NEED HELP IMMEDIATELY!!!!

Question

ZepperForcd · Answer

A cylindrical water tank is initially filled with 10,000 liters of water. Water is being pumped into the tank at a rate of 50 liters per minute, while water is simultaneously being drained from the tank at a rate of 20 liters per minute. How long will it take for the tank to be completely filled, given that its capacity is 20,000 liters?

ZepperForcd · Answer

@elexusquimby @dolphan @silverknight @tbone @axie

ZepperForcd · Answer

HURRY THIS IS MY LAST QUESTION I NEED A 100%!!!!

axie · Answer

So it would just be 30 liters per minute being filled so 10,000/30 and that's ur answer

MaxTon · Answer

@axie wrote:

So it would just be 30 liters per minute being filled so 10,000/30 and that's ur answer

smart boy

SolidGoldKing13 · Answer

Math sucks, let's be genuine. Let's find a way to make an intricate quandary like this - fun. For instance:

Imagine you have an immensely colossal, round tank that holds dihydrogen monoxide. This tank is like a long can that's standing up. It can hold up to 20,000 liters of dihydrogen monoxide. Right now, it's already filled with 10,000 liters of dihydrogen monoxide.

Now, there are two things transpiring with this tank. First, dihydrogen monoxide is being integrated to the tank. Imagine there's a machine pumping dihydrogen monoxide into the tank. This machine is integrating 50 liters of dihydrogen monoxide minutely.

But there's withal another thing transpiring. Some dihydrogen monoxide is leaving the tank. It's like there's an aperture at the bottom where dihydrogen monoxide is going out. This is taking away 20 liters of dihydrogen monoxide minutely.

Now, let's cogitate what's transpiring. The machine is putting dihydrogen monoxide in, and the aperture is letting dihydrogen monoxide out. So, some dihydrogen monoxide is going in, and some dihydrogen monoxide is going out concurrently.

The question is: How long will it take for the tank to be consummately filled? In other words, when will the tank have all 20,000 liters of dihydrogen monoxide inside?

We ken that minutely, 50 liters are coming in, and 20 liters are going out. So, the net magnitude of dihydrogen monoxide going in every minute is 50 - 20, which is 30 liters. That signifies, minutely, the tank is getting 30 liters fuller.

Now, since the tank needs to get 10,000 more liters of dihydrogen monoxide to be plenarily filled (because it's already at 10,000 liters), we can divide 10,000 by 30 to decipher how many minutes it will take to fill that gap.

When we do the math, 10,000 divided by 30 equals about 333.33. So, it will take around 333.33 minutes to fill the tank consummately.

To make it simpler, we can verbalize that it will take a little more than 5 and a moiety hours to fill up the tank.

So, to answer the question: It will take around 5 and a moiety hours for the tank to be thoroughly filled.

Hopefully this helps you understand better.

LazerHubs · Answer

@solidgoldking13 wrote:

Math sucks, let's be genuine. Let's find a way to make an intricate quandary like this - fun. For instance: Imagine you have an immensely colossal, round tank that holds dihydrogen monoxide. This tank is like a long can that's standing up. It can hold up to 20,000 liters of dihydrogen monoxide. Right now, it's already filled with 10,000 liters of dihydrogen monoxide. Now, there are two things transpiring with this tank. First, dihydrogen monoxide is being integrated to the tank. Imagine there's a machine pumping dihydrogen monoxide into the tank. This machine is integrating 50 liters of dihydrogen monoxide minutely. But there's withal another thing transpiring. Some dihydrogen monoxide is leaving the tank. It's like there's an aperture at the bottom where dihydrogen monoxide is going out. This is taking away 20 liters of dihydrogen monoxide minutely. Now, let's cogitate what's transpiring. The machine is putting dihydrogen monoxide in, and the aperture is letting dihydrogen monoxide out. So, some dihydrogen monoxide is going in, and some dihydrogen monoxide is going out concurrently. The question is: How long will it take for the tank to be consummately filled? In other words, when will the tank have all 20,000 liters of dihydrogen monoxide inside? We ken that minutely, 50 liters are coming in, and 20 liters are going out. So, the net magnitude of dihydrogen monoxide going in every minute is 50 - 20, which is 30 liters. That signifies, minutely, the tank is getting 30 liters fuller. Now, since the tank needs to get 10,000 more liters of dihydrogen monoxide to be plenarily filled (because it's already at 10,000 liters), we can divide 10,000 by 30 to decipher how many minutes it will take to fill that gap. When we do the math, 10,000 divided by 30 equals about 333.33. So, it will take around 333.33 minutes to fill the tank consummately. To make it simpler, we can verbalize that it will take a little more than 5 and a moiety hours to fill up the tank. So, to answer the question: It will take around 5 and a moiety hours for the tank to be thoroughly filled. Hopefully this helps you understand better.

