answer:The volume of a solid in the first octant bounded by the planes can be found by determining the intercepts of the plane with the coordinate axes. Given the equation of the plane: {eq}x + frac{y}{2} + frac{z}{4} = 1 {/eq} 1. X-intercept: Set {eq}y = z = 0 {/eq}, then {eq}x = 1 {/eq}. 2. Y-intercept: Set {eq}x = z = 0 {/eq}, then {eq}y = 2 {/eq} (since {eq}y/2 {/eq} equals 1). 3. Z-intercept: Set {eq}x = y = 0 {/eq}, then {eq}z = 4 {/eq} (since {eq}z/4 {/eq} equals 1). The volume {eq}V {/eq} of the solid can be calculated using the formula for the volume of a prism: {eq}V = frac{1}{6} times text{X-intercept} times text{Y-intercept} times text{Z-intercept} {/eq}. Plugging in the intercepts, we have: {eq}V = frac{1}{6} times 1 times 2 times 4 {/eq} {eq}V = frac{4}{3} {/eq} Hence, the volume of the solid in the first octant is {eq}frac{4}{3} {/eq} cubic units.

answer:Let {eq}M {/eq} and {eq}N {/eq} are the two complementary angles. So angle M + angle N = 90^{circ} tag{Eq.1} Given that the measure of an angle is five times the measure of its complementary angle. So angle M = 5 angle N tag{Eq.2} From equation (1) and (2)- begin{align} 5 angle N + angle N &= 90^{circ} [0.3cm] 6 angle N &= 90^{circ} [0.3cm] angle N &= frac{90}{6} [0.3cm] angle N &= 15^{circ} [0.3cm] end{align} and begin{align} angle M &= 5 cdot 15^{circ} [0.3cm] angle M &= 75^{circ} [0.3cm] end{align} Therefore, the measure of the larger angle is {eq}75^{circ} {/eq}.

answer:The equation of a plane in 3D space passing through three points P_1(x_1, y_1, z_1), P_2(x_2, y_2, z_2), and P_3(x_3, y_3, z_3) can be found using the following steps: 1. Calculate the vectors vec{v_1} = P_2 - P_1 and vec{v_2} = P_3 - P_1. 2. Compute the cross product of vec{v_1} and vec{v_2} to get the normal vector vec{n} of the plane. 3. Use the point-normal form of the plane equation n_x(x - x_1) + n_y(y - y_1) + n_z(z - z_1) = 0 with any of the three points. Let's calculate: 1. vec{v_1} = (2, -1, -3) - (4, 5, 2) = (-2, -6, -5) 2. vec{v_2} = (-1, -2, 2) - (4, 5, 2) = (-5, -7, 0) 3. vec{n} = vec{v_1} times vec{v_2} = begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} -2 & -6 & -5 -5 & -7 & 0 end{vmatrix} = (0 cdot (-7) - (-5) cdot (-6))mathbf{i} - ((-2) cdot 0 - (-5) cdot (-5))mathbf{j} + ((-2) cdot (-7) - (-6) cdot (-5))mathbf{k} = (-30)mathbf{i} - (-25)mathbf{j} - (-2)mathbf{k} = -30mathbf{i} + 25mathbf{j} + 2mathbf{k} 4. Using point P_1(4, 5, 2) and the normal vector vec{n} = (-30, 25, 2), we have the equation: -30(x - 4) + 25(y - 5) + 2(z - 2) = 0 Simplifying, we get the equation of the plane: -30x + 120 + 25y - 125 + 2z - 4 = 0 -30x + 25y + 2z = 1 Dividing by -1 to make the coefficient of x positive: 30x - 25y - 2z = -1 Thus, the equation of the plane is: 30x - 25y - 2z + 1 = 0

answer:Balancing redox reactions involves ensuring that the number of electrons lost in the oxidation half-reaction equals the number gained in the reduction half-reaction. 1. The oxidation half-reaction for platinum is: ce{Pt -> Pt^{2+} + 2e-} Here, platinum loses 2 electrons. 2. The reduction half-reaction for hydrogen is: ce{2H+ + 2e- -> H2} In this case, 2 hydrogen ions gain 2 electrons to form hydrogen gas. To balance the overall redox equation, we combine these half-reactions, making sure the number of electrons lost equals the number gained: ce{Pt + 2H+ -> Pt^{2+} + H2} The '2' coefficient is correctly placed in front of ce{H+} to balance the charges, but no additional '2' is needed in front of ce{Pt2+} since the charge is already balanced. The net equation is now balanced, and you can verify this principle using the oxidation state method or other methods covered in chemistry resources, such as the one found at: http://www.chemguide.co.uk/inorganic/redox/equations.html