Solving a system of equations, we will see that there are 6000 kg of coal on the pile.
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How many kilograms of coal are in the pile of coal?</h3>
Let's say that there are N kilograms of coal, and it is planned to burn it totally in D days, these are our variables.
We know that:
- If the factory burns 1500 kilograms per day, it will finish burning the coal one day ahead of planned.
- If it burns 1000 kilograms per day, it will take an additional day than planned.
Then we can write the system of equations:
(1500)*(D - 1) = N
(1000)*(D + 1) = N
Because N is already isolated in both sides, we can write this as:
(1500)*(D - 1) = N = (1000)*(D + 1)
Then we can solve for D:
(1500)*(D - 1) = (1000)*(D + 1)
1500*D - 1500 = 1000*D + 1000
500*D = 2500
D = 2500/500 = 5
Now that we know the value of D, we can find N by replacing it in one of the two equations:
(1500)*(D - 1) = N
(1500)*(5 - 1) = N
(1500)*4= N = 6000
This means that there are 6000 kilograms of coal on the pile.
If you want to learn more about systems of equations:
brainly.com/question/13729904
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The temperature would be about 54.
X's cancel 5y = 20 y= 4 X=0
Answer:
4
Step-by-step explanation:
The perimeter of each square window is 4 times the length of one side, so is 20 feet.
80 feet can be divided into 80/20 = 4 sections of 20 feet each.
Shelly has enough lights to surround 4 windows.
Answer:
x = 0
y = -12
Step-by-step explanation:
Plug in what y is equal to into the second equation
-3(x - 12) = 2x + 36
Distribute the -3
-3x + 36 = 2x + 36
-36 -36
-3x = 2x
They cannot be equal so x = 0
Plug in 0 into the y equation
y = 0 - 12
y = -12
