explain how to do linear functions?
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Which function, Subsitution, Linear COmbination, Graph?
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What you do with extra-polation is assuming that the data is developping along a linear function. So you'll be searching for the function y = c * x where the sum of (your found y-data minus the y-value from the function) is at its minimum.\nSay you found:\n(x,y) = (7,3); (23,8); (26,7); (30,11)\n\nThe function values for these x-values would be:\n(x,y) = (7,7c); (23,23c); (26,26c); (30,30c)\n\nAnd the sum of the differences:\nT = (3-7c) + (8-23c) + (7-26c) + (11-30c) =>\nT = 29 - 86c\n\nT is at its minimum as T = 0, so:\n29 - 86c = 0 =>\nc = 29/86\n\nSo the line you would be looking for is y = 29/86 * x
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A linear equation looks like a line when it is graphed.  The standard form of the equation is y=mx+b where m is the slope, and b is the value of y at the point where the line crosses the y-axis.\n\nA non-linear equation does not look like a line.  It may look like a curve.  A non-linear equation could possibly be a parabola, a circle, a wave, a hyperbola, etc... but not a line!
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Base on the Mathematics for STPM, Penerbit Fajar Bakti SDN.BHD. ,written by Ong Beng Sim...From the cphater 5,Coordinate Geometry, the actual standard form is ax/2+y/2=1.\n\nTo change the form:\n\nfind the axis-y & axis-x first,\n\nwhen y=0, x=2/a (axis-x)\nwhen x=0, y=2   (axis-y)\n\nBase on the standard form of linear funtion is x/a+y/b=1, where, a is axis-x and b is axis-y.\n\nso, substitute the axis-x & axis-y into the standard form of linear function...\n\nTherefore, the standard form of the linear funtion for y=-ax+2 is ax/2+y/2=1
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Homework.\n\nShow it first for intervals. Then extend to simple functions. Then use the fact thats imple functions are dense in L1.
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Call  Y of x like Y(x) \n\nY(x) = Y(x-1)+ (3*4^x)-1(1/9)x\nY(x) = Y(x-2)+ (3*4^(x-1))-1(1/9)(x-1) + (3*4^x)-1(1/9)x\nY(x) = Y(x-3)+ (3*4^(x-2))-1(1/9)(x-2) (3*4^(x-1))-1(1/9)(x-1) + (3*4^x)-1(1/9)x\n...\nY(x) = Y(1) + 3(4^x+...+4^2) - (1/9)(x+...+2)\nY(x) = Y(1) + 3*4[4^(x+1)-1]/[4-1] - [x(x+1)/2-1]/9\nY(x) = (sqrt3)/4 + 4^(x+2) - 4 - [x(x+1)/2-1]/9\nY(x) = 16 * 4^x - x(x+1)/18 - (37/9 - (sqrt3)/4)
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Case 1: f(x) := x-5*y; (y is parameter)\n\nL{f(x)} = int(exp(-s*x)*f(x),x=0..infinity);\nL{f(x)} = int(exp(-s*x)*(x-5*y),x=0..infinity);\nL{f(x)} = 1/s^2-5*y/s;\n\nCase 2: f(y) := x-5*y; (x is parameter)\n\nL{f(y)} = int(exp(-s*y)*f(y),y=0..infinity);\nL{f(y)} = int(exp(-s*y)*(x-5*y),y=0..infinity);\nL{f(y)} = x/s-5/s^2;
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About Me\nMember since: October 03, 2006\nTotal points: 100 (Level 1)\nPoints earned this week:\nTotal answers:\nBest answers:\n\nI need math help again... PLEASE HELP! I m desperately in need.?\n\nAnswer PARTB and PART C in the space provided. Evaluation of your solution to each problem will be based on the following.\n\n*A correct mathematical method for solving the problem is shown\n\n*The final answer is accurate and a check of the answer is shown where asked for by the question.\n\nDetermine if the equation is a linear equation, and if it is, write in standard form.\n\na. 5x+1=3y\n\nb. y=2x   x=y/2\n\nc. 4(2y-x)=1  8y-4x=1\n\nd. 25x-2y+2=0\n\ne. 2x-5y/2=7x\n\n2. a. f(x)= x (exponent 2)+2x+1 f(0),f(2)\n\nb. f(x)=2x-5 f(3),f(-3)\n\nc. f(x)=(-3x)exponent 2 f(0),f(-1)\n\nd. f(x)=2x(exponent 3)+4 f(0),f(-1)\n\ne. f={(1,5),(3,7),(5,9),(7,11)} f(3),F(7)
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Let y = u/v   where u=some function and v =some funtion\n\nthen dy/dx = (v*du/dx - u*dv/dx)/ (v^2)     **Quotient Rule**\n\nIn this case u = 6x^2-1 and v = 2*sqrt(x)\n\nSo you just differentiate and plug into the equation for dy/dx
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Definition:  A function is a one-to-one function if and only if each second element corresponds to one and only one first element. (each x and y value is used only once)\n\nUse the horizontal line test to determine if a function is a one-to-one function.\n\nIf ANY horizontal line intersects your original function in ONLY ONE location, your function will be a one-to-one function and its inverse will also be a function.\n\nTIP:\n\nGraph your function to see if it passes the horizontal line test.\nYou can do the horizontal line test with a pencil.  If the pencil crosses the graph of your function in more than one place (more than one point), then it will not be one-to-one.  If the pencil crosses the graph of your function at only one location, then it will be one-to-one.  \n\nTry it and get back to me.\n\nGuido
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Assuming you mean over R, solve the equation s*t + r*sin(t) = 0 for r and s for all t that is not the trivial solution (r,s)=(0,0). \nIn other words, if f(t)=sin(t) and g(t)=t were linearly dependent, then there is some non-zero slope s, s*g(t) = s*t and some non-zero amplitude r, r*f(t) = r*sin(t) such that the graphs of these functions completely coincide for all t. Since f is non-linear and g is, this is nonsense and the functions are linearly independent.
