url
string | fetch_time
int64 | content_mime_type
string | warc_filename
string | warc_record_offset
int32 | warc_record_length
int32 | text
string | token_count
int32 | char_count
int32 | metadata
string | score
float64 | int_score
int64 | crawl
string | snapshot_type
string | language
string | language_score
float64 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.hackmath.net/en/example/990?tag_id=77
| 1,558,580,607,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-22/segments/1558232257002.33/warc/CC-MAIN-20190523023545-20190523045545-00211.warc.gz
| 798,931,806 | 6,577 |
# Rectangle
Calculate the length of the side GN and diagonal QN of rectangle QGNH when given: |HN| = 25 cm and angle ∠ QGH = 28 degrees.
Result
|GN| = 13.29 cm
|QN| = 28.31 cm
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this example are needed these knowledge from mathematics:
Pythagorean theorem is the base for the right triangle calculator. See also our trigonometric triangle calculator.
## Next similar examples:
1. Rectangle
The rectangle is 11 cm long and 45 cm wide. Determine the radius of the circle circumscribing rectangle.
2. Cincinnati
A map is placed on a coordinate grid. Cincinnati located at (5,4) and San Diego is located at (-10, -3). How far apart is Cincinnati from San Diego on the map? Round to the nearest tenth.
3. Lighthouse
Marcel (point J) lies in the grass and sees the top of the tent (point T) and behind it the top of the lighthouse (P). | TT '| = 1.2m, | PP '| = 36m, | JT '| = 5m. Marcel lies 15 meters away from the sea (M). Calculate the lighthouse distance from the sea.
4. Right Δ
Right triangle has length of one leg 28 cm and length of the hypotenuse 53 cm. Calculate the height of the triangle.
5. Trapezoid MO
The rectangular trapezoid ABCD with right angle at point B, |AC| = 12, |CD| = 8, diagonals are perpendicular to each other. Calculate the perimeter and area of the trapezoid.
6. Triangle SAS
Calculate area and perimeter of the triangle, if the two sides are 51 cm and 110 cm long and angle them clamped is 130°.
7. Axial section
Axial section of the cone is equilateral triangle with area 208 dm2. Calculate volume of the cone.
8. Medians 2:1
Median to side b (tb) in triangle ABC is 12 cm long. a. What is the distance of the center of gravity T from the vertex B? b, Find the distance between T and the side b.
9. Garden
Area of square garden is 6/4 of triangle garden with sides 56 m, 35 m and 35 m. How many meters of fencing need to fence a square garden?
10. Cable car
Find the elevation difference of the cable car when it rises by 67 per mille and the rope length is 930 m.
11. Circle chord
What is the length d of the chord circle of diameter 36 m, if distance from the center circle is 16 m?
12. Isosceles triangle 8
If the rate of the sides an isosceles triangle is 7:6:7, find the base angle correct to the nearest degree.
13. The sides 2
The sides of a trapezoid are in the ratio 2:5:8:5. The trapezoid’s area is 245. Find the height and the perimeter of the trapezoid.
14. Tetrahedral pyramid
What is the surface of a regular tetrahedral (four-sided) pyramid if the base edge a=7 and height v=6?
15. Find the 9
Find the missing angle in the triangle and then name triangle. Angles are: 95, 2x+15, x+3
16. Trigonometric functions
In right triangle is: ? Determine the value of s and c: ? ?
17. Find parameters
Find parameters of the circle in the plane - coordinates of center and radius: ?
| 818 | 2,971 |
{"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}
| 4.1875 | 4 |
CC-MAIN-2019-22
|
latest
|
en
| 0.867257 |
http://gmatclub.com/forum/ps-half-circle-22425.html
| 1,485,134,242,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560281659.81/warc/CC-MAIN-20170116095121-00575-ip-10-171-10-70.ec2.internal.warc.gz
| 121,283,483 | 45,873 |
PS # half circle : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 22 Jan 2017, 17:17
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# PS # half circle
Author Message
SVP
Joined: 28 May 2005
Posts: 1723
Location: Dhaka
Followers: 7
Kudos [?]: 327 [0], given: 0
### Show Tags
02 Nov 2005, 13:26
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 1 sessions
### HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Attachments
half circle.jpg [ 9.14 KiB | Viewed 1163 times ]
_________________
hey ya......
Manager
Joined: 20 Mar 2005
Posts: 201
Location: Colombia, South America
Followers: 1
Kudos [?]: 16 [0], given: 0
### Show Tags
02 Nov 2005, 13:39
it looks like it is sqrt(3) as the angle POQ is 90 degrees and therefore it is a reflection.
another way to think about it
what is the result of rotating a radius 90 degrees along the circumference.
in that case what is the effect in the cosine?
make sense?
Current Student
Joined: 28 Dec 2004
Posts: 3384
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 15
Kudos [?]: 282 [0], given: 2
### Show Tags
02 Nov 2005, 13:49
can you repost the question..not sure what we are asking for s, t?
Manager
Joined: 04 Oct 2005
Posts: 246
Followers: 1
Kudos [?]: 75 [0], given: 0
### Show Tags
02 Nov 2005, 13:56
I get 2sqrt2 - sqrt3... which is neither of the choices...
We know that the radii must be the same for both. We have two triangles...
one with legs of length 1 and sqrt3.... a similar triangle, thus the 3rd length, or the radius, must be 2.
The second triangle with the 90┬░ angle is then a 45-45-90. So the line connecting points P and Q should have a length of sqrt2.
Deduct the length of the x coordinate from the first point, gives you the x coordinate of the second.
:EDIT:
just realized, that this only works when the line PQ is parallel to the x axis... which must not be the case.
Intern
Joined: 22 Mar 2005
Posts: 39
Location: Houston
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
02 Nov 2005, 17:29
Wow, this is a great one. Don't let the diagram fool you here.
First, as was previously mentioned you can see that the radius is 2.
Now calculate the distance between the two points by drawing a line between them. Obviously the distance between them is equal to the hypotenuse of the triangle with the two radii as its legs.
So distance between them = sqrt(2^2+2^2) = sqrt(8)
Now the distance formula between to points says that:
DF = (s+sqrt(3))^2+(t-1)^2 had better equal sqrt(8)^2 = 8 as well.
We need one more equation that relates s and t and we are ready to go.
That equation of course is that s^2+t^2 = 2^2 (eqn of the circle)
Now solve for t^2 = 4 - s^2 and t = sqrt(4-s^2)
Now go back to the distance formula above (DF), multipy it out and then plug in the expressions above for t^2 and t. Now solve for s and you will see that s = 1 and t = sqrt(3).
So the drawing is playing a little trick as the y axis isn't really bisecting the 90 deg angle. As a quick check if it were the other way around and s = sqrt(3) and y = 1, then the distance between the points would simply be sqrt(3) + sqrt(3) = 2sqrt(3) which does not equal sqrt(8) (or 2sqrt(2)).
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19
Kudos [?]: 292 [0], given: 0
### Show Tags
02 Nov 2005, 21:23
first of all, we should find the function of the two lines OP and OQ
from the coordinates provided we can easily find:
OP: y=-1/sqrt3 *x
OQ: y= t/s*x
Becoz OP and OQ are perpendicular --> t/s* ( -1/sqrt3)= -1
---> t/s= sqrt3--> t= s*sqrt3-->t^2= 3*s^2
We have:
--> s^2+3*s^2= 4 ---> s=- or + 1
becoz Q belongs to the II portion of the coordination --> s=1
Last edited by laxieqv on 02 Nov 2005, 21:30, edited 1 time in total.
Current Student
Joined: 29 Jan 2005
Posts: 5238
Followers: 25
Kudos [?]: 378 [0], given: 0
### Show Tags
02 Nov 2005, 21:30
S = sqrt3 by "eyeballing" and under 10 seconds
OA?
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19
Kudos [?]: 292 [0], given: 0
### Show Tags
02 Nov 2005, 21:34
first of all, we should find the function of the two lines OP and OQ
from the coordinates provided we can easily find:
OP: y=-1/sqrt3 *x
OQ: y= t/s*x
Becoz OP and OQ are perpendicular --> t/s* ( -1/sqrt3)= -1
---> t/s= sqrt3--> t= s*sqrt3-->t^2= 3*s^2
We have:
--> s^2+3*s^2= 4 ---> s=- or + 1
becoz Q belongs to the II portion of the coordination --> s=1
SVP
Joined: 28 May 2005
Posts: 1723
Location: Dhaka
Followers: 7
Kudos [?]: 327 [0], given: 0
### Show Tags
03 Nov 2005, 11:29
good job guys... OA is 1.
how long did it take you guys to do it... for me it took over 4 minutes... is it okay?
_________________
hey ya......
03 Nov 2005, 11:29
Display posts from previous: Sort by
| 1,707 | 5,416 |
{"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}
| 3.546875 | 4 |
CC-MAIN-2017-04
|
latest
|
en
| 0.866644 |
http://gmatclub.com/forum/at-a-certain-health-club-30-of-the-members-use-both-the-32273.html?fl=similar
| 1,485,280,106,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560285001.96/warc/CC-MAIN-20170116095125-00122-ip-10-171-10-70.ec2.internal.warc.gz
| 116,689,371 | 39,244 |
At a certain health club, 30% of the members use both the : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 24 Jan 2017, 09:48
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# At a certain health club, 30% of the members use both the
Author Message
Intern
Joined: 17 Jul 2006
Posts: 28
Followers: 0
Kudos [?]: 0 [0], given: 0
At a certain health club, 30% of the members use both the [#permalink]
### Show Tags
22 Jul 2006, 10:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
At a certain health club, 30% of the members use both the pool and sauna, but 40% of the members who use the pool do not use the sauna. What percent of the members of the health club use the pool?
Senior Manager
Joined: 22 May 2006
Posts: 371
Location: Rancho Palos Verdes
Followers: 1
Kudos [?]: 110 [0], given: 0
### Show Tags
22 Jul 2006, 11:06
cejismundo wrote:
At a certain health club, 30% of the members use both the pool and sauna, but 40% of the members who use the pool do not use the sauna. What percent of the members of the health club use the pool?
Let total = T, use pool = P, use sauna = S
P&S : 0.3T
40% of the members who use the pool do not use the sauna means that "60% of the members who use pool use the sauna."
0.6P = 0.3T
P = 1/2T
thus, 50% use the pool
_________________
The only thing that matters is what you believe.
Director
Joined: 28 Dec 2005
Posts: 755
Followers: 1
Kudos [?]: 14 [0], given: 0
### Show Tags
22 Jul 2006, 16:17
cejismundo wrote:
At a certain health club, 30% of the members use both the pool and sauna, but 40% of the members who use the pool do not use the sauna. What percent of the members of the health club use the pool?
Damn, took me too long to solve this, I am so out of practice.
The problem is actually pretty simple.
Let total number of members be X, set of members using the pool be P, and set of members using the sauna be S.
Then PS = 0.3X
PS=0.6P
P=PS/0.6=0.3X/0.6=X/2
Therefore 1/2 the members use the pool, as a percentage its 50%
Re: percentage problem [#permalink] 22 Jul 2006, 16:17
Display posts from previous: Sort by
| 784 | 2,767 |
{"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}
| 4.03125 | 4 |
CC-MAIN-2017-04
|
latest
|
en
| 0.899025 |
https://www.educationquizzes.com/us/middle-school-6th-7th-and-8th-grade/math/business-math-02-calculating-interest-per-year-continued-practice/
| 1,719,015,820,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00677.warc.gz
| 655,354,647 | 10,514 |
US UKIndia
Every Question Helps You Learn
Leading Streak Today
Your Streak Today
Leading Streak Today
Your Streak Today
Calculating interest usually involves money!
# Business Math 02 - Calculating Interest Per Year - Continued Practice
This Math quiz is called 'Business Math 02 - Calculating Interest Per Year - Continued Practice' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14.
It costs only \$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us
Calculation of interest is all around us: savings accounts, car loans, mortgages, credit cards, student loans... Because interest will always be with you, it is important for you to really grasp how it will play a part in your everyday life. In the Business Math quiz entitled “Calculating Interest Per Year”, you revisited the basic steps of how to calculate simple yearly interest. This quiz will continue that practice exercise. By now, you should be able to calculate these pretty quickly but don’t go too fast. Remember to find the right answer you will need to follow the simple formula:
1.
\$216.49 at 12.5% for 3 years.
\$81.61
\$81.81
\$81.11
\$81.18
Working the problem:
\$216.49 at 12.5% for 3 years.
I = PRT
216.49 x 0.125 x 3
216.49 x 0.125 = 27.06
27.06 x 3 = 81.18
I = \$81.18
Answer (d) is the correct answer
2.
\$10.00 at 26% for 2 years.
\$5.20
\$15.20
\$52.00
\$2.50
Working the problem:
\$10.00 at 26% for 2 years.
I = PRT
10 x 0.26 x 2
10 x 0.26 = 2.60
2.60 x 2 = 5.20
I = \$5.20
Answer (a) is the correct answer
3.
\$39,333.33 at 6.23% for 17 years.
\$14,657.99
\$41,657.99
\$16,457.99
\$46,157.99
Working the problem:
\$39,333.33 at 6.23% for 17 years.
I = PRT
39,333.33 x 0.0623 x 17
39,333.33 x 0.0623 = 2,450.47
2,450.47 x 17 = 41,657.99
I = \$41,657.99
Answer (b) is the correct answer
4.
\$900.00 at 3.85% for 10 years.
\$365.40
\$356.40
\$346.50
\$364.60
Working the problem:
\$900.00 at 3.85% for 10 years.
I = PRT
900 x 0.0385 x 10
900 x 0.0385 = 34.65
34.65 x 10 = 346.50
I = \$346.50
Answer (c) is the correct answer
5.
\$5,000.00 at 9% for 15 years.
\$7,650.00
\$67,500.00
\$6,750.00
\$76,500.00
Working the problem:
\$5,000.00 at 9% for 15 years.
I = PRT
5,000 x 0.09 x 15
5,000 x 0.09 = 450
450 x 15 = 6,750
I = \$6,750.00
Answer (c) is the correct answer
6.
\$3,214.00 at 5.65% for 12 years.
\$2,179.09
\$21,790.90
\$1,297.09
\$1,297.90
Working the problem:
\$3,21400 at 5.65% for 12 years.
I = PRT
3214 x 0.0565 x 12
3214 x 0.0565 = 181.59
181.59 x 12 = 2,179.09
I = \$2,179.09
Answer (a) is the correct answer
7.
\$34,900,500.00 at 2.97% for 30 years.
\$310,963,440.00
\$310,963.44
\$3,109,634.40
\$31,096,344.00
Working the problem:
\$34,900,500.00 at 2.97% for 30 years.
I = PRT
34,900,500 x 0.0297 x 30
34,900,500 x 0.0297 = 1,036,544.80
1,036,544.80 x 30 = 31,096,344
I = \$31,096,344.00
Answer (d) is the correct answer
8.
\$12,250.00 at 5.465% for 5 years.
\$33,473.00
\$3,347.30
\$334.73
\$34,373.00
Working the problem:
\$12,250.00 at 5.465% for 5 years.
I = PRT
12,250 x 0.05465 x 5
12,250 x 0.05465 = 669.46
669.46 x 5 = 3,347.30
I = \$3,347.30
Answer (b) is the correct answer
9.
\$2,950,000.00 at 4.05% for 28 years.
\$345,350.00
\$3,435,300.00
\$334,530.00
\$3,345,300.00
Working the problem:
\$2,950,000.00 at 4.05% for 28 years.
I = PRT
2,950,000 x 0.0405 x 28
2,950,000 x 0.0405 = 119,475
119,475 x 28 = 3,345,300
I = \$3,345,300.00
Answer (d) is the correct answer
10.
\$2,450.19 at 2.37% for 7 years.
\$460.49
\$406.49
\$640.49
\$46.49
Working the problem:
\$2,450.19 at 2.37% for 7 years.
I = PRT
2,450.19 x 0.0237 x 7
2,450.19 x 0.0237 = 58.07
58.07 x 7 = 406.49
I = \$406.49
Answer (b) is the correct answer
Author: Christine G. Broome
© Copyright 2016-2024 - Education Quizzes
Work Innovate Ltd - Design | Development | Marketing
| 1,603 | 3,994 |
{"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}
| 4.75 | 5 |
CC-MAIN-2024-26
|
latest
|
en
| 0.907299 |
http://125-problems.univ-mlv.fr/problem34.php
| 1,695,554,815,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233506632.31/warc/CC-MAIN-20230924091344-20230924121344-00233.warc.gz
| 48,286 | 6,247 |
# Problem 34: Worst Case of the Boyer-Moore Algorithm
$$\def\lcs{\mathit{lcs}} \def\per{\mathit{per}} \def\tbsuff{\mathit{good\text{-}suff}} \def\dd{\dot{}\dot{}}$$
a Boyer-Moore string matching is based on a technique that leads to the fastest searching algorithms for fixed patterns. Its main feature is to scan the pattern backward when aligned with a factor of the searched text. A typical pattern preprocessing is shown in Problem 33.
When locating a fixed pattern $x$ of length $m$ in a text $y$ of length $n$, in the generic situation $x$ is aligned with a factor (the window) of $y$ ending at position $j$ (see picture). The algorithm computes the longest common suffix ($\lcs$) between $x$ and the factor of $y$; after possibly reporting an occurrence, it slides the window towards the end of $y$ based on the preprocessing and on information collected during the scan, without missing an occurrence of $x$. Algorithm BM implements the method with table $\tbsuff$ of Problem 33.
BM$(x,y \mbox{ non-empty words}, m,n \mbox{ their lengths})$
\begin{algorithmic}
\STATE $j\leftarrow m-1$
\WHILE{$j \lt n$}
\STATE $i\leftarrow m-1-|lcs(x,y[j-m+1..j])|$
\IF{$i \lt 0$}
\STATE report an occurrence of $x$ at position $j-m+1$ on $y$
\STATE $j\leftarrow j+per(x)$
\COMMENT{$per(x)=good\text{-}suff[0]$}
\ELSE
\STATE $j\leftarrow j+good\text{-}suff[i]$
\ENDIF
\ENDWHILE
\end{algorithmic}
After position $j$ on $y$ is treated, if an occurrence of $x$ is found the algorithm slides naturally the window at distance $\per(x)$. If no occurrence is found, the distance $\tbsuff[i]$ depends on the factor $bu$ of $y$ (it depends on $au$ of $x$ in Problem 33). Value $\per(x)$ and array $\tbsuff$ are preprocessed before the search.
Give examples of a non-periodic pattern and of a text $y$ for which Algorithm BM performs close to $3|y|$ letter comparisons at line 3 for computing the longest common suffix.
## References
• R. S. Boyer and J. S. Moore. A fast string searching algorithm. Commun. ACM., 20(10):762-772, 1977.
• R. Cole. Tight bounds on the complexity of the Boyer-Moore string matching algorithm. SIAM J. Comput., 23(5):1075-1091, 1994.
• M. Crochemore, C. Hancart, and T. Lecroq. Algorithms on Strings. Cambridge University Press, 2007. 392 pages.
• M. Crochemore and W. Rytter. Text algorithms. Oxford University Press, 1994. 412 pages.
• M. Crochemore and W. Rytter. Jewels of Stringology. World Scientific Publishing, Hong-Kong, 2002. 310 pages.
• D. Gusfield. Algorithms on Strings, Trees and Sequences: Computer Science and Computational Biology. Cambridge University Press, Cambridge, 1997.
• D. E. Knuth, J. H. Morris Jr., and V. R. Pratt. Fast pattern matching in strings. SIAM J. Comput., 6(2):323-350, 1977.
• B. Smyth. Computing Patterns in Strings. Pearson Education Limited, Harlow, England, 2003. 423 pages.
| 848 | 2,844 |
{"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": 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}
| 3.625 | 4 |
CC-MAIN-2023-40
|
longest
|
en
| 0.755688 |
https://math.stackexchange.com/questions/83046/maximum-of-the-variance-function-for-given-set-of-bounded-numbers
| 1,701,936,837,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100650.21/warc/CC-MAIN-20231207054219-20231207084219-00008.warc.gz
| 430,923,761 | 39,965 |
# Maximum of the Variance Function for Given Set of Bounded Numbers
Let $$\boldsymbol{x}$$ be a vector of $$n$$ numbers in the range $$\left[0, c \right]$$, where $$c$$ is a positive real number.
What's is the maximum of the variance function of this $$n$$ numbers?
Maximum in the meaning what spread of the number will maximize the variance?
What would be a tighter bound for other assumptions on the spread of the numbers.
The variance of the vector $$\boldsymbol{x}$$ is given by:
$$\operatorname{var} (\boldsymbol{x}) = \frac{1}{n} \sum_{i = 1}^{n} {\left( {x}_{i} - \overline{\mathbf{x}} \right )}^2$$
Where the mean $$\overline{\boldsymbol{x}}$$ is given by:
$$\overline{\boldsymbol{x}} = \frac{1}{n} \sum_{i = 1}^{n} {x}_{i}$$
• What is the variance function of a set? The maximum with respect to what? Please try to rephrase your question using more standard terminology. Nov 17, 2011 at 13:44
• If $X$ takes on values in an interval of length $c$, then its variance is bounded above by $c^2/4$ with equality when half the probability mass is at one end of the interval and half at the other (assuming the interval is closed at both ends; else we have strict inequality) Nov 17, 2011 at 13:47
• @DilipSarwate, Does it work just like Entropy? What if I have only 3 samples in the range [0 1], which variance could they achieve at most? It seems you neglected the factor of how many samples there are.
– Royi
Nov 17, 2011 at 14:24
• @joriki, I've updated the question. Thank You.
– Royi
Nov 17, 2011 at 14:25
• "What if I have only 3 samples in the range [0 1], which variance could they achieve at most?" The maximum is still bounded above by $c^2/4$ which is $1/4$ in this instance. The bound cannot be attained with equality in this instance but that does not invalidate the bound. For odd $n$ and $c = 1$, putting one point at the center (mean) and the rest at the end points gives $\frac{1}{4}\times \frac{n-1}{n}$ which, when $n$ is large, is close enough to the upper bound of $\frac{1}{4}$for gummint purposes. Nov 17, 2011 at 15:50
Since $x_i \leq c$, $\displaystyle \sum_i x_i^2 = \sum_i x_i\cdot x_i \leq \sum_i c\cdot x_i = cn\bar{x}.$ Note also that $0 \leq \bar{x} \leq c$. Then, \begin{align*} n\cdot \text{var}(\mathbf{x}) &= \sum_i (x_i - \bar{x})^2= \sum_i x_i^2 - 2x_i\bar{x} + \bar{x}^2\\ &= \sum_i x_i^2 - 2\bar{x}\sum_i x_i + n\bar{x}^2= \sum_i x_i^2 - n\bar{x}^2\\ &\leq cn\bar{x} - n\bar{x}^2 = n\bar{x}(c-\bar{x}) \end{align*} and thus $$\text{var}(\mathbf{x}) \leq \bar{x}(c-\bar{x}) \leq \frac{c^2}{4}.$$
Added note: (second edit)
The result $\text{var}(X) \leq \frac{c^2}{4}$ also applies to random variables taking on values in $[0,c]$, and, as my first comment on the question says, putting half the mass at $0$ and the other half at $c$ gives the maximal variance of $c^2/4$. For the vector $\mathbf x$, if $n$ is even, the maximal variance $c^2/4$ occurs when $n/2$ of the $x_i$ have value $0$ and the rest have value $c$. Someone else posted an answer -- it has since been deleted -- which said the same thing and added that if $n$ is odd, the variance is maximized when $(n+1)/2$ of the $x_i$ have value $0$ and $(n-1)/2$ have value $c$, or vice versa. This gives a variance of $(c^2/4)\cdot(n^2-1)/n^2$ which is slightly smaller than $c^2/4$. Putting the "extra" point at $c/2$ instead of at an endpoint gives a slightly smaller variance of $(c^2/4)\cdot(n-1)/n$, but both choices have variance approaching $c^2/4$ asymptotically as $n \to \infty$.
• Could you explain the last part of your derivation? I don't understand where $\bar{x}(c - \bar{x}) \lt \frac{c^2}{4}$ comes from. Feb 18, 2017 at 15:08
• I had a google and found a paper called "A Better Bound on the Variance" by Rajendra Bhatia and Chandler Davis which seemed helpful but it relied on the assumption that $\forall x\in \mathbb{R}$, $(M - m)^2 \geq 4(M -x)(x - m)$ where $M \geq m$ and [m, M] are the bounds individual values used to compute the arithmetic mean. That assumption would help here too but I don't understand where it comes from. Feb 18, 2017 at 15:14
• $\bar x \in [0,c]$ and $\bar{x}(c-\bar{x})$ has maximum value $\frac{c^2}{4}$ when $\bar{x}=\frac c2$. With regard to the alleged better bound, if the range is $[m,M]$ instead of $[0,c]$, the variance works out to be $(M-\bar{x})(\bar{x}-m)$ and this is claimed to be smaller than $\frac{(M-m)^2}{4}$. Pardon me if I fail to be greatly impressed by the better bound that Google found for you. Feb 18, 2017 at 19:42
• @delcypher Hi, you can treat x bar as the variable, and use completing the square to find the maximum of the univariate quadratic polynomial Mar 19, 2018 at 22:02
An easier derivation can be done as follows (From [1]):
For any constant $$c$$, we have, $$$$E[(X-c)^2] = E[X^2] - 2E[X]c + c^2$$$$ The above quadratic is minimized when $$c=E[X]$$. It follows that, $$$$\sigma^2 = E[(X-E[X])^2] \leq E[(X-c)^2], \text{for all } c$$$$ By letting $$c = (a+b)/2$$, we obtain, $$$$\sigma^2 \leq E\left[\left(X-\frac{a+b}{2}\right)^2\right] = E[(X-a)(X-b)] + \frac{(b-a)^2}{4} \leq \frac{(b-a)^2}{4}$$$$
since for $$x$$ in $$[a,b]$$, $$(x-a)(x-b)<0$$
Further, the bound could be very conservative. However, in the absence of any other information about $$X$$, it can not be improved.
[1] D. P. Bertsekas and J. N. Tsitsiklis, Introduction to Probability. Athena Scientific, 2002.
| 1,770 | 5,378 |
{"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": 19, "wp-katex-eq": 0, "align": 0, "equation": 3, "x-ck12": 0, "texerror": 0}
| 3.921875 | 4 |
CC-MAIN-2023-50
|
longest
|
en
| 0.904361 |
https://math.answers.com/basic-math/How_fractions_are_equivalent_to_1_3
| 1,669,621,320,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710488.2/warc/CC-MAIN-20221128070816-20221128100816-00371.warc.gz
| 434,482,120 | 47,524 |
0
# How fractions are equivalent to 1 3?
Wiki User
2012-08-23 08:43:28
2/6, 3/9, 4/12, 5/15,...
To get the equivalent fractions of 1/3:
Multiply 1/3 by 2/2, 3/3, 4/4, 5/5,...
1/3 * 2/2 = 2/6
1/3 * 3/3 = 3/9
1/3 * 4/4 = 4/12
1/3 * 5/5 = 5/15
Equivalent fractions of 1/3 = 2/6, 3/9, 4/12, 5/15,...
Wiki User
2012-08-23 08:43:28
Study guides
96 cards
➡️
See all cards
4.04
104 Reviews
| 206 | 396 |
{"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}
| 3.796875 | 4 |
CC-MAIN-2022-49
|
latest
|
en
| 0.757141 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=8543
| 1,369,078,173,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368699201808/warc/CC-MAIN-20130516101321-00026-ip-10-60-113-184.ec2.internal.warc.gz
| 314,457,279 | 784 |
```Question 17686
Hi
Here's the answer:
Consider n be the 1st integer.
The next in the series is n+2.
then the next will be n+4.
thus n+n+2+n+4= 639
3n+6=639
3n=633
n=633/3=211
the integers are 211, 213, 215:)
```
| 88 | 215 |
{"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}
| 3.671875 | 4 |
CC-MAIN-2013-20
|
latest
|
en
| 0.860358 |
http://web.eecs.utk.edu/~jplank/plank/classes/cs202/Notes/Trees/index.html
| 1,686,084,740,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224653071.58/warc/CC-MAIN-20230606182640-20230606212640-00343.warc.gz
| 55,177,938 | 14,456 |
## CS202 Lecture Notes - Basic Trees & Binary Search Trees
• James S. Plank
• Directory: /home/jplank/cs202/Notes/Trees
• Original notes: 2005.
• Last Revision: Tue Apr 11 00:26:13 EDT 2023
Topcoder problems that are good for trees:
A tree is a basic linked data structure that is richer than a list. For tree basics, please read the following parts of the Wikipedia page on trees:
• The first sentence. Maybe. You can just look at the picture to the right, and then skip straight to the Terminology section. Sometimes it seems like the people who write wikipedia pages have nothing to do but try to confuse their readers with pedantic definitions for the sake of completeness. Clearly, they do not teach CS202 or its ilk, or if they do, I'd try to take it from another university. I guess you get what you pay for.
```- Node
- Edge
- Binary
- Complete
- Parent
- Child
- Ancestor
- Descendent
- Root
- Leaf
- Internal
- External
- Height of node or tree
- Depth of node
- Degree
```
• The Traversal Methods section.
Trees are often a very natural way to represent hierarchies. For example, the mathematical expression:
(5a + 10b) / (14a4 - 28b)
is very clearly represented by the following tree:
Using tree terminology, the external, (or leaf) nodes are either variables or values, and the internal nodes are operators.
### Tree Traversals
We typically talk about three tree traversal methods:
### Pre-Order
To visit a node:
• Perform an action (like Print)
• Recursively visit children in order.
Example on the above tree:
Print "/"
Print "+"
Print "*"
Print "5"
Print "a"
Print "*"
Print "10"
Print "b"
Print "-"
Print "*"
Print "14"
Print "exp"
Print "a"
Print "4"
Print "*"
Print "28"
Print "b"
### Post-Order
To visit a node:
• Recursively visit children in order.
• Perform an action
Example on the above tree:
Print "5"
Print "a"
Print "*"
Print "10"
Print "b"
Print "*"
Print "+"
Print "14"
Print "a"
Print "4"
Print "exp"
Print "*"
Print "28"
Print "b"
Print "*"
Print "-"
Print "/"
### In-Order (Binary only)
To visit a node:
• Recursively visit left child.
• Perform an action
• Recursively visit right child.
Example on the above tree:
Print "5"
Print "*"
Print "a"
Print "+"
Print "10"
Print "*"
Print "b"
Print "/"
Print "14"
Print "*"
Print "a"
Print "exp"
Print "4"
Print "-"
Print "28"
Print "*"
Print "b"
Example Question: "Which of the above traversals would allow you to evaluate the expression of the above tree?"
Answer: A post-order traversal -- for each operator in an internal node, you need to first evaluate each of its children before you can perform the operation. For example, you need to evaluate the left subtree of "/" and then the right subtree before you divide the two answers to get the final evaluation of the expression.
In class, I'll go over http://web.eecs.utk.edu/~jplank/topcoder-writeups/Leetcode/Minimum-Fuel-Cost-Capital/index.html, which is a Leetcode problem featuring a postorder traversal, and http://web.eecs.utk.edu/~jplank/topcoder-writeups/Leetcode/Sum-Root-To-Leaf-Numbers/index.html, which is a Leetcode problem featuring a preorder traversal.
### Binary Search Trees
Binary search trees are an exceptionally important type of tree. As their name implies, they are binary trees, where each node has a value, a left child and a right child. The important property of a binary search tree is the following:
• If a node has a left child, then the left child is the root of a binary search tree whose maximum value is less than or equal to the node's value.
• If a node has a right child, then the right child is the root of a binary search tree whose minimum value is greater than or equal to the node's value.
Here are some examples of binary search trees that hold strings:
As demonstrated by the second two trees above, there can be more than one binary search tree that corresponds to the same data.
Here are two examples of trees that are not binary search trees:
Daisy, Luigi and Luther are all greater than Binky. The tree is not binary.
Binary search trees have nice properties. For example, you can sort the data by performing an in-order traversal. You can also find a piece of data simply by traversing a single path from the root to the data, or to where the data would be. For example, in the tree:
If I want to find Luigi, what I do is start at the root of the tree, and compare it to Luigi. If it equals Luigi, then I'm done. If not, and Luigi is less than the node's value, then I recursively continue the process on the tree's left child. If Luigi is greater than the node's value, then I instead recursively continue the process on the tree's right child. If I ever get to a point where the node has no child for me to search on, I can conclude that the value is not in the tree.
Continuing the example of finding Luigi, I would:
• I start at Binky. Since Luigi is greater than Binky, I'll continue searching on Binky's right child.
• I'm next at Fred. Again, Luigi is greater than Fred, so I continue searching on Fred's right child.
• Now I'm at Luther. Luigi is less than Luther, so I continue searching on Luther's left child.
• I have found Luigi.
Similarly, suppose I try to find Calista:
• I start at Binky. Since Calista is greater than Binky, I'll continue searching on Binky's right child.
• I'm next at Fred. Calista is less than Fred, so I continue searching on Fred's left child.
• Now I'm at Daisy. Calista is less than Daisy, so I continue searching on Daisy's left child.
• However, Daisy has no left child. I can conclude that Calista is not in the search tree.
### Binary Search Tree Operations: Find, Insert, Delete
I have just described how to find a value in a binary search tree. Insertion is a pretty simple matter too. To insert a value s, assume that s is not in the tree, and find where s should be. Create the node and put it there.
For example, suppose that we want to insert Calista into the above tree. We try to find the string as we did above. We fail at Daisy's left child, which doesn't exist. Therefore, we create a node for Calista and insert it as Daisy's left child:
To insert duplicate values, do the same procedure, only if you find the value, you continue searching either on the left or the right child, as if you didn't find the key. For example, if you wanted to insert Binky into the tree again, you would either put it as Binky's left child or Calista's left child:
Deletion is the trickiest. To delete a node, you must consider three cases. Let's consider the tree below as an example:
Case 1: The node has no children (it's a leaf node). You can simply delete it. I won't draw an example, but you should see very easily that deleting Calista, Luigi or Waluigi just removes them from the tree.
Case 2: The node has just one child. To delete the node, replace it with that child. I draw two examples below:
After deleting Binky. After deleting Daisy.
Case 3: The node has two children. In this case, you find the node in the tree whose value is the greatest value less than (or equal to) the node's value. That will be the rightmost node in the subtree rooted by the left child. That node will not have a right child. First, delete it. Then use it to replace the node that you are deleting.
Alternatively, you can replace it with the leftmost node in the tree rooted by the node's right child. Both will work.
For example, let's delete Fred from the tree below:
Since Fred has two children, we find the rightmost node in the tree rooted by Fred's left child. That is the node Fiona. We first delete Fiona:
And then we replace Fred with Fiona:
A second example is easier, but sometimes confusing -- suppose we want to delete Luther. Since Luther has two children, we find the rightmost node in the tree rooted by Luther's left child. Since there is only one node in that tree, that's the one we delete: Luigi. We then replace Luther with Luigi:
### Implementation -- API and overview
The file include/bstree.hpp contains a simple binary search tree API. Scan it quickly, but don't go into to much detail. I'll do that with you after the specification:
```#include #include namespace CS202 { /* These are the nodes of a tree. The keys are strings, and the vals are generic pointers. Please see the lecture notes for a more thorough explanation of what a (void *) is. */ class BSTNode { friend class BSTree; protected: BSTNode *left; BSTNode *right; BSTNode *parent; std::string key; void *val; }; class BSTree { public: /* Constructor, copy constructor, assignment overlead, destructor. I am only implementing the constructor and destructor here in the notes. You will implement the other two in your lab. */ BSTree(); BSTree(const BSTree &t); BSTree& operator= (const BSTree &t); ~BSTree(); void Clear(); // Turns the tree into an empty tree. void Print() const; // These are obvious. size_t Size() const; bool Empty() const; bool Insert(const std::string &key, void *val); // Insert the key and val. Returns success (duplicates are not allowed. void *Find(const std::string &key) const; // Return the val associated with the key. Returns NULL if key not found. bool Delete(const std::string &key); // Delete the node with the key. Returns whether there was such a node. std::vector Ordered_Keys() const; // Return a vector of sorted keys std::vector Ordered_Vals() const; // Return a vector of the vals, sorted by the keys. /* You'll write these as part of your lab. */ int Depth(const std::string &key) const; // Distance from a node to the root. Returns -1 if the key is not in the tree. int Height() const; // Returns the depth of the node with maximum depth, plus one. protected: BSTNode *sentinel; // Like the dlists, there is a sentinel. Its right points to the root. size_t size; // Size of the tree void recursive_inorder_print(int level, const BSTNode *n) const; // A helper for Print() void recursive_destroy(BSTNode *n); // A helper for Clear() void make_val_vector(const BSTNode *n, std::vector &v) const; // A helper for Ordered_Vals() /* You'll write these as part of your lab. */ int recursive_find_height(const BSTNode *n) const; // A helper for Height() void make_key_vector(const BSTNode *n, std::vector &v) const; // A helper for Ordered_Keys() BSTNode *make_balanced_tree(const std::vector &sorted_keys, // A helper for the copy constructor and assignment overload. const std::vector &vals, size_t first_index, size_t num_indices) const; }; }; ```
First off, you'll note I'm using a "namespace". It's about time that I introduce you to namespaces. They are pretty simple -- when you define a class (or struct), global variable, or procedure within a namespace, then when someone else wants to use it, they need to either:
• State that they are using the namespace.
• State that they are using the class, variable or procedure from the namespace.
• Prepend the namespace and two colons whenever they use a class, variable or procedure.
In this instance, I'm defining the classes BSTNode and BSTree within the namespace CS202. If I want to use, for example, the BSTree class, then I must do one of the following:
• Put "using namespace CS202" at the top of my code.
• Put "using CS202::BSTree" at the top of my code.
• Use "CS202::BSTree" wherever I would normally just use "BSTree".
For example, in src/bstree_tester.cpp, I say "using CS202::BSTree", and then when I declare pointers to the variables, I say
```BSTree *t1, *t2, *tmp; ```
If I didn't put the "using" statement in there, I'd have to declare the pointers as:
```CS202::BSTree *t1, *t2, *tmp; ```
Namespaces are nice because they help you avoid conflicts in naming. I won't go crazy with them, but I do think they are something you should see an know about at this point in your programming careers.
Now, the BSTree API is a little like the Dlist API in the previous lecture, with some extra bells and whistles. Let's first go over the methods:
• The constructor, destructor, and methods Size() and Empty() are straightforward enough not to need explanation.
• Insert(key, val) inserts the given key and val pair into the tree. The val is of type (void *), which you probably haven't seen before. It is used extensively in C (and I use it a lot in C++ too, but I suspect I'm in the minority). It is basically a catch-all that stands for "any pointer type." You can store any pointer into the tree by "casting" it to a (void *). For example, let's suppose you have a pointer p that you have declared as:
```Person *p; ```
And suppose you want to insert the key "Fred" and the val p into a tree t. Then you would do it as follows:
```t.Insert("Fred", (void *) p); ```
The "(void *) p" part tells the compiler "I know you want a (void *) and I'm giving you a (Person *). It's all good." If you don't put the "(void *)" in, the compiler will yell at you.
I use the (void *) for flexibility -- it allows me to store any pointer in the val field. The STL does the same thing using templates, and it is superior to a (void *) in most respects. I will teach you templates later in the class. Using a (void *) is an "old school" C trick, which I'm using here for convenience.
Insert() returns whether the key was inserted. We're going to disallow inserting duplicate keys, so Insert() will return false if it gets a duplicate key.
• Find(key) returns the val associated with the key, if it's in the tree. The val is a (void *), and to use it, you cast it back to a pointer of the type that you inserted. For example, if you inserted a (Person *) with the Insert() command above, you'd use Find() as follows:
```p = (Person *) Find("Fred"); ```
If the key is not found, then Find() returns NULL. Does that mean that you can't store NULL in the tree? You tell me.
• Depth(key) returns the distance that a key is from the root of the tree. It returns -1 if the key is not in the tree.
• Height() returns the depth of the node with the maximum depth in the tree, plus 1. An empty tree has a height of zero. A tree with one node has a height of one.
• Delete(key) deletes the node with the given key. It returns true or false, depending on whether the key was in the tree.
• Print() prints all the keys by using a reverse inorder traversal. Each key is preceded by two spaces times the key's depth in the tree so that you can see the structure of the tree. I'll go over these calls in the examples below.
• Ordered_Keys() returns a sorted vector of the keys in the tree.
• Ordered_Vals() returns a vector of the vals in the tree, in the same order as Ordered_Keys().
• The copy constructor and assignment overload are special. Not only do they make a copy of the tree, but the copy will be balanced. In other words, the middle key will be the root of the tree. The middle key of the left subtree will be the key of the left child of the root, and the The middle key of the right subtree will be the key of the right child of the root. An so on. If a tree has an even number of elements, the "middle" element is (number-of-elements)/2. So, if a tree has four keys, the "middle" one is key #2 in Ordered_Keys(). Again, I'll go over examples below.
• There are helper methods in the protected part of the class. You'll see them in use below.
In this explanation, I do not implement Height() Depth(), Ordered_Keys(), the copy constructor or the assignment overload. Those tasks are for you in your lab.
### The testing program: Overview and simple commands
There is a testing program in bstree_test.cpp. It's a standard command line tool for managing a tree of Persons:
```class Person { public: string name; string phone; string ssn; void Print() const; }; ```
You run it with a prompt on the command line ("-" for no prompt), and it accepts lines of commands on standard input. You can see the commands when you enter a question mark:
```UNIX> echo '?' | bin/bstree_tester
usage: bstree_tester prompt(- for empty) -- commands on stdin.
commands:
INSERT name phone ssn - Insert the person into the tree.
FIND name - Find the person and print them out.
DELETE person - Delete the person.
PRINT - Print the keys using the Print() method.
EMPTY - Print whether the tree is empty.
SIZE - Print the tree's size.
HEIGHT - Print the tree's height.
DEPTH name - Print the depth of the node whose key is name (-1 if not there).
KEYS - Print the keys using the Ordered_Keys() method.
VALS - Print the vals using the Ordered_Vals() method.
PRINT_COPY - Call the copy constructor and call its Print() method.
REBALANCE - Turn the tree into a balanced tree by calling the assignment overload.
CLEAR - Clear the tree back to an empty tree.
DESTROY - Call the destructor and remake an empty tree.
QUIT - Quit.
? - Print commands.
UNIX>
```
Here I'll show some INSERT, FIND, DELETE, KEYS and VALS commands:
```UNIX> bin/bstree_tester 'BST>'
BST> INSERT Jim-Plank 865-123-4567 111-11-1111 # I'll insert three people
BST> INSERT Harvey-Plank 026-631-5520 826-96-9094
BST> INSERT Abba 462-055-3150 827-30-6292
BST> FIND Abba # Find returns the Person, which we print.
Abba 462-055-3150 827-30-6292
BST> KEYS # KEYS returns a vector of sorted keys, which we print
Abba
Harvey-Plank
Jim-Plank
BST> VALS # VALS returns the vals pointers, in the order of the keys.
Abba 462-055-3150 827-30-6292 # bstree_tester.cpp prints these vals.
Harvey-Plank 026-631-5520 826-96-9094
Jim-Plank 865-123-4567 111-11-1111
BST> FIND Fred # FIND returns NULL when you can't find something.
BST> DELETE Abba # Here, we delete "Abba".
BST> VALS
Harvey-Plank 026-631-5520 826-96-9094
Jim-Plank 865-123-4567 111-11-1111
BST> FIND Abba
BST> QUIT
UNIX>
```
Take a look at how the code that inserts people into the tree. Each line of input is put into the vector sv of the individual words on a line. This shows how we cast to a (void *) when we call Insert():
```/* With the INSERT command, we create a Person, and then insert it into the tree with the name as its key, and a pointer to the person (cast as a (void *)) as its val. */ } else if (sv[0] == "INSERT") { if (sv.size() != 4) { cout << "usage: INSERT name phone ssn" << endl; } else { p = new Person; p->name = sv[1]; p->phone = sv[2]; p->ssn = sv[3]; if (!t1->Insert(p->name, (void *) p)) { /* Here's where we cast to a (void *) */ cout << "Insert " << p->name << " failed." << endl; delete p; } } ```
And below, I show how Find() returns a (void *), but it is cast to a (Person *) for printing.
```/* The Find() method returns a (void *), so I must typecast it to a (Person *) */ } else if (sv[0] == "FIND") { if (sv.size() != 2) { cout << "usage: FIND key" << endl; } else { p = (Person *) t1->Find(sv[1]); /* Here is where we typecast the (void *) to a (Person *). */ if (p == NULL) { cout << "Not found.\n"; } else { p->Print(); } } ```
And finally, take a look at how src/bstree_tester.cpp gets all of the vals in the tree by calling Ordered_Vals(). This is an array of (void *)'s, so each of these must be typecast to a (Person *) to print:
```/* The VALS command calls Ordered_Vals() to get a vector of (void *)'s. We typecast each to a (Person *) and then print the person. */ } else if (sv[0] == "VALS") { vals = t1->Ordered_Vals(); for (i = 0; i < vals.size(); i++) { p = (Person *) vals[i]; p->Print(); } ```
### The Print() method and the PRINT command
The Print() method prints the tree with a special format. The tree is printed in a reverse inorder traversal, and each node of the tree is indented with 2d spaces, if the node's depth is d. Here's an example:
```UNIX> bin/bstree_tester '--------->'
---------> INSERT Binky 944-867-2246 165-79-8849
---------> INSERT Fred 026-631-5520 826-96-9094
---------> INSERT Luther 462-055-3150 827-30-6292
---------> INSERT Waluigi 193-149-4333 106-62-2934
---------> INSERT Daisy 257-554-8530 481-12-6340
---------> INSERT Luigi 018-992-9715 512-23-5507
---------> INSERT Ernie 808-602-6582 702-11-9340
---------> INSERT Calista 457-440-4397 076-91-9105
---------> PRINT # This is the same tree as above
Waluigi
Luther
Luigi
Fred
Ernie
Daisy
Calista
Binky
---------> INSERT Alvin 345-654-3434 242-55-4444
---------> PRINT # Alvin will be inserted as Binky's left child.
Waluigi
Luther
Luigi
Fred
Ernie
Daisy
Calista
Binky
Alvin
---------> DELETE Luigi # Luigi has no children, so deleting him is easy
---------> PRINT
Waluigi
Luther
Fred
Ernie
Daisy
Calista
Binky
Alvin
---------> DELETE Luther # Luther has one child, so we replace him with his child (Waluigi)
---------> PRINT
Waluigi
Fred
Ernie
Daisy
Calista
Binky
Alvin
---------> DELETE Fred # Fred has two children, so we find the node with the
---------> PRINT # greatest key in his left subtree, which is Ernie,
Waluigi # replace Fred's key and val with Ernie's key and val,
Ernie # and delete Ernie.
Daisy
Calista
Binky
Alvin
---------> DEPTH Binky # Binky is the root, which has a depth of 0
0
---------> DEPTH Daisy # Daisy is two edges from the root.
2
---------> DEPTH Calista # And Caliste is three.
3
---------> HEIGHT # Calista has the greatest depth in the tree, so the
4 # height of the tree is Calista's depth plus one.
---------> QUIT
UNIX>
```
### The implementation in src/bstree.cpp
The code that implements the BSTree class is split into two files:
1. src/bstree_notes.cpp implements a lot of the methods of the class.
2. src/bstree_lab.cpp has dummy implementations. You are to write these implementations in Lab A.
Ok, let's look at the interesting parts of the code is in src/bstree_notes.cpp. First, here's the definition of a node of the tree:
```class BSTNode { friend class BSTree; protected: BSTNode *left; BSTNode *right; BSTNode *parent; string key; void *val; }; ```
Besides storing a key and a val, each BSTNode contains a pointer to its left child, right child and parent. As with the Dlist, we are going to have a sentinel node that simplifies the code. The only part of the sentinel that we're going to use is its right pointer. That is going to point to the root of the tree. Rather than point to NULL, pointers that should point to nothing will point to the sentinel. Thus, when we create the following tree:
```UNIX> bstree_test -
INSERT Fred 026-631-5520 826-96-9094
INSERT Binky 944-867-2246 165-79-8849
INSERT Luigi 462-055-3150 827-30-6292
QUIT
UNIX>
```
It is going to have the following representation:
The constructor, size() and Empty() are straightforward as usual. The empty tree has sentinel->right point to sentinel:
```BSTree::BSTree() { sentinel = new BSTNode; sentinel->parent = NULL; sentinel->left = NULL; sentinel->right = sentinel; sentinel->key = "---SENTINEL---"; // This helps with debugging. sentinel->val = NULL; size = 0; } ``` ```bool BSTree::Size() const { return size; } ``` ```bool BSTree::Empty() const { return (size == 0); } ```
The implementation of Find() is a simple while loop that either finds the key or returns NULL when the sentinel has been reached:
```/* This is a standard search on a binary search tree. */ void *BSTree::Find(const string &key) const { BSTNode *n; n = sentinel->right; while (1) { if (n == sentinel) return NULL; if (key == n->key) return n->val; n = (key < n->key) ? n->left : n->right; } } ```
Insert() starts similarly to Find(). If it actually finds the key, then it returns false. Otherwise, the important thing is finding the parent of the newly created node. Once the parent is found, the new node is created, its left and right pointers are set to be the sentinel, and its parent pointer is set to be its parent. At this point, all of its pointers are correct -- the only thing that needs to be fixed is the parent -- the new node is either the parent's left or right child. Once that is figured out, the pointer is set, and we return true:
```bool BSTree::Insert(const string &key, void *val) { BSTNode *parent; BSTNode *n; parent = sentinel; n = sentinel->right; /* Find where the key should go. If you find the key, return false. */ while (n != sentinel) { if (n->key == key) return false; parent = n; n = (key < n->key) ? n->left : n->right; } /* At this point, parent is the node that will be the parent of the new node. Create the new node, and hook it in. */ n = new BSTNode; n->key = key; n->val = val; n->parent = parent; n->left = sentinel; n->right = sentinel; /* Use the correct pointer in the parent to point to the new node. */ if (parent == sentinel) { sentinel->right = n; } else if (key < parent->key) { parent->left = n; } else { parent->right = n; } /* Increment the size and return success. */ size++; return true; } ```
The tough code is Delete(). The first thing that we have to do is find the node. Once we do, we consider the three cases that are described above. If the node has no left child, then we replace the node with its right child. The way we "replace" the node is we set the parent's link to that node to equal the right child. Otherwise, if the node has no right child, then we replace it with its left child.
Those are the easy cases. The hard case is when the node has two children. In that case, we find the maximum node whose key is less than the node's key (There are no duplicates, which simplifies matters). This node is in the variable mlc. We recursively delete mlc, and then replace n's key and val with mlc's key and val:
```bool BSTree::Delete(const string &key) { BSTNode *n, *parent, *mlc; string tmpkey; void *tmpval; /* Try to find the key -- if you can't, return false. */ n = sentinel->right; while (n != sentinel && key != n->key) { n = (key < n->key) ? n->left : n->right; } if (n == sentinel) return false; /* We go through three cases for deletion. */ parent = n->parent; /* Case 1 - I have no left child. Replace me with my right child. Note that this handles the case of having no children, too. */ if (n->left == sentinel) { if (n == parent->left) { parent->left = n->right; } else { parent->right = n->right; } if (n->right != sentinel) n->right->parent = parent; delete n; size--; /* Case 2 - I have no right child. Replace me with my left child. */ } else if (n->right == sentinel) { if (n == parent->left) { parent->left = n->left; } else { parent->right = n->left; } n->left->parent = parent; delete n; size--; /* If I have two children, then find the node "before" me in the tree. That node will have no right child, so I can recursively delete it. When I'm done, I'll replace the key and val of n with the key and val of the deleted node. You'll note that the recursive call updates the size, so you don't have to do it here. */ } else { for (mlc = n->left; mlc->right != sentinel; mlc = mlc->right) ; tmpkey = mlc->key; tmpval = mlc->val; Delete(tmpkey); n->key = tmpkey; n->val = tmpval; } return true; } ```
We need to be really careful making that recursive call. If we set n->key equal to tmpkey before making the recursive call, we'd delete the wrong node. Also, since the recursive call deletes mlc, we can't use it following the recursive call -- this is why we stored mlc->key and mlc->val in tmpkey and tmpval. Finally, since the recursive call decreases the size, we don't do it here.
The last calls are the traversals. We do these recursively with protected methods (we don't want to let others make these calls -- they are just for us). Start with Print() -- it calls recursive_inorder_print() on the root of the tree (in sentinel->right). Recursive_inorder_print() does an in-order traversal in reverse order -- it calls itself recursively on its right child, then it prints the node, and finally it calls recursively on its left child. It always stops when it reaches the sentinel.
```/* Print() simply calls recursive_inorder_print() on the root node of the tree. */ void BSTree::Print() const { recursive_inorder_print(0, sentinel->right); } /* This does an inorder traversal in reverse order. The "Action" is printing "level" spaces, and then the key. You increment the level by two when you make recursive calls. */ void BSTree::recursive_inorder_print(int level, const BSTNode *n) const { if (n == sentinel) return; recursive_inorder_print(level+2, n->right); printf("%*s%s\n", level, "", n->key.c_str()); recursive_inorder_print(level+2, n->left); } ```
The other two recursive traversals are Ordered_Vals and Clear(). The first performs an in-order traversal, pushing vals onto a vector, while the second performs a post-order traversal, deleting nodes of the tree. The post-order traversal is necessary, because we can't use n->left or n->right after we've deleted n. Granted, we could store them in temporary pointers, delete n and then recursively delete the temporary pointers, but the post-order traversal is easier.
