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Nitella [24]
3 years ago
7

Need help on this question!!!!!

Mathematics
1 answer:
den301095 [7]3 years ago
7 0
I think it would be 114 because if you subtract the wall from the playground sides (it says the walls are 4) you would get 21 on one side and 36 on the other. once you add those you would get 114. hope this helps!
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Gas is escaping from a spherical balloon at the rate of 12 ft3/hr. At what rate (in feet per hour) is the radius of the balloon
bija089 [108]

Answer:

This is the rate at which the radius of the balloon is changing when the volume is 300 ft^3 \frac{dr}{dt}=-\frac{3}{225^{\frac{2}{3}}\pi ^{\frac{1}{3}}} \:\frac{ft}{h}  \approx -0.05537 \:\frac{ft}{h}

Step-by-step explanation:

Let r be the radius and V the volume.

We know that the gas is escaping from a spherical balloon at the rate of \frac{dV}{dt}=-12\:\frac{ft^3}{h} because the volume is decreasing, and we want to find \frac{dr}{dt}

The two variables are related by the equation

V=\frac{4}{3}\pi r^3

taking the derivative of the equation, we get

\frac{d}{dt}V=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4}{3}\pi (3r^2)\frac{dr}{dt} \\\\\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}

With the help of the formula for the volume of a sphere and the information given, we find r  

V=\frac{4}{3}\pi r^3\\\\300=\frac{4}{3}\pi r^3\\\\r^3=\frac{225}{\pi }\\\\r=\sqrt[3]{\frac{225}{\pi }}

Substitute the values we know and solve for \frac{dr}{dt}

\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=-\frac{12}{4\pi (\sqrt[3]{\frac{225}{\pi }})^2} \\\\\frac{dr}{dt}=-\frac{3}{\pi \left(\sqrt[3]{\frac{225}{\pi }}\right)^2}\\\\\frac{dr}{dt}=-\frac{3}{\pi \frac{225^{\frac{2}{3}}}{\pi ^{\frac{2}{3}}}}\\\\\frac{dr}{dt}=-\frac{3}{225^{\frac{2}{3}}\pi ^{\frac{1}{3}}} \approx -0.05537 \:\frac{ft}{h}

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4 years ago
Use a graph to find x and y values that make both y=-x+3 and y=2x−5 true.
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Answer:

(8/3,1/3)

Step-by-step explanation:

Two attachments with two methods.

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Point 1, f(x) -4 (x-3) 2+1
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Arlecino [84]

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x = 2, y = 0

Step-by-step explanation:

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If f⃗ =grad(2x2+3y3), find ∫c where c is the quarter of the circle x2+y2=1 in the first quadrant, oriented counterclockwise.
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Presumably, the path starts on the horizontal axis at (1, 0) and ends on the vertical axis at (0, 1). Since we know the vector field is the gradient of a scalar function, we can apply the fundamental theorem of calculus (aka gradient theorem):

\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(0,1)-f(1,0)=3-2=1
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