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Answers by Richard P.

For a fixed value of d,  this set is over determined.   There are three equations but only two unknowns.   One approach is to solve the first two equations for x and y.   This can be done by standard methods.   The result is:       ...

-2tantheta=2√3 (answer)

This is easy with a graphing calculator such as the TI-84 (in degree mode).       theta =  arctan(-√3) = -60 deg.      Other approaches involve using a unit circle chart  or     Remembering that sin(60) =  √3/2...

The solution set is just x = -1/2.   The value x = -1/2 can be easily checked to be a solution.   A way to see that it is the only solution is to use a graphing calculator such as the TI-84.  Put abs(x) into Y1 and abs(x+1) into Y2.  

f(x)=x^2-x-6 (answer)

The function f is a parabola.    The standard quadratic form is  a x2 + b x +c.    So for this case a =1,  b= -1, c = -6    The x coordinate of the vertex is -b/2a  =   1/2.  The y coordinate of the vertex can...

Quadratics (answer)

Since this is a two dimensional system (x,y),   gradient just means slope.   Call the slope of a tangent line m.  The equation of the tangent line is then y = m x +1   Call a candidate point of tangency  (x0,y0).   Then we require     ...

Introduce a Legrange multiplier, λ.    The function to be studied is then:   h(x,y,z) =  xyz + λ (z - exp(-x2-y2))    Denote the partial  derivatives with subscripts ( for example  hx is the partial derivative of h wrt x) Then the conditions...