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Ede4ka [16]
3 years ago
11

Complete the statement, f( ) = -5. Blank 1: Plz answer asap!

Mathematics
1 answer:
tankabanditka [31]3 years ago
4 0
It’s 12 that’s the simple answer
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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}

7 0
4 years ago
On a treasure map, the last clue is that the treasure is buried is buried a certain number of feet away from a tree. the distanc
Trava [24]
The lcm of 20 and 8 is 40 feet.
3 0
3 years ago
Can someone can help me
frozen [14]
Sure with what? u should have asked ur question
3 0
3 years ago
Read 2 more answers
20 POINTS!!! Use the quadratic formula above to solve for h(t) = -4.9t^2 + 8t + 1 where h is the height of the ball in meters an
Elina [12.6K]

Answer:

Two solutions: -0.12 and 1.75.

Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}. Assuming that the x² term is a, the x term is b, and the constant is c, we can plug the values into the equation.

\begin{array}{*{20}c}{\frac{{ - 8 \pm \sqrt {8^2 - 4\cdot-4.9\cdot1} }}{{2\cdot-4.9}}} \end{array}

\begin{array}{*{20}c}{\frac{{ - 8 \pm \sqrt {64 + 19.6} }}{{-9.8}}} \end{array}

\begin{array}{*{20}c}{\frac{{ - 8 \pm \sqrt {83.6} }}{{-9.8}}} \end{array}

\begin{array}{*{20}c}{\frac{{ - 8 \pm \sqrt {9.14} }}{{-9.8}}} \end{array}

\frac{-8 + 9.14}{-9.8} = -0.12

\frac{-8-9.14}{-9.8} =1.75

Hope this helped!

3 0
3 years ago
The following points lie on a certain line.
natali 33 [55]

y = 2x - 7

should be it

5 0
2 years ago
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