What is 1.44 divided by 0.4

Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min, and its coarseness is such that it forms a pile in the shap

11Alexandr11 [23.1K]

Answer:

0.25 feet per minute

Step-by-step explanation:

Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min. Since we are told that the shape formed is a cone, the rate of change of the volume of the cone.

$\dfrac{dV}{dt}=20$ ft^3/min$

$\text{Volume of a cone}=\dfrac{1}{3}\pi r^2 h$

Since base diameter = Height of the Cone

Radius of the Cone = h/2

Therefore,

$\text{Volume of the cone}=\dfrac{\pi h}{3} (\dfrac{h}{2}) ^2 \\V=\dfrac{\pi h^3}{12}$

$\text{Rate of Change of the Volume}, \dfrac{dV}{dt}=\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}$

Therefore: $\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}=20$

We want to determine how fast is the height of the pile is increasing when the pile is 10 feet high.

$h=10$ feet$\\\\\dfrac{3\pi *10^2}{12}\dfrac{dh}{dt}=20\\25\pi \dfrac{dh}{dt}=20\\ \dfrac{dh}{dt}= \dfrac{20}{25\pi}\\ \dfrac{dh}{dt}=0.25$ feet per minute (to two decimal places.)$

When the pile is 10 feet high, the height of the pile is increasing at a rate of 0.25 feet per minute

5 0

3 years ago

A trapezoidal prism undergoes a dilation. The surface area of the pre-image is 35 m². The surface area of the image is 315 m². T

Westkost [7]

The original object is called pre-image, and its dilated version is called image. The height of the image for the considered case is 17.4 m

<h3>How does dilation affect length, area, and volume of an object?</h3>

Suppose a figure (pre image) is dilated (dilated image) by scale factor of k.

So, if a side of the figure is of length L units, and that of its similar figure is of M units, then:

L = k × M

where 'k' will be called as scale factor.

The linear things grow linearly like length, height etc.

The quantities which are squares or multiple of linear things twice grow by square of scale factor. Thus, we need to multiply or divide by k² to get each other corresponding quantity from their similar figures' quantities.

So area of first figure = k² × area of second figure

Similarly,

Volume of first figure = k³ × volume of second figure.

It is because we will need to multiply 3 linear quantities to get volume, which results in k getting multiplied 3 times, thus, cubed.

For this case, we're given that;

Surface area of pre-image = 35 m²
Surface area of dilated image = 315 m²
Height of pre-image = 5.8 m
Height of dilated image.

Let the scale fator of dilation be k.

Then:

So area of pre-image = k² × area of dilated image

Thus, we get:

$35 = k^2 \times 315\\\\k = \sqrt{\dfrac{35}{315}} = 1/3$

(took positive root as scaling is done by non-negative factors as only magnitude matters mostly)

Thus, we get:

Height of pre-image = (1/3) × Height of dilated image (say h)

$5.8 = (1/3) \times h\\h = 3 \times 5.8 = 17.4 \rm \: m$

Thus, the original object is called pre-image, and its dilated version is called image. The height of the image for the considered case is 17.4 m

Learn more about dilation here:

brainly.com/question/3266920

4 0

2 years ago

Read 2 more answers

What is the value of z in the equation 2(4z – 9 – 11) = 166 – 46?

Roman55 [17]

2(4z-9-11)=166-46
8z-18-22=120
8z-40=120
Add 40 from both side
8z-40+40=120+40
8z=160
Divided 8 from both side
8z/8=160/8
z=20. Hope it help!

6 0

2 years ago

Read 2 more answers

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