The metric designators and trade sizes are for identification purposes only and are not actual dimensions, as pointed out in the note in table 300.1(C).
<h3>
What are identification purposes?</h3>
- For identification purposes, an acceptable ID MUST be issued by a Federal, State, County, or Municipal entity.
- For identification purposes, the single document must include the following criteria: photo, name, address (home/employer), date of birth, and be issued by a Federal, State, County, or Municipal entity.
[Refer The table 300.1(C) below.]
Therefore, the metric designators and trade sizes are for identification purposes only and are not actual dimensions, as pointed out in the note in table 300.1(C).
Know more about identification purposes here:
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The correct question is given below:
The metric designators and trade sizes are for identification purposes only and are not actual dimensions as pointed out in the note to Table?
<u>ANSWER: </u>
The values of x and p after solving given equations are 0.445 or
and 260 respectively..
<u>SOLUTION:
</u>
Given, two proportions are
=9 and 
We need to solve the given two proportions.
Let us solve the first proportion,
= 9

On cross-multiplication, we get
4
1 = 9x
4 = 9x
x = 
x = 0.445
Now, let us solve the second proportion 



On cross-multiplication, we get
p = 260
Hence, the values of x and p after solving given equations are 0.445 and 260 respectively.
Check the picture below.
where is the -16t² coming from? that's Earth's gravity pull in feet.
![\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{30}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{6}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+30t+6 \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20~~~~~~%5Ctextit%7Bin%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cend%7Barray%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Cstackrel%7B30%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B6%7D%7B%5Ctextit%7Binitial%20height%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20h%28t%29%3D-16t%5E2%2B30t%2B6%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

![\bf \left(-\cfrac{30}{2(-16)}~~,~~6-\cfrac{30^2}{4(-16)} \right)\implies \left( \cfrac{30}{32}~,~6+\cfrac{225}{16} \right)\implies \left(\cfrac{15}{16}~,~\cfrac{321}{16} \right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{\stackrel{\textit{how many}}{\textit{seconds it took}}}{0.9375}~~,~~\stackrel{\stackrel{\textit{how many feet}}{\textit{up it went}}}{20.0625})~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28-%5Ccfrac%7B30%7D%7B2%28-16%29%7D~~%2C~~6-%5Ccfrac%7B30%5E2%7D%7B4%28-16%29%7D%20%5Cright%29%5Cimplies%20%5Cleft%28%20%5Ccfrac%7B30%7D%7B32%7D~%2C~6%2B%5Ccfrac%7B225%7D%7B16%7D%20%5Cright%29%5Cimplies%20%5Cleft%28%5Ccfrac%7B15%7D%7B16%7D~%2C~%5Ccfrac%7B321%7D%7B16%7D%20%5Cright%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%28%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bhow%20many%7D%7D%7B%5Ctextit%7Bseconds%20it%20took%7D%7D%7D%7B0.9375%7D~~%2C~~%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bhow%20many%20feet%7D%7D%7B%5Ctextit%7Bup%20it%20went%7D%7D%7D%7B20.0625%7D%29~%5Chfill)
This should be your answer
2 less than1/3 idfk.......