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  • How to fix this 7% where 6% , 5% works great.

    - by Stackfan
    Case 1 (discount 6%): Subtotal: 750.00 Discount: 45.00 Handling cost: 24.32 21% VAT: 0.00 Total (this is the amount you will deposit): 729.32 Case 2 (discount 7%): Subtotal: 1250.00 Discount: 87.50 Handling cost: 39.88 21% VAT: 0.00 Total (this is the amount you will deposit): 1202.38 Where i am applying this formula: (729.32 - 0.35) / 1.034/ 0.94 = 750.00 (<<--- CORRECT ) ? (1202.38 - 0.35) / 1.034/ 0.93 = 1250.01 (<<--- My problem why not 1250.00) ? How to correct the 7% formula to get exactly 1250.00 ? Instead of fraction error.

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  • All non-prime factorings

    - by Tony Veijalainen
    Let's say we have numbers factors, for example 1260: >>> factors(1260) [2, 2, 3, 3, 5, 7] Which would be best way to do in Python combinations with every subproduct possible from these numbers, ie all factorings, not only prime factoring, with sum of factors less than max_sum? If I do combinations from the prime factors, I have to refactor remaining part of the product as I do not know the remaining part not in combination. Example results would be: [4, 3, 3, 5, 7] (one replacement) [3, 6, 14, 5] (two replacements)

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  • Fastest way to find the rotation of a vector

    - by kriss
    I have two 2D vectors, say u and v, defined by cartesian coordinates. Imagine that vectors are needles of a clock. I'm looking for the fastest way to find out, using python, if v is after or before u (or in other words find out in wich half plane is v, regarding to position of u). For the purpose of the problem if vectors are aligned answer should be before. It seems easy using some trigonometry, but I believe there should be a faster way using coordinates only. My test case: def after(u, v): """code here""" after((4,2), (6, 1)) : True after((4,2), (3, 3)) : False after((4,2), (2, 1)) : False after((4,2), (3, -3)) : True after((4,2), (-2, -5)) : True after((4,2), (-4, -2)) : False

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  • Efficient Multiplication of Varying-Length #s [Conceptual]

    - by Milan Patel
    Write the pseudocode of an algorithm that takes in two arbitrary length numbers (provided as strings), and computes the product of these numbers. Use an efficient procedure for multiplication of large numbers of arbitrary length. Analyze the efficiency of your algorithm. I decided to take the (semi) easy way out and use the Russian Peasant Algorithm. It works like this: a * b = a/2 * 2b if a is even a * b = (a-1)/2 * 2b + a if a is odd My pseudocode is: rpa(x, y){ if x is 1 return y if x is even return rpa(x/2, 2y) if x is odd return rpa((x-1)/2, 2y) + y } I have 3 questions: Is this efficient for arbitrary length numbers? I implemented it in C and tried varying length numbers. The run-time in was near-instant in all cases so it's hard to tell empirically... Can I apply the Master's Theorem to understand the complexity...? a = # subproblems in recursion = 1 (max 1 recursive call across all states) n / b = size of each subproblem = n / 1 - b = 1 (problem doesn't change size...?) f(n^d) = work done outside recursive calls = 1 - d = 0 (the addition when a is odd) a = 1, b^d = 1, a = b^d - complexity is in n^d*log(n) = log(n) this makes sense logically since we are halving the problem at each step, right? What might my professor mean by providing arbitrary length numbers "as strings". Why do that? Many thanks in advance

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  • 6^x = 5 equation, how to solve it?

    - by Tom
    If it would be 6^x = 1 or 6^x = 6 or 6^x = 36 it would be extremely easy, but how to solve this equation: 6^x = 5 I don't need an answer, I want to find out how to solve equations like this one, I need solution. Thanks.

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  • Why isnt int pow(int base, int exponent) in the standard C++ libraries?

    - by Dan O
    I feel like I must just be unable to find it. Is there any reason that the c++ pow function does not implement the "power" function for anything except floats and doubles? I know the implementation is trivial, I just feel like I'm doing work that should be in a standard library. A robust power function (ie handles overflow in some consistent, explicit way) is not fun to write.

