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2 years ago

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A cylinder has a radius of 5x + 4 and a height of 2x + 5. Which polynomial in standard form best describes the total volume of t

he cylinder? Use the formula v=πr²h for the volume of a cylinder

1 answer:

Nana76 [90]2 years ago

6 0

Answer:

Step-by-step explanation:

Volume = pi·(r^2)*h

r = radius = (5x+4)

h = height = (2x+5)

Volume = pi·(r^2)

Volume = pi·(5x+4)^2*(2x+8)

Volume = pi·(25x^2+40x+16)*(2x+8)

Volume = pi·(50x^2+80x+32+200x^2+32x+128)

Volume = pi·(250x^2+112x+160)

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NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)=−4.9t2+121t

Alenkasestr [34]

NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by

$h(t)=-4.9t^2+121t+283$

The sea level is represented by h = 0, therefore, to find the corresponding time when h splashes into the ocean we have to solve for t the following equation:

$h(t)=0\implies-4.9t^2+121t+283=0$

Using the quadratic formula, the solution for our problem is

$\begin{gathered} t=\frac{-(121)\pm\sqrt{(121)^2-4(-4.9)(283)}}{2(-4.9)} \\ =\frac{121\pm142.083778...}{9.8} \\ =\frac{121+142.083778...}{9.8} \\ =26.8452834694... \end{gathered}$

The rocket splashes after 26.845 seconds.

The maximum of this function happens at the root of the derivative. Differentiating our function, we have

$h^{\prime}(t)=-9.8t+121$

The root is

$-9.8t+121=0\implies t=\frac{121}{9.8}=12.347$

Then, the maximum height is

$h(12.347)=1029.99$

1029.99 meters above sea level.

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1 year ago

What is -2d-2 < 3d+8

OleMash [197]

Answer:

d > -2

Step-by-step explanation:

-2d-2 < 3d+8

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<em>Feel free to mark this as brainliest! :D</em>

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2 years ago

Name a number whose square root and cube root are whole numbers

katrin [286]

The whole number is 10 the square root is 9

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3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

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1 year ago

SUPER SIMPLE, Please help, I will mark brainliest

insens350 [35]

Easy it is clearly A

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2 years ago