```/* This simply calls make_val_vector() on the root. That creates the vector rv, so return it. */ vector BSTree::Ordered_Vals() const { vector rv; make_val_vector(sentinel->right, rv); return rv; } /* This does an inorder traversal, which of course visits the nodes in sorted order of the keys. The "action" is pushing the val onto the vector. That means that the vals get pushed in the correct order. */ void BSTree::make_val_vector(const BSTNode *n, vector &v) const { if (n == sentinel) return; make_val_vector(n->left, v); v.push_back(n->val); make_val_vector(n->right, v); } /* Clear simply calls recursive_destroy on the root of the tree. That deletes all of the nodes but the sentinel. It then sets the root of the tree to the sentinel and the size to 0. */ void BSTree::Clear() { recursive_destroy(sentinel->right); sentinel->right = sentinel; size = 0; } /* Recursive destroy deletes all of the nodes of a tree. It does this with a postorder traversal -- deleting the children before deleting the node. */ void BSTree::recursive_destroy(BSTNode *n) { if (n == sentinel) return; recursive_destroy(n->left); recursive_destroy(n->right); delete n; } /* The destructor calls Clear(), which deletes all of the tree but the sentinel node. Therefore, it must also delete the sentinel node. */ BSTree::~BSTree() { Clear(); delete sentinel; } ```
A final comment on the destructor. Should it also delete the val's? The answer is no -- it's good form only to delete what you create with new. What if the user of this data structure didn't create the val's with new, or what if the user is holding them in a second data structure? Then it would be really bad form for the destructor to delete it!
### The Assignment Overload and Copy Constructor
Note: You are going to implement the assignment overload and copy constructor in the lab. The program below assumes that it is implemented correctly, as it is in the lab directory.
The assignment overload and copy constructor are necessary. The reason is that without them, you'd simply copy the size and the sentinel pointer, and now you'd have two trees pointing to the same sentinel. When one of them calls its destructor, then the second one will be pointing to deleted memory - a disaster!
With this data structure, we are making the assignment overload and copy constructor special. Not only do they make a copy of the tree, but they make the new copy balanced. The reason is that it will improve the performance of subsequent Insert(), Find() and Delete() operations. Let's take an example -- in src/bstree_tester.cpp, the PRINT_COPY command calls a procedure called print_copy() with the tree as a parameter. This will call the copy constructor, so that print_copy() gets a copy of the original tree. This copy will be balanced (the original tree will remain unchanged) -- read the inline comments:
```UNIX> bin/bstree_tester '--------->' # We insert 0 through 6 in order,
---------> INSERT 0 0 0 # which means that our tree is a big line.
---------> INSERT 1 0 0
---------> INSERT 2 0 0
---------> INSERT 3 0 0
---------> INSERT 4 0 0
---------> INSERT 5 0 0
---------> INSERT 6 0 0
---------> PRINT
6
5
4
3
2
1
0
---------> PRINT_COPY # PRINT_COPY calls the copy constructor,
6 # which creates a balanced version of the tree.
5 # Since 3 is the middle key, it becomes the root
4 # of the tree. Its left subtree has the values
3 # 0, 1 and 2. Since 1 is the middle of these,
2 # it is the root of the subtree. Similarly, the
1 # right subtree has the values 4, 5, and 6, so
0 # 5 is the root of the subtree.
---------> PRINT
6 # I'm doing this to show you that the original
5 # tree is unmodified. Only the copy was balanced.
4
3
2
1
0
---------> QUIT
UNIX>
```
The assignment overload also balances. In src/bstree_tester.cpp, the command "REBALANCE" calls the assignment overload and sets the main tree pointer to be this new tree. That way, it balances the tree. I'm going to show that below. Also, make note of the fact that when a tree has an even number of elements, then there are two potential "middle" elements. The code here chooses the higher of the two elements. I'll show that below. Your code will have to do this:
```UNIX> bin/bstree_tester '--------->'
---------> INSERT 0 0 0 # I make a big line again, this time with 8 elements
---------> INSERT 1 0 0
---------> INSERT 2 0 0
---------> INSERT 3 0 0
---------> INSERT 4 0 0
---------> INSERT 5 0 0
---------> INSERT 6 0 0
---------> INSERT 7 0 0
---------> PRINT
7
6
5
4
3
2
1
0
---------> REBALANCE
---------> PRINT
7
6
5
4 # Since there are 8 elements in the tree, there are two potential middles. We choose the larger: 4
3
2 # This subtree has four elements, so there are two potential middles. We choose the larger: 2
1 # This subtree has two elements, so there are two potential middles. We choose the larger: 1
0
---------> QUIT
UNIX>
```
Your lab is simple -- implement the following methods:
• Depth()
• Height()
• Ordered_Keys()
• The copy contstructor.
Some notes here:
• Height() should use the protected method recursive_find_height(), which you'll write.
• Ordered_Keys() should use the protected method make_key_vector(), which you'll write.
• The assignment overload should call Ordered_Keys() to get a vector of keys in sorted order, and Ordered_Vals() to get a vector of vals that correspond to the keys. It should then call make_balanced_tree() to create a subtree of the given region of the keys/vals. make_balanced_tree() should be recursive, and should work by creating the tree in a postorder manner.
I'll give an example. Suppose the keys are "0" through "6" as above. Then here are the calls to make_balanced_tree():
• The first call will be make_balanced_tree(keys, vals, 0, 7).
• That will call make_balanced_tree(keys, vals, 0, 3) and make_balanced_tree(keys, vals, 4, 3).
• make_balanced_tree(keys, vals, 0, 3) will call make_balanced_tree(keys, vals, 0, 1) and make_balanced_tree(keys, vals, 2, 1).
• make_balanced_tree(keys, vals, 4, 3) will call make_balanced_tree(keys, vals, 4, 1) and make_balanced_tree(keys, vals, 6, 1).
• The rest of the make_balanced_tree() calls won't make recursive calls, because they won't have any subtrees.
• The copy constructor should call the assignment overload (see the copy constructor for stacks in the Linkd Data Structures lecture.
### Running times
I know that this has been a long lecture -- this is the last section!
For general trees, you should know the following things:
• A tree with n nodes has n-1 edges.
• Therefore the number of edges is O(n).
• All tree traversals are O(n).
For binary search trees, you should know the following things:
• The running times of Find(), Insert(), Delete() and Depth() are all O(h), where h is the height of the tree.
• Binary search trees do not have to be balanced (the see example with the copy constructor above). That means that a binary search tree can easily have a height h which is O(n).
• Therefore the worst case running times of Find(), Insert(), Delete() and Depth() are O(n).
• The worst case of creating a binary search tree with n elements is O(n2). One way to do that is to insert all of the keys in sorted order.
• When a tree is balanced, its height is O(log(n)).
• So on a balanced binary search tree, the running times of Find(), Insert(), Delete() and Depth() are O(log(n)).
• All of the traversals are still O(n), so Height(), Print(), Ordered_Keys(), Ordered_Vals(), Clear(), the destructor, the copy constructor and the assignment overload are all O(n).
| 9,400 | 37,365 |
{"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-2023-23
|
latest
|
en
| 0.870062 |
http://forums.finehomebuilding.com/breaktime/general-discussion/conversion-linear-feet
| 1,529,747,257,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267864957.2/warc/CC-MAIN-20180623093631-20180623113631-00506.warc.gz
| 126,314,635 | 19,638 |
# Conversion to l.inear feet
## Conversion to l.inear feet (post #185376)
Trying to convert from square feet to linear feet.
formula; sq. ft \ width of board ??
360 square feet. covering with 7 1/2 wide planking would be ??
360 \ .075 = 497 lin feet sound right?
siding a cabin with log siding. front and back are 32' x 10'. side is 24' x 10' with a 12 pitch roof. (360 sq. ft. ??)
320 + 320 + 360 + 360 = 1,360 sq ft. \ .625 (7.5 \ 12) = 2,176 linear feet of 2" x 8" log siding ????
thanks, Chuck
### 360 square feet. covering (post #185376, reply #1 of 9)
360 square feet. covering with 7 1/2 wide planking would be ??
360 \ .075 = 497 lin feet sound right?
No, if the visible coverage is 7-1/2", then 576 LF is closer if the room is 10' x 36'. The 7-1/2" board is .625 of a foot. You're running it the long way.
120" / 7.5" = 16 boards, 36 ft long. = 576 LF.
oddly, a room 4' x 90' would be a little different.
48"/7.5 = 6.4 (7) boards, 90' long = 630 LF.
If you were to buy it by coverage-ask for 360 sf worth.
And just in case-
A room 9'x40' would be.......
108"/7.5 = 14.4 boards (15) 40' long = 600 LF.
As far as the siding, I'll let you figure that out. Waste on the gable end, lengths available, etc.
A Great Place for Information, Comraderie, and a Sucker Punch.
Remodeling Contractor just outside the Glass City.
### No, not right at (post #185376, reply #2 of 9)
No, not right at all.
That 7.5" will give you about 6" of coverage when you figure lap and waste, so you need 720 LF at least for the 360 SF of wall
Oh Well,
We did the best we could...
### This is "new math" to (post #185376, reply #3 of 9)
This is "new math" to me...converting square feet into linear ft. Linear feet is simply the length of a board. Your problem involving how many "sticks" of wood siding to use is easy to figure.
You said your walls were 10 ft high....which equals 120 inches. If you are using 7-1/2 inch wide siding, figure out how much you lose due to face overlap. One poster said you will wind up with approx 6 inch face exposure width....if so, 120 inches divided by 6 inches equals 20 pieces. Hence you need 20 pieces of siding to vertically cover this wall from bottom to top.
Now you say your wall is 32 feet long? Well, how long are your siding pieces? If they are 12 feet long, then you need 3 pieces to cover the length of the wall. 3 X 20 rows = 60 pieces of 12 foot long siding....60 X 12 = 720 LF of siding for one wall.
Now, of course that above example allowed for waste...you really only need 2.66 boards , each 12 ft long to cover the 32ft length of the wall. Hence, 2.67X20 = 54 pieces....
54pcs X 12 ft = 648 LF Total
Of course this last set of calculations doesn't allow for waste...always allow for waste.
Davo
Davo
### It's neat that we can discuss (post #185376, reply #4 of 9)
It's neat that we can discuss math on this forum, one provided us by the publishers of Fine Homebuilding magazine.
And there is nothing more important to fine homebuilding than having the right amount of materials at hand to do the job.
Area is easier than volume because there are only two factors, x and y, or length by width, or height by run, or whatever you want to call it.
So let's get started.
Today's challenge is to figure how many boards we need to slap up on walls. We have the measurements of our walls, and we know how wide our boards are, so let's get out our pencils, sharpen them, and begin.
The way I prefer to look at this is to imagine a small area or our wall, one square foot, 12 x 12 inches, and see it as finished with the boards we will apply. Finished with the cladding, I then compute the quantity it took to do that square foot, and that gives me my factor for figuring my total for the job. I'll want to put a factor on top of that for cutting and waste and knots and more, but that is another topic.
We have boards of 7.5 inch width up there, and I'll run through this two ways. One way will figure it with the boards showing their full 7.5 inch width, and the other will figure it with the boards lapped so that only 5.5 inches will show.
If we are seeing 7.5 inches of width when finished, then each 12 inches of height (if boards are laid up horizontally) will have 12 / 7.5 = 1.6 board widths showing. And that is our factor. If we have a total gross area of 1000 square feet to cover and finish, we will need a net lineal footage of 1000 x 1.6 = 1600 lineal feet of 7.5" boards.
It follows that if our exposure is less than the 7.5 inches because we are lapping, and only showing 5.5 inches of width, our factor then becomes 12 / 5.5 = 2.182, and our footage requirement then computes to 2182 lineal feet.
So it helps, this method of "slicing out" that one square foot of finished wall, and then figuring out the number of units it took to get that finish on, and using that as a factor for extending out the total. If necessary, and it can be helpful at times, take pencil to paper and draw your square foot, and then draw on it the finish, with some dimensions. The method for going forward will be looking right at you from your piece of paper.
"A stripe is just as real as a dadgummed flower."
Gene Davis 1920-1985
### Or it might work better (for (post #185376, reply #5 of 9)
Or it might work better (for some) to figure a 10x10 foot area (100 square feet), since then you have less trouble with the fractional board at the top or bottom. Eg, 5.5" inches divided into 120 inches (10 feet) is 21.8 pieces (or simply call it 22, if you find 21.81818181818181 irrational). So 22 times 10 is 220 linear feat in 100 square feet. And of course divide linear feet by length of a stick to get the number of sticks.
As to allowing for waste, I found it to be about right on our house to simply figure total square feet and then only subtract off the door areas, as if the house had no windows. Not a hard-and-fast rule, certainly, but when I went back and did a more accurate accounting I came up with nearly the same number (and we had about as many left-over sticks as we expected).
Of all the preposterous assumptions of humanity over humanity, nothing exceeds most of the criticisms made on the habits of the poor by the well-housed, well-warmed, and well-fed. --Herman Melville
### >>>As to allowing for waste, (post #185376, reply #6 of 9)
>>>As to allowing for waste, I found it to be about right on our house to simply figure total square feet and then only subtract off the door areas
Yes, and don't forget that if you have the luxury of a CAD program, the problem is greatly simplified. Any decent piece of software will calculate the cumulative area of a class of objects.
### The problem is accounting for (post #185376, reply #7 of 9)
The problem is accounting for cutoffs and other waste. If you're lucky windows will improve your use of cutoffs since where two windows are close together you can use short pieces between them. But if the unbroken areas of the walls are the wrong length -- eg, such that you end up cutting off 17" every other board because some dimension is just longer than an even multiple of 18 -- you can end up with a lot of waste.
Of all the preposterous assumptions of humanity over humanity, nothing exceeds most of the criticisms made on the habits of the poor by the well-housed, well-warmed, and well-fed. --Herman Melville
### Gene....... Gene....... (post #185376, reply #8 of 9)
Gene.......
Gene...................
Gene.........................................
Thanks
;)
Oh Well,
We did the best we could...
### ... You're young and alive. (post #185376, reply #9 of 9)
... You're young and alive. Come out of your half dream-dream, and run if you will to the top of the hill ...
Of all the preposterous assumptions of humanity over humanity, nothing exceeds most of the criticisms made on the habits of the poor by the well-housed, well-warmed, and well-fed. --Herman Melville
| 2,138 | 7,979 |
{"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}
| 3.59375 | 4 |
CC-MAIN-2018-26
|
latest
|
en
| 0.903604 |
https://math.stackexchange.com/questions/1189943/solve-int-fracx-sqrtx2-6xdx
| 1,679,489,264,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296943809.76/warc/CC-MAIN-20230322114226-20230322144226-00441.warc.gz
| 450,729,718 | 34,843 |
# Solve $\int\frac{x}{\sqrt{x^2-6x}}dx$
I need to solve the following integral
$$\int\frac{x}{\sqrt{x^2-6x}}dx$$
I started by completing the square,
$$x^2-6x=(x-3)^2-9$$
Then I defined the substitution variables..
$$(x-3)^2=9\sec^2\theta$$ $$(x-3)=3\sec\theta$$ $$dx=3\sec\theta\tan\theta$$ $$\theta=arcsec(\frac{x-3}{3})$$
Here are my solving steps $$\int\frac{x}{\sqrt{(x-3)^2-9}}dx = 3\int\frac{(3\sec\theta+3)\sec\theta\tan\theta}{\sqrt{9(\sec^2\theta-1)}}d\theta$$ $$=\int\frac{(3\sec\theta+3)\sec\theta\tan\theta}{\tan\theta}d\theta$$ $$=\int(3\sec\theta+3)\sec\theta\tan\theta$$ $$=3\int\sec^2\theta\tan\theta d\theta + 3\int\sec\theta\tan\theta d\theta$$ $$u = \sec\theta, du=\sec\theta\tan\theta d\theta$$ $$=3\int udu + 3\sec\theta$$ $$=\frac{3\sec\theta}{2}+3\sec\theta+C$$ $$=\frac{3\sec(arcsec(\frac{x-3}{3}))}{2}+3\sec(arcsec(\frac{x-3}{3}))+C$$ $$=\frac{3(\frac{x-3}{3})}{2}+3(\frac{x-3}{3})$$ $$=\frac{x-3}{2}+x-3+C$$
$$\int\frac{x}{\sqrt{x^2-6x}}dx=\sqrt{x^2-6x}+3\ln\bigg(\frac{x-3}{3}+\frac{\sqrt{x^2-6x}}{3}\bigg)$$
What did I misunderstood?
• When you go from line 12 to 13 you have not simplified $\tan\theta.$ That is, line 13 should read $$\int(3\sec\theta+3)\sec\theta d\theta.$$
– mfl
Mar 14, 2015 at 20:35
• @mfl I was able to solve it with your correction :) Feel free to post that as an answer if you want me to accept. Mar 14, 2015 at 20:54
• It is not necessary. You have solved the problem by yourself. It was only a typo. Good luck.
– mfl
Mar 14, 2015 at 20:55
• @mfl Well, thanks a lot for your hint. Mar 14, 2015 at 20:56
I prefer to apply hyperbolic substitution.
More precisely, if we let $$x - 3 = 3\cosh(y)$$, we arrive at \begin{align*} \int\frac{x}{\sqrt{x^{2} - 6x}}\mathrm{d}x & = \int\frac{x}{\sqrt{(x - 3)^{2} - 9}}\mathrm{d}x\\\\ & = 3\int\frac{(1 + \cosh(y))\sinh(y)}{\sqrt{\cosh^{2}(y) - 1}}\mathrm{d}y\\\\ & = 3\int\mathrm{d}y + 3\int \cosh(y)\mathrm{d}y\\\\ & = 3y + 3\sinh(y) + c\\\\ & = 3\operatorname{arccosh}\left(\frac{x - 3}{3}\right) + 3\sqrt{\left(\frac{x - 3}{3}\right)^{2} - 1} + c\\\\ & = 3\ln\left(\frac{x - 3}{3} + \frac{\sqrt{x^{2} - 6x}}{3}\right) + \sqrt{x^{2} - 6x} + c\\\\ & = 3\ln\left(x - 3 + \sqrt{x^{2} - 6x}\right) + \sqrt{x^{2} - 6x} + C \end{align*}
which coincides with the proposed result because \begin{align*} \operatorname{arccosh}(z) = \ln\left(z + \sqrt{z^{2} - 1}\right) \end{align*}
Hopefully this helps!
• That logarithm should have an absolute value in the final answer, else you're missing the solutions for negative $x$ (Indeed I think the underlying issue here might stem from $\cosh$ not being an injective function on $\mathbb{R}$, so the substitution doesn't give the complete solution). Mar 25, 2022 at 1:48
• @Lorago thank you very much for the observation. Perhaps it suffices to take $x > 6$, since $\cosh(z)\geq 1$ and $x - 3 = 3\cosh(z)$ (we are not allowed to take $x < 0$ because the $\cosh$ function is always non-negative). What do you think about it? Mar 25, 2022 at 3:03
• The point is the integrand is defined for $x>6$ or $x<0$, so both cases must be handled. And indeed, cosh is the problem, though you can still use cosh for the other case (but I don't think you can do both at once). Your derivation is indeed correct for $x>6$, as is implied by $x-3=3\cosh y$). Interesting to have solutions with Euler and hyperbolic substitutions in this question. Mar 25, 2022 at 11:24
I'll give you an answer that is completely free from all this trigonometric nonsense, that I personally think is cleaner and easier. We wish to compute
$$\int \frac{x}{\sqrt{x^2-6x}}\,\mathrm{d}x.$$
Consider first rewriting it as
$$\int \frac{x}{\sqrt{x^2-6x}}\,\mathrm{d}x=\frac{1}{2}\int \frac{2x-6}{\sqrt{x^2-6x}}\,\mathrm{d}x+3\int \frac{\mathrm{d}x}{\sqrt{x^2-6x}}\,.$$
Now for the first integral, letting $$u=x^2-6x$$ we get that
$$\frac{1}{2}\int \frac{2x-6}{\sqrt{x^2-6x}}\,\mathrm{d}x=\int\frac{\mathrm{d}u}{2\sqrt{u}}=\sqrt{u}+A=\sqrt{x^2-6x}+A.$$
This takes care of the first integral. Now for the second integral, consider the Euler substitution
$$\begin{cases} t=x+\sqrt{x^2-6x},\\ \mathrm{d}t=\frac{x+\sqrt{x^2-6x}-3}{\sqrt{x^2-6x}}\mathrm{d}x. \end{cases}$$
This yields that
$$3\int \frac{1}{\sqrt{x^2-6x}}\,\mathrm{d}x=3\int \frac{\mathrm{d}t}{t-3}=3\ln\lvert t-3\rvert +D=3\ln\lvert x+\sqrt{x^2-6x}-3\rvert+B.$$
$$\int \frac{x}{\sqrt{x^2-6x}}\,\mathrm{d}x=\sqrt{x^2-6x}+3\ln\lvert x+\sqrt{x^2-6x}-3\rvert+C.$$
\begin{aligned} \int \frac{x}{\sqrt{x^2-6 x}} d x & \int \frac{x}{x-3} d\left(\sqrt{x^2-6 x}\right) \\ = & \int\left(1+\frac{3}{x-3}\right)\left(\sqrt{x^2-6 x}\right) \\ = & \sqrt{x^2-6 x}+3 \int \frac{d\left(\sqrt{x^2-6 x}\right)}{\sqrt{\left(\sqrt{x^2-6 x}\right)^2+9}} \\ = & \sqrt{x^2-6 x}+3 \sinh^{-1} \left(\frac{\sqrt{x^2-6 x}}{3}\right)+C \end{aligned}
| 1,955 | 4,795 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 28, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.375 | 4 |
CC-MAIN-2023-14
|
latest
|
en
| 0.516219 |
https://stats.libretexts.org/Bookshelves/Applied_Statistics/Book%3A_Business_Statistics_(OpenStax)/07%3A_The_Central_Limit_Theorem/7.00%3A_Introduction_to_the_Central_Limit_Theorem
| 1,656,927,722,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656104364750.74/warc/CC-MAIN-20220704080332-20220704110332-00108.warc.gz
| 590,651,226 | 25,685 |
# 7.0: Introduction to the Central Limit Theorem
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
Why are we so concerned with means? Two reasons are: they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the Central Limit Theorem.
The Central Limit Theorem is one of the most powerful and useful ideas in all of statistics. The Central Limit Theorem is a theorem which means that it is NOT a theory or just somebody's idea of the way things work. As a theorem it ranks with the Pythagorean Theorem, or the theorem that tells us that the sum of the angles of a triangle must add to 180. These are facts of the ways of the world rigorously demonstrated with mathematical precision and logic. As we will see this powerful theorem will determine just what we can, and cannot say, in inferential statistics. The Central Limit Theorem is concerned with drawing finite samples of size $$n$$ from a population with a known mean, $$\mu$$, and a known standard deviation, $$\sigma$$. The conclusion is that if we collect samples of size $$n$$ with a "large enough $$n$$," calculate each sample's mean, and create a histogram (distribution) of those means, then the resulting distribution will tend to have an approximate normal distribution.
The astounding result is that it does not matter what the distribution of the original population is, or whether you even need to know it. The important fact is that the distribution of sample means tend to follow the normal distribution.
The size of the sample, $$n$$, that is required in order to be "large enough" depends on the original population from which the samples are drawn (the sample size should be at least 30 or the data should come from a normal distribution). If the original population is far from normal, then more observations are needed for the sample means. Sampling is done randomly and with replacement in the theoretical model.
7.0: Introduction to the Central Limit Theorem is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
| 827 | 3,176 |
{"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}
| 4.15625 | 4 |
CC-MAIN-2022-27
|
latest
|
en
| 0.68326 |
https://www.studypool.com/discuss/544708/need-help-with-word-problem-6?free
| 1,495,941,257,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463609409.62/warc/CC-MAIN-20170528024152-20170528044152-00044.warc.gz
| 1,201,513,226 | 13,977 |
##### need help with word problem
Algebra Tutor: None Selected Time limit: 1 Day
Alyssa is 3 years older than Gladys. William is twice as old as Alyssa. Altogether their ages total 89. How old is Gladys?
May 18th, 2015
A = 3 + G
W = 2A
A+ G+ W = 89
3 + G + G + 2 (3+G) = 89
3+ 2G + 6 + 2G =89
9 + 4G = 89
4G = 89 -9
4G = 80
G = 80/4
=20 years
Please let me know if you have any questions and best me if you are satisfactory.
May 18th, 2015
...
May 18th, 2015
...
May 18th, 2015
May 28th, 2017
check_circle
| 210 | 523 |
{"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}
| 3.5625 | 4 |
CC-MAIN-2017-22
|
longest
|
en
| 0.941904 |
http://www.slideserve.com/calvin/warm-up
| 1,508,511,950,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187824225.41/warc/CC-MAIN-20171020135519-20171020155519-00251.warc.gz
| 572,821,788 | 12,627 |
Warm-up
1 / 6
# Warm-up - PowerPoint PPT Presentation
Warm-up. Homework out Warm-up Pg. 485 #2-10 evens in algebra book. 9.6 Graphing Quadratic Inequalities. Reminder: when graphing inequalities, you have to SHADE the part of the graph this makes the inequality true. Steps to graphing quadratic inequality. Sketch the graph of the parabola
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about ' Warm-up' - calvin
An Image/Link below is provided (as is) to download presentation
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
### Warm-up
Homework out
Warm-up
Pg. 485
#2-10 evens in algebra book
9.6 Graphing Quadratic Inequalities
• Reminder: when graphing inequalities, you have to SHADE the part of the graph this makes the inequality true.
Steps to graphing quadratic inequality
• Sketch the graph of the parabola
• Find the vertex,
• 1 point to the left,
• 1 point to the right,
• use a dashed parabola for > or <.
• Use a solid parabola for > or <
• Test one point either inside or outside of the parabola. (0,0) is easiest if it is not on the parabola.
• If that point makes the inequality true, shade in the area that contains that point. If it makes the inequality false, shade the opposite area.
Need a piece of graph paper
• Graph y < x2 – 4x + 2
• Graph y > 2x2 – 8x + 2
9.7 Exploring data: comparing models
4 basic types of algebraic models
| 500 | 1,879 |
{"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}
| 4.28125 | 4 |
CC-MAIN-2017-43
|
latest
|
en
| 0.85595 |
https://www.dynamicmath.xyz/velocity-fields/
| 1,717,107,697,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971684053.99/warc/CC-MAIN-20240530210614-20240531000614-00390.warc.gz
| 626,769,790 | 4,312 |
Velocity fields
Interactive simulations
By Juan Carlos Ponce Campuzano, 17/June/2022
Consider the velocity vector field $\mathbf{v}$ of a steady-state fluid flow. The vector $\mathbf{v}(x, y)$ measures the instantaneous velocity of the fluid particles (molecules or atoms) as they pass through the point $(x, y)$. Steady-state means that the velocity at a point $(x, y)$ does not vary in time -even though the individual fluid particles are in motion. If a fluid particle moves along the curve $\mathbf{x}(t) = (x(t), y(t))$, then its velocity at time $t$ is the derivative $\mathbf{v}= \frac{d\mathbf{x}}{dt}$ of its position with respect to $t$. Thus, for a time-independent velocity vector field $\mathbf{v}(x, y) = ( v_1(x, y), v_2(x, y) )$ the fluid particles will move in accordance with an autonomous, first order system of ordinary differential equations $\frac{dx}{dt}= v_1(x, y),\qquad \frac{dy}{dt}= v_2(x, y).$ According to the basic theory of systems of ordinary differential equations, an individual particle's motion $\mathbf{x}(t)$ will be uniquely determined solely by its initial position $\mathbf{x}(0) = \mathbf{x}_0$. In fluid mechanics, the trajectories of particles are known as the streamlines of the flow. The velocity vector $\mathbf{v}$ is everywhere tangent to the streamlines. When the flow is steady, the streamlines do not change in time. Individual fluid particles experience the same motion as they successively pass through a given point in the domain occupied by the fluid.
Examples
Some basic examples of velocity vector fields of steady-state fluids flow are the following:
• Near a wall (Stagnation point): $\mathbf{v}(x,y)=(k\,x,-k\,y)$, with $k\in \mathbb R^{+}$.
• Rigid body rotation: $\mathbf{v}(x,y)=(-k\,y,k\,x)$, where $k\in \mathbb R$ For $k>0$, rotation is anticlockwise, and for $k < 0$ rotation is clockwise.
• Vortex: $\mathbf{v}(x,y)=\left(\dfrac{-k\,y}{x^2+y^2},\dfrac{k\,x}{x^2+y^2}\right)$, with $k\in \mathbb R.$ For $k>0$, rotation is anticlockwise, and for $k < 0$ rotation is clockwise.
• Source & Sink: $\mathbf{v}(x,y)=\left(\dfrac{k\,x}{x^2+y^2},\dfrac{k\,y}{x^2+y^2}\right)$, with $k\in \mathbb R.$ For $k>0$, the flow is a source, and for $k < 0$ the flow is a sink.
Click on the image below to open interactive applet
Finally, if you find this content useful, please consider supporting the work using the links below.
∞ Thanks!
| 705 | 2,405 |
{"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}
| 3.765625 | 4 |
CC-MAIN-2024-22
|
latest
|
en
| 0.801959 |
https://books.google.com.mx/books?id=G942AAAAMAAJ&lr=
| 1,620,463,165,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243988850.21/warc/CC-MAIN-20210508061546-20210508091546-00597.warc.gz
| 167,523,149 | 11,272 |
# The Elements of Euclid for the Use of Schools and Colleges: With Notes, an Appendix, and Exercises. comprising the first six books and portions of the eleventh and twelfth books
Macmillan and Company, 1880 - 400 páginas
### Comentarios de la gente -Escribir un comentario
No encontramos ningún comentario en los lugares habituales.
### Contenido
Book III 71 Book IV 113 Book V 134 Book VI 173 Book XI 220
Book XII 244 Notes on Euclids Elements 250 Appendix 293 Exercises in Euclid 340
### Pasajes populares
Página 225 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Página 284 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Página 73 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a Right Angle; and the straight line which stands on the other is called a Perpendicular to it.
Página 39 - Triangles upon the same base, and between the same parallels, are equal to one another.
Página 10 - THE angles at the base of an isosceles triangle are equal to one another : and, if the equal sides be produced, the angles upon the other side of the base shall be equal.
Página 353 - AB into two parts, so that the rectangle contained by the whole line and one of the parts, shall be equal to the square on the other part.
Página 67 - ... subtending the obtuse angle, is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle, Let ABC be an obtuse-angled triangle, having the obtuse angle ACB; and from the point A, let AD be drawn perpendicular to BC produced.
Página 300 - Describe a circle which shall pass through a given point and touch a given straight line and a given circle.
Página xv - PROPOSITION I. PROBLEM. To describe an equilateral triangle upon a given Jinite straight line. Let AB be the given straight line. It is required to describe an equilateral triangle upon AB, From the centre A, at the distance AB, describe the circle BCD ; (post.
Página 36 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
| 579 | 2,509 |
{"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-2021-21
|
latest
|
en
| 0.764382 |
http://oeis.org/A266821
| 1,571,790,482,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570987826436.88/warc/CC-MAIN-20191022232751-20191023020251-00030.warc.gz
| 144,976,414 | 4,306 |
This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A266821 Expansion of Product_{k>=1} (1 + 3*x^k) / (1 - x^k). 4
1, 4, 8, 24, 44, 88, 176, 312, 544, 924, 1584, 2552, 4136, 6488, 10128, 15632, 23748, 35640, 53080, 78136, 114024, 165552, 237744, 339544, 481248, 678236, 949008, 1321840, 1830376, 2521688, 3456672, 4717208, 6406680, 8666448, 11672464, 15660528, 20934868 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Convolution of A000041 and A032308. In general, for m > 0, if g.f. = Product_{k>=1} ((1 + m*x^k) / (1 - x^k)) then a(n) ~ sqrt(c) * exp(sqrt(2*c*n)) / (4*Pi*sqrt(m+1)*n), where c = 2*Pi^2/3 + log(m)^2 + 2*polylog(2, -1/m). LINKS Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 5001 terms from Vaclav Kotesovec) FORMULA a(n) ~ sqrt(c) * exp(sqrt(2*c*n)) / (8*Pi*n), where c = 2*Pi^2/3 + log(3)^2 + 2*polylog(2, -1/3) = 7.16861897522987077909937377164783326088308015803... . MAPLE b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, add( (t-> b(t, min(t, i-1)))(n-i*j), j=1..n/i)*4 +b(n, i-1))) end: a:= n-> b(n\$2): seq(a(n), n=0..44); # Alois P. Heinz, Aug 28 2019 MATHEMATICA nmax = 40; CoefficientList[Series[Product[(1+3*x^k) / (1-x^k), {k, 1, nmax}], {x, 0, nmax}], x] PROG (PARI) { my(n=40); Vec(prod(k=1, n, 4/(1-x^k) - 3 + O(x*x^n))) } \\ Andrew Howroyd, Dec 22 2017 CROSSREFS Cf. A000041, A015128, A032308, A264686. Column k=4 of A321884. Sequence in context: A212686 A316961 A180002 * A306484 A208901 A319721 Adjacent sequences: A266818 A266819 A266820 * A266822 A266823 A266824 KEYWORD nonn AUTHOR Vaclav Kotesovec, Jan 04 2016 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified October 22 19:53 EDT 2019. Contains 328319 sequences. (Running on oeis4.)
| 847 | 2,046 |
{"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}
| 3.53125 | 4 |
CC-MAIN-2019-43
|
latest
|
en
| 0.424068 |
https://www.physicsforums.com/threads/comparing-non-uniform-data-sets.435950/
| 1,719,233,814,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00519.warc.gz
| 821,374,805 | 17,929 |
# Comparing Non-Uniform Data Sets
• JPierce
In summary, the student is trying to figure out a way to display a summary of the results of multiple reviewers on a set of items. There are problems with scaling the data, selecting a single sample of data, and dealing with large numbers of students.
JPierce
I'm not sure my title is very descriptive, but I tried my best. I also hope I am posting this in the right forum. If not, please let me know. (I thought it might be better posted in the social sciences forum.)
I have a project where I am analyzing the results of multiple reviewers on a set of items. I am unsure as to the proper method of normalizing the data.
In essence, here is the problem:
We have a large stack of assignments turned in by students, with one assignment turned in by each student. Teachers analyzed each assignment according to three criteria, which I will call A, B, and C. Teachers values for each of these criteria on a scale from 1 to 5. However, this scale is not linear, so A=4 is not twice as "big" as A=2.
I simply want to display a summary of the results using (say) a histogram. I have no interest in calculating summary statistics because the numerical values for each criterion are purely denumerable -- that is, A = 2.4 (which could correspond to say grade level) is meaningless.
So far, so good. But some of the assignments were reviewed by up to five teachers. Others were reviewed by only one.
So one assignment turned in by (say) Jimmy may have the following reviews from five individual teachers:
A = 3; B = 1; C = 4
A = 3; B = 2; C = 4
A = 3; B = 1; C = 2
A = 2; B = 3; C = 4
A = 3; B = 2; C = 4
Another assignment turned in by Mary may only have A = 3; B = 1; C = 4 as measured by a single teacher.
So, how do we handle the fact that some assignments have more reviews than others? We could just scale up the number of reviews to a common value. In other words, we could pretend that Mary turned in five identical assignments, that is,
A = 3; B = 1; C = 4
A = 3; B = 1; C = 4
A = 3; B = 1; C = 4
A = 3; B = 1; C = 4
A = 3; B = 1; C = 4
Somehow that doesn't seem quite right. And it would foul up the precision of the results.
Another idea is to whittle down the number of reviewers to 1 for each assignment, but I have no good criteria for selecting the one sample to keep.
Any ideas?
If I have left out important info, just let me know.
Most reasonable would seem that every student gets an A, B and C grade and each grade would be the average of the grades that the different teachers gave.
So that you would have:
Jimmy: A=2.8, B=1.8, C = 3.6
Marry: A=3.0, B=1.0, C=4.0
JPierce said:
I have no interest in calculating summary statistics because the numerical values for each criterion are purely denumerable -- that is, A = 2.4 (which could correspond to say grade level) is meaningless.
Oh, I am not really sure what you mean here, but it seems to indicate that only whole numbers are meaningful. In that case take medians instead of averages. So that you would have:
Jimmy: A=3, B=2, C=4
Marry: A=3, B=1, C=4
JPierce said:
...We have a large stack of assignments turned in by students, with one assignment turned in by each student. Teachers analyzed each assignment according to three criteria, which I will call A, B, and C. Teachers values for each of these criteria on a scale from 1 to 5. However, this scale is not linear, so A=4 is not twice as "big" as A=2.
I simply want to display a summary of the results using (say) a histogram...But some of the assignments were reviewed by up to five teachers. Others were reviewed by only one.
It sounds kind of similar to the Collaborative Filtering problem made famous by the Netflix Prize, but here if the number of students is not too large it should be possible to present all the data with a few charts.
For example the scores for criteria A could be represented by an Nx5 grid: number the students 1 to N according to some overall rating (such as average total score) and make the colour/brightness of cell (i,j) according to how many reviewers gave student i score j. Then repeat for criteria B and C (so all the data is in 3 grid charts or they could be combined into one with RGB colour mix). I don't know if any packages specifically do this type of chart but it should be possible in Excel using conditional formatting.
Thanks for all suggestions.
The number of students involved is very large -- tens of thousands.
I will look into the Netflix problem.
Using the median might work.
If anyone has more suggestions, please offer them. I will continue reading responses.
## 1. How do you compare two non-uniform data sets?
To compare two non-uniform data sets, it is important to first determine the type of data you are working with. If the data is numerical, you can use statistical measures such as mean, median, and mode to compare the central tendencies of the two sets. If the data is categorical, you can use bar graphs or pie charts to visualize and compare the frequencies of each category.
## 2. What is the best way to visualize non-uniform data sets?
The best way to visualize non-uniform data sets depends on the type of data and the purpose of the comparison. For numerical data, histograms, box plots, or scatter plots can be used to show the distribution and outliers. For categorical data, bar graphs, pie charts, or stacked bar graphs can be used to show the frequencies and proportions of each category.
## 3. How do you handle missing data when comparing non-uniform data sets?
Handling missing data in non-uniform data sets can be challenging, as it can affect the accuracy and reliability of the comparison. One approach is to remove the missing data points, but this can reduce the sample size and potentially skew the results. Another approach is to impute the missing values using statistical methods such as mean or median imputation. However, this can also introduce bias in the results and should be done carefully.
## 4. Can non-uniform data sets be compared using statistical tests?
Yes, non-uniform data sets can be compared using statistical tests, but the choice of test depends on the type of data and the research question being addressed. For numerical data, tests such as t-test, ANOVA, or Mann-Whitney U test can be used. For categorical data, chi-square test or Fisher's exact test can be used. It is important to ensure that the assumptions of the chosen statistical test are met before interpreting the results.
## 5. How do you deal with outliers when comparing non-uniform data sets?
Outliers can significantly affect the results when comparing non-uniform data sets. One approach is to remove the outliers, but this should be done carefully after considering the possible reasons for the outliers. Another approach is to transform the data using methods such as logarithmic or square root transformation, which can help to reduce the impact of outliers. It is also important to report the presence of outliers and the method used to handle them in the analysis.
• Set Theory, Logic, Probability, Statistics
Replies
9
Views
493
• Set Theory, Logic, Probability, Statistics
Replies
1
Views
750
• Set Theory, Logic, Probability, Statistics
Replies
14
Views
1K
• Set Theory, Logic, Probability, Statistics
Replies
1
Views
834
• Set Theory, Logic, Probability, Statistics
Replies
6
Views
876
• Set Theory, Logic, Probability, Statistics
Replies
5
Views
1K
• Set Theory, Logic, Probability, Statistics
Replies
9
Views
1K
• Set Theory, Logic, Probability, Statistics
Replies
9
Views
1K
• Set Theory, Logic, Probability, Statistics
Replies
40
Views
7K
• Set Theory, Logic, Probability, Statistics
Replies
2
Views
2K
| 1,857 | 7,700 |
{"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}
| 3.578125 | 4 |
CC-MAIN-2024-26
|
latest
|
en
| 0.973392 |
https://www.omnicalculator.com/statistics/decimal-random-number-generator
| 1,726,303,473,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00019.warc.gz
| 848,180,293 | 35,037 |
# Decimal Random Number Generator
Created by Anna Szczepanek, PhD
Reviewed by Davide Borchia
Last updated: Jan 18, 2024
Omni's decimal random number generator will assist you in your math or programming tasks by producing as many uniformly distributed random decimal numbers as you need and from the range you need. In the article below, we will briefly explain what random and pseudo-random decimals are.
## What is a random number generator?
A random (or rather pseudo-random) number generator is an algorithm that outputs a sequence of numbers. These numbers should follow no discernible pattern, i.e., it should not be possible to predict the next number of the sequence.
Since there's an algorithm behind the generation procedure, these numbers are not truly random yet they are good enough for most applications. What our decimal random number generator produces is in fact a sequence of pseudo-random decimal numbers.
## Decimal vs integer random numbers
A decimal number is a number written in the base-10 system, i.e., the number system that is most commonly used nowadays. Examples of decimal numbers are 0.33, 2, 2.34, etc.
A decimal number consists of a whole number part and a fractional part. For example, 2.34 has 2 as its whole number part and 0.34 as its fractional part. There is no whole number part in 0.33, while 2 has no fractional part. The latter number can also be written as 2.0. It is, in fact, an example of an integer, which is a decimal number with no fractional part.
When it comes to drawing random numbers, an important difference between integers and decimal numbers is that in every range [a,b], there are finitely many integers but infinitely many decimal numbers. If we draw decimal numbers from any continuous distribution, in particular from the uniform distribution, then the probability of getting duplicates is zero.
## How to use this random number generator with decimals?
Our random number generator with decimals is really straightforward to use:
1. Specify the range from which you want to draw.
2. Tell us how many numbers you want.
3. If you wish, you may adjust the precision, i.e., the number of decimal places.
4. Our decimal random number generator can sort the numbers it produced - just switch the Sort option to Yes.
5. Similarly, you can choose between List and Column output.
🙋 Enter 0 as Precision to produce integers with the help of our random decimal number generator.
## FAQ
### How do I generate a random number from a given range?
If you want to generate a random number from the interval [a, b]:
1. Generate a random number from the interval [0, 1].
2. Multiply this number by (b - a).
3. Add a to the result of Step 2.
4. That's it! What you got in Step 3 is a random number from [a,b].
### Can I express every integer as a decimal?
Yes, every integer can be expressed as a decimal number. In fact, there are two ways to do that: for example, the integer 2 can be expressed as 2 and as 1.999… (with the 9 repeating infinitely many times). In computer science, we would also consider 2.0 or 2.000 as different from 2.
### How do I obtain true random numbers?
Random numbers generated by computer algorithms are, in fact, pseudo-random: they do follow some pattern, yet a highly complicated one. To get a true random number, we can exploit various physical systems, in particular quantum systems.
Anna Szczepanek, PhD
How many numbers?
Minimum value
Maximum value
Precision
Sort results
No
Output format
List
People also viewed…
### Ascending order
Sort numbers from least to greatest with this ascending-order calculator.
### Helium balloons
Wondering how many helium balloons it would take to lift you up in the air? Try this helium balloons calculator! 🎈
### Millionaire
This millionaire calculator will help you determine how long it will take for you to reach a 7-figure saving or any financial goal you have. You can use this calculator even if you are just starting to save or even if you already have savings.
### Post test probability
The post-test probability calculator allows you to calculate pre-test probability, likelihood ratios, and finally, the post-test probability itself.
| 919 | 4,187 |
{"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}
| 4.53125 | 5 |
CC-MAIN-2024-38
|
latest
|
en
| 0.910435 |
https://math.stackexchange.com/questions/540950/rotation-matrix-inverse-using-gauss-jordan-elimination
| 1,568,669,263,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-39/segments/1568514572934.73/warc/CC-MAIN-20190916200355-20190916222355-00116.warc.gz
| 588,468,207 | 32,566 |
# Rotation Matrix inverse using Gauss-Jordan elimination
I'd like to calculate the inverse of a rotation matrix, let take the simplest case which is a $2$ x $2$ rotation matrix:
$R =\begin{bmatrix} \cos \theta & -\sin \theta \\[0.3em] \sin \theta & \cos \theta \end{bmatrix}$
I know that the inverse is the following
$R^{-1} =\begin{bmatrix} \cos \theta & \sin \theta \\[0.3em] -\sin \theta & \cos \theta \end{bmatrix}$
and I know that I can calculate it using the transpose method as such:
$R^{-1}=R^T$
but I fail to calculate the inverse using $Gauss-Jordan$ elimination, that is I don't know how to substract $\cos \theta$ from $\sin \theta$ in the second row. It all gets a bit complicated; I've looked around and nobody has a full step method using $G.-J.$ only the solution or the transpose method. Could someone provide me a full-step solution using $G.-J.$?
$$\begin{bmatrix} \cos t&-\sin t&|&1&0\\ \sin t&\cos t&|&0&1 \end{bmatrix} \xrightarrow{\frac1{\cos t}R1} \begin{bmatrix} 1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ \sin t&\cos t&|&0&1 \end{bmatrix} \xrightarrow{R2-\sin t\,R1} \begin{bmatrix} 1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ 0&\cos t+\frac{\sin^2t}{\cos t}&|&-\frac{\sin t}{\cos t}&1 \end{bmatrix}= \begin{bmatrix} 1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ 0&\frac1{\cos t}&|&-\frac{\sin t}{\cos t}&1 \end{bmatrix} \xrightarrow{\cos t\,R2} \begin{bmatrix} 1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ 0&1&|&-\sin t&\cos t \end{bmatrix} \xrightarrow{R1+\frac{\sin t}{\cos t}R2} \begin{bmatrix} 1 &0&|&\frac1{\cos t}-\frac{\sin^2t}{\cos t}&\sin t\\ 0&1&|&-\sin t&\cos t \end{bmatrix} =\begin{bmatrix} 1 &0&|&\cos t&\sin t\\ 0&1&|&-\sin t&\cos t \end{bmatrix}$$
This is a terrible method to calculate the inverse of any $2\times 2$ matrix.
Edit: of course this does not work when $\cos t=0$; but this is a much easier case: you simply divide by $\sin t$ and permute the rows.
• "Exercise for the interested reader: work out the case $\cos t =0$." :-) – Agustí Roig Oct 27 '13 at 1:27
• That's actually quite good! Thanks for giving a detailed and concise method, although I understand why it is terrible. I could not figure out how do that because I was thinking in a different way. – jtimz Oct 27 '13 at 1:30
• @Agustí: Good point :D – Martin Argerami Oct 27 '13 at 1:30
To simplify notation let $s = \sin (\theta)$, $c = \cos(\theta)$.
If $\theta \neq \pm\pi/2, \$ ($c = 0$ here), then
$R =$ \begin{align*} &\begin{bmatrix} c & -s \\ s & c \end{bmatrix} \sim \\ &\begin{bmatrix} c^2 & -sc \\ s^2 & sc \end{bmatrix} \sim \\ (c^2 + s^2 = 1, right?) &\begin{bmatrix} 1 & 0 \\ s^2 & sc \end{bmatrix} \sim \\ &\begin{bmatrix} 1 & 0 \\ s/c & 1 \end{bmatrix} \end{align*} then finish it off!
If $\theta = \pm \pi/2$, then $\theta \neq 0$ or $\pi$, so perform the similar set of operations first swapping the two rows and ending with a divide-by-$\sin$.
QED
Allow $\theta$ to remain symbolic until the operation is complete. I'm assuming if you had real numbers to plug in then taking the inverse would be trivial.
\begin{bmatrix} c &-s &| &1 &0\\ s & c &| &0 & 1\\ \end{bmatrix} Multiply top line by c \begin{bmatrix} c^2 &-sc &| &c &0\\ s & c &| &0 & 1\\ \end{bmatrix} Bottom line by s \begin{bmatrix} c^2 &-sc &| &c &0\\ s^2 & sc &| &0 & s\\ \end{bmatrix} Add bottom line to top \begin{bmatrix} c^2+s^2 &sc-sc &| &c &s\\ s^2 & sc &| &0 & s\\ \end{bmatrix} Reduce \begin{bmatrix} 1 & 0 &| &c &s\\ s^2 & sc &| &0 & s\\ \end{bmatrix} Multiply top line by $-s^2$ and add to bottom line \begin{bmatrix} 1 & 0 &| &c &s\\ s^2-s^2 & sc &| &-cs^2 & s-s^3\\ \end{bmatrix} Reduce \begin{bmatrix} 1 & 0 &| &c &s\\ 0 & sc &| &-cs^2 & s-s^3\\ \end{bmatrix} Divide bottom line by $s$ \begin{bmatrix} 1 & 0 &| &c &s\\ 0 & c &| &-cs & 1-s^2\\ \end{bmatrix} Note that Pythagorean Theorem gives $c^2 = 1-s^2$ \begin{bmatrix} 1 & 0 &| &c &s\\ 0 & c &| &-cs & c^2\\ \end{bmatrix} Divide bottom line by $c$ \begin{bmatrix} 1 & 0 &| &c &s\\ 0 & 1 &| &-s & c\\ \end{bmatrix} Again, if you want to make specific cases for theta $(=\pi/2,0,etc.)$, then just plug those specific values of $\theta$ in. Notice I did not carry any fractions through; every time I divided by a trig term if fully canceled terms in the numerators.
| 1,625 | 4,253 |
{"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}
| 3.703125 | 4 |
CC-MAIN-2019-39
|
latest
|
en
| 0.55743 |
http://www.onlinemath4all.com/gmat-mock-test-2.html
| 1,516,166,459,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-05/segments/1516084886815.20/warc/CC-MAIN-20180117043259-20180117063259-00359.warc.gz
| 515,170,442 | 10,123 |
# GMAT Mock Test 2
## About the topic GMAT Mock Test 2
In this online GMAT mock test 2,we give the questions whose qualities are in high standard and practicing these questions will definitely make the students to reach their goals.
Example Quiz 1
1. Find the square root of 123454321.
(A) 111111 (B) 11111 (C) 1111 (D) 111
2. Find the time taken by a train 100m long running at a speed of 60 kmph to cross another train of length 80 m running at a speed of 48 kmph in the same direction.
(A) 54 sec (B) 58 sec (C) 62 sec (D) 66 sec
3. A contractor decided to complete the work in 90 days and employed 50 men at the beginning and 20 men additionally after 20 days and got the work completed as per schedule. If he had not employed the additional men,how many extra days would he have needed to complete the work?
(A) 26 (B) 27 (C) 28 (D) 29
4. A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of the total time; to cover the remaining distance in the remaining time, his speed(in kmph) must be
(A) 11 (B) 12 (C) 13 (D) 14
5. John's age after six years will be three-seventh of his father's age. Ten years ago the ratio of their ages was 1 : 5. What is John's father's age at present?
(A) 47 yrs (B) 48 yrs (C) 49 yrs (D) 50 yrs
6. In John’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother doesn't agree with John and he thinks that John's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of John?
(A) 62 kg (B) 65 kg (C) 67 kg (D) 69 kg
7. At his usual rowing rate Michael can travel 12 miles downstream in a certain river in six hours less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for his 24 miles round trip, the downstream 12 miles would then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour?
(A) 12⁄3 mph (B) 22⁄3 mph (C) 31⁄3 mph (D) 51⁄3 mph
8. The monthly incomes of two persons are in the ratio 4:5 and their monthly expenditures are in the ratio 7:9. If each saves \$50 per month, find the monthly income of the second person.
(A) \$ 500 (B) \$ 450 (C) \$ 550 (D) \$ 600
9. A milk vender has two cans of milk. The first contains 25% water and the rest milk. The second contains 50% water and the rest milk. How many liters of milk should he mix from each of the cans so as to get 12 liters of milk such that that the ratio of water to milk is 3:5?
