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ladessa [460]
3 years ago
12

What is the domain of the function

Mathematics
1 answer:
IRINA_888 [86]3 years ago
6 0
\sqrt{5+x} is defined as long as 5+x\ge0, or x\ge-5. However, \dfrac1x is undefined when x=0. This point is contained in the set \{x|x\ge-5\}, so we need to exclude it. We can write the domain in several ways.

\{x|x\ge-5~\land~x\neq0\}
-5\le x
[-5,0)\cup(0,\infty)
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5/16t=-9
Svetradugi [14.3K]
I'm going to assume that the equation is \frac{5}{16}t = -9 where the 't' is not in the denominator

If that assumption is correct, then we do two things to solve for t. First, we multiply both sides by 16. This undoes the division of 16 on the left side

\frac{5}{16}t = -9

16*\frac{5}{16}t = 16*(-9)

5t = -144

-------------------------------

Then we divide both sides by 5 to undo the multiplication of 5 (note: 5t means 5*t or "5 times t")

5t = -144

\frac{5t}{5} = \frac{-144}{5}

t = -\frac{144}{5}

-------------------------------

So here's what all the steps look like:

\frac{5}{16}t = -9

16*\frac{5}{16}t = 16*(-9)

5t = -144

\frac{5t}{5} = \frac{-144}{5}

t = -\frac{144}{5}

note: -144/5 converts to the decimal value of -28.8 so t = -144/5 is the same as t = -28.8
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