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Word 2010 has a particular template that it uses when you first open the program. This template contains default styles, no text and you also use it when you create new, blank documents. As you customize styles, marcos and other elements these settings are automatically stored in this template. What is the name of this template file?
A
Standard.dotx
B
Normal.dotx
C
Empty.dotx
D
Template.dotx
Correct Answer:
Normal.dotx
Simplify the value of $$\frac{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} - {\text{3}} \times 0.9 \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}}}}{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} - 0.9 \times {\text{0}}{\text{.2}} - {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}} - 0.3 \times 0.9}} = ?$$
A
1.4
B
0.054
C
0.8
D
1.0
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
A
-2
B
-1
C
0
D
1
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 x
0
and f(x
0
) are 19.78 and 500.01, respectively. The corresponding error in the central difference estimate for h = 0.02 is approximately
A
1.3 × 10<sup>-4</sup>
B
3.0 × 10<sup>-4</sup>
C
4.5 × 10<sup>-4</sup>
D
9.0 × 10<sup>-4</sup>
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
A
x<sup>2</sup>y + y<sup>2</sup>z + z<sup>2</sup>x
B
2xy + 2yz + 2zx
C
x + y + z
D
0
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 K
A
and K
C
are the adsorption equilibrium constants. The rate controlling step for this reaction is
A
Absorption of A
B
Surface reaction between absorbed A and absorbed B
C
Surface reaction between absorbed A and B in the gas phase
D
Surface reaction between A in the gas phase and absorbed B
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 $${{\text{x}}_1}\left( 0 \right)
A
[ <p><span>A. $${{\text{x}}_{1{\text{f}}}} <p><span>B. $${{\text{x}}_{2{\text{f}}}}
B
$${{\text{x}}_{1{\text{f}}}} = {{\text{x}}_{2{\text{f}}}}
C
$${{\text{x}}_{1{\text{f}}}} = {{\text{x}}_{2{\text{f}}}} = \infty $$
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
A
$${\text{y}} = \left( {{{\text{C}}_1} - {{\text{C}}_2}{\text{x}}} \right){{\text{e}}^{\text{x}}} + {{\text{C}}_3}\cos {\text{x}} + {{\text{C}}_4}\sin {\text{x}}$$
B
$${\text{y}} = \left( {{{\text{C}}_1} + {{\text{C}}_2}{\text{x}}} \right){{\text{e}}^{\text{x}}} - {{\text{C}}_2}\cos {\text{x}} + {{\text{C}}_4}\sin {\text{x}}$$
C
$${\text{y}} = \left( {{{\text{C}}_1} + {{\text{C}}_2}{\text{x}}} \right){{\text{e}}^{\text{x}}} + {{\text{C}}_3}\cos {\text{x}} + {{\text{C}}_4}\sin {\text{x}}$$
D
$${\text{y}} = \left( {{{\text{C}}_1} + {{\text{C}}_2}{\text{x}}} \right){{\text{e}}^{\text{x}}} + {{\text{C}}_3}\cos {\text{x}} - {{\text{C}}_4}\sin {\text{x}}$$
The solution of the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} = 0$$ with boundary conditions
$${\text{i}}{\text{.}}\,\frac{{{\text{dy}}}}{{{\text{dx}}}} = 1{\text{ at x}} = 0;\,{\text{ii}}{\text{.}}\,\frac{{{\text{dy}}}}{{{\text{dx}}}} = 1{\text{ at x}} = 1{\text{ is}}$$
A
y = 1
B
y = x
C
y = x + c where C is an arbitrary constant
D
y = C<sub>1</sub>x + C<sub>2</sub> where C<sub>1</sub> and C<sub>2</sub> are arbitraryconstants
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 $$\int_{ - \infty }^{\text{a}} {\frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ - 1}}{2}{{\left( {\frac{{{\text{x}} - {\text{a}}}}{{\text{b}}}} \right)}^2}}}{\text{dx}}} $$
A
1
B
0.5
C
$$\pi $$
D
$$\frac{\pi }{2}$$
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 -
A
$$\frac{{a - b}}{a}$$
B
$$\frac{{a + b}}{a}$$
C
$$\frac{{a + b}}{b}$$
D
None of these