It is given that \((\frac {x}{3}+1, \frac {y-2}{3}) = (\frac {5}{3}, \frac {1}{3})\).
Since the ordered pairs are equal, the corresponding elements will also be equal.
\((\frac {x}{3}+1, \frac {y-2}{3}) = (\frac {5}{3}, \frac {1}{3})\)
Therefore,
\(\frac {x}{3}+1=\frac {5}{3}\)
⇒ \(\frac{x}{3} = \frac{5}{3} -1,\frac{y-2}{3} = \frac{1}{3}\)
⇒\(\frac{x}{3} = \frac{2}{3}, y = \frac{1}{3}+\frac{2}{3}\)
⇒ x=2, y=1
If the domain of the function $ f(x) = \log_7(1 - \log_4(x^2 - 9x + 18)) $ is $ (\alpha, \beta) \cup (\gamma, \delta) $, then $ \alpha + \beta + \gamma + \delta $ is equal to
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
Let $ A = \{0, 1, 2, 3, 4, 5, 6\} $ and $ R_1 = \{(x, y): \max(x, y) \in \{3, 4 \}$. Consider the two statements:
Statement 1: Total number of elements in $ R_1 $ is 18.
Statement 2: $ R $ is symmetric but not reflexive and transitive.
Draw the Lewis structures for the following molecules and ions: \(H_2S\), \(SiCl_4\), \(BeF_2\), \(CO_3^{2-}\) , \(HCOOH\)
| λ (nm) | 500 | 450 | 400 |
|---|---|---|---|
| v × 10–5(cm s–1) | 2.55 | 4.35 | 5.35 |