Melody

ok Sorry.

This time I will close it again.

You are asking multiple questions and it seems you are trying to disguise this fact by posting as 2 different usernames AND a guest.

One person one name. 

Go back and ask again under your earlier name.

I don't have a clue. Sorry.

How about doing some of your own homework?

There are 2 questions there.  Ask one on a new post and work the other out from any answer that you get.

I am closing this post.

Solve x^2=5   are either of the answers in the rage x<=-2?

Solve |2x+1|=5 is the answer/s in the required range?

Solve 3x+8=5  is the answer/s in the required range?

1) Find or state all the initial conditions.

What is the height?

the vertical speed and the horizontal speed?  

and the acceleration in vertical and horizontal directions?

To find the initial vertical and horizontal speeds.

Draw a right-angled triangle.  The right angle and 30 degrees will be at the base

The hypotenuse will be 70feet/sec

Use trig to find the initial vertical and horizontal speeds.

When you have done that get back with the worked answers (or explain your confusion)

I'd help if I could but I do not have a clue.....

I was going to say that there are 4 choices for one end then 3 for the other end then 4! for the rest (assuming all the books are different)

So that is 4*3*4! = 12*24 = 288

Just as EP found  

This is presented as a test...

Here is my answer:

I don't understand electricity.  This is based solely on the info given.

\(I=\frac{V}{R}\\ R\propto L\\ I=\frac{V}{kL}\\ find\;\;k\\ When \;\;V=50\;\;I=200,\;\;L=2\\ 200=\frac{50}{2k}\\ k=0.125\\so\\ I=\frac{V}{0.125L}\\~\\ Find\; I\;\;when\;V=300\;and\;L=1\\ I=\frac{300}{0.125*1}\\ I=2400\;milliamps \)

LaTex:

I=\frac{V}{R}\\
R\propto L\\
I=\frac{V}{kL}\\
find\;\;k\\
When \;\;V=50\;\;I=200,\;\;L=2\\
200=\frac{50}{2k}\\
k=0.125\\so\\
I=\frac{V}{0.125L}\\~\\
Find\; I\;\;when\;V=300\;and\;L=1\\
I=\frac{300}{0.125*1}\\
I=2400\;milliamps