Solutions
... will make things simpler. This equation now reads 0 = x1 + 3x2 So if we let x2 be any non-zero real number and let x1 = −3x2 , then we see that this equation is satisfied. In particular, if we choose (x1 , x2 ) = (−3, 1) we find an explicit example. We then see that the values x1 = −3 x2 = 1 x4 = 0 ...
... will make things simpler. This equation now reads 0 = x1 + 3x2 So if we let x2 be any non-zero real number and let x1 = −3x2 , then we see that this equation is satisfied. In particular, if we choose (x1 , x2 ) = (−3, 1) we find an explicit example. We then see that the values x1 = −3 x2 = 1 x4 = 0 ...
Full text
... Cassels [1] proved that there are no rationals that satisfy the conditions of (1). Cassels also shows that this problem was expressed by Mordell [3], in equivalent, if not exact form. Additionally, Cassels has compiled an excellent bibliography that demonstrates that the "Mnich" problem has its root ...
... Cassels [1] proved that there are no rationals that satisfy the conditions of (1). Cassels also shows that this problem was expressed by Mordell [3], in equivalent, if not exact form. Additionally, Cassels has compiled an excellent bibliography that demonstrates that the "Mnich" problem has its root ...
solution
... The exponent n can be either even or odd. Assume n is even, then n = 2k where k is an integer. Thus 2n = 22k = 4k. Four to any power has either 4 or 6 as a unit digit (this is obvious when we try a few examples), so the exponent n can be any even integer. Can n be odd? Assume n = 2k + 1. Then 2n = 2 ...
... The exponent n can be either even or odd. Assume n is even, then n = 2k where k is an integer. Thus 2n = 22k = 4k. Four to any power has either 4 or 6 as a unit digit (this is obvious when we try a few examples), so the exponent n can be any even integer. Can n be odd? Assume n = 2k + 1. Then 2n = 2 ...
Lesson 4.1: Solving Two
... Word Problem #2 • Diane sold 9 decorated flowers that cost the same amount each plus a dozen roses for $28. All together she sold $73 in flowers. How much was each decorated flower? ...
... Word Problem #2 • Diane sold 9 decorated flowers that cost the same amount each plus a dozen roses for $28. All together she sold $73 in flowers. How much was each decorated flower? ...
section 6.1
... accurate results when solving systems of equations. It is most often used when a variable is alone on one side of an equation or when it is easy to solve for a variable. One equation is used to express one variable in terms of the other, then it is substituted in the other equation. ...
... accurate results when solving systems of equations. It is most often used when a variable is alone on one side of an equation or when it is easy to solve for a variable. One equation is used to express one variable in terms of the other, then it is substituted in the other equation. ...
Partial differential equation
In mathematics, a partial differential equation (PDE) is a differential equation that contains unknown multivariable functions and their partial derivatives. (A special case are ordinary differential equations (ODEs), which deal with functions of a single variable and their derivatives.) PDEs are used to formulate problems involving functions of several variables, and are either solved by hand, or used to create a relevant computer model.PDEs can be used to describe a wide variety of phenomena such as sound, heat, electrostatics, electrodynamics, fluid flow, elasticity, or quantum mechanics. These seemingly distinct physical phenomena can be formalised similarly in terms of PDEs. Just as ordinary differential equations often model one-dimensional dynamical systems, partial differential equations often model multidimensional systems. PDEs find their generalisation in stochastic partial differential equations.