If the length of the major axis of an ellipse with the equation `5x^(2)+24y^(2)=40` is j and the length of the minor axis of the ellipse is n, then what is the value of `j + n` ?

Options

A. `2.58`
B. `5.66`
C. `6.95`
D. `8.24`

Explanation

To solve the problem, we need to find the lengths of the major and minor axes of the ellipse given by the equation \(5x^2 + 24y^2 = 40\). We will then compute \(j + n\), where \(j\) is the length of the major axis and \(n\) is the length of the minor axis.

### Step-by-Step Solution:

1. **Rewrite the equation in standard form**:
The given equation is \(5x^2 + 24y^2 = 40\). We need to divide the entire equation by 40 to express it in the standard form of an ellipse:
\[ \frac{5x^2}{40} + \frac{24y^2}{40} = 1 \]
This simplifies to:
\[ \frac{x^2}{8} + \frac{y^2}{\frac{5}{3}} = 1 \]

2. **Identify \(a^2\) and \(b^2\)**:
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can identify:
\[ a^2 = 8 \quad \text{and} \quad b^2 = \frac{5}{3} \]

3. **Calculate \(a\) and \(b\)**:
Now, we take the square roots to find \(a\) and \(b\):
\[ a = \sqrt{8} = 2\sqrt{2} \quad \text{and} \quad b = \sqrt{\frac{5}{3}} = \frac{\sqrt{15}}{3} \]

4. **Determine the lengths of the axes**:
The lengths of the major and minor axes are given by:
\[ \text{Length of major axis } (j) = 2a = 2 \times 2\sqrt{2} = 4\sqrt{2} \]
\[ \text{Length of minor axis } (n) = 2b = 2 \times \frac{\sqrt{15}}{3} = \frac{2\sqrt{15}}{3} \]

5. **Calculate \(j + n\)**:
Now we need to find \(j + n\):
\[ j + n = 4\sqrt{2} + \frac{2\sqrt{15}}{3} \]

6. **Approximate the values**:
To find the numerical value, we can approximate:
\[ \sqrt{2} \approx 1.414 \implies 4\sqrt{2} \approx 4 \times 1.414 \approx 5.656 \]
\[ \sqrt{15} \approx 3.873 \implies \frac{2\sqrt{15}}{3} \approx \frac{2 \times 3.873}{3} \approx 2.582 \]
Adding these approximations:
\[ j + n \approx 5.656 + 2.582 \approx 8.238 \]

7. **Final result**:
Therefore, the value of \(j + n\) is approximately \(8.24\).

### Conclusion:
The value of \(j + n\) is \(8.24\).