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The complex number z = x + iy satisfies the condition ∣ z + 1 ∣ = 1, then z lies on
Choices (d) is correct. Given ∣ z + 1 ∣ = 1 ⇒ ∣ z -(- 1 + 0i ∣ = 1 ⇒ The distance of the point z from the point (-1, 0) is constant and is equal to 1. Thus, z lies on a circle with centre (-1, 0) and radius 1. This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex NRead more
Choices (d) is correct.
Given ∣ z + 1 ∣ = 1 ⇒ ∣ z -(- 1 + 0i ∣ = 1
⇒ The distance of the point z from the point (-1, 0) is constant and is equal to 1.
Thus, z lies on a circle with centre (-1, 0) and radius 1.
This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding.
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What does ∣ z ∣ = 4 represent on a plane where z = 4 represent on a plane where z = x + iy, a complex number?
Choice (c) is correct. ∣ z ∣ = 4 ⇒ ∣ x + iy ∣ = 4 ⇒ √x² + y² = 4 ⇒ x² + y² = 16 This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding. For more please visit here: https://www.tiwariacademRead more
Choice (c) is correct.
∣ z ∣ = 4 ⇒ ∣ x + iy ∣ = 4
⇒ √x² + y² = 4 ⇒ x² + y² = 16
This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding.
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The maximum value of Z = 3x + 4y subject to the constraints x ≥ 0, y ≥ 0 and x + y ≤ 1 is
To solve for the maximum value of Z = 3x + 4y subject to the constraints: 1. x ≥ 0 2. y ≥ 0 3. x + y ≤ 1 We plot the constraints first and determine the feasible region. The feasible region is bounded by the points: - (0, 0) where x = 0 and y = 0 - (1, 0) where x + y = 1 and y = 0 - (0, 1) where x +Read more
To solve for the maximum value of Z = 3x + 4y subject to the constraints:
1. x ≥ 0
2. y ≥ 0
3. x + y ≤ 1
We plot the constraints first and determine the feasible region.
The feasible region is bounded by the points:
– (0, 0) where x = 0 and y = 0
– (1, 0) where x + y = 1 and y = 0
– (0, 1) where x + y = 1 and x = 0
Then we calculate Z = 3x + 4y at all the corner points:
– At (0, 0), Z = 3(0) + 4(0) = 0
– At (1, 0), Z = 3(1) + 4(0) = 3
– At (0, 1), Z = 3(0) + 4(1) = 4
The maximum value for Z is 4 and it occurs at (0, 1).
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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/
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