFmlhhJh6m05vmpRbIHQVcw
Helper I
Hey @Seward12533 have you used What If analyses in Power BI?
Lets say now that I am able to display only those companies which are falling behind in market share, would we be able to analyse those clients by asking power bi to display how much revenue would shift towards my company if my Market share increased 5%?
Solution Sage
Isnt that just .05*[Revenue] since that represents the total available market?
Solution Sage
For the what if you could also create a disconnected sliver of percentage amounts and then use SELECTEDVALUE to harvest the user selection and then use that as the percentage.
Since this is a numeric number you could use the New Parametrr button on the Modeling Tab to create both the lookup table and harvest measure for you automatically.
Helper I
@Seward12533 this is perfect, I added this Share Measure into the filter and applied the amounts I wanted to see. Thank you very much for this
Announcements
#### Launching new user group features
Learn how to create your own user groups today! | 608 | 2,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | latest | en | 0.932932 |
https://math.stackexchange.com/questions/1811434/relation-between-the-frattini-property-and-pronormal-subgroups-of-a-solvable-gro | 1,561,382,909,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999539.60/warc/CC-MAIN-20190624130856-20190624152856-00415.warc.gz | 513,714,922 | 35,299 | # Relation between the Frattini Property and Pronormal subgroups of a solvable group
A subgroup $H$ of $G$ is said to satisfy the Frattini Property if for any subgroup $K$ and $L$ such that $H\leq K \unlhd L$ implies that $L \leq N_L(H)K$
A subgroup is $H$ is pronormal in $G$ if for each $g \in G$, there exists $x \in \langle H, H^g \rangle$ such that $H^x = H^g$
A theorem characterising pronormal subgroups of soluble groups was proved by T. Peng which stated that: if $G$ is soluble group, $H$ is pronormal in $G$ $\iff$ H satisfies the Frattini Property
The $\Rightarrow$ direction is true in general since for any $g\in G$, $\langle H, H^g \rangle \leq H^{\langle g \rangle}$ and using my previous question Frattini Property of a subgroup
For the $\Leftarrow$ direction, solvability of the group will be needed. Would induction on $|G|$ be the way of solving this implication?
• I think you already asked this question before, or at least a very similar one. That Frattini property seems to be a rather involved, very specific property and it seems to be not many around here can handle all the info. Try to look for some help perhaps in MathOverflow, – DonAntonio Jun 3 '16 at 19:14
• You can only use induction on $|G|$ for a finite group $G$, and you do not appear to be assuming that $G$ is finite. – Derek Holt Jun 3 '16 at 19:23
• In any case, if it is a published result then why don't you try and read the proof there? – Derek Holt Jun 3 '16 at 19:29
• I don't have access to the paper which is in an Oxford Journal. My university which is the University of Kwa-Zulu Natal in South Africa also does not have library access to this paper. And yes, I missed out the crucial info of $G$ being finite – R Maharaj Jun 3 '16 at 19:47
• @Joanpemo Thank you. I did not know that another Math forum apart from this one existed. – R Maharaj Jun 3 '16 at 19:52 | 528 | 1,870 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-26 | latest | en | 0.925477 |
http://pub12.bravenet.com/forum/static/show.php?usernum=1014493548&frmid=6526&msgid=930896&cmd=show | 1,718,563,843,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861670.48/warc/CC-MAIN-20240616172129-20240616202129-00240.warc.gz | 23,740,337 | 7,127 | ### The Logic Forum Discussion Area
Logic
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enthymeme
I picked up a textbook called "An Introduction" to Logic by Creighton.
I've come early on to this syllogism:
A: All students of the University are over 16 years of age.
B: Baby1nut is a student at the university.
C: Therefore Baby1nut is more than 16 years of age. (so true!)
The author states that this is an Enthymeme. He writes: "The statement He must be over 16 years of age, for he attends the university is an incomplete syllogism . The conclusion as will be readily seen, stands first. "
I really don't see the difference between this, and A: All men are mortal
B. Baby1nut is a man
C: Baby1nut is mortal
Can anyone kindly help out a confused old guy? Many advance thanks!!
Something about you (optional) old guy trying too think weller
Re: enthymeme
Hi, Baby1nut.
The statements as you present them are both right.
"All students of the University are over 16 years of age. Baby1nut is a student at the university. Therefore Baby1nut is more than 16 years of age." - this is a valid syllogism.
Creighton's statement that "He must be over 16 years of age, for he attends the university is an incomplete syllogism . The conclusion as will be readily seen, stands first." is indeed an enthymeme, i.e. an abridged syllogism (the major premise "All students of the University are over 16 years of age" being left tacit).
The underlying argument of the enthymeme is one and the same with your full syllogism.
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