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The following Boolean expression $$Y = A \cdot \overline B \cdot \overline C \cdot \overline D + \overline A \cdot B \cdot \overline C \cdot D + \overline A \cdot \overline B \cdot \overline C \cdot D + \overline A \cdot \overline B \cdot \overline C \cdot D + \overline A \cdot B \cdot C \cdot D + A \cdot \overline B \cdot \overline C \cdot D$$                 can be simplified to
Simplify the value of $$\frac{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} - {\text{3}} \times 0.9 \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}}}}{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} - 0.9 \times {\text{0}}{\text{.2}} - {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}} - 0.3 \times 0.9}} = ?$$
The Boolean expression Y = $$\overline {{\text{ABC}}} $$ D + $$\overline {{\text{AB}}} $$ CD + $$\overline {\text{A}} $$B$$\overline {\text{C}} $$ D + $$\overline {\text{A}} $$BCD reduces to
Consider the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}\left( {\text{t}} \right)}}{{{\text{d}}{{\text{t}}^2}}} + 2\frac{{{\text{dy}}\left( {\text{t}} \right)}}{{{\text{dt}}}} + {\text{y}}\left( {\text{t}} \right) = \delta \left( {\text{t}} \right)$$      with $${\left. {{\text{y}}\left( {\text{t}} \right)} \right|_{{\text{t}} = 0}} = - 2$$   and $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}} = 0.$$
The numerical value of $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}}$$   is
ABCD is a parallelogram and the diagonals \(\overline {AC} \) and \(\overline {BD} \) intersect at 'O'. If E, F, G and H are the mid-point of \(\overline {AO}\), \(\overline {DO}\), \(\overline {CO}\) and \(\overline {BO} \) respectively, then what is the ratio of (EF + FG + GH + HE) and (AD + DC + CB + BA)?
ΔABC is an isosceles triangle and $$\overline {AB} $$  = $$\overline {AC} $$  = 2a unit, $$\overline {BC} $$  = a unit. Draw $$\overline {AD} $$ ⊥ $$\overline {BC} $$ , and find the length of $$\overline {AD} $$
Let the average temperatures in Centigrade (C) and Fahrenheit (F) be $$\overline C $$ and $$\overline F $$. If $$\overline C $$ and $$\overline F $$ are related to $$F = \frac{9}{2}C + 32,$$   then $$\overline F $$ and $$\overline C $$ have the relation
The error in $${\left. {\frac{{\text{d}}}{{{\text{dx}}}}{\text{f}}\left( {\text{x}} \right)} \right|_{{\text{x}} = {{\text{x}}_0}}}$$   for a continuous function estimated with h = 0.03 using the central difference formula $${\left. {\frac{{\text{d}}}{{{\text{dx}}}}{\text{f}}\left( {\text{x}} \right)} \right|_{{\text{x}} = {{\text{x}}_0}}} = \frac{{{\text{f}}\left( {{{\text{x}}_0} + {\text{h}}} \right) - {\text{f}}\left( {{{\text{x}}_0} - {\text{h}}} \right)}}{{2{\text{h}}}},$$        is 2 × 10-3. The values of x0 and f(x0) are 19.78 and 500.01, respectively. The corresponding error in the central difference estimate for h = 0.02 is approximately
Consider a system governed by the following equations:
$$\frac{{{\text{d}}{{\text{x}}_1}\left( {\text{t}} \right)}}{{{\text{dt}}}} = {{\text{x}}_2}\left( {\text{t}} \right) - {{\text{x}}_1}\left( {\text{t}} \right);\,\frac{{{\text{d}}{{\text{x}}_2}\left( {\text{t}} \right)}}{{{\text{dt}}}} = {{\text{x}}_1}\left( {\text{t}} \right) - {{\text{x}}_2}\left( {\text{t}} \right)$$
The initial conditions are such that $${{\text{x}}_1}\left( 0 \right)
The rate of the heterogenous catalytic reaction A(g) + B(g) → C(g) is given by $$ - {{\text{r}}_{\text{A}}} = \frac{{{\text{k}}.{{\text{K}}_{\text{A}}}.{{\text{p}}_{\text{A}}}.{{\text{P}}_{\text{B}}}}}{{1 + {{\text{K}}_{\text{A}}}.{{\text{P}}_{\text{A}}} + {{\text{K}}_{\text{C}}}.{{\text{p}}_{\text{C}}}}},$$      where KA and KC are the adsorption equilibrium constants. The rate controlling step for this reaction is