Is a non linear equation only equal to zero or can another number be as well?
Q. This is an example of the equation in which the question was asked. An example of a non-linear equation is a quadratic equation of the form Ax 2 + bx + c = 0 To get two inputs you will have to equations like this y = x 2 y = 8 x 2
Asked by Charlee B - Wed Sep 3 13:11:31 2008 - - 2 Answers - 0 Comments
A. -c is the other number. ax^2+bx=-c is the same as ax^2+bx+c=0
Answered by David F - Wed Sep 3 13:20:32 2008
Q. This is an example of the equation in which the question was asked. An example of a non-linear equation is a quadratic equation of the form Ax 2 + bx + c = 0 To get two inputs you will have to equations like this y = x 2 y = 8 x 2
Asked by Charlee B - Wed Sep 3 13:11:31 2008 - - 2 Answers - 0 Comments
A. -c is the other number. ax^2+bx=-c is the same as ax^2+bx+c=0
Answered by David F - Wed Sep 3 13:20:32 2008
Non-linear examples?
Q. Hey guys i need help to find one example for non-linear regression(equation or data) please help
Asked by miss_pink_blue - Tue Mar 4 23:49:20 2008 - - 1 Answers - 0 Comments
A. Double click on the following websites:
Answered by arro.mark - Wed Mar 5 02:30:24 2008
Q. Hey guys i need help to find one example for non-linear regression(equation or data) please help
Asked by miss_pink_blue - Tue Mar 4 23:49:20 2008 - - 1 Answers - 0 Comments
A. Double click on the following websites:
Answered by arro.mark - Wed Mar 5 02:30:24 2008
URGENT! Help!!! Can anyone give example of non linear equation (relationship) ???
Q. It can be shapes, patterns, or events in our daily life. Please give little explanation why it is non linear. Thank you.
Asked by Lavender - Tue Jul 15 22:17:59 2008 - - 2 Answers - 0 Comments
A. y^2=4ax x^2+y^2=1
Answered by visram007 - Tue Jul 15 22:21:41 2008
Q. It can be shapes, patterns, or events in our daily life. Please give little explanation why it is non linear. Thank you.
Asked by Lavender - Tue Jul 15 22:17:59 2008 - - 2 Answers - 0 Comments
A. y^2=4ax x^2+y^2=1
Answered by visram007 - Tue Jul 15 22:21:41 2008
give an example of a non linear?
Q. i don't get the difference and how 2 make a linear equation. So could u make 2 linear and 2 non linear equations.
Asked by iloveboyz123456789 - Thu Sep 4 17:56:33 2008 - - 1 Answers - 0 Comments
Q. i don't get the difference and how 2 make a linear equation. So could u make 2 linear and 2 non linear equations.
Asked by iloveboyz123456789 - Thu Sep 4 17:56:33 2008 - - 1 Answers - 0 Comments
Quadratic equations are just one kind of non-linear pattern.?
Q. When graphed, they describe a parabola or arc. In other words, quadratic equations describe something that arches up to a peak and then returns to its original level or something which drops to a low point and then rises. The rate of change on one side of the peak or low point is the same as on the other side and forms a smooth arch-like shape. A quadratic equation that describes a real world parabolic pattern is another kind of model. Can you think of examples of processes that could be modeled with quadratic equations? Here's an example to get you started. The arches in a cathedral can generally be described by a quadratic equation. Although they are static, the curvature of the arch follows the same pattern on both sides of the… [cont.]