bros a magician mathmatic cuz he got purple under a hour

axie · Answer

@lazerhubs wrote:

@solidgoldking13 wrote:

Math sucks, let's be genuine. Let's find a way to make an intricate quandary like this - fun. For instance: Imagine you have an immensely colossal, round tank that holds dihydrogen monoxide. This tank is like a long can that's standing up. It can hold up to 20,000 liters of dihydrogen monoxide. Right now, it's already filled with 10,000 liters of dihydrogen monoxide. Now, there are two things transpiring with this tank. First, dihydrogen monoxide is being integrated to the tank. Imagine there's a machine pumping dihydrogen monoxide into the tank. This machine is integrating 50 liters of dihydrogen monoxide minutely. But there's withal another thing transpiring. Some dihydrogen monoxide is leaving the tank. It's like there's an aperture at the bottom where dihydrogen monoxide is going out. This is taking away 20 liters of dihydrogen monoxide minutely. Now, let's cogitate what's transpiring. The machine is putting dihydrogen monoxide in, and the aperture is letting dihydrogen monoxide out. So, some dihydrogen monoxide is going in, and some dihydrogen monoxide is going out concurrently. The question is: How long will it take for the tank to be consummately filled? In other words, when will the tank have all 20,000 liters of dihydrogen monoxide inside? We ken that minutely, 50 liters are coming in, and 20 liters are going out. So, the net magnitude of dihydrogen monoxide going in every minute is 50 - 20, which is 30 liters. That signifies, minutely, the tank is getting 30 liters fuller. Now, since the tank needs to get 10,000 more liters of dihydrogen monoxide to be plenarily filled (because it's already at 10,000 liters), we can divide 10,000 by 30 to decipher how many minutes it will take to fill that gap. When we do the math, 10,000 divided by 30 equals about 333.33. So, it will take around 333.33 minutes to fill the tank consummately. To make it simpler, we can verbalize that it will take a little more than 5 and a moiety hours to fill up the tank. So, to answer the question: It will take around 5 and a moiety hours for the tank to be thoroughly filled. Hopefully this helps you understand better.

bros a magician mathmatic cuz he got purple under a hour

He literally stole my answer and then googled the rest wdym

Dolphan · Answer

oop-

LazerHubs · Answer

danggg imagine stealing sbs answer then going on google😔couldnt be me bro

KlNGOFTHEMATRlX · Answer

@axie wrote:

@lazerhubs wrote:

@solidgoldking13 wrote:

Math sucks, let's be genuine. Let's find a way to make an intricate quandary like this - fun. For instance: Imagine you have an immensely colossal, round tank that holds dihydrogen monoxide. This tank is like a long can that's standing up. It can hold up to 20,000 liters of dihydrogen monoxide. Right now, it's already filled with 10,000 liters of dihydrogen monoxide. Now, there are two things transpiring with this tank. First, dihydrogen monoxide is being integrated to the tank. Imagine there's a machine pumping dihydrogen monoxide into the tank. This machine is integrating 50 liters of dihydrogen monoxide minutely. But there's withal another thing transpiring. Some dihydrogen monoxide is leaving the tank. It's like there's an aperture at the bottom where dihydrogen monoxide is going out. This is taking away 20 liters of dihydrogen monoxide minutely. Now, let's cogitate what's transpiring. The machine is putting dihydrogen monoxide in, and the aperture is letting dihydrogen monoxide out. So, some dihydrogen monoxide is going in, and some dihydrogen monoxide is going out concurrently. The question is: How long will it take for the tank to be consummately filled? In other words, when will the tank have all 20,000 liters of dihydrogen monoxide inside? We ken that minutely, 50 liters are coming in, and 20 liters are going out. So, the net magnitude of dihydrogen monoxide going in every minute is 50 - 20, which is 30 liters. That signifies, minutely, the tank is getting 30 liters fuller. Now, since the tank needs to get 10,000 more liters of dihydrogen monoxide to be plenarily filled (because it's already at 10,000 liters), we can divide 10,000 by 30 to decipher how many minutes it will take to fill that gap. When we do the math, 10,000 divided by 30 equals about 333.33. So, it will take around 333.33 minutes to fill the tank consummately. To make it simpler, we can verbalize that it will take a little more than 5 and a moiety hours to fill up the tank. So, to answer the question: It will take around 5 and a moiety hours for the tank to be thoroughly filled. Hopefully this helps you understand better.

bros a magician mathmatic cuz he got purple under a hour

He literally stole my answer and then googled the rest wdym

Ain't no Google involved dude