Sure, let's go through each sum step-by-step to determine which one is an irrational number.
### Sum (a): \(2.5 + 3\)
First, we add the numbers:
[tex]\[ 2.5 + 3 = 5.5 \][/tex]
The number 5.5 is a finite decimal, hence it is a rational number.
### Sum (b): \(\sqrt{4} + 5\)
First, we need to evaluate \(\sqrt{4}\):
[tex]\[ \sqrt{4} = 2 \][/tex]
So the sum is:
[tex]\[ 2 + 5 = 7 \][/tex]
The number 7 is an integer and, therefore, a rational number.
### Sum (c): \( \frac{10}{3} + \frac{21}{5} \)
First, we convert the mixed fractions to improper fractions:
[tex]\[ 3 \frac{1}{3} = \frac{10}{3} \][/tex]
[tex]\[ 4 \frac{1}{5} = \frac{21}{5} \][/tex]
We then add these fractions:
[tex]\[ \frac{10}{3} + \frac{21}{5} \][/tex]
To add these fractions, we need a common denominator:
[tex]\[ \frac{10}{3} = \frac{50}{15} \][/tex]
[tex]\[ \frac{21}{5} = \frac{63}{15} \][/tex]
So the sum is:
[tex]\[ \frac{50}{15} + \frac{63}{15} = \frac{113}{15} \][/tex]
Converting this back into decimal:
[tex]\[ \frac{113}{15} = 7.533333\ldots \][/tex]
The number 7.533333\ldots is a finite or repeating decimal, hence it is a rational number.
### Sum (d): \(\sqrt{7} + \sqrt{7}\)
Finally, we add:
[tex]\[ \sqrt{7} + \sqrt{7} = 2\sqrt{7} \][/tex]
Since \(\sqrt{7}\) is an irrational number (square roots of non-perfect squares are irrational), multiplying it by 2 does not change its irrationality. Therefore, \(2\sqrt{7}\) is also an irrational number.
### Conclusion:
From the calculations above, the sum in option (d), \(\sqrt{7} + \sqrt{7}\), which gives \(2\sqrt{7}\), is the irrational number.
- Option (a): \(2.5 + 3 = 5.5\) (rational)
- Option (b): \(\sqrt{4} + 5 = 7\) (rational)
- Option (c): \(3\frac{1}{3} + 4\frac{1}{5} = \frac{113}{15} = 7.533333\ldots\) (rational)
- Option (d): \(\sqrt{7} + \sqrt{7} = 2\sqrt{7}\) (irrational)
Thus, Sum (d): [tex]\(\sqrt{7} + \sqrt{7}\)[/tex] is an irrational number.