Asked by QUEEN ESHA - Fri Jul 17 09:46:16 2009 - - 1 Answers - 0 Comments
A. ballistic missiles have trajectories that follow a parabolic shape. (a simpler illustration: when you throw a ball in a certain angle, the ball follows a parabolic path)
Answered by Alam Ko Iyan - Fri Jul 17 11:45:34 2009
Q. When graphed, they describe a parabola or arc. In other words, quadratic equations describe something that arches up to a peak and then returns to its original level or something which drops to a low point and then rises. The rate of change on one side of the peak or low point is the same as on the other side and forms a smooth arch-like shape. A quadratic equation that describes a real world parabolic pattern is another kind of model. Can you think of examples of processes that could be modeled with quadratic equations? Here's an example to get you started. The arches in a cathedral can generally be described by a quadratic equation. Although they are static, the curvature of the arch follows the same pattern on both sides of the… [cont.]
Asked by QUEEN ESHA - Fri Jul 17 09:46:16 2009 - - 1 Answers - 0 Comments
A. ballistic missiles have trajectories that follow a parabolic shape. (a simpler illustration: when you throw a ball in a certain angle, the ball follows a parabolic path)
Answered by Alam Ko Iyan - Fri Jul 17 11:45:34 2009
Explain in simple terms "functions in algebra", please?
Q. are all linear equations functions, give an example of a non linear function
Asked by Manuel P - Tue Aug 18 14:06:12 2009 - - 3 Answers - 0 Comments
A. A function is a relation that assigns to each input number exactly one output number. The input is called the domain of the function. The output is called the range. Not all functions are linear. Some examples of non-linear functions would be: quadratic function (parabola) y = ax^2 + bx + c cubic function: y = x^3 and so on... Look in your text book. It will have definitions of the terms and examples. QED
Answered by the mathemagician - Tue Aug 18 14:14:18 2009
Q. are all linear equations functions, give an example of a non linear function
Asked by Manuel P - Tue Aug 18 14:06:12 2009 - - 3 Answers - 0 Comments
A. A function is a relation that assigns to each input number exactly one output number. The input is called the domain of the function. The output is called the range. Not all functions are linear. Some examples of non-linear functions would be: quadratic function (parabola) y = ax^2 + bx + c cubic function: y = x^3 and so on... Look in your text book. It will have definitions of the terms and examples. QED
Answered by the mathemagician - Tue Aug 18 14:14:18 2009
Is there any explanation for the non-linear increase of kinetic energy as an object accelerates from gravity?
Q. I've been taking AP Physics for two years, so I'd appreciate an intelligent answer to this, and not something along the lines of "cuz the earth haz mass," or otherwise mindless comments. How is it that gravity causes a constant rate of acceleration? Likewise, why does friction cause a constant rate of deceleration to take place (a disproportionally large amount of energy lost when the object is going fast, relative to a slower speed)? If we plug in an object's velocity at three points in time (due to gravity) into the kinetic energy equation, we end up with the following: At one second: (1/2)m(9.8^2)=x At two seconds: (1/2)m(19.6^2)=4x At three seconds: (1/2)m(29.4^2)=16x By contrast, a car with a certain amount of horsepower, for… [cont.]
Asked by Andre - Sat Apr 4 21:26:35 2009 - - 2 Answers - 0 Comments
Q. I've been taking AP Physics for two years, so I'd appreciate an intelligent answer to this, and not something along the lines of "cuz the earth haz mass," or otherwise mindless comments. How is it that gravity causes a constant rate of acceleration? Likewise, why does friction cause a constant rate of deceleration to take place (a disproportionally large amount of energy lost when the object is going fast, relative to a slower speed)? If we plug in an object's velocity at three points in time (due to gravity) into the kinetic energy equation, we end up with the following: At one second: (1/2)m(9.8^2)=x At two seconds: (1/2)m(19.6^2)=4x At three seconds: (1/2)m(29.4^2)=16x By contrast, a car with a certain amount of horsepower, for… [cont.]
Asked by Andre - Sat Apr 4 21:26:35 2009 - - 2 Answers - 0 Comments
Question about second order linear differential equations?