(A) (5,7) (B) (6,6) (C) (7,5) (D) (8,4)
10. The production of wheat was increased by 20% from the year 1994 to 1995. It was further increased by 25% from 1995 to 1996. The percentage change in the production of wheat from 1994 to 1996 was
(A) 49% (B) 50% (C) 51% (D) 52%
Explanation
Question No.1 Question No.2 Question No.3 Question No.4 Question No.5 Question No.6 Question No.7 Question No.8 Question No.9 Question No.10 jQuery UI Dialog functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} In the given number, we have the first five natural numbers in ascending order up to 5. After 5, we have the first four natural numbers in descending order. Whenever we have a number like this and we want to find square root, we have to replace each digit by 1, up to the digit where we have the first n natural natural numbers in ascending order. So, in our number 123454321, we have to replace each digit by 1 up to 5. That is the square root of 123454321. Hence the square root of 123454321 is 11111. jQuery UI Dialog functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} Total distance covered to cross each other = 100 + 80 = 180 m (when they run in opposite direction or same direction) Relative speed of the two trains = 60-18 = 12 kmph (running in the same direction) = 12X5/18 = 10/3 m/sec Time taken to cross each other = Distance/Speed = 180/(10/3) seconds =180X3/10 seconds =54 seconds Hence, time taken to cross each other = 54 seconds. jQuery UI Dialog functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} Given Information: The work has to completed in 90 days (as per schedule) Total no. of men appointed initially = 50 50 men worked 20 days and completed a part of the work The remaining work is completed by 70 men (50+20=70) in 70 days (90-20=70) If the remaining work is completed by 50 men, no. of days taken by them = (70X70)/50 = 98 days. Hence, extra days needed = 98-70 = 28 days. jQuery UI Dialog functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} From the given information, Already 3 km distance (one half of the distance 6km) has been covered in 30 minutes or 1/2 hr. (two third of the total time 45 minutes) Remaining distance = 3 km Time available = 15 minutes or 1/4 hr Speed required = Distance / Time = 3/(1/4) kmph = 3X4 kmph = 12 kmph Hence, speed required to cover remaining distance is 12 kmph jQuery UI Dialog functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} Ten years ago, age of John and his father are x and 5x Then, present age of John and his father are (x+10) and (5x+10) John's age after six years will be three-seventh of his father's age (x+10+6) = 3/7(5x+10+6) (x+16) = 3/7(5x+16) 7(x+16) = 3(5x+16) Solving the above equation, we get x=8 Present age of John's father = 5x+10 plug x = 8 Present age of John's father = 5(8)+10= 50 yrs jQuery UI Dialog functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} Let "x" be John's weight According to John, we have 65 < x < 72 According to his brother, we have 60 < x < 70 According to his mother, we have x≤68 The values of "x" which satisfy all the above three conditions are 66, 67 and 68 Average of the above three values = (66+67+68)/3 = 201/3 = 67 kg Hence average of different probable weights of John is 67 kg jQuery UI Dialog functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} Let "x" mph be speed in still water and "y" mph be speed of the current. Then, speed downstream = (x+y) mph 12/(x-y) - 12(x+y) = 6 By simplification, we get x2 = y2+4y ----(1) If the speed is doubled, speed in still water = 2x and speed of the current is same "y". Speed down stream = (2x+y) mph and speed up stream = (2x-y) mph 12/(2x-y) - 12(2x+y) = 1 By simplification, we get x2 = (24y+y2)/4 ----(2) Solving (1) and (2), we get y = 8/3 = 22⁄3 mph Hence speed of the current = 22⁄3 mph jQuery UI Dialog functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} From the given information, Income of the 1st person = 4x Income of the 2nd person = 5x Expenditure of the 1st person = 4x-50 Expenditure of the 2nd person = 5x-50 (Because each saves \$50 per month and Expenditure = Income - Savings) Expenditure ratio = 7:9 (4x-50):(5x-50) = 7:9 9(4x-50) = 7(5x-50) 36x-450 = 35x-350 x = 100 Hence income of the second person = 5X100 = \$500 jQuery UI Dialog functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} Let the cost price of 1 liter pure milk be \$1 Milk in 1 liter mixture in the 1st can = 3/4 (that is 75%) Milk in 1 liter mixture in the 2nd can = 1/2 (that is 50%) Milk in 1 liter mixture in the final mix = 5/8 (from the given ratio w:m = 3:5) C.P of 1 liter mixture in the 1st can (c) = \$3/4 C.P of 1 liter mixture in the 2nd can (d) = \$1/2 C.P of 1 liter mixture in the final mix (m) = \$5/8 Rule to find the ratio for producing mixture = (d-m):(m-c) (d-m):(m-c) = 1/2-5/8:5/8-3/4 = 1/8:1/8 = 1:1 The above found ratio 1:1 says that equal quantity of milk should be taken from each of the cans. Since he wants to get 12 liters of milk, he should take 6 liters of milk from each of the cans. The correct answer is option (B) (6,6). jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} Let 100 tons be the production of wheat in 1994 In 1994 ===> 100 tons In 1994-1995 ===> 120 tons (20% increment) In 1995-1996 ===> 150 tons (further 25% increment) When we look in to the above calculations, it is very clear that the production of wheat has been increased by 50 tons in 1996 from 100 tons in 1994 Percentage change = (50/100)X100 % = 50% Hence, the percentage change in the production of wheat from 1994 to 1996 was 50%
| 2,779 | 9,349 |
{"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}
| 4.125 | 4 |
CC-MAIN-2018-05
|
latest
|
en
| 0.944497 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=154041
| 1,369,436,452,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368705195219/warc/CC-MAIN-20130516115315-00003-ip-10-60-113-184.ec2.internal.warc.gz
| 309,947,061 | 851 |
```Question 204102
use L for length and W for width:
Perimeter is: 2*L + 2*W = 28.
Length is 2 meters more than twice the width: L = 2*W+2.
Use the above for length in the perimeter:
2*(2*W+2) + 2*W = 28.
*[invoke explain_simplification "2*(2*W+2) + 2*W = 28." ]
So, width = 4.
Since length = 2*W+2, length = 2*4+2 = 10.
Answer: Width = 4, length = 10.```
| 145 | 362 |
{"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}
| 3.671875 | 4 |
CC-MAIN-2013-20
|
latest
|
en
| 0.662429 |
https://www.hackmath.net/en/example/2931
| 1,558,510,024,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-22/segments/1558232256764.75/warc/CC-MAIN-20190522063112-20190522085112-00320.warc.gz
| 801,618,733 | 7,063 |
# Numbers at ratio
The two numbers are in a ratio 3:2. If we each increase by 5 would be at a ratio of 4:3. What is the sum of original numbers?
Result
s = 25
#### Solution:
2a=3b
3(a+5)=4(b+5)
s = a+b
2a-3b = 0
3a-4b = 5
a+b-s = 0
a = 15
b = 10
s = 25
Calculated by our linear equations calculator.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this example are needed these knowledge from mathematics:
Do you have a system of equations and looking for calculator system of linear equations?
## Next similar examples:
1. Sheep and cows
There are only sheep and cows on the farm. Sheep is eight more than cows. The number of cows is half the number of sheep. How many animals live on the farm?
2. Flowers 2
Cha cruz has a garden. The ratio roses to tulips is 2 : 5, the ratio of roses to orchids is 7 : 6. Cha cruz wonders what the ratio of tulips to orchids is. If Cha cruz has 183 plants, how many of each kind are there?
3. Unknown amount of money
Damian and Denis split an unknown amount in the ratio of 5:4 . Damian got six euros more than Denis. Calculate an unknown amount. Determine how much money got Damian and how Denis.
4. Golden ratio
Divide line of length 14 cm into two sections that the ratio of shorter to greater is same as ratio of greater section to whole length of the line.
5. Three workshops
There are 2743 people working in three workshops. In the second workshop works 140 people more than in the first and in third works 4.2 times more than the second one. How many people work in each workshop?
6. Factory and divisions
The factory consists of three auxiliary divisions total 2,406 employees. The second division has 76 employees less than 1st division and 3rd division has 212 employees more than the 2nd. How many employees has each division?
7. 925 USD
Four classmates saved an annual total 925 USD. The second save twice as the first, third 35 USD more than the second and fourth 10 USD less than the first. How USD save each of them?
8. Nine books
Nine books are to be bought by a student. Art books cost \$6.00 each and biology books cost \$6.50 each . If the total amount spent was \$56.00, how many of each book was bought?
9. Theorem prove
We want to prove the sentense: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started?
10. Potatoes
For three days the store sold 1400 kg of potatoes. The first day they sold 100 kilograms of potatoes less than the second day, the third-day three-fifths of what they sold the first day. How many kgs of potatoes sold every day?
11. Elimination method
Solve system of linear equations by elimination method: 5/2x + 3/5y= 4/15 1/2x + 2/5y= 2/15
12. Legs
Cancer has 5 pairs of legs. The insect has 6 legs. 60 animals have a total of 500 legs. How much more are cancers than insects?
13. Three unknowns
Solve the system of linear equations with three unknowns: A + B + C = 14 B - A - C = 4 2A - B + C = 0
14. Linsys2
Solve two equations with two unknowns: 400x+120y=147.2 350x+200y=144
15. Family
Family has 4 children. Ondra is 3 years older than Matthew and Karlos 5 years older than the youngest Jane. We know that they are together 30 years and 3 years ago they were together 19 years. Determine how old the children are.
16. Linear system
Solve a set of two equations of two unknowns: 1.5x+1.2y=0.6 0.8x-0.2y=2
17. Three friends
Danica, Lenka and Dalibor have altogether 96 kg. Lenka weighs 75% more than Dalibor and Danica weighs 6 kg more than Dalibor. Determine the weight of Danice, Lenka and Dalibor.
| 1,014 | 3,632 |
{"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-2019-22
|
latest
|
en
| 0.948317 |
https://www.javatpoint.com/os-partitioning-algorithms
| 1,675,557,894,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500158.5/warc/CC-MAIN-20230205000727-20230205030727-00103.warc.gz
| 837,428,867 | 17,121 |
# Partitioning Algorithms
There are various algorithms which are implemented by the Operating System in order to find out the holes in the linked list and allocate them to the processes.
The explanation about each of the algorithm is given below.
1. First Fit Algorithm
First Fit algorithm scans the linked list and whenever it finds the first big enough hole to store a process, it stops scanning and load the process into that hole. This procedure produces two partitions. Out of them, one partition will be a hole while the other partition will store the process.
First Fit algorithm maintains the linked list according to the increasing order of starting index. This is the simplest to implement among all the algorithms and produces bigger holes as compare to the other algorithms.
2. Next Fit Algorithm
Next Fit algorithm is similar to First Fit algorithm except the fact that, Next fit scans the linked list from the node where it previously allocated a hole.
Next fit doesn't scan the whole list, it starts scanning the list from the next node. The idea behind the next fit is the fact that the list has been scanned once therefore the probability of finding the hole is larger in the remaining part of the list.
Experiments over the algorithm have shown that the next fit is not better then the first fit. So it is not being used these days in most of the cases.
3. Best Fit Algorithm
The Best Fit algorithm tries to find out the smallest hole possible in the list that can accommodate the size requirement of the process.
Using Best Fit has some disadvantages.
1. 1. It is slower because it scans the entire list every time and tries to find out the smallest hole which can satisfy the requirement the process.
2. Due to the fact that the difference between the whole size and the process size is very small, the holes produced will be as small as it cannot be used to load any process and therefore it remains useless.
Despite of the fact that the name of the algorithm is best fit, It is not the best algorithm among all.
4. Worst Fit Algorithm
The worst fit algorithm scans the entire list every time and tries to find out the biggest hole in the list which can fulfill the requirement of the process.
Despite of the fact that this algorithm produces the larger holes to load the other processes, this is not the better approach due to the fact that it is slower because it searches the entire list every time again and again.
5. Quick Fit Algorithm
The quick fit algorithm suggestsmaintaining the different lists of frequently used sizes. Although, it is not practically suggestible because the procedure takes so much time to create the different lists and then expending the holes to load a process.
The first fit algorithm is the best algorithm among all because
1. It takes lesser time compare to the other algorithms.
2. It produces bigger holes that can be used to load other processes later on.
3. It is easiest to implement.
| 584 | 2,966 |
{"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}
| 3.578125 | 4 |
CC-MAIN-2023-06
|
longest
|
en
| 0.935824 |
https://www.coursehero.com/file/5934110/Dynamics-Hibbeler-CH12-7/
| 1,498,462,699,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128320685.63/warc/CC-MAIN-20170626064746-20170626084746-00361.warc.gz
| 860,271,035 | 182,114 |
Dynamics_Hibbeler_CH12_7
# Dynamics_Hibbeler_CH12_7 - CURVILINEAR MOTION NORMAL AND...
This preview shows pages 1–6. Sign up to view the full content.
CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS Today’s Objectives : Students will be able to: 1. Determine the normal and tangential components of velocity and acceleration of a particle traveling along a curved path. In-Class Activities : Check Homework Reading Quiz Applications Normal and Tangential Components of Velocity and Acceleration Special Cases of Motion Concept Quiz Group Problem Solving Attention Quiz
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
READING QUIZ 1. If a particle moves along a curve with a constant speed, then its tangential component of acceleration is A) positive. B) negative. C) zero. D) constant. 2. The normal component of acceleration represents A) the time rate of change in the magnitude of the velocity. B) the time rate of change in the direction of the velocity. C) magnitude of the velocity. D) direction of the total acceleration.
APPLICATIONS Cars traveling along a clover-leaf interchange experience an acceleration due to a change in velocity as well as due to a change in direction of the velocity. If the car’s speed is increasing at a known rate as it travels along a curve, how can we determine the magnitude and direction of its total acceleration? Why would you care about the total acceleration of the car?
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
APPLICATIONS (continued) A roller coaster travels down a hill for which the path can be approximated by a function y = f(x). The roller coaster starts from rest and increases its speed at a constant rate. How can we determine its velocity and acceleration at the bottom? Why would we want to know these values?
NORMAL AND TANGENTIAL COMPONENTS (Section 12.7) When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian. When the path of motion is known , normal (n) and tangential (t) coordinates are often used. In the n-t coordinate system, the origin is located on the particle (the origin moves with the particle ). The
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
## This note was uploaded on 08/29/2010 for the course EGN 3321 taught by Professor Nohra during the Spring '10 term at University of South Florida.
### Page1 / 19
Dynamics_Hibbeler_CH12_7 - CURVILINEAR MOTION NORMAL AND...
This preview shows document pages 1 - 6. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online
| 619 | 2,827 |
{"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-2017-26
|
latest
|
en
| 0.896826 |
https://socratic.org/questions/how-do-you-find-the-derivative-of-x-8-x-23-1-2-27x-6-4x-6-8
| 1,611,732,584,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610704821253.82/warc/CC-MAIN-20210127055122-20210127085122-00237.warc.gz
| 569,568,295 | 6,038 |
# How do you find the derivative of (x^8(x-23)^(1/2))/(27x^6(4x-6)^8)?
Dec 8, 2017
=(x^8(432x^5(4x-6)^8(x-23)^(1/2)+(x-23)^(-1/2)-2x^4(x-23)^(1/2)(54x^5(3(4x-6)^8+16x(4x-6)^7))))/(2(27x^6(4x-8)^8)^2
#### Explanation:
$y = \frac{{x}^{8} {\left(x - 23\right)}^{\frac{1}{2}}}{27 {x}^{6} {\left(4 x - 6\right)}^{8}} = f \frac{x}{g} \left(x\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{f {\left(x\right)}^{2}}$
$f \left(x\right) = {x}^{8} {\left(x - 23\right)}^{\frac{1}{2}} = h \left(x\right) j \left(x\right)$
$f ' \left(x\right) = h ' \left(x\right) j \left(x\right) + h \left(x\right) j ' \left(x\right)$
$h \left(x\right) = {x}^{8}$
$h ' \left(x\right) = 8 {x}^{7}$
$j \left(x\right) = {\left(x - 23\right)}^{\frac{1}{2}}$
$j ' \left(x\right) = {\left(x - 23\right)}^{- \frac{1}{2}} / 2$
$f ' \left(x\right) = 8 {x}^{7} {\left(x - 23\right)}^{\frac{1}{2}} + \frac{{x}^{8} {\left(x - 23\right)}^{- \frac{1}{2}}}{2} = \frac{16 {x}^{7} {\left(x - 23\right)}^{\frac{1}{2}} + {x}^{8} {\left(x - 23\right)}^{- \frac{1}{2}}}{2} = \frac{16 {x}^{7} {\left(x - 23\right)}^{\frac{1}{2}} + {x}^{8} {\left(x - 23\right)}^{- \frac{1}{2}}}{2}$
$g \left(x\right) = 27 {x}^{6} {\left(4 x - 6\right)}^{8} = a \left(x\right) s \left(x\right)$
$g ' \left(x\right) = a ' \left(x\right) s \left(x\right) + a \left(x\right) s ' \left(x\right)$
$a \left(x\right) = 27 {x}^{6}$
$a ' \left(x\right) = 162 {x}^{5}$
$s \left(x\right) = {\left(4 x - 6\right)}^{8}$
$s ' \left(x\right) = 4 \cdot 8 \cdot {\left(4 x - 6\right)}^{7} = 32 {\left(4 x - 6\right)}^{7}$
$g ' \left(x\right) = 162 {x}^{5} {\left(4 x - 6\right)}^{8} + 27 {x}^{6} 32 {\left(4 x - 6\right)}^{7} = 162 {x}^{5} {\left(4 x - 6\right)}^{8} + 864 {x}^{6} {\left(4 x - 6\right)}^{7} = 54 {x}^{5} \left(3 {\left(4 x - 6\right)}^{8} + 16 x {\left(4 x - 6\right)}^{7}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{27 {x}^{6} {\left(4 x - 6\right)}^{8} \left(\frac{16 {x}^{7} {\left(x - 23\right)}^{\frac{1}{2}} + {x}^{8} {\left(x - 23\right)}^{- \frac{1}{2}}}{2}\right) - {x}^{8} {\left(x - 23\right)}^{\frac{1}{2}} \left(54 {x}^{5} \left(3 {\left(4 x - 6\right)}^{8} + 16 x {\left(4 x - 6\right)}^{7}\right)\right)}{27 {x}^{6} {\left(4 x - 6\right)}^{8}} ^ 2$
$= \frac{\left(\frac{27 {x}^{6} {\left(4 x - 6\right)}^{8} 16 {x}^{7} {\left(x - 23\right)}^{\frac{1}{2}} + {x}^{8} {\left(x - 23\right)}^{- \frac{1}{2}}}{2}\right) - {x}^{8} {\left(x - 23\right)}^{\frac{1}{2}} \left(54 {x}^{5} \left(3 {\left(4 x - 6\right)}^{8} + 16 x {\left(4 x - 6\right)}^{7}\right)\right)}{27 {x}^{6} {\left(4 x - 6\right)}^{8}} ^ 2$
$= \frac{27 {x}^{6} {\left(4 x - 6\right)}^{8} 16 {x}^{7} {\left(x - 23\right)}^{\frac{1}{2}} + {x}^{8} {\left(x - 23\right)}^{- \frac{1}{2}} - 2 {x}^{8} {\left(x - 23\right)}^{\frac{1}{2}} \left(54 {x}^{5} \left(3 {\left(4 x - 6\right)}^{8} + 16 x {\left(4 x - 6\right)}^{7}\right)\right)}{2 {\left(27 {x}^{6} {\left(4 x - 6\right)}^{8}\right)}^{2}}$
=(x^8(432x^5(4x-6)^8(x-23)^(1/2)+(x-23)^(-1/2)-2x^4(x-23)^(1/2)(54x^5(3(4x-6)^8+16x(4x-6)^7))))/(2(27x^6(4x-8)^8)^2
| 1,604 | 3,108 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 21, "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}
| 4.375 | 4 |
CC-MAIN-2021-04
|
latest
|
en
| 0.266525 |
https://netlib.org/lapack/explore-html-3.6.1/d1/d3c/group__double__eig_ga13302edd654451c2fd44502a47bc3740.html
| 1,722,873,878,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00571.warc.gz
| 338,220,931 | 9,407 |
LAPACK 3.6.1 LAPACK: Linear Algebra PACKage
subroutine dlatm4 ( integer ITYPE, integer N, integer NZ1, integer NZ2, integer ISIGN, double precision AMAGN, double precision RCOND, double precision TRIANG, integer IDIST, integer, dimension( 4 ) ISEED, double precision, dimension( lda, * ) A, integer LDA )
DLATM4
Purpose:
``` DLATM4 generates basic square matrices, which may later be
multiplied by others in order to produce test matrices. It is
intended mainly to be used to test the generalized eigenvalue
routines.
It first generates the diagonal and (possibly) subdiagonal,
according to the value of ITYPE, NZ1, NZ2, ISIGN, AMAGN, and RCOND.
It then fills in the upper triangle with random numbers, if TRIANG is
non-zero.```
Parameters
[in] ITYPE ``` ITYPE is INTEGER The "type" of matrix on the diagonal and sub-diagonal. If ITYPE < 0, then type abs(ITYPE) is generated and then swapped end for end (A(I,J) := A'(N-J,N-I).) See also the description of AMAGN and ISIGN. Special types: = 0: the zero matrix. = 1: the identity. = 2: a transposed Jordan block. = 3: If N is odd, then a k+1 x k+1 transposed Jordan block followed by a k x k identity block, where k=(N-1)/2. If N is even, then k=(N-2)/2, and a zero diagonal entry is tacked onto the end. Diagonal types. The diagonal consists of NZ1 zeros, then k=N-NZ1-NZ2 nonzeros. The subdiagonal is zero. ITYPE specifies the nonzero diagonal entries as follows: = 4: 1, ..., k = 5: 1, RCOND, ..., RCOND = 6: 1, ..., 1, RCOND = 7: 1, a, a^2, ..., a^(k-1)=RCOND = 8: 1, 1-d, 1-2*d, ..., 1-(k-1)*d=RCOND = 9: random numbers chosen from (RCOND,1) = 10: random numbers with distribution IDIST (see DLARND.)``` [in] N ``` N is INTEGER The order of the matrix.``` [in] NZ1 ``` NZ1 is INTEGER If abs(ITYPE) > 3, then the first NZ1 diagonal entries will be zero.``` [in] NZ2 ``` NZ2 is INTEGER If abs(ITYPE) > 3, then the last NZ2 diagonal entries will be zero.``` [in] ISIGN ``` ISIGN is INTEGER = 0: The sign of the diagonal and subdiagonal entries will be left unchanged. = 1: The diagonal and subdiagonal entries will have their sign changed at random. = 2: If ITYPE is 2 or 3, then the same as ISIGN=1. Otherwise, with probability 0.5, odd-even pairs of diagonal entries A(2*j-1,2*j-1), A(2*j,2*j) will be converted to a 2x2 block by pre- and post-multiplying by distinct random orthogonal rotations. The remaining diagonal entries will have their sign changed at random.``` [in] AMAGN ``` AMAGN is DOUBLE PRECISION The diagonal and subdiagonal entries will be multiplied by AMAGN.``` [in] RCOND ``` RCOND is DOUBLE PRECISION If abs(ITYPE) > 4, then the smallest diagonal entry will be entry will be RCOND. RCOND must be between 0 and 1.``` [in] TRIANG ``` TRIANG is DOUBLE PRECISION The entries above the diagonal will be random numbers with magnitude bounded by TRIANG (i.e., random numbers multiplied by TRIANG.)``` [in] IDIST ``` IDIST is INTEGER Specifies the type of distribution to be used to generate a random matrix. = 1: UNIFORM( 0, 1 ) = 2: UNIFORM( -1, 1 ) = 3: NORMAL ( 0, 1 )``` [in,out] ISEED ``` ISEED is INTEGER array, dimension (4) On entry ISEED specifies the seed of the random number generator. The values of ISEED are changed on exit, and can be used in the next call to DLATM4 to continue the same random number sequence. Note: ISEED(4) should be odd, for the random number generator used at present.``` [out] A ``` A is DOUBLE PRECISION array, dimension (LDA, N) Array to be computed.``` [in] LDA ``` LDA is INTEGER Leading dimension of A. Must be at least 1 and at least N.```
Date
November 2011
Definition at line 177 of file dlatm4.f.
177 *
178 * -- LAPACK test routine (version 3.4.0) --
179 * -- LAPACK is a software package provided by Univ. of Tennessee, --
180 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
181 * November 2011
182 *
183 * .. Scalar Arguments ..
184 INTEGER idist, isign, itype, lda, n, nz1, nz2
185 DOUBLE PRECISION amagn, rcond, triang
186 * ..
187 * .. Array Arguments ..
188 INTEGER iseed( 4 )
189 DOUBLE PRECISION a( lda, * )
190 * ..
191 *
192 * =====================================================================
193 *
194 * .. Parameters ..
195 DOUBLE PRECISION zero, one, two
196 parameter ( zero = 0.0d0, one = 1.0d0, two = 2.0d0 )
197 DOUBLE PRECISION half
198 parameter ( half = one / two )
199 * ..
200 * .. Local Scalars ..
201 INTEGER i, ioff, isdb, isde, jc, jd, jr, k, kbeg, kend,
202 \$ klen
203 DOUBLE PRECISION alpha, cl, cr, safmin, sl, sr, sv1, sv2, temp
204 * ..
205 * .. External Functions ..
206 DOUBLE PRECISION dlamch, dlaran, dlarnd
207 EXTERNAL dlamch, dlaran, dlarnd
208 * ..
209 * .. External Subroutines ..
210 EXTERNAL dlaset
211 * ..
212 * .. Intrinsic Functions ..
213 INTRINSIC abs, dble, exp, log, max, min, mod, sqrt
214 * ..
215 * .. Executable Statements ..
216 *
217 IF( n.LE.0 )
218 \$ RETURN
219 CALL dlaset( 'Full', n, n, zero, zero, a, lda )
220 *
221 * Insure a correct ISEED
222 *
223 IF( mod( iseed( 4 ), 2 ).NE.1 )
224 \$ iseed( 4 ) = iseed( 4 ) + 1
225 *
226 * Compute diagonal and subdiagonal according to ITYPE, NZ1, NZ2,
227 * and RCOND
228 *
229 IF( itype.NE.0 ) THEN
230 IF( abs( itype ).GE.4 ) THEN
231 kbeg = max( 1, min( n, nz1+1 ) )
232 kend = max( kbeg, min( n, n-nz2 ) )
233 klen = kend + 1 - kbeg
234 ELSE
235 kbeg = 1
236 kend = n
237 klen = n
238 END IF
239 isdb = 1
240 isde = 0
241 GO TO ( 10, 30, 50, 80, 100, 120, 140, 160,
242 \$ 180, 200 )abs( itype )
243 *
244 * abs(ITYPE) = 1: Identity
245 *
246 10 CONTINUE
247 DO 20 jd = 1, n
248 a( jd, jd ) = one
249 20 CONTINUE
250 GO TO 220
251 *
252 * abs(ITYPE) = 2: Transposed Jordan block
253 *
254 30 CONTINUE
255 DO 40 jd = 1, n - 1
256 a( jd+1, jd ) = one
257 40 CONTINUE
258 isdb = 1
259 isde = n - 1
260 GO TO 220
261 *
262 * abs(ITYPE) = 3: Transposed Jordan block, followed by the
263 * identity.
264 *
265 50 CONTINUE
266 k = ( n-1 ) / 2
267 DO 60 jd = 1, k
268 a( jd+1, jd ) = one
269 60 CONTINUE
270 isdb = 1
271 isde = k
272 DO 70 jd = k + 2, 2*k + 1
273 a( jd, jd ) = one
274 70 CONTINUE
275 GO TO 220
276 *
277 * abs(ITYPE) = 4: 1,...,k
278 *
279 80 CONTINUE
280 DO 90 jd = kbeg, kend
281 a( jd, jd ) = dble( jd-nz1 )
282 90 CONTINUE
283 GO TO 220
284 *
285 * abs(ITYPE) = 5: One large D value:
286 *
287 100 CONTINUE
288 DO 110 jd = kbeg + 1, kend
289 a( jd, jd ) = rcond
290 110 CONTINUE
291 a( kbeg, kbeg ) = one
292 GO TO 220
293 *
294 * abs(ITYPE) = 6: One small D value:
295 *
296 120 CONTINUE
297 DO 130 jd = kbeg, kend - 1
298 a( jd, jd ) = one
299 130 CONTINUE
300 a( kend, kend ) = rcond
301 GO TO 220
302 *
303 * abs(ITYPE) = 7: Exponentially distributed D values:
304 *
305 140 CONTINUE
306 a( kbeg, kbeg ) = one
307 IF( klen.GT.1 ) THEN
308 alpha = rcond**( one / dble( klen-1 ) )
309 DO 150 i = 2, klen
310 a( nz1+i, nz1+i ) = alpha**dble( i-1 )
311 150 CONTINUE
312 END IF
313 GO TO 220
314 *
315 * abs(ITYPE) = 8: Arithmetically distributed D values:
316 *
317 160 CONTINUE
318 a( kbeg, kbeg ) = one
319 IF( klen.GT.1 ) THEN
320 alpha = ( one-rcond ) / dble( klen-1 )
321 DO 170 i = 2, klen
322 a( nz1+i, nz1+i ) = dble( klen-i )*alpha + rcond
323 170 CONTINUE
324 END IF
325 GO TO 220
326 *
327 * abs(ITYPE) = 9: Randomly distributed D values on ( RCOND, 1):
328 *
329 180 CONTINUE
330 alpha = log( rcond )
331 DO 190 jd = kbeg, kend
332 a( jd, jd ) = exp( alpha*dlaran( iseed ) )
333 190 CONTINUE
334 GO TO 220
335 *
336 * abs(ITYPE) = 10: Randomly distributed D values from DIST
337 *
338 200 CONTINUE
339 DO 210 jd = kbeg, kend
340 a( jd, jd ) = dlarnd( idist, iseed )
341 210 CONTINUE
342 *
343 220 CONTINUE
344 *
345 * Scale by AMAGN
346 *
347 DO 230 jd = kbeg, kend
348 a( jd, jd ) = amagn*dble( a( jd, jd ) )
349 230 CONTINUE
350 DO 240 jd = isdb, isde
351 a( jd+1, jd ) = amagn*dble( a( jd+1, jd ) )
352 240 CONTINUE
353 *
354 * If ISIGN = 1 or 2, assign random signs to diagonal and
355 * subdiagonal
356 *
357 IF( isign.GT.0 ) THEN
358 DO 250 jd = kbeg, kend
359 IF( dble( a( jd, jd ) ).NE.zero ) THEN
360 IF( dlaran( iseed ).GT.half )
361 \$ a( jd, jd ) = -a( jd, jd )
362 END IF
363 250 CONTINUE
364 DO 260 jd = isdb, isde
365 IF( dble( a( jd+1, jd ) ).NE.zero ) THEN
366 IF( dlaran( iseed ).GT.half )
367 \$ a( jd+1, jd ) = -a( jd+1, jd )
368 END IF
369 260 CONTINUE
370 END IF
371 *
372 * Reverse if ITYPE < 0
373 *
374 IF( itype.LT.0 ) THEN
375 DO 270 jd = kbeg, ( kbeg+kend-1 ) / 2
376 temp = a( jd, jd )
377 a( jd, jd ) = a( kbeg+kend-jd, kbeg+kend-jd )
378 a( kbeg+kend-jd, kbeg+kend-jd ) = temp
379 270 CONTINUE
380 DO 280 jd = 1, ( n-1 ) / 2
381 temp = a( jd+1, jd )
382 a( jd+1, jd ) = a( n+1-jd, n-jd )
383 a( n+1-jd, n-jd ) = temp
384 280 CONTINUE
385 END IF
386 *
387 * If ISIGN = 2, and no subdiagonals already, then apply
388 * random rotations to make 2x2 blocks.
389 *
390 IF( isign.EQ.2 .AND. itype.NE.2 .AND. itype.NE.3 ) THEN
391 safmin = dlamch( 'S' )
392 DO 290 jd = kbeg, kend - 1, 2
393 IF( dlaran( iseed ).GT.half ) THEN
394 *
395 * Rotation on left.
396 *
397 cl = two*dlaran( iseed ) - one
398 sl = two*dlaran( iseed ) - one
399 temp = one / max( safmin, sqrt( cl**2+sl**2 ) )
400 cl = cl*temp
401 sl = sl*temp
402 *
403 * Rotation on right.
404 *
405 cr = two*dlaran( iseed ) - one
406 sr = two*dlaran( iseed ) - one
407 temp = one / max( safmin, sqrt( cr**2+sr**2 ) )
408 cr = cr*temp
409 sr = sr*temp
410 *
411 * Apply
412 *
413 sv1 = a( jd, jd )
414 sv2 = a( jd+1, jd+1 )
415 a( jd, jd ) = cl*cr*sv1 + sl*sr*sv2
416 a( jd+1, jd ) = -sl*cr*sv1 + cl*sr*sv2
417 a( jd, jd+1 ) = -cl*sr*sv1 + sl*cr*sv2
418 a( jd+1, jd+1 ) = sl*sr*sv1 + cl*cr*sv2
419 END IF
420 290 CONTINUE
421 END IF
422 *
423 END IF
424 *
425 * Fill in upper triangle (except for 2x2 blocks)
426 *
427 IF( triang.NE.zero ) THEN
428 IF( isign.NE.2 .OR. itype.EQ.2 .OR. itype.EQ.3 ) THEN
429 ioff = 1
430 ELSE
431 ioff = 2
432 DO 300 jr = 1, n - 1
433 IF( a( jr+1, jr ).EQ.zero )
434 \$ a( jr, jr+1 ) = triang*dlarnd( idist, iseed )
435 300 CONTINUE
436 END IF
437 *
438 DO 320 jc = 2, n
439 DO 310 jr = 1, jc - ioff
440 a( jr, jc ) = triang*dlarnd( idist, iseed )
441 310 CONTINUE
442 320 CONTINUE
443 END IF
444 *
445 RETURN
446 *
447 * End of DLATM4
448 *
subroutine dlaset(UPLO, M, N, ALPHA, BETA, A, LDA)
DLASET initializes the off-diagonal elements and the diagonal elements of a matrix to given values...
Definition: dlaset.f:112
double precision function dlamch(CMACH)
DLAMCH
Definition: dlamch.f:65
double precision function dlarnd(IDIST, ISEED)
DLARND
Definition: dlarnd.f:75
double precision function dlaran(ISEED)
DLARAN
Definition: dlaran.f:69
Here is the call graph for this function:
Here is the caller graph for this function:
| 4,083 | 10,809 |
{"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}
| 3.65625 | 4 |
CC-MAIN-2024-33
|
latest
|
en
| 0.769726 |
https://malcolmmackillop.com/how-to-solve-ohm-s-law/
| 1,716,322,088,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058512.80/warc/CC-MAIN-20240521183800-20240521213800-00728.warc.gz
| 334,919,259 | 13,323 |
# How To Solve Ohm’s Law5 min read
Ohm’s law states that the current through a conductor between two points is directly proportional to the potential difference between those points. In mathematical terms, this is expressed as I = V/R, where I is the current in amps, V is the potential difference in volts, and R is the resistance in ohms.
This law is perhaps the most fundamental of all the laws of electricity, and is the basis for all other electrical calculations. It can be used to solve a wide range of problems, from calculating current and voltage in a circuit, to finding the resistance of a resistor.
To solve a problem using Ohm’s law, you first need to identify the three variables in the equation. Then, you can use the appropriate equation to solve for the desired variable.
For example, if you are given a circuit with a voltage of 12 volts and a resistance of 2 ohms, you would solve for the current in the circuit by using the equation I = V/R, which would give you a value of 6 amps.
Similarly, if you are given a resistor with a value of 1 ohm and a current of 5 amps, you would solve for the voltage across the resistor by using the equation V = I R, which would give you a value of 50 volts.
Ohm’s law is a very versatile equation, and can be used to solve a wide variety of problems. By understanding how to use Ohm’s law, you can effectively solve any electrical problem you may encounter.
## How do you calculate Ohm’s law?
Ohm’s law is a fundamental law of electricity that states that the current through a conductor is directly proportional to the potential difference between its ends. In other words, the current is proportional to the voltage.
To calculate the current, you need to know the voltage and the resistance. The resistance is the opposition to the flow of current, and is measured in ohms.
The current can be calculated using the following equation:
I = V/R
For example, if you have a 12-volt battery and a resistor with a resistance of 2 ohms, the current would be 6 amps (12 volts / 2 ohms).
## What are the 3 formulas in ohms law?
What are the three formulas in ohms law?
The three formulas in ohms law are Voltage (V) = Current (I) x Resistance (R), Power (P) = Voltage (V) x Current (I), and Energy (E) = Power (P) x Time (t).
The voltage (V) is the electrical potential difference between two points in a circuit. The current (I) is the flow of electricity through a conductor. The resistance (R) is the opposition to the current flow. The power (P) is the rate at which energy is converted from one form to another. The energy (E) is the amount of work done or power (P) used multiplied by the time (t) it is applied.
## What is Ohm’s law answer?
Ohm’s law is a law that states that the current through a conductor is directly proportional to the voltage across the conductor, and inversely proportional to the resistance of the conductor. In other words, the current is proportional to the voltage divided by the resistance.
Read also How Does Martial Law Work
The law is named after Georg Simon Ohm, who published it in 1827.
## How do you calculate resistance using Ohm’s law?
Calculating Resistance
The resistor is probably the most common electronic component. A resistor is a component in an electronic circuit that limits the flow of current. The resistance is measured in ohms. The calculation of resistance is simple using Ohm’s law.
V=IR
V is the voltage in volts. I is the current in amps. R is the resistance in ohms.
To calculate the resistance, you need to know the voltage and the current. You can use a multimeter to measure the voltage and current. The resistance is the voltage divided by the current.
For example, if the voltage is 5 volts and the current is 2 amps, the resistance is 5 divided by 2, or 2.5 ohms.
## How do I calculate resistance?
There are a few different ways that you can calculate resistance. One way is to use the formula R = V / I, where R is the resistance, V is the voltage, and I is the current. This formula is used to calculate the resistance of a resistor. Another way to calculate resistance is to use the formula R = ρL / A, where ρ is the resistivity of the material, L is the length of the resistor, and A is the cross-sectional area. This formula is used to calculate the resistance of a wire.
## What is ohm’s law answer?
Ohm’s law is a fundamental law of electricity that states that the current through a conductor is directly proportional to the voltage applied to it, and inversely proportional to the resistance of the conductor. In other words, if you increase the voltage, the current will increase proportionally, and if you decrease the voltage, the current will decrease proportionally.
Read also How To Determine Rate Law From Elementary Steps
Ohm’s law can be mathematically expressed as:
I = V / R
Where I is the current in amps, V is the voltage in volts, and R is the resistance in ohms.
This law is named after German physicist Georg Simon Ohm, who discovered it in 1827. It is one of the most important laws in electrical engineering, and is essential for understanding how electricity behaves in circuits.
## What is Ohm’s law in simple?
In electricity, Ohm’s law is a fundamental law that states that the current through a conductor between two points is directly proportional to the voltage across the two points. Mathematically, Ohm’s law is expressed as:
I = V / R
where
I is the current in amps
V is the voltage in volts
R is the resistance in ohms
| 1,247 | 5,519 |
{"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}
| 4.625 | 5 |
CC-MAIN-2024-22
|
latest
|
en
| 0.960306 |
https://www.latestquality.com/conduct-measurement-system-analysis-msa/
| 1,723,077,214,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640713903.39/warc/CC-MAIN-20240808000606-20240808030606-00088.warc.gz
| 670,595,867 | 19,440 |
Firstly, let me explain to you what measurement system analysis (MSA) is. In short then measurement system analysis (MSA) is a mathematical and experimental process which determines how much variation the measurement system contribute to the overall process variability.
## Measurement System Analysis Terminology
There are a few terminologies that are used in the measurement system analysis. Below I am going to explain all the common terminologies so that it is easier for you to understand the measurement system analysis.
### Discrimination
There is a smallest detectable addition between two measured values. When you use a gauge, there is a minimum count or least value which you can easily measure with the related gauge.
### Accuracy
You need to understand the fact that there is a difference between the observed average and the true average. Usually, the true average might be obtained by applying a more precise measure instrument. The very minute when there appears a difference between the average value and the true average, it suggests that your system isn’t accurate.
### Stability
Here we look at the difference that can be seen in the average of the two sets of measurements at least, which is measured over the time. Like when an object is measured at time (T1) and after time (T2) the average values are completely different.
## Elements of MSA
The key element in the measurement system analysis is the measurement itself. There are a total of 5 parameters to examine in an MSA. They are:
• Bias: It is also denoted as accuracy, in simple terms, it is a measure of the distance that is between the average value of all the quantities and the “True” or “Actual” value of the part.
• Linearity: It is a degree of the consistency of Bias over the whole range of the measurement tool. The amount of Bias also changes over the entire range of use.
• Stability: It refers to the ability of a measurement method which produces the very same values over a period of time when it is measured with the same sample.
• Repeatability: It assesses whether the same evaluator can measure the similar part numerous times with the same measurement tool and get the same value.
• Reproducibility: It assesses whether the dissimilar evaluators can measure the similar part with a similar measurement tool and as a result obtain the same value.
In the measurement system analysis, we rely on the data as much as we can. Along with that, we use various techniques to comprehend the difference within measuring devices. For example the variation that is brought into the measurement system by environment and people. There are a few useful ways for the evaluation of how much error is being contributed by the gauge and how much error is coming from an individual. For an MSA you identify numerous types of errors and how every error can be removed and avoided.
## Measurement Error
This error means that there is a difference between the true value and the value that is measured. It mainly depends on two things.
• The kind of tool that we are using
• The person using the tool
That is why whenever you use an instrument, always keep in mind that there can be possibilities of a measurement error that is present there.
Now you might ask how much measurement error is really acceptable.
Well according to AIAG (2002), a common rule of thumb for the measurement system acceptability is:
• Under 10 percent error is acceptable.
• The errors that range between 10 and 30 percent mean that the system is satisfactory, along with depending on the significance of application, cost of repair, cost of the measurement tool and other factors.
Errors that exceed 30 percent are considered to be unacceptable. It means that you should make improvements on your measuring system.
## Measurement Precision and Accuracy
The main purpose of measurement system analysis is to make the measurement system successful by evaluating the precision, accuracy, and stability.
• All the measurements are denoted accurate if they tend to revolve around the actual value of the sample that is being calculated. The accuracy is achievable when the calculated value has little difference from the actual value.
• The measurements are accurate if they vary from each other by only a small value.
## Use Consistent Measurements
Imagine that there is a call center and it has auditors that evaluate the quality of every phone companion’s call, and all the phone associates are then graded on their quality, which also impacts their paycheck. If every auditor is inspected to a dissimilar standard, then without a doubt, there will be a lot of difference in the measurement technique. Therefore it is crucial to use consistent measurements.
So there you have all the input on the Measurement Systems Analysis. I hope that the article has helped you in understanding what an MSA is and how it is conducted.
| 962 | 4,901 |
{"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}
| 3.671875 | 4 |
CC-MAIN-2024-33
|
latest
|
en
| 0.950467 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-5-polynomials-and-factoring-5-5-factoring-sums-or-differences-of-cubes-5-5-exercise-set-page-338/3
| 1,561,383,109,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-26/segments/1560627999539.60/warc/CC-MAIN-20190624130856-20190624152856-00258.warc.gz
| 764,537,014 | 13,507 |
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
Both terms of the given expression, $9x^4-25 ,$ are perfect squares and their coefficients have $\text{ different signs }.$ Hence, the given expression is $\text{ a difference of squares }.$
| 59 | 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": 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}
| 3.5625 | 4 |
CC-MAIN-2019-26
|
latest
|
en
| 0.8555 |
http://mathhelpforum.com/calculus/147766-optimization-problem.html
| 1,505,882,412,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-39/segments/1505818686169.5/warc/CC-MAIN-20170920033426-20170920053426-00640.warc.gz
| 224,504,451 | 10,666 |
1. ## Optimization Problem
Hi, I'm having problems with the following question. It's probably super easy... but I just keep getting the wrong answer.
A showroom for a car dealership is to be built in the shape of a rectangle with brick on the back and sides, and glass on the front. The floor of the showroom is to have an area of 500m^2.
a) If a brick wall costs $1200/m while a glass wall costs$600/m, what dimensions would minimize the cost of the showroom?
When I try to solve for one of the variables I get, 2400L+1800W=500 Is this the right way to start it?
Any help would be appreciated
2. Originally Posted by sam314159265
Hi, I'm having problems with the following question. It's probably super easy... but I just keep getting the wrong answer.
A showroom for a car dealership is to be built in the shape of a rectangle with brick on the back and sides, and glass on the front. The floor of the showroom is to have an area of 500m^2.
a) If a brick wall costs $1200/m while a glass wall costs$600/m, what dimensions would minimize the cost of the showroom?
When I try to solve for one of the variables I get, 2400L+1800W=500 Is this the right way to start it?
Any help would be appreciated
What you want is to minimize $S = 1200(2L+W) + 600W$ subject to $LW = 500$ .
3. So should I set the equation equal to zero? I tried that and got W=-3L/4, but when I plug that into A=LW and take the derivative to solve for W, I got W=8/3.... I'm not entirely sure what I'm doing wrong
4. Originally Posted by sam314159265
So should I set the equation equal to zero? I tried that and got W=-3L/4, but when I plug that into A=LW and take the derivative to solve for W, I got W=8/3.... I'm not entirely sure what I'm doing wrong
Using $LW = 500$, eliminate either $L$ or $W$ in $S$. This then gives a function of one variable. Now use calculus.
5. This is probably going to sound dumb, but what is "S"? Thanks for helping me btw
| 516 | 1,933 |
{"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": 6, "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}
| 4.21875 | 4 |
CC-MAIN-2017-39
|
longest
|
en
| 0.956214 |
https://betterlesson.com/lesson/601089/solving-quadrilateral-problems
| 1,539,693,461,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-43/segments/1539583510749.55/warc/CC-MAIN-20181016113811-20181016135311-00491.warc.gz
| 639,525,384 | 19,322 |
11 teachers like this lesson
Print Lesson
## Objective
SWBAT apply knowledge of attributes of quadrilaterals to solve problems.
#### Big Idea
Increased opportunities for practice working with quadrilaterals provides students with time to develop a conceptual understanding.
## Review & Refresh
10 minutes
At the beginning of this lesson, students review the types of quadrilaterals and their attributes. To facilitate this discussion I project the same screen that was used for classifying quadrilaterals in the previous lesson (Quadrilateral Properties & Attributes), review the concept of an inclusive definition, and ask students to make statements about each of the quadrilateral types.
Expectations:
Types of quadrilaterals: Students are able to list trapezoid, rhombus, square, rectangle, parallelogram
Attributes of each: Students are aware of the unique vs. inclusive attributes (ex: a rectangle has 2 sets of parallel lines and 4 right angles this includes a square. A square has the same attributes as a rectangle, but also has 4 equal sides).
When students are sharing, I listen carefully for language that is not precise (ex: a parallelogram is a rectangle with slanted sides). This is common kid language, but it is not accurate. I really spend time helping students understand that the side of the angles is an attribute, not a slanted side.
This review and refresh is intended to be a brief way to warm-up the students thinking. The lesson is designed to allow students time to work with these shapes and their attributes in small groups.
## Small Group Independent Practice
20 minutes
Students are given an opportunity to solve various problems involving the properties of quadrilaterals. These problems include finding the measurement of missing angles, naming quadrilaterals, finding area and perimeter, and solving problems with area and perimeter.
These questions were taken from the website mathscore.com, a helpful resource for both teachers and students. The word problems offer various levels of complexity for area and perimeter problems.
Students work in small groups to make sense of each of the problems and then solve them.
As students work, I correct their responses to the basic skills questions.
At the end of class, we share strategies and solutions for solving the area and perimeter word problems. For me, getting the answer is not enough. I want students to know and understanding how they arrived at an answer, so they can then explain it to others. A useful strategy for solving area and perimeter problems is a part/part whole diagram. This is explained in the reflection.
| 528 | 2,639 |
{"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}
| 4.34375 | 4 |
CC-MAIN-2018-43
|
longest
|
en
| 0.927343 |
https://www.coursehero.com/file/1657853/ch8soln/
| 1,495,570,425,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463607649.26/warc/CC-MAIN-20170523183134-20170523203134-00255.warc.gz
| 846,347,832 | 227,178 |
# ch8soln - Chapter 8 Digital Control Problems and Solutions...
This preview shows pages 1–5. Sign up to view the full content.
Chapter 8 Digital Control Problems and Solutions 1. The z -transform of a discrete-time & lter h ( k )ata1 Hz sample rate is H ( z )= 1+(1 / 2) z 1 [1 (1 / 2) z 1 ][1 + (1 / 3) z 1 ] . (a) Let u ( k )and y ( k ) be the discrete input and output of this & lter. Find ad i f erence equation relating u ( k y ( k ). (b) Find the natural frequency and damping coe cient of the & lter&s poles (c) Is the & lter stable? Solution: (a) Find a di f erence equation : H ( z Y ( z ) U ( z ) = / 2) z 1 [1 (1 / 2) z 1 ][1+(1 / 3) z 1 ] = Y ( z ) 1 6 z 1 Y ( z ) 1 6 z 2 y ( z U ( z )+ 1 2 z 1 U ( z ) = y ( k ) 1 6 y ( k 1) 1 6 y ( k 2) = u ( k 1 2 u ( k 1) (b) Two poles at z =1 / 2and z = 1 / 3inz-p lane . z = e sT = s = 0 . 693 T and s = 1 . 10 + 3 . 14 j T in s-plane, where T is the sampling period. Since the sample rate is 1 Hz, T sec. For z = 1 2 , ω n = 0 . 693 T =0 . 693 rad/sec, ζ . 0 z = 1 3 , ω n = 3 . 33 T =3 . 33 rad/sec, ζ . 330 547
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
548 CHAPTER 8. DIGITAL CONTROL (c) Yes, both poles are inside the unit circle. 2. Use the z -transform to solve the di f erence equation y ( k ) 3 y ( k 1) + 2 y ( k 2) = 2 u ( k 1) 2 u ( k 2) , where u ( k )= k, k 0 , 0 ,k< 0 , y ( k )=0 ,k < 0 . Solution: Y ( z ) U ( z ) = 2( z 1 z 2 ) 1 3 z 1 2 z 2 = 2 z 2 u ( k kk 0 0 k< 0 & = U ( z z ( z 1) 2 Y ( z 2 z 2 & z ( z 1) 2 = 2 z z 2 2 z z 1 2 z ( z 1) 2 Taking the inverse z-transform from Table 8.1, y ( k )=2(2 k 1 k )( k 0) 3. The one-sided z -transform is de & ned as F ( z X 0 f ( k ) z k . (a) Show that the one-sided transform of f ( k +1)is Z{ f ( k +1) } = zF ( z ) zf (0). (b) Use the one-sided transform to solve for the transforms of the Fi- bonacci numbers generated by the di f erence equation u ( k +2) = u ( k +1)+ u ( k ). Let u (0) = u (1) = 1. [ Hint: You will need to & nd a general expression for the transform of f ( k +2)intermsofthe transform of f ( k )]. (c) Compute the pole locations of the transform of the Fibonacci num- bers. (d) Compute the inverse transform of the Fibonacci numbers. (e) Show that, if u ( k )rep re sen t sthe k th Fibonacci number, then the ratio u ( k /u ( k ) will approach (1 + 5) / 2. This is the golden rat iova luedsoh igh lybytheGreeks . Solution:
549 (a) Z{ f () k +1) } = P k =0 f ( k z 1 = P j =1 f ( j ) z 1( j 1) ,k +1= j = z P 0 f ( j ) z 1 zf (0) = zF ( z ) (0) (b) u ( k +2) u ( k u ( k )=0 We have : f ( k } = z 2 F ( z ) z 2 f (0) (1) Taking the z-transform, z 2 U ( z ) z 2 u (0) zu (1) [ zU ( z ) (0)] U ( z = ( z 2 z 1) U ( z )=( z 2 z ) u (0) + (1) Since u (0) = u (1) = 1, we have : U ( z )= z 2 z 2 z 1 (c) The poles are at : z = 1 5 2 =1 . 618 , 0 . 618 , α 1 , α 2 (d) (i) By long division : 1+ z 1 +2 z 2 +3 z 3 + ••• 1 z 1 z 2 )1 1 z 1 z 2 z 1 + z 2 z 1 z 2 z 3 2 z 2 + z 3 2 z 2 2 z 3 2 z 4 3 z 3 z 4 u ( k )=1 , 1 , 2 , 3 , 5 ,
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
550 CHAPTER 8. DIGITAL CONTROL (ii) By partial fraction expansion : U ( z )= 1 1 z 1 z 2 = 1 (1 α 1 z 1 )(1 α 2 z 1 ) = & α 1 α 1 α 2 1 α 1 z 1 + & α 2 α 2 α 1 1 α 2 z 1 u ( k α 1 α 1 α 2 α k 1 + α 2 α 2 α 1 α k 2 = ˆ 5+ 5 10 1+ 5 2 !
This is the end of the preview. Sign up to access the rest of the document.
## This note was uploaded on 03/17/2009 for the course MEEM 4700 taught by Professor Staff during the Spring '08 term at Michigan Technological University.
### Page1 / 70
ch8soln - Chapter 8 Digital Control Problems and Solutions...
This preview shows document pages 1 - 5. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online
| 1,581 | 3,824 |
{"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}
| 3.9375 | 4 |
CC-MAIN-2017-22
|
longest
|
en
| 0.705069 |
https://math.stackexchange.com/questions/786046/infinite-product-prod-limits-k-1-infty-left1-fracx2k2-pi2-right/786231
| 1,558,956,996,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-22/segments/1558232262369.94/warc/CC-MAIN-20190527105804-20190527131804-00203.warc.gz
| 556,531,573 | 37,909 |
# Infinite Product $\prod\limits_{k=1}^\infty\left({1-\frac{x^2}{k^2\pi^2}}\right)$
I've been looking at proofs of Euler's sine expansion, that is $$\frac{\sin x}{x}=\prod_{k=1}^\infty\left({1-\frac{x^2}{k^2\pi^2}}\right)$$
All the proofs seem to rely on complex analysis and Fourier series.
Is there any more elementary proof?
I first prove that for any $n$ that is a power of $2$ and for any $x\in\mathbb R$, $$\sin(x)=n p_n(x)\sin\left(\frac x n\right)\cos\left(\frac x n\right),~p_n(x)=\prod_{k=1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\tag{1}$$ using the following more familiar identity $$\sin(x)=2\sin\frac x 2\sin\frac{\pi+x}2$$ we can easily arrive to the following identity that is valid for $n$ equal to any power of $2$, \begin{align}\sin(x)&=2^{n-1}\prod_{k=0}^{n-1}\sin\frac{k\pi+x}n\\ &=2^{n-1}\sin\frac x n\sin\frac{\frac n 2\pi+x}n\prod_{k=1}^{\frac n 2-1}\sin\frac{k\pi+x}n\sin\frac{k\pi-x}n\\ &=2^{n-1}\sin\frac x n\cos\frac x n\prod_{k=1}^{\frac n 2-1}\left(\sin^2\frac{k\pi}n-\sin^2\frac x n\right)\tag{2}\end{align} By considering what happens as $x\to0$ in the recent formula, we can obtain that for $n$ a power of $2$ $$n=2^{n-1}\prod_{k=1}^{\frac n 2-1}\sin^2\frac{k\pi}n\tag{3}$$ and then I replace $(3)$ in $(2)$ and prove $(1)$.
