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tatyana61 [14]
3 years ago
8

A cold water pipe can fill a swimming pool in 4 hours, and a hot water pipe in 6 hours. If cold water pipe is turned on and hot

water pipe is turned on 3 hours later then how long will it take them to fill the pool together?
Mathematics
2 answers:
nirvana33 [79]3 years ago
8 0

Answer:

0.6 hours

Step-by-step explanation:

Let's say the swimming pool is 12,000 gallons.

The flow rate of the cold water pipe is:

12000 gal / 4 hr = 3000 gal/hr

The flow rate of the hot water pipe is:

12000 gal / 6 hr = 2000 gal/hr

If the cold water pipe is turned on for three hours, the volume of water is:

3000 gal/hr × 3 hr = 9000 gal

That leaves 3000 gallons.  The time to fill the rest of the pool with both the cold and hot water pipes is:

3000 gal / (3000 gal/hr + 2000 gal/hr) = 0.6 hr

It takes 0.6 hours (36 minutes).

Romashka [77]3 years ago
5 0

the answer:

0.9

just flip the nine upside down, you have the REAL answer. ;)

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Find e^cos(2+3i) as a complex number expressed in Cartesian form.
ozzi

Answer:

The complex number e^{\cos(2+31)} = \exp(\cos(2+3i)) has Cartesian form

\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2).

Step-by-step explanation:

First, we need to recall the definition of \cos z when z is a complex number:

\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}.

Then,

\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}. (I)

Now, recall the definition of the complex exponential:

e^{z}=e^{x+iy} = e^x(\cos y +i\sin y).

So,

e^{2i-3} = e^{-3}(\cos 2+i\sin 2)

e^{-2i+3} = e^{3}(\cos 2-i\sin 2) (we use that \sin(-y)=-\sin y).

Thus,

e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)

Now we group conveniently in the above expression:

e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2.

Now, substituting this equality in (I) we get

\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2.

Thus,

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right].

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3 years ago
Se cocina una pizza de 20 cm de radio, en su interior se coloca al azar 12 piezas de salami
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Answer:

The area of cheese is 388π cm.

Step-by-step explanation:

Given that,

Radius of pizza = 20 cm

Number of pieces = 12

Radius of circle = 1 cm

We need to calculate the area of pizza

Using formula of area

A=\pi\times r^2

Put the value into the formula

A=\pi\times20^2

A=400\pi\ cm^2

We need to calculate the area of circle

Using formula of area

A'=\pi\times r^2

Put the value into the formula

A'=\pi\times1^2

A'=\pi\ cm^2

Now, we multiply the result by 12 because they are 12 salamis.

So. the area of salamis will be

A'=12\pi\ cm^2

We need to calculate the area of cheese

The total area of ​​the pizza and subtract the area of ​​the salamis

A''=A-A'

Put the value in to the formula

A''=400\pi-12\pi

A=388\pi\ cm^2

Hence,  The area of cheese is 388π cm.

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n200080 [17]

Answer: 10 + 6 + 2.00x + 1.50x = Answer

Step-by-step explanation:

3 0
3 years ago
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