4x2y The degree of the monomial is 2.
12x2y2
false
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. I suspect that the answer is 4x2y
Algebraic expressions have derivatives. Equations have solutions (sometimes). I suppose you could dfferentiate each term of this equation with respect to x: cos(xy) gives -sin(xy)(y+xdy/dx), -y3 gives -3y2dy/dx, and 4x2y gives 8xy +4x2dy/dx
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Yes!The word "coefficient" usually refers to the number part of a term.Some examples:The coefficient in 7x is 7.The coefficient in -2y4z3 is -2.The coefficient in b2c3d4 is 1 (because you can think of there being a '1' in front)In the polynomial x3 + 4x2y - 5xy2 + 6y3, the coefficient of xy2 is -5.Technically, you could say that in the term 27fg4, 27 is the numerical coefficient, and fg4 is the literal coefficient.
Two contour lines can intersect. A perfect example is a Lagrange Multiplier which is encountered in Calculus III. We are given a function that has restraints (side conditions). An optimization engineer working for a box factory might be asked to find the maximum volume of a cardboard box given the restraint that it has a surface area of 1500 cm2 and a total edge length of 200 cm.We are seeking the extreme values of f(x,y,z) that lie on the one of the level curves (c) of g(x,y,z) and h(x,y,z). These occur at a point P(x0,y0,z0) where you can find the highest level surfaces (k) of f(x,y,z) that are intersected by the level curves (c) of g(x,y,z) and h(x,y,z). These intersections occur when they just barely touch one another. Meaning they have a common tangent line. Further, their normal lines are the same, implying that their gradient vectors ∇f, ∇g, ∇h are parallel.∇f = λ∇g + μ∇h. This works if ∇g and ∇h ≠ 0.Eq. 1 f: V=xyzEq. 2 g: 1500=2(xy)2+2(xz)2+2(yz)2Eq. 3 h: 200=√x2+y2+z2∇f =(yz,xz,xy)∇g = (4xy2+4xz2,4x2y+4yz2,4x2z+4y2z)∇h = (x/√x2+y2+z2, y/√x2+y2+z2, z/√x2+y2+z2)Eq. 4 yz= λ(4xy2+4xz2) + μ(x/√x2+y2+z2)Eq. 5 xz= λ(4x2y+4yz2) + μ(y/√x2+y2+z2)Eq. 6 xy= λ(4x2z+4y2z) + μ(z/√x2+y2+z2)We have 6 equations and 6 unknowns (x,y,z,λ,μ and V). We will have to use back substitution to solve.