Select the solutions to the following inequality:

Mathematics

1 answer:

Elena L [17]3 years ago

8 0

X+2> _ 5, x >_5-2, x>_3

-x-2>_5, x<_ -2-5, x<_-7

the last is the answer

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the volume of a gas varies inversely as the pressure upon it. if the volume of the gas is 160 cubic centimeters at a pressure of

marin [14]

It was stated in the problem that a given volume of a is inversely proportional with the pressure of the system. It means that the as the volume increases, the pressure would decrease and as the volume decreases, the pressure would increase. We would express it as:

V α 1/P

To change it to an equality, we introduce a proportionality constant, k. We do as follows:

V = k/P

So, to determine what is asked, we need to first calculate for the value of k.

V = k/P
At V = 160 m^3 P = 108 cmHg
160 = k / (108)
k = 17280

Thus, at P = 87 cmHg
V = 17280 / 87
V = 198.62 m^3

4 0

3 years ago

Could someone help? :)

cluponka [151]

The value missed is 1 min / 60 seconds

<h3>What is unit conversion?</h3>

Units is the conversion between different units of measurement for the same quantity, typically through multiplicative conversion factors.

To convert 720 seconds to hour,

First convert second to minutes divide by 60 and then again convert minutes to hour again divide by 60.

So,

720 seconds * ( 1 minutes / 60 seconds) * ( 1 hour / 60 minutes)

= 12 hours.

Learn more about this concept here:

brainly.com/question/4736731

#SPJ1

8 0

2 years ago

The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of

Lilit [14]

Check the picture below.

$\bf (\stackrel{a}{1}~,~\stackrel{b}{-1})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c = \sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{1^2+(-1)^2}\implies c=\sqrt{2} \\\\[-0.35em] ~\dotfill$

$\bf sin(\theta ) \implies \cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{-1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies -\cfrac{\sqrt{2}}{2}$

$\bf cos(\theta ) \implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{\sqrt{2}}{2} \\\\\\ tan(\theta ) = \cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{1}}\implies tan(\theta ) = -1$