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Mathematics: Post your doubts here!
I am getting 2.28m from two different methods :confused: which book is this question from? -
Mathematics: Post your doubts here!
Final energy - Initial energy = Work done by engine - Work done by friction This one is correct^. Basically the formula is 'Work done by Engine - Work done by Force = Change in Mechanical Energy'. -
Mathematics: Post your doubts here!
What's the answer to this question? -
Mathematics: Post your doubts here!
If it is projected vertically upwards and no other force is mentioned, then the only force acting would be the Gravitational Force which means that the Acceleration will be equal to -10 m/s^2. As far as the initial velocity is concerned, it'll either be mentioned that the object is projected... -
Mathematics: Post your doubts here!
y = 3x^3 - 6z^2 + 4x + 2 Differentiate with respect to 'x'. dy/dx = 9x^2 -12x + 4 dy/dx = 9x^2 - 6x - 6x + 4 dy/dx = 3x (3x - 2) - 2 (3x - 2) dy/dx = (3x-2)^2 No matter what value of 'x' you put in ' (3x-2)^2 ', the gradient will always be positive. -
Mathematics: Post your doubts here!
Q8(ii): First of all, we'll draw an argand diagram representing the inequality 'arg(z − u) = (pi)/4'. Next, we'll draw a perpendicular bisector starting from the origin to the line of angle so that it cuts the line of angle at 90°. This is how the diagram will look like (the red lines... -
Mathematics: Post your doubts here!
Q3(i): AB = OB - OA AB = (2)(-2)(11) - (-1)(2)(5) AB = (3)(-4)(6) '(3)(-4)(6)' is the direction of the plane. 3x - 4y + 6z = k The plane passes through 'b' therefore you can find the value of 'k' using the coordinates of 'b'. 3x - 4y + 6z = k 3(2) - 4(-2) + 6(11) = k 80 = k Put back the... -
Mathematics: Post your doubts here!
arg (u/u*) = 2 arg (u) arg (u/u*): u/u* = 3/5 + 4i/5 arg (u/u*) = tan^-1 [(4/5)/(3/5)] arg (u/u*) = tan^-1 (4/3) arg u: u = 2 + 1 arg u = tan^-1 (1/2) arg (u/u*) = 2 arg (u) tan^-1 (4/3) = 2 tan^-1 (1/2) Hence, proved. -
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Mathematics: Post your doubts here!
(i): x = a cos^3 t dx = -3a cos^2 t sin t y = a sin^3 t dy = 3a sin^2 t cos t dy/dx = (3a sin^2 t cos t)/(-3a cos^2 t sin t) dy/dx = - sin t / cos t (ii): x = a cos^3 t y = a sin^3 t m = - sin t / cos t y - y1 = m ( x - x1) y - a sin^3 t = (- sin t / cos t) ( x - a cos^3 t) y cos t -... -
Mathematics: Post your doubts here!
Please remove the old link and put this one in place of Sketching Diagrams in the first post. -
Mathematics: Post your doubts here!
This's how we'll do Q2, P4 Nov. 06. Part (i): We'll be needing to resolve the forces vertically in order to find the magnitude of the frictional force. Note that the vertical component of the 5N force and the frictional force act in the same direction (upwards) while as the weight... -
Mathematics: Post your doubts here!
To represent the inequality |z-1-i√3|≤ 1, draw a circle of radius 1 with center at '1 , √3'. To represent the inequality arg z ≤ pi/3, make an angle of 60° with the x-axis by placing your protractor at (0,0). This will help you... -
Mathematics: Post your doubts here!
That angle is usually mentioned in the question. Is there any specific mechanics question about this rod-ring thing? -
Mathematics: Post your doubts here!
I don't think that they've asked us to explain why is the angle twice the angle of arg z; In Q7(iii) N03 P3, they've asked us to find the greatest value of arg s for points on this locus. Are you referring to the same question? -
Mathematics: Post your doubts here!
Using the formula of binomial expansion. (1 + x)^n = 1 + [(x)( n )]/1 + [(x)^2]( n )(n-1)]/[(1)(2)] + [(x)^3]( n )(n-1)(n-2)]/[(1)(2)(3)] +[(x)^4]( n )(n-1)(n-2)(n-3)]/[(1)(2)(3)(4)] ........ -
Mathematics: Post your doubts here!
r = 2i - j + 4k + s ( i + j - k) As PQ is perpendicular to 'l', the dot product of PQ and 'l' will be equal to 0. PQ . l = 0 Now we've to find the direction vector of PQ. PQ = OQ - OP PQ = (2i +0j - k) - OP OP lies on 'l' therefore OP = 2i - j + 4k + s ( i + j - k). OP = 2i - j + 4k... -
Mathematics: Post your doubts here!
(1+ax)^6 1 + (ax)(6) + [(ax)^2(6)(5)]/2 + [(ax)^3(6)(5)(4)]/6 The coefficient of 'x' is -30. 6ax = -30x 6a = -30 a = -5 To find the coefficient of 'x^3', substitute the value of 'a = -5' in the equation ' [(ax)^3(6)(5)(4)]/6''. [(ax)^3(6)(5)(4)]/6 [(5x)^3(6)(5)(4)]/6 -2500x^3...