The equation e^(sin x) – e(–sin x) – 4 = 0 has -Maths 9th

The equation e^(sin x) – e(–sin x) – 4 = 0 has -Maths 9th
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Last Answer : Let the roots of the equation x3 – ax2 + bx – c = 0 be (α – 1), α, (α + 1) ∴ S2 = (α – 1)α + α(α + 1) + (α + 1) ( ... ; 1 = b ⇒ 3α2 – 1 = b ∴ Minimum value of b = – 1, when α = 0.

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A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Description : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

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A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Description : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

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According to D' Alembert's principle, m (d 2 x/ dt 2 ) + c (dx/dt) + Kx =0 is the differential equation for damped free vibrations having single degree of freedom. What will be the solution to this differential equation if the system is critically damped? A x = (A + Bt) e – ωt B x = X e – ξωt (sin ω d t + Φ) C x = (A – Bt) e – ωt D x = X e – ξωt (cos ω d t + Φ)

Description : According to D' Alembert's principle, m (d 2 x/ dt 2 ) + c (dx/dt) + Kx =0 is the differential equation for damped free vibrations having single degree of freedom. What will be the solution to this differential equation if the system is ... Φ) C x = (A - Bt) e - ωt D x = X e - ξωt (cos ω d t + Φ)

Last Answer : A x = (A + Bt) e – ωt

According to D' Alembert's principle, m (d 2 x/ dt 2 ) + c (dx/dt) + Kx =0 is the differential equation for damped free vibrations having single degree of freedom. What will be the solution to this differential equation if the system is critically damped? A. x = (A + Bt) e – ωt B. x = X e – ξωt (sin ω d t + Φ) C. x = (A – Bt) e – ωt D. x = X e – ξωt (cos ω d t + Φ

Description : According to D' Alembert's principle, m (d 2 x/ dt 2 ) + c (dx/dt) + Kx =0 is the differential equation for damped free vibrations having single degree of freedom. What will be the solution to this differential equation if the system is ... ) C. x = (A - Bt) e - ωt D. x = X e - ξωt (cos ω d t + Φ

Last Answer : A. x = (A + Bt) e – ωt

According to D' Alembert's principle, m (d 2 x/ dt 2 ) + c (dx/dt) + Kx =0 is the A differential equation for damped free vibrations having single degree of freedom. What will be the solution to this differential equation if the system is critically damped? ( A ) x = (A + Bt) e – ωt (B)x = X e – ξωt (sin ω d t + Φ) (C)x = (A – Bt) e – ωt ( D )x = X e – ξωt (cos ω d t + Φ

Description : According to D' Alembert's principle, m (d 2 x/ dt 2 ) + c (dx/dt) + Kx =0 is the A differential equation for damped free vibrations having single degree of freedom. What will be the solution to this differential equation if the system is ... (C)x = (A - Bt) e - ωt ( D )x = X e - ξωt (cos ω d t + Φ

Last Answer : ( A ) x = (A + Bt) e – ωt

According to D' Alembert's principle, m (d 2 x/ dt 2 ) + c (dx/dt) + Kx =0 is the differential equati damped free vibrations having single degree of freedom. What will be the solution to this differ equation if the system is critically damped? a. x = (A + Bt) e – ωt b. x = X e – ξωt (sin ω d t + Φ) c. x = (A – Bt) e – ωt d. x = X e – ξωt (cos ω d t + Φ)

Description : According to D' Alembert's principle, m (d 2 x/ dt 2 ) + c (dx/dt) + Kx =0 is the differential equati damped free vibrations having single degree of freedom. What will be the solution to this differ equation if the system is critically ... c. x = (A - Bt) e - ωt d. x = X e - ξωt (cos ω d t + Φ)

Last Answer : a. x = (A + Bt) e – ωt

The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th

Description : The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th

Last Answer : (a) In natural numbers, there is only one pair i.e., (1, 1) which satisfy the given equation but in positive real numbers, real numbers and rational numbers there are many pairs to satisfy the given linear equation.

The equation x = 7, in two variables can be written as -Maths 9th

Description : The equation x = 7, in two variables can be written as -Maths 9th

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x = 5 and y = 2 is a solution of the linear equation. -Maths 9th

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Last Answer : (c) (a) Take x + 2y, on putting x=5 and y = 2, we get 5 + 2(2) = 5+ 4 = 9≠7. So, (5, 2) is not a solution of x + 2y = 7 (b) Take 5x + 2y, on putting x = 5 and y = 2, we get 5x 5 + 2 x2 =25+ 4 = ... on putting x = 5 and y = 2, we get 5 5 + 2=25 + 2 =27 ≠7 So, (5, 2) is not a solution of 5x + y = 7.

The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

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Last Answer : (c) Since, the graph of linear equation 2x + 3y = 6 meets the X-axis. So, we put y = 0 in 2x + 3y = 6 ⇒ 2x + 3(0) = 6 = 2x + 0 = 6 ⇒ x = 6/2 ⇒ x = 3 Hence, the coordinate on X-axis is (3, 0).

The graph of the linear equation y = x passes through the point. -Maths 9th

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Last Answer : (c) The linear equation y = x has same value of x and y-coordinates are same. Therefore, the point (1,1) must lie on the line y = x.

Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

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Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

How many solutions of the equation 2x + 1 = x – 3 are there on the Cartesian plane? -Maths 9th

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Last Answer : 2x + 1 = x - 3 2x-x = -3-1 ∴ x = - 4 ..(i) and it can be written as 1.x + 0. y = - 4 ..(ii) (i) Number line represent the all real values of x on the X ... the equation x + 4 = 0 represent a straight line parallel to Y-axis and infinitely many points lie on a line in the cartesian plane.

Find the solution of the linear equation x+2y = 8 which represents a point on -Maths 9th

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Last Answer : We have, x + 2y = 8 ,..(i) (i) When the point is on the X-axis, then put y = 0 in Eq. (i), we get x+2 (0)=8 ⇒ x = 8 Hence, the required point is (8, 0). (ii) When the point is on the Y-axis, then put x = 0 in Eq. (i), we get 0 + 2y = 8 ⇒ y = 8/2 = 4 Hence, the required point is (0, 4).

For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

Description : For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

Last Answer : The given linear equation is 2x + cy= 8. …(i) Now, by condition, x and y-coordinate of given linear equation are same, i.e., x = y. Put y = x in Eq. (i), we get

Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

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Last Answer : Linear equation