For the hyperbola (x^2)/(9) - (y^2)/(16) = 1, the asymptotes have equations:

a = 3, b = 4, so asymptotes: y = ± (4)/(3)x.

Let f: ℝ → ℝ, f(x) = 3x - 6. The inverse f^-1 is:

y = 3x - 6 ⇒ x = (y + 6)/(3), so f^-1(y) = (y + 6)/(3).

Problems & worked solutions

RO M1

For the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$ , the asymptotes have equations:

Show answer & worked solution

$a = 3$ , $b = 4$ , so asymptotes: $y = \pm \frac{4}{3} x$ .

RO M1

Let $f : R \to R$ , $f (x) = 3 x - 6$ . The inverse $f^{- 1}$ is:

Show answer & worked solution

$y = 3 x - 6 \Rightarrow x = \frac{y + 6}{3}$ , so $f^{- 1} (y) = \frac{y + 6}{3}$ .

RO M1

At a meeting, every pair of people shakes hands exactly once. If there are $45$ handshakes total, how many people are there?

Show answer & worked solution

$\frac{n (n - 1)}{2} = 45 \Rightarrow n (n - 1) = 90 \Rightarrow n = 10$ .

Problems & worked solutions

RO M1

For the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$ , the asymptotes have equations:

Show answer & worked solution

$a = 3$ , $b = 4$ , so asymptotes: $y = \pm \frac{4}{3} x$ .

RO M1

Let $f : R \to R$ , $f (x) = 3 x - 6$ . The inverse $f^{- 1}$ is:

Show answer & worked solution

$y = 3 x - 6 \Rightarrow x = \frac{y + 6}{3}$ , so $f^{- 1} (y) = \frac{y + 6}{3}$ .

RO M1

At a meeting, every pair of people shakes hands exactly once. If there are $45$ handshakes total, how many people are there?

Show answer & worked solution

$\frac{n (n - 1)}{2} = 45 \Rightarrow n (n - 1) = 90 \Rightarrow n = 10$ .