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Melody
Nom d'utilisateur
Melody
But
118730
Membership
Stats
Questions
900
Réponses
33648
921 Questions
34316 Answers
+1
1
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+118730
Mathigon website
I just discovered a site called mathigon
https://mathigon.org/dashboard
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Melody
2 avr. 2022
-8
610
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+118730
Reporting problems.
There are a number of people having problems at the moment.
If you lodge an error can you please give all relevant information that you can think of.
Where you are in the world - the city or the country - is often very important information.
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Melody
16 mars 2022
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8
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5
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How to make an empty rectangle in Latex
I just worked out how to make an empty box of various horizonal size easily in LaTex.
This is not a recommended method, but other methods that I have tried here have not worked.
This coding is really simple. Just put letters inside
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Melody
15 mars 2022
-3
873
3
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HAPPY ANNIVERSARY CHRIS! (CPhill)
I just realized that you became a member here exactly 8 years ago today!
It has been great to have you here all that time
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Melody
13 mars 2022
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850
3
+118730
WELCOME BACK CPHILL !
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Melody
11 févr. 2022
-5
719
2
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Very interesting question. No high level maths required.
This question was posted a couple of days ago.
https://web2.0calc.com/questions/geometry_73561
I think it is really interesting. It looks much more difficult than it really needs to be.
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Melody
1 févr. 2022
-7
888
11
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Members have a better experience. Their questions get more attention.
Almost everyone is posting as a guest.
If you post as a member your questions will have a far greater chance of a timely and helpful response.
If you are a genuine learner and interact with us, you will become known and answerers will do
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Melody
9 janv. 2022
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1
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HAPPY BIRTHDAY CHRIS!
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Melody
20 déc. 2021
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Some guidelines for question askers.
Please note:
1) Ask only one question per post.
2) When asking a question talk about what you have tried for yourself. Then people will take you more seriously and will be more able and willing to help.
3) If you get an answer
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Melody
4 nov. 2021
-9
1093
1
+118730
Remeniscing - a little historical humour.
I just found this old post that some of you, who have been around for a while, might find amusing.
https://web2.0calc.com/questions/look-at-the-new-leader-board
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Melody
29 oct. 2021
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#1
+118730
0
What is the domain of (fg)(x)?
Your question doesn't make sense so I assume this notation is new to you.
You probably mean the domain f(g(x)) and you got it from a different question.
The domain is all the values that x is allowed to be.
(The range would be all the values that y is allowed to be.)
I hope this helps. You can ask questions like this on the back of other posts if you want to.
Melody
5 déc. 2013
#4
+118730
0
This is the graph of y=sin
-1
x
It looks like it finishes early, I don't know why, but the end points are (-1,-pi/2) and (1, pi/2)
http://gyazo.com/7f6851adf451e3505295f474e55dfd33
I am going to add more but there are mistakes in my last post so I will fix them first and I will tell you when they are fixed
It was actually the graphing on desmos that showed me my error, which is exactly what Demos is great for.
this is a graph of both on the same grid, I am short of time so I must get this post finished.
http://gyazo.com/a8e1bcc4426b67d5c773a77bd39fc6ca
Melody
5 déc. 2013
#3
+118730
0
Given that p(x)= sin^-1 x and q(x)= x+7, what is the domain of p(q(x))?
The instruction Special has given you is good but what if you didn't have access to the Demos calculator?
By the way, the Demos calculator is fabulous and I use it all the time. But mostly it should be used to check your answer or give you guidance, not to answer the question for you.
With your question you must first look at whether q(x) has put any restrictions on the domain. And no, it has not, q(x) is defined for all real x.
Now look at p(q(x))
p(q(x)) = sin-1(x+7) also written as p(q(x)) = asin(x+7)
Now, I think, AshB, that you are supposed to know what y=sin
-1
x looks like and therefore that it is restricted to -1 < x < 1
So, looking at your problem
-1 <= x+7 <= 1
-1 -7 <= x <= 1 - 7
So the domain of p(q(x)) will be -8 <= x <= -6 ( the errors that I speak about in the next post have been fixed)
I might have a play with demos and see if I can get you a graph or 2
Melody
5 déc. 2013
#4
+118730
0
Don't worry Special. We all do it sometimes. I like seeing all your posts.
Melody
5 déc. 2013
#1
+118730
0
What do you mean, is there more to this question.
Melody
4 déc. 2013
#1
+118730
0
tan (x) sin (x) + sec (x) cos2 (x)
I am assuming you mean
tan (x) sin (x) + sec (x) cos
2
(x)
= sinx/cosx * sinx + 1/cosx * cos
2
(x)
= sin
2
x/cosx + cosx
= (sin
2
x + cos
2
x) / cosx
= 1 / cosx
= sec x
Melody
4 déc. 2013
#1
+118730
0
how i can use cos^-1 ?
It is the inverse of cos or arccos
You use it to find an angle.
Melody
4 déc. 2013
#1
+118730
0
A=900(1+.10/2)^2*2
what is wrong with YOUR calculator?
A=1984.5
Melody
4 déc. 2013
#1
+118730
0
x= [1/2(log13.2/log3)]
this is just a straight calculator question.
1.17430729 to 9 significant figures.
Melody
4 déc. 2013
#2
+118730
0
http://www.betterhealth.vic.gov.au/bhcv2/bhcarticles.nsf/pages/Baby_due_date
Maybe you will find this interesting.
Melody
4 déc. 2013
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