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lakkis [162]
3 years ago
8

A farmer bought a number of pigs for $200. However, 6 of them died before he could sell the rest at a profit of 5 per pig. His t

otal profit was $47. How many pigs did he originally buy?
Mathematics
1 answer:
Vesna [10]3 years ago
4 0

Answer:

40 pigs

Step-by-step explanation:

6*5=30

200-30=170

170/5=34

34+6=40 pigs

to check your answer

40*5=200

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Answer:

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Step-by-step explanation:

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3 years ago
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Drag each tile to the correct box.
ch4aika [34]

Answer:

1. 50 cm

2. 550 mm

3. 5000 mm

4. 5000 cm

5. 5000 m

Step-by-step explanation:

We can convert all the measures in same unit to arrange them one by one.

5000 cm:

As there are 10 millimeters in 1 cm,

5000 cm = 5000*10mm

=> 50000 mm

5000 meters:

As there are 100 cms in one meter

5000 m = 5000*100 cm

=> 500000 cm

Now to convert in millimeter

500000*10= 5000000 mm

50 cm:

50*10 = 500 mm

Then there is 550 mm and 5000 mm

So, the order from least to greatest is:

1. 500 mm which is equivalent of 50 cm

2. 550 mm

3. 5000 mm

4. 50000 mm which was equivalent of 5000 cm

5. 5000000 mm which was equivalent of 5000 m ..

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3 years ago
What is 345 written as a percent?<br> F. 0.0093%. <br> G. 0.93%<br> H. 9.3%<br> I. 93%
zysi [14]

Answer:

34500%

Step-by-step explanation:

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3 years ago
A giant tank in a shape of an inverted cone is filled with oil. the height of the tank is 1.5 metre and its radius is 1 metre. t
skad [1K]

The given height of the cylinder of 1.5 m, and radius of 1 m, and the rate

of dripping of 110 cm³/s gives the following values.

1) The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ <u>9.34 × 10⁻⁵ m/s</u>

2) The rate of change of the oil's height when the height is 20 cm is h' ≈ <u>1.97 × 10⁻³ m/s</u>

3) The rate the oil radius is changing when the radius is 10 cm is approximately <u>0.175 m/s</u>

<h3>How can the rate of change of the radius & height be found?</h3>

The given parameters are;

Height of the tank, h = 1.5 m

Radius of the tank, r = 1 m

Rate at which the oil is dripping from the tank = 110 cm³/s = 0.00011 m³/s

1) \hspace{0.15 cm}V = \frac{1}{3} \cdot \pi \cdot r^2 \cdot h

From the shape of the tank, we have;

\dfrac{h}{r} = \dfrac{1.5}{1}

Which gives;

h = 1.5·r

V = \mathbf{\frac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)}

\dfrac{d}{dr} V =\dfrac{d}{dr}  \left( \dfrac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)\right) = \dfrac{3}{2} \cdot \pi  \cdot r^2

\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}

\dfrac{dr}{dt} = \mathbf{\dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dr} }}

\dfrac{dV}{dt} = 0.00011

Which gives;

\dfrac{dr}{dt} = \mathbf{ \dfrac{0.00011 }{\dfrac{3}{2} \cdot \pi  \cdot r^2}}

When r = 0.5 m, we have;

\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi  \times 0.5^2} \approx  9.34 \times 10^{-5}

The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ <u>9.34 × 10⁻⁵ m/s</u>

2) When the height is 20 cm, we have;

h = 1.5·r

r = \dfrac{h}{1.5}

V = \mathbf{\frac{1}{3} \cdot \pi \cdot \left(\dfrac{h}{1.5} \right) ^2 \cdot h}

r = 20 cm ÷ 1.5 = 13.\overline3 cm = 0.1\overline3 m

Which gives;

\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi  \times 0.1 \overline{3}^2} \approx  \mathbf{1.313 \times 10^{-3}}

\dfrac{d}{dh} V = \dfrac{d}{dh}  \left(\dfrac{4}{27} \cdot \pi  \cdot h^3 \right) = \dfrac{4 \cdot \pi  \cdot h^2}{9}

\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}

\dfrac{dh}{dt} = \dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dh} }<em />

\dfrac{dh}{dt} = \mathbf{\dfrac{0.00011}{\dfrac{4 \cdot \pi  \cdot h^2}{9}}}

When the height is 20 cm = 0.2 m, we have;

\dfrac{dh}{dt} = \dfrac{0.00011}{\dfrac{4 \times \pi  \times 0.2^2}{9}} \approx \mathbf{1.97 \times 10^{-3}}

The rate of change of the oil's height when the height is 20 cm is h' ≈ <u>1.97 × 10⁻³ m/s</u>

3) The volume of the slick, V = π·r²·h

Where;

h = The height of the slick = 0.1 cm = 0.001 m

Therefore;

V = 0.001·π·r²

\dfrac{dV}{dr} = \mathbf{ 0.002 \cdot \pi \cdot r}

\dfrac{dr}{dt} = \mathbf{\dfrac{0.00011 }{0.002 \cdot \pi  \cdot r}}

When the radius is 10 cm = 0.1 m, we have;

\dfrac{dr}{dt} = \dfrac{0.00011 }{0.002 \times \pi  \times 0.1} \approx \mathbf{0.175}

The rate the oil radius is changing when the radius is 10 cm is approximately <u>0.175 m</u>

Learn more about the rules of differentiation here:

brainly.com/question/20433457

brainly.com/question/13502804

3 0
3 years ago
How far could a train travel in 1 hour if it's average speed is 0.79 miles per hour
elena-s [515]

If its average speed is .79 miles in an hour, then it goes .79 miles in an hour.

3 0
3 years ago
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