answer:1. Convert the mixed fractions for the dimensions of the rectangle into improper fractions: [ 60 frac{1}{2} text{ cm} = frac{121}{2} text{ cm} ] [ 47 frac{2}{3} text{ cm} = frac{143}{3} text{ cm} ] 2. Suppose that the artist uses squares of side length ( s ), and that he uses ( M ) squares along the length, and ( N ) along the width. Then we must have: [ M s = 60 frac{1}{2} text{ cm} quad text{and} quad N s = 47 frac{2}{3} text{ cm} ] 3. The total number of squares covering the rectangle is ( M times N ), and we don't need to know ( s ). Dividing the two equations cancels ( s ): [ frac{M s}{N s} = frac{60 frac{1}{2}}{47 frac{2}{3}} implies frac{M}{N} = frac{frac{121}{2}}{frac{143}{3}} = frac{121}{2} cdot frac{3}{143} = frac{121 times 3}{2 times 143} = frac{363}{286} ] 4. Simplify the fraction ( frac{363}{286} ): [ frac{363}{286} = frac{11 times 33}{11 times 26} = frac{33}{26} ] 5. We want positive integers ( M ) and ( N ) such that ( frac{M}{N} = frac{33}{26} ). Since ( frac{33}{26} ) is in its simplest form, the smallest integers ( M ) and ( N ) that work are ( M = 33 ) and ( N = 26 ). 6. Calculate the total number of squares: [ M times N = 33 times 26 = 858 ] # Conclusion [ boxed{858} ]

answer:1. **Applying the Cosine Law in Triangle BKC**: Using the cosine law in triangle BKC, we need to find cos angle C: [ BC = 2, quad KC = 1, quad BK = frac{3 sqrt{2}}{2} ] The cosine law states: [ BK^2 = BC^2 + KC^2 - 2 cdot BC cdot KC cdot cos(angle C) ] Plug in the given values: [ left( frac{3 sqrt{2}}{2} right)^2 = 2^2 + 1^2 - 2 cdot 2 cdot 1 cdot cos(angle C) ] Simplify: [ frac{18}{4} = 4 + 1 - 4 cos(angle C) ] Combine like terms: [ frac{9}{2} = 5 - 4 cos(angle C) ] Solve for cos(angle C): [ 4 cos(angle C) = 5 - frac{9}{2} ] Further simplification: [ 4 cos(angle C) = frac{10}{2} - frac{9}{2} ] [ 4 cos(angle C) = frac{1}{2} ] [ cos(angle C) = frac{1}{8} ] 2. **Using the Angle Bisector Property**: According to the angle bisector theorem, the ratio AB : AK = BC : KC: [ frac{AB}{AK} = frac{BC}{KC} = 2 ] Let AK = x. Then AB = 2x. 3. **Applying the Cosine Law in Triangle ABC**: Using the cosine law in triangle ABC, where AC = AK + KC = x + 1: [ AB^2 = AC^2 + BC^2 - 2 cdot AC cdot BC cdot cos(angle C) ] Plug in the values: [ (2x)^2 = (x + 1)^2 + 4 - 2 cdot (x + 1) cdot 2 cdot frac{1}{8} ] Simplify the equation: [ 4x^2 = (x + 1)^2 + 4 - frac{1}{2} cdot (x + 1) ] Expand and simplify: [ 4x^2 = x^2 + 2x + 1 + 4 - frac{x}{2} - frac{1}{2} ] Combine like terms: [ 4x^2 = x^2 + 2x + 1 + 4 - frac{x}{2} - frac{1}{2} ] Simplify further: [ 4x^2 = x^2 + frac{4x}{2} + frac{7}{2} - frac{x}{2} = x^2 + frac{3x}{2} + frac{7}{2} ] Multiply through by 2 to clear the fraction: [ 8x^2 = 2x^2 + 3x + 7 ] 4. **Solve the Equation for x**: Move all terms to one side: [ 8x^2 - 2x^2 - 3x - 7 = 0 ] Combine like terms: [ 6x^2 - 3x - 7 = 0 ] Solve the quadratic equation for x: [ x = frac{3}{2} ] 5. **Calculate the Area of triangle ABC**: Using the area formula for a triangle and the property sin(theta) = sqrt{1 - cos^2(theta)}, find sin(angle C): [ sin(angle C) = sqrt{1 - left( frac{1}{8} right)^2} = sqrt{1 - frac{1}{64}} = sqrt{ frac{63}{64}} = frac{sqrt{63}}{8} ] Area: [ S_{triangle ABC} = frac{1}{2} BC cdot AC cdot sin(angle C) ] Substitute BC = 2, AC = x + 1 = frac{5}{2}, and sin(angle C) = frac{sqrt{63}}{8}: [ S_{triangle ABC} = frac{1}{2} cdot 2 cdot frac{5}{2} cdot frac{sqrt{63}}{8} = frac{5 sqrt{63}}{16} ] # Conclusion: [ boxed{frac{15 sqrt{7}}{16}} ]

answer:He already has 247 nails and found another 144, so he has: 247 + 144 = 391 nails He needs to buy 109 more nails to reach his total, so: 391 + 109 = 500 nails Therefore, the total number of nails he needs is boxed{500} .

answer:Let mathbf{v} = begin{pmatrix} x y z end{pmatrix}. Then [begin{pmatrix} 3 -4 7 end{pmatrix} times begin{pmatrix} x y z end{pmatrix} = begin{pmatrix} -4z - 7y 7x - 3z -4x + 3y end{pmatrix}.] Expressing the cross product result as a matrix-vector product, [begin{pmatrix} -4z - 7y 7x - 3z -4x + 3y end{pmatrix} = begin{pmatrix} 0 & -7 & -4 7 & 0 & -3 -4 & 3 & 0 end{pmatrix} begin{pmatrix} x y z end{pmatrix}.] Thus, [mathbf{M} = boxed{begin{pmatrix} 0 & -7 & -4 7 & 0 & -3 -4 & 3 & 0 end{pmatrix}}.]