Melody

Thanks Alan,

I always forget how this is done   

I mucked around quite a lot with the second one before I got it.  (Maybe there are some shortcut techniques)

anyway:

It should help you a lot if you work out the value of        \((1-\sqrt3i)^3\)

I will look at the send one a little later.   

That is not quite right.  it is (-4i)^50 = (-4)^50 * i^50 = 4^50 * i^50

You need to work out what i^50

once again you do it with a pattern

hint:

i^1=

i^2=

i^3=

i^4=

i^5=

what is the pattern?

i^ 50 = ?

\((-\sqrt2+i\sqrt2)^{100}=((-\sqrt2+i\sqrt2)^{2})^{50}\\~\\ (-\sqrt2+i\sqrt2)^{2}=2-2-4i = -4i \)

i have just looked at the first one.

Hint:

\((-\sqrt2+i\sqrt2)^{100}=((-\sqrt2+i\sqrt2)^{2})^{50}\)

Nice work Solviet.

Divide everything by  -24

then complete the squares to get    (x-h)^2 + (y-k)^2 >= r^2

Hi Solveit  

I let   z=x+yi

and this is the graph I got.

I don;t intend to do all the working but I will say that

\(|z+4+3i|\\ =|(x+4)+(y+3)i|\\ =(x+4)^2+(y+3)^2\\ etc\)

see if you can do it from there,     

Thank you for finally putting some effort into your presentation.

A mistake that you have made is thinking that    V0 is 70

The initial velocity has both a vertical velocity element and a horizontal velocity element.

The vertical velocity element is   V0 is  70sin(theta)  where theta is the initial angle of the trajectory (the angle of elevation)

So 

\(h(t)= 1/2 gt^2+v_0t+h_0\\ h(t)=-16t^2+(70sin\theta)t+2\)

You are right, this is a concave down parabola and you can find the maximum height by finding h at the centre point.

Centre = -b/2a   from the quadratic formula

\(t=\frac{-70sin\theta}{-32}=\frac{35sin\theta}{16}\\~\\ h(\frac{35sin\theta}{16})=-16(\frac{35^2sin^2\theta}{16*16})+70sin\theta*\frac{35sin\theta}{16}+2\\ h(\frac{35sin\theta}{16})=-(\frac{35^2sin^2\theta}{16})+\frac{70*35sin^2\theta}{16}+2\\ h(\frac{35sin\theta}{16})=(\frac{35^2sin^2\theta}{16})+2\\ h(\frac{35sin\theta}{16})=(\frac{35sin\theta}{4})^2+2\\ \)

This is the maximum height that will be reached.  You cannot give it to the nearest foot because you have not been given the initial angle. 

And you have not been given the initial vertical velocity.  If the question is worded incorrectly and 70 was supposed to be the initial vertical velocity then the logic behind what you have done looks fine.

Please note:  All answers provided need to be checked for logic as well as careless errors.

LaTex:

t=\frac{-70sin\theta}{-32}=\frac{35sin\theta}{16}\\~\\
h(\frac{35sin\theta}{16})=-16(\frac{35^2sin^2\theta}{16*16})+70sin\theta*\frac{35sin\theta}{16}+2\\
h(\frac{35sin\theta}{16})=-(\frac{35^2sin^2\theta}{16})+\frac{70*35sin^2\theta}{16}+2\\
h(\frac{35sin\theta}{16})=(\frac{35^2sin^2\theta}{16})+2\\
h(\frac{35sin\theta}{16})=(\frac{35sin\theta}{4})^2+2\\

You are concerned about YOUR time being wasted?

You are asking for free help and you don't want to waste your time presenting yourself in your best light?

That is like going to a job interview without a resume and DEMANDING that you be given the job!

When you are asked for your resume, you can say:

"Oh, I didn't bother writing it, you might not have given me the job and I didn't want to waste my time."