Now I choose $m<\frac n 2-1$ and break up $p_n(x)$ as follows $$p_n(x)=\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\tag{4}$$ Since for $0<\theta<\frac{\pi}2$, $\sin(\theta)>\frac2{\pi}\theta$, we have $$\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<\frac{x^2}{4k^2}$$ and if I choose $m$ and $n$ large enough such that $\frac{x^2}{4m^2}<1$, then $$0<1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<1,k=m+1,...,\frac n 2-1$$ which implies that $$0<\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)<1$$ Therefore from $(4)$ we have $$p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)$$ Also we have $$\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\ge1-\sum_{k=m+1}^{\frac n 2-1}\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\ge1-\frac{x^2}4\sum_{k=m+1}^\infty\frac1{k^2}=1-s_m$$ and once again by $(4)$, we get $$p_n(x)\ge(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)$$ To summarize, I have shown that $$(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\le p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\tag{5}$$ Now taking $n\to\infty$ in $(5)$ I arrive to $$(1-s_m)\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\le\frac{\sin(x)}x\le\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)$$ After rearrangement and then taking absolute values, we get $$\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|$$ but since we have $$\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le\prod_{k=1}^m\left(1+\frac{x^2}{k^2\pi^2}\right)\le\prod_{k=1}^me^{\frac{x^2}{k^2\pi^2}}\le e^{\sum_{k=1}^\infty\frac{x^2}{k^2\pi^2}}=e^{\ell}$$ I can write $$\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|e^{\ell}$$ Finally since $\lim_{m\to\infty}s_m=0$, this inequality implies our goal, indeed $$\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)=\frac{\sin(x)}x$$
• Thanks, I figured it out. It's just flipped. – Superbus May 8 '14 at 14:40
• Shouldn't (6) be $\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<\frac{n^2}{4k^2}$? That's what I get using the inequality. – Superbus May 8 '14 at 15:08
• Note that $\sin^2\frac x n<\frac{x^2}{n^2}$. – user91500 May 8 '14 at 15:11
• Nice proof, but I'm not fully convinced by (8) and (9). What if $$\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)$$ is negative? And what if $1 - s_m$ is negative? – user85798 May 12 '14 at 11:30
• @Superbus In the following paper, there is another similar proof jstor.org/stable/2690668?origin=JSTOR-pdf – user91500 May 14 '14 at 16:10
Here's a proof using elementary (albeit not very well known) trigonometric integrals.
We may assume $0 < x \leq 1/2$ because the ratio of $\sin(\pi x) / (\pi x)$ to $\prod_{k=1}^\infty \big(1 - (x/k)^2\bigl)$ is readily seen to be an even function of period $1$.
For $n \geq 0$ and $c \in {\bf R}$ define $$I_n(c) := \int_0^{\pi/2} \cos^n t \, \cos ct \, dt.$$ Then $I_0(0) = \pi/2$ and $I_0(2x) = \sin(\pi x) / (2x)$. Hence $$\frac{I_0(2x)}{I_0(0)} = \frac{\sin \pi x}{\pi x}.$$ I claim that also $$\frac{I_0(2x)}{I_0(0)} = \prod_{k=1}^\infty \left( 1 - \frac{x^2}{k^2} \right),$$ which will prove the desired product formula.
The key is a recursion for the ratios $I_n(2x) / I_n(0)$ obtained from the following integration by parts:
Lemma. For $n \geq 2$, $$(n^2 - c^2) I_n(c) = (n^2-n) I_{n-2}(c).$$
Proof: Integrate by parts twice to find $$c^2 I_n(c) = c \int_0^{\pi/2} \cos^n t \, d(\sin ct) = n c \int_0^{\pi/2} \cos^{n-1} t \, \sin t \, \sin ct \, dt$$ $$= -n \int_0^{\pi/2} \cos^{n-1} t \, \sin t \, d(\cos ct) = n \int_0^{\pi/2} (\cos^n t - (n-1) \cos^{n-2} t \sin^2 t) \, \cos ct \, dt.$$ But $\sin^2 t = 1 - \cos^2 t$, so $$c^2 I_n(c) = n \int_0^{\pi/2} (n \cos^n t - (n-1) \cos^{n-2} t) \, \cos ct \, dt = n^2 I_n(c) - (n^2-n) I_{n-2}(c),$$ and the recursion follows. $\Box$
Therefore $$\frac{I_{n-2}(c)}{I_{n-2}(0)} = \frac{n^2-c^2}{n^2} \frac{I_n(c)}{I_n(0)},$$ whence by induction on $m=1,2,3,\ldots$ $$\frac{I_0(c)}{I_0(0)} = \prod_{k=1}^m \frac{(2k)^2-c^2}{(2k)^2} \cdot \frac{I_{2m}(c)}{I_{2m}(0)}.$$ Taking $c=2x$ we find $$\frac{\sin \pi x}{\pi x} = \prod_{k=1}^m \frac{k^2-x^2}{k^2} \cdot \frac{I_{2m}(2x)}{I_{2m}(0)} = \prod_{k=1}^m \left( 1 - \frac{x^2}{k^2} \right) \cdot \frac{I_{2m}(2x)}{I_{2m}(0)},$$ and it remains to prove that $I_n(2x) / I_n(0) \rightarrow 1$ as $n \rightarrow \infty$. But this follows from the inequalities $$I_n(0) > I_n(2x) > I_{n+2}(0)$$ (the second inequality is where we use $x \leq 1/2$: we need $\cos 2xt \leq \cos^2 x$), since by our Lemma $I_{n+2}(0) / I_n(0) = (n+1)/(n+2)$, which approaches $1$ as $n \rightarrow \infty$. Taking $m \rightarrow \infty$ we deduce $$\frac{\sin \pi x}{\pi x} = \prod_{k=1}^\infty \left( 1 - \frac{x^2}{k^2} \right),$$ QED.
I noticed this argument some years back (see for instance #8 of this problem set from 2000). Naturally the formula for integrals $I_n(c)$ is known (even for non-integer $n$, see integral 3.631#9 in Gradshteyn and Ryzhik), but I don't know if and where it was previously noticed that one can use these integrals to prove the product formula for the sine.
I think this is the most motivated elementary proof.
For $z \in \mathbb C$ and $n \in \mathbb N_{> 0}$, let:
$$f_n \left({z}\right) = \frac 1 2 \left[{\left({1 + \frac z n}\right)^n - \left({1 - \frac z n}\right)^n }\right]$$
Then $f_n\left({z}\right) = 0$ if and only if:
\begin{align} &&\left({1 + \frac z n}\right)^n & = \left({1 - \frac z n}\right)^n \\ & \iff & 1 + \frac z n & = \left({1 - \frac z n}\right) e^{2 \pi i \frac k n} \\ & \iff & z & = n \frac {e^{2 \pi i \frac k n} - 1} {e^{2 \pi i \frac k n} + 1} \\ &&& = n i \tan \left({\frac {k \pi} n }\right) \end{align}
Let $n = 2 m + 1$.
Then the roots of $f_{2 m + 1} \left({z}\right)$ are $\left({2 m + 1}\right) i \tan \left({\dfrac {k \pi} {2 m + 1}}\right)$ for $- m \le k \le m$.
Observe that $f_{2m + 1} \left({z}\right)$ is a polynomial of degree $2 m + 1$.
Then for some constant $C$, we have:
\begin{align} f_{2 m + 1} \left({z}\right) & = C z \prod_{\substack {k \mathop = - m \\ k \mathop \ne 0} }^m \left({1 - \frac z {\left({2 m + 1}\right) i \tan \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)\\ & = C z \prod_{k \mathop = 1}^m \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) \end{align}
It can be seen from the binomial theorem that the coefficient of $z$ in $f_n \left({z}\right)$ is $1$.
Hence $C = 1$, and we obtain:
$$f_{2 m + 1} \left({z}\right) = z \prod_{k \mathop = 1}^m \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)$$
First we consider $z = x$ where $x$ is a non-negative real number.
Let $l < m$.
Then:
$$x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) \le f_{2 m + 1} \left({x}\right)$$
Taking the limit as $m \to \infty$ we have:
\begin{align} & &\lim_{m \to \infty} x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} \left({\frac {k \pi / \left({2 m + 1}\right)} {\tan \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)^2 }\right) & \le \frac 1 2 \left({e^x - e^{- x} }\right)\\ & \implies &x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} }\right) & \le \sinh x \end{align}
By the inequality $\tan \theta \ge \theta$ for $0 \le \theta < \dfrac {\pi} 2$ we have:
$$f_{2 l + 1} \left({x}\right) \le x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} }\right) \le \sinh x$$
Taking the limit as $l \to \infty$ we have by Squeeze Theorem:
$$\quad x \prod_{k \mathop = 1}^\infty \left({1 + \frac {x^2} {k^2 \pi^2} }\right) = \sinh x \tag{1}$$
Now let $1 < l < m$.
We have:
\begin{align} &\left \vert{f_{2 m + 1} \left({z}\right) - z \prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)}\right \vert \\ & =\left \vert{z}\right \vert \left \vert{\prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)}\right \vert \cdot \left \vert{\prod_{k \mathop = l + 1}^m \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) - 1}\right \vert\\ & \le \left \vert{z}\right \vert \left[{\prod_{k \mathop = 1}^l \left({1 + \frac {\left \vert{z}\right \vert^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)}\right] \cdot \left[{\prod_{k \mathop = l + 1}^m \left({1 + \frac {\left \vert{z}\right \vert^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) - 1}\right] \\ & = f_{2 m + 1} \left({\left \vert{z}\right \vert}\right) - \left \vert{z}\right \vert \prod_{k \mathop = 1}^l \left({1 + \frac {\left \vert{z}\right \vert^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) \end{align}
Taking the limit as $m \to \infty$ we have:
$$\left \vert{\sinh z - z \prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {k^2 \pi^2} }\right)}\right \vert \le \sinh {\left \vert{z}\right \vert} - \left \vert{z}\right \vert \prod_{k \mathop = 1}^l \left({1 + \frac {\left \vert{z}\right \vert^2} {k^2 \pi^2} }\right)$$
Now take the limit as $l \to \infty$.
By $(1)$ and Squeeze Theorem, we have:
$$\sinh z = z \prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {k^2 \pi^2} }\right)$$
Finally, substituting $z \mapsto i z$, we obtain:
$$\sin z = z \prod_{k \mathop = 1}^l \left({1 - \frac {z^2} {k^2 \pi^2} }\right)$$
In case you want an extremely simple proof:
Let's rewrite the product formula as follows:
$$\sin x = x\prod_{k \mathop = 1}^\infty \left({1 - \frac{x^2}{\pi^2 k^2} }\right)$$
Note that the right-hand side of the equation is a polynomial of a degree of $\infty$, factored into a product of other smaller polynomials of a degree of 2.
Now, the equation holds if the roots of the polynomial and the roots of the sine function coincide. Roots of the sine function:
$$x=0,\pm\pi,\pm2\pi,...$$
In similar manner, you can write down a product formula for the $\cos x$ if you know that the roots of the cosine are
$$\pm\frac{2k-1}{2}\pi;\;k=0,1,2,...$$
etc.
Remark:
"First of all, the result, and then rigor if needed."
• This is all the proof I need. – Kainui Sep 22 '14 at 16:01
• This isn't really a proof, but it gives a good intuition why it works. – Rufflewind Sep 27 '14 at 8:59
| 5,260 | 12,124 |
{"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": 5, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.890625 | 4 |
CC-MAIN-2019-22
|
latest
|
en
| 0.578131 |
https://moment-g.com/how-many-tablespoons-in-7-ounces/
| 1,656,230,950,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656103037649.11/warc/CC-MAIN-20220626071255-20220626101255-00516.warc.gz
| 454,432,283 | 5,028 |
## 7 Grams come Tablespoons (7g come tbsp) basic Conversion
Learn exactly how to fast and easily convert 7 Grams come Tablespoons. So, how many tablespoons in 7 grams?7 Grams (g) is equal to 0.467 Tablespoons (tbsp)or 7 g = 0.467 tbsp
## 7 Grams to Tablespoons conversion Calculator
You are watching: How many tablespoons in 7 ounces
## How to convert 7 g to tbsp
It’s basic to transform grams to tablespoons. Because that the basic equation just divide the grams by 15 to transform them to tablespoons.7g to tbsp calculation:Conversion factortbsp = g ÷ 15 7 Grams to Tablespoons conversion Equation7 g ÷ 15 = 0.467 tbsp
## Common Grams come Tablespoon Conversions
GramsTablespoonsGramsTablespoons1 g0.067 or 1/15 tbsp20 g1.333 tbsp2 g0.133 or 2/15 tbsp30 g2 tbsp3 g0.2 or 1/5 tbsp40 g2.667 tbsp4 g0.267 or 4/15 tbsp50 g3.333 tbsp5 g0.333 or 1/3 tbsp60 g4 tbsp6 g0.4 or 2/5 tbsp70 g4.667 tbsp7 g0.467 or 7/15 tbsp80 g5.333 tbsp8 g0.533 or 8 /15 tbsp90 g5 tbsp9 g0.6 or 3/15 tbsp100 g6.667 tbsp10 g0.667 or 2/3 tbsp1000 g66.667 tbsp
## Ingredient Gram come Tablespoon Conversions
Not every gram come tablespoon conversions room the same, and depend on the specific ingredients density. This counter is not always clear, because a gram is a unit that weight, vice versa, a tablespoon is a unit of volume.For instance let’s look at a tablespoon that sugar. 12.6 grams that sugar would be equal to 1 tablespoon, rather of the general 15 grams every tablespoon. The included sugar is much less dense, so it takes less grams to equal a tablespoon.Here is a perform of usual ingredients v measurement in volume, and their gram to tablespoon counter tables:Ingredient1 Tablespoon (tbsp) =Water14.79 gSugar12.6 gHoney21 gFlour7.83 gMilk15.3 gButter14.19 gBaking Powder13.32 g
Check the end this connect for conversions of food preparation ingredients.
## Convert Tablespoons to Grams
It’s likewise easy to convert tablespoons come grams. For the general equation simply multiply the tablespoons through 15 to convert them to grams.tbsp come g calculation:Conversion factor15 g = 1 tbsp * 15 Example Tablespoons come Grams switch Equation7tbsp * 15 = 105 g
## Convert 7 Grams to various other Units
Do you desire to convert 7 grams into an additional unit? right here is a beneficial table because that converting 7 grams into other units:Unit7 Grams (g) =Micrograms (mcg)7,000,000 mcgMilligram (mg)7,000 mgKilogram (kg)0.007 kgOunce (oz)0.247 ozPound (lb)0.015 lbTeaspoon (tsp)1.4 Tsp
## What is a Gram?
The gram is a unit of mass in the generally used metric mechanism of measurement. The official definition is the a gram is one thousandth that the worldwide Systems of units (SI) base unit because that mass, which is the kilogram.The abbreviated symbol because that a gram is “g”. Example 56 grams is the very same as 56 g.See the dictionary meaning here.
## What is a Tablespoon?
The tablespoon is provided as a measure up of volume, most frequently used together a measure up in food preparation recipes.
See more: How Many Earths Can Fit In Vy Canis Majoris The Largest Class 6 Physics Cbse
Most typically it is 1/16 the a cup or 3 teaspoons.The abbreviated symbol because that a tablespoon is “tbsp”. Examples 52 tablespoons is the exact same as 52 tbsp.See the dictionary meaning here.
| 922 | 3,304 |
{"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}
| 3.90625 | 4 |
CC-MAIN-2022-27
|
latest
|
en
| 0.773074 |
https://www.open.edu/openlearn/mod/oucontent/view.php?id=81976§ion=3.1
| 1,726,408,224,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00401.warc.gz
| 870,059,998 | 21,880 |
### Become an OU student
Everyday maths 1
Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available.
# 3.1 Weighing things
It’s useful to have an idea of how much things weigh. It can help you to work out the weight of fruit or vegetables to buy in a market, for example, or whether your suitcase will be within the weight limit for a flight.
Try estimating the weight of something before you weigh it. It will help you to get used to measures of weight.
Hint: Remember to use appropriate units. Give the weight of small things in grams and of heavy things in kilograms.
Take a look at the example below before having a go at the activity.
## Example: Weighing an apple
1. Which metric unit would you use to weigh an apple?
2. Estimate how much an apple weighs and then weigh one.
3. How much would 20 of these apples weigh? Would you use the same units?
### Method
1. An apple is quite small, so it should be weighed in grams.
2. How much did you estimate that an apple weighs? A reasonable estimate would be 100 g.
When we weighed an apple, it was 130 g.
3. Twenty apples would weigh:
• 130 × 20 = 2,600 g
Remember the metric conversion diagram? To convert from grams to kilograms, you need to divide the figure in grams by 1,000. So the weight of the apples in kilograms is:
• 2,600 g ÷ 1,000 = 2.6 kg
Now try the following activity. Remember to check your answers once you have completed the questions.
## Activity 7: Weighing things
1. How much do ten teabags weigh? Estimate and then weigh them.
2. How heavy is a bottle of sauce? How much would a case of 10 bottles weigh?
Hint: The weight shown on the label is the weight of the sauce – it doesn’t include the weight of the bottle or jar that the sauce comes in. So for an accurate measurement, you need to weigh the bottle rather than read the label!
3. How heavy is a book?
### Discussion
Our suggestions are shown in the table below. Your estimates and measured weights might be different, but they should be roughly similar.
Item Estimated weight Actual weight
Ten teabags 25 g 30 g
Bottle of sauce 500 g 450 g
Book 900 g 720 g
A case of ten bottles of sauce would weigh:
• 450 × 10 = 4,500 g
As previously noted, 1,000 g = 1 kg, so 4,500 g = 4.5 kg – which is how you would more usually express this weight.
If your book weighed more than ours, you might have given its weight in kilograms. If you chose a small book, it may have weighed a lot less.
| 628 | 2,544 |
{"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-2024-38
|
latest
|
en
| 0.934321 |
https://www.enotes.com/homework-help/calculate-limit-1-n-f-1-n-f-2-n-f-1-f-x-arctgx-n-423601
| 1,516,297,446,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-05/segments/1516084887535.40/warc/CC-MAIN-20180118171050-20180118191050-00571.warc.gz
| 903,080,279 | 10,146 |
# calculate in limit 1/n(f(1/n)+f(2/n)+____+f(1)) if f(x)=arctgx in n --> infinite simbol? use antiderivative
sciencesolve | Certified Educator
You may replace the limit by Riemann integral, over the interval `[0,1]` , such that:
`lim_(n->oo)(1/n)(f(1/n) + f(2/n) + ... + f(n/n)) = int_0^1 f(x) dx`
The problem provides the equation of the function, such that:
`f(x) = arctan x`
`int_0^1 f(x) dx = int_0^1 arctan x dx`
You need to use integration by parts, such that:
`arctan x = u => 1/(1 + x^2)dx = du`
`1dx = dv => v = x`
`int_0^1 arctan x dx = x*arctan x|_0^1 - int_0^1 x/(1 + x^2)dx`
You should come up with the substitution, such that:
`1 + x^2 = y => 2x dx = dy => x dx = (dy)/2`
`int_0^1 x/(1 + x^2)dx = int_1^2 (dy)/(2y) `
`int_1^2 (dy)/(2y) = (1/2)ln y|_1^2 => int_1^2 (dy)/(2y) = (1/2)(ln 2 - ln 1)`
`int_1^2 (dy)/(2y) = (1/2)*ln 2`
`int_0^1 arctan x dx = 1*arctan 1 - 0*arctan 0 - (1/2)*ln 2`
`int_0^1 arctan x dx = pi/4 - (1/2)*ln 2`
Hence, evaluating the given limit, using Riemann integral, yields `lim_(n->oo)(1/n)(f(1/n) + f(2/n) + ... + f(n/n)) = pi/4 - (1/2)*ln 2.`
pramodpandey | Student
`lim_(n->oo){(1/n)[f(1/n)+f(2/n)+..........+f(1)]}`
`=lim_(n->oo){(1/n)``[f(0)+f(1/n)+f(2/n)+..........+f(n/n)-f(0)]}`
``
`` `=lim_(n->oo){(1/n)[f(0)+f(1/n)+f(2/n)+..........+f(n/n)]}`
`-lim_(n->oo) (1/n)f(0)`
`=int_0^1tan^(-1)(x)dx-lim_(x->oo)(1/n)*0`
`=(xtan^(-1)x)_0^1-(1/2)(log(1+x^2))_0^1`
`=pi/4-(1/2)log(2)`
``
``
``
| 706 | 1,461 |
{"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}
| 4.21875 | 4 |
CC-MAIN-2018-05
|
latest
|
en
| 0.555523 |
https://www.jiskha.com/questions/532556/how-many-possible-outcomes-are-there-for-15-coins-tossed-simultaneously
| 1,611,548,708,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703564029.59/warc/CC-MAIN-20210125030118-20210125060118-00328.warc.gz
| 825,976,597 | 4,524 |
# math
How many possible outcomes are there for 15 coins tossed simultaneously?
1. 👍
2. 👎
3. 👁
1. 450
1. 👍
2. 👎
## Similar Questions
1. ### Math: Probablity
A penny, a nickel, and a dime are flipped at the same time. Each coin can come out either heads (H) or tails (T). a. What name is given to the act of flipping the coins? b. There are eight elements in the sample space (i.e.,
2. ### math - probability
a) What is the probability of obtaining 2 Heads when a coin is tossed twice? b) What is the probability of obtaining 1 Head when a coin is tossed twice? Keep in mind, the coins are not tossed simultaneously.
3. ### college- practical statistics
Five coins are tossed 3200 times find the frequencies of distribution of heads & tails & tabulate the results. Calculate the mean number of success & the standard deviation.
4. ### math
find the mean,variance and standard deviation for the number of heads when 10 coins are tossed.
1. ### statistic****help*****
if three coins are tossed, what is the number of equally likely outcomes? 3 4 6 8 9
2. ### Statistics
In a gambling game a man is paid \$5 if he gets all heads or all tails when 3 coins are tossed, and he pays out of \$3 if either 1 or 2 heads show. what is his expected gain or loss of the 3 coins?
3. ### math
Two coins are tossed, find the probability that two heads are obtained. Note: Each coin has two possible outcome H(heads), and T (tails).
4. ### Math
5 coins are tossed. How many outcomes are possible?
1. ### Gr 12 Data
Sarah tosses 5 coins, one after the other. a) How many different outcomes are possible? b) Draw a tree diagram to illustrate the different possibilities. c) In how many ways will the first coin turn up heads and the last coin turn
2. ### Prealgebra/probability
A coin is tossed three times. Use a tree diagram to find the number of possible outcomes that could produce exactly two heads. I don't know how to write out a tree diagram on here, but I think this one is heads -> heads, tails ->
3. ### Advanced Math
The probability of getting 2 heads and 1 tail when three coins are tossed is 3 in 8. Find the odds of not getting 2 heads and 1 tail. ANSWER: 5:8? Three Coins are tossed. Find the probability that exactly 2 coins show heads if the
4. ### Probabilities (Math)
An experiment consists of tossing a coin 14 times. (a) How many different outcomes are possible? (b) How many different outcomes have exactly 9 heads? (c) How many different outcomes have at least 2 heads? (d) How many different
| 639 | 2,524 |
{"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}
| 3.78125 | 4 |
CC-MAIN-2021-04
|
latest
|
en
| 0.914143 |
https://e-learnteach.com/square-root-of-negative-9/
| 1,685,660,908,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224648209.30/warc/CC-MAIN-20230601211701-20230602001701-00102.warc.gz
| 264,171,322 | 14,472 |
# Square root of negative 9
The square root of -9 is 3i. Here i stands for iota which means square root of -1. It is the basics of irrational numbers.?-9 = ? If you enter SQRT(-9) into a spreadsheet on your computer, you will get an error message saying something like “argument must be greater. -9 × -9 = 81. The inverse operation of squaring negative negative nine is extracting the square root of negative -9, explained here.The square root of -9 is 3i. (i is the square root of negative one, an imaginary number.) But if you are talking about real numbers, Although the principal negative square root of negative nine is only one of the two square roots, the term square root of -9 usually refers to the.
View this answer now! It’s completely free.
## square root of negative 8
Find the square root of -8. ? Square Root Calculator’ with step-by-step solution using the Babylonian Method or Hero’s Method. Square Root Table.-8 is also a valid square root of 64, and when you’re solving an equation like x2 = 64, there will be two solutions for that reason.Although the principal negative square root of negative eight is only one of the two square roots, the term square root of -8 usually refers to the. 6 plus or minus the square root of negative 8. Natural Language Math Input. Use Math Input Mode to directly enter textbook math notation.What is the Square of Negative -8?
## square root of negative 2
Square root calculator and perfect square calculator. Find the square root, or the two roots, including the principal root, of positive and negative real. You would get (i square root 2) because you can not square root a negative number, the answer is imaginary. Tmdkiller: Square root is NOT. There is a convention in mathematics that ?52 is ?25, not 25. The square is evaluated before the minus. The square root of ?25 can be defined. is not a problem since (-2) (-2) (-2) = -8, making the answer -2. In cube root problems, it is possible to multiply a negative value times itself three. Example 2: How to Find the Square Root of a Negative Number There is no real square root of -1, but we assume that in some universe there is.
## square root of negative 7
-7 × -7 = 49. The inverse operation of squaring negative negative seven is extracting the square root of negative -7, explained here.Simplify – square root of 7. ??7 – 7. The result can be shown in multiple forms. Exact Form: ??7 – 7. Decimal Form: ?2.64575131
– 2.64575131
The square root of -7 is the number y such that y² = -7. Here you can learn all about it in addition to a calculator you will like.The square root of 7 is expressed as ?7 in the radical form and as (7)½ or (7)0.5 in the exponent form. The square root of 7 rounded up to 8 decimal places is. Square root of -7 is neither a rational nor an irrational. It is an imaginary number. Rational and Irrational numbers both together form real numbers.
## square root of negative 3
A negative square root and a positive square root make a positive number. square root of both sides of the equation as shown in Figure 3.=i(1.73205) where i is the square root of -1 in electrical and electronic engineering the the square root of -1 is generally referred to as. The square root of 3 is the positive real number that, when multiplied by itself, gives the number 3. It is denoted mathematically as ?3 or 31/2.2. I have no idea what you want to know. · 3. Note that ?52?25 but rather ?25. · There is a convention in mathematics that ?52 is ?25, not 25. that need the square root of a negative number. numbers become most useful when combined with real numbers to make complex numbers like 3+5i or 6?4i.
| 906 | 3,652 |
{"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}
| 4.09375 | 4 |
CC-MAIN-2023-23
|
latest
|
en
| 0.876974 |
https://metanumbers.com/318413
| 1,638,242,430,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964358903.73/warc/CC-MAIN-20211130015517-20211130045517-00515.warc.gz
| 450,699,915 | 7,375 |
# 318413 (number)
318,413 (three hundred eighteen thousand four hundred thirteen) is an odd six-digits composite number following 318412 and preceding 318414. In scientific notation, it is written as 3.18413 × 105. The sum of its digits is 20. It has a total of 2 prime factors and 4 positive divisors. There are 316,128 positive integers (up to 318413) that are relatively prime to 318413.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 20
• Digital Root 2
## Name
Short name 318 thousand 413 three hundred eighteen thousand four hundred thirteen
## Notation
Scientific notation 3.18413 × 105 318.413 × 103
## Prime Factorization of 318413
Prime Factorization 149 × 2137
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 318413 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 318,413 is 149 × 2137. Since it has a total of 2 prime factors, 318,413 is a composite number.
## Divisors of 318413
4 divisors
Even divisors 0 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 320700 Sum of all the positive divisors of n s(n) 2287 Sum of the proper positive divisors of n A(n) 80175 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 564.281 Returns the nth root of the product of n divisors H(n) 3.97147 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 318,413 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 318,413) is 320,700, the average is 80,175.
## Other Arithmetic Functions (n = 318413)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 316128 Total number of positive integers not greater than n that are coprime to n λ(n) 79032 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 27409 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 316,128 positive integers (less than 318,413) that are coprime with 318,413. And there are approximately 27,409 prime numbers less than or equal to 318,413.
## Divisibility of 318413
m n mod m 2 3 4 5 6 7 8 9 1 2 1 3 5 4 5 2
318,413 is not divisible by any number less than or equal to 9.
## Classification of 318413
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (318413)
Base System Value
2 Binary 1001101101111001101
3 Ternary 121011210002
4 Quaternary 1031233031
5 Quinary 40142123
6 Senary 10454045
8 Octal 1155715
10 Decimal 318413
12 Duodecimal 134325
20 Vigesimal 1jg0d
36 Base36 6tot
## Basic calculations (n = 318413)
### Multiplication
n×y
n×2 636826 955239 1273652 1592065
### Division
n÷y
n÷2 159206 106138 79603.2 63682.6
### Exponentiation
ny
n2 101386838569 32282887429270997 10279291035016465967761 3273059896332697978192683293
### Nth Root
y√n
2√n 564.281 68.2858 23.7546 12.6066
## 318413 as geometric shapes
### Circle
Diameter 636826 2.00065e+06 3.18516e+11
### Sphere
Volume 1.35226e+17 1.27406e+12 2.00065e+06
### Square
Length = n
Perimeter 1.27365e+06 1.01387e+11 450304
### Cube
Length = n
Surface area 6.08321e+11 3.22829e+16 551507
### Equilateral Triangle
Length = n
Perimeter 955239 4.39018e+10 275754
### Triangular Pyramid
Length = n
Surface area 1.75607e+11 3.80457e+15 259983
## Cryptographic Hash Functions
md5 3c3d933a53f7bfed9dca8f482a4909ac 41f2918c20256838d8083be3d3c4646ca6ba0c35 8812f7781582b850149e8a6b3f221e9518e5fa01891bf02c00f6b3babe9d1281 390a37fb84ad635083acd03d40d1ba52a3bb32ddc7187d8e945762d047208dfcf2186fd74eeea2e049a6a413f22ba03ea48eed1a9b7b1376e7a98f56f2d66aba af3273a37da34988b38c5313e720426b9bf5ea78
| 1,468 | 4,250 |
{"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}
| 3.75 | 4 |
CC-MAIN-2021-49
|
latest
|
en
| 0.801283 |
https://www.physicsforums.com/threads/relationships-between-the-roots.857078/
| 1,527,503,836,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794872766.96/warc/CC-MAIN-20180528091637-20180528111637-00312.warc.gz
| 808,071,380 | 16,406 |
# Relationships between the roots
1. Feb 13, 2016
### Bruno Tolentino
A quadratic equation in this format $$x² - 2 A x + B² = 0$$
can be modified and expressed like: x² - 2 (u) x + (u² - v²) = 0.
The roots x1 and x2 are therefore: $$x_1 = x_1(u,v) = u + v$$ $$x_2 = x_2(u,v) = u - v$$ Or: $$x_1 = x_1(a, b) = \frac{a+b}{2} + \frac{|a-b|}{2}$$ $$x_2 = x_2(a, b) = \frac{a+b}{2} - \frac{|a-b|}{2}$$
Until here, these relationships are known. No problem! The problem begins when I ask: "how are these relationships for the cubic equation?"
I know that analog cubic equation is $$x^3 - 3 A x^2 + 3 B^2 x - C^3 = 0$$ and that
$$A = \frac{a+b+c}{3}$$ $$B = \sqrt[2]{\frac{ca + ab + bc}{3}}$$ $$C = \sqrt[3]{abc}$$
a, b and c are the roots. The roots can be called too of x1, x2 and x3. But, by analogy with the relationships above, how I can to express x1, x2 and x3 in terms of a linear combination of a, b and c?
$$x_1 = x_1(a, b, c) = ?$$ $$x_2 = x_2(a, b, c) = ?$$ $$x_3 = x_3(a, b, c) = ?$$
And how express x1, x2 and x3 in terms of u, v and w?
$$x_1 = x_1(u, v, w) = ?$$ $$x_2 = x_2(u, v, w) = ?$$ $$x_3 = x_3(u, v, w) = ?$$
2. Feb 13, 2016
### pasmith
3. Feb 13, 2016
### mathwonk
I usually look at these this way:
To solve a quadratic equation X^2 -bX + c = 0, first assume the solutions are X=r and X=s, and note that then (by the root-factor theorem), the equation factors as X^2 -bX + c = (X-r)(X-s) = X^2 -(r+s)X + rs, so that we must have b = r+s and c = rs. Thus we know in particular the sum b of the roots. Hence if we only knew the difference, say d = r-s, of the roots, we would be done, since then we would have 2r = b+d and 2s = b-d.
So we assume the roots are expressed as a sum, namely r = b/2 + d/2, and s = b/2 - d/2. Then c = rs = b^2/4 - d^2/4, so d^2 = b^2-4c.
Thus 2r = b + sqrt(b^2-4c) and 2s = b - sqrt(b^2-4c), the usual formula, if you notice we had a minus sign in the linear term of our original equation.
CUBIC FORMULA:
To solve a cubic equation, we start with a simplified one of form X^3 -3bX + c = 0, and again assume we want to find X as a sum X = (p+q). Plugging in gives (p+q)^3 = 3b(p+q) + c, and expanding gives p^3 + 3p^2q + 3pq^2 + q^3 = p^3 + q^3 + 3pq(p+q) =
3b(p+q) + c, and for this to hold means that pq = b, and c = p^3+q^3. Cubing the first of these gives p^3q^3 = b^3, and p^3+q^3 = c. Since we know b and c, we know both the sum and the product of the cubes p^3 and q^3. Can we find p^3 and q^3 from this? If so, then we could take cube roots and find p and q, and finally add them and get our root X = p+q.
Just recall in a quadratic equation of form X^2 - BX + C, that B and C are precisely the sum and product of the desired roots, and we can find those roots from B and C. I.e. we can find any two numbers when we know their sum and product, by solving a quadratic.
Since p^3+q^3 = c and p^3q^3 = b^3, the numbers p^3 and q^3, which can be used to give a solution X = p+q of our cubic, are solutions of the quadratic equation
t^2 -ct + b^3 = 0
e.g. to solve X^3 = 9X + 28, we have b = 3, c = 28, and so we solve t^2 -28t + 27 = 0. Here B^2-4C = 676, whose square root is 26, so we get t = (1/2)( 28 ± 26) = {27, 1}, for p^3 and q^3, so p,q are 1 and 3, and hence X=1+3 = 4 solves the cubic. Of course if we know about complex numbers, there are two more cube roots of 1 and 27, and we get two more complex roots. (Only two more because b = pq, so we must always have q = b/p, i.e. the choice of the cube root q is determined by the choice of p.)
Finally, one can translate the variable in any cubic equation to change it into one with zero quadratic term, so this process works in general.
4. Feb 14, 2016
| 1,360 | 3,679 |
{"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}
| 4.28125 | 4 |
CC-MAIN-2018-22
|
latest
|
en
| 0.884103 |
https://www.coursehero.com/file/55258595/MAT1033-Test-3-Reviewpdf/
| 1,642,767,625,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00252.warc.gz
| 758,907,580 | 54,280 |
# MAT1033_-_Test_3_Review.pdf - Review for Test 3 Division of...
• 4
This preview shows page 1 - 2 out of 4 pages.
Review for Test 3 * Division of Polynomials *Factoring *Solve Equations by Factoring *Rational Expressions: Simplify, Multiply, Divide, Add, and Subtract * Complex Fractions MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Divide. 1) (20x 3 - 36x 2 + 31x - 12) ÷ ( - 5x + 4) 1) A) - 4x 2 + 4x - 3 B) x 2 + 4x - 3 C) x 2 - 4x + 3 D) - 4x 2 - 3 2) ( - 8x 3 + 18x 2 - 11x + 21) ÷ ( - 2x + 3) 2) A) 4x 2 - 3x + 1 B) 4x 2 - 3x + 1 + 21 - 2x + 3 C) 4x 2 - 3x + 1 + 18 - 2x + 3 D) x 2 + 1 + - 3 - 2x + 3 3) (36x 3 - 25x) ÷ (6x - 1) 3) A) 6x 2 + x + 36 B) 6x 2 + x + 36 + 36 6x - 1 C) 6x 2 + x - 4 - 4 6x - 1 D) 6x 2 + x - 36 Factor completely, or state that the polynomial is prime. 4) x 3 - 36x 4) A) x(x - 6) 2 B) x(x + 6)(x - 6) C) (x 2 + 6)(x - 6) D) prime Factor the trinomial, or state that the trinomial is prime. 5) 3x 2 - 14x - 5 5) A) (3x - 1)(x + 5) B) (3x + 1)(x - 5) C) (3x - 5)(x + 1) D) prime 6) x 2 + 4x - 21 6) A) (x - 7)(x + 3) B) (x - 7)(x + 1) C) (x + 7)(x - 3) D) prime
Solve.
| 605 | 1,162 |
{"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}
| 3.96875 | 4 |
CC-MAIN-2022-05
|
latest
|
en
| 0.423098 |
https://zh.wikipedia.org/zh-tw/%E6%8B%89%E6%A0%BC%E6%9C%97%E6%97%A5%E6%8F%92%E5%80%BC%E6%B3%95
| 1,527,310,251,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794867309.73/warc/CC-MAIN-20180526033945-20180526053945-00168.warc.gz
| 884,735,384 | 28,887 |
# 拉格朗日插值法
## 定義
${\displaystyle (x_{0},y_{0}),\ldots ,(x_{k},y_{k})}$
${\displaystyle L(x):=\sum _{j=0}^{k}y_{j}\ell _{j}(x)}$
${\displaystyle \ell _{j}(x):=\prod _{i=0,\,i\neq j}^{k}{\frac {x-x_{i}}{x_{j}-x_{i}}}={\frac {(x-x_{0})}{(x_{j}-x_{0})}}\cdots {\frac {(x-x_{j-1})}{(x_{j}-x_{j-1})}}{\frac {(x-x_{j+1})}{(x_{j}-x_{j+1})}}\cdots {\frac {(x-x_{k})}{(x_{j}-x_{k})}}.}$[3]
## 範例
• ${\displaystyle f(4)=10}$
• ${\displaystyle f(5)=5.25}$
• ${\displaystyle f(6)=1}$
${\displaystyle \ell _{0}(x)={\frac {(x-5)(x-6)}{(4-5)(4-6)}}}$
${\displaystyle \ell _{1}(x)={\frac {(x-4)(x-6)}{(5-4)(5-6)}}}$
${\displaystyle \ell _{2}(x)={\frac {(x-4)(x-5)}{(6-4)(6-5)}}}$
${\displaystyle p(x)=f(4)\ell _{0}(x)+f(5)\ell _{1}(x)+f(6)\ell _{2}(x)}$
${\displaystyle .\,\,\,\,\,\,\,\,\,\,=10\cdot {\frac {(x-5)(x-6)}{(4-5)(4-6)}}+5.25\cdot {\frac {(x-4)(x-6)}{(5-4)(5-6)}}+1\cdot {\frac {(x-4)(x-5)}{(6-4)(6-5)}}}$
${\displaystyle .\,\,\,\,\,\,\,\,\,\,={\frac {1}{4}}(x^{2}-28x+136)}$
## 證明
### 存在性
${\displaystyle L(x_{j})=\sum _{i=0}^{k}y_{i}\ell _{i}(x_{j})=0+0+\cdots +y_{j}+\cdots +0=y_{j}}$
${\displaystyle (x-x_{0})\cdots (x-x_{j-1})(x-x_{j+1})\cdots (x-x_{k})}$
${\displaystyle \ell _{j}(x):=\prod _{i=0,\,i\neq j}^{k}{\frac {x-x_{i}}{x_{j}-x_{i}}}={\frac {(x-x_{0})}{(x_{j}-x_{0})}}\cdots {\frac {(x-x_{j-1})}{(x_{j}-x_{j-1})}}{\frac {(x-x_{j+1})}{(x_{j}-x_{j+1})}}\cdots {\frac {(x-x_{k})}{(x_{j}-x_{k})}}}$
## 幾何性質
${\displaystyle P=\lambda _{0}\ell _{0}+\lambda _{1}\ell _{1}+\cdots +\lambda _{n}\ell _{n}=0}$
${\displaystyle \lambda _{0}=\lambda _{1}=\cdots =\lambda _{n}=0}$
## 重心拉格朗日插值法
${\displaystyle \ell (x)=(x-x_{0})(x-x_{1})\cdots (x-x_{k})}$
${\displaystyle \ell _{j}(x)={\frac {\ell (x)}{x-x_{j}}}{\frac {1}{\prod _{i=0,i\neq j}^{k}(x_{j}-x_{i})}}}$
${\displaystyle w_{j}={\frac {1}{\prod _{i=0,i\neq j}^{k}(x_{j}-x_{i})}}}$
${\displaystyle \ell _{j}(x)=\ell (x){\frac {w_{j}}{x-x_{j}}}}$
${\displaystyle L(x)=\ell (x)\sum _{j=0}^{k}{\frac {w_{j}}{x-x_{j}}}y_{j}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)}$
${\displaystyle \forall x,\,g(x)=\ell (x)\sum _{j=0}^{k}{\frac {w_{j}}{x-x_{j}}}}$
${\displaystyle L(x)={\frac {\sum _{j=0}^{k}{\frac {w_{j}}{x-x_{j}}}y_{j}}{\sum _{j=0}^{k}{\frac {w_{j}}{x-x_{j}}}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)}$[7]
## 參考文獻
### 引用
1. ^ E. Waring. Problems Concerning Interpolations. Philosophical Transactions of the Royal Society of London. 1779, 69: 59–67.
2. ^ （英文）E. Meijering. A chronology of interpolation: From ancient astronomy to modern signal and image processing,. Proceedings of the IEEE: 323.
3. ^ （英文）Julius Orion Smith III. Lagrange_Interpolation. Center for Computer Research in Music and Acoustics (CCRMA), Stanford University.
4. ^ 馮有前，《數值分析》，第63頁
5. ^ 李慶揚，《數值分析》第4版，第31頁
6. ^ 馮有前，《數值分析》，第64頁
7. Jean-Paul Berrut, Lloyd N. Trefethen. Barycentric Lagrange Interpolation (PDF). SIAM Review. 2004, 46 (3): 501–517. doi:10.1137/S0036144502417715.[失效連結]
8. ^ 王兆清，李淑萍，唐炳濤. 一維重心型插值：公式、算法和應用. 山東建築大學學報. 2007, 22 (5): 447–453.
9. ^ NICHOLAS J. HIGHAM. The numerical stability of barycentric Lagrange Interpolation (PDF). IMA Journal of Numerical Analysis. 2004, 24 (4): 547–556.
| 1,556 | 3,206 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 80, "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}
| 4.03125 | 4 |
CC-MAIN-2018-22
|
latest
|
en
| 0.370728 |
https://www.physicsforums.com/threads/height-differences-from-relative-uncertainty-of-gravimeters.915837/
| 1,532,071,977,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676591543.63/warc/CC-MAIN-20180720061052-20180720081052-00329.warc.gz
| 953,152,714 | 15,348 |
# Homework Help: Height differences from relative uncertainty of gravimeters
Tags:
1. May 27, 2017
1. The problem statement, all variables and given/known data
The best relative gravimeters have a relative uncertainty of 10-12, that corresponds to a height difference of 3 µm.
2. Relevant equations
g∝(1/r2)
The local gravitational acceleration g outside the Earth is proportional to 1/r2, which means (Δg)/g = -2 (Δr)/r. With (Δg)/g = 10-12 we get Δr = 0.5 * 10-12 r where r is the radius of Earth.
3. The attempt at a solution
I've tried to get to the equation I highlighted in red, but I fail. I know it has derived from taking derivatives, but my derivatives result in different equations.
g∝1/r2 → g=k(1/r2) → ?
2. May 27, 2017
### BvU
The rule is derived from $f(x+\Delta x) = f(x) + f'(x)\Delta x\ + {\mathcal O} (\Delta x)^2$ so you get $$\Delta g = g(x+\Delta x ) - g(x) \approx g'(x) \Delta x \Rightarrow \\ = {-2\over x^3} \Delta x \Rightarrow {\Delta g\over g} = -2 {\Delta x\over x}$$
We usually report the standard deviation which is positive.
3. May 27, 2017
| 337 | 1,082 |
{"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}
| 3.5625 | 4 |
CC-MAIN-2018-30
|
latest
|
en
| 0.855154 |
http://www.ck12.org/physics/Circular-Motion/?difficulty=basic&by=community
| 1,485,143,469,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560281746.82/warc/CC-MAIN-20170116095121-00477-ip-10-171-10-70.ec2.internal.warc.gz
| 387,562,287 | 17,773 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Circular Motion
## The force and acceleration vectors point to the center of the circle.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Circular Motion
Students will learn that in circular motion there is always an acceleration (and hence a force) that points to the center of the circle defined by the objects motion.
MEMORY METER
This indicates how strong in your memory this concept is
0
## Circular Motion
Discusses circular motion and centripetal acceleration.
MEMORY METER
This indicates how strong in your memory this concept is
0
## Circular Motion
Students will learn that in circular motion there is always an acceleration (and hence a force) that points to the center of the circle defined by the objects motion.
MEMORY METER
This indicates how strong in your memory this concept is
0
## Nov 14: A- Circular Motion
Discusses circular motion and centripetal acceleration.
MEMORY METER
This indicates how strong in your memory this concept is
0
## Circular Motion
Students will learn that in circular motion there is always an acceleration (and hence a force) that points to the center of the circle defined by the objects motion.
MEMORY METER
This indicates how strong in your memory this concept is
0
## Unit 7.5 Circular Motion
Students will learn that in circular motion there is always an acceleration (and hence a force) that points to the center of the circle defined by the objects motion.
MEMORY METER
This indicates how strong in your memory this concept is
0
• Real World Application
## Roller Coasters
by Mark Holmstrom //
Even though roller coaster don't go in perfect circles, they are great examples of circular motion!
MEMORY METER
This indicates how strong in your memory this concept is
0
• Real World Application
## Circular Motion
by High Tech High Charter School //at grade
Explore the circular motion of a cyclist following a curved path.
MEMORY METER
This indicates how strong in your memory this concept is
0
| 541 | 2,598 |
{"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}
| 3.578125 | 4 |
CC-MAIN-2017-04
|
longest
|
en
| 0.873287 |
https://socratic.org/questions/what-are-the-mean-and-standard-deviation-of-a-binomial-probability-distribution--3
| 1,657,138,408,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656104676086.90/warc/CC-MAIN-20220706182237-20220706212237-00210.warc.gz
| 554,497,342 | 6,039 |
# What are the mean and standard deviation of a binomial probability distribution with n=15 and p=3/7 ?
Jan 11, 2016
Mean $= 6.43$
SD $= 1.92$
#### Explanation:
Given-
$n = 15$
$p = \frac{3}{7}$
Calculate -
$q = 1 - p = 1 - \frac{3}{7} = \frac{4}{7}$
$M e a n = n p = 15 \times \frac{3}{7} = 6.43$
$S D = \sqrt{n p q} = \sqrt{15 \times \frac{3}{7} \times \frac{4}{7}} = \sqrt{36.7} = 1.92$
| 172 | 398 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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}
| 4.3125 | 4 |
CC-MAIN-2022-27
|
latest
|
en
| 0.538245 |
https://matheducators.stackexchange.com/questions/11491/grading-a-simple-rounding-exercise
| 1,726,555,311,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651739.72/warc/CC-MAIN-20240917040428-20240917070428-00151.warc.gz
| 354,624,756 | 42,595 |
# Grading a Simple Rounding Exercise
A student is given the question:
"Round off each of the following numbers, correct to two significant figures.
• 32.4892
• 8.2673
• 475.98
• 0.078625"
There are two marks for each.
• 32.0000
• 8.3000
• 480.0000
• 0.07900
I believe that $32.0000=32$ so the student has correctly rounded... although I concede they might have somewhat missed the point.
• It seems like they understand rounding and the concept of two sig figs (since all of their answers are rounded correctly and they all only contain 2 sig figs), so they got the point of the assignment. they just wrote their answers wacky and with tons of unnecessary 0's. I would deduct 1 to 2 points overall, imo. Commented Oct 7, 2016 at 12:22
• I think the issue here is with the problem statement. The student correctly solved the task you have given, I would give 100% of points for correctness. (Although I might deduce some clarity points if you have such a thing.) Commented Oct 7, 2016 at 12:52
• @celeriko 1 or 2 points out of how many? Commented Oct 7, 2016 at 13:09
• They show no understanding of the meaning of rounding. 32.4892 does not round to 32.0000. 2 or 3 off out of the 8. Commented Oct 9, 2016 at 3:44
• My husband, an engineer, has drilled into me the difference in his work of a measurement of 3.2 and 3.200. I in turn have passed this on to my students. However, many elementary school teachers don't emphasize this or even understand it. My question is, were the students taught that there is a reason to drop those extra zeroes? Commented Oct 9, 2016 at 16:10
What is the whole point of rounding values? "For brevity, we approximate long decimals by finding the nearest specified place value." (This is copied directly from the slide in my first-day college statistics lecture.) If the expression didn't get shorter, then the student really has missed the whole point, and needs correction. Since it was the same basic error repeated, I would tend to take off the value of one single item (1/4 of the total value for this sequence; i.e, 2 points per your comment).
Secondarily, while mathematically it's true that $32.0000 = 32$, in a practical scientific context that carries different, distinct information. The first expression is saying "this measurement is accurate to the ten-thousandths place", whereas the second is saying "we are only confident of the precision to the units place".
This latter issue is something that I'm not sure my students ever really understand; they may just be going through the motions by rote, which doesn't make me super happy, but we don't have extra time for investigation of instrumentation accuracy.
• Compelling arguments...rounding 475 to 500 doesn't make it shorter. Commented Oct 7, 2016 at 18:48
• @JpMcCarthy: Although for most use cases with n > 1 it makes it similarly shorter in scientific notation, e,g., rounding 475,621 to one significant digit allows us to write $5 \times 10^6$. Commented Oct 7, 2016 at 22:09
• Yes I thought of this afterwards. Of course we don't have great cause to round 475 in this fashion either. Commented Oct 8, 2016 at 7:26
• Why not? Calculations with 500 are much easier in most cases and rounding is often used in order to get a quick approximation of the final result. Commented Oct 8, 2016 at 15:23
• @Jp: 500 also has less cognitive load than 475, can be read more quickly.
– user797
Commented Nov 7, 2016 at 14:53
$32.4892$ rounded off to two significant digits is $32$. The student's answer i.e. $32.0000$ contains six significant digits and an absolute error $|\epsilon_a|=0.5×10^{-4}$
| 953 | 3,602 |
{"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}
| 3.96875 | 4 |
CC-MAIN-2024-38
|
latest
|
en
| 0.963973 |
http://www.physicsforums.com/showthread.php?t=192610
| 1,369,214,582,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368701562534/warc/CC-MAIN-20130516105242-00091-ip-10-60-113-184.ec2.internal.warc.gz
| 661,334,835 | 7,723 |
## Enthalpy of Combustion -- Alkanes
Write a balanced equation for the combustion of CH4(g) (methane)
(its reaction with O2(g) forming the products CO2(g) and H2O(l)).
Given the following standard heats of formation:
$$\Delta$$Hf° of CO2(g) is -393.5 kJ/mol
$$\Delta$$Hf° of H2O(l) is -286 kJ/mol
$$\Delta$$Hf° of CH4(g) is -74.8 kJ/mol
Calculate the difference, $$\Delta$$H-$$\Delta$$E=$$\Delta$$(PV) for the combustion reaction of 1 mole of methane.
(Assume standard state conditions and 298 K for all reactants and products.)
-----------------------------------------------------------------------------------
The ideal gas law must be used.
The incremental change in volume due to liquids is neglected (cf the volume of 1 mole of H2O(l) is 18 mL, whereas the volume of 1 mol of a gas at 298 K is about 24,450 mL.
Be careful to account for the total change in number of moles of gas molecules (be sure the signs and units are correct too).
Note that certain alkanes are gases in the standard state.
-----------------------------------------------------------------------------------
My balanced equation was CH4 (g) + 2O2 (g) ---> CO2 (g) + 2H2O (l)
it says use PV=nRT but i was confused about the change in moles like is it just the change in moles of the GASES so molesfinal - molesininitial = 1 - 3 = -2 ????
and then
(-2)(0.08206)(298) = the difference....???
but i think thats wrong!
| 381 | 1,399 |
{"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}
| 3.5 | 4 |
CC-MAIN-2013-20
|
latest
|
en
| 0.82369 |
https://www.educationquizzes.com/ks1/maths/year-1-calculation-multiplication-using-arrays/
| 1,716,745,219,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00841.warc.gz
| 635,925,338 | 9,603 |
UK USIndia
Every Question Helps You Learn
2 groups of 4 is the same as 4 groups of 2.
# Year 1 Calculation - Multiplication Using Arrays
This quiz addresses the requirements of the National Curriculum KS1 Maths and Numeracy for children aged 5 and 6 in year 1. Specifically this quiz is aimed at the section dealing with multiplication using arrays.
Multiplication using arrays is a way of organising the numbers in a visual way in order to understand how repeated addition is the same as multiplication. Using an array to visualise a problem also highlights how the order of the multiplication can be reversed to give the same answer. For example, 2 x 4 gives the same answer as 4 x 2.
Question 1
2 + 5
4 x 5
5 + 5
5 x 5
2 x 5 is 2 groups of 5, or 5 + 5
Question 2
3 + 3 + 3
6 + 6 + 6
3 x 3
6 x 6
3 x 6 is the same as 3 groups of 6, or 6 + 6 + 6
Question 3
30
13
26
15
5 groups of 3, or 3 + 3 + 3 + 3 + 3 = 15
Question 4
3 + 3 + 3 + 3 + 3 + 3
6 + 6 + 6 + 6 + 6 + 6
3 x 3
6 x 6
6 x 3 is the same as 6 groups of 3, or 3 + 3 + 3 + 3 + 3 + 3
Question 5
18
9
99
81
3 x 3 = 9. 9 plus 9 = 18
Question 6
6
8
4
3
2 groups of 3, or 3 + 3 = 6
Question 7
18
12
8
14
3 groups of 4, or 4 + 4 + 4 = 12
Question 8
2 + 2 + 2 + 2 + 2
4 x 5
5 + 5 + 5 + 5 + 5
5 x 5
5 x 2 is 5 groups of 2, or 2 + 2 + 2 + 2 + 2
Question 9
2 + 2
4 x 2
4 + 4 + 4
2 x 2
2 groups of 4 is the same as 4 groups of 2
Question 10
72
12
27
14
2 groups of 7, or 7 groups of 2, makes 14
You can find more about this topic by visiting BBC Bitesize - How to multiply using an array
Author: Angela Smith
| 626 | 1,561 |
{"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-2024-22
|
latest
|
en
| 0.842628 |
https://www.thestudentroom.co.uk/showthread.php?t=5364450
| 1,571,260,886,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570986669546.24/warc/CC-MAIN-20191016190431-20191016213931-00517.warc.gz
| 1,112,740,281 | 44,173 |
# 2018 Unofficial Markscheme Edexcel AS Further Maths Core Pure Paper 1Watch
Announcements
Poll: How many marks do you think you got?
