45. Which of the following are correct?

A. L.C.M [tex][tex]$(\sqrt{28}, \sqrt{63}, 3 \sqrt{175}, 2 \sqrt{700})$[/tex][/tex] is [tex][tex]$60 \sqrt{7}$[/tex][/tex]
B. L.C.M [tex][tex]$\{2 \sqrt{7}, 3 \sqrt{11}, 5 \sqrt{2}\}$[/tex][/tex] does not exist
C. L.C.M [tex][tex]$(5, 10, 3 \pi) = 30 \pi$[/tex][/tex]
D. L.C.M [tex][tex]$\left\{\frac{2}{5}, \frac{3}{4}, \frac{1}{2}\right\} = 6$[/tex][/tex]



Answer :

To determine which statements are correct, we need to verify each one individually. Here, we can assume the steps and mathematical verifications necessary to validate the given options. Below is the step-by-step robustness of each statement:

### A) L.C.M [tex]$(\sqrt{28}, \sqrt{63}, 3 \sqrt{175}, 2 \sqrt{700})$[/tex] is [tex]$60 \sqrt{7}$[/tex]
Let's break down the terms:
- [tex]\(\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}\)[/tex]
- [tex]\(\sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7}\)[/tex]
- [tex]\(3 \sqrt{175} = 3 \sqrt{25 \times 7} = 15\sqrt{7}\)[/tex]
- [tex]\(2 \sqrt{700} = 2 \sqrt{100 \times 7} = 20\sqrt{7}\)[/tex]

Finding the LCM of [tex]\(2\sqrt{7}\)[/tex], [tex]\(3\sqrt{7}\)[/tex], [tex]\(15\sqrt{7}\)[/tex], and [tex]\(20\sqrt{7}\)[/tex]:
1. Extracting the [tex]\(\sqrt{7}\)[/tex] common factor, we look for the LCM of the coefficients: [tex]\(2, 3, 15, 20\)[/tex].
2. The LCM of the set [tex]\(2, 3, 15, 20\)[/tex] is found as [tex]\( \text{LCM}(2, 3, 15, 20) = 60 \)[/tex].
3. So, the LCM of [tex]\(\sqrt{28}, \sqrt{63}, 3 \sqrt{175}, 2 \sqrt{700}\)[/tex] is [tex]\( 60\sqrt{7} \)[/tex].

Thus, the statement is correct. Calculating this directly matches the given LCM of [tex]\(60\sqrt{7}\)[/tex] which is correct.

### B) L.C.M [tex]$\{2 \sqrt{7}, 3 \sqrt{11}, 5 \sqrt{2}\}$[/tex] does not exist
In this case, we need to consider:
- [tex]\(2\sqrt{7}\)[/tex]
- [tex]\(3\sqrt{11}\)[/tex]
- [tex]\(5\sqrt{2}\)[/tex]

The LCM of irrational numbers involving different radicals ([tex]\(\sqrt{7}, \sqrt{11}, \sqrt{2}\)[/tex]) is not straightforward. These numbers cannot be combined to form a single rational multiple. Therefore, the LCM of irrational numbers with different bases essentially does not exist.

Thus, this statement is correct.

### C) L.C.M [tex]$(5,10,3\pi)=30\pi$[/tex]
Consider the given set:
- Constants: [tex]\(5,\ 10\)[/tex]
- Irrational number: [tex]\(3\pi\)[/tex]

First find the LCM of the constants [tex]\(5,\ 10\)[/tex]:
- The LCM of [tex]\(5\)[/tex] and [tex]\(10\)[/tex] is [tex]\(10\)[/tex].

Since there’s a multiple of [tex]\(3\pi\)[/tex] involved, the resulting LCM should encompass both [tex]\(10\)[/tex] and [tex]\(3\pi\)[/tex].
Combining these:
- [tex]\(10\)[/tex] has no effect over [tex]\(\pi\)[/tex] so the LCM should then be [tex]\(10\)[/tex] times [tex]\(3\pi\)[/tex]:
[tex]\[ \text{LCM}(5, 10, 3\pi) = 10 \cdot 3\pi = 30\pi \][/tex]

Thus, the statement is correct.

### D) L.C.M [tex]$\left\{\frac{2}{5}, \frac{3}{4}, \frac{1}{2}\right\}=6$[/tex]
Consider the given fractions:
- [tex]\(\frac{2}{5}, \frac{3}{4}, \frac{1}{2}\)[/tex]

To find the LCM of fractions:
[tex]\[ \text{LCM} \left(\frac{a}{b}, \frac{c}{d}, \frac{e}{f} \right) = \frac{\text{LCM}(a, c, e)}{\text{GCD}(b, d, f)} \][/tex]

Here, we need to calculate:
1. Numerators LCM:
[tex]\[ \text{LCM}(2, 3, 1) = 6 \][/tex]

2. Denominators GCD:
[tex]\[ \text{GCD}(5, 4, 2) = 1 \][/tex]

Thus:
[tex]\[ \text{LCM}\left( \frac{2}{5}, \frac{3}{4}, \frac{1}{2} \right) = \frac{6}{1} = 6 \][/tex]

Thus, the statement is correct.

### Conclusion:
Summarizing, the correct answers to the respective statements are:
- (A) Correct
- (B) Correct
- (C) Correct
- (D) Correct

All the statements A, B, C, and D are correct, signifying that the respective correct options provide evident results validated by defined mathematical principles, as detailed above.

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