32 Simplify without using calculator: $$ 3.2 .16 \sqrt{80}+2 \sqrt{20}-\sqrt{125} $$ (3) $$ 3.2 .2 \sqrt{\frac{\sqrt{27}+\sqrt{48}+\sqrt{75}}{\sqrt{27}}} $$ (4) [13] QUESTION 4 4.1 The formula for the area of a ring is given as $$ A=\pi\left(R^{2}-r^{2} ight) $$ 4.1.1 Make $$ R $$ the subject of the formula for the given equation. (3) 4.1.2 Determine the value of R if $$ A=120 \mathrm{~mm}^{2} $$ and $$ r=10 \mathrm{~mm} $$. (2) 4.2 Calculate the $$ 16^{ ext {th }} $$ term of an arithmetic sequence if the $$ 5^{ ext {th }} $$ term is 6 and $$ 12^{ ext {th }} $$ term is 41 . (6) 4.3 A theatre has 60 seats in the first row, 68 seats in the second row and 76 seats in the third row and so forth in the increasing pattern. If the theatre has 22 rows, how many seats are in the theatre? (3) 4.4 The value of the of a certain term $$ \left(T_{n} ight) $$, of an arithmetic sequence is 64 . Calculate the position of the term if the first term is -24 and constant difference is 8 . (3)

Question

32 Simplify without using calculator: $$ 3.2 .16 \sqrt{80}+2 \sqrt{20}-\sqrt{125} $$ (3) $$ 3.2 .2 \sqrt{\frac{\sqrt{27}+\sqrt{48}+\sqrt{75}}{\sqrt{27}}} $$ (4) [13] QUESTION 4 4.1 The formula for the area of a ring is given as $$ A=\pi\left(R^{2}-r^{2}ight) $$ 4.1.1 Make $$ R $$ the subject of the formula for the given equation. (3) 4.1.2 Determine the value of R if $$ A=120 \mathrm{~mm}^{2} $$ and $$ r=10 \mathrm{~mm} $$. (2) 4.2 Calculate the $$ 16^{	ext {th }} $$ term of an arithmetic sequence if the $$ 5^{	ext {th }} $$ term is 6 and $$ 12^{	ext {th }} $$ term is 41 . (6) 4.3 A theatre has 60 seats in the first row, 68 seats in the second row and 76 seats in the third row and so forth in the increasing pattern. If the theatre has 22 rows, how many seats are in the theatre? (3) 4.4 The value of the of a certain term $$ \left(T_{n}ight) $$, of an arithmetic sequence is 64 . Calculate the position of the term if the first term is -24 and constant difference is 8 . (3)

UpStudy AI · Accepted Answer

**Problem 3.2 (first expression)**

Simplify
$$
3 \cdot 2 \cdot 16\sqrt{80}+2\sqrt{20}-\sqrt{125}.
$$

1. Write the product:
   $$
   3 \cdot 2 \cdot 16\sqrt{80} = 96\sqrt{80}.
   $$
2. Simplify the radicals:
   - Since $$80=16\cdot5$$, 
     $$
     \sqrt{80}=\sqrt{16\cdot5}=4\sqrt{5}.
     $$
   - Since $$20=4\cdot5$$, 
     $$
     \sqrt{20}=\sqrt{4\cdot5}=2\sqrt{5}.
     $$
   - Since $$125=25\cdot5$$,
     $$
     \sqrt{125}=\sqrt{25\cdot5}=5\sqrt{5}.
     $$
3. Substitute these back:
   $$
   96\sqrt{80} = 96\cdot4\sqrt{5}=384\sqrt{5},
   $$
   $$
   2\sqrt{20}=2\cdot2\sqrt{5}=4\sqrt{5}.
   $$
4. Combine like terms:
   $$
   384\sqrt{5}+4\sqrt{5}-5\sqrt{5}=(384+4-5)\sqrt{5}=383\sqrt{5}.
   $$

**Answer:** $$383\sqrt{5}$$

---

**Problem 3.2 (second expression)**

Simplify
$$
3 \cdot 2 \cdot 2\sqrt{\frac{\sqrt{27}+\sqrt{48}+\sqrt{75}}{\sqrt{27}}}.
$$

1. First, simplify the numerator inside the square root:
   - $$\sqrt{27}=3\sqrt{3}$$ (since $$27=9\cdot3$$),
   - $$\sqrt{48}=4\sqrt{3}$$ (since $$48=16\cdot3$$),
   - $$\sqrt{75}=5\sqrt{3}$$ (since $$75=25\cdot3$$).
   