Q. For (a d y/dx + b dy/dx + cy = 0) ["a","b" and "c" being constant coefficients] My book says the general solution is y = Au + Bv [y=u and y=v being distinct solutions of the differential equation; "A" and "B" are non-zero constants] Question: Why is the general solution y = Au + Bv and how did that result from y=u and y=v? I know that substituting y = Au + Bv and its following values of dy/dx and d y/dx into (a d y/dx + b dy/dx + cy = 0) would give 0 proving that y = Au + Bv is a general solution but my question is about how did we originally acquire this equation ( y = Au + Bv ). --- My book mentions examples of finding the functions u and v in specific cases trying a solution of y = e^(mx) ["m" being a constant] in each example and… [cont.]
Asked by Clark - Sat Oct 11 09:57:18 2008 - - 2 Answers - 0 Comments
A. first question: if u is a solution to the differential equation then a u'' + b u' + c u = 0 moreover, a v'' + b v' + c v = 0 now, note the expression y = A u + B v entering that into the differential equation, we have a (Au'' + Bv'') + b (Au' + Bv') + c (A u + B v) = A(au'' + bu' + cu) + B(av'' + bv' + cv) = 0 thus y = Au + Bv is also a solution to the differential eqn . second question: y = e^(mx) was specifically chosen because its higher powers are simply multiples of the original function namely, the nth order derivative = m^n e^(mx) thus when y = e^(mx) is entered into the differential equation a (m^2 e^(mx)) + b(m e^(mx)) + c(e^(mx)) = e^(mx) (am^2 + bm + c) [and we know that one factor above becomes zero .] Edit: There… [cont.]
Answered by Alam Ko Iyan - Sat Oct 11 10:13:57 2008
Q. For (a d y/dx + b dy/dx + cy = 0) ["a","b" and "c" being constant coefficients] My book says the general solution is y = Au + Bv [y=u and y=v being distinct solutions of the differential equation; "A" and "B" are non-zero constants] Question: Why is the general solution y = Au + Bv and how did that result from y=u and y=v? I know that substituting y = Au + Bv and its following values of dy/dx and d y/dx into (a d y/dx + b dy/dx + cy = 0) would give 0 proving that y = Au + Bv is a general solution but my question is about how did we originally acquire this equation ( y = Au + Bv ). --- My book mentions examples of finding the functions u and v in specific cases trying a solution of y = e^(mx) ["m" being a constant] in each example and… [cont.]
Asked by Clark - Sat Oct 11 09:57:18 2008 - - 2 Answers - 0 Comments
A. first question: if u is a solution to the differential equation then a u'' + b u' + c u = 0 moreover, a v'' + b v' + c v = 0 now, note the expression y = A u + B v entering that into the differential equation, we have a (Au'' + Bv'') + b (Au' + Bv') + c (A u + B v) = A(au'' + bu' + cu) + B(av'' + bv' + cv) = 0 thus y = Au + Bv is also a solution to the differential eqn . second question: y = e^(mx) was specifically chosen because its higher powers are simply multiples of the original function namely, the nth order derivative = m^n e^(mx) thus when y = e^(mx) is entered into the differential equation a (m^2 e^(mx)) + b(m e^(mx)) + c(e^(mx)) = e^(mx) (am^2 + bm + c) [and we know that one factor above becomes zero .] Edit: There… [cont.]
Answered by Alam Ko Iyan - Sat Oct 11 10:13:57 2008
HELP! :( need help plotting non-linear graphs!?
Q. I have exams in a couple of weeks and I have no idea how to do these! I have questions such as Plot the graphs for these equations y=(x+2)^2(x-1) y=3-^x y=x^2-x-6 I don't need to solve those exact questions but I do need to know how to do questions like these. I am slowly teaching myself but am getting stuck really quickly! When I go to plot just a basic curve, I don't really know where to start. E.g. for y=3x^2, how do I know what x values to start plotting at? Sometimes the y values come back really large, like say 36, which is too big for my maths book! Just an example is y = 3x^2, if I do say 3 * 3^2, I get 27 - way too big for me to plot on my maths book! Do I always start plotting at x = 0? Or do I start somewhere else? What… [cont.]
