$\tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)$ $=\tan ^{-1} \tan \frac{\pi}{4}+\cos ^{-1}\left(-\cos \frac{\pi}{3}\right)+\sin ^{-1}\left(-\sin...
Find the principal values of the following:$\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$
$\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ Assume, $y=\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ BBYJU'S $\cos y=-\frac{1}{\sqrt{2}}$ $\cos y=-\cos \frac{\pi}{4}$ $\cos y=\cos...
Maximise Z = x + y, subject to
The area dictated by the requirements equation is given beneath There is no attainable area and in this manner, z has no greatest worth.
Maximise Z = − x + 2y, subject to the constraints:
The achievable district dictated by the limitations, NCERT Solutions Mathematics Class 12 Chapter 12 - 19 is given beneath Here, it very well may be seen that the doable district is unbounded. The...
Minimise and Maximise Z = x + 2y subject to
The plausible district dictated by the limitations, x + 2y ≥ 100, 2x − y ≤ 0, 2x + y ≤ 200, x ≥ 0, and y ≥ 0, is given beneath A (0, 50), B (20, 40), C (50, 100) and D (0, 200) are the corner points...
. Minimise and Maximise Z = 5x + 10y subject to:
The achievable district dictated by the imperatives, x + 2y ≤ 120, x + y ≥ 60, x − 2y ≥ 0, x ≥ 0, and y ≥ 0, is given beneath A (60, 0), B (120, 0), C (60, 30), and D (40, 20) are the corner points...
Minimise Z = x + 2y subject to:
The achievable area controlled by the limitations, 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, and y ≥ 0, is given beneath A (6, 0) and B (0, 3) are the corner points of the achievable area The upsides of Z at...
Maximise Z = 3x + 2y subject to:
The doable area controlled by the requirements, x + 2y ≤ 10, 3x + y ≤ 15, x ≥ 0, and y ≥ 0, is given underneath A (5, 0), B (4, 3), C (0, 5) and D (0, 0) are the corner points of the doable area....
Minimise Z = 3x + 5y such that:
The doable area dictated by the arrangement of requirements, x + 3y ≥ 3, x + y ≥ 2, and x, y ≥ 0 is given underneath It tends to be seen that the plausible district is unbounded. The corner points...
Maximise Z = 5x + 3y subject to:
The practical locale controlled by the arrangement of requirements, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are given underneath O (0, 0), A (2, 0), B (0, 3) and C (20/19, 45/19) are the...
Minimise Z = −3x + 4y subject to.
The feasible region determined by the system of constraints,is given below O (0, 0), A (4, 0), B (2, 3) and C (0, 4) are the corner points of the feasible region The values of Z at these corner...
Maximise Z = 3x + 4y Subject to the constraints:
Solution: The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0, is given below O (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The...