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  • simple rank formula

    - by Graham
    I'm looking for a mathmatical ranking formula. Sample is 2008 2009 2010 A 5 6 4 B 6 7 5 C 7 8 2 I want to add a rank column for each period code field rank 2008 2009 2010 2008 2009 2010 B 6 7 5 2 1 1 A 5 6 4 3 2 2 C 7 2 2 1 3 3 please do not reply with methods that loop thru the rows and columns, incrementing the rank value as it goes, that's easy. I'm looking for a formula much like finding the percent total (item / total). I know i've seen this before but an havning a tough time locating it. Thanks in advance!

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  • getting the value of a filter at an arbitrary time

    - by Andiih
    Context: I'm trying to improve the values returned by the iPhone CLLocationManager, although this is a more generally applicable problem. The key is that CLLocationManger returns data on current velocity as and when it feels like it, rather than at a fixed sample rate. I'd like to use a feedback equation to improve accuracy v=(k*v)+(1-k)*currentVelocity where currentVelocity is the speed returned by didUpdateToLocation:fromLocation: and v is the output velocity (and also used for the feedback element). Because of the "as and when" nature of didUpdateToLocation:fromLocation: I could calculate the time interval since it was last called, and do something like for (i=0;i<timeintervalsincelastcalled;i++) v=(k*v)+(1-k)*currentVelocity which would work, but is wasteful of cycles. Especially as I probably want timeintervalsincelastcalled to be measured as 10ths of a second. Is there a way to solve this without the loop ? i.e. rework (integrate?) the formula so I put an interval into the equation and get the same answer as I would have by iteration ?

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  • How to implement a grapher in C#

    - by iansinke
    So I'm writing a graphing calculator. So far I have a semi-functional grapher, however, I'm having a hard time getting a good balance between accurate graphs and smooth looking curves. The current implementation (semi-pseudo-code) looks something like this: for (float i = GraphXMin; i <= GraphXMax; i++) { PointF P = new PointF(i, EvaluateFunction(Function, i) ListOfPoints.Add(P) } Graphics.DrawCurve(ListOfPoints) The problem with this is since it only adds a point at every integer value, graphs end up distorted when their turning points don't fall on integers (e.g. sin(x)^2). I tried incrementing i by something smaller (like 0.1), which works, but the graph looks very rough. I am using C# and GDI+. I have SmoothingMethod set to AntiAlias, so that's not the problem, as you can see from the first graph. Is there some sort of issue with drawing curves with a lot of points? Should the points perhaps be positioned exactly on pixels? I'm sure some of you have worked on something very similar before, so any suggestions? While you're at it, do you have any suggestions for graphing functions with asymptotes? e.g. 1/x^2 P.S. I'm not looking for a library that does all this - I want to write it myself.

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  • Move an object in the direction of a bezier curve?

    - by Sent1nel
    I have an object with which I would like to make follow a bezier curve and am a little lost right now as to how to make it do that based on time rather than the points that make up the curve. .::Current System::. Each object in my scene graph is made from position, rotation and scale vectors. These vectors are used to form their corresponding matrices: scale, rotation and translation. Which are then multiplied in that order to form the local transform matrix. A world transform (Usually the identity matrix) is then multiplied against the local matrix transform. class CObject { public: // Local transform functions Matrix4f GetLocalTransform() const; void SetPosition(const Vector3f& pos); void SetRotation(const Vector3f& rot); void SetScale(const Vector3f& scale); // Local transform Matrix4f m_local; Vector3f m_localPostion; Vector3f m_localRotation; // rotation in degrees (xrot, yrot, zrot) Vector3f m_localScale; } Matrix4f CObject::GetLocalTransform() { Matrix4f out(Matrix4f::IDENTITY); Matrix4f scale(), rotation(), translation(); scale.SetScale(m_localScale); rotation.SetRotationDegrees(m_localRotation); translation.SetTranslation(m_localTranslation); out = scale * rotation * translation; } The big question I have are 1) How do I orientate my object to face the tangent of the Bezier curve? 2) How do I move that object along the curve without just setting objects position to that of a point on the bezier cuve? Heres an overview of the function thus far void CNodeControllerPieceWise::AnimateNode(CObject* pSpatial, double deltaTime) { // Get object latest pos. Vector3f posDelta = pSpatial->GetWorldTransform().GetTranslation(); // Get postion on curve Vector3f pos = curve.GetPosition(m_t); // Get tangent of curve Vector3f tangent = curve.GetFirstDerivative(m_t); } Edit: sorry its not very clear. I've been working on this for ages and its making my brain turn to mush. I want the object to be attached to the curve and face the direction of the curve. As for movement, I want to object to follow the curve based on the time this way it creates smooth movement throughout the curve.

