We can see that for any integer $x$ , there will always be a $[x]$ , so there will always be a $[x]+1$ . So, we don't need to worry about the domain impacting the range.

$[x]$ can be any integer from $-\infty$ to $\infty$ , and as $\infty+1=\infty$ (at least in terms of a function's range and domain), the range of $[x]+1$ is equal to the range of $[x]$ , which is $(-\infty, \infty)$ .

So, the answer is $\boxed{(-\infty, \infty)}$ and we're done!

6 0

3 years ago

Read 2 more answers

Find the product. (0.5n^5)^2(10n^7)^3

never [62]

Answer:

250{n}^{31}

Step-by-step explanation:

$(0.5 {n}^{5} )^{2} (10 {n}^{7} )^{3} \\ = (0.5)^{2}{n}^{5 \times 2} \times (10)^{3} {n}^{7 \times 3} \\ = 0.25{n}^{10} \times 1000{n}^{21} \\ = 0.25 \times 1000 {n}^{10 + 21} \\ = 250{n}^{31}$

3 0

3 years ago

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Help please! Use the fact that (a+b)^2 = a^2 + 2ab + b^2 to find the value of 81^2. [Hint: (80 + 1)^2]

snow_lady [41]

Answer is below..................

4 0

3 years ago

For the sentence y=5x-2, what are the missing y values in this table

ANTONII [103]

Y=5x-2 so you have to solve for x
Make y = 0
Then
Y=5x-2
+2. +2
2=5x
Divide by five
X= 2/5
Then
Y=5x-2
Put x into the equation
Y=5(2/5)-2
Y=2-2
Y=0
Done I hope that help (＾∇＾)

7 0

3 years ago