   So,
   $$
   \sqrt{27}+\sqrt{48}+\sqrt{75}=3\sqrt{3}+4\sqrt{3}+5\sqrt{3}=12\sqrt{3}.
   $$
2. Compute the fraction:
   $$
   \frac{12\sqrt{3}}{\sqrt{27}}.
   $$
   But
   $$
   \sqrt{27}=3\sqrt{3},
   $$
   hence,
   $$
   \frac{12\sqrt{3}}{3\sqrt{3}}=4.
   $$
3. Now the whole expression becomes:
   $$
   3\cdot2\cdot2\sqrt{4}.
   $$
4. Since $$\sqrt{4}=2$$:
   $$
   3\cdot2\cdot2\cdot2=24.
   $$

**Answer:** $$24$$

---

**Problem 4.1.1**

Given the formula for the area of a ring,
$$
A=\pi\left(R^2-r^2\right),
$$
we solve for $$R$$.

1. Divide both sides by $$\pi$$:
   $$
   \frac{A}{\pi}=R^2-r^2.
   $$
2. Add $$r^2$$ to both sides:
   $$
   R^2=\frac{A}{\pi}+r^2.
   $$
3. Take the square root:
   $$
   R=\sqrt{\frac{A}{\pi}+r^2}.
   $$

**Answer:** $$R=\sqrt{\frac{A}{\pi}+r^2}$$

---

**Problem 4.1.2**

Determine $$R$$ for $$A=120~\mathrm{mm}^2$$ and $$r=10~\mathrm{mm}$$.

Substitute into the formula:
$$
R=\sqrt{\frac{120}{\pi}+10^2}=\sqrt{\frac{120}{\pi}+100}.
$$

**Answer:** $$R=\sqrt{\frac{120}{\pi}+100}~\mathrm{mm}$$

---

**Problem 4.2**

In an arithmetic sequence, the $$n^{\text{th}}$$ term is
$$
T_n=a+(n-1)d.
$$
Given:
$$
T_5=a+4d=6,\quad T_{12}=a+11d=41.
$$

1. Subtract the first equation from the second:
   $$
   (a+11d)-(a+4d)=7d=41-6=35,
   $$
   so,
   $$
   d=5.
   $$
2. Substitute $$d=5$$ into $$T_5$$:
   $$
   a+4(5)=6\quad\Longrightarrow\quad a+20=6,
   $$
   hence,
   $$
   a=-14.
   $$
3. Now find the $$16^{\text{th}}$$ term:
   $$
   T_{16}=a+15d=-14+15\cdot5=-14+75=61.
   $$

**Answer:** $$T_{16}=61$$

---

**Problem 4.3**

The seats in the theatre form an arithmetic sequence with first term $$a_1=60$$ and common difference $$d=68-60=8$$. The theatre has 22 rows.

The sum of the first $$n$$ terms is:
$$
S_n=\frac{n}{2}\left(2a_1+(n-1)d\right).
$$
For $$n=22$$:
$$
S_{22}=\frac{22}{2}\left(2\cdot60+(22-1)8\right)=11\left(120+21\cdot8\right).
$$
Calculate inside:
$$
21\cdot8=168,\quad 120+168=288.
$$
Thus,
$$
S_{22}=11\cdot288=3168.
$$

**Answer:** $$3168$$ seats

---

**Problem 4.4**

An arithmetic sequence has first term $$a=-24$$ and common difference $$d=8$$. It is given that one of the terms is $$64$$; find its position $$n$$.

The $$n^{\text{th}}$$ term is:
$$
T_n=-24+(n-1)8=64.
$$
Solve for $$n$$:
$$
8(n-1)=64+24=88,
$$
$$
n-1=\frac{88}{8}=11,
$$
$$
n=11+1=12.
$$

**Answer:** The term $$64$$ is the $$12^{\text{th}}$$ term.
**Simplified Answers:** 1. **Problem 3.2 (First Expression):** $$ 3 \cdot 2 \cdot 16\sqrt{80} + 2\sqrt{20} - \sqrt{125} = 383\sqrt{5} $$ 2. **Problem 3.2 (Second Expression):** $$ 3 \cdot 2 \cdot 2\sqrt{\frac{\sqrt{27} + \sqrt{48} + \sqrt{75}}{\sqrt{27}}} = 24 $$ 3. **Problem 4.1.1:** $$ R = \sqrt{\frac{A}{\pi} + r^2} $$ 4. **Problem 4.1.2:** $$ R = \sqrt{\frac{120}{\pi} + 100} \, \text{mm} $$ 5. **Problem 4.2:** $$ T_{16} = 61 $$ 6. **Problem 4.3:** $$ 3168 \, \text{seats} $$ 7. **Problem 4.4:** $$ \text{The term } 64 \text{ is the } 12^{\text{th}} \text{ term.} $$

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