Inscription
S'identifier
Nom d'utilisateur
Mot de passe
S'identifier
oublié votre mot de passe?
Accueil
Forum
+0
Formulaire
mathématiques
Aide
nombres complexes
calcul différentiel
calcul différentiel
équations
Graphes de fonction
Algèbre linéaire
Théorie des nombres
Porcentajes
Standard fonctions
Probabilité
Trigonométrie
Conversion des unités
Unités
Sur
Mentions légales
politique de confidentialité
conditions d'utilisation
Crédits
Google+
Facebook
Email du contact
CPhill
Nom d'utilisateur
CPhill
But
130060
Membership
Stats
Questions
56
Réponses
43246
56 Questions
43663 Answers
+16
29
7
+130060
Blocked Posts
I just wrote to the site manager to let him know that this is happening....I'll try to "unblock" as many posts as possible ( but realize that I can't stay on here all day )
CPhi
lire plus ..
●
●
●
●
●
●
●
CPhill
7 août 2023
+18
1
1854
32
+130060
HAPPY BIRTHDAY , MELODY !!!
I baked you a cake.....be sure to share !!!!!
lire plus ..
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
CPhill
16 avr. 2022
+25
3624
21
+130060
HAPPY BIRTHDAY, MELODY !!!!
lire plus ..
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
CPhill
16 avr. 2021
+57
4488
4
+130060
Gotta take a break folks !!!!
I'll come back a little later and try to answer as many questions as I can [ or as many as I know how to do !!! ]
lire plus ..
●
●
●
●
CPhill
26 mars 2020
+46
1
5458
14
+130060
HAPPY BIRTHDAY, HECTICTAR !!!
She is one of the best "answerers" on this site [ even though she IS from Alabama....well.....everyone can't be perfect !!!! (Hehehe!!) ]
lire plus ..
●
●
●
●
●
●
●
●
●
●
●
●
●
●
CPhill
14 oct. 2019
+54
2
4313
1
+130060
'Waiting For Moderation'
I see that this has been happening quite a bit.....
If you "right click" on the image and select "Copy Image Address".....you can open up a new browser window and paste this address into it.....hit "Enter"
lire plus ..
●
CPhill
21 oct. 2018
+84
2
4497
18
+130060
4th Anniversary For Me !!!
Been on this site for 4 years now.....my thanks to Melody (and Andrew) for recommending me as a "mod".....I've learned quite a bit and "met" a lot of good people on here.....I've even managed (somehow) to help some of them....!!!
lire plus ..
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
CPhill
13 mars 2018
+26
1
2937
38
+130060
Hey, guys.....I'm going to be gone for awhile....If you can hang on....I'll answer as many questions as I know how in just a bit....
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
CPhill
6 mars 2018
+32
1
2435
7
+130060
I'm Out Until This Afternoon...You Kids Have Fun !!!
●
●
●
●
●
●
●
CPhill
13 févr. 2018
+48
2
4181
8
+130060
HAPPY BIRTHDAY, HECTICTAR !!!!
You are a real asset to the forum.....!!!!!
Here's your cake.....enjoy.....!!!!
lire plus ..
●
●
●
●
●
●
●
●
CPhill
14 oct. 2017
+31
3
3978
2
+130060
Going To The Picnic
This one isn't too difficult....give it a try.....!!!!!
lire plus ..
●
●
CPhill
25 sept. 2017
+29
2
6914
7
+130060
How Much Do They Weigh ????
This is another problem from American puzzle-maker Sam Lloyd.....
Five little girls taken in couples weigh 129 pounds, 125 pounds, 124 pounds, 123 pounds, 122 pounds, 121 pounds, 120 pound, 118 pounds, 116
lire plus ..
●
●
●
●
●
●
●
CPhill
23 sept. 2017
+26
2
2921
5
+130060
The Width Of The River.....
Here is a problem I've always liked...it was devised by one of the great American "puzzleists" of all time, Sam Lloyd
It has been posted on the forum before, but not in a while.....maybe we have some new people who would like
lire plus ..
●
●
●
●
●
CPhill
20 sept. 2017
+31
2
3232
10
+130060
ReCaptcha Verification
I don't know if others are having this problem or not......when using Opera or Firefox I had to do the ReCaptcha thing when submitting answers.......however.......using Google Chrome seems to solve this problem - for me anyway....
Comments???
lire plus ..
●
●
●
●
●
●
●
●
●
●
CPhill
19 mars 2017
+35
1
3786
6
+130060
Happy Pi Day..(3.14)....Everybody.....!!!!!
lire plus ..
hectictar
●
●
●
●
●
●
CPhill
14 mars 2017
«
dernier
4
3
..
