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tankabanditka [31]
3 years ago
7

Harry drives a delivery van and has several deliveries to make. He is in Horseshoe Bay and has a delivery to make in Highcliffe.

What is the approximate distance (to the nearest kilometre), by road, between Horseshoe Bay and Highcliffe?
Mathematics
2 answers:
GREYUIT [131]3 years ago
7 0

Answer: 56 km

Step-by-step explanation: We want to find the distance, by road, between Horseshoe Bay and Higcliffe (Both in the UK)

The average distance between both places, is 35 miles. where a mile is 1.6 kilometers. so 35 miles = 35*1.6km = 56km

So the distance between Horseshoe Bay and Highcliffe is 56km

iVinArrow [24]3 years ago
6 0

Harry drives a delivery van and has several deliveries to make. He is in Horseshoe Bay and has a delivery to make in Highcliffe. What is the approximatedistance (to the nearest kilometre), by road, between Horseshoe Bay and Highcliffe? km

Your answer is 2 km nnnm

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an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a
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the rate of change of the water depth when the water depth is 10 ft is;  \mathbf{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft  is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the  rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at  a time t and r = radius of the cone shaped at a time t

Then the similar triangles  ΔOCD and ΔOAB is as follows:

\dfrac{h}{r}= \dfrac{20}{8}    ( similar triangle property)

\dfrac{h}{r}= \dfrac{5}{2}

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h = 2.5r

r = \dfrac{h}{2.5}

The volume of the water in the tank is represented by the equation:

V = \dfrac{1}{3} \pi r^2 h

V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h

V = \dfrac{1}{18.75} \pi \ h^3

The rate of change of the water depth  is :

\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

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\dfrac{dv}{dt}= - 4  \ ft^3/sec

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-4 = \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

the rate of change of the water at depth h = 10 ft is:

-4 = \dfrac{ 100 \ \pi }{6.25}\  \dfrac{dh}{dt}

100 \pi \dfrac{dh}{dt}  = -4 \times 6.25

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\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi}

Thus, the rate of change of the water depth when the water depth is 10 ft is;  \mathtt{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

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