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#26 Re: Help Me ! » Infinite lightbulb. » 2006-07-14 23:22:32
In my A-level maths I learn about sequences and series. In geometric progression (a series of number where each previous number is multiplied by a ratio) we have what's called "Sums to infinity". For example, take the series:
1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...
We can see the sum of the series:
1st term, s1 = 1
2nd, s2 = 1 1/2
3rd, s3 = 1 3/4
4th, s4 = 1 7/8
5th, s5 = 1 15/16
6th, s6 = 1 31/32
And so on.
The sum will get ever close to 2. If we use the formula for the sum S to n terms, first term a = 1, ratio r = 1/2
Sn = a(1-r^n)/1-r
Sn = (1(1-(1/2)^n)) / (1 - 1/2) = (1-(1/2)^n) / (1/2) = 2 - (1/2)^n-1
The difference between Sn and 2 is the term (1/2)^n-1. As n gets very large, this term gets very small. We can make Sn as close to 2 as we like by making the value of n very large. Thus the limit of the sum of the series of n tends to infinity 2, we write:
S∞ = 2
Now this really confuses me. When I try to apply this rule anywhere else it is wrong yet we have a whole module on sums to infinity that states that you do eventally reach S = 2!
#27 Re: Help Me ! » Radian Mode » 2006-07-11 23:56:56
Thanks George 
For any with the same question, you need to press the mode button twice and radian mode is the second option there. There are 3 pages of modes but I could not figure this out 
#28 Re: Help Me ! » Radian Mode » 2006-07-05 05:55:48
because the angle in radians is 1.6 when the radius is 1? I was unsure of this but thanks for pointing it out 
Funny:
2M+1.6 = 27.3
Is this a coincidence?
#29 Help Me ! » Radian Mode » 2006-07-04 22:07:54
- RickyOswaldIOW
- Replies: 5
Does anyone know how I can switch my calculator to radian mode? It's a Casio x-83WA.
If no one can then maybe you can tell me if I've worked this out correctly?
(The image might not show up for a while, not sure why).
I need to work out the perimeter of the shaded area.
I started by taking the triangle and spliting it down the middle to make 2 right angled triangles, angle 0.8 and Hypotenuse 9 (call the triangle ABC where Hypotenuse is BC).
AC = 9Sin0.8 = M
Therefore, the length of the chord from the image is 2M. Since the arc is 1.6
2M+1.6 = Perimeter
I cannot find M since I cannot figure out how to put my calculator into radian mode but I looked at the answer at the back of the book.
2M+1.6 = 27.3
2M = 25.7
M = 12.85
I know from a previous example that 10Sin0.8 = 7.174 so I'm guessing that my answer is wrong because of the difference.
#30 Re: Help Me ! » Sector Perimeter » 2006-06-29 02:50:16
Aha! Of course
The perimeter stretches along the two straight edges of the unshaded sector and not just round the arc of the shaded area.
#31 Re: Help Me ! » Sector Perimeter » 2006-06-29 02:45:19
I could look at it this way:
The first sector has angle Pi/2 and radius 20. s=rθ=20*Pi/2=10Pi
Since Pi/2 = 90 degrees which is 1/4 of the full circle, the shaded sector is the other 3/4 of the circle. 1/4 of the circle = 10Pi and thus 3/4 of it would = 30Pi, 94.2.
#32 Re: Help Me ! » Sector Perimeter » 2006-06-29 02:40:55
40? Hum... I can see that the first sector (with the angle Pi/2) has an arc length of 10Pi and the other sector has an arc of 30Pi, that adds up to 40Pi but I do not see how it is related?
P.S. This is the second question I've ever done of this module so I do not know much about it
#33 Re: Help Me ! » Sector Perimeter » 2006-06-29 02:24:24
Do I need to set my calculator to radian mode? How do I do this? It's a casio fx-83wa.
#34 Help Me ! » Sector Perimeter » 2006-06-29 02:21:48
- RickyOswaldIOW
- Replies: 7
I have a question here asking for the Perimeter of a shaded section of a circle:
The full circle is shown with center O. One sector is labled with a radius of 20 and angle Pi/2 (90 degrees). The rest of the circle is shaded and that is the part I need to work out the perimeter of.
I know that a full circle is 2Pi radians (360 degrees) and since the unshaded sector is Pi/2 radians I can deduce
2Pi - Pi/2 = 3Pi/2 (270 degrees)
Now I just use the regular formula for working out the arc length which is
s = rθ
(let θ = the angle in radians)
s = 20 * 3Pi/2 = 60Pi/2 = 94.2 (3.s.f)
Is this correct? The answers at the back of the book states134.2!
#35 Re: Help Me ! » How do you find X? » 2006-05-29 02:00:15
Are you studying high/advance level maths lothy or are you a school student?
