Answer:
In each shelves there are 64/4 books I.e. 16.
Given:


To find:
The quadrant of the terminal side of
and find the value of
.
Solution:
We know that,
In Quadrant I, all trigonometric ratios are positive.
In Quadrant II: Only sin and cosec are positive.
In Quadrant III: Only tan and cot are positive.
In Quadrant IV: Only cos and sec are positive.
It is given that,


Here cos is positive and sine is negative. So,
must be lies in Quadrant IV.
We know that,



It is only negative because
lies in Quadrant IV. So,

After substituting
, we get





Therefore, the correct option is B.
The area for the perimeter of a rectangle is:
<em>P = 2l + 2w</em>
In this problem, we are given that
P = 155 and
l = 45.5. To solve for area, we must find w. To do this, plug in the given info and isolate the w like so:

Now we know
w = 32. Knowing all the necessary info, we can use the area formula, <em>A=lw</em>, to solve for A.

The area is
1,456 sq ft. Hope this helps!
Answer:
<h2>see the answers ⤵️</h2>
Step-by-step explanation:
<h3>to understand the solving steps you need to know about</h3>
<h3>adding or subtracting with redical experience is nothing but algebraic addition or subtraction</h3><h3>let's solve:</h3>


Answer:
tetrahedral
Step-by-step explanation:
According to the valence shell electron pair repulsion theory (VSEPR) the shape of a molecule is dependent on the number of electron pairs on the valence shell of the central atom in the molecule.
The predicted electron pair geometry may sometimes differ from the molecular geometry due to the presence of lone pairs and multiple bonds.
If we consider each nitrogen atom in N2 independently, we will notice that each nitrogen atom has four regions of electron density. Hence the electron pair geometry is tetrahedral.