QUESTION 4
Round off the following numbers to the decimal places indicated:
4.1.1 4,3768534 ( 4 decimal places
4.1.2 decimal place
Given:
4.2.1 Determine the value of if ; ;
(Leave your answer in the simplest surd form)
4.2.2 State whether is rational or irrational.
4.2.3 Between which TWO consecutive integers does lie?
Write the following recurring decimals in fraction form:
4.3.1

Ask by Sandoval Ortiz. in South Africa

Jan 24,2025

Question

QUESTION 4
Round off the following numbers to the decimal places indicated:
4.1.1 4,3768534 ( 4 decimal places
4.1.2 decimal place
Given:
4.2.1 Determine the value of if ; ;
(Leave your answer in the simplest surd form)
4.2.2 State whether is rational or irrational.
4.2.3 Between which TWO consecutive integers does lie?
Write the following recurring decimals in fraction form:
4.3.1

Ask by Sandoval Ortiz. in South Africa

Jan 24,2025

UpStudy AI · Accepted Answer

Certainly! Let's address each part of **QUESTION 4** step by step.

---

### **4.1.1 Rounding 4.3768534 to 4 Decimal Places**

**Number:** 4.3768534  
**Decimal Places to Round:** 4

**Steps:**
1. Identify the fourth decimal place: **4.3768**.
2. Look at the fifth decimal place to decide whether to round up or stay: The fifth digit is **5**.
3. Since the fifth digit is **5** or more, round up the fourth decimal place by 1.

**Rounded Number:**  
**4.3769**

---

### **4.1.2 Rounding √7.24357 to 1 Decimal Place**

**Expression:** √7.24357  
**Decimal Places to Round:** 1

**Steps:**
1. Calculate the square root of 7.24357.  
   - √7 ≈ 2.6458  
   - √7.24357 ≈ **2.692** (approximate value).
2. Round to 1 decimal place:
   - The first decimal place is **6**.
   - The second decimal place is **9** (since it's ≥5, round up).

**Rounded Value:**  
**2.7**

---

### **4.2.1 Determining the Value of $$ p = \sqrt{b^2 + 4ac} $$ with $$ a = 2 $$, $$ b = 6 $$, $$ c = 2 $$**

**Given:**  
$$ a = 2 $$  
$$ b = 6 $$  
$$ c = 2 $$

**Calculation:**
$$
p = \sqrt{b^2 + 4ac} = \sqrt{6^2 + 4 \times 2 \times 2} = \sqrt{36 + 16} = \sqrt{52}
$$
Simplify $$ \sqrt{52} $$:
$$
\sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}
$$

**Value of $$ p $$:**  
$$ p = 2\sqrt{13} $$

---

### **4.2.2 Determining Whether $$ p $$ is Rational or Irrational**

**Expression:**  
$$ p = 2\sqrt{13} $$

**Analysis:**
- $$ \sqrt{13} $$ is an **irrational** number because 13 is not a perfect square.
- Multiplying an irrational number by a rational number (2) still results in an **irrational** number.

**Conclusion:**  
$$ p $$ is **Irrational**.

---

### **4.2.3 Identifying the Two Consecutive Integers Between Which $$ p $$ Lies**

**Expression:**  
$$ p = 2\sqrt{13} $$

**Steps:**
1. Approximate $$ \sqrt{13} $$:
   - $$ \sqrt{9} = 3 $$
   - $$ \sqrt{16} = 4 $$
   - So, $$ \sqrt{13} $$ is between **3** and **4**.
   - More precisely, $$ \sqrt{13} \approx 3.6055 $$.
2. Calculate $$ p $$:
   $$
   p = 2 \times 3.6055 \approx 7.211
   $$
3. Identify the consecutive integers:
   - **7** and **8**.

**Conclusion:**  
$$ p $$ lies between **7** and **8**.

---

### **4.3.1 Writing Recurring Decimals in Fraction Form**

*It appears there might be some formatting issues with the recurring decimals you intended to provide in part 4.3.1. However, I can demonstrate how to convert a recurring decimal to a fraction using an example.*

**Example:** Convert $$ 0.\overline{6} $$ to a fraction.

**Steps:**
1. Let $$ x = 0.\overline{6} $$.
2. Multiply both sides by 10 to shift the decimal point:
   $$
   10x = 6.\overline{6}
   $$
3. Subtract the original equation from this new equation:
   $$
   10x - x = 6.\overline{6} - 0.\overline{6}
   $$
   $$
   9x = 6
   $$
4. Solve for $$ x $$:
   $$
   x = \frac{6}{9} = \frac{2}{3}
   $$

**Conclusion:**  
$$ 0.\overline{6} = \frac{2}{3} $$

*If you provide the specific recurring decimals for parts 4.3.1 and 4.3.2, I can offer precise conversions.*

---

Feel free to provide the specific recurring decimals for **4.3.1** and **4.3.2**, and I'll be happy to assist further!
**Simplified Answers for QUESTION 4:** 1. **4.1.1** Rounded to 4 decimal places: **4.3769** 2. **4.1.2** Rounded to 1 decimal place: **2.7** 3. **4.2.1** $$ p = 2\sqrt{13} $$ 4. **4.2.2** $$ p $$ is **Irrational** 5. **4.2.3** $$ p $$ lies between **7** and **8** 6. **4.3.1** Example: $$ 0.\overline{6} = \frac{2}{3} $$ (Please provide specific decimals for complete solutions.)

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