1. Rewrite $\mathrm{\mid }5x+3\mathrm{\mid }<9$ without using the absolute sign:

The inequality $\mathrm{\mid }5x+3\mathrm{\mid }<9$ means that the expression inside the absolute value lies between $-9$ and $9$.

$$-9<5x+3<9$$

Subtract $3$ from all parts:

$$-12<5x<6$$

Divide through by $5$

$$-\frac{12}{5}<x<\frac{6}{5}$$

2. In how many ways can the letters of the word "EXAMINATION" be arranged?

The word "EXAMINATION" consists of 11 letters, with the following repetitions:

• E: 1 time
• X: 1 time
• A: 2 times
• M: 1 time
• I: 2 times
• N: 2 times
• T: 1 time
• O: 1 time

The formula for the number of arrangements of letters in a word with repetitions is:

$$\frac{n!}{p_1!\cdot p_2!\cdot \cdots \cdot p_k!}$$

where $n$ is the total number of letters, and $p_1,p_2,\dots$ are the frequencies of the repeated letters.

$$\text{Number of arrangements}=\frac{11!}{2!\cdot 2!\cdot 2!}=\frac{39916800}{8}=4989600$$

So, there are 4,989,600 ways to arrange the letters of "EXAMINATION."

3. Find the range and domain of the function $f(x)=5x-2$

• Domain: $f(x)<span\; style="font-family:\; sans-serif;\; text-align:\; var(--bs-body-text-align);">is\; a\; linear\; function,\; so\; it\; is\; defined\; for\; all\; real\; numbers.</span>$

$$\text{Domain: }(-\mathrm{\infty },\mathrm{\infty })$$

• Range: Since $f(x)$ is a linear function with a slope of $5$, it can produce all real values as outputs.

$$\text{Range: }(-\mathrm{\infty },\mathrm{\infty })$$

4. Find the derivative of $y=\sqrt{2x-4}$

Rewrite $y$ as $y=(2x-4{)}^{1\mathrm{/}2}$ Using the chain rule:

$$\frac{dy}{dx}=\frac{1}{2}(2x-4{)}^{-1\mathrm{/}2}\cdot \frac{d}{dx}(2x-4)$$
$$\frac{dy}{dx}=\frac{1}{2}(2x-4{)}^{-1\mathrm{/}2}\cdot 2$$
$$\frac{dy}{dx}=\frac{1}{\sqrt{2x-4}}$$

5. Find $\frac{dy}{dx}$ if $x^2+2xy+7y^3=0$

Differentiate both sides with respect to $\mathrm{x<span\; class="katex"\; style="color:\; var(--body-font-color);\; font-family:\; sans-serif;\; text-align:\; var(--bs-body-text-align);"><span\; style="color:\; var(--body-font-color);\; text-align:\; var(--bs-body-text-align);">(using\; implicit\; differentiation):</span></span>}$

Combine like terms:

$$2x+2y+2x\frac{dy}{dx}+21y^2\frac{dy}{dx}=0$$

Factor out $\frac{dy}{dx}$:

$$2x+2y+\frac{dy}{dx}(2x+21y^2)=0$$

Solve for $\frac{dy}{dx}$:

In how many ways can a futsal team of 6 girls be made from 10 girls?

Case (i): Two girls are always selected
If 2 girls are always selected, we need to choose 4 more girls from the remaining 8 girls.

$$\binom{8}{4} = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} = 70$$

So, there are 70 ways to form the team.

Case (ii): Two girls are never selected
If 2 girls are never selected, we need to choose all 6 girls from the remaining 8 girls.

$$\binom{8}{6} = \binom{8}{2} = \frac{8 \cdot 7}{2 \cdot 1} = 28$$

So, there are 28 ways to form the team.

Discuss the continuity of the function at

The function is defined as:

$$f(x) = \begin{cases} x^2 + 1 & \text{for } x \leq 1 \\ 2x^2 + 1 & \text{for } x > 1 \end{cases}$$

For continuity at $x = 1$, the left-hand limit ($LHL$), right-hand limit ($RHL$), and the value of the function ($f(1)<span\; style="color:\; var(--body-font-color);\; font-family:\; sans-serif;\; text-align:\; var(--bs-body-text-align);">)\; must\; all\; be\; equal.</span>$

1. Left-hand limit (LHL):

$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 1) = 1^2 + 1 = 2$$

2. Right-hand limit (RHL):

$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x^2 + 1) = 2(1^2) + 1 = 3$$

3. Value of $f(1)<strong\; style="font-family:\; sans-serif;\; text-align:\; var(--bs-body-text-align);">:</strong><span\; style="font-family:\; sans-serif;\; text-align:\; var(--bs-body-text-align);">\; For </span>$$x \leq 1$, $f(1)=1^2+1=\mathrm{2<span\; style="font-family:\; sans-serif;\; text-align:\; var(--bs-body-text-align);">.</span>}$

Since $LHL = 2$, $RHL=\mathrm{3<span\; style="color:\; var(--body-font-color);\; font-family:\; sans-serif;\; text-align:\; var(--bs-body-text-align);"> </span>}$$LHL \neq RHL$, the function is not continuous at $x = 1$.

Find the equation with roots A$+$$B$ and 2$\mathrm{/}$$($$A$$+$$B$$)$:

Given equation:

$$3x^2 - 6x + 5 = 0$$

Sum of roots ($A + B$) = $\frac{-b}{a} = \frac{6}{3} = 2$
Product of roots ($A \cdot B$) = $\frac{c}{a} = \frac{5}{3}$

New roots are $A+B = 2$ and $\frac{2}{A+B} = \frac{2}{2} = 1$.
The equation with these roots is:

$$x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$$
$$x^2 - (2 + 1)x + (2 \cdot 1) = 0$$
$$x^2 - 3x + 2 = 0$$