Asked by Ater Atra Atrum - Tue Nov 4 03:40:44 2008 - - 2 Answers - 0 Comments
A. Check the domain of x, and y for what it's valid, i refered nonlinear equation chapter full in this 20 minutes for answering your question. No-linear equetion needs further support of domain defination for plotting the graph. Define the domain of the equation you will have your answer. there must be some poles and zeros take out that and then try, answer will be same. one more thing try one variable constant increment.. like x and find respective Y value 1 2 3 4
Answered by Pramey - Tue Nov 4 04:02:44 2008
Q. I have exams in a couple of weeks and I have no idea how to do these! I have questions such as Plot the graphs for these equations y=(x+2)^2(x-1) y=3-^x y=x^2-x-6 I don't need to solve those exact questions but I do need to know how to do questions like these. I am slowly teaching myself but am getting stuck really quickly! When I go to plot just a basic curve, I don't really know where to start. E.g. for y=3x^2, how do I know what x values to start plotting at? Sometimes the y values come back really large, like say 36, which is too big for my maths book! Just an example is y = 3x^2, if I do say 3 * 3^2, I get 27 - way too big for me to plot on my maths book! Do I always start plotting at x = 0? Or do I start somewhere else? What… [cont.]
Asked by Ater Atra Atrum - Tue Nov 4 03:40:44 2008 - - 2 Answers - 0 Comments
A. Check the domain of x, and y for what it's valid, i refered nonlinear equation chapter full in this 20 minutes for answering your question. No-linear equetion needs further support of domain defination for plotting the graph. Define the domain of the equation you will have your answer. there must be some poles and zeros take out that and then try, answer will be same. one more thing try one variable constant increment.. like x and find respective Y value 1 2 3 4
Answered by Pramey - Tue Nov 4 04:02:44 2008
What is the best method for solving a Differential Equation?
Q. If I am given an non -homogeneous Differential Equation, how can I know what the best method for solving it is? In other words how can I tell whether I should use the Method of Undetermined Coefficients, the Laplace Transform, or Variation of Parameters?? Does this have something to do with linear independence? for example: y'' - 4y' + 3y = 4e^2x + xsin(2x) You don't have to solve it just an idea of the problem
Asked by Cheesburger 0 Doome - Thu Apr 17 20:12:15 2008 - - 2 Answers - 0 Comments
A. this depends on which technique you are most able to answer... of course with Laplace Transforms... you must be familiar with many of the transformations to apply that... in the method of undetermined coefficients ... you have the term xsin(2x) on the right side... that alone causes 4 unknowns... thus you might need 5 unknowns ... but you are only solving for system of linear equations... if you are fast in solving that ... it might work for you... in variation of parameters... you must be able to integrate the expression to arrive at the answer... considering that the homogenous part will give you exponential answers... the Wronskian is at most in exponential ... and you will be integrating exponential and trigonometric... i think… [cont.]
Answered by Alam Ko Iyan - Thu Apr 17 20:26:13 2008
Q. If I am given an non -homogeneous Differential Equation, how can I know what the best method for solving it is? In other words how can I tell whether I should use the Method of Undetermined Coefficients, the Laplace Transform, or Variation of Parameters?? Does this have something to do with linear independence? for example: y'' - 4y' + 3y = 4e^2x + xsin(2x) You don't have to solve it just an idea of the problem
Asked by Cheesburger 0 Doome - Thu Apr 17 20:12:15 2008 - - 2 Answers - 0 Comments
A. this depends on which technique you are most able to answer... of course with Laplace Transforms... you must be familiar with many of the transformations to apply that... in the method of undetermined coefficients ... you have the term xsin(2x) on the right side... that alone causes 4 unknowns... thus you might need 5 unknowns ... but you are only solving for system of linear equations... if you are fast in solving that ... it might work for you... in variation of parameters... you must be able to integrate the expression to arrive at the answer... considering that the homogenous part will give you exponential answers... the Wronskian is at most in exponential ... and you will be integrating exponential and trigonometric... i think… [cont.]
