Top answers

The point A lies on the curve y=5(x^2)+9x , The tangent to the curve at A is parralel to the line 2y-x=3. Find an equation to this tangent at A.

y-(any value of the function)=0.5(x-(any value in the domain)) e.g. Point with point A being (1,14) The answer would be y-14=0.5(x-1)
RA
Answered by Rehman A. • Maths tutor
4275 Views

Express cos(x) + (1/2)sin(x) in terms of a single resultant sinusoidal wave of the form Rsin(x+a)

cos(x) + (1/2)sin(x) : Rsin(x + a) = R{sin(x)cos(a) + cos(x)sin(a)} = (1/2)sin(x) + (1)cos(x) (comparing coeffs.) Therefore Rcos(a) = 1/2 and separately Rsin(a) = 1 So tan(a) = 2 and R^2 = 5/4. Answer: sqrt(...
HT
Answered by Hakkihan T. • Maths tutor
8203 Views

Find the general solution, in degrees, of the equation 2 sin(3x+45°)= 1

A general way of solving these equations is getting them to the form sin(y)=k . In this case, to do so, we have to divide by 2 and then put y=3x+45° We then get sin(y)=1/2 . You should know which angles have...
CG
9327 Views

The quadratic equation 2x^2+ 6x+7 has roots a and b. Write down the value of a+b and the value of ab.

A general method would be to compute a and b with the general solution formula for quadratic equations. However, a special property of quadratic equations can make this a lot easier. In fact, given the equat...
CG
6291 Views

find dy/dx of the equation y=ln(x)2x^2

Here it is necessary to use the chain rule to solve the derivative. If we equate our equation in terms of the following notation: ln(x)='u'and 2x^2='v' and use the chain rule formula dy/dx=udv/dx+vdu/dx we c...
PG
Answered by Pierce G. • Maths tutor
4590 Views