0-10 (0)
0%
11-20 (1)
3.45%
21-30 (1)
3.45%
31-40 (5)
17.24%
41-50 (3)
10.34%
51-60 (6)
20.69%
61-70 (8)
27.59%
71-80 (5)
17.24%
#1
3a (black lines aren't part of the half-line)
3b
lw-1-1il = 3
(x-1)^2 + (y-1)^2 = 9
tan (pi/4) = y / (x-2)
y = x-2
(x-1)^2 + (x-3)^2 = 9
2x^2 - 8x + 1 = 0
x = 7.5 + 2 * 14^0.5
x^2 = 19.5 + 4 * 14^0.5
y = 0.5 * 14^0.5
y^2 = 3.5
x^2 + y^2 = 11 + 2 * 14^0.5
4b
7
We know that the real root 3 must be accompanied by two imaginary roots or a triangle won't be formed in the diagram.
The coefficient of z^2 is 1, so the sum of roots is -1.
3 + x + iy + x - iy = -1
2x = -4
x = -2
So the two other roots have a real part of -2.
3 - (-2) gives a base of 5.
The triangle has area 35.
So it's height is 14.
This means the imaginary part of the two other roots is 7 and -7 respectively.
Hence the roots are -2 + 7i, -2 -7i and 3.
If anyone remembers any other questions post the solution and I'll paste it into the OP.
1
1 year ago
#2
0
1 year ago
#3
Note to self to come back and add what I remember later... Could probably manage the prove by induction questions (if I can remember the matrix)!
The first two questions were nice, unfortunately I can't remember the numbers for them
Questions ~~(will check when I complete them tomorrow)
5 - 8 ^n 4n - 8n
2 1 ? 1-4n
Is divisible by 21
Posted from TSR Mobile
0
1 year ago
#4
for 3b you forgot to square the absolute value of the complex number
0
1 year ago
#5
(Original post by IWantIPods)
I got the hard part of the question right then got the easy part wrong , the last clearing up step
0
1 year ago
#6
Disclaimer:
These are my own answers. Whether they are right or not, I have no way to tell. Also, I don’t particularly remember the order of the questions so it’s going to be weird. I may have messed up the parts as well.
1
(a) This was a find the inverse q. I can’t remember this exactly but the determinant was -69 and the numbers were very very weird.
(b)
(c) The 3 planes meet at a single point.
Questions 2-6
Sorry. I can’t remember the order of the questions at all but I do have some idea of what the answers are…
Summations q:
(a) Show that q. Hope everyone got this
(b) n=29
Transformations q:
(a) Rotation 120 degrees anticlockwise about the origin
(b) or something like that. I think it was a reflection in the line y=-x?
(c) You had to multiply the matrices from (b) and (a) together. It had to be that order.
(d) k was something like It was something like that. It had an integer and an irrational number.
Vectors q:
(a) Can’t remember
(b) angle: 7 point something degrees, shortest distance: 3 hundred and something cm
Complex Numbers q:
(a) It was a shade the region q
(b) |w| was I remember it was something like that.
7
p = 41, q = -159
8
Proof by Induction q. I’m hoping everyone was able to do this.
9
(a) , [tex]b=\frac{274}{15}
(b) 268pi
(c) 1 mark q about something insignificant.
(d) Something along the lines of there was a 12% difference so the model is unreliable.
PS. Again, these were just my answers. Whether they are right, I don’t know. If you guys have anything you disagree with, please comment.
0
1 year ago
#7
The vectors question was 7.58 if you used sine, and 7.60 if you used cosine. I imagine they'll be looking for something which rounds to 7.6
I got 375cm for the pipe, but I know there were a few other answers floating around
Part c) in question 9 was a limitation on the model (eg. assumes bottle is full/has no air in etc.)
Question 2 was the roots question, I can't remember what the numbers were though...I think the squared coefficient was -19, but I might be misremembering.
0
1 year ago
#8
Ques 1. M=
(2 1 -3
4 -2 1
3 5 -2)
and i found the inverse by finding the cofactors of minors. Changing the signs. Transpose and dividing by determinant (although I know a shortcut but couldnt risk the marks).
(1/69) ( 1 13 5
-11 -5 14
-26 7 8)
part b had matrix M as three equations which had to be solved simultaneously.
2x+y-3z = -4
4x-2y+z = 1
3x+5y-2z = 5
I just multiplied the answer to part a by (-4,1,5) to get (2,1,3).
0
1 year ago
#9
Does anyone remember how much marks each question was worth?
I remembered some of them and I've tried estimating the ones I couldn't, which gave me something like this:
Q1 Matrices (5 marks)
Q2 Roots of Polynomials (5 marks)
Q3 Complex Number / Argand Diagram (12 marks)
Q4 Summations proof and the Cosine Summation (12 marks)
Q5 Vectors (12 marks)
Q6 Matrix transformations (5 marks)
Q7 Roots on Argand diagram create a triangle with are of 35 (6 marks)
Q8 Proof by induction (12 marks)
Q9 Volumes of Revolution (11 marks)
P.S. The order of the questions is probably wrong, can't really remember much about questions 3-7
0
1 year ago
#10
How is the proof by induction done? Year 12 student who will take the AS next year. I can get down to f(k+1) -f(k) but do not know where to go from there. Is it really 21?
0
1 year ago
#11
(Original post by TurkishMathsHelp)
How is the proof by induction done? Year 12 student who will take the AS next year. I can get down to f(k+1) -f(k) but do not know where to go from there. Is it really 21?
Yes it is If I'm remembering right it was actually f(k+1)-4f(k)
Posted from TSR Mobile
0
1 year ago
#12
(Original post by Lemur14)
Yes it is If I'm remembering right it was actually f(k+1)-4f(k)
Posted from TSR Mobile
So it would be f(k+1)-4f(k)= 16(4^k) + 5(5^2k) - 16 (4^k) + 4/5 (5^2k)
That gives 21(5^k-1) oh okay thanks.
Also if anyone can explain it please, the summation formula but the cos part of it. I do not understand it and my teachers only give the textbook for work. I used the classwiz to get the number but would like to know how to do the expanation. Sorry, probably sound stupid
0
1 year ago
#13
(Original post by TurkishMathsHelp)
So it would be f(k+1)-4f(k)= 16(4^k) + 5(5^2k) - 16 (4^k) + 4/5 (5^2k)
That gives 21(5^k-1) oh okay thanks.
Also if anyone can explain it please, the summation formula but the cos part of it. I do not understand it and my teachers only give the textbook for work. I used the classwiz to get the number but would like to know how to do the expanation. Sorry, probably sound stupid
I can't remember what it was but that sounds about right. If it simplifies to that I'm sure it's fine
I don't think there was a summation formula for the cos part. You had to go through and substitute the values in and work out the total like that as you did.
Posted from TSR Mobile
0
1 year ago
#14
I don’t understand the cosine part of the summations question, why is it 130(2 2 2 2 2 2 2)
0
1 year ago
#15
(Original post by StugglinAlevels)
I don’t understand the cosine part of the summations question, why is it 130(2 2 2 2 2 2 2)
Hey sorry this is really late for everyone but the summations with the cosine was a weird questions. The TLDR is that for the sigma rcos(rpi/2) you subbed in 1,2,3,4,5,6,7,8....28 in for the r's. Now if you started this you can notice a pattern (the cos is in radians by the way) it goes , 0,-4,0,6,0,-8,0,10... you notice that after each set of 4 the difference between the sum of the numbers is 2 (0+-4+0+6=2). Because you know it goes up until 28 you know there will be 28/4 sets=7. Because the sum of each set of 4 is 2. the total is 14 because there is 2x7 two's in the brackets.
0
1 year ago
#16
Hi there, do you guys have an unofficial mark scheme for the further applied as Edexcel 2018 paper?
0
1 year ago
#17
Hi there, do you have an unofficial mark scheme for the further applied paper? Thanks - Henry
0
10 months ago
#18
could anyone show how you got from the third line to the fourth line on the series question where n=29 please
0
X
new posts
Back
to top
Latest
My Feed
### Oops, nobody has postedin the last few hours.
Why not re-start the conversation?
see more
### See more of what you like onThe Student Room
You can personalise what you see on TSR. Tell us a little about yourself to get started.
### University open days
• The University of Law
The Bar Series: Applications and Interviews - London Bloomsbury campus Postgraduate
Thu, 17 Oct '19
• Cardiff Metropolitan University
Sat, 19 Oct '19
• Coventry University
Sat, 19 Oct '19
### Poll
Join the discussion
#### Why wouldn't you turn to teachers if you were being bullied?
They might tell my parents (2)
3.23%
They might tell the bully (8)
12.9%
I don't think they'd understand (13)
20.97%
It might lead to more bullying (22)
35.48%
There's nothing they could do (17)
27.42%
| 2,641 | 8,586 |
{"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}
| 3.953125 | 4 |
CC-MAIN-2019-43
|
latest
|
en
| 0.901125 |
https://www.askiitians.com/forums/General-Physics/9/22062/kinematics.htm
| 1,713,289,414,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817103.42/warc/CC-MAIN-20240416155952-20240416185952-00739.warc.gz
| 601,451,717 | 43,942 |
# The time of flight of a projectile is 10 sec.its horizontal range is 100m.Calculate angle and velocity of projection
Chetan Mandayam Nayakar
312 Points
13 years ago
v2sin(2theta)=100g, vsin(theta)=5g,thus v2=25g2/sin2(theta), thus 50g2cot(theta)=100g, theta=arccot(1/5), v=10√26 m/sec
509 Points
13 years ago
time of flight=2usin@/g=10........1
horizontal range=u^2sin2@ /g=100........2
dividing 1by2
usin2@/2sin@=10 or
ucos@=10....3
divide 1 by 3
we get tan@=5 ,cos@=1/sqrt26
on putting value of @ in 3 we get
u=10sqrt26
Ashwin Sheoran
44 Points
5 years ago
Let θ be the angle of projection (with the horizontal) and ‘u’ be the velocity of projection. Given time of flight is, t = 10 s.
Horizontal range, x = 100 m
=> u cosθ × 10 = 100
=> u cosθ = 10 …………………..(1)
=> u2 cos2θ = 100 ………………..(2)
Now, considering the vertical motion of the projectile,
v = u + at
=> -u sinθ = u sinθ – gt (since, initial vertical velocity = u sinθ, final vertical velocity after time t is –u sinθ)
=> u sinθ = gt/2 = (10 × 10)/2 = 50 ………………..(3)
=> u2 sin2θ = 2500 ………………………………….(4)
(2) + (4) => u2 = 2600
=> u = 50.99 m/s
(3)/(1) => tanθ = 5
=> θ = 78.690
| 440 | 1,167 |
{"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}
| 3.9375 | 4 |
CC-MAIN-2024-18
|
latest
|
en
| 0.659385 |
https://www.diagrampedia.com/how-to-calculate-gain-of-a-circuit/
| 1,713,024,664,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816820.63/warc/CC-MAIN-20240413144933-20240413174933-00157.warc.gz
| 713,190,302 | 12,389 |
# How To Calculate Gain Of A Circuit
By | January 8, 2023
Understanding how to calculate gain of a circuit is essential for anyone looking to optimize their electrical systems. Whether you’re designing a commercial building, a residential house, or a computer system, understanding the basics of gain calculation can help you make more efficient and cost-effective decisions.
Gain in an electrical circuit is defined as the ratio between the output power and input power. This ratio can be expressed either as a positive or negative number, depending on the design of the circuit. In other words, if the circuit amplifies the signal, the gain is positive; if the circuit attenuates the signal, the gain is negative.
To calculate the gain of a circuit, the first step is to identify the input and output power levels. Input power is the power supplied to the circuit, while output power is the power delivered by the circuit. The next step is to determine the voltage and current values at both points. Once these values are known, the gain can be calculated using the formula:
Gain = (Vout x Iout) / (Vin x Iin)
Gain is usually stated in decibels (dB). A gain of 0 dB means that the circuit does not amplify or attenuate the signal. A gain of 10 dB means that the output power is ten times greater than the input power. Conversely, a gain of -10 dB indicates that the output power is one-tenth of the input power.
In some cases, the gain of a particular circuit can also be calculated by measuring the voltage gain and the current gain separately. Voltage gain is the ratio of the output voltage and the input voltage. Similarly, the current gain is the ratio of the output current and the input current. Both of these ratios can be used to calculate the overall gain of the circuit.
It’s important to note that gain is not always the same across the entire frequency range of a circuit. Many circuits have different gains at different frequencies, so it’s important to take this into account when calculating the gain. Additionally, the gain of the circuit may also be affected by external factors such as temperature and humidity.
By understanding how to calculate gain of a circuit, you can more accurately predict the performance of your electrical systems and make better decisions about their design. With the right knowledge and tools, you can ensure that your designs are efficient and cost-effective.
Amplifier Gain Instrumentationtools
From The Q And A Nuts Volts Magazine
Voltage Amplifier Gain Formula Examples What Is Study Com
How To Determine Power Gain And Voltage In Rf Systems Analog Technical Articles Ti E2e Support Forums
Amplifier Gain Amplifiers And Active Devices Electronics Textbook
Op Amp Open Loop Gain
How To Calculate The Midband Gain Of A Transistor Circuit
Voltage Amplifier Gain Formula Examples What Is Study Com
Amplifier Gain Decibel Or Db Electrical4u
Amplifier Gain Amplifiers And Active Devices Electronics Textbook
Solved Calculate The Voltage Gain For Circuit Shown In Chegg Com
Op Amp Gain Calculator
Solved C Calculate The Amplifier Gain Of Circuit Below Chegg Com
What Are Open Loop And Closed Gains Of An Op Amp Toshiba Electronic Devices Storage Corporation Europe Emea
Solved Assuming λ 0 Calculate The Voltage Gain Of Circuit Shown In 1 Answer Transtutors
Solved Calculate The Voltage Gain For This Circuit 16v 90 Chegg Com
Solved 1 Calculate The Closed Loop Voltage Gain Of Chegg Com
Solved Check Point 4 Calculate The Overall Voltage Gain Of Chegg Com
Amplifier Gain Amplifiers And Active Devices Electronics Textbook
| 748 | 3,594 |
{"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}
| 4.03125 | 4 |
CC-MAIN-2024-18
|
latest
|
en
| 0.91815 |
https://www.enotes.com/homework-help/what-does-2-tons-equal-pounds-477359
| 1,498,455,021,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128320679.64/warc/CC-MAIN-20170626050425-20170626070425-00339.warc.gz
| 875,350,671 | 12,785 |
# what does 2 tons equal in pounds
electreto05 | College Teacher | (Level 1) Assistant Educator
Posted on
To convert tons to pounds, first we convert tons in kilograms knowing that:
1 ton(metric) = 1000 kg
then:
2 tons(metric) = 2 x 1000 kg = 2000 kg
Now we convert kilograms into pounds, knowing that:
1 kg = 2.204622621848776 pounds
then:
2000 kg = 2000 x 2.204622621848776 = 4409.245243697552 pounds
So that:
2 tons(metric) ≈ 4409.25 pounds
rsarvar1a | Student, Grade 9 | (Level 1) Honors
Posted on
I'll assume that the two units that you are looking at here are the imperial units of the pound and the short ton.
First, we want to find the conversion for short tons to pounds, which is pretty easy to find. A reliable source tells us that:
1 short ton = 2000 pounds
So, all you have to do is apply this conversion. If the number of tons is 2, and the number of pounds per ton is 2000, you can simply multiply 2 by 2000 to get the answer. So:
(Number of tons) * 2000 (Pounds in a ton) = 4000 (pounds in two tons)
| 310 | 1,036 |
{"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-2017-26
|
longest
|
en
| 0.873174 |
https://classroomsecrets.co.uk/mixed-age-year-5-and-6-perimeter-and-area-step-3-resource-pack/
| 1,716,714,350,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058872.68/warc/CC-MAIN-20240526074314-20240526104314-00168.warc.gz
| 133,897,533 | 79,091 |
Mixed Age Year 5 and 6 Perimeter and Area Step 3 Resource Pack – Classroom Secrets | Classroom Secrets
All › Mixed Age Year 5 and 6 Perimeter and Area Step 3 Resource Pack
# Mixed Age Year 5 and 6 Perimeter and Area Step 3 Resource Pack
## Step 3: Mixed Age Year 5 and 6 Perimeter and Area Resource Pack
Mixed Age Year 5 and 6 Perimeter and Area Step 3 Resource Pack includes a teaching PowerPoint and differentiated varied fluency and reasoning and problem solving resources for this step which covers Year 5 Calculate Perimeter for Autumn Block 5.
### What's included in the Pack?
This Mixed Age Year 5 and 6 Perimeter and Area Step 3 pack includes:
• Mixed Age Year 5 and 6 Perimeter and Area Step 3 Teaching PowerPoint with examples for both year groups.
• Year 5 Calculate Perimeter Varied Fluency with answers.
• Year 5 Calculate Perimeter Reasoning and Problem Solving with answers.
#### National Curriculum Objectives
Mathematics Year 5: (5M7a) Measure and calculate the perimeter of composite rectilinear shapes in centimetres and metres
Differentiation for Year 5 Calculate Perimeter:
Varied Fluency
Developing Questions to support calculating the perimeter of simple rectilinear shapes. Includes whole centimetres and metres, with no conversion of units.
Expected Questions to support calculating the perimeter of composite rectilinear shapes. Includes half and quarter lengths (represented as decimals or fractions) with no conversion of units.
Greater Depth Questions to support calculating the perimeter of composite rectilinear shapes. Includes half and quarter lengths (represented as decimals or fractions) with some conversion of units.
Reasoning and Problem Solving
Questions 1, 4 and 7 (Problem Solving)
Developing Investigate the perimeter of a square using the given information. Whole numbers only and no conversion of units.
Expected Investigate the perimeter of a shape using the given information. Using knowledge of all table facts and whole numbers.
Greater Depth Investigate the perimeter of a shape using the given information. using knowledge of all table facts and decimal numbers.
Questions 2, 5 and 8 (Problem Solving)
Developing Calculate a new perimeter when a shape is altered from square to rectangle, using whole numbers.
Expected Calculate a new perimeter when a shape is altered, using 2-digits and some decimals.
Greater Depth Calculate a new perimeter when a shape is altered using multiplication and including decimals and fractions.
Questions 3, 6 and 9 (Reasoning)
Developing Use knowledge of calculating the perimeter of a shape to explain why an answer is or is not possible. Includes whole numbers with no conversion of units.
Expected Use knowledge of calculating the perimeter of a shape to explain why an answer is or is not possible. Includes some fractions and decimals with no conversion of units.
Greater Depth Use knowledge of calculating the perimeter of a shape to explain why an answer is or is not possible. Includes fractions, decimals and some conversion of units.
| 629 | 3,042 |
{"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-2024-22
|
latest
|
en
| 0.839502 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-with-applications-10th-edition/chapter-12-sequences-and-series-12-7-l-hospital-s-rule-12-7-exercises-page-659/17
| 1,585,601,535,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00359.warc.gz
| 988,838,744 | 12,507 |
## Calculus with Applications (10th Edition)
$$\frac{1}{{12}}$$
\eqalign{ & \mathop {\lim }\limits_{x \to 8} \frac{{\root 3 \of x - 2}}{{x - 8}} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}8{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 8} \left( {\root 3 \of x - 2} \right) = \root 3 \of 8 - 2 = 2 - 2 = 0 \cr & \mathop {\lim }\limits_{x \to 8} \left( {x - 8} \right) = 8 - 8 = 0 \cr & \cr & {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr & {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr & {\text{denominator}}{\text{}} \cr & {\text{for }}\root 3 \of x - 2 \to {D_x}\left( {{x^{1/3}} - 2} \right) = \frac{1}{3}{x^{ - 2/3}} - 0 = \frac{1}{{3\root 3 \of {{x^2}} }} \cr & {\text{for }}\left( {x - 8} \right) \to {D_x}\left( {x - 8} \right) = 1 \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 8} \frac{{\root 3 \of x - 2}}{{x - 8}} = \mathop {\lim }\limits_{x \to 8} \frac{1}{{3\root 3 \of {{x^2}} }} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{1}{{3\root 3 \of {{{\left( 8 \right)}^2}} }} \cr & = \frac{1}{{3\left( 4 \right)}} \cr & = \frac{1}{{12}} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 8} \frac{{\root 3 \of x - 2}}{{x - 8}} = \frac{1}{{12}} \cr}
| 627 | 1,505 |
{"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}
| 4.75 | 5 |
CC-MAIN-2020-16
|
latest
|
en
| 0.467456 |
https://psiberg.com/maxwell-boltzmann-distribution/
| 1,686,132,758,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224653631.71/warc/CC-MAIN-20230607074914-20230607104914-00750.warc.gz
| 505,673,902 | 58,328 |
The Maxwell-Boltzmann distribution is the probability distribution that describes the velocities distribution of gas particles in a container. The particles of a gas are identical but distinguishable.
In the 1880s, this theory was developed by James Clerk Maxwell and Ludwig Boltzmann. It is based on the kinetic theory of gases which states that gas molecules are constantly in motion and are colliding with each other at various energies.
The Maxwell-Boltzmann distribution of their energies at a given temperature can be graphed using a fraction of particles on the y-axis against kinetic energy on the x-axis. Moreover, the distribution can be used to calculate various statistical properties of a gas, such as the mean velocity, activation energy, and the effect of catalyst.
In the above graph, activation energy is labeled as Ea. It is the minimum amount of energy required for colliding particles to react. Vrms is the average root mean square velocity of all the particle present in the system.
Outline
## What is Maxwell Boltzmann’s distribution law?
Maxwell Boltzman’s law states that the continuously colliding particles of an ideal gas are distributed in different fractions or groups, having a certain amount of energy. This statistical distribution of the particles is shown in the figure below:
### Three postulates of Maxwell-Boltzman distribution law
1. All particles present in the system are distinguished.
2. Number particles are associated with each energy state.
3. The total number of particles in the entire system is constant.
## Factors affecting Maxwell Boltzmann distribution curve
The Boltzmann distribution curve displays the fraction of particles with motions. So, the changes in conditions may alter its representation. The most prominent effect is caused by two factors: temperature and catalyst.
### Effect of temperature on Boltzmann distribution curve
The increase in temperature increases the kinetic energy, so the number of collisions raises the probability of successful collisions.
For instance, If we heat the gas to a higher temperature, the peak will shift towards the right due to an increase in kinetic energy. As the graph moves to the right, the height of the curve peak decrease. Since the area under the curve representing the total number of particles in the container is conserved, therefore upon cooling, the graph becomes taller and narrower while on heating it becomes wider and shorter.
In the graph mentioned below, the peak of the curve at T2 is slightly steeper than the T1. Moreover, the shaded area shows the particles with greater energy than the activation energy. Thus, increasing the temperature increases the rate of reaction.
The general graph which represents the curves obtained at various temperatures is given below:
## Effect of catalyst
In a chemical reaction, the reactant particle requires energy to get converted to the product. This minimum energy is known as activation energy. So, the particles with energy greater than Ea can react. If a catalyst is added to the system, it provides an alternative pathway to the reactant with lower activation energy. Now the molecules under the shaded regions A and B can react to form a product at a given temperature.
## Experimental evidence
Boltzmann emphasized the discrete nature of the distribution of energy levels. This was later verified by Max Plank’s quantum theory. After that, Maxwell Boltzmann’s distribution was subjected to verification due to its practical importance. Some of the experiments are explained below:
### Stern experiment
The Stern experiment is the most famous experiment for this purpose. The principle of this experiment is the heating of silver and the emission of atomic silver. The platinum wire is silver-doped is used in it. Upon heating, the emitted silver atoms whose speed has to measured is pass from the slit and the ranges of particle distribution to their respective velocity. However, this experiment has an accuracy of 85%.
Zartman KO experiment
Zartman and KO conduct the second experiment in 1930. The apparatus consists of an oval vessel with a narrow opening. A stream of bismuth atoms is passed through two parallel slits. Bismuth atoms collide with a drum-shaped screen. When the drum is stationary, the particles collide in the same spot, but when it moves quickly, the particles disperse throughout the drum. A microphotometer can be used to measure the thickness of deposited atoms, which is then used to calculate the velocity distribution. Furthermore, it confirms Maxwell Boltzmann’s distribution law with a 1% agreement of uncertainty.
## Applications of Boltzmann law
Maxwell’s distribution law has many applications in thermodynamic systems of physics and chemical synthesis. It gives the basis for the authenticity of the kinetic molecular of gases. The Maxwell-Boltzmann velocity distribution function is valid even in non-ideal gases.
Note that, it is perfectly applicable where long-range interactions between the particles of the system are absent or feeble.
Some of its famous applications are mentioned below:
• The average kinetic energy of a system
• Activation energy
• Suitability of the catalyst
• Specific heat of solids and liquids
• Statistical thermodynamics in classical limits
Related resources
## Concepts Berg
What does the area under a Maxwell-Boltzmann distribution represent?
The area under the curve represents the number of molecules present in the sample. The molecules of the system remain unchanged during changes in temperature.
What is the Maxwell-Boltzmann distribution?
Maxwell-Boltzmann distribution is the law that describes the probability of finding the particles at different temperatures in a system.
What is James Clerk Maxwell most famous for?
James Clark Maxwell is famous for his work on energy distribution and the theory of electromagnetic radiation.
How does the Maxwell-Boltzmann Distribution work?
It works as the population or ratio of particles over the total number of particles. The graph can describe the distribution of molecules at various temperatures. Moreover, The effect of temperature and the addition of a catalyst can be assumed by using the Maxwell-Boltzmann distribution law.
What are the properties of Maxwell Boltzmann’s statistics?
The maxwell Boltzmann statistics gives insight into the distribution of particles at several states. The statistical approach is only applicable at the classical level because here, the particle density is not negligible.
What is the Maxwell-Boltzmann distribution curve?
It is the representation of particle distribution at different temperatures. It is used to derive the average kinetic energy of the particle, root means square value and the effect of the catalyst on activation energy.
What are the limitations of the Maxwell-Boltzmann distribution law?
This law is only applicable when the particle size is within classical limits. It does explain the photoelectric effect and black body radiation.
Why is the Boltzmann distribution important?
It describes the distribution of ideal gas particles at different temperatures in a given state. Moreover, This distribution is further used to calculate the energy of activation and temperature ranges required for a chemical reaction.
What is a good explanation of the Maxwell-Boltzmann distribution?
It is simply explained as a graph plotted between the number of particles at the y-axis against temperature ranges at the x-axis.
References
| 1,420 | 7,518 |
{"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}
| 3.5625 | 4 |
CC-MAIN-2023-23
|
longest
|
en
| 0.914055 |
https://www.mindxmaster.com/2015/12/very-important-tech-mahindra-campus.html
| 1,524,511,826,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125946165.56/warc/CC-MAIN-20180423184427-20180423204427-00600.warc.gz
| 849,284,359 | 100,123 |
Very important tech mahindra campus questions 2016
These are best tech mahindra campus questions that were asked in the test. You can go through them and see if are able to solve them this the largest collection consists of arithmetic section, logical sections, Programming and other important questions.
These are best tech mahindra campus questions that were asked in the test. You can go through them and see if are able to solve them this the largest collection consists of arithmetic section, logical sections, Programming and other important questions.
ARITHMETIC SECTION
This section consists of 29 problems. The questions are simple though time consuming.
1. If a boat is moving in upstream with velocity of 14 km/hr and goes downstream with a velocity of 40
km/hr, then what is the speed of the stream ?
(a) 13 km/hr
(b) 26 km/hr
(c) 34 km/hr
(d) none of these
Ans. A
2. Find the value of ( 0.75 * 0.75 * 0.75 - 0.001 ) / ( 0.75 * 0.75 - 0.075 + 0.01)
(a) 0.845
(b) 1.908
(c) 2.312
(d) 0.001
Ans. A
3. A can have a piece of work done in 8 days, B can work three times faster than the A, C can work
five times faster than A. How many days will they take to do the work together ?
(a) 3 days
(b) 8/9 days
(c) 4 days
(d) can't say
Ans. B
4. A car travels a certain distance taking 7 hrs in forward journey, during the return journey increased
speed 12km/hr takes the times 5 hrs. What is the distance traveled.
(a) 210 kms
(b) 30 kms
(c) 20 kms
(c) none of these
Ans. B
5. Instead of multiplying a number by 7, the number is divided by 7. What is the percentage of error
obtained ?
6. Find (7x + 4y ) / (x-2y) if x/2y = 3/2 ?
(a) 6
(b) 8
(c) 7
(d) data insufficient
Ans. C
7. A man buys 12 lts of liquid which contains 20% of the liquid and the rest is water. He then mixes it
with 10 lts of another mixture with 30% of liquid. What is the % of water in the new mixture?
8. If a man buys 1 lt of milk for Rs.12 and mixes it with 20% water and sells it for Rs.15, then what is
the percentage of gain?
9. Pipe A can fill a tank in 30 mins and Pipe B can fill it in 28 mins. If 3/4th of the tank is filled by Pipe
B alone and both are opened, how much time is required by both the pipes to fill the tank completely ?
10. If on an item a company gives 25% discount, they earn 25% profit. If they now give 10% discount
then what is the profit percentage.
(a) 40%
(b) 55%
(c) 35%
(d) 30%
Ans. D
11. A certain number of men can finish a piece of work in 10 days. If however there were 10 men less
it will take 10 days more for the work to be finished. How many men were there originally?
(a) 110 men
(b) 130 men
(c) 100 men
(d) none of these
Ans. A
12. In simple interest what sum amounts of Rs.1120/- in 4 years and Rs.1200/- in 5 years ?
(a) Rs. 500
(b) Rs. 600
(c) Rs. 800
(d) Rs. 900
Ans. C
13. If a sum of money compound annually amounts of thrice itself in 3 years. In how many years
will it become 9 times itself.
(a) 6
(b) 8
(c) 10
(d) 12
Ans A
14. Two trains move in the same direction at 50 kmph and 32 kmph respectively. A man in the slower
train
observes the 15 seconds elapse before the faster train completely passes by him.
What is the length of faster train ?
(a) 100m
(b) 75m
(c) 120m
(d) 50m
Ans B
15. How many mashes are there in 1 squrare meter of wire gauge if each mesh
is 8mm long and 5mm wide ?
(a) 2500
(b) 25000
(c) 250
(d) 250000
Ans B
16. x% of y is y% of ?
(a) x/y
(b) 2y
(c) x
(d) can't be determined
Ans. C
17. The price of sugar increases by 20%, by what % should a housewife reduce the consumption of
sugar so that expenditure on sugar can be same as before ?
(a) 15%
(b) 16.66%
(c) 12%
(d) 9%
Ans B
18. A man spends half of his salary on household expenses, 1/4th for rent, 1/5th for travel expenses, the
man deposits the rest in a bank. If his monthly deposits in the bank amount 50, what is his monthly
salary ?
(a) Rs.500
(b) Rs.1500
(c) Rs.1000
(d) Rs. 900
Ans C
20. The population of a city increases @ 4% p.a. There is an additional annual increase of 4% of the
population due to the influx of job seekers, find the % increase in population after 2 years ?
21. The ratio of the number of boys and girls in a school is 3:2 Out of these 10% the boys and 25% of
girls are scholarship holders. % of students who are not scholarship holders.?
22. 15 men take 21 days of 8 hrs. each to do a piece of work. How many days of 6 hrs. each would it
take for 21 women if 3 women do as much work as 2 men?
(a) 30
(b) 20
(c) 19
(d) 29
Ans. A
23. A cylinder is 6 cms in diameter and 6 cms in height. If spheres of the same size are made from the
material obtained, what is the diameter of each sphere?
(a) 5 cms
(b) 2 cms
(c) 3 cms
(d) 4 cms
Ans C
24. A rectangular plank (2)1/2 meters wide can be placed so that it is on either side of the diagonal of a
square shown below.(Figure is not available)What is the area of the plank?
Ans :7*(2)1/2
25. The difference b/w the compound interest payble half yearly and the simple interest on a
certain sum lent out at 10% p.a for 1 year is Rs 25. What is the sum?
(a) Rs. 15000
(b) Rs. 12000
(c) Rs. 10000
(d) none of these
Ans C
26. What is the smallest number by which 2880 must be divided in order to make it into a
perfect square ?
(a) 3
(b) 4
(c) 5
(d) 6
Ans. C
27. A father is 30 years older than his son however he will be only thrice as old as the son after 5 years
what is father's present age ?
(a) 40 yrs
(b) 30 yrs
(c) 50 yrs
(d) none of these
Ans. A
28. An article sold at a profit of 20% if both the cost price and selling price would be Rs.20/- the profit
would be 10% more. What is the cost price of that article?
29. If an item costs Rs.3 in '99 and Rs.203 in '00.What is the % increase in price?
(a) 200/3 %
(b) 200/6 %
(c) 100%
(d) none of these
Ans. A
Tech mahindra questions
Programming questions (15) C-Test Paper
1. #include
* What is wrong in the following problem
main() {
int i,j;
j = 10;
i = j++ - j++;
printf("%d %d", i,j);
} ans: 0, 12
2.#include
* What is the output of the following problem
main() {
int j;
for(j=0;j<3;j++)
foo();
}
foo() {
static int i = 10;
i+=10;
printf("%dn",i);
}
/* Out put is (***since static int is used i value is retained between
* 20 function calls )
* 30
* 40
*/
3.#include
#include
#include
/* This is asked in PCS Bombay walk-in-interview
* What is wrong in the following code
*/
main()
{ char *
c;
c = "Hello";
printf("%sn", c);
}
/*ans:- Hello, The code is successfully running */
4. #include
/* This problem is given in PCS BOMBAY walk-in-interview.
* What is the final value of i and how many times loop is
* Executed ?
*/
main()
{ int i,j,k,l,lc=0;
/* the input is given as 1234 567 */
printf("Enter the number string:<1234 567 n");
scanf("%2d%d%1d",&i,&j,&k);
for(;k;k--,i++)
for(l=0;l
printf("%d %dn",i,l);}
printf("LOOPS= %dn", lc-1);
} /* Ans: i
=
16, and loop is executed for 169 times */
5.#include
/* This is given in PCS Bombay walk-in-interview */
/* What is the output of the following program */
main() {
union {
int a;
int b;
int c;
} u,v;
u.a = 10;
u.b = 20;
printf("%d %d n",u.a,u.b);
} /* Ans :
The latest value assigned to any of the union member
will be present in the union members so answer is
20 20
*/
6.#include
main()
{
float i, j;
scanf("%f %f", &i, &j);
printf("%.2f %.3f", i, j);
}
/Ans:- 123.34 3. 234 */
7.#include
/* This is given in PCS Bombay walk-in-interview
* What is the out put of the following problem ?
*/
main()
{ char *
str =
"
12345";
printf("%c %c %cn", *str, *(str++), *(str++));
} /* Ans: It is not 1
2
3
* But it is 3 2 1 Why ??
*/
8.#include
/* This problem is asked in PCS Bombay Walk-in-interview
* Write a macro statement to find maximum of a,b
*/
#define max(a,b) (ab)?a:b
main()
{ int a,b;
a=3;
b=4;
printf("%d",max(a,b));
} /* Ans is very simple the coding is just for testing it
and output is 4 */
~
9.#include
/* This problem is asked in PCS Bombay
* What is the output of the following coding
*/
main()
{ int len=4;
char *st="12345678";
st = st -len;
printf("%cn",*st);
} /* Ans :
It will print some junk value */
~
10.#include
main()
{
func(1);
}
func(int i){
static char *str ={ "One","Two","Three","Four"};
printf("%sn",str[i++]);
return;
} /* Ans:- it will give warning because str is pointer to the char but
it is initialized with more values
if it is not considered then the answer is Two */
11.
#include
main()
{ int i;
for (i=1;i<100; i++)
printf("%d %0xn",i,i);
} /* Ans:- i
is from 1
to 99 for the first format,
for the second format 1to9, ato f, 10 to 19,1ato1f, 20 to 29, etc */
12.#include
/* This problem is asked in PCS Bombay walk-in-interview
* In the following code please write the syntax for
* assing a value of 10 to field x of s and id_no 101 of s
*/
struct {
int x;
int y;
union {
int id_no;
char *name;
}b;
}s,*st;
main()
{s
t = &s;
st-x=10;
st-b.id_no = 101;
printf("%d %dn",s..x,s.b.id_no);
} /* Ans: The answer is st-x=10;
* st-b.id_no=101;
*/
13.#include
/* This problem was asked in PCS Bombay in a walk-in-interview
* Write a recursive function that calculates
* n * (n-1) * (n-2) * ....... 2 * 1
*/
main() {
int factorial(int n);
int i,ans;
printf("n Enter a Number:");
scanf("%d",&i);
ans = factorial(i);
printf("nFactorial by recursion = %dn", ans);
} int factorial(int n)
{ if (
n
<= 1) return (
1);
else
return ( n * factorial(n-1));
}
~
14.#include
/* This problem is asked in PCS Bombay walk-in-interview
* What is the output of the following problem
*/
main(){
int j,ans;
j = 4;
ans = count(4);
printf("%dn",ans);
} int count(int i)
{ if (
i
<
0) return(i);
else
return( count(i-2) + count(i-1));
}
/* It is showing -18 as an answer */
15.#include
main()
{ int i=4;
if(i=0)
printf("statement 1");
else
printf("statement 2");
} /* statement 2 */
This is pcsb paper.
1. #include
* What is wrong in the following problem
main() {
int i,j;
j = 10;
i = j++ - j++;
printf("%d %d", i,j);
} ans: 0, 12
2.#include
* What is the output of the following problem
main() {
int j;
for(j=0;j<3;j++)
foo();
}
foo() {
static int i = 10;
i+=10;
printf("%dn",i);
}
/* Out put is (***since static int is used i value is retained between
* 20 function calls )
* 30
* 40
*/
3.#include
#include
#include
/* This is asked in PCS Bombay walk-in-interview
* What is wrong in the following code
*/
main()
{ char *
c;
c = "Hello";
printf("%sn", c);
} /*ans:- Hello, The code is successfully running */
4. #include
/* This problem is given in PCS BOMBAY walk-in-interview.
* What is the final value of i and how many times loop is
* Executed ?
*/
main()
{ int i,j,k,l,lc=0;
/* the input is given as 1234 567 */
printf("Enter the number string:<1234 567 n");
scanf("%2d%d%1d",&i,&j,&k);
for(;k;k--,i++)
for(l=0;l
printf("%d %dn",i,l);}
printf("LOOPS= %dn", lc-1);
} /* Ans: i
=
16, and loop is executed for 169 times */
5.#include
/* This is given in PCS Bombay walk-in-interview */
/* What is the output of the following program */
main() {
union {
int a;
int b;
int c;
} u,v;
u.a = 10;
u.b = 20;
printf("%d %d n",u.a,u.b);
} /* Ans :
The latest value assigned to any of the union member
will be present in the union members so answer is
20 20
*/
6.#include
main()
{
float i, j;
scanf("%f %f", &i, &j);
printf("%.2f %.3f", i, j);
}
/Ans:- 123.34 3. 234 */
7.#include
/* This is given in PCS Bombay walk-in-interview
* What is the out put of the following problem ?
*/
main()
{ char *
str =
"
12345";
printf("%c %c %cn", *str, *(str++), *(str++));
} /* Ans: It is not 1
2
3
* But it is 3 2 1 Why ??
*/
8.#include
/* This problem is asked in PCS Bombay Walk-in-interview
* Write a macro statement to find maximum of a,b
*/
#define max(a,b) (ab)?a:b
main()
{ int a,b;
a=3;
b=4;
printf("%d",max(a,b));
} /* Ans is very simple the coding is just for testing it
and output is 4 */
~
9.#include
/* This problem is asked in PCS Bombay
* What is the output of the following coding
*/
main()
{ int len=4;
char *st="12345678";
for(i=0; i<6; i++)
st = st -len;
printf("%cn",*st);
} /* Ans :
It will print some junk value */
~
10.#include
main()
{
func(1);
}
func(int i){
static char *str ={ "One","Two","Three","Four"};
printf("%sn",str[i++]);
return;
} /* Ans:- it will give warning because str is pointer to the char but
it is initialized with more values
if it is not considered then the answer is Two */
11.
#include
main()
{ int i;
for (i=1;i<100; i++)
printf("%d %0xn",i,i);
} /* Ans:- i
is from 1
to 99 for the first format,
for the second format 1to9, ato f, 10 to 19,1ato1f, 20 to 29, etc */
12.#include
/* This problem is asked in PCS Bombay walk-in-interview
* In the following code please write the syntax for
* assing a value of 10 to field x of s and id_no 101 of s
*/
struct {
int x;
int y;
union {
int id_no;
char *name;
}b;
}s,*st;
main()
{s
t = &s;
st-x=10;
st-b.id_no = 101;
printf("%d %dn",s.x,s.b.id_no);
} /* Ans: The answer is st-x=10;
* st-b.id_no=101;
*/
13.#include
/* This problem was asked in PCS Bombay in a walk-in-interview
* Write a recursive function that calculates
* n * (n-1) * (n-2) * ....... 2 * 1
*/
main() {
int factorial(int n);
int i,ans;
printf("n Enter a Number:");
scanf("%d",&i);
ans = factorial(i);
printf("nFactorial by recursion = %dn", ans);
} int factorial(int n)
{ if (
n
<= 1) return (
1);
else
return ( n * factorial(n-1));
}
~
14.#include
/* This problem is asked in PCS Bombay walk-in-interview
* What is the output of the following problem
*/
main(){
int j,ans;
j = 4;
ans = count(4);
printf("%dn",ans);
} int count(int i)
{ if (
i
<
0) return(i);
else
return( count(i-2) + count(i-1));
}
/* It is showing -18 as an answer */
15.#include
main()
{ int i=4;
if(i=0)
printf("statement 1");
else
printf("statement 2");
} /* statement 2
*/
LOGICAL SECTION
Directions: For questions 30-39 fill the missing number or letter in the given series
30. a, c, e, g, _
(a) h
(b) i
(c) d
(d) j
Ans. B
31. a, e, i, m, q, u, _, _
(a) y,c
(b) b,f
(c) g,i
(d) none
Ans. A
32. ay , bz , cw , dx ,__
(a) gu
(b) ev
(c) fv
(d) eu
Ans. D
33. 1, 2, 3, 5, 7, 11, __
(a) 15
(b) 9
(c) 13
(d) 12
Ans. C (series of prime numbers)
34. kp , lo , mn , __
(a) nm
(b) np
(c) op
(d) pq
Ans. A
35. R,M,__,F,D,__
(a) I, C
(b) A, Q
(c) L, N
(d) B, Q
Ans. A
36. ___, ayw, gec, mki, sqo
(a) awx
(b) usq
(c) prs
(d) lmn
Ans. B
37. 1, 3, 4, 8, 15, 27, __
(a) 60
(b) 59
(c) 43
(d) 50
Ans D
38. 0, 2, 3, 5, 8, 10, 15, 17, 24, 26,__
(a) 45
(b) 55
(c) 35
(d) 48
Ans. C
39. 2, 5, 9, 19, 37,__
(a) 64
(b) 55
(c) 75
(d) 40
Ans C
Directions for questions 40 to 45: Select the alternative that logically follows form the two given statements.
40. All scientists are fools. All fools are literates.
(a)All literates are scientists
(b) All scientists are literates
(c) No scientists are literates
(d) Both (a) and (b) are correct
Ans. B
41. No apple is an orange. All bananas are oranges.
(a) All apples are oranges
(b) Some apples are oranges
(c) No apple is a banana
(d) None of the above
Ans. A
42. All pens are elephants. Some elephants are cats.
(a) Some pens are cats
(b) No pens are cats
(c) All pens are cats
(d) None of the above
Ans. D
43. All shares are debentures.No debentures are deposits.
(a) All shares are deposits
(b) Some shares are deposits
(c) No shares are deposits
(d) None of the above
Ans. C
44. Many fathers are brothers. All brothers are priests.
(a) No father is a priest
(b) Many fathers are not priests
(c) Many fathers are priests
(d) Both (b) and (c)
Ans. B
45. Some green are blue. No blue are white.
(a) No green are white
(b) Some green are white
(c) No green are white
(d) None of the above
Ans. B
46. If the word "CODING" is represented as DPEJOH , then the word "CURFEW" can be represented?
(a) dvsgfx
(b) dvshfx
(c) dgshfx
(d) dtsgfy
Ans. A
47. If in a certain code "RANGE" is coded as 12345 and "RANDOM" is coded as 123678, then the
code for the word "MANGO" would be
(a) 82357
(b) 84563
(c) 82346
(d) 82543
Ans. D
Directions for questions 48-50:The questions are based on the following data
In a class of 150 students 55 speak English;85 speak Telugu and 30 speak neither English nor Telugu
48. How many speak both English and Telugu?
(a) 10
(b) 15
(c) 20
(d) 12
Ans. C
49.How many speak only Telugu?
(a) 55
(b) 45
(c) 65
(d) none of the above
Ans.C
50.How many speak at least one of the two languages?
(a) 120
(b) 100
(c) 250
(d) 50
Ans. A
Other Direct important questions
1. A can have a piece of work done in 8 days, B can work three times faster than the A, C can work five times faster than A. How many days will they take to do the work together ?
2. Of a mutual instrument vibrate 6,8 & 12 intervals respectively. If all three vibrate together what is the time interval before all vibrate together again? Ans : 1/2 sec HCF of DR.
3. Certain no. of men can finish a piece of work in 10 days. If however there were 10 men less it will take 10 days more for the work to be finished. How many men were there originally. Ans : 110 men
4. In simple interest what sum amounts of Rs.1120/- in 4 years and Rs.1200/- in 5 years. Ans : Rs.800/-
==>Sum of money at compound interest amounts of thrice itself in 3 years. In how many years will it take 9 times itself. Ans : 6
5. Two trains in the same direction at 50 & 32 kmph respectively. A man in the slower train observes the 15 seconds elapse before the faster train completely passes him. What is the length of faster train ? Ans : 75m
==>How many mashes are there in a sq. m of wire gauge. Each mesh being 8mm long X 5mm width Ans : 25000
6. x% of y is equal to y% of ? Ans : x
7. The price of sugar increases by 20%, by what % house-wife should reduce the consumption of sugar so that expenditure on sugar can be same as before Ans : 16.66
8. A man spending half of his salary for house hold expenses, 1/4th for rent, 1/5th for travel expenses, a man deposits the rest in a bank. If his monthly deposits in the bank amount 50. What is his monthly salary ? Ans : 1000
9. The population of a city increases @ 4% p.a. That is an additioanl annual increase of 4% of the population due to this influx of job seekers, the % increase in population after 2 years is
10.The ratio of no. of boys & girls in a school is 3:2 Out of these ?% the boys & 25% of girls are scholarship holders. % of students who are not scholarship holders.?
11.15 Men take 21 days of 8 hrs. each to do a piece of work. How many days of 6 hrs. each would do if 21 women take. If 3 women do as much work of 2 men. Ans : 30
12.A cylinder ingot 6cms in diameter and 6 cms in height is and spheres all of the same size are made from the material obtained.what is the diameter of each sphere? Ans :3cms
14.The difference b/w the compound interest payble half yearly and the simple interest on a certain sum cont out at 10% p.a for 1 year is Rs 25 what is the sum Ans:10,000
15.what is the smallest n0 by which 2880 must be divided in order to make it a perfect square ?
a)3 b)4 c)5 d) 6 e)8 Ans: 3
16.A father is 30 times more than his son however he will be only thrice as old as the son what is father's present age ?Ans : 40
17.An article sold at a profit of 20% if both the c.p & s.p were tobe Rs.20/- the profit would be 10% more. What is the c.p of that article? Ans : 1% loss
18.How can a process be started automatically in UNIX ? ANS : Through CRONTAB { options were crontab, inittab, and two more }
19.What is the meaning of run level 6 ? ANS. Reboot
20.How can you determine which file has generated the CORE ?
22.3 to 4 questions on conditional operator :?:
23.Write a macro for sqaring no.
24.Trees -3 noded tree ( 4 to 5 questions fundamentals) Maximum possible no.of arrnging these nodes
25.Arrange the nodes in depth first order breadth first order
26.Linked lists Q) Given two statments
27.Allocating memory dynamiccaly
28.Arrays Tree the above both and find the mistake
29.Pointers (7 to 8 questions) Schaum series Pointer to functions, to arrays 4 statements ->meaning,syntax for another 4 statements
30.Booting-def(When you on the system the process that takes place is ------
31.Type of global variable can be accessible from any where in the working environment ( external
global variable)
Name
ltr
item
MindxMaster: Very important tech mahindra campus questions 2016
Very important tech mahindra campus questions 2016
These are best tech mahindra campus questions that were asked in the test. You can go through them and see if are able to solve them this the largest collection consists of arithmetic section, logical sections, Programming and other important questions.
https://2.bp.blogspot.com/-rD8E-9dkRmI/Vk8GmA7WvRI/AAAAAAAAA3Q/I-z7_Z80CKM/s400/20151120164836.jpg
https://2.bp.blogspot.com/-rD8E-9dkRmI/Vk8GmA7WvRI/AAAAAAAAA3Q/I-z7_Z80CKM/s72-c/20151120164836.jpg
MindxMaster
https://www.mindxmaster.com/2015/12/very-important-tech-mahindra-campus.html
https://www.mindxmaster.com/
https://www.mindxmaster.com/
https://www.mindxmaster.com/2015/12/very-important-tech-mahindra-campus.html
true
5332415103371288268
UTF-8
| 6,728 | 20,798 |
{"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}
| 3.9375 | 4 |
CC-MAIN-2018-17
|
longest
|
en
| 0.920474 |
https://www.physicsforums.com/threads/basic-addition-question.637666/
| 1,660,768,444,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882573104.24/warc/CC-MAIN-20220817183340-20220817213340-00333.warc.gz
| 797,401,466 | 22,284 |
g4143
Hello, mathematicians,
I have a very simple question. Well its simple to state. Can you arrive at all the higher level mathematics with plain addition? Here's an example of what I mean. Can you calculate things like trigonometry and logarithms with just addition?
I ask this question because I've been following and participating in a debate which claims that you can achieve all the higher level mathematics functionality with just addition.
Myself, I fall on the side of basic addition can achieve all the higher level functions and I'm curious what a professional mathematicians would say.
Thank-you for any replies.
Number Nine
I'm at a loss as to how addition would give you category theory, so I'll say "no". Much of higher mathematics doesn't even concern itself with numbers.
Hey g4143 and welcome to the forums.
You do actually raise indirectly an interesting point about the duality between multiplication and addition, which is seen the logarithmic function where log(xy) = log(x) + log(y) so in this sense, there is actually a bridge to relate multiplication purely to addition.
However with this said, even if you are able to eventually reduce all log(xy) down to summations where x and y are real numbers (or even complex ones), you might want to think about exponentiation: log(x^y) = y*log(x) and although you could do a double log to convert to an addition, this raises an issues of how to deal with this.
I am basically being generous in the above assumption and am assuming you have a way to eventually break any product of real numbers into an addition which you can calculate quantities without resorting to addition, and basically I think this is going to involve a lot of abstract symbolic relations between things to get around the fact that calculating explicitly the logarithm requires a power series that has powers (i.e. repeated multiplication) if you want to do a specific finite computation.
The analogue that I am making can be seen with what people do with the sine and cosine functions: we know for example some exact expressions for sine and cosine (like for various fractions of pi) and we can use all the various sine and cosine expressions to get exact expressions for a particular quantity without needing to actually use the power series and I think if you wanted to do the above, you would need to do the exact same thing but for logarithms.
Interesting enoughly, there is a direct connection with exponentiation and the trig functions, so there is actually a bridge that you can build and see if you can cross.
Problem is though, that we don't really have a way currently that is known of, to just get the exact value for the sine of any number on the real line in exact form, so if you want to do your thing, you need to solve this problem first (I still think it's actually possible to do this, but it's going to be rather difficult).
But this would only look at being able to express arithmetic operations in terms of only two operations and a lot of this would require a completely complex symbolic framework.
The other thing you need to look at, as hinted above by Number Nine, is the abstract stuff that does not necessarily correlate to arithmetic in a direct way like sets and the rest of the other abstract things built on general sets.
Sets have a duality too but it deals with intersections and unions, and sets don't work like numbers: they are a completely different way of looking at things because sets do not have rank or order like numbers do. You can introduce relations and things on various sets but it is by no means required to do so.
So since you have these sets that do not have any notion of rank, or comparison in the arithmetic sense, you can't really apply that kind of thinking to all the stuff based on sets which is pretty much all of mathematics at some level.
I suggest you take a look at the set framework and the two operations of intersections and unions to get a feel of the ground-work for modern abstract mathematics and then consider that in light of your question.
jimgram
Consider that in machine language (I.E. hexadecimal or binary) only addition and subtraction are possible. So at the root of all computer programs, the initial machine only adds and subtracts
Consider that in machine language (I.E. hexadecimal or binary) only addition and subtraction are possible. So at the root of all computer programs, the initial machine only adds and subtracts
That is an interesting point.
In mathematics though, the variability and nature of information is a little bit more subtle: we aren't dealing with things that have something that corresponds to a numeric quantity.
