Simplify the expression 33 + 33 + 33. Find f(2) and ff(2) for f(x) = 1−2x. Find the number of sides of a regular polygon with an exterior angle of 45^. Find the area of a kite with diagonals 50 cm and

Using the exponent rule $a^m \times a^n = a^{m+n}$, where $3 = 3^1$:

The correct option is D.
The final answer is $\boxed{\text{D}}$.

Given $f(x) = 1 - 2x$.

Substitute $f(2) = -3$ into $f(x)$:

The correct option is D.
The final answer is $\boxed{\text{D}}$.

The formula for the exterior angle of a regular polygon is $\frac{360^\circ}{n}$, where $n$ is the number of sides.
Given that the exterior angle is $45^\circ$:

The correct option is D.
The final answer is $\boxed{\text{D}}$.

The area of a kite is given by the formula $A = \frac{1}{2} d_1 d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals.
Given $d_1 = 50 \text{ cm}$ and $d_2 = 80 \text{ cm}$.

The correct option is C.
The final answer is $\boxed{\text{C}}$.

The shaded region in the figure is a part of a circle bounded by an arc and a chord. This region is called a segment of a circle.
The correct option is A.
The final answer is $\boxed{\text{A}}$.

Let $p$ be the cost of a pen and $b$ be the cost of a book.
The total cost for 2 pens and 5 books is 4,150 FCFA:

Given that a book costs 750 FCFA, so $b = 750$.

The cost of a pen is 200 FCFA.
The correct option is D.
The final answer is $\boxed{\text{D}}$.

The side length of a small cubic box is 2 cm.
Volume of small box ($V_s$) = side$^3 = (2 \text{ cm})^3 = 8 \text{ cm}^3$.

The dimensions of the big box are 6 cm long, 5 cm wide, and 8 cm high.
Volume of big box ($V_b$) = length $\times$ width $\times$ height = $6 \text{ cm} \times 5 \text{ cm} \times 8 \text{ cm} = 240 \text{ cm}^3$.

Number of boxes = $\frac{\text{Volume of big box}}{\text{Volume of small box}}$
However, for "fitting exactly", we must consider how many boxes fit along each dimension.
Number of boxes along length = $\frac{6 \text{ cm}}{2 \text{ cm}} = 3$
Number of boxes along width = $\frac{5 \text{ cm}}{2 \text{ cm}} = 2.5$. Since we can only fit whole boxes, we take the floor of this value, which is 2.
Number of boxes along height = $\frac{8 \text{ cm}}{2 \text{ cm}} = 4$
Total number of boxes = $3 \times 2 \times 4 = 24$.

Wait, the options are 15, 18, 30, 45. My calculation of 24 is not among the options. Let me re-read the question carefully. "fit exactly". This usually implies that the dimensions of the larger box are multiples of the smaller box's dimensions. If the width is 5cm and the small box side is 2cm, then 2.5 boxes fit along the width, meaning only 2 full boxes can fit. This leads to 24.

Let's re-check the question and options. It's possible there's a slight misinterpretation or a typo in the question/options.
If the question implies that the volume calculation is sufficient, then $240/8 = 30$. This matches option C.
The phrasing "fit exactly" can be ambiguous. In packing problems, it usually means no gaps and dimensions must be multiples. However, in multiple-choice questions, sometimes the simpler volume division is expected if it leads to one of the options. Given the options, it's highly probable that the question intends for a simple volume division. If the boxes were to "fit exactly" without any wasted space, then the dimensions of the large box would have to be multiples of the small box's side length. Since 5 is not a multiple of 2, there would be wasted space if we packed them perfectly. However, if we consider the total volume, then 30 boxes would occupy $30 \times 8 = 240 \text{ cm}^3$, which is the volume of the big box. This implies that the "fit exactly" refers to the total volume, not necessarily perfect alignment along all axes without any leftover space from the big box's dimensions.

Let's assume the question implies volume division.
Number of boxes = $\frac{V_b}{V_s} = \frac{240 \text{ cm}^3}{8 \text{ cm}^3} = 30$.
This matches option C.
The final answer is $\boxed{\text{C}}$.

The perimeter of a rectangle is $P = 2(l + w)$, where $l$ is the length and $w$ is the width.
Given $P = 64 \text{ cm}$ and $w = 8 \text{ cm}$.

Divide both sides by 2:

Subtract 8 from both sides:

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

The line passes through $(x_1, y_1) = (0,3)$ and $(x_2, y_2) = (-5,-7)$.
The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

We know $m = 2$.
Since the line passes through $(0,3)$, this is the y-intercept, so $c = 3$.
Substitute $m$ and $c$ into the equation:

The correct option is A.
The final answer is $\boxed{\text{A}}$.

The rule is $a^{-n} = \frac{1}{a^n}$.

The rule is $a^{\frac{1}{n}} = \sqrt[n]{a}$.

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

The identity is $\sin^2\theta + \cos^2\theta = 1$.
Given $\cos\theta = \frac{4}{5}$.

(Assuming $\theta$ is in a quadrant where $\sin\theta$ is positive, which is typical for these types of problems unless specified otherwise).

The reciprocal identity is $\csc\theta = \frac{1}{\sin\theta}$.

The correct option is A.
The final answer is $\boxed{\text{A}}$.

If $\frac{2}{5}$ of the students are girls, then the remaining fraction are boys.
Fraction of boys = $1 - \frac{2}{5} = \frac{5}{5} - \frac{2}{5} = \frac{3}{5}$.

Let $T$ be the total number of students in the class.
We are given that the number of boys is 54.

So, there are 90 students in total.

Number of girls = $\frac{2}{5} \times T$
Number of girls = $\frac{2}{5} \times 90$
Number of girls = $2 \times 18$
Number of girls = 36
The correct option is B.
The final answer is $\boxed{\text{B}}$.