
At line y = –2, we have (5, –2) (6, –2) (1, –2) (0, –2)
⇒ 4 points
At line y = –1, we have (4, –1) (5, –1) (6, –1) (2, –1) (1, –1) (0, –1)
⇒ 6 points
At line y = 0, we have (0, 0) (1, 0) (2, 0) (3, 0) (4, 0) (5, 0) (6, 0)
⇒ 7 points
At line y = 1, we have (1, 1), (2, 1), (3, 1), (4, 1), (5, 1) i.e. 5 points
Similarly,
At line y = –5, we have 5 points
At line y = –4, we have 7 points
At line y = –3, we have 6 points
Then, the total integral points = 2(5 + 7 + 6) + 4 = 40
So, the answer is 40.
Let \(S=\left\{ z\in\mathbb{C}:\left|\frac{z-6i}{z-2i}\right|=1 \text{ and } \left|\frac{z-8+2i}{z+2i}\right|=\frac{3}{5} \right\}.\)
Then $\sum_{z\in S}|z|^2$ is equal to
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
A Complex Number is written in the form
a + ib
where,
The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.