2
1
»
#2
+130060
+3
The diameter is 2
To see this, note that the circumference is given by C = 2πR, or C=2Rπ. But 2R = D, where D is the diameter.
Therefore, C=Dπ
So 6.28 = Dπ
And if we let π = 3.14
Then, 6.28 = D * 3.14
So D = 6.28 / 3.14 = 2
CPhill
26 mars 2014
#1
+130060
+2
Example:
Twice a number = 2n
So, 5 less than twice would be
2n - 5
CPhill
25 mars 2014
#3
+130060
+2
Take a look at this website, Stu. I've found that his explanations are pretty good. He also gives some examples.
http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeIntro.aspx
If you have any specific questions, post them to the forum.
Hope this helps.
CPhill
25 mars 2014
#2
+130060
+2
Although I'm not the most "technical" user of the calculator, it will add and subtract fractions (and reduce them to simplest form). It will also convert an "unknown" decimal to a fraction.
Examples
108/357
9/93+8/102
0.1752055660974067
I'm not sure what you might be after, but if you have more specific questions, post them to the forum and we'll see if we can answer them.
CPhill
25 mars 2014
#1
+130060
+2
First, let sin(x) = x
So
3SIN(x)^2-7SIN(x)-3=0
= 3x^2 - 7x - 3 = 0
Solving, we have.....
3x^2 - 7x - 3 = 0
This implies that sin(x) = (7 - √85) / 6 and sin(x) = (7 + √85) / 6
The second answer is impossible, since it implies that sin(x) > 1
Solving for the first, we have
asin(((7-sqrt(85))/6))
Hope this helps
CPhill
25 mars 2014
#6
+130060
+2
Bertie
Your method of substituting cos(pi/3) for 1/2 and sin (pi/3) for (√3)/2 is quite ingenious!!!!
I like that
MUCH BETTER
!!!
CPhill
25 mars 2014
#3
+130060
+2
sqrt(3)cos(x) - sin(x)= 1
Let's see if this works
(add sin(x) to both sides and square both sides)
3(cos(x)^2 = 1 + 2sin(x) +sin(x)^2
(using cos(x)^2 = 1 - sin(x)^2, we have)
3*(1 - sin(x)^2) = 1 + 2sin(x) +sin(x)^2
(Simplify)
4*sin(x)^2 + 2*sin(x) - 2 = 0
(Divide through by 2)
2*sin(x)^2 = sin(x) -1 = 0
(Factor)
(2sin(x) -1) (sin(x) +1 ) = 0
Solving, we get x = (pi/6 + 2*pi*n , 3*pi/2 +2*pi*n)
Another 2nd quad angle, 5*pi/6, solves (2sin(x) -1 = 0), but it doesn't "work in the original problem) (due to "squaring," we get an extraneous solution??)
I think that may be it.....but check my work and method!!
CPhill
25 mars 2014
#1
+130060
+2
This is an interesting one........
For all real numbers x,
√(x)^2 = IxI = x if {x ≥ 0 and -x if x < 0}
So, √(-1)^2 = I-1I = -(-1) = 1
Now, many people assume that the result could be "-1" - due to following faulty "proof."
Step 1 √(-1)^2 =
Step 2 √(-1) * √(-1) =
Step 3 ( i * i ) =
Step 4 ( -1 )
The problem comes in the transition from Step 1 to Step 2.
In general, √(a * a) = √(a) * √(a) only if a ≥ 0. (Thus, we can't "split" the radicals in this manner.)
I think that's it !!!
CPhill
22 mars 2014
#1
+130060
+2
How do you Solve this?
2(x+3) - 2x^2(x+3) = 0
In my answer sheet they rewrite it to (x+3) * (2-2x^2) = 0
Im unsure what happens to the other (x+3)
They (almost) took the greatest common factor, (x + 3), out of both terms......... !!!!
We can also take a "2" out !!!
Doing this, we get
2 * (x + 3) * (1 - x^2) = 0
We can get rid of the 2 by dividing through, so this becomes
(x + 3) * (1 - x^2) = 0
To solve, set each factor = 0
The first one is easy
x + 3 = 0
Subtracting 3 from both sides, we get
x = -3 and that's one solution
Now, setting the other factor to 0 we get
1 - x^2 = 0
This can be factored as
(1 - x) * (1 + x) = 0
And setting these two things to 0 (as before), we get the other two solutions of 1 and -1
Hope this helps
CPhill
22 mars 2014
#1
+130060
+2
how do i do a problem like this 1/2(w-4)??
I assume you just want to "distribute" the 1/2 over the terms in the parentheses.
This would just give us
1/2w - 2
CPhill
21 mars 2014
«
dernier
4365
4364
..
9
8
7
6
5
..
2
1
»