Q: (i) How many patterns does he need to form the smallest square? (ii) What are the dimensions of the largest square that he can form?

I can’t give the numeric answers yet — I need the full problem statement (how many patterns he has and the size/shape of each pattern). Explanation / what I need Please tell me whether the "patterns" are identical tiles, and give their dimensions (or area), and whether the supply is limited (how many patterns he actually has). Without that, there are several common interpretations; below I show the standard methods so you can apply them once you supply the missing data. If the question is: "How many identical rectangular patterns (each of size a×b) are needed to make the smallest square?" Let the area of one pattern be $m=a\\times b$. You need the smallest integer $N$ such that $N\\cdot m$ is a perfect square. Factor $m=\\prod p_i^{e_i}$. For each prime $p_i$ with odd exponent $e_i$, include one extra $p_i$ in $N$ to make the exponent even. Equivalently, write $m=u^2\\cdot t$ where $t$ is squarefree; then the minimal $N$ is $t$. Example: if a pattern is $2\\times3$ so $m=6=2\\cdot3$, then $t=6$ and the smallest $N$ is $6$ (since $6\\cdot6=36$, a perfect square). The resulting square will have area $36$ and side length $6$ (if units are unit squares). If the question is: "Given a finite number $T$ of identical patterns, what is the largest square (by number of patterns) he can form?" The largest number of whole patterns you can arrange into a perfect square count is the largest perfect square ≤ $T$, i.e. $\lfloor\\sqrt{T}\\rfloor^2$ patterns. The side length (in patterns) is $\lfloor\\sqrt{T}\\rfloor$. Example: if he has $T=20$ patterns, the largest square uses $\\lfloor\\sqrt{20}\\rfloor^2=4^2=16$ patterns, arranged as a $4\\times4$ square of patterns. If you paste the exact problem (pattern dimensions and how many he has), I’ll compute (i) the minimal number needed for the smallest square and (ii) the dimensions of the largest square he can form, with full step-by-step working.

Q: mr.lim gave 3600 to his wife an two children altogether. his wife received 500 more than his son. his son received twice as much as his daughter. how much did mr. lim's wife receive?

The correct answer is: $1740$ Explanation Let the daughter's share be $d$, the son's share be $s$, and the wife's share be $w$. We are given $s = 2d$, $w = s + 500$, and $w + s + d = 3600$. Steps: Write the total with the given relations: $$ (s+500) + s + d = 3600 $$ Substitute $d = \frac{s}{2}$ (since $s = 2d$): $$ (s+500) + s + \frac{s}{2} = 3600 $$ Combine terms: $$ \frac{5}{2}s + 500 = 3600 $$ Solve for $s$: $$ \frac{5}{2}s = 3100 \quad\Rightarrow\quad s = \frac{3100 \times 2}{5} = 1240 $$ Find the wife's amount: $$ w = s + 500 = 1240 + 500 = 1740 $$ Therefore, Mr. Lim's wife received $1740$.

Q: A radio station is giving away tickets to a play. They plan to give away tickets for seats that cost $10 and $20. They want to give away at least 20 tickets. The total cost of all the tickets they give away can be no more than $280.

All integer solutions (let $x$ = number of $10 tickets$, $y$ = number of $20 tickets$) are: For $y=0,1,\dots,8$, any integer $x$ with $20-y \le x \le 28-2y$. There are 45 possible combinations. The maximum number of tickets is 28 (all $10$-dollar tickets); the maximum number of $20$-dollar tickets is 8 (with $x=12$). Explanation Write the constraints: At least 20 tickets: $x+y \ge 20$. Cost no more than \$280: $10x+20y \le 280$. $x,y$ are nonnegative integers. Simplify the cost inequality by dividing by 10: $x+2y \le 28$. Combine inequalities to bound $y$: Subtract $(x+y \ge 20)$ from $(x+2y \le 28)$ to get $y \le 8$. So $y=0,1,\dots,8$. For each fixed integer $y$ in that range, solve for $x$: From $x+y \ge 20$ we get $x \ge 20-y$. From $x+2y \le 28$ we get $x \le 28-2y$. So $20-y \le x \le 28-2y$ (integers). Count solutions: For $y=0$ there are $9$ values ($x=20\ldots28$), for $y=1$ there are $8$ values, ..., for $y=8$ there is $1$ value. Total $9+8+\dots+1 = 45$ combinations. Examples of extremes: Maximum tickets: $x=28,y=0$ gives 28 tickets costing \$280. Maximum $20$-dollar tickets: $y=8,x=12$ gives 20 tickets costing \$280. Minimum cost while meeting at least 20 tickets: $x=20,y=0$ costing \$200.

Q: What is the Plural for Quiz

The plural of "quiz" is quizzes. Explanation Rule: For nouns ending in -s, -x, -z, -ch, or -sh, form the plural by adding -es (e.g., bus → buses, box → boxes, church → churches). For short, one-syllable words ending in a single consonant after a short vowel (like quiz), English orthography typically doubles the final consonant before adding a vowel suffix. That’s why quiz becomes quizzes (with a doubled z). Pronunciation: quizzes is pronounced /ˈkwɪzɪz/. So use quizzes for more than one quiz.

Q: 500 Grams to oz

The correct answer is: $17.63698\text{ oz}$ Explanation Convert grams to ounces using the relation $1\text{ oz} = 28.349523125\text{ g}$ (avoirdupois ounce). Steps: Write the conversion factor: $1\text{ oz} = 28.349523125\text{ g}$, so $1\text{ g} = 0.03527396195\text{ oz}$. Multiply: $$500\text{ g} \times 0.03527396195\text{ oz/g} = 17.636980975\text{ oz}$$ Round as desired: $17.63698\text{ oz}\approx 17.64\text{ oz}$ (to two decimal places). Therefore, 500 grams ≈ $17.63698\text{ oz}$ (about 17.64 oz).

Question

Line s is the perpendicular bisector of JK. If line s intersects JK at point L, which of the following statements must be true? Check all that apply.

Accepted Answer

I don't see the answer choices — but the statements that must be true are:

$JL = LK$ (L is the midpoint of $JK$).  
Line $s$ is perpendicular to $JK$ (so $s \perp JK$ at $L$).  
$J,\;L,\;K$ are collinear (because $L$ lies on segment $JK$).  
Every point on line $s$ is equidistant from $J$ and $K$ (if $P$ is any point on $s$, then $PJ = PK$).

Explanation

By definition a perpendicular bisector both bisects the segment and is perpendicular to it. Bisecting means it meets $JK$ at its midpoint, so $JL = LK$ and $L$ lies on the line containing $J$ and $K$. Perpendicular means $s$ meets $JK$ at a right angle.
The equidistance property is a standard fact: any point on the perpendicular bisector of a segment is the same distance from the segment’s endpoints (can be shown by two congruent right triangles).

If you paste the specific answer choices I can mark which of those match the true statements above.