Can someone with these questions (1 , 3 , 4)

Remark

4 is long (but not hard) and I am in a terrible time bind. I will take a short at doing it but it will be abbreviated.

One

Ax = 70*cos(51.7) = 43.38

Ay = 70*sin(51.7) = 54.93

That makes C the correct answer. To resolve these, Ax is always taken along the +x axis. and Ay is always associated with the y axis. A(x) always uses Cos(x) and Ay = sin(x)

Three

A would be true if Ax and Ay were vectors. They don't seem to be. They look like scalers. Magnitudes don't add that way.

B Theta = tan-1(Ay/Ax) The Statement given is simply not true.

C Tan(Theta) = Ay / Ax Not The given result

D is correct Cos(theta) = The magnitude along the +x axis divided by the hypotenuse which is sqrt(Ax^2 + Ay^2)

Four

The best way to do this is to set up x and y values in a table. I can't do this well, but I'll try.

x                                                            y

12*cos(53) = 7.22                             15*sin(53)= 9.58

15*cos(-40) = 11.49                            15.sin(-40) = -9.64

Sum =            18.69                                             = - 0.08

As you can see most of the magnitude and direction will come from Ax

So the answer is 19 for the magnitude and 0o for the direction. The other choices are taken from theta = 40o. That minus sign makes all the difference.          

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .