In two days, 48,000 gallons of oil gushed out of a well. At what rate did the oil flow from the well?

2 answers:

7 0

48 000 gallons / 2 days = 24 000 gallons per day
24 000 gallons / 24 hours = 1 000 gallons per hour
<span>1 000 gallons / 60 minutes = 16.(6) gallons per minute</span>

NNADVOKAT [17]3 years ago

4 0

1,000 gallons per hour

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1 point) Use Stoke's Theorem to evaluate ∫CF⋅dr∫CF⋅dr where F(x,y,z)=xi+yj+1(x2+y2)kF(x,y,z)=xi+yj+1(x2+y2)k and CC is the bound

Drupady [299]

Answer:

Step-by-step explanation:

Thinking process:

$\int\limits^a_b {cF} \, .dr$

= $\int\limits^a_b {x} \, dx \int\limits^a_b {x} \, dx curlFdS$ by Stoke's Theorem

= $\int\limits^a_b {} \, \int\limits^a_b {} \, < 12y, -12x, 0 > .< z_x,-z_y, 1 > dA\\= \int\limits^a_b {} \, \int\limits^a_b {} \, < 12y, -12x, 0 > .< 2x, 2y, 1 > dA$ since z = $25-x^{2} -y^{2} \\= 0$

3 0

3 years ago

Which equation has the same solution as 2x^2+16x-10=0

dmitriy555 [2]

D
Divide the whole thing by 2
2(X^2+8x-5)
X^2+8x. =5
Half of 8 is 4 sq it so you add 16
X^2+8x+16= 5+16
(X+4)^2=21

8 0

2 years ago

Read 2 more answers

How do you solve his with working

AlexFokin [52]

Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

$\bf \begin{array}{cllll}
\textit{circumference of a circle}\\\\ 
2\pi r
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{arc's length}\\\\
s=\cfrac{\theta r\pi }{180}
\end{array}\\\\
-------------------------------$

$\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+
\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}
\\\\\\
15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888$

b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.

$\bf \begin{array}{cllll}
\textit{area of a circle}\\\\ 
\pi r^2
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{area of a sector of a circle}\\\\
s=\cfrac{\theta r^2\pi }{360}
\end{array}\\\\
-------------------------------$

$\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}
\\\\\\
90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884$