# WRITE THE DEGREE OF EACH POLYNOMIAL GIVEN BELOW: XYZ

Applying the AM-GM is the right strategy, but you need to bởi vì it a bit differently. X^2+4xy+4y^2+4z^2 = x^2+2xy+2xy+4y^2+2z^2+2z^2 ge 6sqrt<6>x^2 cdot 2xy cdot 2xy cdot 4y^2 cdot 2z^2 cdot 2z^2 ...

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fMy; Solution:: We Can write xcdot ycdot z= 3000 = 2^3cdot 3^1cdot 5^3 Now let x=2^x_1cdot 3^y_1cdot 5^z_1 và 2^x_2cdot 3^y_2cdot 5^z_2 and 2^x_3cdot 3^y_3cdot 5^z_3 ...
This is intended as a complement to lớn Alex Bartel's answer. (1) Consider the 2d case given by x = x(t), y = y(t). Here is a graph of a parametrized function with (x(t_0),y(t_0)) (in red) và (x(t_0 + delta), y(t_0 + delta)) ...

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In general, such a line will be parametrized by x = az + 2, y = bz + 5 unless the line lies on the plane z = 0, which is easy to see not lớn be the case. In a more general case, the linear ...
the gradient is indeed the solution. Take a curve, gamma that lies in S. Then Fcirc gamma is zero everywhere. So its derivative is zero. But its derivative is just abla F(gamma(s)). gamma'(s) ...

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What you actually want here is the point-line distance formula for three dimensions . However, the cylinder equation you did get is correct. Using formula 10 in the liên kết I gave, we have r^2=fraclangle x,y,z angle imes(langle x,y,z angle-langle 1,1,1 angle)1^2+1^2+1^2=frac(z-y)^2+(x-z)^2+(y-x)^21^2+1^2+1^2 ...
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left< eginarray l l 2 & 3 \ 5 và 4 endarray ight> left< eginarray l l l 2 & 0 & 3 \ -1 và 1 và 5 endarray ight>

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