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Mathematics: Post your doubts here!

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AoA!
In part (a), you write this function as 5 sin (θ + 53.13).
So, 3 sin θ + 4 cos θ + 7 can be rewritten as 5 sin (θ + 53.13) + 7. The least value of this sin (θ + 53.13) part of the function is -1. Consequently, the least value of the entire function is = 5(-1) + 7 = 2.

Thank u!!
 
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[(1 + sin x) / cos x] + (cos x / (1 + sin x)]
Take the LCM and write it as a single fraction.
{[(1 + sin x)^2 + cos^2 x] / cos x(1 + sin x)}
Expand (1+sin x)^2 so that it becomes 1 + 2 sin x + sin^2 x. Now in the numerator you have
1 + 2 sin x + sin^2 x +cos^2 x. Recall the identity sin^2 x +cos^2 x = 1. Therefore, replace sin^2 x +cos^2 x with 1 in the numerator so that it becomes 1 + 2 sin x + 1. The fraction that we now have is
(2 + 2 sin x) / cos x (1 + sin x). Take '2' common from the numerator so that it becomes 2 ( 1 + sin x) / cos x (1 +sin x). ( 1 + sin x) being common is cancelled out and you are left with 2 / cos x. Thus, LHS = RHS.
 
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You'll first find the co-ordinates of C which'll be (10,0). Now as BA=CD, you can find the coordinates of D by vector move. To go from B to A, you need to move 4 places to the left on the x-axis (-1, 0, 1, 2) and 6 places upwards on the y-axis (9, 10, 11, 12, 13, 14). Now you'll simply add '4' to the x-coordinate of C and '6' to the y-coordinate of C. This'll come out to be (14,6) and thus these are the co-ordinates of D.
 
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May/June 2006 Question One :/ I don't understand the question.
1. A curve has equation y=k/x. Given that the gradient of the curve is -3 when x=2, find the value of the constant k. :/
 
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May/June 2006 Question One :/ I don't understand the question.
1. A curve has equation y=k/x. Given that the gradient of the curve is -3 when x=2, find the value of the constant k. :/

Differentiate 'y=k/x' using the quotient rule. The derivative will come out to be 'dy/dx=(-k) / x^2'. Then substitute the value of x=2 and the gradient (dy/dx)=-3 in this equation and calculate the value of 'k'. The value of 'k' is 12. If you are finding it difficult to differentiate, then it'll be done as follows:

y = k / x
dy / dx = [(0)(x) - (k)(1)] / x^2
dy / dx = (-k) / x^2

Substitute '-3' in the place of dy / dx and '2' in the place of x:

-3 = (-k) / (2)^2
-12 = -k
12=k
 
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a related rates of change question
Two quantities p and q are related by the equation (p-1)(q+2)=k where k is a constant. When p=5 units, q is 7 units and q is changing at the rate of 0.04 units per second. Find the rate at which p is changing.
The answer at the back of the book is 0.018 units per second
The one I'm getting is -0.018 seconds
I don't get it, where is the mistake?
 
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^No! The only thing given in the question was the equation of the curve i.e. y = 1 / (2x + 2) ^2 !
If you have that book, Pure Mathematics 1, you can see the question for yourself! Exercise 16 d, Question 5, part c !
 
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