$$(x + y - 7)^2 + (y + 3x - 13)^2 = 0$$
The value of $(x^3 + y^3)$ is ________ (in integer).
The given equation is: $ (x + y - 7)^2 + (y + 3x - 13)^2 = 0 $ Since the squares of real numbers are always non-negative, the sum of two squares can only be zero if each term is individually zero. Therefore, we have two separate equations:
We need to solve this system of linear equations for the integers \(x\) and \(y\).
The integer solutions are \(x = 3\) and \(y = 4\).
Now, we calculate the value of \(x^3 + y^3\) using the found integer values:
$ x^3 + y^3 = 3^3 + 4^3 $ $ 3^3 = 3 \times 3 \times 3 = 27 $ $ 4^3 = 4 \times 4 \times 4 = 64 $ $ x^3 + y^3 = 27 + 64 $ $ x^3 + y^3 = 91 $The calculated value of \( (x^3 + y^3) \) is 91.
The relationship between two variables $x$ and $y$ is given by $x + py + q = 0$ and is shown in the figure. Find the values of $p$ and $q$.
Note: The figure shown is representative.
The real variables $x, y, z$ and the real constants $p, q, r $ satisfy
$\frac{x}{pq - r^2} = \frac{y}{qr - p^2} = \frac{z}{rp - q^2}$
Given the denominators are non-zero, the value of $px + qy + rz$ is
The complex function
$e^{-\left(\frac{2}{z-1}\right)}$
has __________________
Consider two matrices: $P = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and $Q = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$.
Which of the following statement is/are true?