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Tju [1.3M]
3 years ago
5

A city stores fresh water in a spherical tank. The tank has a radius of 32 feet. One cubic foot is approximately 7.5 gallons.If

the city drains 140,000 gallons of water out of the full tank, how many gallons are left in the tank? Round your answer to the nearest gallon.
Mathematics
1 answer:
Ann [662]3 years ago
3 0

Answer: There will be about 888,915 gallons of water left in the spherical tank

Now, multiply that by 7.5 and you will have 1,028,915 gallons of water in the tank.

Finally, just subtract the 140,000 gallons and we are at our answer.

Hope this helps!

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Use pencil and paper to create a table of values for the equation, x - 2y = 6. Rearrange the equation into the y = mx + b form.
Lunna [17]

Answer:

y = 1/2x -3

x=-2 ⇒ y =  -4

x=-1 ⇒ y =  -3.5

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Step-by-step explanation:

Hi, to answer this question we have to isolate y:

x - 2y = 6

-2y =6-x

y = (6-x)/-2

y = -3+1/2x

y = 1/2x -3  

Now, we have to create a table with the next values (see attachment)

x=-2 ⇒ y = 1/2x -3 = 1/2(-2)-3= -4

x=-1 ⇒ y = 1/2x -3 = 1/2(-1)-3= -3.5

x=0 ⇒ y = 1/2x -3 = 1/2(0)-3= -3

x=1 ⇒ y = 1/2x -3 = 1/2(1)-3= -2.5

x=2 ⇒ y = 1/2x -3 = 1/2(2)-3= -2

Feel free to ask for more if needed or if you did not understand something.

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4 years ago
Find the exact value of cos(sin^-1(-5/13))
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bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}

\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

8 0
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