Answered by Alam Ko Iyan - Thu Apr 17 20:26:13 2008
Linear Combinations: Clarify what constitutes coplanarity and linear dependence?
Q. With the understanding that coplanarity constitutes three vectors existing in the same plane (two vectors can only be either collinear or a basis for 2-D space), is it possible for three vectors to be coplanar without forming a triangle? If no: Is it possible for three vectors to be linearly dependent without being coplanar (without forming a triangle)? (example: two collinear vectors and a third vector expressed in a linearly dependent combination) (NOTE: My textbook gives these equations, which, I think, support the idea that three vectors must form a triangle to be coplanar, but that it is possible for three vectors to be linearly dependent without being coplanar: linear combination for coplanarity: x=au+bv, where x, u, v are vectors & [cont.]
Asked by Raeychaelle - Wed Mar 5 15:02:04 2008 - - 1 Answers - 0 Comments
A. First off, any two vectors are always "coplanar". Now suppose you have 3 vectors (u,v,w) that are linearly dependent: au + bv + cw = 0 then you can re-write this equation as w = (-a/c) u + (-b/c) v that is, w is a linear combination of vectors u and v. Since u and v lie on some plane, linear combinations of u and v cannot leave that plane. Therefore, w also lies on the same plane.
Answered by Ikabob - Thu Mar 6 01:35:37 2008
Q. With the understanding that coplanarity constitutes three vectors existing in the same plane (two vectors can only be either collinear or a basis for 2-D space), is it possible for three vectors to be coplanar without forming a triangle? If no: Is it possible for three vectors to be linearly dependent without being coplanar (without forming a triangle)? (example: two collinear vectors and a third vector expressed in a linearly dependent combination) (NOTE: My textbook gives these equations, which, I think, support the idea that three vectors must form a triangle to be coplanar, but that it is possible for three vectors to be linearly dependent without being coplanar: linear combination for coplanarity: x=au+bv, where x, u, v are vectors & [cont.]
Asked by Raeychaelle - Wed Mar 5 15:02:04 2008 - - 1 Answers - 0 Comments
A. First off, any two vectors are always "coplanar". Now suppose you have 3 vectors (u,v,w) that are linearly dependent: au + bv + cw = 0 then you can re-write this equation as w = (-a/c) u + (-b/c) v that is, w is a linear combination of vectors u and v. Since u and v lie on some plane, linear combinations of u and v cannot leave that plane. Therefore, w also lies on the same plane.
Answered by Ikabob - Thu Mar 6 01:35:37 2008
microecon: optimal choice and demand?
Q. a.Anita s preference for nutella (#1) and books (#2) is given by the utility function u(x1,x2)= a*ln x1 +b*x2. It's an example of quasilinear prefs, where utility is linear in one good and non-linear in the other. Assume prices, income denoted by (p1, p2, y). (i)Anita s MRS? Interpret expression you get. (ii)The prefs form convex set, utility function is smooth &Anita consumes positive amounts of both. Derive her ordinary demand funtions for nutella and books using the substitution method. In other words, find x1* (p1,p2,y) and x2* (p1,p2,y). (Hint: Use budget equation to substitute x2 into utility function &maximize this new utility w/respect to independent variable x1). (iii)What does the ordinary demand of x1 depend on? x2? ( [cont.]
