answer:To find an equation of the plane, we can use the following steps: 1. Find two vectors parallel to the plane. 2. Find a vector perpendicular to the plane by taking the cross product of the two vectors found in step 1. 3. Use the point-normal form of the equation of a plane to write an equation of the plane. Following these steps, we have: 1. Two vectors parallel to the plane are: {eq}overrightarrow{v_1} = langle 4-3, 4-5, 0-(-3) rangle = langle 1, -1, 3 rangle {/eq} {eq}overrightarrow{v_2} = langle 4-4, 5-4, 2-0 rangle = langle 0, 1, 2 rangle {/eq} 2. A vector perpendicular to the plane is: {eq}overrightarrow{n} = overrightarrow{v_1} times overrightarrow{v_2} = begin{vmatrix} mathbf{i}& mathbf{j} & mathbf{k} 1& -1 & 3 0& 1 & 2 end{vmatrix} = (1*3-(-1)(2))mathbf{i}-(0*3-1*2)mathbf{j}+(0*(-1)-1*1)mathbf{k} = langle 5, 2, -1 rangle {/eq} 3. Using the point-normal form of the equation of a plane, we have: {eq}5(x-3) + 2(y-5) - (z+3) = 0 {/eq} Simplifying, we get: {eq}5x - 15 + 2y - 10 - z - 3 = 0 {/eq} {eq}5x + 2y - z = 28 {/eq} Therefore, an equation of the plane containing the three points (3, 5, -3), (4, 4, 0), and (4, 5, 2) in which the coefficient of x is -5 is: {eq}boxed{-5x - 2y + z = -28} {/eq}

answer:(a) Since the image is formed on a screen, it indicates that the lens is convergent, commonly known as a convex lens. A divergent (concave) lens would create a virtual image that cannot be projected on a screen. (b) The magnification (M) is the ratio of the image size to the object size. In this case, the image is twice the size of the object, so the magnification is -2 (negative because the image is inverted). The object distance (d_o) is positive since the object is in front of the lens, and the image distance (d_i) is also positive because the image is on the opposite side of the lens. Thus, the image is inverted. (c) To determine the relationship between the object distance (d_o) and the focal distance (f), we can use the lens equation: [ frac{1}{d_o} + frac{1}{d_i} = frac{1}{f} ] Since the image is real and projected on the screen (d_i > 0), we can rearrange the equation to show that the object distance must be greater than the focal distance: [ d_i = frac{1}{frac{1}{f} - frac{1}{d_o}} > 0 ] This inequality leads to: [ d_o > f ] Hence, the object distance is greater than the lens's focal distance.

answer:The binomial coefficient binom{n}{k} represents the number of ways to choose k elements from a set of n elements, without regard to order. To compute binom{4033}{4031}, we can use the formula: binom{n}{k} = frac{n!}{k!(n-k)!} Here, n=4033 and k=4031. Therefore, binom{4033}{4031} = frac{4033!}{4031!(4033-4031)!} = frac{4033 times 4032}{2 times 1} = 8066 times 2016 = 162693312 So, binom{4033}{4031} = 162693312.

answer:90555 = 3 cdot 5 cdot 6037