answer:The cell membrane acts as a selective barrier, controlling the movement of substances into and out of the cell. It allows the passage of essential nutrients, oxygen, and water while preventing the entry of harmful substances and the loss of vital cellular components. The membrane's selective permeability helps maintain the cell's internal environment by regulating the concentrations of ions, molecules, and other substances, thereby contributing to homeostasis.

answer:To find the equation of the plane, we first determine the vectors formed by the given points. Let's denote the points as P_1 = (-1, 0, -5), P_2 = (-5, 5, -2), and P_3 = (3, 2, -4). Vector vec{P_1P_2} = (-5 - (-1), 5 - 0, -2 - (-5)) = (-4, 5, 3) Vector vec{P_1P_3} = (3 - (-1), 2 - 0, -4 - (-5)) = (4, 2, 1) The normal vector of the plane, vec{n}, can be found by taking the cross product of vec{P_1P_2} and vec{P_1P_3}: vec{n} = vec{P_1P_2} times vec{P_1P_3} = begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} -4 & 5 & 3 4 & 2 & 1 end{vmatrix} = (1 cdot 2 - 3 cdot 2) mathbf{i} - (3 cdot 4 - 1 cdot 4) mathbf{j} + (4 cdot 5 - (-4) cdot 2) mathbf{k} = (-4, -8, 28) Now, we use one of the points, say P_1, and the normal vector vec{n} to write the equation of the plane: -4(x + 1) - 8(y - 0) + 28(z + 5) = 0 Simplifying, we get the equation of the plane: -4x - 8y + 28z + 144 = 0 Therefore, the equation of the plane is: x + 2y - 7z - 36 = 0

answer:- The cevian is a line segment that connects a vertex of a triangle to a point on the opposite side. The coordinates of the point on the opposite side are given by the formula left( frac{a(b^2+c^2-a^2)}{2bc}, frac{a^2+b^2-c^2}{2b} right), where a, b, and c are the lengths of the sides of the triangle. - The median is a line segment that connects a vertex of a triangle to the midpoint of the opposite side. The coordinates of the midpoint of the opposite side are given by the formula left( frac{b+c}{2}, frac{a}{2} right), where a, b, and c are the lengths of the sides of the triangle. - The symmedian is a line segment that connects a vertex of a triangle to the centroid of the triangle. The centroid is the point of intersection of the medians of a triangle. The coordinates of the centroid are given by the formula left( frac{a+b+c}{3}, frac{a^2+b^2+c^2}{3(a+b+c)} right), where a, b, and c are the lengths of the sides of the triangle. - The altitude is a line segment that connects a vertex of a triangle to the base of the triangle and is perpendicular to the base. The coordinates of the point on the base are given by the formula left( a cos (C), a sin (C) right), where a is the length of the side opposite angle C. Cevian: left( begin{array}{cc} 3 sqrt{12 sin (21 {}^{circ})+13} & 0 frac{9 (2 sin (21 {}^{circ})+3)}{2 sqrt{12 sin (21 {}^{circ})+13}} & frac{9 cos (21 {}^{circ})}{sqrt{12 sin (21 {}^{circ})+13}} end{array} right) Median: left( begin{array}{cc} frac{3}{2} (6+9 cos (111 {}^{circ})) & 0 frac{9 sin (111 {}^{circ})}{2} & frac{9 cos (111 {}^{circ})}{2} end{array} right) Symmedian: left( begin{array}{cc} frac{1}{3} (6+9 cos (111 {}^{circ})+9 cos (222 {}^{circ})) & 0 frac{9 sin (111 {}^{circ})+9 sin (222 {}^{circ})}{3} & frac{9 cos (111 {}^{circ})+9 cos (222 {}^{circ})}{3} end{array} right) Altitude: left( begin{array}{cc} 6 cos (111 {}^{circ}) & 0 6 sin (111 {}^{circ}) & 6 cos (111 {}^{circ}) end{array} right)

answer:To find the eigenvectors, we first need to determine the eigenvalues. We solve the characteristic equation: left| begin{matrix} 9 - lambda & -7 -9 & 5 - lambda end{matrix} right| = 0 This simplifies to: lambda^2 - 14lambda - 18 = 0 Factoring or using the quadratic formula, we get: lambda = 7 pm sqrt{67} Now we find the eigenvectors for each eigenvalue: For lambda = 7 + sqrt{67}: begin{bmatrix} 2 - sqrt{67} & -7 -9 & -2 - sqrt{67} end{bmatrix} begin{bmatrix} v_1 v_2 end{bmatrix} = begin{bmatrix} 0 0 end{bmatrix} Reducing the system, we get the eigenvector: v = begin{bmatrix} frac{1}{9} left(-2 - sqrt{67}right) 1 end{bmatrix} For lambda = 7 - sqrt{67}: begin{bmatrix} 2 + sqrt{67} & -7 -9 & -2 + sqrt{67} end{bmatrix} begin{bmatrix} v_1 v_2 end{bmatrix} = begin{bmatrix} 0 0 end{bmatrix} Reducing the system, we find the eigenvector: v = begin{bmatrix} frac{7}{2 + sqrt{67}} 1 end{bmatrix} Hence, the eigenvectors are: 1. For lambda = 7 + sqrt{67}: v = begin{bmatrix} frac{1}{9} left(-2 - sqrt{67}right) 1 end{bmatrix} 2. For lambda = 7 - sqrt{67}: v = begin{bmatrix} frac{7}{2 + sqrt{67}} 1 end{bmatrix}