To solve the problem, we need to find the relationship between the number of elements in sets \( A \) and \( B \), and subsequently use those values to calculate the distance between the points \( P(m, n) \) and \( Q(-2, -3) \).
Now, we need to find the distance between points \( P(6, 3) \) and \( Q(-2, -3) \) using the distance formula:
The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
\(d = \sqrt{((-2) - 6)^2 + ((-3) - 3)^2}\) \(d = \sqrt{(-8)^2 + (-6)^2}\) \(d = \sqrt{64 + 36}\) \(d = \sqrt{100}\) \(d = 10\)
Thus, the distance between point \( P(m, n) \) and point \( Q(-2, -3) \) is 10. Therefore, the correct answer is 10.
The total number of subsets of a set with \(m\) elements is \(2^m\) and for a set with \(n\) elements is \(2^n\). Given:
\(2^m = 2^n + 56.\)
Rearranging:
\(2^m - 2^n = 56.\)
Factoring the left side:
\(2^n (2^{m-n} - 1) = 56.\)
Since \(56 = 2^3 \times 7\), we set \(2^n = 8 \implies n = 3\) and
\(2^{m-n} - 1 = 7 \implies 2^{m-n} = 8 \implies m - n = 3.\)
Therefore:
\(m = 6, \quad n = 3.\)
The distance between points \(P(6, 3)\) and \(Q(-2, -3)\) is given by:
\(\text{Distance} = \sqrt{(6 - (-2))^2 + (3 - (-3))^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10.\)
Thus, the correct answer is 10.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,