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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We should know that ⇒ 1 lb = 16 oz
The amount of purchased celery = 12 lb = 12 * 16 = 192 oz
<span>Trim and cut it into pieces, ending up with 5.5 oz of trim.
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</span><span>The yield will be = 192/5.5 = 34.91
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</span><span>Taking the smallest integer number.
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</span><span>So, the yield = 34
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</span><span>which will be equal = 34*5.5 = 187 oz
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So, the yield </span>
percentage =
She made 6 patches of cookies because if she uses 1/8 per batch and 3/4 x2 equals 6/8 , she made 6 batches.
it is definitely the answer B