answer:Let {eq}x {/eq} be the percentage abundance of Si-29. Then, the percentage abundance of Si-28 is {eq}(100-x-3.09)=96.91-x {/eq}, and the percentage abundance of Si-30 is 3.09. Using the formula for average atomic mass, we have: {eq}begin{align*} 28.10;rm amu &=27.97693;rm amu times dfrac{96.91-x}{100}+ 28.97649;rm amu times dfrac{x}{100}+29.97377;rm amu times dfrac{3.09}{100}[0.3cm] Rightarrowqquad x&=boxed{mathbf{5.12%}} end{align*} {/eq}

answer:This problem aims to illustrate the distinction between the union of subspaces and their span. The span of a set of vectors represents the set of all linear combinations of those vectors, while the union simply combines the vectors without filling in the gaps. The span of e_1 and e_2 forms a two-dimensional subspace, which is a plane in mathbb{R}^4. The span of e_3 is a one-dimensional subspace, a line parallel to the z-axis. The union of these two subspaces consists of vectors either in the plane or on the line. This is not a vector space because it does not include all linear combinations of e_1, e_2, and e_3. To find a vector in the span of e_1, e_2, and e_3 but not in their union, consider a vector with non-zero first and second coordinates and zeros elsewhere (denoted as v) and another with a non-zero third coordinate and zeros elsewhere (denoted as w). The sum of v and w will be outside the union but within the span of the three basis vectors. For instance, let v = (1, 1, 0, 0) and w = (0, 0, 1, 0). Then, v + w = (1, 1, 1, 0), which belongs to operatorname{span}(e_1, e_2, e_3) but not to operatorname{span}(e_1, e_2) cup operatorname{span}(e_3), as it lies outside both the xy-plane and the z-axis. Geometrically, the spans can be visualized as follows: (1) The span of e_1, e_2, and e_3 is the entire xyz-hyperplane in mathbb{R}^4. (2) The union of the spans of e_1 and e_2 and e_3 individually is the xy-plane combined with the z-axis. Note that the union does not cover points in the xy-plane that are not on the z-axis, while the span of all three vectors does.

answer:To analyze the object's motion, consider the forces acting on it: 1. Force from the right: 10 newtons (towards the positive direction) 2. Force from the left: 17 newtons (towards the negative direction) 3. Additional force from the right: 7 newtons (towards the positive direction) Initially, the net force is the vector sum of these forces: {eq}begin{align*} F_{net}&=F_{right1} + F_{left} + F_{right2} &=10; text{N} - 17; text{N} + 7; text{N} &=0; text{N} end{align*} {/eq} As the net force is zero, the object will not accelerate but will continue to move with a constant velocity, maintaining its motion to the right. Therefore, the correct answer is: A. The object will maintain its motion to the right.

answer:To expand the characteristic polynomial f(lambda) = det(lambda A + B) of two conics A: XAX^T and B: XBX^T in a form that can be computed by a computer program or MATLAB, we can use the following steps: 1. **Symbolic Representation**: - Define symbolic variables for the elements of matrices A and B. - Use the symbolic toolbox in MATLAB (e.g., `syms`) to represent the symbolic variables. 2. **Matrix Construction**: - Construct the matrices A and B using the symbolic variables. 3. **Characteristic Polynomial**: - Construct the matrix lambda A + B by adding the symbolic variable lambda multiplied by matrix A to matrix B. - Use the `det` function in MATLAB to compute the determinant of the matrix lambda A + B. 4. **Output**: - The output of the `det` function will be a symbolic expression representing the characteristic polynomial f(lambda). 5. **Numerical Evaluation**: - To evaluate the characteristic polynomial for specific values of lambda, substitute the desired values into the symbolic expression and use the `eval` function in MATLAB. Here's an example code in MATLAB: ```matlab syms x; % Symbolic variable for lambda A = eye(3); % Define matrix A as the identity matrix B = [1, 2, 3; 4, 5, 6; 7, 8, 9]; % Define matrix B characteristic_polynomial = det(x * A + B); % Compute the characteristic polynomial disp(characteristic_polynomial); % Display the symbolic expression lambda_value = 2; % Example value for lambda numerical_value = eval(characteristic_polynomial); % Evaluate the polynomial at lambda = 2 disp(numerical_value); % Display the numerical value ``` This code will output the symbolic expression of the characteristic polynomial and the numerical value when lambda = 2.