You also have to consider the nature of the limits of what you can do with addition and subtraction: these operations are built on more fundamental and flexible ones that are the basis for boolean algebra on binary numbers.
If you look at an adder circuit, it is derived from using gates that implement AND and XOR routines for the most basic adder (and an extra term if you are using a full adder as opposed to a half-adder).
If you wanted to, you could write your own adder if only you had the ability to perform the basic binary operations (AND and OR).
The real absolute fundamental instruction of anything that is either binary based or is equivalent in some form to a the standard binary based computer are the logical operators, and you can derive every single instruction (and indeed this is actually done in practice) by using these things in various combinations.
Staff Emeritus
Homework Helper
Find me the area of a circle using basic addition.
Find me $\pi^2$ using basic addition.
Define $\sin(x)$ and $\log(x)$ using basic addition.
Define composition of functions using basic addition.
Describe the union of sets using basic addition.
If you can give a satisfactory answer to the previous five simple problems, then you might have a point.
jimgram
Add $\pi$ to itself $\pi$ times.
Microprocessors consist of registers (a series of flip-flops) and memory registers. Parts of the circuits can be dedicated to (using addition) multiplication, division, squares and roots. Along with tables, other parts are dedicated to complex functions (using the aforementioned sections for addition, multiplication, etc.). Taken alone, a register can only increment or decrement - hence the need to do it extremely fast
g4143
Find me $\pi^2$ using basic addition.
Can you post a solution that solves the above equation so that the answer is in the form (3.141592653589793238462643383279502884197169399....)^2 accurately by any method?
g4143
There is a series expansion formula for this g4143: just slightly adjust the basel problem.
http://en.wikipedia.org/wiki/Basel_problem
My rudimentary math skills are probably apparent by now but can you represent pi^2 as a number in the form 9.86960440109... accurately?
Add $\pi$ to itself $\pi$ times.
What does "∏ times" mean?
OP, do you know what a computable number is?
Last edited:
Staff Emeritus
Homework Helper
Add $\pi$ to itself $\pi$ times.
Please define what it means to add something $\pi$ times.
Last edited:
Staff Emeritus
Homework Helper
Can you post a solution that solves the above equation so that the answer is in the form (3.141592653589793238462643383279502884197169399....)^2 accurately by any method?
I wanted you to define $\pi^2$ using just basic addition. I don't want you to actually find the number. Just define it.
And please don't come with silliness such as adding things $\pi$ times, without defining what that means.
jimgram
If 'x' = 3, then you could find 3^2 by adding x to itself x times (3+3+3=9)
Staff Emeritus
Homework Helper
If 'x' = 3, then you could find 3^2 by adding x to itself x times (3+3+3=9)
Sure, and what if $x=\pi$? It doesn't seem to work well then, does it?
jimgram
There's always a level of accuracy (e.g. number of decimal points) that must be accepted. What's you point?
Staff Emeritus
Homework Helper
There's always a level of accuracy (e.g. number of decimal points) that must be accepted. What's you point?
Not in math. Everything is precise and accurate in math.
Number Nine
More: Categorize, up to homeomorphism, all topological 2-manifolds using only addition.
lamball1
∏^2 = 3∏ + (∏/10) + (4∏/100) + (∏/1000).......
forever onwards
Homework Helper
Okay, if you don't like $\pi$, how would you add 1.5 to itself 1.5 times? micromass's question has nothing to do with "level of accuracy".
Mentor
Consider that in machine language (I.E. hexadecimal or binary) only addition and subtraction are possible. So at the root of all computer programs, the initial machine only adds and subtracts
A bit late, but a couple of other members have dredged this thread up.
Many processors have MUL and DIV operators as part of their instruction sets, so machine code produced by these processors is not limited to only addition and subtraction. In any case, there really is no such thing as "machine language" per se. Each processor type produces its own sort of machine language.
Homework Helper
Gold Member
Many processors have MUL and DIV operators as part of their instruction sets, so machine code produced by these processors is not limited to only addition and subtraction.
Even processors that don't have these will at least have right-shift and left-shift instructions. (i.e. multiply/divide by powers of 2), from which you can somewhat painfully build functions to multiply/divide by arbitrary numbers.
Homework Helper
A bit late, but a couple of other members have dredged this thread up.
Many processors have MUL and DIV operators as part of their instruction sets, so machine code produced by these processors is not limited to only addition and subtraction. In any case, there really is no such thing as "machine language" per se. Each processor type produces its own sort of machine language.
Since the internal coding of numbers in any computer processor only supports a finite subset of the rational numbers, what a processor can do is irrelevant to the mathematical concept of multiplication.
jimgram
Okay, if you don't like $\pi$, how would you add 1.5 to itself 1.5 times? micromass's question has nothing to do with "level of accuracy".
Machines have no particular use of decimal points. You register '15' (1111 binary or 0f hex) 15 times then shift the decimal twice.
Staff Emeritus
Homework Helper
∏^2 = 3∏ + (∏/10) + (4∏/100) + (∏/1000).......
forever onwards
Then you are making use of limits and not only of addition. This is my point.
Mentor
Since the internal coding of numbers in any computer processor only supports a finite subset of the rational numbers, what a processor can do is irrelevant to the mathematical concept of multiplication.
I don't understand your point, unless possibly you were talking about real number multiplication, which I wasn't. The operations I mentioned, MUL and DIV, are Intel X86 opcodes that take integer operands. Floating point arithmetic operations usually use different opcodes, assuming they are actually supported by the processor.
My comment was in rebuttal to what jimgram said, which was that in machine language, only addition and subtraction were possible.
the_wolfman
I think a professional mathematician would say the answer is no. (I'm an engineer).
In math there are often two basic operations: addition and multiplication. From these operations we can define other operations like subtraction, division, exponents, etc.
Now if you consider the real numbers its tempting to say that all operations can be defined from addition. This is because we can define multiplication in terms of addition. A*B is A added to itself B times.
But this is not always so. Consider 2x2 matrices. Can you define matrix multiplication in terms of matrix addition?
Its true that we can define matrix multiplication and addition in terms of real number operations, but when you consider more abstract groups this is not always the case.
There are also groups with a well defined multiplication, but no addition operation. For example Braid groups fall into this category (i think).
lamball1
Then you are making use of limits and not only of addition. This is my point.
But then this also counts for adding pi together? Of course all irrational numbers cannot be completely calculated, but this does not present an argument against the idea that all other operations can be based on simple addition and subtraction.
Number Nine
But then this also counts for adding pi together? Of course all irrational numbers cannot be completely calculated, but this does not present an argument against the idea that all other operations can be based on simple addition and subtraction.
Define function composition using addition. Keep in mind that you'll have to define the notion of a binary operation, as well as the notion of a function between sets, using only addition.
| 2,791 | 12,943 |
{"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}
| 3.578125 | 4 |
CC-MAIN-2022-33
|
latest
|
en
| 0.967114 |
https://www.khanacademy.org/math/cc-sixth-grade-math/x0267d782:cc-6th-rates-and-percentages/cc-6th-percent-word-problems/v/percent-word-problems
| 1,701,558,603,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100452.79/warc/CC-MAIN-20231202203800-20231202233800-00176.warc.gz
| 950,733,667 | 115,260 |
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
Lesson 6: Percent word problems
# Percent word problem: recycling cans
In this math lesson, we learn how to calculate the percentage of recycled cans. Given that 13 out of 20 cans are recycled, we divide 13 by 20 to get 0.65 as a decimal. Multiplying by 100, we find that 65% of cans are recycled, promoting eco-friendly practices and waste reduction. Created by Sal Khan.
## Want to join the conversation?
• this is so hard i still dont under stand this and btw im only in 6th grade and im dumb i vant do it
• Hi!
First of all, you are super smart! Don't be mad at yourself - it will only make you feel worse! Believe that you CAN do it!
I would recommend trying to do this problem this way:
Ok, so let's make the fraction bigger. Twenty times five is 100, but since we have to multiply both parts of the fraction equally, then we multiply 13 by 5 also. This gives us a fraction of 65 over 100, or 65%. Does this make sense?
• Instead of moving the decimal point 2 places to the right, we could have ALSO made 0.65 as 65/100 and since 100 is the same thing as PERCENT, we could write it as 65% too...
• ORRRRRRR you can make it even easier and just do
13/20 since 13 is out of the whole number 20.
I would recommend you write this setup down to make it easier.
so as I said 13/20 because 13 is a part of the whole number which is 20.
soooooo, after that set it up in fraction form.
Let me backtrack a little. so what you're going to do is you're going to set up 13/20 in a fraction form. on the left side before the equal sign. After the = sign, you would put ?/100. why did I choose over a hundred well because that Will be your variable and 100 is a total number that you're going to take from. after that you're going to cross multiply or as you would call the butterfly effect so what you're going to do is going to multiply 13 * 100 fraction that is after equal signs of 13 * 100 and that will give you 1300. Ater that your going to * 20 which is the denominator of 13. the whole number 1300* the variable or? XYZ whatever it doesn't really matter then you're going to do 13* 100 which is 1300. then you multiply 20 by the variable in this case X. after you multiply 20*X, that should give you 20x. so the setup now should be 1300 is equal to or = to 20x.
Since you multiplied now do the opposite, divide. whatever sign you use first do the oppisite of that next. So if you add then subtract, if you times then divide, and vise versa!
now you have to divide with whatever the denominator number is in this case the number would be 20 you have to divide both sides by the denominator which is 20. equal number so that would be 20 so you would divide 1300 divided by 20 that should give you 65. Then 20X / 20, it would be 1 or in this case 1X or just X.
Hopefully, this helped everyone. The butterfly effect is very easy or in other words, cross multiplication.
• it not clear on how you get percent
• multiply the number (from which percentage is going to be find) with 100 and divide the number. like there are 13 cans out of 20 then it will be like this= 13*100/20
=13*10/2
= 13*5
• Wouldn't you be able to just do 13 divided by 2 over 10 and get 65%?
• 13/2 over 10 = 0.65
To get to a percent, you must then multiply by 100
0.65 * 100 = 65%
• When you divide by 10 or 100 etc, you move the digits. So the place value is changing, not the decimal point.
• Hi!
First of all, you are super smart! Don't be mad at yourself - it will only make you feel worse! Believe that you CAN do it!
I would recommend trying to do this problem this way:
Ok, so let's make the fraction bigger. Twenty times five is 100, but since we have to multiply both parts of the fraction equally, then we multiply 13 by 5 also. This gives us a fraction of 65 over 100, or 65%. Does this make sense?
• At I think Sal should have said forget the numerator
• He actually said, "to get the denominator from 20 to 100 you would multiply by 5" so he is correct, the person who transcribed the video just made a little mistake.
• Mr sal can u answer my question his is hard
| 1,117 | 4,272 |
{"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}
| 4.46875 | 4 |
CC-MAIN-2023-50
|
latest
|
en
| 0.95834 |
https://www.traditionaloven.com/tutorials/angle/convert-diameter-part-unit-to-right-angle-90-degree.html
| 1,709,413,038,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00485.warc.gz
| 1,021,876,499 | 17,555 |
Convert Ø dia- part to ∠ , ∠ | diameter part to right angles
# angle units conversion
## Amount: 1 diameter part (Ø dia- part) of angle Equals: 0.011 right angles (∠ , ∠) in angle
Converting diameter part to right angles value in the angle units scale.
TOGGLE : from right angles into diameter parts in the other way around.
## angle from diameter part to right angle conversion results
### Enter a new diameter part number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other angle measuring units - complete list.
How many right angles are in 1 diameter part? The answer is: 1 Ø dia- part equals 0.011 ∠ , ∠
## 0.011 ∠ , ∠ is converted to 1 of what?
The right angles unit number 0.011 ∠ , ∠ converts to 1 Ø dia- part, one diameter part. It is the EQUAL angle value of 1 diameter part but in the right angles angle unit alternative.
Ø dia- part/∠ , ∠ angle conversion result From Symbol Equals Result Symbol 1 Ø dia- part = 0.011 ∠ , ∠
## Conversion chart - diameter parts to right angles
1 diameter part to right angles = 0.011 ∠ , ∠
2 diameter parts to right angles = 0.021 ∠ , ∠
3 diameter parts to right angles = 0.032 ∠ , ∠
4 diameter parts to right angles = 0.042 ∠ , ∠
5 diameter parts to right angles = 0.053 ∠ , ∠
6 diameter parts to right angles = 0.064 ∠ , ∠
7 diameter parts to right angles = 0.074 ∠ , ∠
8 diameter parts to right angles = 0.085 ∠ , ∠
9 diameter parts to right angles = 0.095 ∠ , ∠
10 diameter parts to right angles = 0.11 ∠ , ∠
11 diameter parts to right angles = 0.12 ∠ , ∠
12 diameter parts to right angles = 0.13 ∠ , ∠
13 diameter parts to right angles = 0.14 ∠ , ∠
14 diameter parts to right angles = 0.15 ∠ , ∠
15 diameter parts to right angles = 0.16 ∠ , ∠
Convert angle of diameter part (Ø dia- part) and right angles (∠ , ∠) units in reverse from right angles into diameter parts.
## Angles
This calculator is based on conversion of two angle units. An angle consists of two rays (as in sides of an angle sharing a common vertex or else called the endpoint.) Some belong to rotation measurements - spherical angles measured by arcs' lengths, pointing from the center, plus the radius. For a whole set of multiple units of angle on one page, try that Multiunit converter tool which has built in all angle unit-variations. Page with individual angle units.
# Converter type: angle units
First unit: diameter part (Ø dia- part) is used for measuring angle.
Second: right angle (∠ , ∠) is unit of angle.
QUESTION:
15 Ø dia- part = ? ∠ , ∠
15 Ø dia- part = 0.16 ∠ , ∠
Abbreviation, or prefix, for diameter part is:
Ø dia- part
Abbreviation for right angle is:
∠ , ∠
## Other applications for this angle calculator ...
With the above mentioned two-units calculating service it provides, this angle converter proved to be useful also as a teaching tool:
1. in practicing diameter parts and right angles ( Ø dia- part vs. ∠ , ∠ ) measures exchange.
2. for conversion factors between unit pairs.
3. work with angle's values and properties.
To link to this angle diameter part to right angles online converter simply cut and paste the following.
The link to this tool will appear as: angle from diameter part (Ø dia- part) to right angles (∠ , ∠) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.
| 939 | 3,480 |
{"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}
| 3.578125 | 4 |
CC-MAIN-2024-10
|
latest
|
en
| 0.696847 |
https://www.statlect.com/glossary/critical-value
| 1,726,722,091,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00641.warc.gz
| 906,121,930 | 14,002 |
# Critical value
In a test of hypothesis, a critical value is a number that separates two regions:
1. the critical region, that is, the set of values of the test statistic that lead to a rejection of the null hypothesis;
2. the acceptance region, that is, the set of values for which the null is not rejected.
## Notation
In what follows we are going to use the following notation:
• is the test statistic (e.g., a z-statistic);
• is the critical region;
• is the acceptance region.
Thus, the null hypothesis is rejected ifand it is not rejected if
## Definition
Here is a formal definition.
Definition A critical value is a boundary of the acceptance region .
Let us make an example.
Example If the acceptance region is the intervalthen the critical region isThe critical values are the two boundariesIf, for example, the value of the test statistic is , then belongs to the acceptance region and the null hypothesis is not rejected.
## One-tailed tests
A test is called one-tailed if there is only one critical value .
There are two cases:
1. left-tailed test: the null is rejected only if
2. right-tailed test: the null is rejected only if
The following table summarizes the two cases by using the symbols introduced above.
### How is the critical value determined in left-tailed tests?
Typically, is chosen so as to a achieve a desired size of the test.
Remember that the size is the probability of rejecting the null hypothesis when it is true. Denote it by .
For a left-tailed test, we have
In the majority of practical cases, the test statistic is a continuous random variable. As a consequence, the probability that it takes any specific value is equal to zero. In particular,
Thus, we can writewhere is the cumulative distribution function (cdf) of the test statistic.
All we need to do in order to determine the critical value is to find a that solves the equation
We explain below how to solve it.
### How is the critical value determined in right-tailed tests?
Things are similar for right-tailed tests. In these tests, we have
Thus, we need to solve the equation
### How are the equations solved?
For the most common distributions such as the normal distribution and the t distribution, the equationsand have no analytical solution.
The reason is that the inverse of the cdf is not known in closed form.
So, for example, we cannot compute analytically the solution of the first equation as
However, virtually any calculator or statistical software has pre-built functions that allow us to easily solve these equations numerically.
The (old-fashioned) alternative is to look up the critical value in special tables called statistical tables. See this lecture if you want to know more about these alternatives.
### Example
Let us make an example of a left-tailed test.
Suppose that the size of the test is , which means that we are happy with a 5% probability of incorrectly rejecting the null when it is true.
Suppose that our test statistic has a standard normal distribution.
Then, we need to findwhere is the cumulative distribution function of a standard normal distribution.
We are going to use a free web app called Wolfram Alpha to find .
Here is the result.
Thus, the critical value is .
We are going to reject the null hypothesis if is less than .
## Two-tailed tests
A test is called two-tailed if there are two critical values and and the null hypothesis is rejected only if
We assume without loss of generality that .
Thus, we can add a new line to the table shown in the previous section:
### How do you find the two critical values?
As in the case of a one-tailed test (see above), also in the two-tailed case the critical values are chosen so as to achieve a pre-defined size of the test.
The size can be computed as follows:
By making again the assumption that the test statistic is a continuous random variable, we obtainwhere is the distribution function of .
### How is the equation solved?
Our problem is to solve one equation in two unknowns ( and ).
There are potentially infinite solutions to the problem because one can choose one of the two critical values at will and choose the remaining one so as to solve the equation.
There is no general rule for choosing one specific solution.
Some possibilities are to:
We do not discuss these possibilities here, but we refer the reader to Berger and Casella (2002).
We instead discuss the case in which the test statistic has a symmetric distribution. This is the most relevant case in practice because in many tests has a normal or a Student's t distribution and both of these distributions are symmetric.
### Solution for symmetric distributions (practically relevant)
A distribution is symmetric (around zero) whenfor any number .
We can exploit this fact by making the additional assumption that the two critical values are opposite:
Without loss of generality, we can assumewith .
It follows that the size of the test can be written asand the equation to solve becomes
This is an equation in one unknown () that can be solved using the methods (numeric inversion, tables, etc.) discussed in the previous section on one-tailed tests.
## Summary
Everything we have said thus far is summarized by the following table.
## More details
If you want to read a more detailed exposition of the concept of critical value and of related concepts, go to the lecture entitled Hypothesis testing.
## References
Berger, R. L. and G. Casella (2002) "Statistical inference", Duxbury Advanced Series.
## Keep reading the glossary
Previous entry: Covariance stationary
Next entry: Cross-covariance matrix
| 1,171 | 5,646 |
{"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}
| 4.3125 | 4 |
CC-MAIN-2024-38
|
latest
|
en
| 0.902811 |
https://gmatclub.com/forum/it-is-believed-that-military-strategists-in-the-medieval-150674.html?fl=similar
| 1,495,626,986,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463607813.12/warc/CC-MAIN-20170524112717-20170524132717-00408.warc.gz
| 770,735,305 | 62,492 |
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 24 May 2017, 04:56
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# It is believed that military strategists in the medieval
Author Message
TAGS:
### Hide Tags
Moderator
Joined: 01 Sep 2010
Posts: 3166
Followers: 855
Kudos [?]: 7300 [0], given: 1063
It is believed that military strategists in the medieval [#permalink]
### Show Tags
10 Apr 2013, 14:22
2
This post was
BOOKMARKED
00:00
Difficulty:
5% (low)
Question Stats:
80% (01:00) correct 20% (01:09) wrong based on 105 sessions
### HideShow timer Statistics
It is believed that military strategists in the medieval ages used manned kites surveying enemy troops, so anticipating techniques of modern aerial surveillance.
(A) used manned kites surveying enemy troops, so anticipating techniques of
(B) were using manned kites to survey enemy troops anticipating techniques of
(C) had used manned kites, surveyed enemy troops, and anticipated techniques of
(D) used manned kites to survey enemy troops, a technique anticipating
(E) using manned kites to survey enemy troops, were anticipating techniques of
[Reveal] Spoiler: OA
_________________
If you have any questions
New!
Verbal Forum Moderator
Joined: 16 Jun 2012
Posts: 1132
Location: United States
Followers: 278
Kudos [?]: 3107 [1] , given: 123
Re: It is believed that military strategists in the medieval age [#permalink]
### Show Tags
10 Apr 2013, 16:34
1
KUDOS
It is believed that military strategists in the medieval ages used manned kites surveying enemy troops, so anticipating techniques of modern aerial surveillance.
(A) used manned kites surveying enemy troops, so anticipating techniques of >>> Wrong. "so anticipating" is wrong modifier, hence, coveys wrongmeaning. The strategists did not anticipate techniques. Must be "the techniques used to anticipate something".
(B) were using manned kites to survey enemy troops anticipating techniques of >>> Wrong. Because "anticipating" modifies enemy troops. that's wrong.
(C) had used manned kites, surveyed enemy troops, and anticipated techniques >>> Wrong. Not parallel
(D) used manned kites to survey enemy troops, a technique anticipating >>> Correct. Good structure and correct modifier
(E) using manned kites to survey enemy troops, were anticipating techniques of >>> Wrong. Not parallel.
So, I vote for D.
_________________
Please +1 KUDO if my post helps. Thank you.
"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."
Chris Bangle - Former BMW Chief of Design.
Moderator
Joined: 01 Sep 2010
Posts: 3166
Followers: 855
Kudos [?]: 7300 [0], given: 1063
Re: It is believed that military strategists in the medieval [#permalink]
### Show Tags
11 Apr 2013, 02:42
Concepts Tested – Meaning, Verb Tense
The sentence, in its original form, makes no sense. It appears as though the kites
are doing the surveying on their own. Also the last part of the sentence after the
comma (so anticipating....) makes no logical sense with the rest of the sentence.
So eliminate A
B - Unnecessarily uses the past continuous tense. Also distorts the meaning by
making it appear as though the enemy troops were anticipating something.
C - Makes ‘used’, ‘surveyed’, and ‘anticipated’ parallel but this is not what the
sentence is implying. The manned kites were actually used to do the survey.
Between D and E, E again distorts the meaning by suggesting that the military
strategists were anticipating something. ‘Anticipating’ in the original sentence
implies ‘preceding’ or ‘was a forerunner of’. D gets this meaning across correctly
_________________
Intern
Joined: 06 Mar 2012
Posts: 35
Location: India
GPA: 3.4
Followers: 1
Kudos [?]: 57 [0], given: 12
Re: It is believed that military strategists in the medieval age [#permalink]
### Show Tags
13 Apr 2013, 04:31
pqhai wrote:
It is believed that military strategists in the medieval ages used manned kites surveying enemy troops, so anticipating techniques of modern aerial surveillance.
(A) used manned kites surveying enemy troops, so anticipating techniques of >>> Wrong. "so anticipating" is wrong modifier, hence, coveys wrongmeaning. The strategists did not anticipate techniques. Must be "the techniques used to anticipate something".
(B) were using manned kites to survey enemy troops anticipating techniques of >>> Wrong. Because "anticipating" modifies enemy troops. that's wrong.
(C) had used manned kites, surveyed enemy troops, and anticipated techniques >>> Wrong. Not parallel
(D) used manned kites to survey enemy troops, a technique anticipating >>> Correct. Good structure and correct modifier
(E) using manned kites to survey enemy troops, were anticipating techniques of >>> Wrong. Not parallel.
So, I vote for D.
@carcass - Although I picked D, I had a doubt - a technique after the comma is modifying the entire phrase before the comma or just the enemy troops. Can you please explain.
Senior Manager
Joined: 23 Mar 2011
Posts: 463
Location: India
GPA: 2.5
WE: Operations (Hospitality and Tourism)
Followers: 23
Kudos [?]: 259 [0], given: 59
Re: It is believed that military strategists in the medieval [#permalink]
### Show Tags
13 Apr 2013, 06:51
The sentence has a clear split of verb tense. A simple past "used" and modifier "a technique..." is best in Option D
_________________
"When the going gets tough, the tough gets going!"
Bring ON SOME KUDOS MATES+++
-----------------------------
My GMAT journey begins: http://gmatclub.com/forum/my-gmat-journey-begins-122251.html
GMAT Club Legend
Joined: 01 Oct 2013
Posts: 10371
Followers: 997
Kudos [?]: 224 [0], given: 0
Re: It is believed that military strategists in the medieval [#permalink]
### Show Tags
01 May 2017, 22:33
Hello from the GMAT Club VerbalBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: It is believed that military strategists in the medieval [#permalink] 01 May 2017, 22:33
Similar topics Replies Last post
Similar
Topics:
9 In medieval times, Libya was the principal corridor 10 22 Jul 2014, 23:57
3 What seemed as remarkable as the medieval invention of 9 10 May 2013, 03:53
65 Medieval monks, through their manuscript illuminations, 25 23 Apr 2017, 03:23
13 The combat code of medieval knights resembled that of 11 23 Dec 2015, 18:22
7 It is believed that military strategists in the medieval 5 15 Sep 2015, 22:55
Display posts from previous: Sort by
| 1,849 | 7,482 |
{"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}
| 3.703125 | 4 |
CC-MAIN-2017-22
|
latest
|
en
| 0.87682 |
https://gmatclub.com/forum/the-table-shows-partial-results-of-a-survey-in-which-consumers-were-as-268909.html?sort_by_oldest=true
| 1,579,427,456,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579250594391.21/warc/CC-MAIN-20200119093733-20200119121733-00091.warc.gz
| 473,471,628 | 157,635 |
GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video
It is currently 19 Jan 2020, 02:49
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# The table shows partial results of a survey in which consumers were as
Author Message
TAGS:
### Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 60481
The table shows partial results of a survey in which consumers were as [#permalink]
### Show Tags
25 Jun 2018, 05:11
37
00:00
Difficulty:
45% (medium)
Question Stats:
66% (01:24) correct 34% (01:26) wrong based on 1394 sessions
### HideShow timer Statistics
The table shows partial results of a survey in which consumers were asked to indicate which one of six promotional techniques most influenced their decision to buy a new food product. Of those consumers who indicated one of the four techniques listed, what fraction indicated either coupons or store displays?
A. 2/7
B. 1/3
C. 2/5
D. 4/9
E. 1/2
NEW question from GMAT® Official Guide 2019
(PS10307)
Attachment:
PS10307_f011.jpg [ 31.44 KiB | Viewed 14410 times ]
_________________
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3209
Re: The table shows partial results of a survey in which consumers were as [#permalink]
### Show Tags
25 Jun 2018, 05:19
Solution
Given:
• The table shows partial results of a survey in which consumers were asked to indicate which one of six promotional techniques most influenced their decision to buy a new food product
To find:
• Out of those consumers who indicated one of the four techniques listed, what fraction indicated either coupons or store displays
Approach and Working:
If we assume the total number of consumers who responded is 100, then we can write:
• Total consumers responded to the four mentioned techniques = (35 + 22 + 18 + 15) = 90
• Total consumers responded to either coupons or store displays = (22 + 18) = 40
• Therefore, the required fraction = $$\frac{40}{90} = \frac{4}{9}$$
Hence, the correct answer is option D.
_________________
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4214
Re: The table shows partial results of a survey in which consumers were as [#permalink]
### Show Tags
25 Jun 2018, 08:14
2
Top Contributor
2
Bunuel wrote:
The table shows partial results of a survey in which consumers were asked to indicate which one of six promotional techniques most influenced their decision to buy a new food product. Of those consumers who indicated one of the four techniques listed, what fraction indicated either coupons or store displays?
A. 2/7
B. 1/3
C. 2/5
D. 4/9
E. 1/2
Attachment:
PS10307_f011.jpg
Since we aren't told the number of people involved in the survey, let's make things easier on ourselves and assign a "nice" value to the number of people involved in the survey.
Let's say there were 100 people in the survey.
When we apply this to our table, we see that 35 people chose TV ads, 22 people chose coupons, 18 people chose store displays, and 15 people chose samples.
35 + 22 + 18 + 15 = 90
So, the table shows the results for 90 consumers
What fraction indicated either coupons or store displays?
Among the 90 consumers surveyed, 22 chose coupons, and 18 chose store displays
22 + 18 = 40
Fraction = 40/90
= 4/9
Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2806
Re: The table shows partial results of a survey in which consumers were as [#permalink]
### Show Tags
26 Jun 2018, 18:01
Bunuel wrote:
The table shows partial results of a survey in which consumers were asked to indicate which one of six promotional techniques most influenced their decision to buy a new food product. Of those consumers who indicated one of the four techniques listed, what fraction indicated either coupons or store displays?
A. 2/7
B. 1/3
C. 2/5
D. 4/9
E. 1/2
Notice that you are instructed to use the incomplete table; the percentages add up to only 90%. So first, you will add the percentages for “Coupons” and “Store Displays,” but then you must divide by the total of 90% to get the final answer, as shown here:
(22 + 18)/(35 + 22 + 18 + 15) = 40/90 = 4/9
_________________
# Jeffrey Miller
[email protected]
181 Reviews
5-star rated online GMAT quant
self study course
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.
Director
Joined: 09 Mar 2018
Posts: 991
Location: India
Re: The table shows partial results of a survey in which consumers were as [#permalink]
### Show Tags
15 Feb 2019, 00:49
1
Bunuel wrote:
The table shows partial results of a survey in which consumers were asked to indicate which one of six promotional techniques most influenced their decision to buy a new food product. Of those consumers who indicated one of the four techniques listed, what fraction indicated either coupons or store displays?
A. 2/7
B. 1/3
C. 2/5
D. 4/9
E. 1/2
When we add all the %'s we get 90 as total value
either coupons or store displays = 40/90
D
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.
Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Senior Manager
Joined: 12 Sep 2017
Posts: 316
Re: The table shows partial results of a survey in which consumers were as [#permalink]
### Show Tags
30 Jun 2019, 19:47
Why we should we add the other 10%?
Manager
Joined: 07 Sep 2019
Posts: 116
Location: Viet Nam
Concentration: Marketing, Strategy
WE: Brand Management (Consumer Products)
Re: The table shows partial results of a survey in which consumers were as [#permalink]
### Show Tags
03 Dec 2019, 15:00
Could anyone help explain why the stem says "or" "what fraction indicated either coupons or store displays" then we have a sum (22+18)? I think it should be 22/90 or 18/90.
(22+18)/90 --> stem should be "what fraction indicated coupons and store displays"
_________________
--- One day you will thank yourself for not giving up ---
Intern
Joined: 20 Mar 2013
Posts: 19
GMAT 1: 570 Q46 V23
GMAT 2: 620 Q47 V28
The table shows partial results of a survey in which consumers were as [#permalink]
### Show Tags
23 Dec 2019, 01:00
This question tries to trick you ! Otherwise the maths is pretty simple
You should solve 40/90 and not 40/100
The table shows partial results of a survey in which consumers were as [#permalink] 23 Dec 2019, 01:00
Display posts from previous: Sort by
| 1,848 | 7,079 |
{"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}
| 4.0625 | 4 |
CC-MAIN-2020-05
|
latest
|
en
| 0.915528 |
https://socratic.org/questions/how-do-you-simplify-6-3i-5-4i
| 1,719,199,882,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00841.warc.gz
| 469,578,211 | 5,839 |
# How do you simplify (6-3i)+(5+4i)?
Oct 5, 2016
$11 + i$
#### Explanation:
Summing complex numbers is a piece of cake!
Every complex number has the form $z = \textcolor{red}{a} + \textcolor{b l u e}{b} i$, where $a$ and $b$ are real numbers (the real part and the imaginary part of $z$) and $i$ is the imaginary unit.
To sum two complex numbers, you simply need to sum the real and complex numbers together! So:
(color(red)6-color(blue)3i)+(color(red)5+color(blue)4i) =color(red)6+color(red)5-color(blue)3i+color(blue)4i =color(red)11+(color(blue)(-3+4))i =color(red)11+i
| 195 | 579 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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}
| 4.46875 | 4 |
CC-MAIN-2024-26
|
latest
|
en
| 0.722982 |
https://bridgewinners.com/article/view/should-district-9-have-resigned/?cj=528636
| 1,585,975,545,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585370520039.50/warc/CC-MAIN-20200404042338-20200404072338-00425.warc.gz
| 400,665,866 | 21,002 |
Should District 9 have resigned?
In the finals of the GNT, District 9 (D9) was down to District 21 (D21) by 77 imps after 45 boards, with 15 boards to play. They chose to forfeit the match. There are certainly many reasons to do so, "fatigue", "team fragmentation", "hopelessness", "hey, we have won this too many times already", "time for dinner", "sponsor's whim", etc. It is in their discretion to do so and I do not want to question their decision.
But what I would like to explore is a very simplified model of their likelihood of winning. I am going to use a basic statistical model, looking at the results of the first 45 boards to determine a standard deviation, then periodizing it for the remaining 15 boards to determine their likelihood of recovering 78 imps using the easily criticized assumptions that the distribution of imps can be represented by a normal distribution around a mean and that the volatility of the first 45 boards is representative of the volatility for the remaining 15. It isn't difficult to challenge these assumptions in the abstract, and given that Meckwell with what may be the greatest record for comebacks in the history of the game plays for D9, they have additional problems of credibility in their application in this case.
But you need to start somewhere and the results may prove to be instructional. I have used the board scores in the BBO archives. (There is a one imp discrepancy between the totals and the deal by deal scores. I used the deal scores.) Over 45 boards, D21 averaged +1.73 imps per board. To compute the standard deviation, a measure of volatility, one sums the squares of the difference between the board score and the mean, divides by the number of boards, and takes the square root of the result. I computed a standard deviation of 7.86. (That is 2783/45=61.84, the square root of which is 7.86.)
Traditional statistical analysis would say that 68% of the occurrences should be within one standard deviation of the mean, 95% within two standard deviations, 99% within three. For example, the model would predict that 68% of the time, the score on a board will be between +9.59 (1.73+7.86) and -6.13 (1.73-7.86) imps in favor of D21.
To determine what that means over the 15 remaining boards, one periodizes the result by multiplying the standard deviation by the square root of the number of periods. 7.86 times the square root of 15 equals 30.4. D9 had to win back 78 imps; two standard deviations (95% of the time) is 60.8, three standard deviations (99%) is 91.2. But this is the probability of both tails and we want to look at just one, so the result is halved. In short, the model predicts that approximately 99% of the time, had they played on, D21 would have prevailed.
| 661 | 2,749 |
{"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}
| 3.671875 | 4 |
CC-MAIN-2020-16
|
latest
|
en
| 0.960641 |
https://www.statistics-lab.com/%E6%95%B0%E5%AD%A6%E4%BB%A3%E5%86%99matlab%E4%BB%BF%E7%9C%9F%E4%BB%A3%E5%86%99simulation%E4%BB%A3%E5%81%9Atype-of-dyads/
| 1,702,026,165,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100739.50/warc/CC-MAIN-20231208081124-20231208111124-00498.warc.gz
| 1,084,802,826 | 40,337 |
MATLAB是一个编程和数值计算平台，被数百万工程师和科学家用来分析数据、开发算法和创建模型。
MATLAB主要用于数值运算，但利用为数众多的附加工具箱，它也适合不同领域的应用，例如控制系统设计与分析、影像处理、深度学习、信号处理与通讯、金融建模和分析等。另外还有配套软件包提供可视化开发环境，常用于系统模拟、动态嵌入式系统开发等方面。
statistics-lab™ 为您的留学生涯保驾护航 在代写matlab方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写matlab代写方面经验极为丰富，各种代写matlab相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
The dyad (binary group) is a fundamental kinematic chain with two links $(n=2)$ and three one degree of freedom joints $\left(c_{5}=3\right.$ ). Figure $1.7$ depicts different types of dyads: rotation rotation rotation (dyad RRR) or dyad of type one, Fig. 1.7a; rotation rotation translation (dyad RRT) or dyad of type two, Fig.1.7b; rotation translation rotation (dyad RTR) or dyad of type three, Fig. $1.7 \mathrm{c}$; translation rotation translation (dyad TRT) or dyad of type four, Fig. $1.7 \mathrm{~d}$; rotation translation translation (dyad RTT) or dyad of type five, Fig. 1.7e. The advantage of the group classification of a mechanical system is in its simplicity. The solution of the whole mechanical system can be obtained by adding partial solutions of different fundamental kinematic chains [55-57].
The number of DOF for the mechanism in Fig. $1.8$ is $M=1$. If $M=1$, there is one driver link (one actuator). The rotational link 1 can be selected as the driver link. If the driver link is separated from the mechanism the remaining moving kinematic chain (links $2,3,4,5$ ) has the number of DOF equal to zero. The dyad is the simplest system group with two links and three joints. On the contour diagram, the links 2 and 3 form a dyad and the links 4 and 5 represent another dyad. The mechanism has been decomposed into a driver link (link 1) and two dyads (links 2 and 3, and links 4 and
5). The dyad with the links 2 and 3 is a RTR dyad and the dyad with the links 4 and 5 is also a RTR dyad. The whole mechanism can be symbolized as a R-RTR-RTR mechanism.
For planar mechanisms the two degrees of freedom joints can be substituted and mechanisms with one degree of freedom joints are obtained. The transformed mechanism has to be equivalent with the initial mechanism from a kinematical point of view. The number of degrees of freedom of the transformed mechanism has to be equal to the number of degrees of freedom of the initial mechanism. The relative motion of the links of the transformed mechanism has to be the same as the relative motion of the links of the initial mechanism.
A planar straight link with the end nodes at $A$ and $B$ is considered. The coordinates of the joint $A$ are $\left(x_{A}, y_{A}\right)$ and the coordinates of the joint $B$ are $\left(x_{B}, y_{B}\right)$. The length $A B=l_{A B}$ is constant and the following relation can be written
$$\left(x_{B}-x_{A}\right)^{2}+\left(y_{B}-y_{A}\right)^{2}=A B^{2},$$
The angle of the link $A B$ with the horizontal axis $O x$ is $\phi$ and the slope $m$ of $A B$ is
$$m=\tan \phi=\frac{y_{B}-y_{A}}{x_{B}-x_{A}} .$$
The equation of the straight link is $y=m x+n$, where $x$ and $y$ are the coordinates of any point on this link and $n$ is the intercept of $A B$ with the vertical axis $O y$.
The R-RRT (slider-crank) mechanism shown in Fig. 1.10a has the dimensions: $A B=0.5 \mathrm{~m}$ and $B C=1 \mathrm{~m}$. The driver link 1 makes an angle $\phi=\phi_{1}=60^{\circ}$ with the horizontal axis. The positions of the joints and the angles of the links with the horizontal axis will be calculated. A Cartesian reference frame $x y$ is selected. The joint $A$ is in the origin of the reference frame, $A \equiv O$,
$$x_{A}=0, y_{A}=0 .$$
The coordinates of the joint $B$ are
\begin{aligned} &x_{B}=A B \cos \phi=(0.5) \cos 60^{\circ}=0.250 \mathrm{~m} \ &y_{B}=A B \sin \phi=(0.5) \sin 60^{\circ}=0.433 \mathrm{~m} \end{aligned}
The coordinates of the joint $C$ are $x_{C}$ and $y_{C}$. The joint $C$ is located on the horizontal axis $y_{C}=0$. The length of the segment $B C$ is constant
$$\left(x_{B}-x_{C}\right)^{2}+\left(y_{B}-y_{C}\right)^{2}=B C^{2},$$
or
$$\left(0.250-x_{C}\right)^{2}+(0.433-0)^{2}=1^{2} .$$
The two solutions for $x_{C}$ are:
$$x_{C_{1}}=1.151 \mathrm{~m} \text { and } x_{C_{2}}=-0.651 \mathrm{~m}$$
To determine the position of the joint $C$, for the given angle of the link 1 , an additional condition is needed. For the first quadrant, $0 \leq \phi \leq 90^{\circ}$, one condition can be $x_{C}>$ $x_{B}$. The $x$-coordinate of the joint $C$ is $x_{C}=x_{C_{1}}=1.151 \mathrm{~m}$. The angle of the link 2 (link $B C$ ) with the horizontal is
$$\phi_{2}=\arctan \frac{y_{B}-y_{C}}{x_{B}-x_{C}} .$$
The numerical solutions for $\phi_{2}$ is $154.341$ degrees. The MATLAB code is:
clear; clc; close all;
뭉 Input data
clear; clc; close
ㅇㅎㅇ Input data
$\mathrm{AB}=0.5 ;$ 황 $(\mathrm{m}}$
$A B=0.5 ;$ 핳 $(\mathrm{m})$
## 数学代写|matlab仿真代写simulation代做|Velocity and Acceleration Analysis for Rigid Body
The motion of a rigid body $(R B)$, with respect to a fixed reference frame, is defined by the position, velocity and acceleration of all points of the rigid body. A fixed orthogonal Cartesian reference frame $x_{0} y_{0} z_{0}$ with the constant unit vectors $\mathbf{1}{0}, \mathbf{J}{0}$, and $\mathbf{k}_{0}$, is shown in Fig. 1.12. The body fixed (mobile or rotating) orthogonal Cartesian
reference frame $x y z$ with unit vectors $\mathbf{1}, \mathbf{J}$ and $\mathbf{k}$ is attached to the moving rigid body. The unit vectors $\mathbf{1}{0}, \mathbf{J}{0}$, and $\mathbf{k}{0}$ of the primary reference frame are constant with respect to time and the unit vectors $\mathbf{1}, \mathbf{J}$, and $\mathbf{k}$ are functions of time, $t$. The unit vectors $\mathbf{1}, \mathbf{J}$, and $\mathbf{k}$ of the body-fixed frame of reference rotate with the body-fixed reference frame. The origin $O$ is arbitrary. The position vector of any point $M, \forall M \in(R B)$, with respect to the fixed reference frame $x{0} y_{0} z_{0}$ is $\mathbf{r}{M}=\mathbf{r}{O_{0} M}$ and with respect to the rotating reference frame $O x y z$ is denoted by $\mathbf{r}=\mathbf{r}{O M}$. The position vector of the origin $O$ of the rotating reference frame with respect to the fixed point $O{0}$ is $\mathbf{r}{o}=\mathbf{r}{{ }{0}} o$. The position vector of $M$ is $$\mathbf{r}{M}=\mathbf{r}{O}+\mathbf{r}=\mathbf{r}{O}+x \mathbf{1}+y \mathbf{J}+z \mathbf{k}$$
where $x, y$, and $z$ represent the projections of the vector $\mathbf{r}=\mathbf{r}{O M}$ on the rotating reference frame $$\mathbf{r}=x \mathbf{1}+y \mathbf{j}+z \mathbf{k}$$ The distance between two points of the rigid body $O$ and $M$ is constant, $O \in(R B)$, and $M \in(R B)$. The components $x, y$ and $z$ of the vector $\mathbf{r}$ with respect to the rotating reference frame are constant. The unit vectors $\mathbf{1}, \mathbf{J}$, and $\mathbf{k}$ are time-dependent vector functions. For the unit vectors of an orthogonal Cartesian reference frame, 1, Jand $\mathbf{k}$, there are the following relations $$\mathbf{1} \cdot \mathbf{1}=1, \quad \mathbf{J} \cdot \mathbf{J}=1, \quad \mathbf{k} \cdot \mathbf{k}=1, \quad \mathbf{1} \cdot \mathbf{J}=0, \quad \mathbf{j} \cdot \mathbf{k}=0, \quad \mathbf{k} \cdot \mathbf{1}=0 .$$ The velocity of an arbitrary point $M$ of the rigid body with respect to the fixed reference frame $x{0} y_{0} z_{0}$, is the derivative with respect to time of the position vector $\mathbf{r}{M}$ \begin{aligned} \mathbf{v}{M} &=\frac{d \mathbf{r}{M}}{d t}=\frac{d \mathbf{r}{O_{o} M}}{d t}=\frac{d \mathbf{r}{O}}{d t}+\frac{d \mathbf{r}}{d t} \ &=\mathbf{v}{O}+x \frac{d \mathbf{1}}{d t}+y \frac{d \mathbf{j}}{d t}+z \frac{d \mathbf{k}}{d t}+\frac{d x}{d t} \mathbf{1}+\frac{d y}{d t} \mathbf{j}+\frac{d z}{d t} \mathbf{k} \end{aligned}
where $\mathbf{v}{O}=\dot{\mathbf{r}}{O}$ is the velocity of the origin of the rotating reference frame $O x y z$ with respect to the fixed reference frame. Because all the points in the rigid body maintain their relative position, their velocity relative to the rotating reference frame $x y z$ is zero, i.e., $\dot{x}=\dot{y}=\dot{z}=0$.
The velocity of point $M$ is
$$\mathbf{v}{M}=\mathbf{v}{O}+x \frac{d \mathbf{1}}{d t}+y \frac{d \mathbf{j}}{d t}+z \frac{d \mathbf{k}}{d t}=\mathbf{v}_{O}+x \mathbf{i}+y \mathbf{j}+z \dot{\mathbf{k}}$$
The derivative of the Eq. (1.11) with respect to time gives
$$\frac{d \mathbf{1}}{d t} \cdot \mathbf{1}=0, \quad \frac{d \mathbf{J}}{d t} \cdot \mathbf{J}=0, \quad \frac{d \mathbf{k}}{d t} \cdot \mathbf{k}=0$$
## matlab代写
5）。具有链接 2 和 3 的对子是 RTR 对子，具有链接 4 和 5 的对子也是 RTR 对子。整个机制可以用R-RTR-RTR机制来表示。
(X乙−X一种)2+(是乙−是一种)2=一种乙2,
X一种=0,是一种=0.
X乙=一种乙因φ=(0.5)因60∘=0.250 米 是乙=一种乙罪φ=(0.5)罪60∘=0.433 米
(X乙−XC)2+(是乙−是C)2=乙C2,
(0.250−XC)2+(0.433−0)2=12.
XC1=1.151 米 和 XC2=−0.651 米
φ2=反正切是乙−是CX乙−XC.
clear; cl；关闭所有;
뭉 输入数据
ㅇㅎㅇ 输入数据
## 数学代写|matlab仿真代写simulation代做|Velocity and Acceleration Analysis for Rigid Body
d1d吨⋅1=0,dĴd吨⋅Ĵ=0,dķd吨⋅ķ=0
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
| 3,877 | 9,980 |
{"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}
| 3.609375 | 4 |
CC-MAIN-2023-50
|
latest
|
en
| 0.6359 |
http://blogs.ptc.com/2010/10/14/a-brief-lesson-in-numerical-equality-with-mathcad/
| 1,524,179,746,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125937074.8/warc/CC-MAIN-20180419223925-20180420003925-00207.warc.gz
| 44,247,988 | 24,610 |
# A Brief Lesson in Numerical Equality with Mathcad
After twenty-two years of mathematics teaching, the urge to explore and explain mathematical concepts has become a force of habit. Recently, I was exploring Mathcad’s equality symbols and I came up with an idea for an algebra lesson on numerical inequality that leverages Mathcad’s powerful computational thinking and communication capabilities to dig deeper into the concepts of equality and rational numbers.
Enjoy!
Exploring equality with the Mathcad’s Numeric Evaluation Symbol (=)
Consider the expression:
If we were to estimate its value, we would expect a quantity greater than 5 (25/5), but less than 6.25 (25/4). What results does Mathcad give us?
Using the Mathcad’s numerical evaluation equal sign:
Given our estimate that seems about right. We could satisfy ourselves with that, but let’s check the result. Using Mathcad’s Boolean Equal sign:
Well, this is interesting. The result of 0 that Mathcad gives us let’s us know that this Boolean equation is false. What is going on here?
Perhaps we have a rounding error? Maybe we should extend the evaluation to 6 decimal places?
Well, that did not fix the problem
Are we concerned that our answer is wrong? Probably not. But, what would it mean to be correct?
Using Fractions instead of Decimals
Let’s try something different: Let’s convert the result to a fraction.
Well, this is another surprise! 8,377,962 over 1,571,845!!! Did something go wrong? How would we check:
So, it seems like we are on track. But, let’s use this fraction in a Boolean statement.
It’s still false.
Gee, we are really getting deep into the mud of concept of equality today. Like a field trip, really.
Try a new approach – Use Mathcad’s Symbolic Evaluation Symbol (->)
Let’s use a different equality symbol. Mathcad has a second evaluation operator to supplement the = evaluation symbol. The -> evaluator is the symbolic evaluation symbol. Just to illustrate the difference, here’s an example:
The arrow gives us a symbolic result that is algebraically equivalent to the given expression. The numeric evaluator gives us an error in the first instance because x was not defined. But, after using the assignment operator (:=), both the numeric and the symbolic evaluation symbol return the same result.
Now, what happens if we evaluate our tricky numerical expression with the symbolic evaluation operator.
Wow! The numerator was simplified as 17 + 8 = 25. And, the denominator was rationalized — Mathcad removed the radical number from the denominator by multiplying the top and bottom of the fraction by the square root of 22.
Yet, even this result is not truly equal to the original expression.
So far, in evaluating the original expression we have determined that
• Is between 5 and 6.25
• Is close to 5.33
• But, a more exact estimate is 5.330018
• Two fractional results that estimate the exact value of the expression are
• And, none of these quantities is exactly equal to the original expression in Mathcad.
Are we stuck? Is the exact value unknowable?
Why all the inequality?
By now, we should have realized that the expression we are working with is an irrational number like pi and therefore we will never know the exact answer. But, what does Mathcad think? Is Mathcad teasing us, rather than just telling us that the expression is irrational and all answers are estimates?
No, Mathcad is evaluating the expression as exactly as it can and then assessing our Boolean statements against its best estimate. Let’s see what happens if we maximize the number of decimal places in our result.
Finally, a true statement! But, is this the exact answer? Not entirely, because the quantity is still irrational. But, we are now within one-one-hundred-trillionth of the precise answer. And, that is the default limit on Mathcad’s numeric calculations.
So, even with a powerful computational tool like Mathcad, many of our results are approximations. And this is a grand argument for using tools like Mathcad in secondary mathematics. How else would we dig so deep into the concepts of equality and irrational numbers? How else will we prepare our secondary students for careers in STEM fields where tools like Mathcad are the industry standard for engineering and scientific calculations?
This entry was posted in Education, Mathcad and tagged , , , , . Bookmark the permalink.
## 5 thoughts on “A Brief Lesson in Numerical Equality with Mathcad”
1. Alex says:
LOL mathcad is too smart: 2.4*3=7.19999999999999999
2. Michael Parsons says:
How may we introduce an approximately equal sign (the wavy equal) to state that a result is approximately the same as the exact answer.
1. I assume you want to do this for documentation purposes. To do so, I would create a text box, put in the two variables (or numbers) as math regions inside the text box, and then add the “approximately equal” sign between them by copying from the character map.
1. Michael Parsons says:
Yes, I this is for documentation purposes. A text region would be fine. How do I copy the symbol from the character map?
2. I do it this way, but there are many ways.
1) Windows key + R (this brings up the Run prompt)
2) charmap (this brings up the character map)
3) Pick the font you want. Search for the approximately equal sign.
4) ‘Select’ the character.
5) ‘Copy’ button.
6) Create text box/block in Mathcad.
7) Paste.
Actually, I realized a quicker way would be to just copy it from: http://en.wikipedia.org/wiki/Approximation#Unicode
Then paste into a text box/block in Mathcad.
| 1,219 | 5,579 |
{"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}
| 4.625 | 5 |
CC-MAIN-2018-17
|
latest
|
en
| 0.912475 |
https://wisdomanswer.com/how-do-you-find-the-lcm-of-8-and-32/
| 1,718,689,656,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861747.46/warc/CC-MAIN-20240618042809-20240618072809-00724.warc.gz
| 558,971,839 | 43,318 |
# How do you find the LCM of 8 and 32?
## How do you find the LCM of 8 and 32?
The LCM of 8 and 32 is 32. To find the least common multiple of 8 and 32, we need to find the multiples of 8 and 32 (multiples of 8 = 8, 16, 24, 32; multiples of 32 = 32, 64, 96, 128) and choose the smallest multiple that is exactly divisible by 8 and 32, i.e., 32.
### How do you find the lowest common multiple of 6 and 8?
The LCM of 6 and 8 is 24. To find the least common multiple (LCM) of 6 and 8, we need to find the multiples of 6 and 8 (multiples of 6 = 6, 12, 18, 24; multiples of 8 = 8, 16, 24, 32) and choose the smallest multiple that is exactly divisible by 6 and 8, i.e., 24.
#### What is the least common multiple of 8 20 and 32?
160
The least common multiple of 8, 20 and 32 is 160.
What is the LCM and GCF of 6 and 8?
24
Example 1: Find the GCF of 6 and 8, if their LCM is 24. Therefore, the greatest common factor of 6 and 8 is 2. Example 2: For two numbers, GCF = 2 and LCM = 24. If one number is 8, find the other number.
Which is the least common multiple of 6 and 8?
When we compare the two lists to see what they have in common, we get the answer to “What are the common multiples of 6 and 8?” Since 24 is the first number they have in common, 24 is the least common multiple of 6 and 8.
## How to find the least common multiple in math?
Least Common Multiple is the smallest multiple that two or more numbers have in common. How to find LCM: list out all the multiples of each number and then find the smallest one they have in common. Find the LCM for [Math Processing Error] 8 and [Math Processing Error] 6.
### Which is the least common multiple of 4, 6 and 12?
Common Multiples of 4 : 4,8,12,16,20,24,28,….. Common Multiples of 6: 6,12,18,24,30,36,42….. Common Multiples of 12: 12,24,36,48,60,72,…. From the above-given multiples of 4, 6 and 12, you can see, 12 is the smallest common multiple. Therefore, LCM. of 4, 6 and 12 is 12.
