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Write 9sin(x) + 12 cos(x) in the form Rsin(x+y) and hence solve 9sin(x) + 12 cos(x) = 3

9sin(x) + 12 cos(x) = Rsin(x+y) =R(sin(x)cos(y)+cos(x)sin(y))= (Rcos(y))sin(x) + (Rsin(y))cos(x)Therefore by matching the coefficientsRsin(y)=12, Rcos(y)=9 [1]SoRsin(y)/Rcos(y) = 12/9 = 4/3, therefore tan...

JH
Answered by James H. • Maths tutor
5681 Views

g(x) = e^(x-1) + x - 6 Show that the equation g(x) = 0 can be written as x = ln(6 - x) + 1, where x<6

0 = ex-1+ x - 6 ex-1 = 6-x x-1 = ln (6-x) -> here we have taken the natural log of both sides, but it only shows on one side as the natural log of e is 1.x = ln (6-x) + 1Question ...

SN
Answered by Sumrah N. • Maths tutor
6898 Views

Find the coefficient of x^4 in the expansion of: x(2x^2 - 3x + 1)(3x^2 + x - 4)

Only the terms which will form the x4 term need to be considered:x * 2x2 * x + x * -3x * 3x2 = -7x4Therefore the answer is -7.

FH
Answered by Finn H. • Maths tutor
4593 Views

An object of mass 3kg is held at rest on a rough plane. The plane is inclined at 30º to the horizontal and has a coefficient of friction of 0.2. The object is released, what acceleration does the object move with?

We need to use Newtons law F=ma going down the slope.
We can see that the only forces acting in this direction are the component of the weight and friction, so we have that: F = Wsin30 - μR = 3a

AD
Answered by Asha D. • Maths tutor
3241 Views

Show that (sec(x))^2 /(sec(x)+1)(sec(x)-1) can be written as (cosec(x))^2.

( sec2(x))/((sec(x)+1)(sec(x)-1))Then, by the rule of 'difference of two squares', we know that this equals= (sec2(x))/(sec2(x)-1)= (sec2x/tan2x)s...

RS
Answered by Rishi S. • Maths tutor
10928 Views

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