Convert 1011.11 to base 10.

To convert a binary number with a fractional part to base 10, we sum the products of each digit with the corresponding power of 2.
The given binary number is $1011.11_2$.
The integer part is $1011_2$:

$$1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0$$ $$= 1 \times 8 + 0 \times 4 + 1 \times 2 + 1 \times 1$$ $$= 8 + 0 + 2 + 1 = 11$$

The fractional part is $0.11_2$:

$$1 \times 2^{-1} + 1 \times 2^{-2}$$ $$= 1 \times \frac{1}{2} + 1 \times \frac{1}{4}$$ $$= \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$$

Combining the integer and fractional parts:

$$11 + \frac{3}{4} = 11\frac{3}{4}$$

The final answer is $\boxed{\text{D. } 11\frac{3}{4}}$.

Given that O is the center of the circle and AB // CD.
Also, $\angle ABO = 45^\circ$.
Since OA and OB are radii, $\triangle OAB$ is an isosceles triangle with OA = OB.
Therefore, $\angle OAB = \angle ABO = 45^\circ$.
The sum of angles in $\triangle OAB$ is $180^\circ$:

$$\angle AOB = 180^\circ - (\angle OAB + \angle ABO)$$ $$\angle AOB = 180^\circ - (45^\circ + 45^\circ)$$ $$\angle AOB = 180^\circ - 90^\circ = 90^\circ$$

Since AB // CD, the chords AB and CD are parallel. In a circle, parallel chords are equidistant from the center. This implies that the lengths of the chords are equal, so AB = CD.
If two chords are equal in length, they subtend equal angles at the center.
Therefore, $\angle COD = \angle AOB = 90^\circ$.
In $\triangle OCD$, OC and OD are radii, so OC = OD. This means $\triangle OCD$ is an isosceles triangle.
Thus, $\angle OCD = \angle ODC$.
The sum of angles in $\triangle OCD$ is $180^\circ$:

$$\angle OCD + \angle ODC + \angle COD = 180^\circ$$ $$2 \angle OCD + 90^\circ = 180^\circ$$ $$2 \angle OCD = 180^\circ - 90^\circ$$ $$2 \angle OCD = 90^\circ$$ $$\angle OCD = \frac{90^\circ}{2}$$ $$\angle OCD = 45^\circ$$

The final answer is $\boxed{\text{A. } 45^\circ}$.

Given matrices $X = \begin{pmatrix} 4 & -11 \\ -8 & 5 \end{pmatrix}$ and $Y = \begin{pmatrix} -5 & 2 \\ 0 & 6 \end{pmatrix}$.
We need to evaluate $4X - 3Y$.
First, calculate $4X$:

$$4X = 4 \begin{pmatrix} 4 & -11 \\ -8 & 5 \end{pmatrix} = \begin{pmatrix} 4 \times 4 & 4 \times (-11) \\ 4 \times (-8) & 4 \times 5 \end{pmatrix} = \begin{pmatrix} 16 & -44 \\ -32 & 20 \end{pmatrix}$$

Next, calculate $3Y$:

$$3Y = 3 \begin{pmatrix} -5 & 2 \\ 0 & 6 \end{pmatrix} = \begin{pmatrix} 3 \times (-5) & 3 \times 2 \\ 3 \times 0 & 3 \times 6 \end{pmatrix} = \begin{pmatrix} -15 & 6 \\ 0 & 18 \end{pmatrix}$$

Now, subtract $3Y$ from $4X$:

$$4X - 3Y = \begin{pmatrix} 16 & -44 \\ -32 & 20 \end{pmatrix} - \begin{pmatrix} -15 & 6 \\ 0 & 18 \end{pmatrix}$$ $$= \begin{pmatrix} 16 - (-15) & -44 - 6 \\ -32 - 0 & 20 - 18 \end{pmatrix}$$ $$= \begin{pmatrix} 16 + 15 & -50 \\ -32 & 2 \end{pmatrix}$$ $$= \begin{pmatrix} 31 & -50 \\ -32 & 2 \end{pmatrix}$$

The final answer is $\boxed{\text{A. } \begin{pmatrix} 31 & -50 \\ -32 & 2 \end{pmatrix}}$.

We need to solve the simultaneous equations:

1. $2x + y = 4$
2. $3x - y = 1$ Add equation (1) and equation (2) to eliminate $y$:

$$(2x + y) + (3x - y) = 4 + 1$$ $$5x = 5$$

Divide by 5:

$$x = 1$$

Substitute $x=1$ into equation (1):

$$2(1) + y = 4$$ $$2 + y = 4$$

Subtract 2 from both sides:

$$y = 4 - 2$$ $$y = 2$$

The solution is $x=1$ and $y=2$. The final answer is $\boxed{\text{C. } x=1, y=2}$.

We need to solve the equation $\log_2 2 + \log_x 8 = 4$.
First, evaluate $\log_2 2$. Since $2^1 = 2$, $\log_2 2 = 1$.
Substitute this value into the equation:

$$1 + \log_x 8 = 4$$

Subtract 1 from both sides:

$$\log_x 8 = 3$$

Convert the logarithmic equation to its exponential form. If $\log_b a = c$, then $b^c = a$.

$$x^3 = 8$$

Take the cube root of both sides:

$$x = \sqrt[3]{8}$$ $$x = 2$$

The final answer is $\boxed{\text{B. } 2}$.

The aircraft flies from Q $(25^\circ N, 38^\circ E)$ to R $(72^\circ S, 38^\circ E)$.
Since both points are on the same longitude ($38^\circ E$), the flight is along a line of longitude.
The difference in latitude, $\theta$, is the sum of the absolute values of the latitudes because one is North and the other is South:

$$\theta = 25^\circ + 72^\circ = 97^\circ$$

To use the arc length formula $s = R\theta$, we must convert $\theta$ to radians:

$$\theta = 97^\circ \times \frac{\pi}{180^\circ} \text{ radians}$$

Given $R = 6,400 \text{ km}$ and $\pi = \frac{22}{7}$.
Substitute these values into the formula:

$$s = 6400 \times \frac{97}{180} \times \frac{22}{7}$$ $$s = \frac{6400 \times 97 \times 22}{180 \times 7}$$ $$s = \frac{13657600}{1260}$$ $$s = \frac{1365760}{126}$$ $$s = \frac{682880}{63}$$ $$s \approx 10839.365 \text{ km}$$

Rounding to the nearest kilometer:

$$s \approx 10839 \text{ km}$$

The final answer is $\boxed{\text{D. } 10,839}$.

We need to express the product of 0.016 and 3.54 in standard form.
First, calculate the product:

$$0.016 \times 3.54 = 0.05664$$

Now, express 0.05664 in standard form ($a \times 10^n$, where $1 \le |a| < 10$).
Move the decimal point two places to the right to get 5.664.
Since we moved the decimal point two places to the right, the exponent of 10 will be $-2$.

$$0.05664 = 5.664 \times 10^{-2}$$

The final answer is $\boxed{\text{E. } 5.664 \times 10^{-2}}$.