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Pavel [41]
3 years ago
13

Mr. Phan fills a portion of his pool with water on Monday. Then on Tuesday, he fills the pool with water coming from a hose at a

constant rate. The table shows the number of feet of water in the pool as a function of time on Tuesday. What is the rate of change of the function? Explain your reasoning.
Mathematics
1 answer:
Zanzabum3 years ago
7 0

Answer:

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do it or he will hunt you down and kill u (lets destroy the moderators!!!!!!!!)

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JOIN THE BOB GANG!!!!!!!!!

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
Help with this question quick???
SIZIF [17.4K]
Mars is the correct answer
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The roots of the equation 2x^2+8x-7=-2 are?
olchik [2.2K]
<span>2x^2+8x-7=-2
</span><span>2x^2+8x - 5 = 0

using the formula:- x = [-8 +/- sqrt(8^2 - 4*2*-5) ]  / 4
                                  =  -2 +/- sqrt 104)/2
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3 years ago
Diagram 5 shows a right cylinder with a diameter of 2xcm. Given that the total surface area of the cylinder is 96cm³.Find the ma
Paraphin [41]

Given:

The diameter of the right cylinder is 2x cm.

The total surface area is 96 cm cube.

The radius is calculated as,

\begin{gathered} r=\frac{d}{2} \\ r=\frac{2x}{2} \\ r=x\text{ cm} \end{gathered}

The total surface area is,

\begin{gathered} S=2\pi rh+2\pi(r)^2 \\ 96=2\pi xh+2\pi(x^2) \\ h=\frac{96-2\pi(x^2)}{2\pi x} \end{gathered}

Volume is,

\begin{gathered} V=\pi(r)^2h \\ =\pi(x^2)\frac{96-2\pi(x^2)}{2\pi x} \\ =\frac{x(96-2\pi(x^2)}{2} \end{gathered}

Now, differentiate with respect to x,

\begin{gathered} \frac{dV}{dx}^{}=\frac{d}{dx}(\frac{x(96-2\pi(x^2)}{2}) \\ =\frac{d}{dx}\mleft(x\mleft(-\pi x^2+48\mright)\mright) \\ =\frac{d}{dx}\mleft(x\mright)\mleft(-\pi x^2+48\mright)+\frac{d}{dx}\mleft(-\pi x^2+48\mright)x \\ =1\cdot\mleft(-\pi x^2+48\mright)+\mleft(-2\pi x\mright)x \\ =84-3\pi(x^2)\ldots\ldots\ldots\ldots\text{.}(1) \end{gathered}

Now,

\begin{gathered} \frac{dV}{dx}=0 \\ 84-3\pi(x^2)=0 \\ x^2=\frac{16}{\pi} \\ x=\sqrt[]{\frac{16}{\pi}} \end{gathered}

Now, differentiate (1) with respect to x again,

\begin{gathered} \frac{d^2V}{dx^2}=\frac{d}{dx}(84-3\pi(x^2)) \\ =-6\pi x \\ At\text{ x=}\sqrt[]{\frac{16}{\pi}} \\ \frac{d^2V}{dx^2}=-6\pi\sqrt[]{\frac{16}{\pi}}

Since, the double derivative is negative.

So,\text{ the volume is maximum at }\sqrt[]{\frac{16}{\pi}}

So, the volume becomes,

\begin{gathered} V=\pi(x^2)h \\ V=\pi(\sqrt[]{\frac{16}{\pi}})^2h \\ V=\frac{16h}{\pi} \end{gathered}

Answer: maximum volume of the cylinder is,

6 0
1 year ago
What are the next two numbers? 0.03 0.12 0.21 ... ...
iris [78.8K]
If it is 0.03,0.12,0.12 just add 0.09.
The next two numbers are 0.30,0.39
7 0
3 years ago
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