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3 years ago

7.7.33

1 answer:

7 0

We know that

At sea level, the height is 0 and
the pressure is 98 kilopascals

At 1000 ft, the height is 1000 and
the pressure <span>decreases about 11.41%

</span>i(100%-11.41%)/100----> (0.8859)

therefore
1) at 1000 ft--------> the pressure is 98*(0.8859)----------> 86.82 kilopascals
2) at 2000 ft--------> the pressure is 86.82*(0.8859)-------> 76.91 kilopascals
3) at 3000 ft--------> the pressure is 76.91*(0.8859)-------> 68.14 kilopascals
4) at 4000 ft---------> the pressure is 68.14*(0.8859)-------> 60.36 kilopascals

the answer is
<span>The pressure at an altitude of 4000 m is about </span>60.36 kilopascals

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students can buy up to 10 books at the book fair the cost of the books as shown in the graph what is the domain of the graph

german

The answer is 8 okay

5 0

3 years ago

Read 2 more answers

Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar

Brrunno [24]

The distance between the two points is $d=8.6m$

The polar coordinate of A is $\left(4.47,296.57\right)$

The polar coordinate of B is $\left(4.24,135\right)$

Explanation:

The two points are $A(2,-4)$ and $B(-3,3)$

The distance between two points is given by,

$d=\sqrt{(2+3)^{2}+(-4-3)^{2}}\\d=\sqrt{(5)^{2}+(-7)^{2}}\\d=\sqrt{25+49}\\d=8.6$

Thus, the distance between the two points is $d=8.6m$

The polar coordinates of A can be written as $(Distance, tan^{-1} \frac{y}{x} )$

Distance = $\sqrt{x^{2} +y^{2} }$

Substituting $A(2,-4)$ , we get,

Distance = $\sqrt{2^{2} +(-4)^{2} }=\sqrt{4+16 }=4.47$

$tan^{-1} \frac{y}{x} =tan^{-1} \frac{-4}{2}=-63.43$

To make the angle positive, let us add 360,

$\theta=360-63.43=296.57$

The polar coordinate of A is $\left(4.47,296.57\right)$

Similarly, The polar coordinate of B can be written as $(Distance, tan^{-1} \frac{y}{x} )$

Distance = $\sqrt{x^{2} +y^{2} }$

Substituting $B(-3,3)$ , we get,

Distance = $\sqrt{(3)^{2}+(3)^{2}}=4.24$

$tan^{-1} \frac{y}{x} =tan^{-1} \frac{3}{-3}=-45$

To make the angle positive, let us add 360,

$\theta=180-45=135^{\circ}$

The polar coordinate of B is $\left(4.24,135\right)$

5 0

3 years ago

What is 5.316 - 1.942 btw (show ur work) :)

erastovalidia [21]

Answer:

3.374

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4\\$

$\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}$

$\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}\\$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&11&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&\linethrough{1}&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{3}&7&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}$