The domain of the vector function F––(t)=t−3−−−−√i–+ln(10−2t−−−−−−√)–j+1t−3√k––𝐹_(𝑡)=𝑡−3𝑖_+𝑙𝑛(10−2𝑡)_𝑗+1𝑡−3𝑘_ is given by the interval

The domain of a vector function is the set of all possible values of the variable that will make the function "work" or be defined. In this case, we have three components to consider: t−3−−−−√i–, ln(10−2t−−−−−−√)–j, and 1t−3√k––.

1. For the first component t−3−−−−√i–, the square root function is defined for all values greater than or equal to 0. Therefore, t must be greater than or equal to 3.

2. For the second component ln(10−2t−−−−−−√)–j, the natural logarithm function is defined for all values greater than 0. Therefore, the expression inside the logarithm, 10−2t−−−−−−√, must be greater than 0. Solving this inequality gives us t < 5.

3. For the third component 1t−3√k––, the function is defined for all t ≠ 0. However, since we have a cube root in the denominator, we must also have t ≠ 3 to avoid division by zero.

Taking all these restrictions together, the domain of the vector function is given by the intersection of all these intervals, which is [3, 5).