#### Which is the least common multiple of 6 7 21?
Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60. Multiples of 7: 7, 14, 21, 28, 35, 42, 56, 63. Multiples of 21: 21, 42, 63. Find the smallest number that is on all of the lists. We have it in bold above. So LCM (6, 7, 21) is 42.
| 770 | 2,242 |
{"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}
| 4.8125 | 5 |
CC-MAIN-2024-26
|
latest
|
en
| 0.929324 |
https://de.mathworks.com/matlabcentral/answers/563603-how-to-compute-the-density-of-a-3d-point-cloud
| 1,669,506,295,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446709929.63/warc/CC-MAIN-20221126212945-20221127002945-00328.warc.gz
| 248,016,741 | 30,548 |
# How to compute the density of a 3D point cloud?
228 views (last 30 days)
Ali on 12 Jul 2020
Commented: Ali on 12 Jul 2020
I'm trying to write a program that operates on 3D point clouds (.ply, .pcd), I need to know the density of the given point cloud file and compare it with a threshold or certain percentage to decide for some operations. How can I do that?
I have found this for 3D point clouds on the web:
Two methods can be used to compute the density:
• either 'Precise': the density is estimated by counting for each point the number of neighbors N (inside a sphere of radius R)
• or 'Approximate': the density is simply estimated by determining the distance to the nearest neighbor (which is generally much faster). This distance is considered as being equivalent to the above spherical neighborhood radius R (and N = 1).
However I'm not sure how to put this into matlab code, I guess I should use pcfitplane for sphere fitting and findNearestNeighbors. Would appreciate a step by step example.
Edited: Thiago Henrique Gomes Lobato on 12 Jul 2020
You can just directly apply the definition you gave considering that the density is N/Volume. The easiest is the second one, which would be:
K = 1;
for idx=1:length(AllPoints)
[~,r] = findNearestNeighbors(AllPoints,AllPoints(idx,:),K);
density(idx) = 1/(4*pi*r.^3/3);
end
The first one is a little more complicated but at the same time not so much:
R = 1; % depends on your data
for idx=1:length(AllPoints)
Distances = sqrt( sum( (AllPoints-AllPoints(idx,:)).^2 ,2) );
Ninside = length( find(Distances<=R) );
density(idx) = Ninside/(4*pi*R.^3/3);
end
Ali on 12 Jul 2020
I did try my suggested order and indeed your statement is true, no difference in results. I appreciate your time and help.
### Categories
Find more on Point Cloud Processing in Help Center and File Exchange
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by
| 497 | 1,980 |
{"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}
| 3.578125 | 4 |
CC-MAIN-2022-49
|
latest
|
en
| 0.913937 |
https://offices-www-office.com/qa/how-many-zeros-are-there-in-45-lakhs.html
| 1,620,953,538,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243989616.38/warc/CC-MAIN-20210513234920-20210514024920-00502.warc.gz
| 478,060,328 | 7,584 |
# How Many Zeros Are There In 45 Lakhs?
## How can I write 1 lakh in English?
A lakh (/læk, lɑːk/; abbreviated L; sometimes written Lac or Lacs) is a unit in the Indian numbering system equal to one hundred thousand (100,000; scientific notation: 105).
In the Indian 2,2,3 convention of digit grouping, it is written as 1,00,000..
## How much is 37 lakhs?
As you can see, 37 lakh is the same as 3.7 million.
## How do you write 10 lakhs in numbers?
10 Lakhs = 1 Million = 1 followed by 6 Zeros = 1,000,000….As per the Indian currency system:1-Ones.10-Tens.100-Hundred.1000-Thousand.10,000-Ten thousand.1,00,000-One Lakh.10,00,000-Ten.1,00,00,000-One.More items…•
## How do you write 25 lakhs in numbers?
As you can see, 25 lakh is the same as 2.5 million.
## How do you write 15 lakh in numbers?
Note: We found that some people call it 15 lakhs or 15 lac, but the correct way to say it is 15 lakh (lakh without the trailing “s”).
## How do you write 35 lakhs?
Note: We found that some people call it 35 lakhs or 35 lac, but the correct way to say it is 35 lakh (lakh without the trailing “s”).
## How many dollars is 5 lakhs?
Convert Indian Rupee to US DollarINRUSD1000 INR13.4261 USD5000 INR67.1305 USD10000 INR134.261 USD50000 INR671.305 USD7 more rows•Nov 2, 2020
## How do you write 20 lakhs?
Note: We found that some people call it 20 lakhs or 20 lac, but the correct way to say it is 20 lakh (lakh without the trailing “s”).
## How do you write 1 crore 50 lakhs in numbers?
To write the 1 crore 50 lakhs in numerical form we just have to replace the first zero with a five, because 50 lakhs is the exact half value of 1 crore. So the numerical value will be; 15000000.
## How do you write 50 lakhs in numbers?
Note: We found that some people call it 50 lakhs or 50 lac, but the correct way to say it is 50 lakh (lakh without the trailing “s”).
## How much lakhs is 60?
As you can see, 60 lakh is the same as 6 million.
## How many zeros are there in 40 lakhs?
6As you can see, 40 lakh is the same as 4 million. How many zeros in 40 lakh? When we count the trailing zeros in 40 lakh above, we see that the answer is 6.
## How much is 3 lakhs?
As you can see, 3 lakh is the same as 0.3 million.
## How do you write 17 lakhs?
Note: We found that some people call it 17 lakhs or 17 lac, but the correct way to say it is 17 lakh (lakh without the trailing “s”).
## How do you write 13 lakhs?
Note: We found that some people call it 13 lakhs or 13 lac, but the correct way to say it is 13 lakh (lakh without the trailing “s”).
## Is 60 lakhs a good salary in India?
It’s a very good salary for someone who is 30 years old and you should be proud of that but how you make the most of it is by investing a large chunk of that income maintaining a balanced life style.
## How do you write 5 lakhs?
Let’s take a 6-digit number 5,46,783. According to the Indian place value system, it is written as Five lakh, forty-six thousand, seven hundred and eighty-three.
## How do you write 24 lakhs?
Note: We found that some people call it 24 lakhs or 24 lac, but the correct way to say it is 24 lakh (lakh without the trailing “s”).
| 945 | 3,157 |
{"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}
| 3.9375 | 4 |
CC-MAIN-2021-21
|
longest
|
en
| 0.892747 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-and-trigonometry-10th-edition/prerequisites-review-exercises-page-63/66
| 1,675,595,520,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500251.38/warc/CC-MAIN-20230205094841-20230205124841-00574.warc.gz
| 804,353,033 | 12,154 |
Algebra and Trigonometry 10th Edition
$-2y^2+11y-8$
$8y-[2y^2-(3y-8)]=8y-[2y^2-3y-(-8)]=8y-[2y^2-3y+8]=8y-2y^2-(-3y)-8=8y-2y^2+3y-8=(8+3)y-2y^2-8=-2y^2+11y-8$
| 116 | 159 |
{"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 | 4 |
CC-MAIN-2023-06
|
latest
|
en
| 0.506138 |
http://www.mathskey.com/question2answer/39238/multiply-the-following-polynomials
| 1,544,857,373,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-51/segments/1544376826800.31/warc/CC-MAIN-20181215061532-20181215083532-00289.warc.gz
| 429,894,864 | 10,906 |
# Multiply the following polynomials.
Multiply the following polynomials.
(3x^2 + 4)(x - 3)(3x2 - 4) = (3x^2 + 4)(3x^2 - 4)(x - 3)
= [ (3x^2)^2 - (4)^2 ] (x - 3)
= [ 9x^4 - 16 ] (x - 3)
= (9x^4 - 16)x - 3(9x^4 - 16)
= (9x^5 - 16x) - (27x^4 - 48)
= 9x^5 - 16x - 27x^4 + 48
= 9x^5 - 27x^4 - 16x + 48
(3x^2 + 4)(x - 3)(3x2 - 4) = 9x^5 - 27x^4 - 16x + 48.
| 246 | 383 |
{"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}
| 3.84375 | 4 |
CC-MAIN-2018-51
|
longest
|
en
| 0.118589 |
http://kullabs.com/classes/subjects/units/lessons/notes/note-detail/1475
| 1,544,919,829,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00128.warc.gz
| 163,909,302 | 12,330 |
Note on Energy stored in a Stretched Wire, Poisson's Ratio and Elastic after Effect
• Note
• Things to remember
Energy stored in a Stretched Wire
In order to stretch a wire, certain work must be done against restoring force. This work done is stored in the form of potential energy (U) of the stretched wire.
Suppose a wire having length 'L' and cross-section area (A) is being deformed by a deforming force 'F' . Then the small work done 'dw' to elongated in by 'dl' is given by
$$dw = F dl \dots (i)$$
Hence, the total work done 'w' to elongated the wire by length 'l' is obtained by integrating dw between the limits o and 'l'.
Therefore,
$$W = \int dW$$
$$W = \int _o^l F dl \dots {ii}$$
Here 'F' depends upon the elongation. So when the elongation is 'l' the definition of young's Modulus (Y) gives
$$Y =\frac {stress}{strain} = \frac {F/A}{l/L} = frac {FL}{Al}$$
$$\therefore F = \frac {YAl}{L} \dots (iii)$$
Using this in equation (ii) and integrating, we get
$$dW = \int_0^l \frac {YAl}{L} dl$$
$$= \frac {YA}{L} \left [\frac l2 \right ]_0^l$$
$$= \frac {YA}{L} \left [\frac {l^2 - 0^2}{2} \right ]$$
$$= \frac 12 \frac {YA}{L} l^2$$
$$= \frac 12 \left (\frac {YAl}{L} \right ) l$$
$$= \frac 12 F \times l$$
$$\therefore W= \frac 12F \times l$$
$$\boxed {= \frac 12 \text {force} \times \text {elongation}}$$
This work done is equal to the potential energy stored (U).
Energy stored (U) = work done (w)
$$U = \frac 12\text {force} \times \text {elongation}$$
Energy Density (U)
The energy stored per unit volume of the stretched wire is called its energy density.
$$\text {Energy density} (U) =\frac {\text {enrgy stored}}{\text {volume}}$$
$$= \frac {Y F\times l}{A \times L}$$
$$=\frac 12 \frac FA \times \frac lL$$
$$= \frac 12 \text {stress} \times \text {strain}$$
$$\therefore Energy Density = \frac 12 \text {stress} \times \text {strain}$$
Poisson's Ratio
When a deforming force is applied at the free end of a suspended wire, the length of the wire increases and its dimension decreases. Thus there occurs longitudinal strain as well as lateral strain. The change in dimensions per unit original dimension along the force is applied is called longitudinal strain. It is denoted by α, while the change in dimension in the per unit dimension in perpendicular direction is called lateral strain and is denoted by β. Experimentally, it has been found that within the elastic limit the lateral strain β is directly proportional to the longitudinal strainα, i.e
$$\beta \propto \alpha$$
$$\text {or} \beta = \sigma \alpha$$
$$\text {or} \sigma = \frac {\beta}{\alpha}$$
whereσ is proportionality constant which is known as Poisson's ratio and is defined as the ratio of lateral strain to the longitudinal strain within elastic limit.
$$\text {poisson's ratio,} \:\sigma = \frac {\beta}{\alpha} = \frac {\text {llateral strain}}{\text {longitudinal strain}}$$
Let L, D be the original length and diameter respectively of the wire, and dL and d be the slight increase in length and a corresponding slight decrease in diameter of the wire as shown in the figure.
$$\text {Longitudional strain,} \:\alpha = \frac {dL}{L}$$
$$\text {Lateral strain,} \:\beta = \frac {d}{D}$$
Therefore, Poisson 's ratio for the material of the wire is given by
$$\sigma = \frac {\beta}{\alpha} \frac {d/D}{dL/L}$$
$$\boxed {\therefore\sigma = \frac dD \times \frac {L}{dL}}$$
Poisson's ratio is a unitless and dimensionless quantity.
Elastic after Effect
We know that on removing the deforming forces from the elastic bodies, they regain their original configuration. Some elastic bodies regain their original configuration immediately, but a few others take some tome to recover their original configurations. The temporary delay in regaining the original configuration by an elastic body after removal of a deforming force is called elastic after effect.
Elastic Fatigue
When a body is repeating deformed and released, its elasticity is lost. This is known as elastic fatigue. Due to elastic fatigue it is possible to break an iron nail after deforming repeatedly.
Elastic Hysteresis
Due to elastic after-effect the strain produced in an elastic lags behind the stress to which it is subjected and this effect is known as elastic hysteresis.
When a body is subjected to the road cycle of increasing and decreasing stress then the strain produced is larger when the body is unloaded then when it is loaded.
In order to stretch a wire, certain work must be done against restoring force. This work done is stored in the form of potential energy (U) of the stretched wire.
The energy stored per unit volume of the stretched wire is called its energy density.
The change in dimensions per unit original dimension along the force is applied is called longitudinal strain.
The change in dimension in the per unit dimension in perpendicular direction is called lateral strain and is denoted by β.
The temporary delay in regaining the original configuration by an elastic body after removal of a deforming force is called elastic after effect.
When a body is repeating deformed and released, its elasticity is lost. This is known as elastic fatigue.
Due to elastic after effect the strain produced in an elastic lags behind the stress to which it is subjected and this effect is known as elastic hysteresis.
.
0%
| 1,402 | 5,355 |
{"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}
| 4.125 | 4 |
CC-MAIN-2018-51
|
latest
|
en
| 0.789391 |
https://techauthority.tech/2022/01/30/best-time-to-buy-and-sell-stock-ii-interview-question-on-arrays-python/?wp-story-load-in-fullscreen=true&wp-story-play-on-load=true
| 1,679,613,743,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296945218.30/warc/CC-MAIN-20230323225049-20230324015049-00737.warc.gz
| 619,007,027 | 33,453 |
This is an array interview question that is solved using Python Language.
Below, you are given an integer array prices where prices[i] is the price of a given stock on the ith day.
On each day, you may decide to buy and/or sell your stock. You can only get to hold at most one share of the stock at any time. However, you can buy it again immediately and get to sell it on the same day.
Now, find and return the maximum profit you can achieve.
Input: prices = [7,1,5,3,6,4] Output: 7 Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4. Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3. Total profit is 4 + 3 = 7.
Now, we have to come up with an algorithm to find the maximum profit. So if the given prices are [7,1,5,3,6,4], then the result will be 7, as we will buy on day 2 (price 1), then sell on day 3, (price 5), so profit is 5 – 1 = 4. Now but on day 4 (price 3), and sell on day 5 (price 6) so profit is 6 – 3 = 3.
Example Let us see the following implementation to get better understanding:
Solution 1:
class Solution(object):
def maxProfit(self, prices):
if not prices:
return 0
n = len(prices)
dp = [0 for i in range(n)]
ymin = prices[0]
for i in range(1,n):
ymin = min(ymin,prices[i])
dp[i] = max(dp[i],prices[i]-ymin)
zmax = [0 for i in range(n)]
zmax[-1] =prices[-1]
tempp = 0
for i in range(n-2,-1,-1):
zmax[i] = max(zmax[i+1],prices[i])
zmin = [prices[-1],n]
for i in range(n-2,-1,-1):
tempp = max(tempp,zmax[i+1]-prices[i+1])
obj = Solution()
print(obj.maxProfit([7,1,5,3,6,4]))
Solution 2:
class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
| 562 | 1,677 |
{"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}
| 4.1875 | 4 |
CC-MAIN-2023-14
|
latest
|
en
| 0.823059 |
https://www.coursehero.com/file/5997075/S-72-2410-solution-2/
| 1,490,722,067,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-13/segments/1490218189802.18/warc/CC-MAIN-20170322212949-00230-ip-10-233-31-227.ec2.internal.warc.gz
| 887,478,567 | 22,707 |
S-72_2410_solution_2
# S-72_2410_solution_2 - S-72.2410 Information Theory Haanp...
This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: S-72.2410 Information Theory Haanp aa & Linja-aho Homework 2, solutions 2009 Homework 2, solutions Deadline: November 16th, 16:00 The box for returning exercises is in the E-wing, 2nd floor corridor. 1. Let X and Y be two independent integer-valued random variables. Let X be uniformly distributed over { 1 , 2 ,..., 8 } , and let Pr { Y = k } = 2- k ,k = 1 , 2 , 3 ,... (a) Find H ( X ) (b) Find H ( Y ) (c) Find H ( X + Y,X- Y ). Solution: (a) For uniform distribution H ( X ) = 8(- 1 8 log 1 8 ) = log 8 = 3 (b) H ( Y ) =- k =1 1 2 k log 1 2 k = 1 2 k k log 2 = k =0 k ( 1 2 ) k = 1 2 (1- 1 2 ) 2 = 2 (c) Because ( X,Y ) ( X + Y,X- Y ) is a one-to-one -transformation, H ( X + Y,X- Y ) = H ( X,Y ). Because X and Y are independent, H ( X,Y ) = H ( X ) + H ( Y ) = 3 + 2 = 5. 2. A deck of n cards in order 1 , 2 ,...,n is provided. One card is removed at random then replaced at random. What is the entropy of the resulting deck? Solution: There are n 2 possible ways to shuffle the deck as described, and all have the same probability. But the solution is not that simple, because some shuffling actions result in same deck. There are basically three ways to shuffle the deck: The card is removed and put back in the same place. In this case,The card is removed and put back in the same place....
View Full Document
## This note was uploaded on 10/27/2010 for the course ECE 221 taught by Professor Sd during the Spring '10 term at Huston-Tillotson.
### Page1 / 4
S-72_2410_solution_2 - S-72.2410 Information Theory Haanp...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online
| 631 | 2,028 |
{"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}
| 3.875 | 4 |
CC-MAIN-2017-13
|
longest
|
en
| 0.872265 |
https://www.nagwa.com/en/videos/134102908257/
| 1,586,468,224,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585371876625.96/warc/CC-MAIN-20200409185507-20200409220007-00485.warc.gz
| 1,022,475,555 | 5,325 |
# Video: US-SAT05S4-Q04-134102908257
If 5(π + π) = 7, what is the value of π + π?
01:28
### Video Transcript
If five multiplied by π plus π is equal to seven, what is the value of π plus π?
Weβre told in the question that five multiplied by π plus π is equal to seven. In order to calculate the value of π plus π, we need to divide both sides by five. On the left-hand side, five divided by five is equal to one. This means that weβre left with π plus π. On the right-hand side, weβre left with seven-fifths or seven over five. If five multiplied by π plus π is equal to seven, then the value of π plus π is seven-fifths.
We could convert this answer into a decimal, as the line in a fraction means divide. We need to divide seven by five. This is equal to 1.4. We could also write our top heavy or improper fraction as a mixed number. Seven-fifths is the same as one and two-fifths. This is because seven divided by five is equal to one remainder two. The three answers seven-fifths, 1.4, and one and two-fifths are all correct values of π plus π.
| 350 | 1,120 |
{"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}
| 4.3125 | 4 |
CC-MAIN-2020-16
|
latest
|
en
| 0.934657 |
http://www.questionpaperz.in/venn-diagram-questions-answers/
| 1,521,503,761,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-13/segments/1521257647244.44/warc/CC-MAIN-20180319234034-20180320014034-00346.warc.gz
| 456,311,486 | 12,836 |
Verbal reasoning is the understanding of concepts which are based on general sense things. Verbal Reasoning is part of every competitive exam nowadays and the candidates need to prepare it very well to get all its questions solved. This section is very scoring part of exam and so preparation is very important. Verbal reasoning testes are often used for the entrance of candidates in various schools universities etc. These are the aptitude tests which includes various parts. This post is based on the logical Venn diagrams of verbal reasoning.
Logical Venn diagrams represents the mathematical or logical sets as circles or closed curves. Venn diagram shows all possible logical relations between finite collections of different sets. These diagram test the students’ theory and working towards diagrams. Venn diagrams are the important topic in verbal reasoning as this topic includes the diagrams. The main aim of this is to test the ability about the relation between some items of a group by diagrams.
Some examples of the Venn diagrams are as follows:
Question 1: In an organization of pollution control board, engineers are represented by a circle, legal experts by a square and environmentalist by a triangle. Who is most represented in the board as shown in the following figure?
1. Environmentalists
2. Legal Experts
3. Engineers with legal background
4. Environmentalists with Engineering background
Answer: D (Environmentalists with Engineering background is most represented in the board)
Question 2: Study the diagram and identify the people who can speak only one language.
1. L+M+O
2. K+J+I
3. K
4. I
These are the various types of questions included in Venn diagrams. To solve these questions various tricks and tips including with various formulas are adopted. Without these tricks and formulas the solution to these questions cannot be found. So candidates need to prepare all these questions in detail then only they can attempt the question in exam without wastage of time. Venn diagrams are easy if you know all the tricks than everybody easily solve the problems. We all know that these are mathematical based and we know that maths involves lot of practice. So, candidates should start preparing these chapters than only they can score good marks in exams. We have provided you all the information related to logical Venn diagrams.
| 459 | 2,365 |
{"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}
| 4.03125 | 4 |
CC-MAIN-2018-13
|
latest
|
en
| 0.960335 |
https://gmatclub.com/forum/is-x-y-1-a-n-x-a-n-y-2-3n-1-is-odd-open-1053.html
| 1,498,686,165,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128323801.5/warc/CC-MAIN-20170628204133-20170628224133-00099.warc.gz
| 753,067,071 | 48,246 |
It is currently 28 Jun 2017, 14:42
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Is x > y? 1) (a^n)*x > (a^n)*y 2) 3n + 1 is odd open
Author Message
Manager
Joined: 28 Feb 2003
Posts: 100
Is x > y? 1) (a^n)*x > (a^n)*y 2) 3n + 1 is odd open [#permalink]
### Show Tags
29 May 2003, 05:52
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Is x > y?
1) (a^n)*x > (a^n)*y
2) 3n + 1 is odd
open to debate......i hope i don't screw up on this one like the last one
Manager
Joined: 25 May 2003
Posts: 54
### Show Tags
29 May 2003, 13:22
am i missing something here?? seems suspiciously easy...A?
Manager
Joined: 12 Mar 2003
Posts: 59
### Show Tags
29 May 2003, 19:44
Ans : C
1) Insuff, since a and n can be -ve or +ve
2) Insuff, n is even - nothing about x and y
Combined : n is even, and if a is -ve or +ve a^n can be divided out.
=> x>y
Suff
Intern
Joined: 25 May 2003
Posts: 17
Location: Thailand
### Show Tags
29 May 2003, 21:40
In my view, I vote for C
supposed that X and Y are positive 5 and 2 repectively.
Clearly, the statement (I) doesn't tell the values of a^n. If a^n is negative value, X may be less than Y. On the contrary, if a^n is positive, X will be more than Y. Thus, Insufficient.
The statment II only show the value of n which can be negative or positve,the value of a and the value of x and y. Therefore. Insufficient
Combined them together, Thus, the answer is C because a^n always be positive and therefore, X always be more than Y
Manager
Joined: 28 Feb 2003
Posts: 100
### Show Tags
30 May 2003, 02:41
I also had C in mind
to prove that x > y, you have to prove that the common number that is multiplied on both sides is POSITIVE.
From 1, a^n can be negative if n is odd and "a" is negative..
however if n is even than a^n will AlWAYS be positive....
therefore both together are sufficient
Manager
Joined: 08 Apr 2003
Posts: 150
### Show Tags
30 May 2003, 02:50
Hi bhavesh,
I went on the same lines of reasoning as you gave. I put in C. But, another thing crossed my mind.
The question is x > y ?? So it can be YES or NO, correct
So, even if you say if a is negative and n is odd, you are multiplying the same number on both sides. So they get cancelled anyhow and even if the greater than sign reverses to say x < y , you still have an answer.
Am i right???
Manager
Joined: 28 Feb 2003
Posts: 100
### Show Tags
30 May 2003, 02:54
if "a" is positive, than x>y...the answer is YES
however if "a" is negative and "n" is odd, than a^n will be negative and in this case x<y.......answer is NO
therefore you cannot say for sure YES or NO.....
depends on sign of "a" and value of "n"
Founder
Joined: 04 Dec 2002
Posts: 15153
Location: United States (WA)
GMAT 1: 750 Q49 V42
### Show Tags
06 Jun 2003, 20:25
neema wrote:
Is it C?
you are still a student and you are getting into MBA?
you feel like you are ready?
or you are a different kind of student ?
Senior Manager
Joined: 11 Nov 2003
Posts: 356
Location: Illinois
### Show Tags
11 Feb 2004, 11:58
This one is a very old one.
Quote:
Is x > y?
1) (a^n)*x > (a^n)*y
2) 3n + 1 is odd
open to debate......i hope i don't screw up on this one like the last one
The agreed upon answer for this question by all the members who responded to this question in the past is C.
I have following questios.
(1) Should the second statement be given as 3n + 1 is positive odd integer ?
If the word positive is not included then we can take
3n + 1 = -1 => n -2/3 = 0.66666..... (Is this odd or even? )
If the word integer is not included,
Can we consider 3n + 1 = 5.5 ? (This is odd number, correct?)
=> n= 4.5/3 => 1.5 (This should be odd)
I think if these conditions are not included, then the answer should be E.
Finally, please let me know (-1)^1.2 = ?
I thought that 1.2 is considered even number so answer should be 1. But the calculator says invalid input function.
(1.2 here is decimal number = 12/10)[/b]
Senior Manager
Joined: 11 Nov 2003
Posts: 356
Location: Illinois
### Show Tags
11 Feb 2004, 12:05
Quote:
Finally, please let me know (-1)^1.2 = ?
I thought that 1.2 is considered even number so answer should be 1. But the calculator says invalid input function.
(1.2 here is decimal number = 12/10)
But my other questions remain.
11 Feb 2004, 12:05
Display posts from previous: Sort by
| 1,511 | 5,013 |
{"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-2017-26
|
latest
|
en
| 0.895119 |
https://comojogaruno.com/how-many-ways-can-8-people-sit-around-a-table-if-two-insist-on-sitting-next-to-each-other
| 1,701,790,495,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100551.2/warc/CC-MAIN-20231205140836-20231205170836-00072.warc.gz
| 223,262,224 | 11,026 |
# How many ways can 8 people sit around a table if two insist on sitting next to each other *?
The formula for the typical scenario is (n-1)!. The n becomes 7 in this case, since the two people form a block (act as one). Then, multiply by their own permutations (2!). So, 6!2!, if there are no placeholders (name cards on the table). If this is the case, we are not just looking for the order, but actual seats. With placeholders, we also multiply by the number of people, which makes the formula equivalent to n!. So, in this case it would be 7!2!. Also, we usually assume clockwise is different than anticlockwise, so we refrain from dividing by 2. what is the answer in1in =_____cm? Can anyone help me please for 50 points!! [-3,-2], [-2,0], [-1,2],[ 0,4][ 1,-6],[2,8] function or not? monthly payments of 12,000 for 9 years that will start 6 months from now.what is the period of deferral? Malolos in one of the oldest towns in the region that is known in making Ensaimadas, Inipit and Pastillas. Aside from those sweets, the said town is a … lso popular in making Kakanin like Suman Yangit, Puto, Kutsinta, Ginatan Bilo Bilo, Sapin Sapin, Suman Murwekos, Sweetened Macapuno and Kalamay. S Situation Ka Ditha, one of the makers of Kakanin in Barangay Balite, City of Malolos, males Sumang Murwekos, a rice cake she learned from her aunt, Nana Elena. She sells Sumang Murwekos to the barrio for additional income. As anyone who sells things can tell you, deciding on an appropriate price is very important. If your price is too low, you will not make much of a profit. If your price is too high, you will also probably not make much of a profit because fewer people will buy what you are selling. The best price is the price that leads to the maximum profit. Suppose Ka Ditha's profit y can be found by y = x²+24x 60, where x represents the price of each Suman Murwekos. What price should she charge to receive the maximum profit?
| 515 | 1,938 |
{"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}
| 3.75 | 4 |
CC-MAIN-2023-50
|
latest
|
en
| 0.941705 |
http://kb.tableau.com/articles/howto/how-to-change-military-time-into-standard-time
| 1,524,755,397,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125948285.62/warc/CC-MAIN-20180426144615-20180426164615-00302.warc.gz
| 172,170,666 | 51,009 |
KNOWLEDGE BASE
## How to Change Military Time Into Standard 12-Hour Time Format
Published: 04 May 2017
### Question
How to convert a military time into a standard 12-hour format.
### Environment
• Tableau Desktop
To convert a 24-hour integer-based time format, that may contain either three values such as 735 or four values such as 1348, follow the steps outlined below.
### Step 1
First, separate the hours from the minutes while ensuring that we do not add extra values due to the different lengths possible for the integer value.
• Create the following calculation titled ‘Scheduled Trimmed Mil Hour’ in the example using the following formula.
IF
`LEN(STR([Scheduled Departure Time])) = 4 THEN LEFT(STR([Scheduled Departure Time]),2)ELSEIFLEN(STR([Scheduled Departure Time])) = 3 THEN LEFT(STR([Scheduled Departure Time]),1)`
`END`
### Step 2
Now we need to separate the minutes from the hours.
• Create the following calculation titled ‘Scheduled Minutes’ in the example using the following formula.
`RIGHT(STR([Scheduled Departure Time]),2)`
### Step 3
Next we need to determine if the time is AM or PM.
• Create the following calculation titled ‘Scheduled AM/PM’ in the example using the following formula:
`IF INT([Scheduled Trimmed Mil Hour]) > 12 THEN 'PM' ELSE 'AM'` `END`
### Step 4
Now we need to incorporate all these values into one single field.
• Create the following calculation titled ‘Scheduled Standard Time’ in the example using the following formula.
`STR([Scheduled Standard Hour])+':'+[Scheduled Minutes]+ ' ' + [Scheduled AM/PM]`
### Step 5
Lastly we need to have the new field recognized as an actual time field by Tableau Desktop.
• Create the following calculation titled ‘Scheduled Time Date Parsed’ in the example using the following formula.
`DATEPARSE('hh:mm a', [Scheduled Standard Time])`
| 431 | 1,842 |
{"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-2018-17
|
latest
|
en
| 0.759177 |
https://protonstalk.com/statistics/assumed-mean-method/
| 1,702,135,971,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100912.91/warc/CC-MAIN-20231209134916-20231209164916-00368.warc.gz
| 523,150,727 | 55,213 |
# Assumed Mean Method
The assumed mean method is a technique used in statistics to calculate the arithmetic mean. It is a useful shortcut method to calculate the mean from a set of data.
Index
## Assumed Mean Method Formula
Here, we present the assumed mean method formula for different kinds of data.
### Discrete Data
Let us consider a collection of $$\boldsymbol{N}$$ values whose mean we have to find. Each value is denoted by $$x_i$$. We make a guess and choose an approximate mean, $$x_0$$, roughly in the middle of the data. This is called the assumed mean.
Then, we calculate deviations from the assumed mean, defined as:
$$d_i = x_i – x_0$$
Now, we add up these deviations for all values in the data.
$$A = \sum_{i=1}^N d_i$$
This is divided by the total number of observations $$N$$. This value $$D$$ is the difference between the actual mean and the assumed mean.
$$D = \frac{A}{N}$$
Finally, the actual mean is obtained as follows.
$$\bar{x} = x_0 + D$$
### Grouped Data
The process is very similar in grouped data. The only major differences are as follows:
1. For each class, the assumed mean is subtracted from the class mark (midpoint) of class interval.
2. The individual deviations $$d_i$$ is multiplied by the frequency $$f_i$$ of the class intervals.
Thus, the assumed mean formulae become,
$$d_i = x_i – x_0$$
Where, $$x_i$$ are class marks.
$$A = \sum_{i=1}^N f_i d_i$$
Where, $$f_i$$ is class frequency.
$$D = \frac{A}{N}$$
Finally,
$$\bar{x} = x_0 + D$$
## Applications of Assumed Mean Method
This method can be used to make the calculation of mean much faster by hand. If the assumed mean is chosen well, the deviations tend to be small and almost cancel out, making the addition easier.
The process used for grouped data also speeds up the procedure very much, providing accurate results.
## Example Questions
Question 1. Find the mean of this set of values by assumed mean method.
91, 48, 9, 37, 6, 42, 23, 45, 63, 84, 88, 29, 28, 10, 8
Solution. Let us assume mean $$x_0 = 40$$. Then the deviations $$x_i$$ from mean become,
$$51, 8, -31, -3, -34, 2, -17, 5, 23, 44, 48, -11, -12, -30, -32$$
Now, we have their sum $$A = 11$$.
Now, the difference $$D$$ between actual mean and assumed mean is
$$D = \frac{A}{N}$$
So, $$D = \frac{11}{15} = 0.733$$.
Finally, the actual mean is given by,
$$\bar{x} = x_0 + D = 40 + 0.733 = 40.733$$
So the mean of the above data is $$\bar{x} = 40.733$$. A good choice of assumed mean simplified the process significantly.
Question 2. Find the mean of the following data, using classes of width 10.
66,75,79,56,61,77,92,75,78,51,82,85,59,66,91,98,90,73,71,83,85,91,77,70
Solution. Let us group the data into the following classes: $$50-60, 60-70, 70-80, 80-90, 90-100$$. Then we have can arrange the required values in a table as follows. Let us assume a mean of 75.
Thus, we get $$D = \frac{A}{N} = \frac{50}{24}$$. So, $$D \approx 2.1$$.
We then have the actual mean as follows.
\begin{align} \bar{x} & = x_0 + D \\ & = 75 + \frac{50}{24} \\ & \approx 75 + 2.1 \\ \bar{x} & = 77.1 \end{align}
## FAQs
How do you calculate the assumed mean?
We calculate the mean by the assumed mean method for a series of $$N$$ values $$x_i$$ as follows.
1. Assume a mean $$x_0$$.
2. Calculate deviations $$d_i = x_i – x_0$$ from assumed mean, for each $$x_i$$.
3. Sum the deviations for all values. $$A = \sum_{i=1}^N d_i$$.
4. Calculate $$D = \frac{A}{N}$$.
Find actual mean $$\bar{x}$$ as follows: $$\bar{x} = x_0 + D$$.
How can you find the mean for grouped data by the assumed mean method?
The process is the same as the method used for discrete data, with the only differences being:
1. The values $$x_i$$ become the class-marks for each class interval.
2. The deviations $$d_i$$ are multiplied by the frequency $$f_i$$ of each class interval before being added.
In short,
$$\bar{x} = x_0 + \frac{\sum f_i d_i}{N}$$
Where, $$N = \sum f_i$$
What is the difference between actual mean and assumed mean?
The actual mean $$\bar{x}$$ is the real arithmetic mean of the given data.
The assumed mean $$x_0$$ is a guess of the approximate value for the mean. It is used in the assumed mean method to simplify calculations.
How can you find standard deviation by the assumed mean method?
We define, $$B = \sum_{i=1}^N {d_i}^2$$.
Then, we have standard deviation $$\sigma$$ as,
$$\sigma = \sqrt{\frac{B-ND^2}{N}}$$
Where, $$N$$ is number of observations, $$D$$ is difference between $$\bar{x}$$ and $$x_0$$ as defined above.
Scroll to Top
Scroll to Top
| 1,380 | 4,546 |
{"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
|
longest
|
en
| 0.883037 |
https://www.coloradoplc.org/resources?page=3
| 1,503,475,003,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-34/segments/1502886117911.49/warc/CC-MAIN-20170823074634-20170823094634-00081.warc.gz
| 887,647,928 | 10,251 |
# Resource Finder
This tool allows you to search for materials across all groups on the site.
Resource Overview Type Content Area Group
Algebra I - Finding Your Roots - High School - Unit 4
Inquiry Questions (Engaging- Debatable):
What is the best way to model simple projectile motion? (MA10-GR.HS-S.2-GLE.1-EO.c.v.1)
How would shooting a basketball on the moon be different from Earth?
Curriculum Guide, Units Math Colorado's Sample Instructional Units Resource Bank
Algebra I - Graphing Linear Equations - Unit 3
Essential Questions:
1. A linear function consists of a slope and a y-intercept, with the exception of vertical lines.
2. There are multiple methods of graphing a linear equation.
3. Linear patterns can be modeled from scatter plots through use of line-of-best fit.
4. Real-world situations can be modeled with linear equations.
Lines that are either parallel or perpendicular can be determined by identifying their slopes
Curriculum Guide, Units Math Strasburg School District
Algebra I - Home on the Range – Part 1 - Unit 1
Inquiry Questions (Engaging- Debatable):
Can all real-world situations be modeled with a function?
MA10-GR.HS-S.2-GLE.2-EO.b
Curriculum Guide, Units Math Colorado's Sample Instructional Units Resource Bank
Algebra I - Home on the Range – Part 2- Unit 5
Inquiry Questions (Engaging- Debatable):
How does the same melody in a different key relate to a function being translated or scaled? (MA10-GR.HS-S.2-GLE.1-EO.e.i)
Curriculum Guide, Units Math Colorado's Sample Instructional Units Resource Bank
Algebra I - One and Two Variable Inequalities - Unit 5
Essential Questions:
1. “Properties of solving” can be used to produce equivalent inequalities.
2. One-variable inequalities can be graphed on a number line, whereas two-variable inequalities can be graphed on the coordinate plane.
3. Compound and absolute value inequalities can be rewritten and solved as two separate inequalities.
4. Graphing solutions to inequalities in two-variables can be represented as half-planes
Curriculum Guide, Units Math Strasburg School District
Algebra I - Planning for High School Mathematics - All Units
Six (6) Algebra 1 Units in a single curriculum planning document and Curriculum at a Glance overview
Colorado's District Sample Curriculum Project By Content Area
Curriculum Guide, Units Math Colorado's Sample Instructional Units Resource Bank
Algebra I - Power to the Variable - Instructional Unit 2
Enhanced and Updated - March 31, 2014 - This updated instructional unit includes learning experiences, teacher and student resources, assessment ideas, and differentiation options. The storyboard document provides a roadmap of teaching activities and a possible performance assessment for the unit.
Inquiry Questions (Engaging- Debatable):
What are the parameters that affect gas mileage in a car and how would you model them? (MA10-GR.HS-S.2-GLE.2-EO.b.i)
What are the consequences of a population that grows exponentially?
Curriculum Guide, Units Math Colorado's Sample Instructional Units Resource Bank
Algebra I - Poynomials and Factoring - Unit 9
Essential Questions:
1. Polynomials can be simplified using properties of addition, subtraction, and multiplication.
2. Polynomials can be factored into a product of polynomials.
3. A product of polynomials can be expanded.
4. Graphs of polynomial equations can be utilized to find solutions.
Curriculum Guide, Units Math Strasburg School District
Algebra I - Properties of Exponents - Unit 7
Essential Questions:
1. Use properties of exponents to simplify exponential expressions.
2. Rational exponents can be simplified using properties of exponents and be rewritten in radical form.
3. Expressions in scientific notation can be simplified using properties of exponents.
4. Real-life situations can be represented using exponential expressions.
Curriculum Guide, Units Math Strasburg School District
Algebra I - Real Numbers and Expressions
Essential Questions: 1) Real number expressions are simplified by correct use of order of operations. 2) Real-world situations are represented by translating verbal terms into symbolic notation? 3) Equations and inequalities have a solution or solutions that will make them true statements, whereas expressions can have different outcomes depending upon the substituted input. 4) Data can be summarized graphically and by measures of central tendency and spread.
Curriculum Guide, Units Math Strasburg School District
Algebra I - Statistics Lie – Find Out How - Unit 3
Inquiry Questions (Engaging- Debatable):
Most people who die of lung cancer have an ashtray at home. Do ashtrays cause cancer?
What makes a statistic believable? What makes a statistic accurate? Is there a difference between the two?
What makes data meaningful or actionable? 3.1.IQ.1
Curriculum Guide, Units Math Colorado's Sample Instructional Units Resource Bank
Algebra I - Writing Linear Equations - Unit 4
Essential Questions:
1. A linear equation can be written when given a slope and the y-intercept, a slope and a point, or two points.
2. When given a graph of a line, an equation can be determined.
3. Knowledge of parallel or perpendicular lines can aid in developing linear equations.
A line-of-best-fit can be utilized to determine a model for real-world situations.
Curriculum Guide, Units Math Strasburg School District
Algebra I Units - ECBOCES
Nine Algebra I units covering: Review of Real Numbers, Linear Equations & Inequalities, Functions, Linear Functions, Systems of Equations, Polynomials, Quadratic Equations & Functions, Statistics, and Exponential & Radical Functions
Curriculum Guide, Lesson Plan, Units Math East Central BOCES
Algebra II - Functions and Equations - Unit 1
Essential Questions:
1. Equations, inequalities, and formulas are solved by using inverse operations.
2. Functions model relationships between two or more variables.
3. Transformations include translating and stretching/shrinking the graph horizontally or vertically.
Curriculum Guide, Units Math Strasburg School District
Algebra II - Systems of Equations - Unit 2
Essential Questions:
1. Systems of equations can be solved by graphing, substitution, elimination, and matrices.
2. The solution to a system of inequalities is called a feasible region.
3. Systems of equations are used to solve real world problems in finance and business.
Curriculum Guide, Units Math Strasburg School District
| 1,419 | 6,477 |
{"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}
| 4.09375 | 4 |
CC-MAIN-2017-34
|
longest
|
en
| 0.842531 |
https://www.thestudentroom.co.uk/showthread.php?t=3934967
| 1,527,264,088,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794867094.5/warc/CC-MAIN-20180525141236-20180525161236-00469.warc.gz
| 838,362,321 | 39,681 |
You are Here: Home >< Physics
# Calculating increase in diameter after a force is exerted watch
1. Question is:
A nylon cylinder of diameter 69.9mm is placed inside a steel tube having a slightly larger inner diameter. The nylon cylinder is then compressed by an axial force of 80 kN.
What shall be the minimal value of the inner diameter of the steel tube to avoid lateral contact stresses? Compression on inner walls of steel tube due to lateral expansion of nylon bar must be avoided.
Take E = 3.1GPa and Poisson’s ratio 0.4 for nylon. Required accuracy 0.01 mm.
My working so far, taking L as 1m:
Axial Strain = P/EA
= 80kN/(3.1GPa*0.25π0.0699)
= 6.72x10-4
ΔL=Axial Strain*L
= 6.72x10-4*1
= 6.72x10-4
To find increase in diameter:
Δd=v*d*Strain
=0.4*0.0669*6.72x10-4
=1.798x10-5
=0.01798mm
Is this correct?
2. I also cant figure out how I should be working out:
A steel bar of a square cross-section 67 mm on a side, is subject to an axial tensile force of 113 kN.
Young's modulus (modulus of elasticity) is 200 GPa for the steel.
Determine the strain in the bar with the required accuracy of 5%.
### Related university courses
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: March 6, 2016
Today on TSR
### Exam Jam 2018
Join thousands of students this half term
### Solo - A Star Wars Story
Poll
The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.
Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE
| 497 | 1,740 |
{"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}
| 3.78125 | 4 |
CC-MAIN-2018-22
|
latest
|
en
| 0.877133 |
http://math.stackexchange.com/questions/87349/exact-solutions-to-the-equation-sinx-n-cosx
| 1,469,633,576,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469257826908.63/warc/CC-MAIN-20160723071026-00142-ip-10-185-27-174.ec2.internal.warc.gz
| 167,345,743 | 17,488 |
# Exact solutions to the equation $\sin(x) = n \cos(x)$
Let $n$ be a positive integer. Can we precisely solve the equation $$\sin(x) = n\cos(x)$$ in $x$?
For $n=1$, we get $x=\pi/4$.
-
To add to the answer by @Fredrik, note that you don't lose any solutions by dividing by $\cos$ since both sides cannot be zero simultaneously. – S.D. Dec 1 '11 at 11:30
The set of solutions is $x_n+\pi\mathbb Z$, for some $x_n$ in $[\pi/4,\pi/2)$ depending on $n\geqslant1$, such that $x_1=\pi/4$, $x_n\lt x_{n+1}$ for every $n$ and $x_n=\pi/2-1/n+o(1/n)$ when $n\to\infty$. – Did Dec 1 '11 at 13:23
By the way: 18 minutes. – Did Dec 1 '11 at 13:26
The only rational values of $\tan(x)$ or $\cot(x)$ for rational $x$ in degrees are $0$ and $\pm 1$. See for instance
| 280 | 754 |
{"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}
| 3.921875 | 4 |
CC-MAIN-2016-30
|
latest
|
en
| 0.816629 |
http://oeis.org/A017714
| 1,575,920,197,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540521378.25/warc/CC-MAIN-20191209173528-20191209201528-00445.warc.gz
| 108,286,784 | 4,311 |
This site is supported by donations to The OEIS Foundation.
Please make a donation to keep the OEIS running. We are now in our 55th year. In the past year we added 12000 new sequences and reached 8000 citations (which often say "discovered thanks to the OEIS"). We need to raise money to hire someone to manage submissions, which would reduce the load on our editors and speed up editing. Other ways to donate
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A017714 Binomial coefficients C(n,50). 4
1, 51, 1326, 23426, 316251, 3478761, 32468436, 264385836, 1916797311, 12565671261, 75394027566, 418094152866, 2160153123141, 10468434365991, 47855699958816, 207374699821536, 855420636763836, 3371363686069236, 12736262814039336, 46252743903616536 (list; graph; refs; listen; history; text; internal format)
OFFSET 50,2 LINKS G. C. Greubel, Table of n, a(n) for n = 50..10000 FORMULA From G. C. Greubel, Nov 03 2018: (Start) G.f.: x^50/(1-x)^51. E.g.f.: x^50*exp(x)/50!. (End) MATHEMATICA Table[Binomial[n, 50], {n, 50, 5!}] (* Vladimir Joseph Stephan Orlovsky, Sep 25 2008 *) PROG (Sage) [binomial(n, 50) for n in range(50, 68)] # Zerinvary Lajos, May 23 2009 (Python) A017714_list, m = [], [1]*51 for _ in range(10**2): A017714_list.append(m[-1]) for i in range(50): m[i+1] += m[i] # Chai Wah Wu, Jan 24 2016 (PARI) for(n=50, 80, print1(binomial(n, 50), ", ")) \\ G. C. Greubel, Nov 03 2018 (MAGMA) [Binomial(n, 50): n in [50..80]]; // G. C. Greubel, Nov 03 2018 CROSSREFS Sequence in context: A172639 A231648 A017767 * A202044 A238319 A014943 Adjacent sequences: A017711 A017712 A017713 * A017715 A017716 A017717 KEYWORD nonn,changed AUTHOR STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified December 9 13:50 EST 2019. Contains 329877 sequences. (Running on oeis4.)
| 696 | 2,035 |
{"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}
| 3.578125 | 4 |
CC-MAIN-2019-51
|
latest
|
en
| 0.65959 |
http://www.ask.com/web?q=How+Do+You+Divide+Fractions%3F&o=2603&l=dir&qsrc=3139&gc=1
| 1,488,304,922,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-09/segments/1487501174167.41/warc/CC-MAIN-20170219104614-00599-ip-10-171-10-108.ec2.internal.warc.gz
| 292,721,705 | 15,106 |
Web Results
To divide fractions together, first flip the second fraction, turning it into its reciprocal. Multiply the numerators and multiply the denominators to get your answer. Reduce if necessary.
If given 1/2 รท 1/6, you would change 1/6 into its reciprocal, 6/1. Then you would multiply 1/2 X 6/1 to get your final answer.
## Dividing fractions: 3/5 ÷ 1/2 (video) | Khan Academy
Learn to divide two fraction (3/5 ÷ 1/2). The answer will be a mixed number.
## Dividing Fractions - Math is Fun
www.mathsisfun.com/fractions_division.html
Step 1. Turn the second fraction (the one you want to divide by) upside down (this is now a reciprocal). Step 2. Multiply the first fraction by that reciprocal. Step 3.
## Dividing Fractions | CoolMath4Kids
www.coolmath4kids.com/math-help/fractions/dividing-fractions
There's a really cool trick for these... FLIP and MULTIPLY! Check it out: That's it -- then GO FOR IT! Done! Look at another one: 6/11 divided by 1/2. Head on over ...
## Divide Two Fractions - WebMath
www.webmath.com/divfract.html
This page will show you how to divide two fractions. There are three combinations of this. 1) Dividing two "normal" fractions, 2) Dividing a mixed number by a ...
## Fractions - Dividing fractions - In Depth - Math.com
www.math.com/school/subject1/lessons/S1U4L5DP.html
Dividing by fractions is just like multiplying fractions, except for one additional step. To divide any number by a fraction: First step: Find the reciprocal of the ...
## How to Divide Fractions by Fractions: 12 Steps (with Pictures)
www.wikihow.com/Divide-Fractions-by-Fractions
How to Divide Fractions by Fractions. Dividing a fraction by a fraction might seem confusing at first, but it is really very simple. All you need to do is flip, multiply ...
| 476 | 1,782 |
{"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}
| 4.28125 | 4 |
CC-MAIN-2017-09
|
latest
|
en
| 0.850464 |
https://pinoycravings.com/bakeing/how-much-rice-do-i-cook-to-make-2-cups.html
| 1,660,012,257,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00398.warc.gz
| 430,639,968 | 18,596 |
# How much rice do I cook to make 2 cups?
Contents
RICE CONVERSIONS & EQUIVALENTS 1 cup uncooked white rice, or wild rice, yields 3 cups of cooked rice. 1 cup brown whole grain rice, yields 4 cups of cooked rice.. 1 cup long grain rice, yields 3 cups cooked rice. 1 cup dry pre-cooked instant rice, yields 2 cups cooked rice.
## How much rice do I need for 2 cups?
Measure one cup of long grain white rice into a cup and level it off. One cup of dry rice will make enough cooked rice for two to three adult servings. (Or two adults and two small children.) The cool thing about this recipe is it is proportional.
## Does 1 cup of dry rice make 2 cups of cooked rice?
Measure a quarter cup of uncooked rice per person, or a half cup per person for more generous servings. One cup of uncooked rice will yield approximately three cups cooked.
## How do you yield 2 cups of cooked rice?
Rice to Liquid Ratio
The most common ratio for rice to liquid is 1 to 2 or 1 cup rice to 2 cups of water which will yield 3 cups cooked rice. The formula is simple: 1-2-3. So if you were cooking 2 cups of rice, you would cook it with 4 cups of liquid to yield 6 cups of cooked rice.
## Is it 1 cup of rice to 2 cups of water?
The basic water to white rice ratio is 2 cups water to 1 cup rice. You can easily, double and even triple the recipe; just make sure you are using a pot large enough to hold the rice as it cooks and expands.
## Is 2 cups of rice a lot?
This one that talks about serving sizes specifically and recommends 1/2 cup uncooked per person or less: When it comes to rice the norm seems to be about ½ cup (90g) per person, although some people prefer to use a bit less – about 1/3 cup (60g) per person.
## How do I measure a cup of rice?
Put 16 tablespoons of rice into a clear glass cup. Put a piece of tape at the top of the rice level. Now you do have a measuring cup (sort of). You can measure your rice using any container as long as you keep the proportion of rice to water the same.
## How much does 1/3 cup dry rice make cooked?
1/3 cup dry makes roughly 1 cup cooked RightRice. So, the full bag will be roughly 4 cups of cooked RightRice. 5 of 8 found this helpful.
## How much water do you need for 4 cups of rice?
How much water do I use for 4 cups of rice? To make 4 cups of dry rice, you will need 6- 7 cups of water.
## How much water do I use for one and a half cups of rice?
The biggest mistake most people make which results in gluey rice is using the wrong rice to water ratio. The correct rice to water ratio is 1 : 1.5 (1 cup of rice to 1.5 cups of water).
THIS IS IMPORTANT: How do you cook Uncle Ben's 10 Minute Rice?
## Is 1 cup of rice a cup of water?
The correct rice to water ratio is 1:2. You will need 1 cup of rice and 2 cups of water (or any relative portion).
## How do you calculate cooked rice?
Whether it be white, brown, or wild – rice gains volume and weight after cooking. Different rices absorb different amounts of water, but for ease of estimation use the “3X rule” – 1 cup of dry rice (185 g) will give you about 3 cups of cooked rice (555 g) for the same macronutrient content.
| 822 | 3,146 |
{"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}
| 3.578125 | 4 |
CC-MAIN-2022-33
|
latest
|
en
| 0.951386 |
https://techblog.izotope.com/2015/09/
| 1,582,066,060,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875143815.23/warc/CC-MAIN-20200218210853-20200219000853-00541.warc.gz
| 595,930,700 | 16,176 |
## How Loud is Disaster Area?
Tedd Terry
Disaster Area, the “loudest band in the galaxy” from Douglas Adams’ The Restaurant at the End of the Universe, is remarkably loud.
So loud that their audience prefers to listen to them in a concrete bunker 37 miles away from the stage. So loud that Disaster Area play their instruments remotely from a spaceship that’s orbiting a different planet. So loud that they have trouble booking shows because their PA violates ordnance regulations.
If their audience prefers a reasonable OSHA-approved listening level of 83 dB SPL, and their listening bunker has one-foot thick concrete walls, how loud is it at ground zero of the Disaster Area stage if the concert took place on Earth?
We’ve done some napkin math to figure it out, so try it yourself and then compare numbers after the cut!
## How do we measure how loud something is?
Almost everyone is familiar with the decibel as an indicator of volume: more is louder. A large enough number means it hurts and listening to it long enough causes some permanent damage that makes it hard to hear quiet things for a while (or forever!). However, a decibel is much more than just an indicator of loudness: it’s an important measurement of distance between two values.
Decibels are in common usage in many fields to express ratios of difference that are otherwise cumbersome to throw around. Decibels have a few properties that make them really useful to us:
• They use a logarithmic scale, meaning that each doubling of amplitude corresponds to an increase of about 6 dB.
• They represent a ratio, so they’re always specified in terms of some reference value.
Since decibels measure a change in loudness, there are a few different ways of measuring them for different applications.
For example, the amplitude of sound in the real world is measured in dB SPL (sound pressure level). The amplitude of audio data in a computer is measured in dBFS (full scale). In dB SPL, 0 dB means 0.00002 Pascals: a tiny amount of air pressure. In dBFS, 0 dB means the largest sample value that can possibly be expressed by the wordlength of your bit depth, or a sample value of 1 if floating point is used.