Asked by musicluver - Wed Jun 11 15:32:14 2008 - - 2 Answers - 0 Comments
A. ii) this is how we did it in my class: start with income function Y=p1x1+p2x2 x2=(Y-p1x1)/p2 then substitute that into the utility function: U=a ln x1 + b((Y-p1x1)/p2) then take the derivative: dU/dx1=a/x1 - (b*p1)/p2 foc: a/x1=(b*p1)/p2 x1=(a*p2)/(b*p1) that's your demand curve for good 1 then substitute that into the x1 function x2=(Y-p1(a*p2/b))/p2 iii) you should be able to do that now that you have the demand curve iv) take the derivative of the demand functions with respect to the price of the good v) take the derivative of the demand functions with respect to the price of the other good EDIT: for part iv, to find whether it's an ordinary good, you take the derivative of the demand curve because you want to know what will happen to… [cont.]
Answered by Danajaan - Wed Jun 11 15:51:10 2008
Q. a.Anita s preference for nutella (#1) and books (#2) is given by the utility function u(x1,x2)= a*ln x1 +b*x2. It's an example of quasilinear prefs, where utility is linear in one good and non-linear in the other. Assume prices, income denoted by (p1, p2, y). (i)Anita s MRS? Interpret expression you get. (ii)The prefs form convex set, utility function is smooth &Anita consumes positive amounts of both. Derive her ordinary demand funtions for nutella and books using the substitution method. In other words, find x1* (p1,p2,y) and x2* (p1,p2,y). (Hint: Use budget equation to substitute x2 into utility function &maximize this new utility w/respect to independent variable x1). (iii)What does the ordinary demand of x1 depend on? x2? ( [cont.]
Asked by musicluver - Wed Jun 11 15:32:14 2008 - - 2 Answers - 0 Comments
A. ii) this is how we did it in my class: start with income function Y=p1x1+p2x2 x2=(Y-p1x1)/p2 then substitute that into the utility function: U=a ln x1 + b((Y-p1x1)/p2) then take the derivative: dU/dx1=a/x1 - (b*p1)/p2 foc: a/x1=(b*p1)/p2 x1=(a*p2)/(b*p1) that's your demand curve for good 1 then substitute that into the x1 function x2=(Y-p1(a*p2/b))/p2 iii) you should be able to do that now that you have the demand curve iv) take the derivative of the demand functions with respect to the price of the good v) take the derivative of the demand functions with respect to the price of the other good EDIT: for part iv, to find whether it's an ordinary good, you take the derivative of the demand curve because you want to know what will happen to… [cont.]
Answered by Danajaan - Wed Jun 11 15:51:10 2008
Math Help Best Answer 10 points?
Q. This is an practice math problem that i need to put in a paper, but the problem is i cant actually find a formula or a way that we actually do this in real life any help? Share a real-world application involving either a non-linear polynomial or a rational expression or a radical expression.it must include a specific equation, function, or formula and you must clearly explain what it does / how it works. Your example needn't be complicated, just practical
Asked by marky w - Mon Oct 20 17:51:51 2008 - - 6 Answers - 0 Comments
A. Real world problem: what's a better deal: a 10" pizza for $8.00 or a 20" pizza for $25.00 Nonlinear polynomial: unit price = price/area = price / ( r ) Lower unit price is the better deal. 25/(10) < 8/(5) ... Answer, the 20" pizza is a better deal.
Answered by intc_escapee - Mon Oct 20 18:13:14 2008
Q. This is an practice math problem that i need to put in a paper, but the problem is i cant actually find a formula or a way that we actually do this in real life any help? Share a real-world application involving either a non-linear polynomial or a rational expression or a radical expression.it must include a specific equation, function, or formula and you must clearly explain what it does / how it works. Your example needn't be complicated, just practical
Asked by marky w - Mon Oct 20 17:51:51 2008 - - 6 Answers - 0 Comments
A. Real world problem: what's a better deal: a 10" pizza for $8.00 or a 20" pizza for $25.00 Nonlinear polynomial: unit price = price/area = price / ( r ) Lower unit price is the better deal. 25/(10) < 8/(5) ... Answer, the 20" pizza is a better deal.
Answered by intc_escapee - Mon Oct 20 18:13:14 2008
From Yahoo Answer Search: 'examples of non linear equations'
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