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Yes. In the first one, since there is a 2/(x+2)^2, this can only take on positive values (and moreover can take on any positive value), because of the square in the denominator (and positive constant numerator). This causes the range to become (2, infinity). With the second one, the 2/(x+2) can...

ℝ \ {2} (You forgot a backslash.)

Essentially. (Make sure to first express y in terms of x for the 2y + mx + 8 = 0.)

Yes (for y = 2/[(x+2)2] + 2).

Yes

Yeah that's the range. And it's because the graph doesn't touch 2 that we put an open bracket around the 2.

Horizontal line through the point (0, c). (Where the x-axis is the horizontal axis and the y-axis is the vertical axis, as usual.)

You said the range is all real numbers excluding 2. But this is not correct. Can you see why?

The domain is correct, but the range is not. (And the 2 and -2 should be written in curly brackets, like {-2}.)

Who said that? You still generally need to do well in a subject to get a high ATAR with it. It's just that you may not need such a high raw mark to do so if it has good scaling.

Re: Several Variable Calculus You ended up with the negative of your previous answer because you used the oppositely oriented normal.

Why are so many BOS users seemingly transitioning to this "one rival site" (no prizes for guessing which site it is, I think)? Is it because they are finding they can't get their questions answered on BOS but can on the rival site?

Yes. What Drongoski gave is the natural domain (assuming we are dealing with functions of a real variable). You can read up about this here: https://en.wikipedia.org/wiki/Domain_of_a_function#Natural_domain .

$\noindent Assuming this was the case, the third term is $ar^{2} = 54$ and the sixth term is $ar^{5} = 11\tfrac{83}{125}$. This implies that $r^{3} = \frac{ar^{5}}{ar^{3}} = \frac{11\tfrac{83}{125}}{54} = \frac{27}{125}$. Taking cube roots, we have $\boxed{r = \frac{3}{5}}$. Now$...

$\noindent Possibly, the original poster meant $11\tfrac{83}{125}$.$

$\noindent The no. of total possible hands of five cards is $\binom{52}{5}$. The no. of hands that give a royal flush is just $4$ (one way from each suit). Hence the probability of getting a royal flush is $\frac{4}{\binom{52}{5}} = \frac{1}{649740}$.$ $\noindent An expression $\binom{n}{k}$...

For Q2: a) Note that the number made will be greater than 500 if and only if the first digit is chosen to be either 5, 8, or 9. There are four choices for the first digit altogether (1, 5, 8 or 9), and precisely three of them result in a number larger than 500 (5, 8 or 9). Hence the answer is 3...

Re: Several Variable Calculus Imagine it as polar coordinates where the radius is e^x and angle is y ==> radius > 0 (and can attain any positive value) and angle can be anything ==> any point except (0, 0) is in the image (since any nonzero point can be expressed in polar coordinates with a...

Domain

$\noindent It would be $(-\infty, \infty)$. In other words, $\mathbb{R}$.$