Class work Without using a Ealculator, determine between whieh tho integers the following irrational numbers lie D) $$ \sqrt{50} $$ $$ = $$ $$ = $$ $$ herefore $$ (2) $$ \sqrt{29} $$ $$ = $$ $$ = $$ $$ herefore $$ $$ 3 \sqrt[3]{45} $$ $$ = $$ $$ = $$ - $$ \quad $$ - (1) $$ -\sqrt{54} $$ $$ = $$ $$ = $$

Question

Class work Without using a Ealculator, determine between whieh tho integers the following irrational numbers lie D) $$ \sqrt{50} $$ $$ = $$ $$ = $$ $$ 	herefore $$ (2) $$ \sqrt{29} $$ $$ = $$ $$ = $$ $$ 	herefore $$ $$ 3 \sqrt[3]{45} $$ $$ = $$ $$ = $$ - $$ \quad $$ - (1) $$ -\sqrt{54} $$ $$ = $$ $$ = $$

UpStudy AI · Accepted Answer

Sure, let's determine between which two integers each of the given irrational numbers lie without using a calculator.

---

### (1) $$ -\sqrt{54} $$

**Step 1:** Find the nearest perfect squares around 54.

- $$ 7^2 = 49 $$
- $$ 8^2 = 64 $$

**Step 2:** Determine where 54 falls.

- $$ 49 < 54 < 64 $$

**Step 3:** Apply square roots.

- $$ 7 < \sqrt{54} < 8 $$

**Step 4:** Since we have a negative sign:

- $$ -8 < -\sqrt{54} < -7 $$

**Conclusion:**

$$
-\sqrt{54} \quad \text{lies between} \quad -8 \quad \text{and} \quad -7
$$

---

### (2) $$ \sqrt{29} $$

**Step 1:** Find the nearest perfect squares around 29.

- $$ 5^2 = 25 $$
- $$ 6^2 = 36 $$

**Step 2:** Determine where 29 falls.

- $$ 25 < 29 < 36 $$

**Step 3:** Apply square roots.

- $$ 5 < \sqrt{29} < 6 $$

**Conclusion:**

$$
\sqrt{29} \quad \text{lies between} \quad 5 \quad \text{and} \quad 6
$$

---

### (D) $$ \sqrt{50} $$

**Step 1:** Find the nearest perfect squares around 50.

- $$ 7^2 = 49 $$
- $$ 8^2 = 64 $$

**Step 2:** Determine where 50 falls.

- $$ 49 < 50 < 64 $$

**Step 3:** Apply square roots.

- $$ 7 < \sqrt{50} < 8 $$

**Conclusion:**

$$
\sqrt{50} \quad \text{lies between} \quad 7 \quad \text{and} \quad 8
$$

---

### (3) $$ 3\sqrt[3]{45} $$

**Step 1:** Find the nearest perfect cubes around 45.

- $$ 3^3 = 27 $$
- $$ 4^3 = 64 $$

**Step 2:** Determine where 45 falls.

- $$ 27 < 45 < 64 $$

**Step 3:** Apply cube roots.

- $$ 3 < \sqrt[3]{45} < 4 $$

**Step 4:** Multiply by 3.

- $$ 3 \times 3 = 9 $$
- $$ 3 \times 4 = 12 $$

- So, $$ 9 < 3\sqrt[3]{45} < 12 $$

**Refined Estimate:**

For a more precise interval:

- $$ \sqrt[3]{45} $$ is slightly greater than $$ 3.5 $$ (since $$ 3.5^3 = 42.875 $$ and $$ 3.6^3 = 46.656 $$)
- Therefore, $$ 3.5 < \sqrt[3]{45} < 3.6 $$
- Multiplying by 3: $$ 10.5 < 3\sqrt[3]{45} < 10.8 $$

**Conclusion:**

$$
3\sqrt[3]{45} \quad \text{lies between} \quad 10 \quad \text{and} \quad 11
$$

---

### Summary

1. **$$ -\sqrt{54} $$** lies between **-8** and **-7**.
2. **$$ \sqrt{29} $$** lies between **5** and **6**.
3. **$$ \sqrt{50} $$** lies between **7** and **8**.
4. **$$ 3\sqrt[3]{45} $$** lies between **10** and **11**.

---

Feel free to ask if you need further explanations or assistance!
$$ -\sqrt{54} \quad \text{lies between} \quad -8 \quad \text{and} \quad -7 $$ $$ \sqrt{29} \quad \text{lies between} \quad 5 \quad \text{and} \quad 6 $$ $$ \sqrt{50} \quad \text{lies between} \quad 7 \quad \text{and} \quad 8 $$ $$ 3\sqrt[3]{45} \quad \text{lies between} \quad 10 \quad \text{and} \quad 11 $$

Class work
Without using a Ealculator, determine between whieh tho integers the following irrational numbers lie
D)

(2)

-
(1)

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Class work
Without using a Ealculator, determine between whieh tho integers the following irrational numbers lie
D)

(2)

-
(1)