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  • How to turn this simple 10 digit hex number back into 8 digits?

    - by Babil
    The algorithm to convert input 8 digit hex number into 10 digit are following: Given that the 8 digit number is: '12 34 56 78' x1 = 1 * 16^8 * 2^3 x2 = 2 * 16^7 * 2^2 x3 = 3 * 16^6 * 2^1 x4 = 4 * 16^4 * 2^4 x5 = 5 * 16^3 * 2^3 x6 = 6 * 16^2 * 2^2 x7 = 7 * 16^1 * 2^1 x8 = 8 * 16^0 * 2^0 Final 10 digit hex is: = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 = '08 86 42 98 E8' The problem is - how to go back to 8 digit hex from a given 10 digit hex (for example: 08 86 42 98 E8 to 12 34 56 78) Some sample input and output are following: input output 11 11 11 11 08 42 10 84 21 22 22 33 33 10 84 21 8C 63 AB CD 12 34 52 D8 D0 88 64 45 78 96 32 21 4E 84 98 62 FF FF FF FF 7B DE F7 BD EF

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  • Android: Find coordinates of a certain point X meters from my location moving towards the point I am

    - by Aidan
    Hi Guys, I'm constructing a geolocation based application and I'm trying to figure out a way to make my application realise when a user is facing the direction of the given location (a particular long / lat co-ord). I've done some Googling and checked the SDK but can't really find anything for such a thing. Does anyone know of a way? Example. Point A = Phones current location. Point B = A's orientation in relation to true north + 45 + max distance towards the direction your facing, Point C = A's orientation in relation to true north - 45 + max distance towards the direction your facing. So now you have a triangle constructed. pretty schweet huh? yeah.. I think so.. So now that I have my fancy Triangle I use something called Barycentric Coordinates ( http://en.wikipedia.org/wiki/Barycentric_coordinates_(mathematics) ). This will allow me to test another point and see if it is in the triangle. If it is, it means we're facing it AND it's within the right distance. So it should be displayed on screen. If I'm facing 90 degrees from true north. The distance it travels should be that direction. 90 degrees from true north. It should not be 100 degrees or something from true north! But the problem is I haven't yet figured out how I make the device realise it must go "out" the direction it is facing.

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  • C++ integer floor function

    - by Thomas
    I want to implement greatest integer function. int x = 5/3; My question is with greater numbers could there be a loss of precision as 5/3 would produce a double? EDIT: Greatest integer function is integer less than or equal to X. Example: 4.5 = 4 4 = 4 3.2 = 3 3 = 3 What I want to know is 5/3 going to produce a double? Because if so I will have loss of precision when converting to int. Hope this makes sense. RE-EDIT: What OP calls 'greatest integer' most of us call 'floor'.

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  • Android debug mode not working waiting for device

    - by Blue42
    I'm trying to get my android phone working to run apps from inside an IDE. Got it working on windows no problem, currently working in fedora (18) a lot though so wanted to get it working with that. Got IntelliJ and android sdk installed, problem is when I try to run the default hello world app it wont work it just says waiting for device.. Ran adb devices got List of devices attached ???????????? no permissions Leads me to believe the driver isn't installed? Phone I'm using is a HTC Sensation. Does anyone know what I can do to try and resolve it? The HTC web page doesn't offer me drivers to install. Also noticed in /etc/udev/rules.d/..android.rules there is nothing about Sensation. Seems it recognises my nexus 7 though.. Edit: Tried my nexus.. got List of devices attached 901839238298923 offline So it doesn't even work with that.. confusing.. Any help would be appreciated. Thanks, B