For real world audio, SPL is measured with a sound pressure level meter (as one would expect), which is basically a calibrated microphone with some readout and parameters for how often a measurement is taken and whether or not some frequencies are filtered out before measuring. There’s a lot of pretty good SPL meter apps available for phones, which makes it pretty easy to measure your subway ride and e-mail yourself a CSV so you can nerd out on the data.
### dB SPL Reference
Because the decibel is all about the reference, we can build some meaningful perceptual references for dB SPL values. Here are some common world dB values (many of which are cribbed from this handy list):
• 0 dB SPL: threshold of hearing, a mosquito 10 feet away, the eardrum moves less than 1/100 the length of an air molecule
• 10 dB SPL: absolute silence, AT&T-Bell Laboratory “Quiet Room”
• 40 dB SPL: whispered conversation
• 60 dB SPL: normal conversation
• 83 dB SPL: average loudness in a THX-certified cinema program; perception of frequencies is equalized around this loudness (this is where you perceive bass and treble about as well as you perceive mids)
• 85 dB SPL: permanent hearing damage after eight hours of exposure
• 110 dB SPL: average loudness at a typical rock concert or construction site
• 120 dB SPL: threshold of pain, permanent hearing damage after less than a minute of exposure
This general range is why we say that human hearing has a dynamic range of 120 dB. We can go even farther and say that the effective dynamic range of our perception is even lower (probably about 60 dB) because we can’t hear stuff buried by everyday noise like HVACs or wind, and we don’t like listening to loud things for very long (or, at least, we probably shouldn’t). This is probably something to consider, given a recent trend toward placing an importance on high bit depth audio delivery formats, where the dynamic range can be 144 dB or above. In our DAWs we have an insane amount of dynamic range available at the mix stage: about 196 dB!
The “threshold of pain” bit is somewhat subjective, but really this is around the ceiling of where it’s safe for humans to experience sound without protection. The point is, sound is pressure, and levels past this threshold make bad things happen to your hearing forever. Of course, that doesn’t stop rock and roll:
• 130 dB SPL: Front row of an AC/DC concert
• 150 dB SPL: The Who’s sound system in 1976: measured at 120 dB 50m away from the speaker, audible 100 miles away
• 154 dB SPL: The loudest sound system on Earth (used to test if spaceflight vehicles can withstand the forces of takeoff)
• 155 dB SPL: Loud enough to blur vision and cause difficulty breathing
• 194 dB SPL: 1 ATM (14.7 PSI) of pressure, rupturing of eardrums and probably some other distressing events as well
194 dB is pretty much the ceiling for audio on Earth. Past this point, pressure essentially clips, which would probably sound awesome except for you’d be dead.
While we’re talking relative loudness, it’s interesting to note that a sound needs to be at least roughly 6 dB louder than whatever it’s competing with to be perceived, and probably 12 dB louder to hear it well. If your subway ride has a noise floor of 80 dB SPL (it probably does), you’re rocking your jams between 86 and 92 dB SPL to hear them. This phenomenon is known as auditory masking, and a more sophisticated model is exploited by audio codecs like MP3 to discard sound that cannot be perceived.
90dB SPL, by the way, is about where it becomes really important to pay attention to your sound level exposure: permanent hearing damage can occur after just two hours , and safe exposure time halves for every additional 3dB.
## OK so how loud is Disaster Area?
The cultured audience of Disaster Area fans sipping Pan Galactic Gargle Blasters inside their listening bunker are enjoying the noise at a refined, safe loudness of 83 dB SPL.
### The bunker’s walls
The mass of the bunker’s concrete walls absorbs much of the energy present outside the structure. We can quantify this property as transmission loss, which varies by building material and density.
We take a typical density of concrete (2400 kg/m3), make it 1 foot thick (.3 meters) and get a mass of 720 kg/m2. Eyeball that value on this handy chart and we get a transmission loss of about 54 dB. If the level in the room is 83 dB SPL, we can add the transmission loss to get the level outside the Disaster Area bunker: 137 dB SPL.
### Between the PA and the bunker
Sound halves in power every doubling of a distance (more or less: there are some caveats here about point sources and line sources and positioning against reflective surfaces but for our purposes we’ll consider the Disaster Area PA an unencumbered point source in space). A 2x change in acoustic power is about 6 dB.
There are 37 miles between the PA and the bunker, which is pretty close to the beautifully round value of 60,000 meters. There’s about 15 doublings between the stage and our Disaster Area concertgoers (log2 60000 ≈ 15.87).
15 doublings × 6 dB is 90 dB. Add that to the level just outside the bunker wall and we get the level at the stage: 227 dB SPL, louder than Earth’s atmosphere can support.
### What happens above 194 dB SPL?
Is it meaningful to measure SPL above 194 dB? We’ve already seen that we can use decibels to compare the ranges of acoustic and digital systems. We can abuse decibels a little further to make observations about the power of natural disasters, explosions, and other phenomena we can observe with sound (but would like very much to be standing far away from). This gives us a context for comparing a Disaster Area concert on Earth to some other similarly catastrophic event, like the eruption of Krakatoa or the yield of Ivy Mike.
We’re not really dealing with sound at this point: sound is the pushing and pulling of pressure waves and our only reference for the kind of noise Disaster Area makes is the shockwaves resulting from heavy ordnance. We have to consider Disaster Area’s stage as the source of an explosion and estimate the shockwave force of the event.
This means that, above 194 dB SPL, instead of losing 6 dB per doubling of distance, we look at how explosions behave, where the shockwave loses around 18 dB (at maximum — there’s some transfer here but we’re keeping it simple for napkin math).
So, considering that:
• There are 10 doublings between bunker and stage until we hit around 194 dB (the “threshold of atmosphere”): 10 × 6 dB = 60 dB
• There are 5 remaining doublings between that point and the stage. 5 × 18 dB = 90 dB
• We don’t care about the fancy pants math or reality, really, because we’re talking about an imaginary space band that competes with rocket launches for loudness.
The total level difference between bunker and stage is 150 dB, which brings Disaster Area to a very pretty 287 dB SPL.
For comparison:
• 194 dB SPL: 1 ATM (14.7 PSI) of pressure, rupturing of eardrums
• 200 dB SPL: Instantaneous human death from pressure waves
• 210 dB SPL: Explosion of 1 ton of TNT
• 220 dB SPL: Saturn V rocket launch
• 286 dB SPL: Eruption of Mount St. Helens, which knocked down trees for 16 miles around it and blew out windows ~200 miles away in the Seattle-Tacoma area
• 287 dB SPL: Disaster Area concert (stage)
• 310 dB SPL: Eruption of Krakatoa, which created an anti-node of pressure on the other side of the planet
That’s our best rough calculation. Did you get another number? Got an acoustic insight we’ve missed? Leave a comment and let us know!
| 2,198 | 9,762 |
{"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}
| 3.8125 | 4 |
CC-MAIN-2020-10
|
longest
|
en
| 0.940502 |
https://www.physicsforums.com/threads/hamilton-method.404960/
| 1,544,558,494,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-51/segments/1544376823702.46/warc/CC-MAIN-20181211194359-20181211215859-00616.warc.gz
| 980,216,535 | 14,661 |
Homework Help: Hamilton method
1. May 22, 2010
1. The problem statement, all variables and given/known data
[PLAIN]http://img411.imageshack.us/img411/4412/sssa.jpg [Broken]
2. Relevant equations
3. The attempt at a solution
Actually I have very basic knowledge of university physics and math, so the only things I've done are calculating Hamilton equations (I hope correctly)
q'=p2+p1
p'=-(q2+q1)
and f3 as I guess it was poisson brackets
f3=4q2p2-4q1p1
Now I don't have enough theoretical mechanics knowledge to move on, any help would be very welcome. Thanks!
Last edited by a moderator: May 4, 2017
2. May 22, 2010
vela
Staff Emeritus
You didn't get the equations right. Hamilton's equations are
$$\dot{p_i} &=& -\frac{\partial H}{\partial q_i}$$
$$\dot{q_i} &=& \frac{\partial H}{\partial p_i}$$
So you should get four equations in total.
3. May 23, 2010
Ok, so I get four equations (thank you for pointing it out):
q'1=p2
q'2=p1
p'1=-q2
p'2=-q1
Now, should I integrate them, and if yes, what should I do next to get solution?
4. May 23, 2010
vela
Staff Emeritus
The problem now is that you have differential equations involving two functions, q1 and p2 for example. You want to combine equations so that you get a differential equation that involves only one function, which you can then solve.
5. May 23, 2010
I'm not sure if I understand you correctly, but ok, this is what I get, but still can't see the whole idea
p1=-p''1
p2=-p''2
q1=-q''1
q2=-q''2
6. May 23, 2010
vela
Staff Emeritus
Those are differential equations you can solve. You actually only need two of them. If you find q1, for instance, you can use one of the original equations, p2=q1', to solve for p2.
7. May 23, 2010
I can't :D Well I think q1 and q2 should be trigonometric functions (cosines) while p1 and p2 (-sines), but I thought that q and p are some kind coordinate and moment functions
8. May 26, 2010
vela
Staff Emeritus
You should review how to solve basic differential equations. Mechanics is already difficult enough to learn on its own, but not having a good grounding in mathematics makes it even more so.
9. May 27, 2010
Yeah, I know, I actually more needed than wanted to solve this problem. I eventually managed to integrate equations (yeah, it turned out to be easy task after one glance into math book), but the easier way to show that functions were of the same system, was just to calculate poisson brackets of hamilton function and all other functions
{H;f1}={H,f2}={H,f3}=0
Anyway, thanks for help :)
10. May 27, 2010
vela
Staff Emeritus
Ah, of course. It's been so long since I've taken classical mechanics I had forgotten all about that.
| 771 | 2,677 |
{"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}
| 3.9375 | 4 |
CC-MAIN-2018-51
|
latest
|
en
| 0.928478 |
https://possiblywrong.wordpress.com/2011/07/02/representing-permutations/
| 1,628,100,811,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046154897.82/warc/CC-MAIN-20210804174229-20210804204229-00620.warc.gz
| 442,089,761 | 19,000 |
## Representing Permutations
Often, a mathematical puzzle or problem is interesting because the answer is interesting. The recent card tricks and other prisoner puzzles discussed lately are of this type. There is a solution when it seems none exists, or we can succeed with a much higher probability than intuition would suggest. As Peter Winkler puts it, these are “puzzles you think you must not have heard correctly.”
Other problems, like the one presented last week, are interesting because the method of solution is interesting. The actual answer may not be terribly surprising, but the process of arriving at the answer involves something surprisingly simple, elegant, and/or useful in other applications. (Dick Lipton at Gödel’s Lost Letter has an interesting article about historical examples of this in mathematics and computer science, where “the proof is more important than the result.”)
In last week’s problem, we were asked for the probability of successfully unlocking all of $n$ boxes, given that we may initially select and break into $k$ of them. The answer is rather unsurprising; in fact, it is what you might guess the answer to be without any thought: the probability is $k/n$. But I like this problem anyway, because showing that the probability is $k/n$ involves a very simple yet handy idea, that of the canonical cycle representation of a permutation.
First, we can represent the random distribution of keys into boxes by a randomly selected permutation $\pi \in S_n$, where box $i$ contains the key that unlocks box $\pi(i)$. There are several different ways to write such a permutation. For example, we can simply list the image elements $\langle\pi(1), \pi(2), ..., \pi(n)\rangle$ in order.
Another representation that is more convenient for this problem is to decompose the permutation into cycles, where each element is mapped to the element appearing immediately after it in the same cycle (wrapping around if necessary). Using parentheses to group elements in a cycle, the following notations represent the same permutation:
$(2, 4, 5)(1)(6, 3) = \langle 1, 4, 6, 5, 2, 3\rangle$
Cycles are important in this prisoner puzzle. If we assume WLOG that we break open boxes 1 through $k$, then we can unlock all of the remaining boxes if and only if every cycle in $\pi$ contains an element in ${1, 2, ..., k}$. We can find the desired probability by counting the number of permutations that have this property.
There are two disadvantages of cycle representations of a permutation. First, we need the parentheses to indicate which elements belong to the same cycle. Thinking like a computer scientist, if we want to store a permutation as a simple linear array of $n$ elements, then the cycle representation requires storing additional information about where the parentheses go.
The second disadvantage is that cycle representations are not unique. It does not matter in what order we write the cycles; nor does it matter which element we write first in any particular cycle. For example, all of the following represent the same permutation:
$(2, 4, 5)(1)(6, 3) = (1)(2, 4, 5)(3, 6) = (3, 6)(2, 4, 5)(1)$
The canonical cycle representation simply fixes a particular ordering of cycles and elements within each cycle, by writing each cycle with its least element first, and ordering the cycles from left to right in decreasing order of first element. For example, the last of the three representations above is the “canonical” one. (It is worth pointing out that an equally workable “canonical” representation would be to write each cycle with its largest element first, and order the cycles from left to right in increasing order of first element.)
This representation solves both of our problems: it is unique… and it does not require the parentheses! That is, each of the $n!$ possible lists of elements corresponds to a unique permutation whose canonical cycle representation has that same ordering of elements.
Coming back to the puzzle again, we can now restate the desired property of a permutation very simply in terms of its canonical cycle representation. We can unlock all of the remaining boxes if and only if the first element in the canonical cycle representation of the corresponding permutation is at most $k$. The number of permutations with this property is $k(n-1)!$, and so the probability is $k/n$.
This entry was posted in Uncategorized. Bookmark the permalink.
This site uses Akismet to reduce spam. Learn how your comment data is processed.
| 989 | 4,529 |
{"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": 18, "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}
| 4.09375 | 4 |
CC-MAIN-2021-31
|
latest
|
en
| 0.922993 |
https://rbseguide.com/class-8-maths-chapter-7-ex-7-1-english-medium/
| 1,725,882,511,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651098.19/warc/CC-MAIN-20240909103148-20240909133148-00025.warc.gz
| 449,399,466 | 12,867 |
# RBSE Solutions for Class 8 Maths Chapter 7 Construction of Quadrilaterals Ex 7.1
RBSE Solutions for Class 8 Maths Chapter 7 Construction of Quadrilaterals Ex 7.1 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 7 Construction of Quadrilaterals Exercise 7.1.
Board RBSE Textbook SIERT, Rajasthan Class Class 8 Subject Maths Chapter Chapter 7 Chapter Name Construction of Quadrilaterals Exercise Exercise 7.1 Number of Questions 5 Category RBSE Solutions
## Rajasthan Board RBSE Class 8 Maths Chapter 7 Construction of Quadrilaterals Ex 7.1
Question 1.
Construct a quadrilateral ABCD in which AB = 4.0 cm., BC = 6.0 cm., CD = DA = 5.2 cm. and AC = 8.0 cm.
Solution
Draw a rough sketch of the given measurements.
Steps of Construction
1. First of all we draw a line segment AB of length 4.0 cm.
2. Draw a arc of length 6.0 cm(RBSESolutions.com)from point B and an arc of radius 8.0 cm from point A. Both arcs cuts each other at point C. Join AC and BC. So, we obtained a triangle ABC.
3. Two arcs of radius 5.2 cm. each, are drawn from points A and C intersecting each other at point D. Join AD and CD.
Thus, we obtained a required quadrilateral ABCD.
Question 2.
Construct a quadrilateral JUMP in which JU = 3.5 cm., UM = 4.0 cm., MP = 5.0 cm., PJ = 4.5 cm. and PU = 6.5 cm.
Solution
Draw a rough sketch of the(RBSESolutions.com)given measurements :
Steps of Construction
1. First, we draw a line segment of length JU = 3.5 cm.
2. Draw an arc of radius 6.5 cm. from point U and an arc of radius 4.5 cm. from point J which cuts the first arc at point P. Join UP and JP. So, we obtained a triangle JUP.
3. Taking P as center draw(RBSESolutions.com)an arc of radius 5.0 cm. draw an arc of radius 4.0 cm. from point U. Both arc intersect each other at M.
4. Join PM and UM.
5. Thus, we obtained a required quadrilateral JUMP.
Question 3.
Construct a parallelogram MORE in which MO = 3.6 cm., OR = 4.2 cm., MR = 6.5 cm. Measure the remaining sides and write them in your notebook.
Solution
Draw a rough sketch of the given(RBSESolutions.com)measurements
Steps of Construction
1. First of all we draw a line segment MO = 3.6 cm.
2. Draw an arc of radius 4.2 cm. from point O and an arc of radius 6.5 cm. from point M which intersect each other at R. Join MR and OR. So, we(RBSESolutions.com)obtained a triangle MOR.
3. Draw an arc of radius 4.2 cm. from point M and an arc of radius 3.6 cm. from R which intersect each other at E. Join ME and ER. Thus, we get a required parallelogram MORE. Measure of remaining sides are RE = 3.6 cm. and ME 4.2 cm.
Question 4.
Construct a rhombus BEST in which BE = 4.5 cm. and ET = 6.0 cm. Then measure the diagonals BS.
Solution
Draw a rough sketch of the given measurements :
Steps of Construction
1. First of all we draw a line segment BE = 4.5 cm.
2. Draw an arc of radius 4.5 cm. from(RBSESolutions.com)point B and an arc of radius 6.5 cm from point E which intersect each other at T. Join BT and ET. So, we obtained a triangle BET.
3. Draw an arc of radius 4.5 cm. from point T and an arc of radius 4.5 from E which intersect each other at S. Join ES and TS.
Thus, we obtained a required rhombus BEST.
The measure of diagonal BS is 6.8 cm.
Question 5.
Construct a quadrilateral PQRS in which PQ = 4.4 cm., QR = 4.0 cm., RS = 6.4 cm., SP = 2.8 cm. and QS = 6.6 cm. Measure the diagonal PR.
Solution
First of all we draw a rough sketch of the given measurements :
Steps of Construction
1. First of all we draw a line of segment PQ = 4.4 cm.
2. Draw an arc of radius 2.8 cm. from point P and an arc of radius 6.6 cm. from point Q which intersect to each other of point S. Join PS and QS. So, we obtained a triangle PQS.
3. Taking S as centre draw an arc(RBSESolutions.com)of radius 6.4 cm. and taking Q as centre draw an arc of radius 4 cm. which intersect to each other at point R. Join RQ and RS.
Thus, we obtained a required quadrilateral PQRS.
Length of diagonal PR = 6 cm.
We hope the given RBSE Solutions for Class 8 Maths Chapter 7 Construction of Quadrilaterals Ex 7.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 7 Construction of Quadrilaterals Exercise 7.1, drop a comment below and we will get back to you at the earliest.
| 1,245 | 4,292 |
{"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}
| 4.75 | 5 |
CC-MAIN-2024-38
|
latest
|
en
| 0.840364 |
https://static.tinkutara.com/question/105815.htm
| 1,716,773,217,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00441.warc.gz
| 471,390,099 | 2,143 |
Integration Questions
Question Number 105815 by I want to learn more last updated on 31/Jul/20
Answered by john santu last updated on 01/Aug/20
$$\frac{{dx}}{{d}\theta}\:=\:\mathrm{1}+\mathrm{sin}\:\theta\:;\:\frac{{dy}}{{d}\theta}\:=\:\mathrm{sin}\:\theta \\$$$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\:\sqrt{\mathrm{sin}\:^{\mathrm{2}} \theta+\left(\mathrm{1}+\mathrm{sin}\:\theta\right)^{\mathrm{2}} }\:{d}\theta\: \\$$$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\:\sqrt{\mathrm{2sin}\:^{\mathrm{2}} \theta+\mathrm{2sin}\:\theta+\mathrm{1}}\:{d}\theta \\$$$$\sqrt{\mathrm{2}\:}\:\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\:\sqrt{\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{sin}\:\theta+\frac{\mathrm{1}}{\mathrm{2}}}\:{d}\theta \\$$$$\sqrt{\mathrm{2}}\:\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\:\sqrt{\left(\mathrm{sin}\:\theta+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\:{d}\theta\: \\$$$$\\$$
Commented by bemath last updated on 01/Aug/20
$${let}\:\mathrm{sin}\:\theta+\frac{\mathrm{1}}{\mathrm{2}}\:=\:{t}\:\Rightarrow{dx}\:=\:\frac{{dt}}{\mathrm{cos}\:\theta} \\$$$$\mathrm{cos}\:\theta\:=\:\sqrt{\mathrm{4}−\left(\mathrm{2}{t}−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{3}−\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}{t}} \\$$$${I}=\sqrt{\mathrm{2}}\:\underset{\mathrm{1}/\mathrm{2}} {\overset{\mathrm{1}/\mathrm{2}} {\int}}\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\:\sqrt{\mathrm{3}−\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}{t}\:}\:{dt}\:=\:\mathrm{0} \\$$
| 615 | 1,569 |
{"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}
| 3.765625 | 4 |
CC-MAIN-2024-22
|
latest
|
en
| 0.205235 |
https://www.teacherspayteachers.com/Product/6th-Grade-Equations-and-Inequalities-Game-6th-Grade-Math-Game-2387159
| 1,484,955,172,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560280888.62/warc/CC-MAIN-20170116095120-00121-ip-10-171-10-70.ec2.internal.warc.gz
| 993,225,325 | 57,441 |
Subjects
Resource Types
Common Core Standards
Product Rating
4.0
File Type
PDF (Acrobat) Document File
3.69 MB | 41 pages
PRODUCT DESCRIPTION
Equations and inequalities have never been this much fun! This 6th Grade I Can Math Game focuses on equations and inequalities with variables, and analyzing relationships between two variable, and provides students with practice in the form of multiple choice or short answer questions. QR codes (optional) make this game even more interactive as students get immediate feedback on their work! This game can also be used for independent practice, a guided math activity, whole group review, or for progress monitoring. The possibilities are endless.
With QR Codes!
SAVE over \$10 when you purchase the “I CAN” math game BUNDLE!!! {CLICK HERE}
Looking for MORE "I Can" Math Games for 6th Grade?
Dividing Fractions
Operations with Whole Numbers & Decimals
Ratios & Unit Rate
Expressions
Rational Numbers
Coordinate Planes
Equations & Inequalities
Area & Volume
Surface Area
Statistics & Probability
What is an “I CAN” game?
It is literally a can that contains a variety of “test like” questions. Each question comes in multiple choice format, or short response format (you pick the format that fits your needs).
Both sets of cards (multiple choice & short answer) come with and without QR codes. It is up to you if you want to incorporate them into your classroom!
This Game Includes….
2 different sized can covers (includes directions for students)
A guide on how to assemble the game
A guide on how to use this product
1 Checklist to help monitor student progress in this skill
40 Practice Questions – 2 formats for each question (multiple choice/written response)
QR codes (optional; for checking answers)
HOW TO USE THIS RESOURCE!!!
As a Group Game:
Place this “I Can” game out as one of your math centers. In groups of 2 or more, students can play this game against one another by seeing who can collect the most cards. To collect a card, students must answer the question correctly. If they check their answer and it is incorrect, another player can attempt to answer the question correctly and keep the card for themselves. If a student pulls an “I Can” card, they can add this to their pile of cards as a bonus and pull another card to solve.
As Independent Practice:
Students will pull a card from the can and solve it. They should record their answers on the “My Answers” sheet. When they are finished, they can check their answers using the answer key or QR code. It is a good idea to offer a reward/incentive for completing the set of cards, and/or mastering a certain percentage.
As a Progress Monitoring Tool:
When students complete this activity independently, have them keep track of their progress using the “Checklist” provided (or you can use the checklist and check their work yourself). You can then use this checklist to see if the student has mastered the focus skill. You can also use this information to help you determine if, and in what area, further instruction is needed.
Standards Covered by this Resource:
CCSS.MATH.CONTENT.6.EE.B.5
Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.
CCSS.MATH.CONTENT.6.EE.B.6
Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.
CCSS.MATH.CONTENT.6.EE.B.7
Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q and x are all nonnegative rational numbers.
CCSS.MATH.CONTENT.6.EE.B.8
Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams.
CCSS.MATH.CONTENT.6.EE.C.9
Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation. For example, in a problem involving motion at constant speed, list and graph ordered pairs of distances and times, and write the equation d = 65t to represent the relationship between distance and time.
Still have questions??? Check out my blog post to see MORE!! {CLICK HERE}
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
Kindergarten
Algebra 1
6th Grade I Can Math Games
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
LOOKING for MORE great RESOURCES???
CLICK HERE to see ALL of my Common Core LANGUAGE ARTS resources!
CLICK HERE to see ALL of my BUNDLES, and SAVE some \$\$!
If you would like to get updates on NEW and CURRENT resources...
FOLLOW me on Teachers Pay Teachers!
FOLLOW the One Stop Teacher Shop BLOG!
If you would like to contact me about pricing on a license for an ENTIRE school, county, or district, please email me at [email protected]
www.teacherspayteachers.com/Store/One-Stop-Teacher-Shop
As such, it is for use in one classroom only. This item is also bound by copyright laws. Redistributing, editing, selling, or posting this item (or any part thereof) on the Internet are all strictly prohibited without first gaining permission from the author. Violations are subject to the penalties of the Digital Millennium Copyright Act. Please contact me if you wish to be granted special permissions!
Total Pages
41
Included
Teaching Duration
N/A
4.0
Overall Quality:
4.0
Accuracy:
4.0
Practicality:
4.0
Thoroughness:
4.0
Creativity:
4.0
Clarity:
4.0
Total:
21 ratings
| 1,407 | 6,120 |
{"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}
| 3.578125 | 4 |
CC-MAIN-2017-04
|
latest
|
en
| 0.911894 |
https://testbook.com/question-answer/0-1-mole-of-hcl-equal-to--5fb7c192e65dbf31caeb7013
| 1,642,459,226,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320300624.10/warc/CC-MAIN-20220117212242-20220118002242-00717.warc.gz
| 626,643,735 | 32,354 |
# 0.1 mole of HCl equal to:
1. 3.65 gm
2. 36.5 gm
3. 18 gm
4. 1.8 gm
Option 1 : 3.65 gm
Free
CT 1: हिन्दी (आदिकाल)
6912
10 Questions 40 Marks 10 Mins
## Detailed Solution
Concept:
Mole Concept -
• The quantity one mole of a substance signifies 6.022 × 1023 number of particles of that substance which may be atoms, molecules, or ions.
• The quantity is a universal constant like Dozen, Gross, etc., and is known as Avogadro number, denoted by NA. after the scientist Amedeo Avogadro.
• Examples- In one mole of H2, there are 6.022 × 1023 molecules of hydrogen, and the number of atoms is 2 × 6.022 × 1023, as one molecule of hydrogen contains two-atom each.
• The mass of one mole of a substance is called its Molar Mass (M) or Atomic mass expressed in grams.
• The volume occupied by a mole of gas is 22.4 L at NTP, called its Molar Volume.
• The no. of moles (n) is calculated as =
The number of particles / Avogadro’s number.
Calculation:
Given:
• The number of moles of HCl = 0.1.
• The number of particles in one mole = 6.022 × 1023.
• The mass of one mole of any substance is its molar mass, so the mass of one mole of HCl = 1 + 35.5 = 36.5g.
• So, the mass of .1 mole =
$${36.5\over 10} = 3.65g$$
Hence, 0.1 moles of HCl equal to 3.65g.
$${X_1} = \;\frac{{no.\;of\;moles\;of\;sustance\;1}}{{total\;number\;of\;moles\;in\;the\;solution}}$$
$$\sum {X_i = 1}$$
| 496 | 1,378 |
{"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}
| 3.828125 | 4 |
CC-MAIN-2022-05
|
latest
|
en
| 0.711531 |
http://www.teach-nology.com/lessons/lsn_pln_view_lessons.php?action=view&cat_id=5&lsn_id=22179
| 1,544,845,212,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-51/segments/1544376826686.8/warc/CC-MAIN-20181215014028-20181215040028-00516.warc.gz
| 475,288,709 | 3,718 |
Lesson Plan : Parallel & Intersecting Line Segments, Rays, and Lines
Teacher Name: Ms. Gibson Grade: Grade 3 Subject: Math
Topic: Plane Geometry and basic Algebra Skills: Parallel & Intersecting Line Segments, Rays, and lines Content: Key Activities: Children model parallel and intersecting pairs of line segments, as well as polygons. They also hurt for real-world examples of parallel and intersecting figures. Key Vocabulary: -parallel -intersect Goals: * Identify line segments, lines, and rays. * Identify parallel and intersecting lines, line segments, and rays. * Model and draw parallel and intersecting pairs of lines, line segments, and rays. * Model geometric figures. Objectives: To guide children as they model and draw polygons parallel and intersecting line segments, rays, and lines. Materials: Vocabulary Chart ***Students can add key terms from this lesson*** -Straightedge -Blank paper Introduction: 1. Begin with a MAD-MINUTE! 2. Have a quick review from yesterdays lesson. ***See if children can represent 3 rays with the straws & twist ties*** 3. Tell the students briefly what we will be going over and what they should expect. Development: 1. Discussing Parallel and Intersecting Line Segments, Rays, and Lines. -Have the students use their straightedge to draw a line at the top and bottom of their blank paper. *After the students have done so, explain to them that they had just drawn parallel lines... -Have some of the students come up to the board and write what they know about line segments and parallel lines. (what are examples of parallel lines inside and outside of the classroom?) -Explain to the class about intersecting lines. ***Write on the board the word and definition.*** *****ACTIVITY***** Practice: 1. Rope Activity! -Have several different lengths of light rope, elastic rope, etc. A. have 3 pairs of students act as points. -Each pair of students will represent a line segment by stretching a rope between them. -Position some of the children so that the lines are both parallel and intersecting. -Ask the students questions and get the students to try a few different things and explore. 2. Geometry Hunt! (Pg. 130) -have the students look around the classroom and in the hallways to try to find parallel lines and intersecting lines. -have the students record them in the appropriate area of their worksheet. Accommodations: Vocabulary chart ***Add key terms to keep th chart growing! Checking For Understanding: Math journal (pg. 130) -if children are able to find examples of both parallel and intersecting lines, along with being able to draw at least on example then they are on track. Closure: Review what we went over and go over all of the vocabulary terms.
| 588 | 2,719 |
{"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-2018-51
|
latest
|
en
| 0.916752 |
https://math.stackexchange.com/questions/1022879/geometric-interpretation-of-the-cross-ratio?noredirect=1&lq=1
| 1,582,883,031,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875147116.85/warc/CC-MAIN-20200228073640-20200228103640-00423.warc.gz
| 439,777,407 | 36,584 |
# Geometric Interpretation of the Cross-Ratio
The cross ratio of 4 points $A,B,C,D$ in the plane is defined by
$$(A,B,C,D) = \frac{AC}{AD} \frac{BD}{BC}$$
And it's a ratio which is preserved under projections, inversions and in general, by Möbius-Transformations.
Although I can see it's utility and power, I cannot see a geometric definition or intuition for the cross ratio. Can someone give me any insights about this?
EDIT: To be more specific, I'd like to express the cross-ratio as the length of some segment constructible with straightedge-and-compass.
Draw the circle passing through $B,C,D$ and the circle passing through $A,C,D$ (or the line for three collinear points). The argument of the cross-ratio $\arg(A,B,C,D)$ is the angle between the two circles* where they meet at $C$.
To work this out, notice that the construction and the answer you get are invariant under Möbius maps, so you can make everything simple before you calculate by putting $D$ at $\infty$, $C$ at $0$, and $B$ at $1$.
To work out what the modulus of the cross-ratio should be, just swap some of the points over and repeat. For example, $(A,C,B,D)=1-(A,B,C,D)$ so $\arg[1-(A,B,C,D)]$ gives the angle at $B$ between the circles passing through $A,B,D$ and $B,C,D$.
There are several other angles that you could measure, but they are all related by various bits of spherical geometry (I find it easiest to think about this stuff on a sphere, by stereographic projection) in such a way that knowing two of them tells you all the rest.
$\ast$ There's some ambiguity in which angle to take, since there are two choices, and what counts as positive or negative. This can be resolved by checking for each circle the order the three marked points come in, and whether the fourth point is inside or outside the circle; details left as an exercise for the reader (translation: I can't be bothered to write them down).
• Thank you very much! That's exactly the kind of solution I was looking for. – Jonas Gomes Jan 31 '15 at 20:01
What you have written is the "classical" cross-ratio of projective geometry, which is defined not for any old 4-tuple of points in the plane but only for a 4-tuple of points that lie on some line. The geometric meaning is that the cross-ratio is a complete "projective" invariant. For example, if you are looking in perspective at 4 collinear points $A,B,C,D$, and if you wish to know whether there is another perspective from which those 4 points are equally spaced in order along a line, then that other perspective exists if and only if the cross ratio of $A,B,C,D$ equals the cross-ratio of four equally spaced points on the line such as $1,2,3,4$, namely $\frac{1-3}{1-4} \frac{2-4}{2-3} = \frac{4}{3}$.
Our mammalian brains know this: if our eye looks at a perspective picture of a line with marks labelled by the integers in order, it won't look like a number line to us unless the cross-ratios of all successive 4-tuples are equal to $\frac{4}{3}$. There's a nice explanation of this in John Stillwell's book "The Four Pillars of Geometry".
But for a general 4-tuple of points in the plane, your formula is not correct. Instead you should think of $A,B,C,D$ as being four points in the complex plane $\mathbb{C}$, and an expression like $AC$ should be replaced by the complex valued difference, and then one uses complex valued multiplication and division to obtain the complex valued cross-ratio formula $$\frac{A-C}{A-D} \cdot \frac{B-D}{B-C}$$
• Thanks for you answer, I should've written more carefully the expression for the cross ratio, indeed. I understand the cross ratio is a complete projective invariant, but I can't find a geometrical interpretation for the number $(A,B,C,D)$. For example, given three points $P,Q,R$ with $P-Q-R$ we can define the geometric mean of $P,Q,R$ by $\sqrt{PQ * QR}$ and a geometric meaning for this would be the height of the right triangle of hypotenuse $PR$ and with projections $PQ$ and $QR$. That's the sort of interpretation I'm looking for. – Jonas Gomes Nov 15 '14 at 21:46
• @JonasGomes: One geometric meaning for the cross-ratio, in somewhat the sense that you are asking, is that there is a unique Mobius transformation that takes $A \to 0$, $B \to \infty$, and $D \to 1$, and that unique Mobius transformation takes $C$ to the cross-ratio. – Lee Mosher Nov 16 '14 at 3:47
• And can you give a more geometrical meaning? Maybe the measure of some segment constructible with edge-and-compass? – Jonas Gomes Nov 16 '14 at 4:48
• @JonasGomes: I don't know how to do anything like that. Cross-ratio is a conformal invariant. The group of conformal transformations (Mobius transformations) is quite a bit larger than, say, the group of Euclidean isometries or even the group of Euclidean similarities. So the invariants of conformal geometry---such as cross-ratio---are expected to be quite a bit less rigid than isometry invariants or similarity invariants. – Lee Mosher Nov 16 '14 at 14:42
Let's consider the cross-ratio in the complex plane so that it looks like $$(A,B,C,D)=\frac{A-C}{A-D}\cdot\frac{B-D}{B-C}$$
Given trhee point $$A,B,C$$ in $$\widehat{\mathbb C}=\mathbb C\cup\{\infty\}$$ there is a unique Moebius transformation $$z\mapsto \frac{az+b}{cz+d}$$ that sends $$A\to 1, D\to 0, C\to\infty$$ and this is exactly $$f(z)= \frac{A-C}{A-D}\cdot\frac{z-D}{z-C}$$
Thus $$(A,B,C,D)$$ as defined above is just $$f(B)$$. In other words, the cross-ratio of four points is the image of one point once three of them are mapped to $$0,1,\infty$$ via a Moebius map.
Thus, any geometric quantity that you can "compute" from the four points $$0,1,\infty,z$$ by means that are invariant under Moebius maps, is the same quantity that you compute for any four points with $$(A,B,C,D)=z$$. For instance, the three angles of the triangle with vertices $$0,1,z$$.
Moreover, if you consider $$\widehat{\mathbb C}$$ as the boundary of the hyperbolic space $$\mathbb H^3$$, then the points $$A,B,C,D$$ are the vertices at infinity of a so-called ideal tetrahedron. The cross-ratio is a complete isometry invariant of geodesic ideal tetrahedra. Thus any metric (hyperbolic metric) quantity you compute (e.g. distances between opposite edges) depends only on the cross-ratio. You can find more about this viewpoint in any book of hyperbolic geometry.
It was my Conjecture.
If a right circular cone in 3-space of semi-vertical angle $\alpha$ has a pencil of four
rays/generators (OA,OB,OC, OD)through its vertex O, then,
the cross-ratio of their projection on an arbitrary plane is a projective invariant, is some trigonometric or other function of $\alpha$.
Earlier thoughts ( Drexel Univ. Math Forum geometry.puzzles) :
| 1,777 | 6,694 |
{"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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.859375 | 4 |
CC-MAIN-2020-10
|
latest
|
en
| 0.944586 |
https://www.gradesaver.com/textbooks/math/trigonometry/CLONE-68cac39a-c5ec-4c26-8565-a44738e90952/chapter-1-trigonometric-functions-section-1-1-angles-1-1-exercises-page-9/91
| 1,720,873,142,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00453.warc.gz
| 654,216,220 | 12,808 |
## Trigonometry (11th Edition) Clone
130$^{\circ}$
Since this number is larger than 360°, to find the smaller coterminal positive angle we need to subtract 360 to get an angle less than 360 850° -360° = 490° 490° -360° = 130$^{\circ}$
| 76 | 235 |
{"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}
| 3.671875 | 4 |
CC-MAIN-2024-30
|
latest
|
en
| 0.765323 |
https://www.techylib.com/en/view/vijay0021/electricity.pdf_2
| 1,618,246,473,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00624.warc.gz
| 973,452,270 | 12,069 |
# ELECTRICITY.pdf
Internet and Web Development
Apr 6, 2021 (6 days and 10 hours ago)
3 views
ELECTRICITY.pdf
https://www.entrancei.com/
https://www.entrancei.com/ncert
-
solutions
-
class
-
10
-
science
https://www.entrancei.com/
https://www.entrancei.com/ncert
-
solutions
-
class
-
10
-
science
ELECTRICITY
Electricity and its Effect (notations)
Physical Quantity
Symbols
SI unit
Voltage (potential difference)
V
Volt (V)
Power
P
Watt (W)
Charge
Q
Coulomb (C)
Work or Energy
W
Joule (J)
Resistance
R
Ohm (
)
Current
I
Ampe
re (A)
Resistivity
Ohm metre (
m)
Laws of electric forces:
(i)
Like charges repel and unlike charges attract each other.
(ii)
Charges of a conductor reside on its outer surface.
Current:
The rate of flow of charges (Q) through a conductor is cal
led current (I) and is given
by.
Current =
Time
charge
or
t
Q
I
=
. The SI unit of current is ampere (A).
1 Ampere
second
1
coulomb
1
=
The current flowing through a circuit is measured by a device called ammeter.
Ammeter is
connected in series with the conductor. The direction of the current is taken as the
direction of the flow of positive charge and opposite to the flow of electrons through the
conductor.
Electric cell:
It is the simplest form of arrangement to maintain a
constant potential
difference between two points.
Electromotive force:
The potential difference at the terminals of cells in an open
circuit is called electromotive force (emf) and is denoted by letter E.
Potential difference is the work done in bringin
g a unit charge from one place to
another.
charge
work
Difference
Potential
=
,
(C)
Coulomb
1
)
(J
Joule
1
(V)
Volt
1
=
Ohms law:
At any constant temperature the current (I) flowing through a conductor is
directly proportional to the potential difference (V) across it. Mathematical
ly,
I
V
vice
-
versa
V
I
or
V = RI
R
V
I
I
V
R
=
=
,
where
R
Resistance,
V
Voltage (P.D.),
I
Current
Symbols of a few commonly used components in Circuit Diagrams
Component
Symbol
Component
Symbol
An electric cell
Electric bulb
Battery of cells
A resistance
https://www.entrancei.com/
https://www.entrancei.com/ncert
-
solutions
-
class
-
10
-
science
Plug key or switch
(open)
or
Variable resistance
(R
heostat)
or
A closed plug or
switch
or
Ammeter
A
+
A wire joint
Voltmeter
V
+
Wires crossing
Galvanometer
G
+
Resistance:
Resistance is a property of a conductor by virtue of which it opposes the
flow of electricity through it. Resistance is measured in Ohms
(
)
. Resistance is a scalar
quantity.
Conductor:
Low
-
resistance material which allows the flow of electric current through
it is called a conductor. All metals are conductors except Hg and Pb etc.
Resistor:
High
-
resistance materials are called resistors. Resistors become hot when
current fl
ows through them (nichrome wire is a typical resistor).
Insulator:
A material which does not allow heat and electricity to pass through it is
called an insulator. Rubber, dry wood etc., are insulators.
Equivalent Resistance:
A single resistance which can
replace a combination of
resistance
s
such that current through the circuit remains the same is called equivalent
resistance.
Law of Combination of Resistances in Series:
When number of resistances are
connected in series, the equivalent resistance is equ
al to the sum of the individual resistances.
3
2
1
V
V
V
V
+
+
=
3
3
2
2
1
1
,
,
,
IR
V
IR
V
IR
V
IR
V
=
=
=
=
3
2
1
IR
IR
IR
IR
+
+
=
n
R
R
R
R
R
+
+
+
=
.....
3
2
1
Things to remember in series connection
(a)
When a number of resistances are connect
ed in series, the equivalent or resultant
resistance is equal to the sum of individual resistances and resultant resistance is greater than
any individual resistance.
https://www.entrancei.com/
https://www.entrancei.com/ncert
-
solutions
-
class
-
10
-
science
(b)
If
n
resistances each of value
R
are connected in series, the equivalent resistance
R
e
is given by:
R
e
=
R
+
R
+
R
..........
n
times
R
e
=
nR
R
e
= Number of resistors × resistance of each resistor
(c)
Equal current flows through each resistance and it is also equal to the total current
in the circuit. This is because there i
s no other path along which the current can flow.
(d)
The potential difference across the ends of the combination is distributed across
the ends of each of the resistances. The potential difference across any one of the resistances
is directly proportiona
l to its resistance.
(e)
The equivalent resistance when used in place of the combination of resistances
produces the same current with the same potential difference applied across its ends.
(f)
When two or more resistances are joined in series, the resul
t is the same as
increasing the length of the conductor. In both cases the resultant resistance is higher.
(g)
In a series combination, the equivalent resistance is greater than the greatest
resistance in the combination.
Law of Combination of Resistance
s in Parallel:
If resistance
.....
,
,
,
3
2
1
R
R
R
etc are
connected
in parallel
then the equivalent resistance (
R
) is given by
3
2
1
I
I
I
I
+
+
=
3
3
2
2
1
1
,
,
,
R
V
I
R
V
I
R
V
I
R
V
I
=
=
=
=
3
2
1
R
V
R
V
R
V
R
V
+
+
=
n
R
R
R
R
R
1
.....
1
1
1
1
3
2
1
+
+
+
=
Things to remember in parallel connection
(a)
When a number of resistances are connected in parallel, the reciprocal of the
equivalent or resultant resistance is equal to the sum of reciprocals of the individual
resistances and is always smaller than th
e individual resistances. This is because there are a
number of paths for the flow of electrons.
https://www.entrancei.com/
https://www.entrancei.com/ncert
-
solutions
-
class
-
10
-
science
(b)
If there are n resistances connected in parallel and each resistance has a value
of
R
n
R
R
R
R
e
..........
1
1
1
1
+
+
=
times
R
n
R
e
=
1
n
R
R
e
=
resistors
of
number
resistor
each
of
Resistance
=
e
R
(c)
The potential difference across each resistance is the same and is equal to the
total potential difference across the combination.
(d)
The main current divides itself and a different current flows th
rough each resistor.
The maximum current flows through the resistor having minimum resistance and vice versa.
(e)
If an equivalent resistance
R
e
is connected in place of combination, it produces
the same current for the same potential difference applied a
cross its ends.
(f)
In a parallel combination, the equivalent resistance is lesser than the least of all
the resistances.
(g)
If two resistances
R
1
and
R
2
are connected in parallel then
2
1
2
1
2
1
1
1
1
R
R
R
R
R
R
R
e
+
=
+
=
s
resistance
two
of
Sum
s
resistance
two
of
Product
2
1
2
1
=
+
=
R
R
R
R
R
e
(h)
If ther
e are
n
resistors each of resistance
R
.
Let
R
S
be the resultant resistance of
series combination and
R
p
be the resultant resistance of parallel combination.
Then,
R
S
= nR
R
p
=
n
R
2
/
n
n
R
nR
R
R
p
S
=
=
.
Electrical
energy:
Capacity of the flowing electricity to do work is called its electrical
energy.
Electrical energy (work) =
R
t
V
Pt
Rt
I
t
I
V
2
2
=
=
=
The SI unit of electrical energy is Joule.
One Joule is the amount of energy
consumed when an electrical appliance
of one watt rating is used for one second. The
commercial (practical) unit of electrical energy is kilowatt
-
hour (kWh).
Power,
R
V
R
I
VI
t
W
P
2
2
=
=
=
=
https://www.entrancei.com/
https://www.entrancei.com/ncert
-
solutions
-
class
-
10
-
science
The SI unit of electric power is watt (W).
The power of a machine doing work at the
rate of 1 Joule p
er second is equal to one watt.
Electrical energy = Electrical power × Time.
Important Formulae:
1.
Coulomb’s law
2
2
1
r
q
q
K
F
=
(
k
is constant of
proportionality
)
1
q
and
2
q
= two electric charges
r
= distance between two electric charges
F
= Force
2.
V
W
Q
Q
V
W
Q
W
V
=
=
=
;
;
.
.
d
p
V
=
W
= work done,
Q
= Quantity of charge transferred
3.
R
V
I
I
V
R
I
R
V
=
=
=
;
;
V
=
pd
;
R
= Resistance,
I
= current.
4.
l
A
R
A
I
R
=
=
;
R
= Resistance;
l
= length;
A
= Area of cross section;
= rho, a constant known as
resistivity
5.
Series combination
n
R
R
R
R
R
+
+
+
=
.....
3
2
1
6.
Parallel combination
n
R
R
R
R
R
1
.....
1
1
1
1
3
2
1
+
+
+
=
For equal resistances
nR
Rs
=
(For series co
nnection)
n
R
Rp
=
(For parallel connection)
2
n
Rp
Rs
=
Rs
= Effective resistance i
n
series
Rp
= Effective resistance i
n
parallel
n
= number of resistors
R
= Resistance of each resistor
7.
Time
consumed
Energy
time
work
Power
;
=
=
=
t
W
P
8
.
W
=
V
×
I
×
t
; Power =
potential difference
× current × time
)
(
2
Rt
I
W
=
=
R
t
V
W
2
9.
P
=
V
×
I
; Power =
potential difference
× current
https://www.entrancei.com/
https://www.entrancei.com/ncert
-
solutions
-
class
-
10
-
science
10.
R
I
P
=
2
; Power =
resistance
(current)
2
11.
R
V
P
2
=
;
resistance
)
difference
(potential
Power
2
=
12.
Electric energy =
P
×
t
; electric energy = power × time
https://www.entrancei.com/
https://www.entrancei.com/ncert
-
solutions
-
class
-
10
-
science
| 2,802 | 9,172 |
{"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}
| 3.8125 | 4 |
CC-MAIN-2021-17
|
latest
|
en
| 0.908845 |
https://gmatclub.com/forum/what-is-the-ratio-of-the-surface-area-of-a-cube-to-the-surfa-60832.html?fl=similar
| 1,495,697,823,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463608004.38/warc/CC-MAIN-20170525063740-20170525083740-00270.warc.gz
| 772,817,906 | 61,561 |
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 25 May 2017, 00:37
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# What is the ratio of the surface area of a cube to the surfa
Author Message
TAGS:
### Hide Tags
SVP
Joined: 04 May 2006
Posts: 1902
Schools: CBS, Kellogg
Followers: 23
Kudos [?]: 1122 [1] , given: 1
What is the ratio of the surface area of a cube to the surfa [#permalink]
### Show Tags
04 Mar 2008, 00:26
1
KUDOS
6
This post was
BOOKMARKED
00:00
Difficulty:
45% (medium)
Question Stats:
54% (01:49) correct 46% (00:41) wrong based on 211 sessions
### HideShow timer Statistics
What is the ratio of the surface area of a cube to the surface area of a rectangular solid identical to the cube in all ways except that its length has been doubled?
A. 1/4
B. 3/8
C. 1/2
D. 3/5
E. 2
OPEN DISCUSSION OF THIS QUESTION IS HERE: what-is-the-ratio-of-the-surface-area-of-a-cube-to-the-surfa-127459.html
[Reveal] Spoiler: OA
_________________
Last edited by Bunuel on 27 Mar 2014, 01:08, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
Manager
Joined: 31 Oct 2007
Posts: 113
Location: Frankfurt, Germany
Followers: 2
Kudos [?]: 65 [1] , given: 0
### Show Tags
05 Mar 2008, 21:41
1
KUDOS
1
This post was
BOOKMARKED
One side surface area of a cube = x*x = x^2
Total 6 sides = 6x^2
As for the rectangular, Height (H) and Width (W) are same as Cube, x. Only Length = 2x.
L x H = 2x * x = 2x^2 ----> 4 sides = 2x^2 * 4 = 8x^2
W * H = x * x = x^2 ------> 2 sides = x^2 * 2 = 2x^2
Total 6 sides = 8x^2 + 2x^2 = 10x^2
Ratio of cube area to rectangular area = 6x^2 / 10x^2 ----> 6/10 ----> 3/5 (E)
_________________
Jimmy Low, Frankfurt, Germany
Blog: http://mytrainmaster.wordpress.com
GMAT Malaysia: http://gmatmalaysia.blogspot.com
Manager
Joined: 14 Jan 2013
Posts: 153
Concentration: Strategy, Technology
GMAT Date: 08-01-2013
GPA: 3.7
WE: Consulting (Consulting)
Followers: 4
Kudos [?]: 285 [0], given: 30
Re: What is the ratio of the surface area of a cube to the [#permalink]
### Show Tags
26 Mar 2014, 17:01
Guys,
Can some one draw this for me....
~M14
_________________
"Where are my Kudos" ............ Good Question = kudos
"Start enjoying all phases" & all Sections
__________________________________________________________________
http://gmatclub.com/forum/collection-of-articles-on-critical-reasoning-159959.html
http://gmatclub.com/forum/percentages-700-800-level-questions-130588.html
http://gmatclub.com/forum/700-to-800-level-quant-question-with-detail-soluition-143321.html
Math Expert
Joined: 02 Sep 2009
Posts: 38859
Followers: 7728
Kudos [?]: 106073 [2] , given: 11607
Re: What is the ratio of the surface area of a cube to the [#permalink]
### Show Tags
27 Mar 2014, 01:10
2
KUDOS
Expert's post
2
This post was
BOOKMARKED
Mountain14 wrote:
Guys,
Can some one draw this for me....
~M14
What is the ratio of the surface area of a cube to the surface area of a rectangular solid identical to the cube in all ways except that its length has been doubled?
A. 1/4
B. 3/8
C. 1/2
D. 3/5
E. 2
30 second approach:
A cube has 6 faces, suppose each has an area of 1, so surface area of the cube will be 6;
A rectangular solid identical to the cube in all ways except that its length has been doubled is just complicated way of saying that the rectangular solid is built with two cubes, hence it has 4+4+2=10 faces of a little cube (just imagine to cubes put one on another) and thus the surface area of the rectangular solid will be 10;
Ratio: 6/10=3/5.
OPEN DISCUSSION OF THIS QUESTION IS HERE: what-is-the-ratio-of-the-surface-area-of-a-cube-to-the-surfa-127459.html
_________________
Re: What is the ratio of the surface area of a cube to the [#permalink] 27 Mar 2014, 01:10
Similar topics Replies Last post
Similar
Topics:
2 If the surface area of a certain cube is 2400, what is its volume? 4 20 Jan 2017, 09:03
2 What is the surface area of a cube whose edges measure (1/6)? 3 16 Nov 2016, 07:32
2 If the surface area of cube A is 4 times the surface area of cube B, 5 23 Sep 2016, 06:15
If a cube has a volume of 125, what is the surface area of one side? 2 09 Sep 2016, 03:55
12 What is the ratio of the surface area of a cube to the surfa 15 28 Feb 2017, 18:45
Display posts from previous: Sort by
# What is the ratio of the surface area of a cube to the surfa
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
| 1,582 | 5,190 |
{"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-2017-22
|
longest
|
en
| 0.880811 |
http://www.algebra.com/algebra/homework/Coordinate-system/Coordinate-system.faq.question.157577.html
| 1,369,555,123,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368706637439/warc/CC-MAIN-20130516121717-00036-ip-10-60-113-184.ec2.internal.warc.gz
| 303,878,998 | 4,688 |
# SOLUTION: The steepness, or grade, of a road is expressed as a percent. If a road rises 3 feet for every 24 horizontal feet, what is the sope of the road? what percent grade is this?
Algebra -> Algebra -> Coordinate-system -> SOLUTION: The steepness, or grade, of a road is expressed as a percent. If a road rises 3 feet for every 24 horizontal feet, what is the sope of the road? what percent grade is this? Log On
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Algebra: Coordinate systems, graph plotting, etc Solvers Lessons Answers archive Quiz In Depth
Question 157577: The steepness, or grade, of a road is expressed as a percent. If a road rises 3 feet for every 24 horizontal feet, what is the sope of the road? what percent grade is this? Answer by Earlsdon(6291) (Show Source): You can put this solution on YOUR website!Remember...slope = rise over run. Rise is 3 ft. and run is 24 ft, so slope = 3/24. Simplify to 1/8 and the decimal equaivalent is 0.125 which, changed to a percent = 12.5% The grade is 12.5%
| 310 | 1,173 |
{"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}
| 3.75 | 4 |
CC-MAIN-2013-20
|
latest
|
en
| 0.917487 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.