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Fofino [41]
3 years ago
7

Help!!!!!!!!!!!!!!!

Mathematics
1 answer:
notsponge [240]3 years ago
7 0

Answer:

I THANK its 1290

Step-by-step explanation:

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The metric designators and trade sizes are for identification purposes only and are not actual dimensions, as pointed out in the
Basile [38]

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:

brainly.com/question/11827791

#SPJ4

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?

5 0
2 years ago
Solve the proportion. a. 4/x = 9 b. 7.5/26 = 75/p
lina2011 [118]

<u>ANSWER: </u>

The values of x and p after solving given equations are 0.445 or \frac{4}{9} and 260 respectively..

<u>SOLUTION: </u>

Given, two proportions are \frac{4}{x} =9 and \frac{7.5}{26} = \frac{75}{p}

We need to solve the given two proportions.

Let us solve the first proportion, \frac{4}{x} = 9

\frac{4}{x} = \frac{9}{1}

On cross-multiplication, we get

4 \times 1 = 9x

4 = 9x

x = \frac{4}{9}

x = 0.445

Now, let us solve the second proportion  \frac{7.5}{26} = \frac{75}{p}

\frac{\frac{75}{10}}{26}=\frac{75}{p}

\frac{75}{10 \times 26} = \frac{75}{p}

\frac{1}{260} = \frac{1}{p}

On cross-multiplication, we get

p = 260

Hence, the values of x and p after solving given equations are 0.445 and 260 respectively.

7 0
3 years ago
A baseball is thrown into the air with an upward velocity of 30 ft/s. its initial height was 6 ft, and its maximum height is 20.
marissa [1.9K]

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

\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+30}t\stackrel{\stackrel{c}{\downarrow }}{+6} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)

\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

7 0
4 years ago
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What Is The Range Of The Function F(x) = 3x - 12 For The Domain {-2,2}
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This should be your answer

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4 years ago
2 less than one-third of the points that the Panthers scored
ValentinkaMS [17]
2 less than1/3 idfk.......
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