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  • Linear Interpolation. How to implement this algorithm in C ? (Python version is given)

    - by psihodelia
    There exists one very good linear interpolation method. It performs linear interpolation requiring at most one multiply per output sample. I found its description in a third edition of Understanding DSP by Lyons. This method involves a special hold buffer. Given a number of samples to be inserted between any two input samples, it produces output points using linear interpolation. Here, I have rewritten this algorithm using Python: temp1, temp2 = 0, 0 iL = 1.0 / L for i in x: hold = [i-temp1] * L temp1 = i for j in hold: temp2 += j y.append(temp2 *iL) where x contains input samples, L is a number of points to be inserted, y will contain output samples. My question is how to implement such algorithm in ANSI C in a most effective way, e.g. is it possible to avoid the second loop? NOTE: presented Python code is just to understand how this algorithm works. UPDATE: here is an example how it works in Python: x=[] y=[] hold=[] num_points=20 points_inbetween = 2 temp1,temp2=0,0 for i in range(num_points): x.append( sin(i*2.0*pi * 0.1) ) L = points_inbetween iL = 1.0/L for i in x: hold = [i-temp1] * L temp1 = i for j in hold: temp2 += j y.append(temp2 * iL) Let's say x=[.... 10, 20, 30 ....]. Then, if L=1, it will produce [... 10, 15, 20, 25, 30 ...]

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  • simple dividing

    - by dontoo
    Lets say I am dividing 2592 / 324 = 8 Practically I must guess that 324 goes 8 times in 2592 without reminder. For example, first I try with 7 ( 7*324 = 2268, reminder 324, so 7 is wrong, so I try with 8 and 8 is correct) Is there anyway to do dividing without guessing( trying ), like multiplication ( all mechanically, digit by digit )?

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  • How to make simple mathematical operations and displaying the result in a textbox

    - by Audel
    Hi, I know is a rather simple question but I just can't find an appropriate example in google or anywhere. I've got this piece int numberOfPlays = int.Parse(textBox2.Text); numberOfPlays = (numberOfPlays++); textBox2.Text = (numberOfPlays.ToString()); MessageBox.Show(numberOfPlays.ToString()); So basically what I want to do is to get the value of the textBox2, make it an integer and then add 1 to it. I can't think of any more details right now, so if i'm not clear enough please ask Thanks in advance

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  • using context resource in gwt 2 hosted mode

    - by rafael
    Hello all, I am moving a web app from gwt 1.5 to gwt 2.0. I am trying to connect to the a database resource I have in my context.xml file.In gwt 1.5 I had set up root.xml in tomcat-conf-gwt-localhost. I have no idea where to set up the resource in GWT 2.0. I tried placing my context.xml file in war-META-INF with no luck. Anyone have an idea where to place the context.xml file to be able to use a jndi database resource in GWT 2.0? Thanks in advanced

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  • Uniforme distance between points

    - by Reonarudo
    Hello, How could I, having a path defined by several points that are not in a uniform distance from each other, redefine along the same path the same number of points but with a uniform distance. I'm trying to do this in Objective-C with NSArrays of CGPoints but so far I haven't had any luck with this. Thank you for any help. EDIT I was wondering if it would help to reduce the number of points, like when detecting if 3 points are collinear we could remove the middle one, but I'm not sure that would help.

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  • Calculate coordinates between startpoint and endpoint

    - by MikeB123
    Given.. 1 - The start-point GPS coordinates, 2 - The end-point GPS coordinates, 3 - The speed at which the object is travelling, 4 - Knowing that the trip trajectory will be a straight line... How can I calculate what my GPS coordinated will be in n minutes' time? That is to say, how can I calculate my position at a given time after the trip has started before the trip has ended?

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  • bcdiv() bcadd() bcsub() with Php

    - by Pieman
    Will this code be 'stressful' for a server? Or is it easy to bcdiv/sub/add to 10000 decimal places? I'm thinking of looping it afew times... Not Sure... $s2 = (bcdiv('1', $test, 10000)); $s = bcsub($s, $s2, 10000); $test += 2; $s3 = (bcdiv('1', $test, 10000)); $s = bcadd($s, $s3, 10000); $test += 2; Any advice? :)

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