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4 years ago

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Water flows at a rate of 6000 cubic inches per minute into a cylindrical tank. The tank has a diameter of 180 inches and a heigh

t of 72 inches. What is the height, in inches, of the water in the tank after 5 minutes? Round your answer to the nearest tenth

1 answer:

I am Lyosha [343]4 years ago

7 0

Answer:

s

Step-by-step explanation:

sffssffsffs

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I need help if you answer this your littearly a pro hacker pls

Anit [1.1K]

Answer:

24 meters cubed

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Step-by-step explanation:

V=Bh

V=6*4

V=24

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3 years ago

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Which answer is the best estimate of the correlation coefficient for the variables in the scatter plot?

Vesnalui [34]

Its Letter B -0.5

negative because its slope is negative (from left to right it's going downward)

so there are 2 negavite numbers in the choices, -0.5 and -0.95.

The nearer the value to 1 the less scatter it is. The farther the value to 1 (means nearer to 0) the more scatter it is. Because the plot in the picture is mostly scattered the value should be nearer to 0. It is negative 0.5.

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3 years ago

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Help plzzz!! Marking most Brainly;)

kolezko [41]

Answer:

D it

Step-by-step explanation:

We have two equations, A quadratic and a linear equation.

The domain of a linear equation and quadratic equation is all real numbers but since it dealing with time we must make sure the roots or x-intercepts is positive.

Looking at the quadratic equation, we can use the discramnt formula to see if all roots are positve.

$b {}^{2} - 4ac$

${ - 6}^{2} - 4(1)(8.75)$

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3 years ago

put the following decimals in order of size, starting with the one of least value 0.125 0.4 0.62 1.05 0.05

xeze [42]

Answer:

0.05, 0.125, 0.4, 0.62, 1.05

Step-by-step explanation:

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3 years ago

Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

<u>apply chain rule</u>

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$ <u />

<u>maximum volume:</u> <u>when h = 40/3</u>

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

<u>minimum volume:</u> <u>when h = 0</u>

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

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