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The linearization is the equation of the line tangent to f at x=a.\nThe point that L(x) is based on is (a,f(a)), which is (4,√13)\nFind the derivative of the function.\nf'(x) = 2x·[1/2√(x²+9)] = x/√(x²+9)\nf'(4) = 4/√13\nThe linearization:\nL(x) = (4/√13)x - 15/√13
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A linear function can refer to two slightly different concepts. In geometery and elementary algebra a linear function is a first degree polynomial mathematical function of the form:\n\nf(x) = m x + c \nwhere m and c are constants.\n\nThe problem with this geometric definition is that functions of the above form - despite their names - do not necessarily satisfy the conditions of a linear map. Therefore, some people refer to functions of the above form as affine functions. If and only if a function is of the above form with c equal to zero, the function satisfies the properties of a linear map, preserving scalar multiplication and vector addition for all points in its domain.\n\nLinear functions always have as domain the set of all real numbers and a range of all real numbers. By the geometric definition, the derivative of linear functions in terms of their independent variable, x, is always the constant m.\n\nLinear functions (according to the geometric definition) can also be written in the form:\n\ny = m x + c \nand plotted on an x,y graph. It forms a straight line, as the name implies.\n\nThe constant m is often called the slope or gradient while c is the y-intercept, which gives the point of intersection between the graph of the function and the y-axis.
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(Ok, by the way, this is not a linear equation. A linear equation is when the equation includes "x" and "y")\n\nBefore you can get the answer, you have to obviously solve it. \nSo for this equation, you want to get the variable alone; to one side. Now looking at the equation, everything isn't simplified. The left side is, you can not do anything with "x" and "-15"...Though, on the right side, there are like terms (expressions/terms that can be simplified)..."x" and "2x" become 3x. See 1x (1x is same as x) plus 2x equals 3x. So now your equation is x-15=3x+1 Now that everything is simplifed, you can solve it. As I said, get the variable to one side. Usually, the variable is on the left side of the equation, but it doesn't really matter which side its on...So to get x alone from the left side, you have to do what to get rid of NEGATIVE 15?...Do the opposite, POSTIVE 15, which then cancel each other out and x will be left alone. However when you add 15 to the left side, you have to add 15 to the right side. So adding 15 to the right side, 3x+1+15= 3x+16...Now the equation should be x=3x+16. So now, you have to get the 3x from the right side to the left side. How do you get sixteen alone? You minus 3x from both sides. The right side will result with having just 16 as the other side will have -2x. The equation is ow -2x=16. Again, get x alone. And how to do that is divide -2 to both sides. So -2x divided by -2 is 1x or just x and 16 divided by -2 is -8. At the end you should have x=-8\n\nHere's the work:\nx-15=x+2x+1\nx-15=3x+1\nx-15+15=3x+1+15\nx=3x+16\nx=3x+16-3x\nx-3x=16\n-2x=16\n-2x/-2=16/-2\nx=-8\n\nAnd the "check" works too(when you plug in -8 as x)\n\nx-15=x+2x+1\n-8-15=-8+2(-8)+1\n-23=-8-16+1\n-23=-23
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the answer is linear function:\nf(n)=2n-1
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To write any equation in "function form", just solve for y (get y by itself on one side of the =), and replace y with f(x).\n\nI'll let you do that.  :)
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Multiply the first equation by 2.\n\nSubtract the second equation from the new first equation.\n\nSolve for y.\n\nTake your value for y and substitute it into the original first equation.\n\nSolve for x.