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What is the default width of text field? Correct Answer 40 characters

The default width of a text field is twenty characters. The character limit of the text field can be extended with the proper syntax by mentioning the size of the text field you want to create. The 20 character limit is for the one-line text input field.

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Simplify the value of $$\frac{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} - {\text{3}} \times 0.9 \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}}}}{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} - 0.9 \times {\text{0}}{\text{.2}} - {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}} - 0.3 \times 0.9}} = ?$$
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
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}}}}},$$      where KA and KC are the adsorption equilibrium constants. The rate controlling step for this reaction is
Let $$\nabla \cdot \left( {{\text{f}}\overrightarrow {\text{v}} } \right) = {{\text{x}}^2}{\text{y}} + {{\text{y}}^2}{\text{z}} + {{\text{z}}^2}{\text{x}},$$      where f and v are scalar and vector fields respectively. If $$\overrightarrow {\text{v}} = {\text{y}}\overrightarrow {\text{i}} + {\text{z}}\overrightarrow {\text{j}} + {\text{x}}\overrightarrow {\text{k}} ,$$     then $$\overrightarrow {\text{v}} \cdot \nabla {\text{f}}$$   is
The general solution of the differential equation, $$\frac{{{{\text{d}}^4}{\text{y}}}}{{{\text{d}}{{\text{x}}^4}}} - 2\frac{{{{\text{d}}^3}{\text{y}}}}{{{\text{d}}{{\text{x}}^3}}} + 2\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - 2\frac{{{\text{dy}}}}{{{\text{dx}}}} + {\text{y}} = 0$$       is
The solution of the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} = 0$$   with boundary conditions
$${\text{i}}{\text{.}}\,\frac{{{\text{dy}}}}{{{\text{dx}}}} = 1{\text{ at x}} = 0;\,{\text{ii}}{\text{.}}\,\frac{{{\text{dy}}}}{{{\text{dx}}}} = 1{\text{ at x}} = 1{\text{ is}}$$
If {x} is a continuous, real valued random variable defined over the interval (-$$\infty $$, +$$\infty $$) and its occurrence is defined by the density function given as:
$${\text{f}}\left( {\text{x}} \right) = \frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ - 1}}{2}{{\left( {\frac{{{\text{x}} - {\text{a}}}}{{\text{b}}}} \right)}^2}}}$$     where 'a' and 'b' are the statistical attributes of the random variable {x}. The value of the integral $$\int_{ - \infty }^{\text{a}} {\frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ - 1}}{2}{{\left( {\frac{{{\text{x}} - {\text{a}}}}{{\text{b}}}} \right)}^2}}}{\text{dx}}} $$
Let $$\frac{{\text{a}}}{{\text{b}}}: - \frac{{\text{b}}}{{\text{a}}} = {\text{x}}:{\text{y}}{\text{.}}$$    If $$\left( {{\text{x - y}}} \right) = $$  $$\left\{ {\frac{{\text{a}}}{{\text{b}}}{\text{ + }}\frac{{\text{b}}}{{\text{a}}}} \right\}{\text{,}}$$   then x is equal to -