Two pumps are filling large vats of liquid. One vat is empty, and the pump is filling it at a rate of 6 gallons per minute. The other is already filled with 10 gallons, and it is continuing to fill at a rate of 4 gallons per minute. Write an equation to show when the two vats will have an equal amount of liquid. Use x as your variable.
1. 6x = 10 + 4x
2. 8x = 40 + 2x
3. 300 - 15x = 280 + 20x
4. 14
5. 5
Mr bot???
Two pumps are filling large vats of liquid. One vat is empty, and the pump is filling it at a rate of 6 gallons per minute. The other is already filled with 10 gallons, and it is continuing to fill at a rate of 4 gallons per minute. Write an equation to show when the two vats will have an equal amount of liquid. Use x as your variable.
Two pumps are filling large vats of liquid. One vat is empty, and the pump is filling it at a rate of 6 gallons per minute. The other is already filled with 10 gallons, and it is continuing to fill at a rate of 4 gallons per minute. Write an equation to show when the two vats will have an equal amount of liquid. Use x as your variable.
To write an equation showing when the two vats will have an equal amount of liquid, we need to consider the rates at which each vat is being filled. Let's analyze the situation:
The first vat starts empty and is being filled at a rate of 6 gallons per minute. Therefore, the amount of liquid in the first vat, in terms of time (t), can be represented as 6t.
The second vat already has 10 gallons and is continuing to fill at a rate of 4 gallons per minute. This means that the amount of liquid in the second vat, in terms of time (t), can be represented as 10 + 4t.
To find when the two vats will have an equal amount of liquid, we need to set the two expressions equal to each other:
6t = 10 + 4t
Now, we can solve for the variable x:
6t - 4t = 10
2t = 10
t = 10/2
t = 5
The variable x represents the time in minutes when the two vats will have an equal amount of liquid. Therefore, the equation is x = 5.
Sure, I can come up with an equation, but let me warn you, it's going to be filled with clownish humor. Brace yourself!
Let's call the time in minutes as x. At time x, the empty vat will be filled with 6x gallons, and the vat with 10 gallons will be filled with 10 + 4x gallons. To find when both vats have an equal amount of liquid, we can set the two expressions equal to each other:
6x = 10 + 4x
Now, let's simplify this equation and solve it with my clown expertise. I'll subtract 4x from both sides to get rid of those clownish x's hanging around in different vats:
6x - 4x = 10 + 4x - 4x
2x = 10
Finally, we divide both sides by 2, because even clowns know how to share:
2x/2 = 10/2
x = 5
So, there you have it! At time x = 5 minutes, those vats will have an equal amount of liquid. Just remember to keep an eye out for any clown fish swimming around in there!
Let's assume that after x minutes, the two vats will have an equal amount of liquid.
The amount of liquid in the first vat after x minutes will be 6x (since it is filling at a rate of 6 gallons per minute).
The amount of liquid in the second vat after x minutes will be 10 + 4x (since it already has 10 gallons and is filling at a rate of 4 gallons per minute).
To find when the two vats will have an equal amount of liquid, we can set the two expressions equal to each other:
6x = 10 + 4x
Simplifying this equation, we get:
2x = 10
x = 5
Therefore, the two vats will have an equal amount of liquid after 5 minutes.