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The number of solutions of the system of in equations x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1 is
To graph the system of inequalities: 1. x + 2y ≤ 3 2. 3x + 4y ≥ 12 3. x ≥ 0 4. y ≥ 1 We draw a graph of the inequalities. For x + 2y ≤ 3, the line x + 2y = 3 crosses the x-axis at (3, 0) and the y-axis at (0, 1.5). - For 3x + 4y ≥ 12, the line 3x + 4y = 12 cuts the x-axis at the point (4, 0) and y-aRead more
To graph the system of inequalities:
1. x + 2y ≤ 3
2. 3x + 4y ≥ 12
3. x ≥ 0
4. y ≥ 1
We draw a graph of the inequalities.
For x + 2y ≤ 3, the line x + 2y = 3 crosses the x-axis at (3, 0) and the y-axis at (0, 1.5).
– For 3x + 4y ≥ 12, the line 3x + 4y = 12 cuts the x-axis at the point (4, 0) and y-axis at (0, 3).
-x ≥ 0 and y ≥ 1 limits the feasible region to the first quadrant above the line y = 1.
When we plot these constraints, we see that the feasible region is empty because the two lines do not intersect within the given constraints.
Thus, the number of solutions is zero.
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Rutherford’s alpha (a-) particle scattering experiment resulted into the discovery of
(c) Rutherford experiment resulted in the discovery of nucleus. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-4/
(c) Rutherford experiment resulted in the discovery of nucleus.
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-4/
See lessCorner points of the feasible region determined by the system of linear constraints are (0,3), (1,1) and (3,0). Let Z = 4x + 5y be the objective function. The minimum value of Z occurs at
To determine the minimum value of Z = 4x + 5y, we input the coordinates for the corner points into the objective function. 1. For the point (0, 3): Z = 4(0) + 5(3) Z = 15 2. For the point (1, 1): Z = 4(1) + 5(1) Z = 9 3. For the point (3, 0): Z = 4(3) + 5(0) Z = 12 The point (1, 1) is where the miniRead more
To determine the minimum value of Z = 4x + 5y, we input the coordinates for the corner points into the objective function.
1. For the point (0, 3):
Z = 4(0) + 5(3)
Z = 15
2. For the point (1, 1):
Z = 4(1) + 5(1)
Z = 9
3. For the point (3, 0):
Z = 4(3) + 5(0)
Z = 12
The point (1, 1) is where the minimum value of Z = 9 occurs.
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Which of the following is not correct?
Choice (c) is correct. As sec θ ≤ - 1 or sec θ ≥ 1, so sec θ cannot be equal to 1/2. This question related to Chapter 3 maths Class 11th NCERT. From the Chapter 3: Trigonometric Functions. Give answer according to your understanding. For more please visit here: https://www.tiwariacademy.com/ncert-sRead more
Choice (c) is correct.
As sec θ ≤ – 1 or sec θ ≥ 1, so sec θ cannot be equal to 1/2.
This question related to Chapter 3 maths Class 11th NCERT. From the Chapter 3: Trigonometric Functions. Give answer according to your understanding.
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Tangent function is negative in
Choice (b) is correct. As tangent function is negative in 2nd and 4th quadrants. This question related to Chapter 3 maths Class 11th NCERT. From the Chapter 3: Trigonometric Functions. Give answer according to your understanding. For more please visit here: https://www.tiwariacademy.com/ncert-solutRead more
Choice (b) is correct. As tangent function is negative in 2nd and 4th quadrants.
This question related to Chapter 3 maths Class 11th NCERT. From the Chapter 3: Trigonometric Functions. Give answer according to your understanding.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-3