Solve the following expression:

[tex]$\beta e^{u^2} e^{l v e}$[/tex]

Which of the following represents a fraction?

A. [tex]$\frac{1}{6} + \frac{3}{10}$[/tex]

B. [tex]$\frac{1}{3} + \frac{5}{12}$[/tex]



Answer :

Certainly! Let's address part b, which involves calculating the sum of the given fractions [tex]\(\frac{1}{3} + \frac{5}{12}\)[/tex]. Below is the detailed step-by-step solution for this problem:

### Step 1: Understand the Problem
We are given two fractions:
- [tex]\(\frac{1}{3}\)[/tex]
- [tex]\(\frac{5}{12}\)[/tex]

Our goal is to find the sum of these two fractions.

### Step 2: Find a Common Denominator
To add fractions, they need to have a common denominator. The denominators in our fractions are 3 and 12. The least common multiple (LCM) of 3 and 12 is 12.

### Step 3: Convert Fractions to Common Denominator
We need to convert both fractions so that they have the denominator of 12.

- For [tex]\(\frac{1}{3}\)[/tex]:
We multiply the numerator and the denominator by 4 to get an equivalent fraction:
[tex]\[ \frac{1 \times 4}{3 \times 4} = \frac{4}{12} \][/tex]

- For [tex]\(\frac{5}{12}\)[/tex]:
The denominator is already 12, so we do not need to change this fraction.

### Step 4: Add the Fractions
Now both fractions are [tex]\(\frac{4}{12}\)[/tex] and [tex]\(\frac{5}{12}\)[/tex]. Since they have a common denominator, we can add the numerators directly:
[tex]\[ \frac{4}{12} + \frac{5}{12} = \frac{4 + 5}{12} = \frac{9}{12} \][/tex]

### Step 5: Simplify the Result (if necessary)
We should check if the resulting fraction can be simplified. The greatest common divisor (GCD) of 9 and 12 is 3. Therefore, we can simplify [tex]\(\frac{9}{12}\)[/tex] by dividing both the numerator and the denominator by 3:
[tex]\[ \frac{9 \div 3}{12 \div 3} = \frac{3}{4} \][/tex]

So, the simplest form of [tex]\(\frac{9}{12}\)[/tex] is [tex]\(\frac{3}{4}\)[/tex].

### Step 6: Conclusion
The sum of the fractions [tex]\(\frac{1}{3} + \frac{5}{12}\)[/tex] is [tex]\(\frac{3}{4}\)[/tex].

So, the answer to part b is [tex]\(\frac{3}{4}\)[/tex].