Q: If xy and yz statement must be true?

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There appear to be 10 terms in the determinant. A determinant can only have a perfect number of terms. So something has gone wrong with the question. 1: x2 plus 1 2: xy 3: xz 4: xy 5: y2 plus 1 6: yz 7: 1 plus x2 plus y2 plus z2 8: xz 9: yz 10: z2 plus 1

y(z+x) + 4(x+z)

With a fixed surface area you can actually maximize the volume of the box without using multi-variable calculus. Here's how: Suppose the dimensions are x, y, z with 2(xy + yz + zx) = C for some constant C. This implies xy + yz + zx = C1 for a different constant C1. We will use what mathematicians call the AM-GM inequality (Arithmetic mean - geometric mean inequality) which says that, for some positive numbers, the arithmetic mean is always greater than or equal to the geometric mean, with equality occurring iff all the numbers are equal. The AM-GM inequality says that (xy + yz + zx)/3 >= (xy*yz*zx)1/3 (couldn't do cube root) ((xy + yz + zx)/3)3/2 >= xyz The left side is equal to a constant (since it has the expression for surface area), so the maximal value of xyz (the volume) is equal to that constant. This happens when x = y = z, i.e. the box is cube-shaped.

10 cm

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The only common factor to all terms is yz. → xy³z² + y²z + xyz = yz(xy²z + y + x)

2(xy+xz+yz)=100 xy+xz+yz=50 or x(y+z)+yz=50 x=2, y=4, z=7

If WXYZ is a square, which statements must be true? Check all that apply: ANSWERS (apex): angle W is supplementary to angle Y. angle W is congruent to angle Y. angle W is a right angle. WXYZ is a parallelogram WX ≅ XY

xy times yz divided by two is the area.2a*a/2 is the area 36a^2=36a=6xy is 12.CHECK:6*12=72half of 72 is 36. Yeah!

There appear to be 10 terms in the determinant. A determinant can only have a perfect number of terms. So something has gone wrong with the question. 1: x2 plus 1 2: xy 3: xz 4: xy 5: y2 plus 1 6: yz 7: 1 plus x2 plus y2 plus z2 8: xz 9: yz 10: z2 plus 1

8.3

y(z+x) + 4(x+z)

If the lengths of the edges are x, y and z units, then the total surface area is 2*(xy + yz + zx).

With a fixed surface area you can actually maximize the volume of the box without using multi-variable calculus. Here's how: Suppose the dimensions are x, y, z with 2(xy + yz + zx) = C for some constant C. This implies xy + yz + zx = C1 for a different constant C1. We will use what mathematicians call the AM-GM inequality (Arithmetic mean - geometric mean inequality) which says that, for some positive numbers, the arithmetic mean is always greater than or equal to the geometric mean, with equality occurring iff all the numbers are equal. The AM-GM inequality says that (xy + yz + zx)/3 >= (xy*yz*zx)1/3 (couldn't do cube root) ((xy + yz + zx)/3)3/2 >= xyz The left side is equal to a constant (since it has the expression for surface area), so the maximal value of xyz (the volume) is equal to that constant. This happens when x = y = z, i.e. the box is cube-shaped.

10 cm

I am assuming that you are in a three dimensional world. Then the three planes of projection would be the xy plane, the xz plane, and the yz plane.