answer:To find the cross product, arrange the vectors in a matrix and compute the determinant for each component. The cross product of begin{pmatrix} 4 3 -5 end{pmatrix} and begin{pmatrix} 2 -1 4 end{pmatrix} is: [ begin{pmatrix} i & j & k 4 & 3 & -5 2 & -1 & 4 end{pmatrix} ] Using the determinant formula, compute each component as follows: - i-component: begin{vmatrix} 3 & -5 -1 & 4 end{vmatrix} = 3(4) - (-5)(-1) = 12 - 5 = 7 - j-component: -begin{vmatrix} 4 & -5 2 & 4 end{vmatrix} = -(4(4) - (-5)(2)) = -(16 + 10) = -26 - k-component: begin{vmatrix} 4 & 3 2 & -1 end{vmatrix} = 4(-1) - 3(2) = -4 - 6 = -10 Thus, the cross product is: [ boxed{begin{pmatrix} 7 -26 -10 end{pmatrix}} ] To verify if the resulting vector is orthogonal to begin{pmatrix} 4 3 -5 end{pmatrix}, compute the dot product: [ (7)(4) + (-26)(3) + (-10)(-5) = 28 - 78 + 50 = 0 ] Conclusion: The cross product of the two vectors is begin{pmatrix} 7 -26 -10 end{pmatrix}, and it is orthogonal to the vector begin{pmatrix} 4 3 -5 end{pmatrix} as the dot product equals zero.

answer:To solve the problem step by step, we follow the given information and perform the calculations accordingly: 1. First, we calculate the total weight of the bags of chips Jasmine buys. Since each bag weighs 20 ounces and she buys 6 bags, the total weight of the bags of chips is: [ 20 , text{ounces/bag} times 6 , text{bags} = 120 , text{ounces} ] 2. Next, we find out how many tins of cookies Jasmine buys. She buys 4 times as many tins as the bags of chips, so: [ 6 , text{bags} times 4 = 24 , text{tins} ] 3. Then, we calculate the total weight of the tins of cookies. Since each tin weighs 9 ounces and she buys 24 tins, the total weight of the tins of cookies is: [ 9 , text{ounces/tin} times 24 , text{tins} = 216 , text{ounces} ] 4. To find the total weight Jasmine has to carry, we add the weight of the bags of chips and the tins of cookies: [ 120 , text{ounces} + 216 , text{ounces} = 336 , text{ounces} ] 5. Finally, since we need to find the total weight in pounds and there are 16 ounces in a pound, we convert the total weight into pounds: [ 336 , text{ounces} div 16 , text{ounces/pound} = 21 , text{pounds} ] Therefore, Jasmine has to carry boxed{21} pounds.

answer:Let's break down the problem step by step: 1. There are 5 classes, and each class uses 2 whiteboards. So, the total number of whiteboards used is 5 classes * 2 whiteboards/class = 10 whiteboards. 2. It costs 100 to use all the whiteboards for one day. 3. Ink costs 50 cents per ml, which is the same as 0.50/ml. Now, we need to find out how many milliliters of ink are used for all the whiteboards in one day. Since 100 is spent on ink, we can calculate the total milliliters of ink used as follows: Total cost / Cost per ml = Total ml of ink used 100 / 0.50/ml = 200 ml So, 200 ml of ink is used for all 10 whiteboards in one day. To find out how many milliliters of ink each whiteboard needs for a day's use, we divide the total ml of ink by the number of whiteboards: Total ml of ink / Number of whiteboards = ml of ink per whiteboard 200 ml / 10 whiteboards = 20 ml per whiteboard Each whiteboard needs boxed{20} ml of ink for a day's use.

answer:We are given three vertices of a parallelogram in the order of traversal as (a, b), (0,0), and (c,d). We are required to find the coordinates of the fourth vertex. 1. Let us denote the coordinates of the vertices of the parallelogram as follows: - A(a, b) - B(0, 0) - C(c, d) 2. We know that in a parallelogram, the diagonals bisect each other. Therefore, the midpoint of the diagonal AC should coincide with the midpoint of the diagonal BD where D is the fourth vertex of the parallelogram we need to determine. 3. Let’s find the midpoint of diagonal AC: [ text{Midpoint of } AC = left( frac{a + c}{2}, frac{b + d}{2} right) ] 4. Let D(x, y) be the fourth vertex. Using the property of the parallelogram, the midpoint of diagonal BD should be the same as the midpoint of diagonal AC. Therefore, the midpoint of BD is: [ text{Midpoint of } BD = left( frac{0 + x}{2}, frac{0 + y}{2} right) = left( frac{x}{2}, frac{y}{2} right) ] 5. Equate the midpoints of the diagonals AC and BD: [ left( frac{a + c}{2}, frac{b + d}{2} right) = left( frac{x}{2}, frac{y}{2} right) ] 6. By comparing the coordinates, we get: [ frac{a + c}{2} = frac{x}{2} quad Rightarrow quad x = a + c ] [ frac{b + d}{2} = frac{y}{2} quad Rightarrow quad y = b + d ] So, the coordinates of the fourth vertex D are (a+c, b+d). # Conclusion The coordinates of the fourth vertex of the parallelogram are: [ boxed{(a+c, b+d)} ]