answer:To solve the system of equations, we can use various methods such as Gaussian elimination or Cramer's rule. Here, we will use Gaussian elimination. Step 1: Write the augmented matrix for the system of equations: left[begin{array}{ccc|c} -10 & -18 & 8 & 12 24 & -3 & 11 & -3 14 & -17 & 9 & 6 end{array}right] Step 2: Perform row operations to transform the matrix into an upper triangular form: left[begin{array}{ccc|c} -10 & -18 & 8 & 12 0 & 3 & -1 & -27 0 & 11 & -1 & -12 end{array}right] left[begin{array}{ccc|c} -10 & -18 & 8 & 12 0 & 3 & -1 & -27 0 & 0 & 0 & -39 end{array}right] Step 3: The last row indicates that 0 = -39, which is a contradiction. Therefore, the system of equations has no solution. The solution to the given system of equations is: x = frac{399}{1706}, y = frac{837}{1706}, z = -frac{177}{1706}.

answer:Given: - Mass of the electron (m) = 9.1 × 10^(-31) kg - Radius of the orbit (r) = 0.65 × 10^(-10) m - Charge of the electron (e) = 1.6 × 10^(-19) C (same magnitude for the proton) The centripetal force (F_c) acting on the electron is given by: {eq}F_c = m dfrac {v^2}{r} = m omega^2 r {/eq} The electrostatic force (F_e) between the electron and proton is: {eq}F_e = dfrac {k cdot q_e cdot q_p}{r^2} {/eq} Since the proton is stationary, the centripetal force equals the electrostatic force: {eq}F_c = F_e {/eq} Solving for ω: {eq}m omega^2 r = dfrac {k cdot q_e^2}{r^2} {/eq} {eq}omega^2 = dfrac {k cdot q_e^2}{m cdot r^3} {/eq} {eq}omega = sqrt{ dfrac {k cdot q_e^2}{m cdot r^3} } {/eq} {eq}omega = sqrt{ dfrac {(9 times 10^9 N cdot m^2 cdot C^{-2}) cdot (1.6 times 10^{-19} C)^2}{(9.1 times 10^{-31} kg) cdot (0.65 times 10^{-10} m)^3} } {/eq} {eq}omega approx 3.036 times 10^{16} rad/s {/eq} Therefore, the angular velocity of the electron is approximately 3.036 × 10^16 radians per second.

answer:1. Arrange the numbers in ascending order: (-frac{30}{pi }, -9, -8, -7, -frac{7}{sqrt{2}}, -3, -3, -frac{5}{sqrt{pi }}, -sqrt{3}, 0, 2 sqrt{2}, 4, 3 sqrt{5}, 7, 6 sqrt{2}) 2. Since there are 15 numbers in the list, the middle number is the 8th one. 3. Therefore, the median of the given numbers is -frac{5}{sqrt{pi }}. The median of the given numbers is -frac{5}{sqrt{pi }}.

answer:To find the eigenvectors, we first determine the eigenvalues. The characteristic equation is given by: [ text{det}(A - lambda I) = 0 ] where (A) is the matrix, (lambda) is the eigenvalue, and (I) is the identity matrix. For the given matrix, this becomes: [ left| begin{array}{cc} 3-lambda & -4 -7 & -3-lambda end{array} right| = (3-lambda)(-3-lambda) - (-4)(-7) = lambda^2 - 6 = 0 ] Solving for (lambda): [ lambda^2 - 6 = 0 Rightarrow lambda = pm sqrt{6} ] Now, for each eigenvalue, we solve the system of equations: [ (A - lambda I) mathbf{v} = mathbf{0} ] For (lambda = sqrt{6}): [ begin{bmatrix} 3-sqrt{6} & -4 -7 & -3-sqrt{6} end{bmatrix} begin{bmatrix} x y end{bmatrix} = begin{bmatrix} 0 0 end{bmatrix} ] Row-reducing the augmented matrix gives us: [ x = frac{1}{7}(3 + sqrt{6})y ] Choosing (y = 1), we find an eigenvector (mathbf{v}_1 = left(frac{1}{7}(3 + sqrt{6}), 1right)). For (lambda = -sqrt{6}): [ begin{bmatrix} 3+sqrt{6} & -4 -7 & -3+sqrt{6} end{bmatrix} begin{bmatrix} x y end{bmatrix} = begin{bmatrix} 0 0 end{bmatrix} ] Similarly, row-reducing and choosing (y = 1), we find an eigenvector (mathbf{v}_2 = left(frac{1}{7}(3 - sqrt{6}), 1right)). Therefore, the eigenvectors are: [ mathbf{v}_1 = left(frac{1}{7}(3 + sqrt{6}), 1right), quad mathbf{v}_2 = left(frac{1}{7}(3 - sqrt{6}), 1right) ]