What is the domain of sqrt(25-(t^2+9)) ?

The domain of sqrt(25-(t^2+9)) is -4<= t<= 4

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domain of sqrt(25-(t^2+9))

domain

25 - (t^{2} + 9)

- 4 \leq t \leq 4

Interval Notation

[- 4, 4]

Solution steps

Find non-negative values for radicals:

- 4 \leq t \leq 4

The function domain

- 4 \leq t \leq 4

f (x) = - x (x + 2) (x - 4)

f (x) = \frac{5}{( x + 1 ) ^{2}}

f (x) = \frac{x + 1}{5 - x}

f (x) = 4 x + 13

y = - \frac{1}{2} x - 3

What is the domain of sqrt(25-(t^2+9)) ?
The domain of sqrt(25-(t^2+9)) is -4<= t<= 4