#36 Re: Help Me ! » How do you find X? » 2006-05-29 01:51:27
number n is converted to a different base
I think I see what you mean by this:
2^2 is a 1 with 2 zeros after it in base 2 so x^x is a 1 with x zeros after it? But lothy did specify
(x power of x)
To solve it exactly, I would agree that it's impossible since we have no values given to work with so we can only approximate. I did not know we could do this previously but what you guys are saying seems to make sense
#37 Re: Help Me ! » How do you find X? » 2006-05-28 14:33:30
I know nothing about log 
2² = 4 and ²√4 = 2
5³ = 125 and ³√125 = 5
so
So the xth root of n is x, I see.
x^x = n and x√n = x
#38 Re: Help Me ! » How do you find X? » 2006-05-28 14:11:17
If you take an equation such as
2y = 4
and you wish to get the value of 'y', you must move the 2 over to the other side of the = sign.
We don't know the value of 'y' to start but we know that 2y equals 4. To get just 1y we must divide the 2y by 2:
2y / 2 = y
When you do somthing to one side of the equation you MUST do the same on the other side. Since we have divided the 'y' by 2, we must also divide the 4 on the other side by 2, so:
2y = 4
(2y)/2 = (4)/2
y = 2
_________________________________________________________________
With your example, we start with x^x. You wish to know the value of x. The opposite of to the power of is the root of. So to get x on its own we must:
x√(x^x) = x
We must again do this same step to the other side of the equation
n = x^x
x√n = x√(x^x)
x√n = x
#39 Re: Help Me ! » Doing math in university? » 2006-05-25 13:48:04
How would one go about advertising themself as a mathematician so that they might be hired to do such jobs?
#40 Re: Help Me ! » Bart Simpson Rocks » 2006-05-19 03:26:24
You don't know the secret mathematical formula for getting rich that only mathematicians know?
#41 Re: Help Me ! » Bart Simpson Rocks » 2006-05-19 02:43:02
krassi: Bart simpson is a member of this forum who think's maths is not cool.
#42 Re: Help Me ! » A question about questions. » 2006-05-19 02:35:53
Those programs are about a year old now which is when I started learning maths. I'd probably write the apps in visual basic 6 (not .net). It's a very easy language to make GUIs in! I've look through that code and it has some very interesting statements such as
if (n*i)/i = nWhat a pointless statement! God knows why it's there :S
P.S. liuv, it is I who added you on MSN!
#43 Re: Help Me ! » A question about questions. » 2006-05-17 12:12:10
3(x-2)^2 + x - 6
3(x-2)(x-2) + x - 6
(3x-6)(x-2) + x - 6
3x^2 - 12x + 12 + x - 6
3x^2 - 11x + 6
(3x - 2)(x - 3)
P.S. My algebra drags.
#44 Re: Help Me ! » A question about questions. » 2006-05-17 12:03:33
Here are some of those old console apps I made the can solve some questions, I found them in my archives.
Be Warned! The code is very messy.
http://www.philoswald.f2s.com/maths1.rar
You'll need to run them from a command line rather than double clicking on the executable or you wont get to read the last line.
#45 Re: Help Me ! » A question about questions. » 2006-05-17 11:57:25
I've made a couple of console based apps in C but they are quite limited so far (and the code is very very messy!). Once I can factorise properly in my head I'll start work on one that can produce quadratic and cubic polynomials.
I'm a bit stuck at the moment on the following for now:
factorise 3(x-2)^2 + x - 6
(3x - 6)^2 + x - 6
(3x - 6)(3x - 6) + x - 6
9x^2 - 36x + 36 + x - 6
9x^2 - 35x + 30
I've tried trial and error to put
-1, -30
-2, -15
-3, -10
-5, -6
into both
(3x )(3x )
(9x )(x )
But cannot for the life of me work it out!
#46 Re: Help Me ! » A question about questions. » 2006-05-16 08:53:01
Well it's definately some new questions
I'm going to write some apps that will generate questions of choise, would mathsisfun be interested in having these apps?
#47 Help Me ! » A question about questions. » 2006-05-16 07:25:15
- RickyOswaldIOW
- Replies: 14
I am currently studying a-level maths and have a few folders and books to learn from as well as a tutor. Once I have learnt a module I like to revise it often but I am finding now that I simply remember the answer to all the questions I have done previously and am not exercising the skill of actually working the answers out. Do you know of any resources with many many different questions or maybe some computer program that can generate questions on a given module?
#48 Re: Help Me ! » Math class is boring!!! » 2006-05-03 14:39:07
In my eyes, maths is a tool. Unless you want to construct somthing it is pretty useless and no fun! People use maths in pretty much everything these days even if it's just simple adding up coins to buy some sweets. I find that if I can figure out what applications the maths I am learning has, it becomes a lot easier to learn - but that's the hard part 
#49 Re: Help Me ! » Curve Sketching » 2006-05-03 04:14:07
-2x² - 7x + 15
-2[x² + 7/2x] + 15
-2[(x + 7/4)² - 49/16] + 15
-2(x + 7/4)² + 18 + 1/16
Sketch the curve of -2x² - 7x + 15:
2(x + 7/4)² = 18 + 1/16
(x + 7/4)² = 9 + 1/32
x + 7/4 = ±√9 + 1/32
x = - 7/4 ±√9 + 1/32
So the curve would cross the x-axis at -4.76 and +1.26 which, as you already know, is wrong!.
#50 Re: Help Me ! » Curve Sketching » 2006-05-03 00:43:29
Ah good point!!! I'll write both answers